{
  "Tag": [
    "AMC",
    "AMC 10",
    "AMC 10 A",
    "symmetry",
    "blogs",
    "number theory",
    "least common multiple"
  ],
  "Problem": "Here's a topic for discussing problems on the 2005 AMC10A. I'll start off.\r\n\r\nEmbarrassingly, I spent 20 minutes on #24 by misreading the instructions and not noticing that n>1. Luckily, there was no choice for 2, because that's what I got first.  :blush:",
  "Solution_1": "#17 might have been the most interesting problem, since it can be solved with a symmetry approach like the ones discussed in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=60]Mr. Rusczyk's blog[/url].  Well, sort of.  Since we have an arithmetic sequence,\r\n\r\nAB + BC + CD + DE + EA = 5(middle segment)\r\n\r\nBut AB + BC + CD + DE + EA = 2(A+B+C+D+E) = 2(3+5+6+7+9) = 60\r\n\r\nSo middle segment = 12",
  "Solution_2": "Man, I was so disappointed after missing #22 about the sets of multiples of 4's and 6's. You had to look for the LCM of those numbers, and my mind ignorantly thought for a moment that LCM (4,6) = 24, so I picked 333, which I thought should have been 334. Right after the test, I realized my mistake.  :(",
  "Solution_3": "[quote=\"keta\"]Man, I was so disappointed after missing #22 about the sets of multiples of 4's and 6's. You had to look for the LCM of those numbers, and my mind ignorantly thought for a moment that LCM (4,6) = 24, so I picked 333, which I thought should have been 334. Right after the test, I realized my mistake.  :([/quote]\r\n\r\ndude lcm of 4 and 6 is 12",
  "Solution_4": "Yeah, he said he realized his mistake right after the test.  Anybody else think #24 was a really nice problem?  The whole difference of squares...and number 8 was just the diagram used to prove the Pythagorean Theorem, which I'm sure most of us are familiar with.",
  "Solution_5": "Most of them were great. I agree that 17 and 24 were cool. Was there any way to do 25 besides $A=\\frac 12 ab\\sin\\theta$? I would have gotten that one had I known the formula before taking the test. Instead I omitted it and our proctor explained the formula afterwards.",
  "Solution_6": "For number 25, you don't need the sine formula... just draw a couple of lines.\r\nConnecting DC, the areas of triangles ADC and DBC have the ratio 19/25, (same height)\r\nand similarly, connecting ED, the areas of triangles AED and EDC have the ratio 14/42.\r\nI almost left 24 blank, but got it in the last 3 minutes... is trial and error the only way to do it?",
  "Solution_7": "[quote=\"sky9073\"]For number 25, you don't need the sine formula... just draw a couple of lines.\nConnecting DC, the areas of triangles ADC and DBC have the ratio 19/25, (same height)\nand similarly, connecting ED, the areas of triangles AED and EDC have the ratio 14/42.\nI almost left 24 blank, but got it in the last 3 minutes... is trial and error the only way to do it?[/quote]\r\n\r\nWell, we know that both n and n+48 have to be squares of prime numbers, and 48 can be expressed as 23+25 or 9+11+13+15 or 3+5+7+9+11+13, so there's only 3 possibilities for n and n+48 being squares. Check all 3 and you'll find that only n=121 and n+48=169 works.",
  "Solution_8": "i personally like the [i]trefoi[/i]l problem\r\nalso the one where we had an arithmetic sequence.\r\nhated octagon problem (forgot that the sqrt(2) was divided by 2... D'OH!) hence I got 7 intead of 7/2",
  "Solution_9": "What I didn't like about 25 was that my friend who isn't really good at math assumed the the triangles were right in 25, and got it right, but I checked, and found that they weren't, and didn't answer it.",
  "Solution_10": "I'm really annoyed at myself about 25 now. I knew the 1/2ab(sinC) formula but I only looked at it in the last 2 minutes because I thought: Wow, #25, must be too hard, let's check my answers first. :(\r\n\r\nKudos to you who got it :D",
  "Solution_11": "i screwed up on reading 21 and thought it was the other way around....\r\ni can't believe I missed 12 for number 15\r\nalso, i marked number 23 wrong even though I had the right answer...\r\nand I couldn't really get an \"actual\" solution to 19 w/o guessing",
  "Solution_12": "does anyone know why that soccer problem is 1/5?? my intuition would tell me it would be 1/2, but apparently not....",
  "Solution_13": "Intuition is often not good enough my friend  :P \r\nThere were 4 cases where B was 2nd and A won.\r\nOne of them consisted of 4 games (ABAA I think).\r\nlet p be the probability of a 5 game occurence with A winning and B winning the second game\r\nso we can write 2p (for the ABAA game) + 3p = 1\r\nso 5p = 1\r\nprobability of a 5 game win that fits the requirements is 1/5\r\n(I put 1/4 on the test...)",
  "Solution_14": "hmm okay.. maybe i dont understand it well b/c i dont really know the prerequisities.... (i didnt take the AMC A, so this is based upon what other ppl told me)\r\n\r\n5 games; Team B won game 2 and you want to find the probability that Team A wins the series?",
  "Solution_15": "no, team B wins game 2, team A wins the series.\r\n\r\nFind probability that team B wins game 1 also.",
  "Solution_16": "[quote=\"yif man12\"][quote=\"sky9073\"]For number 25, you don't need the sine formula... just draw a couple of lines.\nConnecting DC, the areas of triangles ADC and DBC have the ratio 19/25, (same height)\nand similarly, connecting ED, the areas of triangles AED and EDC have the ratio 14/42.\nI almost left 24 blank, but got it in the last 3 minutes... is trial and error the only way to do it?[/quote]\n\nWell, we know that both n and n+48 have to be squares of prime numbers, and 48 can be expressed as 23+25 or 9+11+13+15 or 3+5+7+9+11+13, so there's only 3 possibilities for n and n+48 being squares. Check all 3 and you'll find that only n=121 and n+48=169 works.[/quote]\r\n\r\nAlternatively, let n = k^2 and n+48 = l^2.  Then l^2-k^2 = 48, so (l+k,l-k) = (1,48), (2,24),... (48,1).  Of these, only the pairs where k = 1, 4, or 11 work, so n = 1, 16, or 121.  But n > 1, and 16 isn't the square of a prime, so only n=121 works.",
  "Solution_17": "[quote=\"pkothari13\"]does anyone know why that soccer problem is 1/5?? my intuition would tell me it would be 1/2, but apparently not....[/quote]\r\n\r\nAs for your intuition, it is true that team B has a 1/2 chance of winning game 1.  However, that's not the question - the question is actually 'out of all the cases where team B wins game 2 but team A wins the series, what is the probability that team B will win the first game'.  Since the set of cases where team B wins game 2 but team A wins the series is a different set from the set of all games, it's no longer 1/2 - we have additional conditions that we must satisfy.\r\n\r\nThe four possible outcomes, and the probability that they occur, is\r\nABAA = (1/2)^4 = 1/16\r\nABABA = (1/2)^5 = 1/32\r\nABBAA = 1/32\r\nBBAAA = 1/32\r\n\r\nWe are given that one of these occured (since B won game 2, A won series).  One of these occurs 1/16+1/32+1/32+1/32 = 5/32 of the time, so now we need to multiply all of the probabilities by 32/5 so that they add up to 1 (since we know that one of them must occur):\r\n\r\nABAA => 2/5\r\nABABA => 1/5\r\nABBAA => 1/5\r\nBBAAA => 1/5\r\n\r\nAnd there you have it.",
  "Solution_18": "Even though a \"best of 5\" series usually stops when one team wins 3 games, we can imagine that the series continues for the full 5 games.  For example, the event ABAA splits into two outcomes ABAAA and ABAAB.  That way, it is clear that there are 5 equally-likely outcomes, and we don't have to worry about some outcomes being weighted more than others.\r\n\r\nAdding such fictitious games to \"best of\" series often makes such problems a lot easier.  I recall using that trick in the past, but I can't think of the examples now."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "find integer $a$ such that:$x^2-x+a|x^{13}+x+90$",
  "Solution_1": "[quote=\"amir2\"]find integer $a$ such that:$x^2-x+a|x^{13}+x+90$[/quote]\r\n$x^2-x+a|x^{13}+x+90$\r\nPut $x=0$, we get $a|90$\r\nPut $x=1$, we get $a|92$\r\nPut $x=-1$, we get $a+2|88$\r\nSo $a$ can only be $-1$ or $2$.\r\nWhen $a=-1$, let $x^{13}+x+90=(x^2-x-1)Q(x)$. But when $x^2-x-1=0$, $x^{13}+x+90\\neq 0$. So $a=-1$ does not satisfy the condition.\r\nWhen $a=2$, let $u$ be a root of $x^2-x+2=0$. Then\r\n$u^2=u-2$\r\n$u^3=u^2-2u=-u-2$\r\n$u^6=u^2+4u+4=5u+2$\r\n$u^{12}=25u^2+20u+4=45u-46$\r\n$u^{13}=45u^2-46u=-u-90$ or $u^{13}+u+90=0$.\r\nTherefore $a=2$."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "LaTeX",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Prove that sum and product of 2 algebric integers is an algebric integer.\r\n\r\nProof:Suppose a,b are algebric integers then a satisfies a minimal monic polynomial with integer coefficients i.e. let\r\n$a^n +a_1a^{n-1}+...+a_n=0$.\r\n So, $a^n =-(a_1a^{n-1}+...+a_n)$ \u2026\u2026\u2026\u2026..(i)\r\nThus all the higher powers of a can be written in the form of linear combinations of 1,a,\u2026.a^n-1  \r\nSo this set generates Z[a]. Similarly if b satisfies a minimal monic polynomial of degree m, then 1,b,\u2026,b^m-1 generates Z[b].\r\nNow we claim that Z[a,b] is generated by elements of the form aibj where I=0(1)n-1\r\nJ=0(1)m-1.\r\nTo see this take any element p belonging to Z[a,b]. then p is of the form \r\n $P= \u00e5\u00e5x_ij a_ib_j$ . Since there are only a finitely many terms in the expression, starting from the 1st term check the power of a. If it is less than n, we leave the term unchanged and go to the next term. If the power is greater than n we reduce it by the help of (i).After reducing the power of a from each term we then reduce the power of b by the same method. By these operations each term becomes a product of polynomials where the degree of a & b are atmost n-1 and m-1 respectively. Multiplying all the terms and rearranging them we can write p as a linear combination of aibj . Thus aibj generate Z[a,b] as claimed and as the no of such generators is finite we have Z[a,b] a finitely generated Z-module.\r\n\r\nIs my proof correct?It certainly is easy for a double starred herstein problem",
  "Solution_1": "Please use $\\LaTeX$ correctly (e.g. use a^{n-1} to get $a^{n-1}$, not as you do a^(n-1) ).",
  "Solution_2": "I am very sorry I dont know latex. :blush: \r\nI have a lot of problems with the summation sign.I know its difficult for you to read but plz tell me if my soln is correct or not.",
  "Solution_3": "[quote=\"the game\"]Prove that sum and product of $2$ algebric integers is an algebric integer.\n\nProof:Suppose $a,b$ are algebric integers then a satisfies a minimal monic polynomial with integer coefficients i.e. let\n$a^n +a_1a^{n-1}+...+a_n=0$.\n So, $a^n =-(a_1a^{n-1}+...+a_n)$ \u2026\u2026\u2026\u2026..(i)\nThus all the higher powers of a can be written in the form of linear combinations of $1,a,....a^{n-1}$  \nSo this set generates $\\mathbb{Z}[a]$. Similarly if $b$ satisfies a minimal monic polynomial of degree $m$, then $1,b,...,b^{m-1}$ generates $\\mathbb{Z}[b]$.\nNow we claim that $\\mathbb{Z}[a,b]$ is generated by elements of the form $a^i b^j$ where $i=0...n-1$\n$J=0...m-1$.\nTo see this take any element $p$ belonging to $\\mathbb{Z}[a,b]$. then $p$ is of the form \n $p= \\sum_{i,j} x_{i,j} a^ib^j$ . Since there are only a finitely many terms in the expression, starting from the 1st term check the power of $a$. If it is less than $n$, we leave the term unchanged and go to the next term. If the power is greater than $n$ we reduce it by the help of (i).After reducing the power of $a$ from each term we then reduce the power of $b$ by the same method. By these operations each term becomes a product of polynomials where the degree of $a$ & $b$ are atmost $n-1$ and $m-1$ respectively. Multiplying all the terms and rearranging them we can write $p$ as a linear combination of $a^ib^j$ . Thus $a^ib^j$ generate $\\mathbb{Z}[a,b]$ as claimed and as the no of such generators is finite we have $\\mathbb{Z}[a,b]$ a finitely generated $\\mathbb{Z}$-module.\n\nIs my proof correct?It certainly is easy for a double starred herstein problem[/quote]\r\nThe above should be what you want :) (click on quote to see the plain text for the formulas)\r\n\r\nTo the prove:\r\nThe part \"finetely generated $\\implies$ algebraic integers\" is missing.\r\nThe rest seems ok."
}
{
  "Tag": [],
  "Problem": "Can 10th grader apply RSI?\r\nThanks!",
  "Solution_1": "From what I've heard, you should only apply to RSI in 10th grade if you intend to graduate in 11th grade."
}
{
  "Tag": [
    "invariant",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "This is just beautiful: suppose that $A,B$ are invertible complex matrices which commute to their commutator. Then all eigenvalues of the commutator are roots of the unity.",
  "Solution_1": "I suppose the commutator here is $ABA^{-1}B^{-1}$ and not $AB-BA$.",
  "Solution_2": "Yes, sorry.",
  "Solution_3": "Let $\\lambda$ be an eigenvalue of $[A,B]=ABA^{-1}B^{-1}$. because $A,B$ commute with $[A,B]$, the $\\lambda$-eigenspace of $[A,B]$ is invariant under both $A$ and $B$. By restricting attention to this eigenspace, we may as well assume that $[A,B]=\\lambda$. \r\n\r\nWe now have $AB=\\lambda BA$. Let $\\sigma$ be an eigenvalue of $A$ and let $x$ be a $\\sigma$-eigenvector of $A$. We then have $ABx=\\lambda BAx=\\lambda\\sigma Bx$, i.e. $Bx$ is a $\\lambda\\sigma$-eigenvector of $A$. We repeat this procedure to conclude that for all positive integers $n,\\ \\lambda^{n}\\sigma$ is an eigenvalue of $A$. Since $A$ has finitely many eigenvalues, this can only happen if $\\lambda$ is a root of unity."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "I'm hoping to get a summation symbol like this:\r\n\r\n$ \\displaystyle\\sum_{k\\equal{}0}^n$\r\n\r\nin a fraction like this: $ \\displaystyle\\frac{\\sum_{k\\equal{}0}^n}{\\sum_{j\\equal{}0}^u}$.\r\n\r\nBut I can't figure out how to get the summation sign to render stylistically in a fraction.\r\n\r\nHelp appreciated. Thanks!",
  "Solution_1": "Use \\sum\\limits... \r\n$ \\frac{\\sum\\limits_{k\\equal{}0}^{n}}{\\sum\\limits_{j\\equal{}0}^{u}}$"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "trigonometry",
    "logarithms",
    "complex numbers",
    "calculus computations"
  ],
  "Problem": "Compute the \"antiderivative\" of the function $ln(sinx)$",
  "Solution_1": "There is no elementary antiderivative. The best we can do is to evaluate some definite integrals for that function.",
  "Solution_2": "Starting with the argument I just posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152690]here[/url] but taking the real part instead of the imaginary part, we get this result:\r\n\r\n$\\frac1n\\ln|2\\sin2\\theta|=\\ln|1-e^{i\\theta}|=\\sum_{n=1}^{\\infty}\\frac{\\cos n\\theta}{n}.$\r\n\r\nIntegrating, we get\r\n\r\n$\\int_{0}^{x}\\frac12\\ln|2\\sin2\\theta|\\,d\\theta=\\sum_{n=1}^{\\infty}\\frac{\\sin nx}{n^{2}}.$\r\n\r\nOne could, with sufficient patience, use this to calculate certain definite integrals, such as\r\n\r\n$\\int_{0}^{\\pi}\\ln\\sin x\\,dx.$ Carcul claimed as much."
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Does there exists a real function f such that for all real x the set of solutions of the equation f(t)=x is in bijection with the reals?",
  "Solution_1": "Of course. f(0,abcdef...)=0,ace..., where 0,abcde.. is a decimal representation of a number.",
  "Solution_2": "What a simple example! Thank you. I start to ask some very easy questions, but I like them. I don't know why. \r\n  Another (maybe stupid question): does there exist a continuous function with those properties?",
  "Solution_3": "Yes, it exists, There exists a continious function f which maps the segment to a square. If f(t)=(x(t),y(t)), then x(t) is such a function."
}
{
  "Tag": [
    "algorithm",
    "analytic geometry",
    "graphing lines",
    "slope",
    "trigonometry",
    "ratio"
  ],
  "Problem": "I am looking for some justification as to why when you have a line in Ax+By+C=0 form you divide by  :pm:   :sqrt: A :^2: +B :^2:  to get it into normal form and then the coefficents represent cos :phi: x + sin :phi: y - p =0. \r\n\r\nDoes anyone know where I can find a link that would tell me this?\r\n\r\nI have been trying to do this myself but cannot find justification for the C term.\r\n\r\nThanks!",
  "Solution_1": "What do you mean that you cannot find a justification for the C term?  You need to explain more what you're asking about for us (well, me at least) to understand and help.",
  "Solution_2": "Okay well I am trying to find a justification as to why one preforms a certain algorithim to get from general form of an equation of a line, Ax+By+C=0 to Normal form, xcos( :phi: )+ysin( :phi: )-p=0.\r\n\r\nThe algorithim is that you find sqrt(A^2+B^2), and give it a sign that is opposite of C's sign.  Then you divide all terms of Ax+By+C=0 by this quantity, and now the line is in normal form.\r\n\r\nI have figured out why this is true for the A/ :pm:  :sqrt: A :^2: +B :^2: and B/:pm:  :sqrt: A :^2: +B :^2: but I can't figure out why it works w/the C term.  (FYI I did this by changing Ax+By+C=0 to y=(-A/B)x+(-C/B), since the slope of this line is -A/B I constructed a triangle w/ A*n, B*n, and n*sqrt(A^2+B^2) leg lengths, n is a constant, and formed trig ratios w/  90+ :phi: , to find out why the coefficents were in the normal form now.)",
  "Solution_3": "[quote=\"mrblueskies\"]I have figured out why [b]this[/b] is true ...[/quote]\r\n(emphasis mine)\r\n\r\nWhat is the \"this\" you are refering to?  I understand everything you've written.  I understand the operation you're carrying out.  What I don't understand is what you think is happening to specific coefficients -- what does the \"this\" in that sentence refer to?  \"Normal form\" is a property of the equation of the line; it isn't a property of the individual coefficients.",
  "Solution_4": "\"this\" as in the algorithim, I found out why this algorithim transformes the line so the coefficents of x and y represent cos phi and sine phi.  I cannot however find out why the new C term after preforming the algorithim represents p.",
  "Solution_5": "Hey,\r\n\r\nWell...if you have the line Ax + By + C = 0, dividing by  :pm:  :sqrt: (A^2 + B^2) will give you the hypotenuse of a right triangle with sides A and B. In our case, since the slope is -A/B, say n = arctanA/B. Dividing gives us that sin(n)x + cos(n)y - C/root(A^2 + B^2) = 0. Is that the p you are trying to \"justify\"?",
  "Solution_6": "I don't understand what you think p is -- it's just a number, which is the result of dividing C by :sqrt:(A<sup>2</sup> + B<sup>2</sup>).  What significance are you attributing to it?  All it's doing is telling you that there is some constant left over.",
  "Solution_7": "p is the perpendicular distance of the line from the origin.\r\n\r\nIf the line is at an angle  :theta: to the x-axis (so that tan( :theta: )= -A/B) then the perpendicular distance p  of the line from the origin is given by\r\n\r\np = ycos( :theta: ) - xsin( :theta: )\r\n\r\nif the line crosses the y-axis above the origin, or\r\n\r\np = xsin( :theta: ) - ycos( :theta: )\r\n\r\nif the line crosses the y-axis below the origin, where (x,y) is any point on the line.\r\n\r\nIn the first case, set  :phi: =  :theta: + :pi: /2 and in the second case set :phi: =  :theta: - :pi: /2, and then in both cases you have\r\n\r\np = xcos( :phi: ) + ysin( :phi: )\r\n\r\nand, also in both cases, tan( :phi: ) = -cot( :theta: ) = B/A"
}
{
  "Tag": [
    "integration",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": ":roll: What is the product $1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot ... \\cdot n$?",
  "Solution_1": "$1 \\cdot 2 \\cdot 3 \\cdot ... \\cdot n = n!$ ~ $\\sqrt{2\\pi n} e^{n \\ln(n)-n+\\frac{1}{12n}-...}$",
  "Solution_2": "$\\Gamma(n+1) = \\int_{0}^{\\infty} t^n e^{-t} \\, dt$\r\n\r\nAlternately:\r\n\r\nhttp://mathworld.wolfram.com/StirlingsSeries.html",
  "Solution_3": "For completness sake:\r\n$n! = \\Gamma ( n + 1 )$ for all positive integer $n$.",
  "Solution_4": "[quote=\"sagar1331\"]:roll: What is the product 1*2*3*4*.........n?[/quote]\r\nMy friend says:\r\nEvgeny: $1*2*...*n=n*(n-1)*...*1$.\r\nRoman: It's very silly question. ;)",
  "Solution_5": "I agree with Roman..  :P \r\n\r\n(Sorry for the useless post)\r\n\r\nOh, well, just to make it more valuable : http://mathworld.wolfram.com/Factorial.html\r\n\r\n;)"
}
{
  "Tag": [
    "inequalities",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "inequalities solved"
  ],
  "Problem": "given positive reals a1,a2,... a5 satisfy  \\sum 1/(1+a)=1\r\nProve that  \\sum a/(4+a <sup>2</sup> ) \\leq 1.",
  "Solution_1": "you joined CWMO?\r\nI got a silver this yr..\r\n\r\nThe official solution used a substitution.\r\nanyone wants me to post it?",
  "Solution_2": "West Math Olympiad?",
  "Solution_3": "[quote=\"liyi\"]West Math Olympiad?[/quote]\r\n\r\nyep, something like Western China Mathematical Olympiad",
  "Solution_4": "Very beautiful and quite challenging. Do you see anything similar with Usamo 2003 or Japan 1997?\r\n   Let 1/(1+a_i)=x_i. Then a_1=(x_2+x_3+x_4+x_5)/x_1 and so on. The inequality becomes sum x_1*(x_2+x_3+x_4+x_5)/(4x_1^2+(x_2+x_3+x_4+x_5)^2)<=1. We prove that this is true for all x_i, not necessarily with sum 1. Since it is homogenuous, we can assume that sum x_i=5. It becomes sum x_i*(5-x_i)/(4+(x_i-1)^2)<=5. It is obvious that LHS is at most sum x_i(5-x_i)/4 and from now on everything is clear since sum x_i=5 and we can apply Cauchy ( rest in peace; beautiful inequality had he discovered).",
  "Solution_5": "For $x>0$, we have $\\displaystyle \\frac{x}{4+x^2} \\leq \\frac{1}{20}\\left(1+\\frac{15}{1+x}\\right)$. In fact, this is equivalent to $(x-4)^2(x+4)\\geq0$.\r\n\r\nAdding, we get $\\displaystyle  \\sum_{i=1}^5\\frac{a_i}{4+a_i^2} \\leq \\frac{1}{20}\\sum_{i=1}^5 \\left(1+\\frac{15}{1+a_i}\\right) = 1$.",
  "Solution_6": "[quote=\"fuzzylogic\"]For $x>0$, we have $\\displaystyle \\frac{x}{4+x^2} \\leq \\frac{1}{20}\\left(1+\\frac{15}{1+x}\\right)$. [/quote]\n\nSomebody asked me how I came up with this inequality. I have to give credit to [b]hxtung[/b]. I learned this technique from his post  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180]Poland 1996 inequality[/url].  This technique can be applied to Japan 1997 and USAMO 2003 as well. \n\nFor this particular problem,  I expect that  $\\displaystyle \\frac{x}{4+x^2} \\leq \\frac{ax+b}{1+x}$ for some $a,b$. This is equivalent to \n$f(x) = (ax+b)(4+x^2) - x (1+x) \\geq 0$, where $f(x)$ is a cubic polynomial.\n\nSo how do we find $a,b$? I expect the equality holds when all $a_i$ are equal to 4 (from the given condition). So I expect $f(4)=0$. Further, I expect $(x-4)^2$ divides $f(x)$, otherwise $f(x)\\geq0$ can't be true. So $f'(4)=0$ too. From $f(4)=f'(4)=0$, I found that $a=\\frac{1}{20}$ and $b=\\frac{4}{5}$. The rest is just verify that the inequality is true.\n\n\nThis problem can be extended to $n$ variables as follows:\n[quote]Let $a_1,a_2,\\ldots, a_n > 0$ satisfying $\\displaystyle \\sum_{i=1}^n \\frac{1}{1+a_i}=1$. Prove that\n\\[\n\\sum_{i=1}^n \\frac{a_i}{(n-1)+a_i^2} \\leq 1.\n\\][/quote]\r\n\r\nCheers! :D  :D"
}
{
  "Tag": [
    "ratio",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Numbers $ 1,2,...,64$ are written on a $ 8\\times 8$ board. For every two numbers $ a,b$ with $ a>b$ in the same row or column, the ratio $ \\frac{a}{b}$ is calculated. The characteristic of the board is defined as the least of these ratios. Find the greatest possible value of a characteristic.",
  "Solution_1": "Is the answer 64/7 or am I being stupid?",
  "Solution_2": "It has to be at most $ \\frac{64}{14}$ (there are only 13 numbers less than 14, so 14 appears in the denominator of some ratio).",
  "Solution_3": "Ah, sorry - somehow I completely missed \"or column\" when I read that!\r\n\r\n64/14 is indeed right.",
  "Solution_4": "[hide=\"Ponderings\"]Let the characteristic be $ r$. In the row containing 1, note that the next smallest entry is at least $ r$. The next after that is at least $ r^2$; otherwise, the ratio between the two would be less than $ r$, meaning that $ r$ would not be the characteristic. Then by similar reasoning, the largest entry in the row is at least $ r^7$. To maximize $ r$, you want the largest entry in this row to be as high as possible. Therefore, $ r<\\sqrt[7]{64}$.\n\nYou can do better by considering the maximum among the least entries in each row. This maximum is at least 8, so $ 8r^7<64$.\n\nFurther note that since the entries are integers, you will always be considering the ceiling of the $ r$ powers at each step, so that deflates it even more. As an example, the second smallest entry in the row containing 1 is at least 2. The maximum among second smallest entries in each row is at least 16. Then we have $ 16r^6<64$.\n\nI'm pretty sure this continues to the 7th smallest (or second largest) entry in the row, suggesting the maximum characteristic is less than $ \\frac{64}{56}$.[/hide]\r\n\r\nThis method doesn't lend itself to determine whether characteristics can actually be achieved, though."
}
{
  "Tag": [
    "calculus",
    "integration",
    "algebra",
    "function",
    "domain",
    "Ring Theory",
    "algorithm"
  ],
  "Problem": "Let [b]F [/b]be a field. State under what circumstances, if any, [b]F[/b] is respectively an Euclidean Domain, Principal Ideal Domain, Uniqe Factorization Domain. Justify your answers\r\n\r\nThank you very much in advanced",
  "Solution_1": "What do you mean? If you have a certain field which satisfies the criteria for those domains, it'll be that.\r\n\r\nFor instance construct a norm over your field as follows \r\n$ N(a) \\equal{} 1$ for any $ a \\not \\equal{} 0$\r\n$ N(0) \\equal{} 0$\r\nThen your field is an Euclidean Domain.\r\n\r\nTo be a PID, its really easy to see, since every element in a field is a unit. Hence with the Euclidean Algorithm there won't be any non-zero remainders...",
  "Solution_2": "There's no need to define a Euclidean valuation since a field is vacuously Euclidean: there are no nonzero remainders upon division. Since Euclidean => PID => UFD all else follows. But, of course, it's trivial to directly verify the other claims  too."
}
{
  "Tag": [],
  "Problem": "Roslyn has ten boxes. Six of the boxes contain pencils, three of\nthe boxes contain pens, and two of the boxes contain both pens\nand pencils. How many boxes contain neither pens nor pencils?",
  "Solution_1": "Call the number of boxes with neither pens nor pencils $ x$.  Using [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]PIE[/url], we know that $ 6\\plus{}3\\minus{}2\\plus{}x\\equal{}10\\implies x\\equal{}\\boxed{3}$."
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics",
    "Bulgaria"
  ],
  "Problem": "Find the greatest positive integer $n$ such that we can choose $2007$ different positive integers from $[2\\cdot 10^{n-1},10^{n})$ such that for each two $1\\leq i<j\\leq n$ there exists a positive integer $\\overline{a_{1}a_{2}\\ldots a_{n}}$ from the chosen integers for which $a_{j}\\geq a_{i}+2$.\n\n[i]A. Ivanov, E. Kolev[/i]",
  "Solution_1": "Nobody?  :)",
  "Solution_2": "[hide=\"Does it help?\"]$ 1)$ Notice that for all the $ 2007$ numbers $ a_1 \\ge 2$ \n$ 2)$ If $ n\\equal{}4k$ then a choice covering the maximum amount of pairs is $ 24682468\\cdots2468$ covering $ 6 \\cdot \\frac{k(k\\plus{}1)}{2}$ pairs \n$ 3)$ There are $ \\frac{n(n\\plus{}1)}{2}$ pairs[/hide]",
  "Solution_3": "[quote=\"mszew\"]\n$ 2)$ If $ n \\equal{} 4k$ then a choice covering the maximum amount of pairs is $ 24682468\\cdots2468$ covering $ 6 \\cdot \\frac {k(k \\plus{} 1)}{2}$ pairs \n[/quote]\r\n\r\nUmm... how can you prove this is maximum?\r\n\r\nMoreover, the other $ n$s won't cover this many pairs...",
  "Solution_4": "Any complete solutions?  :roll:",
  "Solution_5": "[url=https://dgrozev.wordpress.com/2022/07/10/packing-balls-method/]A solution and some other stuff[/url] related to this problem."
}
{
  "Tag": [
    "rotation",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "An $m$x$n$ board is divided into unit squares. An L-triomino consists of three unit squares: a central square and two outer squares. An L-triomino is located in the upper left corner of the board, the central square covering the corner square of the board. In a move one may rotate the triomino around the midpoint of one of its outer squares by multiples of 90 degrees. For which $m$ and $n$ is it possible to move the triomino to the lower right corner of the board using a finite number of moves?",
  "Solution_1": "Anybody solved it?"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all integers triples (p,q,r) such that:\r\n$ r|p^q\\plus{}1 , q|r^p\\plus{}1 , p|q^r\\plus{}1$",
  "Solution_1": "An idea:I will not write the whole solution but an effective idea is to consider the multiplicative order (http://en.wikipedia.org/wiki/Multiplicative_order) of,p mod r, r mod q and q mod r. Then using elementary properties you will reach the solution\r\nIf you have questions about my post ask\r\n\r\nSolution\r\n[hide]The solution is (2,3,5) and permutations[/hide]\r\n\r\nDaniel"
}
{
  "Tag": [],
  "Problem": "I'm trying to come up with a formula to calculate the radius for the largest circle whose interior contains exactly n numbers of lattices.  n =0,1,2,3 through 9.  \r\nI can draw the circles with 0, 1, 2, 3 etc lattice points but I can't seem to come up with a formula whereby I can just enter the n value (0 through 9) and it computes the radius.  \r\nI figure there must be a formula perhaps one formula for even numbers for n and one for odd numbers.  The reason I say this is because for the even numbers for n the center point of the cicle is between two lattice points where the odd n's form triangles for the center point.\r\nMayber theres just some simple pattern that explains it.\r\nPLEASE HELP AS THIS IS DRIVING ME CRAZY!",
  "Solution_1": "Can someone please help me on this problem.  \r\n\r\ni know that there must be a formula that will work for n=even and another for n=odd.  I can graph the various points but need to come up with a formula for computing the radius.\r\n\r\nThanks for any help offered,",
  "Solution_2": "What exactly are lattices inside a circle?",
  "Solution_3": "latices, integer coordinates.",
  "Solution_4": "Well I know pretty much nothing about lattice points and such, but here's what seems to be a good website if you haven't seen it: http://mathworld.wolfram.com/PointLattice.html.",
  "Solution_5": "I do not think there will be a \"nice\" formula in the way that you hope.  For each number of lattice points, there is such a radius, but it is unlikely to have a nice expression in terms of n, I'm afraid.",
  "Solution_6": "Are these any particular points, like say the first n along the number line (ok its trivial but just to give an example) or just n random points?"
}
{
  "Tag": [],
  "Problem": "how do you prove that the interior angles of a triangle are concurrent using ceva's theorum?",
  "Solution_1": "Try to look around at the forums to determine the most appropriate place to post questions like this.  I'll move this to a high school forum.",
  "Solution_2": "Please don't post the same thing twice :!:"
}
{
  "Tag": [
    "Putnam",
    "linear algebra",
    "matrix",
    "college contests"
  ],
  "Problem": "Alan and Barbara play a game in which they take turns filling entries of an initially empty $ 2008\\times 2008$ array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?",
  "Solution_1": "Pick two columns.  If Alan plays in one column, Barbara places the same real number in the corresponding position int he other column.  If Alan plays in neither column, Barbara makes an arbitrary move in another column.  Since there are an even number of positions not in either column, when the game ends two columns are identical and Barbara wins.",
  "Solution_2": "Yay!  That's what I said!  (Except much wordilier.  \"Wordilier\" should be a word.)\r\n\r\n(BONUS: The professor who runs the Putnam at my school is named Barbara.)",
  "Solution_3": "Barbara can even pair off all the columns, and force a rank of no more than 1004.",
  "Solution_4": "The natural question people down at my school were asking was this: what if the matrix was $ n \\times n$ and $ n$ was odd? Picking two columns doesn't work because Alan can run Barbara out of the remaining spaces.",
  "Solution_5": "Well it was already given n = 2008, an even. However, whether n is even or odd doesn't matter. \r\n\r\nLet i denote the ith row and let j denote the jth column of the matrix. \r\n\r\nThe solution I put was if Alan fills entry a_ij with a number where 0 < i, j <= 2008, then if Barbara fills out entry a_IJ where I = 2009 - i and J = j with the same number, the matrix will at least have two identical rows, and by the property of the determinants, the determinant of the matrix will be 0. Therefore, Barbara will always win with this strategy. \r\n\r\nEDIT: Looking in my reference, it seems so long as the matrix has two identical rows or columns for any n x n matrix, the determinant will be 0. \r\n\r\nEDIT 2: Will I still get full credit if I justified that \"since at least a pair of rows are identical, the determinant of matrix is 0\" by \"property of determinants\"? Or was I supposed to provide proof to this property as well?",
  "Solution_6": "I made up a first-player win proof... and forgot to check that who moves. This should work for any odd n (on even n a mirroring strategy works for a second-player win).\r\n\r\nSince exchanging rows and columns does not affect the magnitude of the determinant I'll just order them by move first played in. Alan's first move is arbitrary, and Barbara can either play in the same row/column or in a new row, new column. Regardless, Alan plays a number so that the position\r\n\r\nx1 x1\r\n--- x2\r\n\r\nis reached, where x1, x2 are real numbers. Barbara may fill in the -- slot or play in a new row, new column. If she plays in the spot and doesn't fill in x2, then the two rows are LI, so she should play x2 for optimality (we'll get back to this in the end). Regardless of which she does, Alan forces the following position:\r\n\r\nx1 x1 ---\r\nx2 x2 ---\r\n--- --- x3\r\n\r\nNow consider if Barbara plays in any of the four marked spots - Alan can immediately play a new number in the other spot to force LI of the rows/columns, for example\r\n\r\nx1 x1 ---\r\nx2 x2 ---\r\nx4 x5 x3\r\n\r\nAnd so Barbara is forced into yet a new row/column. Alan again copies her move to force:\r\n\r\nx1 x1 --- ---\r\nx2 x2 --- ---\r\n--- --- x3 x3\r\n--- --- --- x4\r\n\r\nand her dilemma continues. Notice when the second x4 is placed, all 8 empty positions are forbidden to her - Alan can force all 4 rows/columns to be LI by picking suitable new numbers. Thus assuming she does not give up LD on rows/columns, eventually the position becomes a 2x2-block diagonal matrix with equal entries on each row, with one remaining row. Alan plays a new number in it (since 4 moves per block), and now Barbara must suicide until Alan forces every single row/column to be LI and wins.\r\n\r\nBack to the possibility that Barbara gives up making two rows LD earlier, to wit\r\n\r\nx1 x1 ---\r\nx2 x3 ---\r\n--- --- x4\r\n\r\nwith x4 being Alan's new move. Now rows 1/2 are LI as are columns 1/2, and suppose Barbara plays in this submatrix, case 1:\r\n\r\nx1 x1 ---\r\nx2 x3 ---\r\nx5 --- x4\r\n\r\nAlan can put a suitably huge number in the bottom row such that columns 1-3 are all LI, the other case is:\r\n\r\nx1 x1 x5\r\nx2 x3 ---\r\n--- --- x4\r\n\r\nwhence Alan places another huge number in the rightmost column. Thus giving up rows cannot improve Barbara's condition and we (on no errors in the above) can conclude it's a first-player win.\r\n\r\nPS: I wrote this on my A2, except I forgot that when the entire matrix was 2x2-block diagonal... Alan moves and suicides next.",
  "Solution_7": "I tried something a bit different.  I had Barbara responding with 0's based on where Alan played.  If Alan played in an m,n spot which wasn't on the boundary, Barbara would play in the m+1, n-1 spot.  Or if there was already a 0 on that diagonal, then play in the m+1, n+1 spot.  On the boundary the idea was similar, broken up into cases to make sure the move happened on the diagonal if possible.  Would this be enough to guarantee that the matrix' determinant was 0?",
  "Solution_8": "Wow, I can't believe I missed this one! And, I thought A1 was way too easy!",
  "Solution_9": "I don't really like this problem.  It reminds me of the kind of problem I write when I have writer's block: something that looks kind of cool, but reduces down to something quite trivial.  I'll take the 10 free points, though.",
  "Solution_10": "[color=red]This proof is wrong\n\n\nFirst:  any 2n*2n matrix can have its determinant defined by the difference in the sums of the left-wise and right-wise diagonal products.  Then, any left-wise and right-wise diagonal intersect in either 2 or 0 elements.\n\nLet the first player place any number in the M_i,j position.  \n\nBy definition, the left-wise and right-wise diagonals that intersect at M_i,j also intersect at M_p(i),p(j), where p(x) = (x+1004)mod2008.  \n\nIf Barbara places the integer \"0\" at M_p(i),p(j), then she makes the diagonal's products equal 0 for both diagonals.  \n\nIn addition, as long as she continues that pattern, every diagonal will eventually have a product of 0, so both sums will equal 0 and the difference of sums will equal 0.[/color]",
  "Solution_11": "[quote=\"collegebookworm\"]First:  any 2n*2n matrix can have its determinant defined by the difference in the sums of the left-wise and right-wise diagonal products. [/quote]\r\nThis is not true. The determinant of a $ 4\\times 4$ matrix has $ 24$ terms, not $ 8$ terms.",
  "Solution_12": "[quote=\"Kent Merryfield\"][quote=\"collegebookworm\"]First:  any 2n*2n matrix can have its determinant defined by the difference in the sums of the left-wise and right-wise diagonal products. [/quote]\nThis is not true. The determinant of a $ 4\\times 4$ matrix has $ 24$ terms, not $ 8$ terms.[/quote]\r\nFrosh's mistake...folly...I guess I shouldn't take Calc 2 then.",
  "Solution_13": "What does that have to do with linear algebra? They're unrelated classes, and there's no shame in getting something wrong when you haven't seen it yet.",
  "Solution_14": "A fellow test-taker at my university told me what he wrote for his answer, which makes sense to me.\r\nWhen Alan places a number in any spot in the array, Barbara places the same number in the column, but one row up or down.  That was she forces the rows to be linearly dependent, and the resulting determinate is zero.  I think that's the gist of what he said.\r\n\r\nI spent a long time thinking about what would make the determinate easier to calculate, for what reason I don't know.  Also, if Alan was particularly dim, Barbara could fill a whole row up with zeros and win the game.",
  "Solution_15": "[quote=\"jaws1216\"]I tried something a bit different.  I had Barbara responding with 0's based on where Alan played.  If Alan played in an m,n spot which wasn't on the boundary, Barbara would play in the m+1, n-1 spot.  Or if there was already a 0 on that diagonal, then play in the m+1, n+1 spot.  On the boundary the idea was similar, broken up into cases to make sure the move happened on the diagonal if possible.  Would this be enough to guarantee that the matrix' determinant was 0?[/quote]\r\nMy gut would say no. Regardless, if you don't know the answer to your question, it means you didn't sufficiently prove it on the exam, thus won't get points (1 if you're very lucky). So it doesn't really affect your score whether that works or not.\r\n\r\n\r\nOn another note, I have a question stemming from this problem:\r\nHow much attention does Barbara actually have to pay? Specifically, how many moves must she make non-arbitrarily? 2008 is easy--just do the two identical rows method. Currently I think that this is also the best we can achieve, but I'm not entirely sure how to prove it. Any thoughts?",
  "Solution_16": "[quote=\"Sly Si\"]I don't really like this problem.  It reminds me of the kind of problem I write when I have writer's block: something that looks kind of cool, but reduces down to something quite trivial.  I'll take the 10 free points, though.[/quote]\r\n\r\nsame, what a POS problem.",
  "Solution_17": "I paired up all the rows by identifying row i with row 2009 - i, except I...accidentally wrote 2008 - i. I later stated that row 1 was paired with row 2008, so it was clearly a typo. How bad is this?",
  "Solution_18": "^uh oh.  I did the same thing",
  "Solution_19": "Here's another quick proof.\r\n\r\nLet Alan place his number, say $ r$, at $ a_{ij}$. Have Barbara place the same number at $ a_{(2008\\minus{}i\\plus{}1)j}$. It is clear that Barbara can always place a number down. After this process, the matrix will be symmetrical vertically. Since $ 2008$ is even, then upon determinant expansion along the first column, the sign of the term with $ a_{i1}$ will be the opposite of $ a_{(2008\\minus{}i\\plus{}1)1}$. The determinants of the submatrices corresponding to these elements will be identical. Thus, the determinant will end up as zero.",
  "Solution_20": "What if 2008 was replaced by 2009?  Who has a winning strategy?",
  "Solution_21": "Try it with 3 before asking about 2009. And look at the 2002 Putnam - it's not exactly the same problem, but ...",
  "Solution_22": "We are discussing this on an internal mailing list.  \r\nAccording to the discussion, B can guarentee the rank of the matrix is no more than ceiling(n/2).",
  "Solution_23": "If $A$ places a real number in row 1, $B$ places $0$ in row 2 underneath $A$. If $A$ places a real number in row 2, $B$ places $0$ in row 1, above $A$. By generalized laplace, the determinant is $0$.\n\nWhy didn't I just make it so two rows are the same? I don't know   :| "
}
{
  "Tag": [
    "trigonometry",
    "algebra",
    "polynomial",
    "absolute value",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "$A \\in \\mathcal{M}_{n}(\\mathbb{Z})$, $\\parallel A\\parallel _{\\infty}=\\max_{\\parallel X\\parallel =1}\\parallel AX\\parallel $, where $\\parallel . \\parallel $ is the euclidian norm.\r\nShow that either $\\parallel A\\parallel _{\\infty}$ could be written $2 \\cos (\\pi /k)$ with $k \\in \\mathbb{N}$, $k\\geqslant 2$, or $\\parallel A\\parallel _{\\infty}\\geqslant 2$.",
  "Solution_1": "Hint: use Kronecker's theorem on polynomials whose roots have modulus 1.",
  "Solution_2": "Nice hint harazi, I hope that won't spoil the fun :lol:",
  "Solution_3": "Surely not, there is still some work to do after that!",
  "Solution_4": "Opps, I think I didn't correctly evaluate the difficulty. First, I thought it is a direct consequence of Kronecker's theorem, but it's not really like that. Ok, here's my proof. Suppose that the norm is smaller than 2 and introduce $B=(0,A,^{t}A, 0)$ notation by blocs, the first two elements being the first line.  It is not difficult to see that $C$ has the same norm as $B$ and thus smaller than 2. The idea is that hopefully, $C$ has all eigenvalues real numbers, because it is symetric. Moreover, any eigenvalue is surely in $(-2,2)$ because its norm is smaller than $2$. Because $C$ has integer coefficients, its characteristic polynomial is monic with integer coefficients and  all roots in $(-2,2)$. A direct application of Kronecker's theorem shows that all of its roots are of the form $2\\cos (2\\pi r)$ for some rational $r$. Because $C$ is diagonalisable in orthonormal basis, it is enough to prove that the maximal absolute value of a root of $P$ must be of the form $2\\cos (\\pi/k)$ for some integer $k$. Of course, using the irreducibility of the cyclotomic polynomial, we infer that if $2\\cos (2\\pi u/v)$ is a root of $P$, so is $2\\cos(2\\pi/v)$. A small case analysis (whether $v$ is odd or not) shows that indeed the maximal modulus of a root of $P$ has the form $2\\cos(\\pi/k)$ and this finishes the proof, except that as usual I forgot the case when $P$ is of the form $X^{2n}$, that is $C$ is nilpotent and symmetric, thus 0 and we find again the desired form. [/code]",
  "Solution_5": "Nice proof,\r\nA detail\r\n[quote=\"harazi\"]... It is not difficult to see that $C$ has the same norm as $B$ and thus smaller than 2....[/quote]\r\nyou meant probably \"It is not difficult to see that $B$ has the same norm as $A$ and thus smaller than 2\", etc. because I don't see any $C$ matrix.",
  "Solution_6": "Yes, sorry, C is so close to B on french keyboard..."
}
{
  "Tag": [
    "function",
    "algebra",
    "difference of squares",
    "special factorizations"
  ],
  "Problem": "Refer to the picture.\n\n\n\nI made this question up because of something my friend showed me.\n\n\n\nI couldn't think of a way to describe this question so I had to draw it with MS Paint.  Below in the drawings are actually all squares although some might look like rectangles.  Heh...\n\n\n\nAnyways, my question is...\n\nIn Square 1, there is 1 square that can be found.\nIn Square 2, there are 5 squares that can be found.\nIn Square 3, there are 14 squares that can be found.\nIn Square 4, there are 30 squares that can be found.\n\na) Make a formula that will be able to solve for Square N where N is any number from 1 to  :inf: .\nb) Solve for Square 50.\n\n\n\nNote: I'm not sure if this question can be solved however...\n\nI think this is an easy question but I couldn't solve it ... as always.\n\n\n\nAnother friend of mine showed me the answer to the formula from a book...(highlight for spoilers)\n\n\n\n[hide]n(n+1)(2n+1)/6[/hide]\n\n\n\nEdit : Typos",
  "Solution_1": "Yep, that formula your friend showed you is correct.\r\n\r\nHere's a question that will help you begin the path to figuring out why that is so:\r\n\r\nWhat is (the number of squares in the 50th square) - (the number of squares in the 49th square)?\r\n(Of course, 50 could be replaced there by any number, and the 49 by a number one less.)",
  "Solution_2": "Thx for your advice.\r\n\r\nI thought about that as well.\r\n\r\nThe difference of squares between the consecutive large squares is always the square of a number.\r\n\r\nFor example, Square 1 has 1 square in total.  Square 2 has 5 squares in total therefore the difference is 4 which is the square of 2.  This pattern repeats.\r\n\r\nBut after that, I can't really figure out how the factors of the formula all get pieced together.",
  "Solution_3": "Okay, that's great -- so in other words, we could describe the number of squares in the [i]n[/i]th diagram as the sum of the first [i]n[/i] square numbers, right?\r\n\r\nNow, one thing you can do is to check that the formula you have obeys that law -- that, starting with any value of n, if you increase n by 1, the function will increase in value by (n + 1)^2.  This is really just an implimentation of the mathematical method known as induction.  See if you can make it work in this case.",
  "Solution_4": "[quote=\"JBL\"]\nNow, one thing you can do is to check that the formula you have obeys that law -- that, starting with any value of n, if you increase n by 1, the function will increase in value by (n + 1)^2.  This is really just an implimentation of the mathematical method known as induction.  See if you can make it work in this case.[/quote]\r\n\r\nWhen I did my work, I tried doing something just like (n+1)^2 and that's where I halted all my steps because I couldn't find any more solutions.",
  "Solution_5": "never mind... JBL already posted it.",
  "Solution_6": "more help?\r\n\r\nthx"
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Suppose $ A$ has a right-inverse $ B$. Then $ AB \\equal{} I$ leads to $ A^{T}AB \\equal{} A^{T}$ or $ B \\equal{} (A^{T}A)^{\\minus{}1}A^{T}$. But that satisfies BA = I; it is a left inverse. What step is not justified? \r\n\r\nCan someone point this one out, I can't seem to see any mistakes. Assuming A is m by n, it seems to work out. \r\n\r\nTHanks",
  "Solution_1": "$ A^T A$ is not necessarily invertible, is it?",
  "Solution_2": "Hi bythecliff,\r\nLet $ A$ be a $ (m,n)$ matrix and $ B$ a $ (n,m)$ matrix.\r\nif $ m\\not\\equal{}n$ you can't have $ AB\\equal{}I_m$ and $ BA\\equal{}I_n$ because $ rank(AB),rank(BA)\\leq{inf(m,n)}$.\r\nIf $ m\\equal{}n$ it's wellknown that $ AB\\equal{}I$ implies $ BA\\equal{}I$: indeed\r\n$ AB\\equal{}I$ implies $ det(A)\\not\\equal{}0$ implies $ A$ is invertible implies $ B\\equal{}A^{\\minus{}1}$."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "search",
    "geometry proposed"
  ],
  "Problem": "The incircle k of triangle ABC touches the sides at points A',B',C'.\r\nFor any point K on k, let d be the sum of the distances from K to the sides of the\r\ntriangle A'B'C'. Prove that KA + KB + KC > 2d.\r\n\r\nI have an algebraic solution, but I am looking for a solution using Erd\u00f6s Mordell",
  "Solution_1": "Let distances from $ K$ to the $ B^{'}C^{'},C^'A^',A^'B^',BC,CA,AB$ be $ x,y,z,a,b,c$. Then we have $ x^2 = bc,y^2 = ca,z^2 = ab$.(this can be proved as follow: Let $ P,Q,R$ be projection of $ K$ on $ AB,B^'C^',CA. \\angle KQP = \\angle KC^'P = \\angle KB^'C^' = \\angle KRQ$ and similarly, $ \\angle KPQ = \\angle KQR$. Hence triangle $ KQP$ and triangle $ KRQ$ are simialr, and we obtain $ x^2 = bc$). By Erd\u00f6s Mordell theorem, $ KA + KB + KC\\geq 2(a + b + c)\\geq 2(\\sqrt {bc} + \\sqrt {ca} + \\sqrt {ab}) = 2(x + y + z) = 2d$. Obviously, Equal isn't hold. Hence $ KA + KB + KC > 2d$.",
  "Solution_2": "Thanks Euler. :)",
  "Solution_3": "Sorry but I can't contain my curiosity:\r\n\r\nWhat are F1 and F2 in your avatar (I am supposing that O and N are the circuncircle and the Nagel point)  :oops:",
  "Solution_4": "[quote=\"m.candales\"]Sorry but I can't contain my curiosity:\n\nWhat are F1 and F2 in your avatar (I am supposing that O and N are the circuncircle and the Nagel point)  :oops:[/quote]\r\n$ F_1,F_2$ is two Fermat point and $ N$ is nine-point center. My avatar is picture of Lester theorem( http://www.mathlinks.ro/viewtopic.php?search_id=298908544&t=16494)"
}
{
  "Tag": [
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "$f: R\\to R$\r\n$fx)= x+m$, for x<=1\r\n        $2mx-1$, for x>1\r\nIf the function is surjective, what is the $m$'s interval ?",
  "Solution_1": "$f$ automatically takes all values in $(-\\infty,m+1]$. We then need that $2mx-1$ for $x>1$ take values in the rest of $\\mathbb{R}$. So the minimum of $2mx-1$ must be $m+1$ or less. When $x\\to1$, $2mx-1\\to 2m-1$, so:\r\n\r\n$2m-1\\leq m+1$\r\n\r\nyielding,\r\n\r\n$m\\leq2$\r\n\r\nHowever, it's obvious that $2mx-1$ must go to $\\infty$ for some $x$ also, which means that $m>0$. Hence,\r\n\r\n$m\\in(0,2]$"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "combinatorial geometry",
    "linear algebra unsolved"
  ],
  "Problem": "Let $ A \\in \\mathcal{M}_n(\\mathbb{C})$, $ H(A)\\equal{}\\{ X^*AX \\quad | \\quad  X \\in \\mathbb{C}^n, X^*X\\equal{}1 \\}$ ($ X^*$ denotes transpose conjugate).\r\nFor which $ n\\in \\mathbb{N}^*$, do we have the following statement:\r\n\r\nIf $ H(A)$ is the convex hull of the eigenvalues of $ A$, then $ A$ is normal.",
  "Solution_1": "Answer : For $ n\\leq 4$ \r\n This year it's really a nice problem at ENS , especially in compare with the other one ,but unfortunately I didn't succeed ... :(",
  "Solution_2": "I have a partial proof: with Facts 1-2 we can prove that the answer is YES if $ n\\leq3$. Unfortunately we cannot conclude if $ n\\equal{}4$.\r\n\r\nThere exists $ P$ unitary matrix s.t. $ P^{\\minus{}1}AP$ is triangular. Thus we may assume that $ A\\equal{}[a_{ij}]$ is upper triangular with $ a_{ii}\\equal{}\\lambda_i$. $ X^*AX\\equal{}\\sum_i{\\lambda_i|x_i|^2}\\plus{}\\sum_{j>i}a_{ij}\\bar{x_i}x_j$ where $ X\\equal{}[x_i]$ satisfies $ \\sum_i{|x_i|^2}\\equal{}1$. Let $ \\mathcal{C}$ be the convex hull of $ (\\lambda_i)$ and $ \\mathcal{B}$ its edge.\r\nThe question is: must $ A$ be a diagonal matrix ?\r\nFact 1: let $ i,j$ s.t. $ \\lambda_i\\not\\equal{}\\lambda_j$ and $ [\\lambda_i,\\lambda_j]\\subset\\mathcal{B}$. \r\nLet $ X$ s.t. $ x_k\\equal{}0$ except $ x_i\\equal{}\\alpha_i/\\sqrt{2},x_j\\equal{}\\alpha_j/\\sqrt{2}$ where $ |\\alpha_i|\\equal{}|\\alpha_j|\\equal{}1$. $ X^*AX\\equal{}\\lambda_i/2\\plus{}\\lambda_j/2\\plus{}\\bar{\\alpha_i}\\alpha_j{a_{ij}}/2$. $ X^*AX\\in\\mathcal{C}$ and $ \\lambda_i/2\\plus{}\\lambda_j/2\\in\\mathcal{B}$. Then the complex number $ \\bar{\\alpha_i}\\alpha_j{a_{ij}}$ is always in the same half plane for all $ \\alpha_i,\\alpha_j$. This fact implies that $ a_{ij}\\equal{}0$.\r\nFact 2: let distinct $ i,j$ s.t. $ \\lambda_i\\equal{}\\lambda_j$. \r\nIf $ \\mathcal{C}\\not\\equal{}\\{\\lambda_i\\}$ then $ \\bar{\\alpha_i}\\alpha_j{a_{ij}}$ is in a fixed angular sector (eventually in a half plane) for all $ \\alpha_i,\\alpha_j$. This implies that $ a_{ij}\\equal{}0$. \r\nIf $ \\mathcal{C}\\equal{}\\{\\lambda_i\\}$ then $ \\bar{\\alpha_i}\\alpha_j{a_{ij}}\\equal{}0$ and $ a_{ij}\\equal{}0$."
}
{
  "Tag": [
    "abstract algebra",
    "function",
    "algebra",
    "polynomial",
    "topology",
    "Ring Theory",
    "superior algebra"
  ],
  "Problem": "At the site http://www.math.uga.edu/~bensondj/ I found this remarkable remark:\r\n\r\n[b]Foundations sometimes matter\nThe projective dimension of $\\mathbb C(x,y,z)$ as a module over $\\mathbb C[x,y,z]$ is:\ntwo if the continuum hypothesis holds and three otherwise.[/b]\r\n\r\nI assume C(x,y,z) is a function field over the complex, and C[x,y,z] is a polynomial ring, so I understand that the former is a module over its subring. What is the \"projective dimension\" here?\r\n\r\nLiMa",
  "Solution_1": "Hi Lima ;-),\r\n\r\nis the general definition unclear?\r\n \r\nhttp://planetmath.org/encyclopedia/ProjectiveDimension.html\r\n \r\nI guess that results of homological algebra may involve foundations by the recursive construction of homotopy equivalences of arbitrary projective resolutions of a module. But I can't imagine what does this matter in the case of finite resolutions  :maybe:. Well, perhaps CH admits the construction of curious modules, which may shorten the projective dimension. :-)",
  "Solution_2": "... after a short google-session ;-)\r\n\r\nCH plays even often a role dealing with projective dimensions. For an example, see\r\n \r\nmpcs.usip.edu/faculty/vas/homepage_sa_UMDja/Semisimple_abstract.pdf\r\n \r\nBut this seems to be quite difficult to understand :/"
}
{
  "Tag": [
    "function",
    "logarithms",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Can anyone give an example of a function $ f \\in L^p(\\Bbb R)$ such that $ f \\notin L^h(\\Bbb R)$ for any $ h > p$? ($ p$ is finite)",
  "Solution_1": "Try $ f(x)\\equal{}|x\\ln^2x|^{\\minus{}1/p}$ for $ x\\in (0,1/2)$ and zero otherwise."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "function",
    "absolute value",
    "complex analysis"
  ],
  "Problem": "MAIN QUESTION:\r\n\r\n\r\nDoes      $ |(cos(R^2 cos(2 \\theta)) \\plus{} i sin(R^2 cos(2 \\theta)))| \\equal{} 1$ UNIFORMLY even though \r\nthe absolute value of  $ e^{i R^2 e^{i 2 \\theta}}$ is NOT equal to 1?\r\n\r\nAnd if so, IN WHAT CASES THEN DO A NUMBER OF THE FORM $ e^{i \\theta}$ actually have an absolute magnitude of 1? A theta value of $ R^2 cos(2 \\theta)$ seems to work (correct me if I'm wrong), but a theta value of $ R^2 e^{i 2 \\theta}$ doesn't seem to work. I'd like to know why (besides the \"you need to simplify it\" explanation - which does not help with my intuition here). \r\n==\r\n\r\nSo I want to evaluate $ \\int_0^{\\infty} e^{i z^2}$. One of the central steps is to evaluate the integral around the contour line $ z \\equal{} R e^{i \\theta}$. So I substitute in $ z \\equal{} R e^{i \\theta}$ so that I get $ e^{i R^2 e^{i 2 \\theta}} \\equal{} e^{i R^2 cos(2 \\theta) \\minus{} R^2 sin(2 \\theta)}.$. I have to show convergence to 0. This equals $ e^{i R^2 cos(2 \\theta)} e^{ \\minus{} R^2 sin(2 \\theta)} \\equal{} (cos(R^2 cos(2 \\theta)) \\plus{} i sin(R^2 cos(2 \\theta))) e^{ \\minus{} R^2 sin(2 \\theta)}$. So my sources say that the absolute value of this becomes $ e^{ \\minus{} R^2 sin(2 \\theta)},$ implying that $ |(cos(R^2 cos(2 \\theta)) \\plus{} i sin(R^2 cos(2 \\theta)))| \\equal{} 1$ throughout the entire interval. Of course, the absolute value of $ e^{i \\theta}$ is 1 as long as the number is of this form. But then, why can't the absolute value of $ e^{i R^2 e^{i 2 \\theta}}$ also be 1 throughout the range? What properties make a the absolute value of a number of the form $ e^{i \\theta}$ NOT converge to 1 for ANY value of $ \\theta$?\r\n\r\n====\r\n\r\nAnother question: Since I have to evaluate       $ \\int_0^{\\infty} e^{i z^2} dz$, where $ dz \\equal{} i e^{i \\theta} d \\theta$ - I have to multiply this equation by that factor. I probably have to do this for the final solution. But since this is bounded by i, and my only objective is to show that this integral converges to 0, this factor of dz should be effectively negligible, right?\r\n\r\n==",
  "Solution_1": "On your first question- of course not. That absolute value relation only holds when the argument of the sines and cosines is pure real. We do have that $ \\sin^2 t\\plus{}\\cos^2 t\\equal{}1$ everywhere, but that's different.\r\n\r\nThe integral you're trying to find is along the $ x$-axis, so there are no extra multiples- it's just $ \\int_0^{\\infty} e^{iz^2}\\,dz$ there.\r\n\r\nOn the circular arc $ z\\equal{}Re^{i\\theta}$, we can estimate\r\n$ \\left|\\int_{Re^{i\\alpha}}^{Re^{i\\beta}}e^{iz^2}\\,dz\\right|\\le \\int_{Re^{i\\alpha}}^{Re^{i\\beta}}\\left|e^{iz^2}\\right|\\,|dz|\\equal{} \\int_{\\alpha}^{\\beta}\\left|\\exp(iR^2e^{2i\\theta})\\right|\\cdot \\left|iRe^{i\\theta}\\right|\\,d\\theta$\r\n$ \\left|\\int_{Re^{i\\alpha}}^{Re^{i\\beta}}e^{iz^2}\\,dz\\right|\\le \\int_{\\alpha}^{\\beta}\\exp\\left(\\minus{}R^2\\cdot\\text{Im}(e^{2i\\theta})\\right)\\cdot R\\,d\\theta$\r\n\r\nNow, that $ \\sin 2\\theta$ comes from the imaginary part. The $ \\alpha$ and $ \\beta$ you want are zero and $ \\frac{\\pi}{2}$- the first quadrant piece.\r\nProving this integral goes to zero is tricky, because the function we're integrating only goes to zero away from the endpoints. Integration by parts is the most elementary way of dealing with this.",
  "Solution_2": "[quote=\"Simfish\"]MAIN QUESTION:\n\n\nDoes      $ |(cos(R^2 cos(2 \\theta)) + i sin(R^2 cos(2 \\theta)))| = 1$ UNIFORMLY even though \nthe absolute value of  $ e^{i R^2 e^{i 2 \\theta}}$ is NOT equal to 1?\n\nAnd if so, IN WHAT CASES THEN DO A NUMBER OF THE FORM $ e^{i \\theta}$ actually have an absolute magnitude of 1? A theta value of $ R^2 cos(2 \\theta)$ seems to work (correct me if I'm wrong), but a theta value of $ R^2 e^{i 2 \\theta}$ doesn't seem to work. I'd like to know why (besides the \"you need to simplify it\" explanation - which does not help with my intuition here). \n[/quote]\r\n\r\nYou forgot the $ e^{ - R^2Sin(2\\theta)}$ part:\r\n\r\n$ e^{iR^2e^{i2\\theta}} = e^{iR^2[Cos(2\\theta) + iSin(2\\theta)]}$\r\n\r\n$ = e^{iR^2Cos(2\\theta) - R^2Sin(2\\theta)}$\r\n\r\n$ = e^{ - R^2Sin(2\\theta)}\\left[Cos(R^2Cos2\\theta) + iSin(R^2Cos2\\theta)\\right]$\r\n\r\nso the absolute value of $ e^{iR^2e^{i2\\theta}}$ is $ |e^{ - R^2Sin(2\\theta)}|$ and that's going to be all kinds of values depending on R and $ \\theta$.\r\n\r\nNow, $ |e^{iw}| = |Cos[w] + iSin[w]|$  which is 1 when $ w$ is real but when w is a complex number it's:\r\n\r\n$ e^{i(x + iy)} = e^{ - y + ix}$ and so:\r\n\r\n$ |e^{i(x + iy)}| = |e^{ - y}|$\r\n\r\nSo now you see when the complex component of w is zero, then the absolute value of $ e^{iw}$ is 1.  In your case, \r\n\r\n$ w = R^2e^{i2\\theta} = R^2[Cos(2\\theta) + iSin(2\\theta)]$\r\n\r\nso the complex component is in general not zero so the absolute value of your expression won't in general be 1.",
  "Solution_3": "[quote=\"jmerry\"]\nProving this integral goes to zero . . . [/quote]\r\n\r\nFirst I'd make the substitution $ \\phi\\equal{}2\\theta$ and obtain:\r\n\r\n$ \\frac{R}{2}\\int_0^{\\pi} Exp[\\minus{}R^2 Sin(\\phi)]d\\phi$\r\n\r\nNow we wish to find an upper bound for this integral.  Note the plot below:\r\n\r\n[img]http://img87.imageshack.us/img87/3495/plotaik6.jpg[/img]\r\n  \r\n\r\nOn Blue:  $ Sin(\\phi)\\geq \\frac{2}{\\pi} \\phi$\r\n\r\nOn Green:   $ Sin(\\phi)\\geq \\minus{}\\frac{2}{\\pi} \\phi\\plus{}2$\r\n\r\nThen I can write:\r\n\r\n$ \\left|\\frac{R}{2}\\int_0^{\\pi} Exp[\\minus{}R^2 Sin(\\phi) d\\phi\\right|\r\n\\leq \\frac{R}{2}\\left(\\left|\\int_0^{\\pi/2} Exp[\\minus{}R^2\\frac{2\\phi}{\\pi}]d\\phi\\right|\\plus{}\r\n\\left|\\int_{\\pi/2}^{\\pi}Exp[\\minus{}R^2(\\minus{}\\frac{2}{\\pi}\\phi\\plus{}2)]d\\phi\\right|\\right)$\r\n\r\nevaluating the integrals I obtain:\r\n\r\n$ \\left|\\int_0^{\\pi/2}R Exp[\\minus{}R^2 Sin(2\\theta)]d\\theta\\right|\\leq \\frac{R}{2}\\left[\\frac{\\pi}{2R^2}(1\\minus{}e^{\\minus{}R^2})\\plus{}\\frac{\\pi}{2R^2}(1\\minus{}e^{\\minus{}R^2})\\right]$\r\n\r\nor:\r\n\r\n$ \\left|\\int_0^{\\pi/2}R Exp[\\minus{}R^2 Sin(2\\theta)]d\\theta\\right|\\leq \\frac{\\pi}{R}\\left(1\\minus{}e^{\\minus{}R^2}\\right)$\r\n\r\nwhich goes to zero as $ R\\to\\infty$"
}
{
  "Tag": [
    "Putnam",
    "Harvard",
    "college",
    "college contests"
  ],
  "Problem": "As I know, of the 5 fellows, there'll be one receiving the putnam scholarship to Harvard, but I can't find past scholars. Anyone knows? and how would they decide who'll receive the scholarship?",
  "Solution_1": "(At least long ago, when I was Putnam scholar)\r\nThe advisors/professors of the 5 Putnam Fellows write recommendations, and one is selected based on that.\r\n\r\nI think admission to graduate study at Harvard is in practice never denied to that winner.  Even before he completes his bachelor's degree: when I was at Harvard, another Putnam scholar began there immediately without completing his bachelor's degree.\r\n\r\nIn fact the fellowship (one year tuition plus generous allowance for living and expenses at Harvard) need not be used for graduate study in mathematics.  It has sometimes been used for undergraduate study (by a student already at Harvard), and it has at least once been used for Harvard Medical School.\r\n\r\nThe volumes published by the MAA on the Putnam have listings of the winners at the back, and the holders of the Putnam Fellowship are noted in the first volume (1938--1964) but not in the later ones.",
  "Solution_2": "[quote=\"transseries\"](At least long ago, when I was Putnam scholar)\nI think admission to graduate study at Harvard is in practice never denied to that winner.  Even before he completes his bachelor's degree: when I was at Harvard, another Putnam scholar began there immediately without completing his bachelor's degree.[/quote]\r\n\r\nWhat?!  You can do that?  Can you go to grad school in math without completing a bachelor's degree first?",
  "Solution_3": "Yes, it occasionally happens, but for obvious reasons it's quite rare.",
  "Solution_4": "[quote=\"sexynsmartjenny\"]What?!  You can do that?  Can you go to grad school in math without completing a bachelor's degree first?[/quote]\r\n\r\nYou can if the graduate school admits you.",
  "Solution_5": "[quote=\"transseries\"](At least long ago, when I was Putnam scholar)\nThe advisors/professors of the 5 Putnam Fellows write recommendations, and one is selected based on that.\n\nI think admission to graduate study at Harvard is in practice never denied to that winner.  Even before he completes his bachelor's degree: when I was at Harvard, another Putnam scholar began there immediately without completing his bachelor's degree.\n\nIn fact the fellowship (one year tuition plus generous allowance for living and expenses at Harvard) need not be used for graduate study in mathematics.  It has sometimes been used for undergraduate study (by a student already at Harvard), and it has at least once been used for Harvard Medical School.\n\nThe volumes published by the MAA on the Putnam have listings of the winners at the back, and the holders of the Putnam Fellowship are noted in the first volume (1938--1964) but not in the later ones.[/quote]\r\n\r\nThank you for the info! I ask about this since I am quite confident to become a fellow this year or next year, and going to harvard is my dream. getting the scholarship helps my admission to harvard much easier.",
  "Solution_6": "[quote=\"ohhotgirl\"][quote=\"transseries\"](At least long ago, when I was Putnam scholar)\nThe advisors/professors of the 5 Putnam Fellows write recommendations, and one is selected based on that.\n\nI think admission to graduate study at Harvard is in practice never denied to that winner.  Even before he completes his bachelor's degree: when I was at Harvard, another Putnam scholar began there immediately without completing his bachelor's degree.\n\nIn fact the fellowship (one year tuition plus generous allowance for living and expenses at Harvard) need not be used for graduate study in mathematics.  It has sometimes been used for undergraduate study (by a student already at Harvard), and it has at least once been used for Harvard Medical School.\n\nThe volumes published by the MAA on the Putnam have listings of the winners at the back, and the holders of the Putnam Fellowship are noted in the first volume (1938--1964) but not in the later ones.[/quote]\n\nThank you for the info! I ask about this since I am quite confident to become a fellow this year or next year, and going to harvard is my dream. getting the scholarship helps my admission to harvard much easier.[/quote]\r\n\r\n\r\n\r\nwow, what is your name?\r\n\r\nIf you have the ability to become a fellow, I don't think there is any difficulty for you to get into Harvard.\r\n\r\nPutnam fellows never need to worry about grad admissions",
  "Solution_7": "[quote=\"EulerFish\"]Putnam fellows never need to worry about grad admissions[/quote]  What basis do you have for making this statement?  (My guess: none whatsoever.)",
  "Solution_8": "[quote=\"JBL\"][quote=\"EulerFish\"]Putnam fellows never need to worry about grad admissions[/quote]  What basis do you have for making this statement?  (My guess: none whatsoever.)[/quote]\r\n\r\nI don't mean that the title \"putnam fellow\" can guarantee admissions. \r\n\r\nWhat I tried to say was that if someone has the ability to be a putnam fellow, he should have made a background strong enough to earn admissions from top grad schools without much difficulty.",
  "Solution_9": "No, it doesn't work that way. The Putnam is a very strong positive factor, but the top schools will care if you screw up otherwise- and people do. Like me.",
  "Solution_10": "[quote=\"jmerry\"]No, it doesn't work that way. The Putnam is a very strong positive factor, but the top schools will care if you screw up otherwise- and people do. Like me.[/quote]\r\nHow did you \"screw up\"?",
  "Solution_11": "[quote=\"ohhotgirl\"]\n\nThank you for the info! I ask about this since I am quite confident to become a fellow this year or next year, and going to harvard is my dream. getting the scholarship helps my admission to harvard much easier.[/quote] \r\n\r\n\r\nDid you take putnam last Saturday?\r\n\r\nIt is amazing to heard someone saying \"confident to become a fellow\"."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "$m, m+1 , ... , m+n$ are consecutive natural numbers and sum of numbers on digits $m$ and $m+n$ are divisible 8, but  sum of numbers on digits other natural numbers are not divisible 8. Find maximum value of $n$.",
  "Solution_1": "$n\\le 15$\r\nLet $d(n)$ be the sum of the digits of n. \r\nNote that if the number last digit of $k$ isn't $9$ we have that \r\n$d(k)+1=d(k+1)(\\mod 8)$.\r\nIf the last $t$ digits of $k$ are $9$ we have that:\r\n$d(k)+1-t\\equiv d(k+1) \\left(\\mod 8\\right)$ because $9=1 (\\mod 8)$\r\nIf the last digit of $m$ is $0$ or $1$ we have that\r\n$d(m)= d(m+8) (\\mod 8)$\r\nIf the last digit of $m$ is $a\\not=0,1$ then $n \\le 10-a+7 \\le 10-2+7 = 15$ Because the first digit  $n+k$ that ends at $0$ is at least $1$ modulo $8$ and one number between $d(n+k), d(n+k+1)...,d(n+k+7)$ is $0$ modulo $8$.\r\nNow we note that if $m=9999992$ the secuence $d(m),d(m+1),...,d(m+15)$ modulo 8 is:\r\n$0,1,2,3,4,5,6,7,1,2,3,4,5,6,7,0$ and we are done."
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all continuous functions $f : \\mathbb{R}\\to\\mathbb{R}$ such that for all $x \\in\\mathbb{ R}$,\r\n\\[f (f (x)) = f (x)+x.\\]",
  "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128763]Done[/url]  ($a=b=1$ satisfy  $a\\not =0,b>0,a+b\\not =1$)",
  "Solution_2": "Reviewing old problems, any solutions here???",
  "Solution_3": "[quote=N.T.TUAN]Find all continuous functions $f : \\mathbb{R}\\to\\mathbb{R}$ such that for all $x \\in\\mathbb{ R}$,\n\\[f (f (x)) = f (x)+x.\\][/quote]\n\nSee http://artofproblemsolving.com/community/c6h1344617p7316804 subcase 7.1.1\n",
  "Solution_4": "Thank you!! "
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a, b, c$ be positive real numbers. Prove that: $ \\frac{a^2}{2a^2\\plus{}ab\\plus{}c^2}\\plus{}\\frac{b^2}{2b^2\\plus{}bc\\plus{}a^2}\\plus{}\\frac{c^2}{2c^2\\plus{}ca\\plus{}b^2} \\le \\frac{3}{4}$  \r\n :maybe:",
  "Solution_1": "[quote=\"nguoivn\"]Let $ a, b, c$ be positive real numbers. Prove that: $ \\frac {a^2}{2a^2 \\plus{} ab \\plus{} c^2} \\plus{} \\frac {b^2}{2b^2 \\plus{} bc \\plus{} a^2} \\plus{} \\frac {c^2}{2c^2 \\plus{} ca \\plus{} b^2} \\le \\frac {3}{4}$  \n :maybe:[/quote]\r\n\r\n\r\nassume $ a \\equal{} max{(a,b,c)}$\r\n\r\nThe following inequality is equivalent to   \r\n\r\n$ (2b^2 \\plus{} 7bc \\plus{} 2ca)(a \\minus{} b)^2(a \\minus{} c)^2 \\plus{} ((12acb^2 \\plus{} 10ac^2b\\minus{} 22b^2c^2)$$ \\plus{} ac^3 \\plus{} b^4 \\plus{} 8b^3c \\plus{} 7ab^3 \\plus{} 4bc^3 \\plus{} 3c^4)(a \\minus{} b)(a \\minus{} c)$\r\n\r\n$ \\plus{} ((3a \\minus{} c)b^3 \\plus{} (15ca \\minus{} 10c^2)b^2 \\plus{} (15ac^2 \\minus{} c^3)b \\plus{} 3ac^3)(a \\minus{} b)^2\\geq 0$\r\n\r\nobvious true."
}
{
  "Tag": [],
  "Problem": "What is the units digit of the prime number $ 2^{44,497}\\minus{}1$?",
  "Solution_1": "The units digit of powers of 2 repeats as follows:\r\n\r\n2,4,8,6,2,4,...\r\n\r\n44497=1 mod 4, so the answer is 2-1=1."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation",
    "ratio",
    "symmetry",
    "reflection",
    "similar triangles"
  ],
  "Problem": "This is Tournament of the Towns 2004 problem (8 points)\r\nBut I think it is easier than that one with geometrical locus.\r\n\r\nAngle AOB and COD are combined with a rotation such that\r\nray OA is combined with a ray OC, OB with OD.\r\nIn them are inscribed two circles which intersect in points E and F.\r\nProve that angle AOE  is equal to the angle DOF.",
  "Solution_1": "Just wondering, had all the countries finished Tournament of Towns yet? I'm not sure how this contest is conducted internationally.",
  "Solution_2": "This is a sketch. Some details are left out.\r\n\r\nI'll assume WLOG (because I don't want to name other points) that $A,B$ are the points of tangency between $OA,OB$ and the circle inscribed in $\\angle AOB$, and the same for $C,D$. In this case we replace the word \"rotation\" in the text with \"spiral similarity\". I'm sure you won't mind :).\r\n\r\nLet $\\mathcal C_1$ be the circle inscribed in $\\angle AOB$ and $\\mathcal C_2$ the other one, and let $r_1,r_2$ be their respective radii, and $p_1,p_2$ the powers of $O$ wrt these two circles (respectively). Also, let $P$ be the second pt of intersection between $OE$ and $\\mathcal C_2$, while $Q$ is the second pt of intersection between $OF$ and $\\mathcal C_1$. Assume for now that we have shown $\\frac{OE}{OQ}=\\frac{OP}{OF}=\\frac{r_1}{r_2}\\ (*)$. In this case we find the triangles $QCF,EBP$ to be similar, and we\u2019re pretty much done.\r\n\r\nIn order to prove $(*)$, let $X$ be the intersection between the tangent at $E$ to $\\mathcal C_1$ and the tangent at $F$ to $\\mathcal C_2$. Let $Y$ be the second intersection between $XF$ and $\\mathcal C_1$, and $Z$ the second intersection between $XE$ and $\\mathcal C_2$. Try to show that $(1)\\ \\frac{XE}{XF}=\\frac{r_1}{r_2}$ and $(2)\\ XEF$ is similar to both $OEQ$ and $OPF$.",
  "Solution_3": "Another solution:\r\n\r\nAgain assume that $A, B, C, D$ are points of the relavant tangencies. Let $O_1, O_2$ be the centres of the two circles respectively ($O_1$ for the circle in $\\angle AOB$).\r\n\r\nSince $\\angle AOB= \\angle COD$, we have $\\angle AOO_1 = DOO_2$. Similar triangles give $\\frac{O_1O}{O_2O} = \\frac{r_1}{r_2}$ (the $r$'s are the radii of the circles)\r\n\r\nLet the internal angle bisector of $\\angle O_1OO_2$ touch $O_1O_2$ at $X$. Then all four ratios\r\n\r\n\\[\r\n\\frac{O_1O}{O_2O}, \\frac{O_1E}{O_2E}, \\frac{O_1X}{O_2X}, \\frac{O_1F}{O_2F}\r\n\\]\r\nare equal to $r_1/r_2$. By the Apollonius circle, the points $O, E, X, F$ are concyclic (or collinear in the trivial case $r_1=r_2$.) By symmetry, $XE=XF$, so $\\angle XOE = \\angle XOF$. After knowing this, the rest is just trivial angle calculations about the point $O$.",
  "Solution_4": "Wow, billzhao you have a very pretty solution, nobody from our country\r\ndidn't solve it in a such way!  Very good idea with Apollonius circle.\r\n\r\n Actually all who solved it used the same idea that Grobber used,\r\nFor example I just explained solution using inversion and symmetry \r\n- similar to Grobber spiral similarity.",
  "Solution_5": "This might come as a shock to you, but I for one was very much thinking about an Apollonius circle :).",
  "Solution_6": "I also solved it wth apollonius circle during the test! gteat billzhao!",
  "Solution_7": "Here's my solution, in the test. I found it quite easy for an 8 points problem. \r\n Let the circles C1 and C2 be tangent at A,B,C and D to the lines. Then consider the inversion with center O and ratio OA.OD . Clearly it takes the circle C1 to a circle equal to C2 but reflected across the angle bisector of AOD, and the same with C2 to C2' equal to C1 but reflected. Suposse E maps to E' . It is obvious now to reflect the inverted figure across the angle bisector of AOD. C2' maps to C2'',  C1' maps to C1'' and E' to E'' . Compare the obtained figure with the original one. It is obvious that C1'' = C2 , C2'' = C1, therefore E'' = F, and we know that the line OE'' (the same as OF) is the reflection of the line OE' (the same as OE), so OF is the reflection across the angle bisector of AOD of the line OE, and the conclussion follows.\r\nhope u could understand what i did.."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "Hi,\r\nI have a deadLine untill Thursday to find out the solutions to the following Questions.\r\nCan Anyone Help me??? please??\r\n\r\n1) A cylinder with Relative Density=0.5 and height=10m is holding vertical with the bottom base touching the surface of water.\r\nIf we leave the cylinder to move vertical towards water, Find the maximum depth in which the upper base of cylinder will reach.\r\n\r\n2) Flow of Water in an experimental tube with diametre 100mm has pressure fall 10.000 Pascal with n(a coefficient)=0.01. Find the length of the tube when Turbulent appears with Re=1000000\r\n\r\n3) Water is flowing in a Venturi Meter of 400mm X 100mm with discharge 72000lit/h and  pipe difference 250mm.\r\nIf the 'narrowing' is 200mm X 50 mm, \r\nfind the coefficient z of the meter.\r\n\r\n4) An Open reservoir fill with water has at it's side walls and in 2meters depth from the surface, Orifice(hole) with diametre=25mm.\r\na) Find the Velocity of Water at the Exit of the Orifice\r\nb) At the Orifice we Adjust a Pipe with diametr=25mm and length=9meters. Find the velocity at the exit of the Pipe.\r\nc) How much % has the Velocity increase or Decrease??\r\nGenerally, take the water flow normal.\r\n\r\n5) A cube with edge of 10 cm and Relative Density p=1.2 is equalising in 2 non-mixable fluids of relative density a=0.6 and b=1.6. \r\nFind the parts of the Cube's Edge that are Immersed for each case.\r\n\r\n\r\nPLEEEEEAAAASSSSEEEEEEE!!!!!",
  "Solution_1": "YO new member.\r\n\r\nFOr the first question,i'm not really getting the wordings.Could u be more clear?\r\n\r\nI think the fourth one has something to do with Toricelli's equation.Directly substitute the values i guess.",
  "Solution_2": "[quote=\"shadysaysurspammed\"]YO new member.\r\n\r\nFOr the first question,i'm not really getting the wordings.Could u be more clear?\r\nquote]\r\n\r\nWell, Excuse my English,\r\nWhat The Problem means is that We have a Cylinder that we are holding above the surface of water (the down part of cylinder touches the water) and the question is :\r\nIf we leave it to fall towards the water , the upper part of cylinder in what depth will reach (I suppose before changing it's vertical direction to some angle). \r\nSo, We have it's 'Relative Density' =0.5 and the cylinder height=10meters.\r\n\r\nFor All I know, the 'Relative Density' =( Weight of Object / weight of moved water )\r\n\r\nI don't know if this is the right Solution but I think something like :\r\n\r\nWeight of Cylinder = Weight of Moved Water   ==>\r\n9,81(gravity) X 0.5 (Relative Density) X 1000  = 9810 X(10m X h)     [where h=what we are looking for].\r\n\r\nso solving this , should be something like : h= 0.05 m\r\n\r\nBut I guess this is wrong cause the Result (h=0.05m) seems too small as a number. If I could guess I'd say h=50meters but who knows!\r\n\r\nThat's why I want your help.!!\r\n\r\nPlease, I've LOST all my notes and books in a fire!!! [and I need answers till Thursday]\r\n\r\nThanks again.",
  "Solution_3": "[quote=\"shadysaysurspammed\"] What starts with \"f\" and ends in \"uck\"? \n\nanswer-Firetruck!!(heheh,i know what u were thinkin:)) [/quote]Really? I thought it was \"Friar Tuck,\" but then that must be because I just watched \"Robin Hood (Prince of Thieves)\" the other night on TV :)"
}
{
  "Tag": [
    "geometry",
    "inradius",
    "inequalities",
    "circumcircle",
    "trigonometry",
    "function",
    "incenter"
  ],
  "Problem": "In any triangle ABC  proove  [tex] \\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{p}{2r} [/tex]\r\n\r\n[i]Edited by Myth[/i]",
  "Solution_1": "What is $s$?",
  "Solution_2": "Semiperimeter , I suppose... :?",
  "Solution_3": "Notation \"p\" is generally accepted for semiperimeter. Am I right?",
  "Solution_4": "yes , s  is semiperimeter",
  "Solution_5": "I'm more accustom to having $s$ as the semiperimeter and $p$ as the perimeter.\r\n\r\nBut that goes along with Darij's [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=138794#138794]comment[/url].",
  "Solution_6": "[quote=\"nickolas\"]In any triangle ABC  proove  [tex] \\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{p}{2r} [/tex]\n\n[i]Edited by Myth[/i][/quote]\r\n\r\nMyth, you hardly made the problem easier to understand. I have always been used to denoting the semiperimeter by s (if my memory doesn't cheat me, this notation goes back to Euler). Could you better just write explicitely what is meant by $m_a$, ..., p, r, as I proposed in another thread? Thanks.\r\n\r\n  Darij",
  "Solution_7": "I hear that $s$ is a semiprimetr for the first time. In all russian books I saw notation $p$ is used (see geometr's bible Prasolov's \"Planimetry\").",
  "Solution_8": "[quote=\"nickolas\"]In any triangle ABC  proove  [tex] \\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{p}{2r} [/tex][/quote]\r\n\r\nAt first, the $\\geq$ sign should actually be an $\\leq$ sign here, and also, I rewrite the problem using the notations I prefer:\r\n\r\n[color=blue]In any triangle ABC with semiperimeter $s=\\frac{a+b+c}{2}$, inradius r and medians $m_a$, $m_b$, $m_c$, we have $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$.[/color]\r\n\r\n[i]Proof.[/i]\r\n\r\n[color=red][b]EDIT:[/b] The proof below is WRONG. Thanks to Mak for pointing this out. In fact, the \"well-known inequalities\" $m_a\\leq\\sqrt{s\\left(s-a\\right)}$, $m_b\\leq\\sqrt{s\\left(s-b\\right)}$, $m_c\\leq\\sqrt{s\\left(s-c\\right)}$ which I used are incorrect; actually, one has $m_a\\geq\\sqrt{s\\left(s-a\\right)}$, $m_b\\geq\\sqrt{s\\left(s-b\\right)}$, $m_c\\geq\\sqrt{s\\left(s-c\\right)}$.\n\nSorry to everybody for the wrong proof. I will try to find a correct one.[/color]\r\n\r\nUsing the well-known inequalities $m_a\\leq\\sqrt{s\\left(s-a\\right)}$, $m_b\\leq\\sqrt{s\\left(s-b\\right)}$, $m_c\\leq\\sqrt{s\\left(s-c\\right)}$, we find\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{\\sqrt{s\\left(s-a\\right)}}{a}+\\frac{\\sqrt{s\\left(s-b\\right)}}{b}+\\frac{\\sqrt{s\\left(s-c\\right)}}{c}$.\r\n\r\nHence, in order to prove the inequality\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$,\r\n\r\nit will be enough to establish the sharper inequality\r\n\r\n$\\frac{\\sqrt{s\\left(s-a\\right)}}{a}+\\frac{\\sqrt{s\\left(s-b\\right)}}{b}+\\frac{\\sqrt{s\\left(s-c\\right)}}{c}\\leq\\frac{s}{2r}$.\r\n\r\nUpon division by $\\sqrt{s}$, this inequality becomes\r\n\r\n$\\frac{\\sqrt{s-a}}{a}+\\frac{\\sqrt{s-b}}{b}+\\frac{\\sqrt{s-c}}{c}\\leq\\frac{s}{2r\\sqrt{s}}$.\r\n\r\nNow, let's transform the right hand side of this inequality: On the one hand,\r\n\r\n(s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s.\r\n\r\nOn the other hand, by a well-known formula for the inradius of a triangle,\r\n\r\n$r=\\sqrt{\\frac{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}{s}}$,\r\n\r\nso that\r\n\r\n$r\\sqrt{s}=\\sqrt{\\frac{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}{s}}\\sqrt{s}=\\sqrt{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}$.\r\n\r\nHence, the fraction on the right hand side of our inequality can be transformed as follows:\r\n\r\n$\\frac{s}{2r\\sqrt{s}}=\\frac{\\left(s-a\\right)+\\left(s-b\\right)+\\left(s-c\\right)}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}$\r\n$=\\frac{\\left(s-a\\right)}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}+\\frac{\\left(s-b\\right)}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}+\\frac{\\left(s-c\\right)}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}$\r\n$=\\frac{\\sqrt{s-a}}{2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}}+\\frac{\\sqrt{s-b}}{2\\sqrt{\\left(s-c\\right)\\left(s-a\\right)}}+\\frac{\\sqrt{s-c}}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)}}$.\r\n\r\nHence, the inequality that we must prove rewrites as\r\n\r\n$\\frac{\\sqrt{s-a}}{a}+\\frac{\\sqrt{s-b}}{b}+\\frac{\\sqrt{s-c}}{c}$\r\n$\\leq\\frac{\\sqrt{s-a}}{2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}}+\\frac{\\sqrt{s-b}}{2\\sqrt{\\left(s-c\\right)\\left(s-a\\right)}}+\\frac{\\sqrt{s-c}}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)}}$.\r\n\r\nBut this follows from the following three inequalities:\r\n\r\n$\\frac{\\sqrt{s-a}}{a}\\leq\\frac{\\sqrt{s-a}}{2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}}$;\r\n$\\frac{\\sqrt{s-b}}{b}\\leq\\frac{\\sqrt{s-b}}{2\\sqrt{\\left(s-c\\right)\\left(s-a\\right)}}$;\r\n$\\frac{\\sqrt{s-c}}{c}\\leq\\frac{\\sqrt{s-c}}{2\\sqrt{\\left(s-a\\right)\\left(s-b\\right)}}$.\r\n\r\nSo, to complete the proof, it remains to verify these three inequalities. We will only verify the first inequality,\r\n\r\n$\\frac{\\sqrt{s-a}}{a}\\leq\\frac{\\sqrt{s-a}}{2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}}$;\r\n\r\nthe other two are just its cyclic permutations.\r\n\r\nThe proof of the inequality $\\frac{\\sqrt{s-a}}{a}\\leq\\frac{\\sqrt{s-a}}{2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}}$ is very easy: First, we divide it by $\\sqrt{s-a}$ and obtain\r\n\r\n$\\frac{1}{a}\\leq\\frac{1}{2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}}$;\r\n\r\nthis is obiviously equivalent to\r\n\r\n$a\\geq 2\\sqrt{\\left(s-b\\right)\\left(s-c\\right)}$.\r\n\r\nNow we square this:\r\n\r\n$a^2\\geq 4\\left(s-b\\right)\\left(s-c\\right)$.\r\n\r\nBut\r\n\r\n$4\\left(s-b\\right)\\left(s-c\\right)=\\left(2\\left(s-b\\right)\\right)\\cdot\\left(2\\left(s-c\\right)\\right)=\\left(c+a-b\\right)\\cdot\\left(a+b-c\\right)$\r\n$=\\left(a-\\left(b-c\\right)\\right)\\cdot\\left(a+\\left(b-c\\right)\\right)=a^2-\\left(b-c\\right)^2$,\r\n\r\nso that $a^2\\geq 4\\left(s-b\\right)\\left(s-c\\right)$ rewrites as $a^2\\geq a^2-\\left(b-c\\right)^2$, what is trivial since $\\left(b-c\\right)^2\\geq 0$.\r\n\r\nProof complete.\r\n\r\nNote that in the initial inequality $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$, equality holds if and only if the triangle ABC is equilateral. This is easy to see from the above proof.\r\n\r\n  Darij",
  "Solution_9": "Thank you  very much  Darij!",
  "Solution_10": "Hey, but the proof is wrong!!\r\n\r\n  darij",
  "Solution_11": "In  fact , this is  a prove  for  the ineq\r\n\r\n[tex] \\frac{w_a}{a}+\\frac{w_b}{b}+\\frac{w_c}{c}\\leq \\frac{s}{2r} [/tex]\r\nwhere  [tex] w_a , w_b , w_c [/tex] are  angle  bisectors  and  [tex]  s  [/tex] is semiperimeter",
  "Solution_12": "[quote=\"darij grinberg\"]Hey, but the proof is wrong!![/quote]\r\nAt least, it is a good edification for long non-conceptual proofs. :?",
  "Solution_13": "One man asked me some days ago to prove $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{3\\sqrt 3}{2}$.\r\n\r\nI proved it. And now I can't find the topic, which contain this inequality, though I remember we discussed it. I want to know what solution was proposed here.\r\nDoes anybody know?",
  "Solution_14": "[quote=\"Myth\"]One man asked me some days ago to prove $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\geq\\frac{3\\sqrt 3}{2}$.\n\nI proved it. And now I can't find the topic, which contain this inequality, though I remember we discussed it. I want to know what solution was proposed here.[/quote]\r\n\r\nHmm, I don't remember whether there was a separate topic for this inequality, but it is a special case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21016]the general geometric inequality[/url] stating that $\\frac{MA}{a}+\\frac{MB}{b}+\\frac{MC}{c}\\geq\\sqrt3$ for every point M in the plane of the triangle ABC (in your case, one has just to take M = centroid of triangle ABC).\r\n\r\nBy the way, are there any advances on the initial inequality $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$? I have showed it for all acute-angled triangles ABC, but I can't prove it for obtuse-angled ones...\r\n\r\n  darij",
  "Solution_15": "[quote=\"darij grinberg\"][color=blue]In any triangle ABC with semiperimeter $s=\\frac{a+b+c}{2}$, inradius r and medians $m_a$, $m_b$, $m_c$, we have $\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$.[/color][/quote]\r\n\r\nAfter giving a wrong proof of this inequality in post #9, I have now succeeded to find a correct one, although I find it quite ugly. Anyway, here it goes:\r\n\r\nFirst let's consider the case when the triangle ABC is acute-angled or right-angled. Let D be the midpoint of its side BC. Then, the segment AD is the A-median of triangle ABC and thus has the length $AD=m_a$. On the other hand, if O is the circumcenter of triangle ABC, then $OD\\perp BC$ (since the point O, being the circumcenter of triangle ABC, must lie on the perpendicular bisector of its side BC). Thus, in the right-angled triangle ODC, we have $OD=OC\\cdot\\cos\\measuredangle DOC$. But since the point O is the circumcenter of triangle ABC, we have < BOC = 2 < BAC = 2A by the central angle theorem, and also we have OA = OB = OC = R, where R is the circumradius of triangle ABC. Thus, the triangle BOC is isosceles with OB = OC, and hence the line OD, being the altitude from its apex O (since $OD\\perp BC$), must bisect its angle < BOC. Hence, $\\measuredangle DOC=\\frac{\\measuredangle BOC}{2}=\\frac{2A}{2}=A$. Thus, $OD=OC\\cdot\\cos\\measuredangle DOC=R\\cdot\\cos A$.\r\n\r\nNow, by the triangle inequality in the (possibly degenerate) triangle AOD, we have\r\n\r\n$AD\\leq OA+OD=R+R\\cdot\\cos A=R\\cdot\\left(1+\\cos A\\right)=R\\cdot 2\\cos^2\\frac{A}{2}=2R\\cos^2\\frac{A}{2}$.\r\n\r\nBy the extended law of sines, $2R=\\frac{a}{\\sin A}$; thus,\r\n\r\n$AD\\leq 2R\\cos^2\\frac{A}{2}=\\frac{a}{\\sin A}\\cdot \\cos^2\\frac{A}{2}=a\\frac{\\cos^2\\frac{A}{2}}{\\sin A}$\r\n$=a\\frac{\\cos^2\\frac{A}{2}}{2\\sin\\frac{A}{2}\\cos\\frac{A}{2}}=\\frac{a}{2}\\cdot\\frac{\\cos\\frac{A}{2}}{\\sin\\frac{A}{2}}=\\frac{a}{2}\\cdot\\cot\\frac{A}{2}$.\r\n\r\nSince $AD=m_a$, this becomes $m_a\\leq\\frac{a}{2}\\cdot\\cot\\frac{A}{2}$. Dividing this by a, we get $\\frac{m_a}{a}\\leq\\frac12\\cdot\\cot\\frac{A}{2}$. By the half-angle laws, $\\tan\\frac{A}{2}=\\frac{r}{s-a}$, so that\r\n\r\n$\\frac{m_a}{a}\\leq\\frac12\\cdot\\cot\\frac{A}{2}=\\frac12:\\tan\\frac{A}{2}=\\frac12:\\frac{r}{s-a}=\\frac{s-a}{2r}$.\r\n\r\nSimilarly, $\\frac{m_b}{b}\\leq\\frac{s-b}{2r}$ and $\\frac{m_c}{c}\\leq\\frac{s-c}{2r}$. Thus,\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s-a}{2r}+\\frac{s-b}{2r}+\\frac{s-c}{2r}=\\frac{3s-\\left(a+b+c\\right)}{2r}=\\frac{3s-2s}{2r}=\\frac{s}{2r}$.\r\n\r\nThus, the inequality in question is proved for the case of an acute-angled or right-angled triangle ABC.\r\n\r\nThe case when the triangle ABC is obtuse-angled is much more tricky. In fact, assume WLOG that A is the obtuse angle. Then, we have $OD=-R\\cdot\\cos A$ rather than $OD=R\\cdot\\cos A$, and the inequality $m_a\\leq\\frac{a}{2}\\cdot\\cot\\frac{A}{2}$ doesn't hold. But rather rough estimates can be applied:\r\n\r\nFirst, we have A > 90\u00b0; in other words, $\\measuredangle CAB>90^{\\circ}$. Thus, the point A lies inside the circle with diameter BC. The center of this circle is the midpoint D of the segment BC, and the radius of this circle is $\\frac{BC}{2}=\\frac{a}{2}$. Thus, $AD<\\frac{a}{2}$. Since $AD=m_a$, this becomes $m_a<\\frac{a}{2}$. Hence, $\\frac{m_a}{a}<\\frac12$.\r\n\r\nFor the next estimate, we will use vectors. If E is the midpoint of the segment CA, then $\\overrightarrow{BE}=\\frac{\\overrightarrow{BC}+\\overrightarrow{BA}}{2}$, so that\r\n\r\n$BE=\\left|\\overrightarrow{BE}\\right|=\\left|\\frac{\\overrightarrow{BC}+\\overrightarrow{BA}}{2}\\right|=\\frac12\\cdot\\left|\\overrightarrow{BC}+\\overrightarrow{BA}\\right|\\leq\\frac12\\cdot\\left(\\left|\\overrightarrow{BC}\\right|+\\left|\\overrightarrow{BA}\\right|\\right)$       (by the triangle inequality)\r\n$=\\frac12\\cdot\\left(BC+BA\\right)=\\frac12\\cdot\\left(a+c\\right)$.\r\n\r\nSince the point E is the midpoint of the side CA of triangle ABC, the segment BE is the B-median of this triangle; thus, its length is $BE=m_b$, and hence this inequality becomes $m_b\\leq\\frac12\\cdot\\left(a+c\\right)$. Similarly, $m_c\\leq\\frac12\\cdot\\left(b+a\\right)$.\r\n\r\nThus,\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}<\\frac12+\\frac{\\frac12\\cdot\\left(a+c\\right)}{b}+\\frac{\\frac12\\cdot\\left(b+a\\right)}{c}$\r\n$=\\frac12\\cdot\\left(1+\\frac{a+c}{b}+\\frac{b+a}{c}\\right)=\\frac12\\cdot\\left(1+\\left(\\frac{a+b+c}{b}-1\\right)+\\left(\\frac{a+b+c}{c}-1\\right)\\right)$\r\n$=\\frac12\\cdot\\left(\\frac{a+b+c}{b}+\\frac{a+b+c}{c}-1\\right)=\\frac12\\cdot\\left(\\frac{2s}{b}+\\frac{2s}{c}-1\\right)=\\frac{s}{b}+\\frac{s}{c}-\\frac12$.\r\n\r\nLet F be the area of triangle ABC; then, by a formula for the area of a triangle, $F=\\frac12bc\\sin A$. Since the angle A is obtuse, sin A < 1, and thus this becomes $F<\\frac12bc$. On the other hand, by another formula for the area of a triangle, F = rs. Thus, $rs<\\frac12bc$. In other words, 2rs < bc. This yields $\\frac{s}{b}<\\frac{c}{2r}$ and $\\frac{s}{c}<\\frac{b}{2r}$. Thus,\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}<\\frac{s}{b}+\\frac{s}{c}-\\frac12<\\frac{c}{2r}+\\frac{b}{2r}-\\frac12=\\frac{c+b}{2r}-\\frac12$\r\n$=\\frac{\\left(a+b+c\\right)-a}{2r}-\\frac12=\\frac{2s-a}{2r}-\\frac12=\\frac{s+\\left(s-a\\right)}{2r}-\\frac12=\\frac{s}{2r}+\\frac{s-a}{2r}-\\frac12$.\r\n\r\nBut since A > 90\u00b0, we have $\\frac{A}{2}>\\frac{90^{\\circ}}{2}=45^{\\circ}$, and thus, since the tangens function is monotonically increasing on the interval [0\u00b0; 90\u00b0[, we conclude that $\\tan\\frac{A}{2}>\\tan 45^{\\circ}=1$. But by the half-angle laws, $\\tan\\frac{A}{2}=\\frac{r}{s-a}$. Hence, $\\frac{r}{s-a}>1$, so that s - a < r. Thus,\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}<\\frac{s}{2r}+\\frac{s-a}{2r}-\\frac12<\\frac{s}{2r}+\\frac{r}{2r}-\\frac12=\\frac{s}{2r}+\\frac12-\\frac12=\\frac{s}{2r}$.\r\n\r\nSo the required inequality is proven for the case of an obtuse-angled triangle ABC, too. This completes the proof of the inequality.\r\n\r\nFinally, let's remark that the inequality we just proved,\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$,\r\n\r\ncan be rewritten in the form\r\n\r\n$m_ah_a+m_bh_b+m_ch_c\\leq s^2$,\r\n\r\nwhere $h_a$, $h_b$, $h_c$ are the three altitudes of triangle ABC. In fact, this is an equivalent transformation, since the area F of the triangle ABC satisfies $F=\\frac12ah_a=\\frac12bh_b=\\frac12ch_c=rs$, so that $a=2F:h_a$, $b=2F:h_b$, $c=2F:h_c$ and $r=F:s$, and thus\r\n\r\n$\\frac{m_a}{a}+\\frac{m_b}{b}+\\frac{m_c}{c}\\leq\\frac{s}{2r}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\frac{m_a}{2F:h_a}+\\frac{m_b}{2F:h_b}+\\frac{m_c}{2F:h_c}\\leq\\frac{s}{2F:s}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ \\frac{m_ah_a}{2F}+\\frac{m_bh_b}{2F}+\\frac{m_ch_c}{2F}\\leq\\frac{s^2}{2F}$\r\n$\\Longleftrightarrow\\ \\ \\ \\ \\ m_ah_a+m_bh_b+m_ch_c\\leq s^2$.\r\n\r\nIn the form $m_ah_a+m_bh_b+m_ch_c\\leq s^2$, the inequality was posted by Manlio at http://www.mathlinks.ro/Forum/viewtopic.php?t=5208 .\r\n\r\n  Darij",
  "Solution_16": "I read this post this noon and have a solution which is considerably longer than Darij 's. However, as this solution is replete with properties, so I post it here.\r\n    In order to prove this, let 's first say a little bit about about a bijection $f$ that maps from the set of all triangles $X$ to itself (the set $X$ doesn't have any two congruent triangles in it ). Let 's map a triangle with sides $a, b, c$ to the triangle with side $\\frac{2}{\\sqrt{3}}m_{a}, \\frac{2}{\\sqrt{3}}m_{b}, \\frac{2}{\\sqrt{3}}m_{c}$. Then of course $f = f^{-1}$. Notice that $f$ preserves the area of triangles.\r\n    Now let me introduce some lemmas that are very useful :\r\n    [b]Lemma 1[/b]  Let $ABC$ be a triangle. Then there exists a triangle $XYZ$ such that $sin^{2}{X}= \\frac{1}{2}sinA(\\sum tg{\\frac{A}{2}}, ...$\r\n    [b]Lemma 2 [/b] Let $XYZ$ be a triangle and $x, y, z$ be real numbers. Then $\\sum yz sin^{2}{X}\\leq \\frac{1}{4}(x+y+z)^{2}$\r\n    From lemma 1, it follows that   \r\n    [b]Lemma 3 [/b] Let $XYZ$ be a triangle and $x, y, z$ be real numbers. Then $\\sum yz sin{X}\\leq \\frac{(x+y+z)^{2}}{2.(\\sum \\tg \\frac{A}{2})}$\r\n    Let 's stop there and progress our proof for a moment. Call $R'$ the circumradius of the image triangle $XYZ$ of $ABC$ under $f$. Then $\\frac{2}{\\sqrt{3}}m_{a}= 2R' sin{X}, ....$. By applying lemma 3 we obtain \r\n    ${m_{a}.bc = \\sqrt{3}R'(\\sum bc.sinX) \\leq \\frac{\\frac{\\sqrt{3}}{2}}R'(a+b+c)^{2}}{\\sum tg{\\frac{X}{2}}}= \\frac{2\\sqrt{3}R'p^{2}}{\\sum tg{\\frac{X}{2}}}$.\r\n    We need to show that    \r\n  $\\frac{2\\sqrt{3}R'p^{2}}{\\sum tg{\\frac{X}{2}}}\\leq abc.\\frac{p}{2r}$\r\n    Notice that $abc = 4RS$, and thus it is equivalent to proving $R' \\leq \\frac{\\sum{tg\\frac{X}{2}}}{\\sqrt{3}}.R$\r\n    Carrying out the bijection $f$, this inequality is equivalent to \r\n       $R \\leq \\frac{\\sum{tg \\frac{A}{2}}}{\\sqrt{3}}.R'$  [color=blue](*) [/color]\r\n    We now go on introducing the lemmas\r\n    [b]Lemma 4 [/b] $4Rm_{a}\\geq b^{2}+c^{2}$\r\n    It is easy to prove if one compares the diameter and the chord formed by the $A$ median of the circumcircle. Therefore it follows that\r\n    [b]Lemma 5[/b]  $2R(m_{a}+m_{b}+m_{c})\\geq a^{2}+b^{2}+c^{2}$\r\n    Lemma 6 is a direct corollary of this inequality if we carry out the bijection $f$\r\n    [b]Lemma 6[/b] $\\frac{m_{a}.m_{b}.m_{c}}{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}\\geq r$, here $r$ is the inradius of $ABC$.\r\n     [b]Lemma 7[/b] $a+2+b^{2}+c^{2}= 2(p^{2}-r^{2}-4Rr)$\r\n     [b]Lemma 8[/b] $\\sum tg{\\frac{A}{2}}= \\frac{4R+r}{p}$\r\n     By replacing $R' = \\frac{2m_{a}.m_{b}.m_{c}}{3\\sqrt{3}S}\\geq \\frac{2(m_{a}^{2}+m_{b}^{2}+m_{c}^{2}). r}{3\\sqrt{3}S}= \\frac{(a^{2}+b^{2}+c^{2}).r}{2\\sqrt{3}S}= \\frac{(p^{2}-r^{2}-4Rr).r}{\\sqrt{3}S}$ and using lemma 4 to the inequality [color=blue](*)[/color], we obtain we need to prove\r\n      $p^{2}(R+r) \\geq 16R^{2}r+r^{3}+8Rr^{2}$ [color=blue](**)[/color]\r\n     And here is the final lemma (which I don't prove here, as a special case of Leibnitz 's theorem)\r\n     [b]Lemma 9 [/b] Let $I, G$ be the incenter and the centroid of $ABC$. Then $IG^{2}= p^{2}+5r^{2}-16Rr \\geq 0$\r\n     [color=blue](**)[/color]  follows immediately as $R \\geq 2r$.\r\n     PS This may be the longest post I ever reply in ML.  :lol:"
}
{
  "Tag": [
    "2nd edition"
  ],
  "Problem": "k is a range?",
  "Solution_1": "hmm I remember saying  no questions about the problems :D I belive that the questions are pretty well put :) read the questions carefully and take your time :)",
  "Solution_2": "suppose I think the question 3 is wrong (the one with BMN isosceles)?  could you please double check that question?   :D",
  "Solution_3": "I've checked the problem and I don't think there is anything wrong with it! :) anyway it's good that you asked :D",
  "Solution_4": "[quote=\"Valentin Vornicu\"]I've checked the problem and I don't think there is anything wrong with it! :) anyway it's good that you asked :D[/quote]\r\n\r\nthen i must be really stupid because I still cannot draw a proper diagram to that question for x /= 1    :(  :(  :(",
  "Solution_5": "... I will answer this question next Monday :D - anyway the problem's fine, I can assure you of that."
}
{
  "Tag": [
    "videos",
    "Princeton",
    "college"
  ],
  "Problem": "Before I start, let me just say I'm not necessarily bad at grammar, or overly obsessed with it (or the SAT, for that matter). There just always seems to be ONE question I get wrong in the writing section every time, and I get frustrated with 48C, 1W, 0O test after test. So here's one more (and hopefully the last):\r\n\r\nA major cause of stress in school is [u]where seniors must manage not only academic requirements and sports schedules, but also[/u] standardized testing and college applications, during the first semester.\r\n\r\n(A) where seniors must manage not only academic requirements and sports schedules, but also\r\n(B) seniors need to manage not only academic requirements and sports schedules, but also\r\n(C) where seniors must manage not only academic requirements and sports schedules, and also\r\n(D) when seniors must manage both academic requirements and sports schedules, but also\r\n(E) the management by seniors of not only academic requirements and sports schedules, but also\r\n\r\n\r\n[hide=\"My reasoning...\"]\nOkay, so I know the correct format is \"not only... but also\", so I can eliminate (C), which says \"and also\".\nFurthermore, \"a major cause\" cannot be a \"where\" or \"when\", so eliminate (A) and (D). I'm down to (B) and (E).\nNow take out all the fillers and look at the skeleton sentence:\n(B) A cause is seniors need to manage not only requirements but also testing.\n(E) A cause is the management of not only requirements but also testing.\nThe management causes the stress, not the seniors, so the answer must be (E)![/hide]\n\n\n[hide=\"But I was wrong!\"]\nThe answer is (B). Can someone please explain?[/hide]",
  "Solution_1": "I think you're parsing (E) wrong.  Your interpretation is that \"the management causes the stress, not the seniors.\"  But if we strip (E) down to a basic copular sentence, we get \"A cause of stress is [the] management\" which is certainly OK.\r\n\r\nMy best guess as to why (E) is wrong is that it's in the passive voice-- except actually it's not, it just looks suspiciously like it.\r\n\r\nI find the lack of an overt complementizer in (B) pretty jarring, but maybe that's just me.",
  "Solution_2": "[quote=\"Osud\"]I think you're parsing (E) wrong.  Your interpretation is that \"the management causes the stress, not the seniors.\"  But if we strip (E) down to a basic copular sentence, we get \"A cause of stress is [the] management\" which is certainly OK.[/quote]\n\nI'm not sure what you mean, I never said (E) was [i]not[/i] okay...\n\n[quote=\"Osud\"]My best guess as to why (E) is wrong is that it's in the passive voice-- except actually it's not, it just looks suspiciously like it.[/quote]\r\n\r\nThe book says that (E) is a \"very awkward passive phrase\". But is (B) any less awkward?",
  "Solution_3": "[quote]The book says that (E) is a \"very awkward passive phrase\".[/quote]\n\nWhat your book says is completely wrong.  (E) is emphatically [b]not[/b] a passive sentence.  \n\nHere's how the passive works.  In converting a sentence from active to passive, a few things happen:\n(1)  The active verb is converted to a form of \"to be\" plus a participle;\n(2)  The direct object of the active verb is promoted to subject position;\n(3)  The subject of the active verb is demoted to an optional \"by\" phrase.\n\nSo, if your active example is \"seniors manage academic requirements and sports schedules,\" then the passive form of the sentence will be \"academic requirements and sports schedules are managed by seniors.\"  We've converted the active verb \"manage\" to the passive \"are managed,\" moved \"academic requirements and sports schedules,\" the direct object of the \"manage,\" to subject position, and moved the subject of the active verb \"manage\" to an optional \"by\" phrase, \"by seniors.\"\n\n\"The management by seniors of academic requirements and sports schedules\" is not a passive phrase.  This is a noun phrase headed by the noun \"management.\"  The fact that this noun phrase has a \"by\" phrase containing what would have been the subject had \"management\" been an active verb does nothing to change the fact that nouns cannot be either active or passive and that the real verb of the sentence, \"is,\" is active, not passive.\n\nAgreed that that construction is very awkward, though.\n\n[quote]I'm not sure what you mean, I never said (E) was not okay... [/quote]\r\nSorry, I wasn't clear about that.  You said in your earlier post that you read the sentence to mean that \"The management causes the stress, not the seniors,\" I responded with an alternate meaning, and by \"OK\" I meant that my parse of the sentence was in line with what you thought the sense of the sentence should have been.  That make sense?",
  "Solution_4": "[quote=\"Osud\"]\n[quote]I'm not sure what you mean, I never said (E) was not okay... [/quote]\nSorry, I wasn't clear about that.  You said in your earlier post that you read the sentence to mean that \"The management causes the stress, not the seniors,\" I responded with an alternate meaning, and by \"OK\" I meant that my parse of the sentence was in line with what you thought the sense of the sentence should have been.  That make sense?[/quote]\r\n\r\nThanks for your explanation, makes sense now!\r\n\r\nAs for the passive stuff, yeah, I see what you mean.. with management as the noun, the phrase is not really passive at all. But do you agree with them that (B) is the correct answer? And if it is, is awkwardness the only thing wrong with (E)? And [i]if it is[/i], isn't (B) awkward too???  :lol:",
  "Solution_5": "I would actually say that (B) is worse than (E), and here's why.  Imagine you have read only this much of the sentence:\r\n\r\n[quote]A major cause of stress in school is[/quote]\n\nWhat do you naturally expect to come next?  A noun phrase, of course.  So then you hit the next word:\n\n[quote]seniors[/quote]\n\nAnd, since it's a noun, you naturally interpret it as the noun phrase that's the complement of the verb.  But wait!  There's more!\n\n[quote]need to manage not only academic requirements[/quote]\r\n\r\nYou start interpreting that as part of a noun phrase headed by \"seniors,\" but that's ungrammatical if all of that stuff goes in the noun phrase.  So you have to go back and interpret the whole thing starting with \"seniors\" as a complementizer phrase introduced by a silent complementizer.  Usually in English silent complementizers are used when omitting an overt complementizer won't create a garden-path sentence (so-called because it leads you down a garden path instead of the 'real' interpretation).  \r\n\r\nEven if grammar books have nothing against garden-path sentences, it is very bad form to write a sentence which many people will parse wrong on the first reading. \r\n\r\nIf it were up to me, I would pick (D).  The \"both...but also\" isn't great, but at least it is easy to read and it's immediately obvious what the sentence means.  \r\n\r\nActually, if it were me, I'd just say \"Managing standardized testing and college applications, on top of academic requirements and sports schedules, is a major cause of stress in school for seniors.\"  Or, if you really wanted to put \"a major cause of stress in school\" at the front of the sentence, I would say, \"It is a major cause of stress for seniors to manage standardized testing and college applications on top of academic requirements and sports schedules.\"  Just delete the prepositional phrase \"in school,\" it's not really adding anything here.  What else would it be, a major source of stress in church?  In video gaming?  In Sub-Saharan Africa? :lol:",
  "Solution_6": "I agree. When I first read the question, I really thought none of the choices made good sentences.\r\n\r\nBut thanks a lot for your input, I feel better about the question now that we've thrown it through a grinder. :)",
  "Solution_7": "Which book was this question from?",
  "Solution_8": "Cracking the SAT 2009 Edition. The Princeton Review.",
  "Solution_9": "For the SAT, I wouldn't trust a test-prep book besides the College Board's own Study Guide.  You can use the other books if you have run out of other material, but don't be surprised to see errors.  In math, I have caught numerous errors by the outside companies.\r\n\r\nDisclosure: I work for a tutoring company.",
  "Solution_10": "If you add \"that\" to (B), it becomes an obvious right choice. \"That\" can be sometimes omitted but here the omission creates a terribly awkward sentence, much worse than (E), so I suspect it was just a typographical error. (E) changes the meaning of the sentence: the cause is not \"management by seniors\" but the \"necessity for seniors to manage\" if you want that type of construction.",
  "Solution_11": "Fedja wrote:\r\n[quote]If you add \"that\" to (B), it becomes an obvious right choice. \"That\" can be sometimes omitted but here the omission creates a terribly awkward sentence, much worse than (E), so I suspect it was just a typographical error.[/quote]\n\nThat's what I was trying to get at with the business about silent vs. overt complementizers, but didn't think to give an example with an overt complementizer... thanks, fedja!   :blush: \n\nbuzzer11 wrote:\n[quote]I agree. When I first read the question, I really thought none of the choices made good sentences.[/quote]\n\nI think your instincts are spot-on, it's just that the SAT requires rote application of a set of prescriptive rules instead of a more holistic approach to style, grammar, and readability. This is what I meant earlier when I said not to take the stuff too seriously-- if you puzzle over something for a while and all of the answers seem a bit off to you, then stick with your instinct.  The SAT is a big deal at the time you take it, but you have a whole life of writing ahead of you.  You shouldn't try to permanently override your instincts about writing and language for the sake of one test.\n\n[quote]But thanks a lot for your input, I feel better about the question now that we've thrown it through a grinder. [/quote]\r\n\r\nSure!  Glad I could help.  (I'd never thought of myself as a grinder before...  :gleam:)",
  "Solution_12": "A, C, and D are automatically eliminated, because you know it shouldn't begin with \"where\" or \"when\".\r\n\r\nThe management is not the cause of stress. The cause of stress is either the things that the seniors have to manage or the fact that they have to manage them.\r\n\r\nHence, the best answer is B.\r\n\r\n\r\nHaving said that, I don't really like Princeton Review. Their practice problems are usually significantly easier than the actual test, and are sometimes weird. Kaplan has the right difficulty, but is riddled with technical errors and typos. Barrons is a bit harder than the real thing. For subject tests in science, Barrons introduces more material than is required for the actual tests, and the questions are not as understanding-of-theory type as the real ones, but perhaps provide the best practice.\r\n\r\nAlso, I don't particularly agree with the above poster. When I took the SAT's, I went off gut feeling for every single writing question, and it worked just fine. In the past, when I tried consciously invoking particular grammar rules, I got bogged up and confused, and ended up not only getting questions wrong, but also not finishing on time.",
  "Solution_13": "Why can't management be the cause of the stress? When I read the question that was what I thought the question was implying. The seniors were stressed because they have to [i]manage[/i] a lot.\r\n\r\nIn terms of the \"gut feeling\" idea, I usually go with strict grammar rules and not my gut and I do fine most of the time. I guess it depends on the person.",
  "Solution_14": "I've reduced B and E by eliminating some adjectives and unnecessary prepositional phrases. I have placed in parentheses what is communicated as the cause of stress.\r\n\r\n(B) A cause of stress is ([implicit*: the fact that] seniors need to manage various tasks).\r\n\r\n(E) A cause of stress is (the management) by seniors of various tasks.\r\n\r\nThe management is not a cause of stress -- management is generally a good thing. Causes of stress might be the tasks themselves, the fact that they have to manage these tasks (as in part B), or inability to manage.\r\n\r\nI totally agree that the assumption of (*) should not be necessary. However, in the SAT's, the first thing you must do is see if the sentence means what it wants you to mean, and then see if it is properly expressed -- it is not just about finding a sentence that is gramatically correct! Sometimes they will try to trick you by making a few gramatically correct responses, only one of which has the desired meaning.",
  "Solution_15": "Oh. When I read (B) I just read \"the cause of stress is --> seniors\" and so I eliminated that one. :(",
  "Solution_16": "And that [i][b]is[/b][/i] a bad mistake ;)",
  "Solution_17": "[quote=\"TZF\"](B) A cause of stress is ([implicit*: the fact that] seniors need to manage various tasks). [/quote]\n\n[quote=\"fedja\"]And that is a bad mistake  :wink: [/quote]\r\n\r\nThen how do we know to interpret (B) implying an object?"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "algebra",
    "binomial theorem",
    "calculus computations"
  ],
  "Problem": "I need help on getting something straightened out.\r\n\r\nFor the integral $ \\int_0^\\frac {1}{2}\\cfrac{\\sqrt {x}}{\\sqrt {1 - x}}\\: dx$, if I make the substitution $ \\sqrt {x} = \\sin{y}$, I get this:\r\n\r\n$ x = \\sin^2y$\r\n$ dx = 2\\sin{y}\\cos{y}$\r\n\r\nand for\r\n\r\n$ \\sqrt {0} = 0 = \\sin{y}$ and $ \\sqrt {\\cfrac{1}{2}} = \\cfrac{1}{\\sqrt {2}} = \\sin{y}$ to hold true,\r\n\r\n$ y = 0$ and $ y = \\cfrac{\\pi}{4}$ respectively.\r\n\r\n\r\nso:\r\n\r\n\\begin{align*} & \\int_0^\\frac {1}{2}\\cfrac{\\sqrt {x}}{\\sqrt {1 - x}}\\: dx \\\\\r\n& = \\int_0^\\frac {\\pi}{4}\\cfrac{\\sin{y}}{\\sqrt {1 - \\sin^2y}}\\cdot{}2\\sin{y}\\cos{y}\\: dy \\\\\r\n& = 2\\int_0^\\frac {\\pi}{4}\\sin^2y\\: dy \\end{align*}\r\n\r\n\r\n\r\nwas the above procedure correct? In particular, did I change the bounds of integration in the correct step?\r\n\r\nMy grasp of this is still kinda shaky.. What is this called anyway? Replacing bounds? It'd be great if someone could tell me under what section of a calculus course I could find more information on this. Thanks! :)",
  "Solution_1": "Seems fine to me :) Usually the corresponding section in Calculus textbooks is called either \"Change or variable\" or \"Finding integrals by substitution\". Just look at the table of contents and see what looks like it.",
  "Solution_2": "we can't find any mistakes in your calculation, undefined 117. :) \r\n\r\nHow about this one?\r\n\r\nQ. 1 $ \\int_0^1 x\\sqrt {1 \\minus{} x}\\ dx$.\r\n\r\nQ.2 Let $ x \\equal{} \\cos ^3 t,\\ y \\equal{} \\sin ^ 3 t\\ \\left(0\\leq t\\leq \\frac {\\pi}{2}\\right)$.\r\n\r\n$ \\int_0^1 y\\ dx$.",
  "Solution_3": "Thanks fedja - I found [url=http://en.wikipedia.org/wiki/Integration_by_substitution]this Wikipedia article[/url] :D\r\n\r\n@kunny: hehe I spent 40 minutes filling up two pages trying to solve that first one by substituting $ x = \\sin^2\\theta$ :rotfl:\r\n\r\nhmm idk is there a way to do it that way? I get $ \\int_0^\\frac {\\pi}{2}2\\sin^3\\theta\\cos^3\\theta\\: d\\theta$, which looks a bit like Q. 2\r\n\r\nokay so\r\n\r\n[hide=\"Q. 1\"]\n$ \\int_0^1x\\sqrt {1 - x}\\: dx$\n\n\n$ u = 1 - x$, so $ x = 1 - u$ and $ du = - 1\\: dx$.\n\n\n\\begin{align*} & - \\int_1^0(1 - u)\\sqrt {u}\\: du \\\\\n& = - \\int_1^0(u^\\frac {1}{2} - u^\\frac {3}{2})\\: du \\\\\n& = - \\int_1^0u^\\frac {1}{2}\\: du + \\int_1^0u^\\frac {3}{2}\\: du \\\\\n& = - \\left[\\cfrac{2}{3}u^\\frac {3}{2}\\right]_1^0 + \\left[\\cfrac{2}{5}u^\\frac {5}{2}\\right]_1^0 \\\\\n& = \\boxed{\\cfrac{4}{15}} \\end{align*}\n\nw00t! :lol:\n[/hide]\r\nI'll need to think a little more about Q. 2 :|",
  "Solution_4": "O.K.  :lol:  Q1 is right.",
  "Solution_5": "Well, You could do the q.1 using that substitution ($ x = sin^2\\theta$), but you won't get the integral you've written. You'll get $ \\int_{0}^{\\pi/2}2\\sin^3\\theta\\cos^2\\theta d\\theta$. And now, you just need to transform it into the form suitable for integration. Of course, it is hard to do it (and quite unnecessary, as your first solution is elegant enough), but here's the form:\r\n\r\n$ \\frac {1}{8} (2\\sin \\theta + \\sin 3\\theta - \\sin 5\\theta)$\r\n\r\nwhich gives you the same answer: $ \\frac {4}{15}$.",
  "Solution_6": "mk I think I'll need a hint for Q. 2 :D\r\n[hide=\"Q. 2\"]\n\\begin{align*} \\int_0^1y\\: dx & = \\int_0^\\frac {\\pi}{2}\\sin^3t\\cdot( - 3\\cos^2t\\sin{t})\\: dt \\\\\n& = - 3\\int_0^\\frac {\\pi}{2}\\sin^4t\\cos^2t\\: dt \\\\\n& = - 3\\int_0^\\frac {\\pi}{2}\\sin^4t(1 - \\sin^2t)\\: dt \\\\\n& = - 3\\left(\\int_0^\\frac {\\pi}{2}\\sin^4t\\: dt - \\int_0^\\frac {\\pi}{2}\\sin^6t\\: dt\\right) \\end{align*}\nerr..\n\\begin{align*} & - 3\\int_0^\\frac {\\pi}{2}(\\sin^2t\\cos{t})^2\\: dt \\\\\n& = - 3\\int_0^\\frac {\\pi}{2}(\\cos{t} - \\cos^3t)^2\\: dt \\end{align*}\nmeh. :|\n[/hide]",
  "Solution_7": "Well, let's stop here:\r\n\r\n$ \\minus{} 3\\left(\\int_0^\\frac {\\pi}{2}\\sin^4t\\: dt \\minus{} \\int_0^\\frac {\\pi}{2}\\sin^6t\\: dt\\right)$\r\n\r\nThese two can be easily evaluated if you know:\r\n\r\n$ \\int\\sin^n {ax}\\;dx \\equal{} \\minus{}\\frac{\\sin^{n\\minus{}1} ax\\cos ax}{na} \\plus{} \\frac{n\\minus{}1}{n}\\int\\sin^{n\\minus{}2} ax\\;dx$\r\n\r\n\r\nBut you can also use this nice result \r\n\r\n[url]http://www.mathlinks.ro/viewtopic.php?t=214671[/url]",
  "Solution_8": "whoa.. that stuff looks crazy beastly, but thanks for reminding me about integration by parts.\r\n\r\nhmm..\r\n\r\nFor $ \\int\\sin^3t( - 3\\cos^2t\\sin{t})\\: dt$\r\n\r\n$ u = \\sin^3t$ and $ dv = - 3\\cos^2t\\sin{t}$\r\n$ du = 3\\sin^2t\\cos{t}$ and $ v = \\cos^3t$\r\n\r\nso\r\n\r\n\\begin{align*} \\int\\sin^3t( - 3\\cos^2t\\sin{t})\\: dt & = \\sin^3t\\cos^3t + 3\\int\\cos^4t\\sin^2t\\: dt \\\\\r\n0 & = (\\sin{t}\\cos{t})^3 + 3\\left(\\int\\cos^4t\\sin^2t\\: dt + \\int\\sin^4t\\cos^2t\\: dt\\right) \\\\\r\n0 & = (\\sin{t}\\cos{t})^3 + 3\\int\\sin^2t\\cos^2t(\\cos^2t + \\sin^2t)\\: dt \\\\\r\n0 & = (\\sin{t}\\cos{t})^3 + 3\\int\\sin^2t\\cos^2t\\: dt \\end{align*}\r\n\r\nheh.. I'm probably going way off-course.",
  "Solution_9": "No need really for this kind of calculations, but I can't guarantee that you won't solve the problem this way :lol:  I thought a bit and found two more ways (well, the first one was the first I used, and the second one is more a showing off than a method):\r\n\r\n1. $ \\sin x\\equal{}\\frac{e^{ix}\\minus{}e^{\\minus{}ix}}{2i}$. Now, if you plug it in those two sine integrals, using binomial theorem reduces you to the easily integrable form shown in 2.\r\n\r\n2. Theoretically, after using addition (multiple angle) trig formulae, it is possible to obtain: \r\n\r\n$ \\frac{1}{32}(2 \\minus{} \\cos2t \\minus{} 2 \\cos4t \\plus{}\\cos6t)$ which is the same you'll get in Method 1 using binomial theorem and proper grouping of elements.",
  "Solution_10": "I think I've seen that form $ \\sin{x}\\equal{}\\cfrac{e^{ix}\\minus{}e^{\\minus{}ix}}{2i}$ before, but sorry I'm not familiar with it. What's it called? Hopefully I can look it up and familiarize myself :)",
  "Solution_11": "Well, I call it the complex form (or, even better-complex definition) of sine. $ x$ doesn't have to be real $ (x\\in\\mathbb{C})$.",
  "Solution_12": "ooh I found this page: [url]http://en.wikipedia.org/wiki/De_Moivre%27s_formula[/url]\r\n\r\nthat explains it in sorta easy terms..\r\n\r\nI'll just need to eventually convince myself that $ e^{ix} = \\cos{x} + i\\sin{x}$\r\n\r\nBut yes! It works! :lol:\r\n\\[ - 3\\left(\\int_0^\\frac {\\pi}{2}\\sin^4t\\: dt - \\int_0^\\frac {\\pi}{2}\\sin^6t\\: dt\\right)\r\n\\]\r\n[hide=\"First integrand\"]\n\\begin{align*} \\sin^4t & = \\left(\\cfrac{e^{ix} - e^{ - ix}}{2i}\\right)^4 \\\\\n& = \\cfrac{1}{8}\\cdot\\cfrac{e^{4ix} + e^{ - 4ix}}{2} - \\cfrac{4}{8}\\cdot\\cfrac{e^{2ix} + e^{ - 2ix}}{2} + \\cfrac{6}{16} \\\\\n& = \\cfrac{1}{8}\\cos{4x} - \\cfrac{1}{2}\\cos{2x} + \\cfrac{3}{8} \\end{align*}\n[/hide]\n[hide=\"Second integrand\"]\nI spent a bit more time on this one because I kept messing up pos/neg signs and the fractions. :P\n\\begin{align*} \\sin^6t & = \\left(\\cfrac{e^{ix} - e^{ - ix}}{2i}\\right)^6 \\\\\n& = - \\cfrac{1}{32}\\cdot\\cfrac{e^{6ix} + e^{ - 6ix}}{2} + \\cfrac{6}{32}\\cdot\\cfrac{e^{4ix} + e^{ - 4ix}}{2} - \\cfrac{15}{32}\\cdot\\cfrac{e^{2ix} + e^{ - 2ix}}{2} + \\cfrac{20}{64} \\\\\n& = - \\cfrac{1}{32}\\cos{6x} + \\cfrac{3}{16}\\cos{4x} - \\cfrac{15}{32}\\cos{2x} + \\cfrac{5}{16} \\end{align*}\n[/hide]\r\n\r\nSo combining the two parts we see that the original expression equals:\r\n\\[ - 3\\left[\\int_0^\\frac {\\pi}{2}\\left(\\cfrac{1}{8}\\cos{4x} - \\cfrac{1}{2}\\cos{2x} + \\cfrac{3}{8}\\right)\\: dx - \\int_0^\\frac {\\pi}{2}\\left( - \\cfrac{1}{32}\\cos{6x} + \\cfrac{3}{16}\\cos{4x} - \\cfrac{15}{32}\\cos{2x} + \\cfrac{5}{16}\\right)\\: dx\\right]\r\n\\]\r\nwhich, as hsiljak said, equals\r\n\\[ - 3\\left[\\int_0^\\frac {\\pi}{2}\\cfrac{1}{32}(2 - \\cos{2x} - 2\\cos{4x} + \\cos{6x})\\: dx\\right]\r\n\\]\r\nThen simply:\r\n\\begin{align*} & - \\cfrac{3}{32}\\left(\\int_0^\\frac {\\pi}{2}2\\: dx - \\int_0^\\frac {\\pi}{2}\\cos{2x}\\: dx - 2\\int_0^\\frac {\\pi}{2}\\cos{4x}\\: dx + \\int_0^\\frac {\\pi}{2}\\cos{6x}\\right) \\\\\r\n& = - \\cfrac{3}{32}\\left(\\left[\\cfrac{}{}2x\\right]_0^\\frac {\\pi}{2} - \\left[\\cfrac{1}{2}\\sin{2x}\\right]_0^\\frac {\\pi}{2} - \\left[\\cfrac{2}{4}\\sin{2x}\\right]_0^\\frac {\\pi}{2} + \\left[\\cfrac{1}{6}\\sin{6x}\\right]_0^\\frac {\\pi}{2}\\right) \\\\\r\n& = \\boxed{ - \\cfrac{3}{32}\\pi} \\end{align*}\r\nyeah! :w00t:\r\n\r\nThanks for the help, hsiljak, and thanks for the problems kunny!\r\n\r\nEDIT: oops, er I accidentally switched variables along the way, but you knew what I meant. :lol:"
}
{
  "Tag": [
    "LaTeX",
    "function"
  ],
  "Problem": "There are a few minor changes in the way the LaTeX code is rendered on the site.  For those of you unfamiliar with LaTeX, read the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX guide[/url] provided by this site to learn more about typesetting mathematics.  \r\n\r\nFirst of all, all the formulas are rendered with the \\displaystyle command from default. You need not add that command anymore. The \\displaystyle command makes fractions, summations, etc., render more legibly, as shown below:\r\n\r\nWith \\displaystyle: $ \\sum_{k \\equal{} 0}^n \\frac 1{k}$ \r\n\r\nWithout \\displaystyle: $ \\textstyle \\sum_{k \\equal{} 0}^n \\frac 1{k}$. \r\n\r\nTo force your posts to be without \\displaystyle you can use the \\textstyle \r\n command (start your LaTeX code with \\textstyle). \r\n\r\nSecond, we encourage all users to use the normal LaTeX syntax, i.e. the dollar signs, instead of the [tex ] and [/tex ] tags (without spaces there). \r\n\r\nThe [tex ] button will still be available and can still be used as before, but after rendering the text, the forum will automatically transform the [tex ] tags into equivalent dollar signs. \r\n\r\nIf you do not edit your post, you will \"feel nothing\". If you do, the [tex ] tags that you had used will be replaced by dollar signs. \r\n\r\nPS I do not know if everyone knows this already, so I will mention it here again: on most browsers the latex code used to produce a block of math text (the whole latex code, including the dollar signs) can be viewed by holding the mouse a couple of seconds still above the picture.",
  "Solution_1": "[quote=\"RC-7th\"]Is there a command for the thing that makes this $\\textstyle \\sum\\limits_{k=0}^n \\frac 1{k}$ different from this \\[\\textstyle \\sum\\limits_{k=0}^n \\frac 1{k}\\] ?[/quote] You probably didn't carefully read the text above. You are using the \\limits command, which automatically acts as a \\displaystyle. Without the \\limits command (which is not necessary!), the code above would be \r\n$ \\textstyle \\sum^n_{k=0} \\frac 1k $.",
  "Solution_2": "The \\limits thing is just a habit. I was wondering if there was a command equivalent of using double dollar signs [code]($$)[/code] on the ends.",
  "Solution_3": "next time, if you want to display dollars use [code ] (dollars) [/code ] (without spaces), and use all the time the function preview, to see if there are any errors in your messages ;) \r\n\r\nas for your question, if you had carefully read the post above, you would have discovered that \"we encourage all users to use the normal LaTeX syntax\", that means any known LaTeX syntax, including double dollars as\r\n\r\n\\[ \\frac 12 \\cdot \\frac 34 = \\frac 38 \\] does. \r\n\r\nPlease read carefully the general annoucements, as they usually contain all the answers you need. :) \r\n\r\nGood luck \"tExING\" :D",
  "Solution_4": "The feature of some web browsers to show LaTeX code when scrolled over that v.v. remarked about on the original post does not work in Mozilla.",
  "Solution_5": "Or Opera..",
  "Solution_6": "Sorry guys, I have just tested it on IE Explore, and Semaca Mobile browser for SE (where it works!). \r\n\r\nIt is curious why it does not work in Opera, or in Mozzila. I will try to see what is wrong ...  :maybe: \r\n\r\nIf anyone knows more about these browsers, you can check the source code yourselves and tell me what exactly is wrong :)  ;)",
  "Solution_7": "We might be able to add netscape to that list (or maybe my version is simply too old :? )",
  "Solution_8": "Ok, I've identified the problem. It simply seems that Opera, Mozzila and Netscape just do not accept the alt tag. Thus, there is only one solution. Tell me if it works ;)",
  "Solution_9": "Works fine here (Opera speaking :)).",
  "Solution_10": "Works with Firefox now. Thanks.",
  "Solution_11": "Lemme try in Firefox:\r\n\r\n\\[\r\n\\frac{1}{2}! = \\frac{1}{2}\\Gamma{}\\left(\\frac{1}{2}\\right) = \\frac{\\sqrt{\\pi}}{2}\r\n\\]",
  "Solution_12": "if you type $$ and $$ at the ends of a sentence, does the code always come out as text in the center of a line and always occupy the whole line?\r\nFor example if you type \"I have $$3$$ dollars\", it will come out as \r\n\r\nI have\r\n                                                                                                                                                                              3\r\ndollars. \r\n\r\n\r\nHow can you aviod this?",
  "Solution_13": "[quote=\"henryyangrui\"]if you type \\[ and \\] at the ends of a sentence, does the code always come out as text in the center of a line and always occupy the whole line?\nFor example if you type \"I have \\[3\\] dollars\", it will come out as \n\nI have\n                                                                                                                                                                              3\ndollars. \n\n\nHow can you aviod this?[/quote]What do you mean? This is how normal $\\LaTeX$ works. If you want inline equations, use a single dolar sign at the beginning and the end of a formula: $\\$ \\frac 12 $\\$ will reveal $ \\frac 12 $ ;)",
  "Solution_14": "Oh, i used double $ instead. Thx."
}
{
  "Tag": [
    "geometry",
    "trapezoid",
    "LaTeX"
  ],
  "Problem": "Hey..ive been bustin my brains with this problem.  It might be easy but im still having troubles\r\n\r\nA equivalent triangle (A,B,C) with 10 cm as sides thus all angles are 60 degrees.\r\nA half-circle passes trough point A and B passing trough the interior of the triangle, yet there is a small portion of the circle outside the triangle.  Find the area of that portion.\r\n\r\nIf you dont understand the question, just tell me. I'll try to do a drawing",
  "Solution_1": "BTW...\r\nall i need is a guideline of how to do it. \r\nI dont need the solution\r\nthx in advance",
  "Solution_2": "edit: oops.",
  "Solution_3": "hmm. i guess this is the picture, right??\r\n[hide=\"hint\"]\nangle $AOM=120$ because $\\triangle MOB$ is equalateral. with side length of $5$.\nnow find $S_{\\widehat {AOM}+\\triangle MOB}$ and subtract if from the area of the half circle[/hide]",
  "Solution_4": "sorry...i seem to have miss explainedthe question.\r\nfirst of all, it is a half circle we are talking about...\r\nsecond of all, i forgot to mention that the diameter of the circle is a side of the triangle so a portion of the circle would not be in the triangle....\r\n\r\n\r\ni could do an image, but we're can i host it???\r\n\r\nEDIT...\r\ni am in grade 10 math, dont over complicate plz",
  "Solution_5": "here's a picture i drew...\r\nfind the area of the shaded part\r\n\r\nBTW...each side is 10 cm",
  "Solution_6": "[hide=\"tostart you off..\"]\ni dont know if i made a correct assumption but anyways:\ndraw a line at top of circle to form a trapezoid at bottom of the equiliateral triangle (in other words, 3 smaller equilateral triangles of sidelength 5. \n[/hide]\n[hide=\"stilldontgetit?\"]\nThe area of an equilateral triangle is \nS^2*sqrt3/4 where s is sidelength, so the area of the three triangles is 75sqrt3/4 \nSince we know the diameter of the circle to be 10 we know that half the circle would be 25pi/2\nthe area o fthe two shaded things and thing at top of circle is 25pi/2-75sqrt3/4 <--\nim assuming they are all the same so u multiply this value ------------------------------| by 2/3\nim not gonna bother to actually calculate it but u get the idea :D[/hide]\r\n\r\nEdit: sry i havent bothered to learn latex yet so i hope you can still read it  :P",
  "Solution_7": "I agree with sheepwarrior, the answer is \\[\\frac{2}{3}\\left(\\frac{25\\pi}{2}-\\frac{75\\sqrt{3}}{4}\\right).\\] Here's a picture:"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "[b]Let ABCD be a convex quadrilateral inscribed in the circle[/b] $w$.[b] I note the [/b][b]intersections[/b]\n\n$S\\in AC\\cap BD,\\ P\\in BB\\cap DD,\\ R\\in AA\\cap CC$. [b]Prove that:[/b]\n\n$1.\\ AB.CD=AD.BC\\Longleftrightarrow R\\in BD\\Longleftrightarrow P\\in AC$.[b] ($XX$ is the[/b] [b]tangent in the point[/b] $X\\in w$[b]).[/b]\n\n[b]2. The point S is the foot of a simedian-line at least in the one from the triangles ABC, ABD, ACD, BCD.[/b]",
  "Solution_1": "P = BB intersect DD ???  I think I am seeing deja vu... :( \r\n\r\ncould you explain?",
  "Solution_2": "Did you read the last sentence, al.M.V.? There, levi explains what he means by $XX$, in general. The post isn't that long, man, you could have read the whole thing.. :)",
  "Solution_3": "al.m.v  ,i think levi mean :tangent ;)",
  "Solution_4": "Yes"
}
{
  "Tag": [
    "geometry",
    "number theory",
    "relatively prime"
  ],
  "Problem": "Prove or give a counterexample that a right angled triangle with integer sides can't have area as a perfect square.",
  "Solution_1": "because one leg is odd the other is even.....and the odd cant be a perfect square without the even containing an odd multiple of it too",
  "Solution_2": "[quote=\"usaha\"]because one leg is odd the other is even.....and the odd cant be a perfect square without the even containing an odd multiple of it too[/quote]\n\n[quote]$\\cdots$ and the odd cant be a perfect square without the even containing an odd multiple of it too[/quote]\r\ncan you elborate more?",
  "Solution_3": "I welcome criticism of this proof as I am trying to become better at this sort of thing.\r\n\r\nProof:\r\n\r\n[hide]\nIt can be shown that all primitive pythagorean triples (primitive in that there is no factor common to the lengths of all three sides) can be described for positive integers $a$ and $b$, where $a>b$, such that the two legs are $a^{2}-b^{2}$ and $2ab$ and the hypotnuse is $a^{2}+b^{2}$. \n\nThus the area of any such triangle is equal to half base times height or:\n$A=\\frac{1}{2}(a^{2}-b^{2})(2ab)=(a+b)(a-b)ab$\n\nConsider non-primitive cases; all such cases can be described as a primitive case with each side multiplied by a common positive integer factor $k$. For such a non-primitive case, the area would be $A=k^{2}(a+b)(a-b)ab$ and for A to be a perfect square, the $k^{2}$ factors and the product $(a+b)(a-b)ab$ must still be a perfect square for the area to be a perfect square. \n\nThus, all we must do is show that the product $(a+b)(a-b)ab$ cannot be a perfect square to cover all primitive and non-primitive cases.\n\nThus the only cases to consider are:\n1) $(a-b)=(a+b)ab$\n2) $(a+b)=(a-b)ab$\n\nSince a and b are both positive integers and a>b, case 1 is clearly rediculous.\n\nManipulating case 2 gives:\n$a-b=\\frac{1}{a}+\\frac{1}{b}$\nSince a and b are positive integers, this could only be possible if a-b=1. Thus:\n$1=\\frac{1}{b+1}+\\frac{1}{b}$\nAnd this clearly does not work for any positive integer b. \n\nThus the product $(a+b)(a-b)ab$ for relatively prime a and b cannot ever be a perfect square, and thus the area of a right triangle with integer sides cannot ever be a perfect square.[/hide]",
  "Solution_4": "[quote=\"usaha\"]and the odd cant be a perfect square without the even containing an odd multiple of it too[/quote]\n\nNot true.  Consider the triple $9, 40, 41$.  \n\n[quote=\"bthings_2000\"]Thus, all we must do is show that the product $(a+b)(a-b)ab$ cannot be a perfect square to cover all primitive and non-primitive cases.\n\nThus the only cases to consider are:\n1) $(a-b)=(a+b)ab$\n2) $(a+b)=(a-b)ab$[/quote]\r\n\r\nI have a problem with this narrowing down to two cases; what about the case where $a+b, a-b, a, b$ are all simultaneously square?",
  "Solution_5": "Hmm, good point t0rajir0u, and I've messed around with that for a while and not figured it out... input from anyone welcome.",
  "Solution_6": "Yea toujorou, you're right.\r\n\r\nWhat i was trying to say is that (this is like a guess) if a leg is a perfect square, the other side must contain an odd power of an odd multiple (so 9, 40, 41 the 40 has only 1 multiple of 5), so when you do the area, it cant be a perfect square.  Also, for like generic cases like 5,12,13, im guessing that since one is even the other is odd, that when you multiply it up there will be an odd number power of an odd multiple, AI dont know what i just said.",
  "Solution_7": "Well, we can reduce it to the case of a primitive triangle (since dividing each side by $g$ divides area by $g^{2}$, so it remains a perfect square), and then one leg is twice a square and the other leg is a square. So we're looking at\r\n\\[4a^{4}+b^{4}=c^{2}\\]\r\nin positive integers.\r\n\r\nProbably factoring the left hand side or using primitive Pythagorean triple formula could help here.",
  "Solution_8": "some treatment available here... external link.\r\n[hide][url]http://mcraefamily.com/MathHelp/PythagTriangleAreaNotSquare.htm[/url][/hide]"
}
{
  "Tag": [
    "function",
    "inequalities",
    "logarithms",
    "induction",
    "rearrangement inequality",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $f(x)=ax^{2}+bx+x$ $(a,b,c>0)$; $a+b+c=1$, $x_{1}>0,x_{2}>0,...,x_{n}>0$, $x_{1}x_{2}...x_{n}=1$\r\nProve that:\r\n$f(x_{1})f(x_{2})f(x_{3})...f(x_{n}) \\geq 1$",
  "Solution_1": "$f(x)f(1/x)=a^{2}+b^{2}+c^{2}+ab(x+1/x)+ac(x^{2}+1/x^{2})+bc(x+1/x)=1+(ab+bc)\\frac{(x-1)^{2}}{x}+ac\\frac{(x^{2}-1)^{2}}{x^{2}}\\ge 1$.\r\nWe have $f'(x)\\ge 0$ and $f(1)=1$. It give your result.",
  "Solution_2": "It's not clear to me how you get the result from what you've written.\r\n\r\nHow about the following:\r\n\r\nFor $x < 1 < y$, \r\n$f(x)f(y)=a^{2}+b^{2}+c^{2}+ab(x^{2}y+x y^{2})+ac(x^{2}+y^{2})+bc(x+y) \\geq a^{2}+b^{2}+c^{2}+ab(x^{2}y^{2}+x y)+ac(x^{2}y^{2}+1)+bc(xy+1) = f(xy) f(1)$,\r\n\r\nusing the rearrangement inequality for each term.  So if the $x_{i}$ are not identically 1, we can convert a non-1 value to 1 while decreasing the product of the $f(x_{i})$.  Eventually all the $x_{i}$ are equal to 1, and the product of the $f(x_{i})$ is exactly 1.",
  "Solution_3": "$f(xy)f(1)-f(x)f(y)=(abxy+bc)(x-1)(y-1)+ac(x^{2}-1)(y^{2}-1)$.\r\nLet $x_{1}\\le x_{2}\\le ....x_{k}\\le 1\\le x_{k+1}\\le...\\le x_{n}$. Let $\\sigma(1)=1,y_{1}=x_{1},\\sigma(k+1)=n+1-\\sigma(k), \\ if \\ y_{k}<1, \\ sigma(k+1)=\\sigma(k)-1 \\ else \\ y_{k+1}=y_{k}*x_{\\sigma(k+1)}.$ Then\r\n$f(y_{k+1})\\lef(y_{k})f(x_{\\sigma(k)})$. Therefore\r\n$1\\le \\pmod_{k}f(x_{k}).$",
  "Solution_4": "I'm afraid I'm not following this either.\r\n\r\nFor example, if $y_{1}> 1$, then $\\sigma(2) = 0$.",
  "Solution_5": "If $y_{k}>1$ we find little $x_{m}$ don't used before. If $y_{k}<1$ we find biggest $x_{m}$ don't used before. Define $m=\\sigma (k+1)$. \r\nIt give $(y_{k+1}=y_{k}x_{\\sigma (k+1)}, \\ (y_{k}-1)(x_{\\sigma (k+1)}-1)\\le 0 \\ \\forall k .$",
  "Solution_6": "is there any solutions else? I dont like Rust's Solution  :wink:",
  "Solution_7": "I just did this problem in a practice test (1990 All-Soviet Olympiad, I think?)\r\n\r\nSay $e^{a_{i}}= x_{i}$. Let $g(x) = \\log (a e^{2x}+b e^{x}+c)$. Then $g''(x) > 0$ for $x > 0$ (this is easy to confirm, unless I made an algebra error). Then we have $\\sum a_{i}= 0$. We need to show $\\sum g(a_{i}) = 0$. But Karimata gives $\\sum g(a_{i}) \\geq \\sum g(0)$, and since $a+b+c = 1$, $g(0) = \\log(1) = 0$, and we have the desired result.\r\n\r\nDeedlit: Maybe I am missing something, but you appear to be using rearrangement backwards? Certainly $x+y \\leq x y+1$ is not a true statement.",
  "Solution_8": "[quote=\"hieuchuoi@\"]is there any solutions else? I dont like Rust's Solution  :wink:[/quote]\n\nWas there a problem with mine?\n\n[quote]Deedlit: Maybe I am missing something, but you appear to be using rearrangement backwards? Certainly x+y \\leq x y+1 is not a true statement.[/quote]\r\n\r\nWhoops, yes the inequality is the wrong way.  I'll fix it.",
  "Solution_9": "A nice solution  :D \r\nUse this lemma:\r\n$f(x).f(y) \\geq f(\\sqrt{xy})^{2}$\r\nThe rest is for you  :roll: !",
  "Solution_10": "It is nice. But we need $f(x_{1})f(x_{2})...f(x_{n})\\ge(f((x_{1}...x_{n})^{1/n}))^{n}.$\r\nFor example consider $n=3.$",
  "Solution_11": "Yes, we easy have it by induction",
  "Solution_12": "Explain it for $n=3$.",
  "Solution_13": "sorry, i was wrong, and this is two new solutions\r\n1) you can apply Holder ineq  :D \r\n2) you can prove with n is odd, n is even and we have conclusion\r\nif n is odd, we add $a_{n+1}=1$  :D",
  "Solution_14": "It is easy to prove that $f(x)f(y)\\geq (f(\\sqrt{xy}))^{2}$\r\nNow suppose that $f(x_{1})...f(x_{n})\\geq (f(\\sqrt[n]{x_{1}..x_{n}}))^{n}$ \r\n\r\nis true for $n=2^{k}$ and prove it for $n=2^{k+1}$  :wink:",
  "Solution_15": "no, it's true for any positive integer $n$  :wink:",
  "Solution_16": "I can prove it forall n. But it is not trivial for $n\\not =2^{k}$, for example for n=3.",
  "Solution_17": "Well, the easy way would be to take the logarithm and apply Jensen's inequality, which would make it essentially the same as 101101101's proof.",
  "Solution_18": "[quote=\"hieuchuoi@\"]no, it's true for any positive integer $n$  :wink:[/quote]\r\n\r\nWho said that ?  :wink: \r\nAfter the proof for $n=2^{k}$  suppose $n$ is arbitary and $x_{1}..x_{n}=1$  . Let $k$ a positive integer such that $2^{k-1}<n\\leq 2^{k}$ . Let us add ,if necessary, $x_{n+1}=x_{n+2}=...=x_{2^{k}}=1$. Since $f(x_{n+1})=....=f(x_{2^{k}})=1$ we may write \r\n$f(x_{1})....f(n)=f(x_{1})...f(x_{2^{k}})\\geq (f(\\sqrt[2^{k}]{x_{1}...x_{2^{k}}}))^{2^{k}}=1$  \r\n\r\n :wink:"
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "Found this somewhere.\r\n\r\nSolve without a calculator. Explain.\r\n\r\n(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)\r\n----------------------------------------------------------------------------\r\n(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)",
  "Solution_1": "http://www.mathlinks.ro/Wiki/index.php/Sophie_Germain_Identity\r\n\r\nhttp://www.mathlinks.ro/Wiki/index.php/1987_AIME_Problems/Problem_14"
}
{
  "Tag": [
    "number theory",
    "prime numbers",
    "number theory unsolved"
  ],
  "Problem": "I need a help with this problem (again on set N-[1]):\r\n\r\nLet's divide all prime numbers into two infinite sets A,B so that each prime number is element of exactly one of sets A,B. Now let each composite number that is product of some primes only from A be element of A too and also let each composite number that is product of some primes only from B is element of B too. Let's denote the set of the remaining composite numbers (such that they aren't elements of A or B) as C. Prove that for each natural number n at least one of sets A,B or C contains n consecutive natural numbers.\r\n\r\nThanks for all your replies...",
  "Solution_1": "I think it is very simple.\r\nSuppose $p_1$, ..., $p_n\\in A$ and $q_1$, ..., $q_n\\in B$. Then due to Chinese residue theorem we can find $m$ s.t. $p_iq_i\\,|\\,m+i$, $i=1..n$. So $C$ contains $n$ consecutive numbers.",
  "Solution_2": "Aargh...too late...\r\n\r\nPierre.",
  "Solution_3": "Welcome to my club!"
}
{
  "Tag": [
    "integration"
  ],
  "Problem": "A variable force F acts on a body which is free to move.The displacement of the body is proportional to $t^3$ where t is time.The power delivered by F to the body will be proprtional to what power of $t$?",
  "Solution_1": "[color=blue]I think the answer is $t^4$[/color]",
  "Solution_2": "0\r\n\r\n$f = ct$\r\n$a =\\frac{ct}{m}$\r\n$v = \\frac{1}{m}\\frac{ct^2}{2}$\r\n$d = \\frac{1}{m}\\frac{ct^3}{3}$\r\n\r\n$d$ is proportional to $t^3$\r\n\r\n$p = \\frac{1}{t}\\int f(t)dx$\r\n$p = \\frac{ctd}{t}$\r\n$p$ is proportional to $t^0$",
  "Solution_3": "$x = k t^3$\r\n\r\n$v = 3 k t^2$\r\n\r\n$a = 6 k t$\r\n\r\n$F = 6 k m t$\r\n\r\n$P = Fv = 18 k t ^3$\r\n\r\nSo the power is proportional to $t^3$.",
  "Solution_4": "durt is correct :)",
  "Solution_5": "isnt power work over time not force times velocity?",
  "Solution_6": "[quote=\"oooooh ya\"]isnt power work over time not force times velocity?[/quote]\r\n\r\nyeah,my bad",
  "Solution_7": "$work = \\int f dx$\r\n\r\n$dist = vt$ if v is constant, not for a variable acceleration/force\r\n\r\n[quote=\"phucnv87\"]I think the answer is $t^4$[/quote]\r\npost your work so we can see if it is",
  "Solution_8": "$x = k t^3$\r\n$v = 3 k t^2$\r\n$a = 6 k t$\r\n$F = 6 k m t$\r\n$dx = 3 k t^2 dt$\r\n$W = \\int F dx$\r\n$W = \\int 18 k^2 m t^3 dt$\r\n$P = W/t$\r\n$P = \\frac{9}{2} k^2 m t^4/t$\r\n$P = \\frac{9}{2} k^2 m t^3$ so it is $t^3$",
  "Solution_9": "It seems we all seem to be getting different answers. I forgot to include $k^2$ and $m$ in my solution. Annie is on the right track until she says$P = W / t$. Its actually $P = \\frac{dW}{dt} = \\frac{d}{dt} \\left( \\frac{9}{2} k^2 m t^4 \\right) = 18 k^2 m t^3$.",
  "Solution_10": "ooh I want to play too\r\n\r\n$P = \\frac{dW}{dt} = \\frac{d}{dt} (F \\cdot x) = F \\frac{dx}{dt} + x \\frac{dF}{dt} = ma \\frac{dx}{dt} + x m\\frac{da}{dt} = m \\left( \\frac{dx}{dt} \\frac{d^2x}{dt^2} + x \\frac{d^3x}{dt^3} \\right)$. Let $x = kt^3$. Then the above is $m (3kt^2 \\cdot 6kt + kt^3 \\cdot 6k) = 24mk^2t^3$.",
  "Solution_11": "[b]MysticTerminator[/b], note that $\\text dW = F \\text dx$, not $\\text dW = \\text d(Fx)$  ;)  \r\n\r\nThe answer is $P = 18mk^2t^3$...",
  "Solution_12": "I was wondering why the formula looked different ...  ;)"
}
{
  "Tag": [
    "limit",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all $a,b,c \\in R$ such that $[na]\\cdot [n^{2}b]=[n^{3}c]$ for all $n \\in \\mathbb N$.\r\n\r\nMy solution isn't short so i want the other solution :)  :)\r\n\r\n[i]Edited by Myth[/i]",
  "Solution_1": "Please, don't call topics like \"nice but easy\"!\r\n\r\nRead http://www.mathlinks.ro/viewtopic.php?p=140098",
  "Solution_2": "We have $ab=\\lim_{n\\to\\infty} \\frac{[na]\\cdot [n^2b]}{n^3}=\\lim_{n\\to\\infty}\\frac{[n^3c]}{n^3}=c$, so $c=ab$.\r\n\r\nConsider the case $a,b>0$. Let $a=a_0,a_1a_2a_3...$ and $b=b_0,b_1b_2b_3...$ (binary representation).\r\nChoose large $n$ s.t. $2^na>10$ and $2^{2n}b>10$. Then $[2^na]\\cdot [2^{2n}b]=\\overline{a_0a_1...a_n}\\cdot \\overline{b_0b_1...b_{2n}}$ and $2^{3n}ab>\\overline{a_0a_1...a_n}\\cdot \\overline{b_0b_1...b_{2n}}+10\\cdot 0,a_{n+1}a_{n+2}...+10\\cdot 0,b_{2n+1}b_{2n+2}...$. It follows that $10\\cdot 0,a_{n+1}a_{n+2}...<1$ and $10\\cdot 0,b_{2n+1}a_{2n+2}...<1$, i.e. $a_{n+1}=0$ and $b_{2n+1}=b_{2n+2}=0$, so $a_n=b_n=0$ for all large $n$. And we can choose arbitrary base instead of 2. Therefore $a$ and $b$ are integer.\r\n\r\nI applied the same analysis for cases $(a>0,b<0)$, $(a<0,b>0)$ and $(a<0,b<0)$. However, to write it here is to long."
}
{
  "Tag": [],
  "Problem": "Exista vreun sir real marginit $\\left( x_{n}\\right)_{n \\geq 1}$ astfel incat $\\left| \\left( x_{i}-x_{j}\\right) \\left( i-j \\right) \\right| \\geq 1, \\, \\forall i \\neq j$?",
  "Solution_1": "Asta e problema 6 de la olimpiada sovietica 1978.\r\nUn exemplu de astfel de \u015fir este $x_{n}=4\\{n\\sqrt{2}\\}.$"
}
{
  "Tag": [
    "algebra solved",
    "algebra"
  ],
  "Problem": "if a+b+c=2 then prove that:\r\n \r\n3(a^3+b^3+c^3)+10(ab+bc+ca)>=16",
  "Solution_1": "after homogenisation it becomes $a^3+b^3+c^3+3abc\\geq ab(a+b)+bc(b+c)+ac(a+c)$. \r\nYou may suppose $a\\geq b\\geq c$.\r\nIf $c\\geq 0$, it is true by Schur.\r\nIf $c<0$; put x=a+b. Then the conclusion follows rapidly from x>2"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be a finite group of order $ n.$  Suppose $ G$ has a simple subgroup of order $ >n/3$ and also has a proper normal subgroup of order $ 3.$  Then $ n\\equal{}6.$",
  "Solution_1": "Suppose that $ H\\leq G$ is simple, $ N\\lhd G$, $ |N|\\equal{}3$, $ |H| > \\frac{n}{3}$.\r\n\r\nIf $ H \\equal{} G$, then $ G$ is simple so $ N\\equal{}G$ but then $ N$ is not a proper subgroup, contradiction.\r\n\r\nSo $ |H|\\equal{}\\frac{n}{2}$.  Let $ N \\equal{} \\langle x \\rangle$, then the cosets $ H$, $ xH$, $ x^2 H$ cannot be distinct, so we have $ x\\in H$.  Hence $ N \\lhd H$, so $ N\\equal{}H$ and $ 3 \\equal{} \\frac{n}{2}$."
}
{
  "Tag": [],
  "Problem": "I'm having trouble with this problem. Can someone help me, not necessarily giving me the answer but hints would be nice.  :)  thanks!\r\n[b]\nFor each of n=84 and n=88, find the smallest integer multiple of n whose base 10 representation consists entirely of 6's and 7's.[/b]",
  "Solution_1": "[hide]\nWell, you should prime factorize them, and then look at each factor by itself.\nFor example, 88 is 11*8, so you should see how you can arrange the last 3 digits to make it divisible by 8, then see which of these arrangements you can add a digit or two to to make it divisible by 11 and have it be the smallest.\n\nAnd, 84 is 7 * 4 * 3, so you look at arrangements of the last 2 digits, and you see it must be 76. So, obviously there are have to be 3 7s. (to be divisible by 3), so see how you can add 2 more 7's and as few 6s as possible to the left of the 76.\n[/hide]",
  "Solution_2": "wow! that makes a lot of sense, i never would have thought of that...thanks a lot klebian!!! :)"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "For each $n$, denote \r\n\\[ P_n(X)=nX^{n-1}+(n-1)X^{n-2}+...+2X+1.  \\]\r\nFind all $n$ such that the polynomial $P_n(X)$ is irreducible over $Q$.",
  "Solution_1": "I can prove that it's prime for $n$ prime... Valentin Vornicu has a similar problem, for $n=2004$.\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=75053&postorder=asc",
  "Solution_2": "As I assume everyone is well aware $P_n(X)=(X+X^2+...+X^n)'$ so I made a few computations and got that $P_n(X)=\\frac{X^{n+1}-2X^n+1}{(X-1)^2}$ (assuming I made the computations correct)... now what?!... does this help?",
  "Solution_3": "[quote=\"spix\"]As I assume everyone is well aware $P_n(X)=(X+X^2+...+X^n)'$ so I made a few computations and got that $P_n(X)=\\frac{X^{n+1}-2X^n+1}{(X-1)^2}$ (assuming I made the computations correct)... now what?!... does this help?[/quote]\r\nMaybe, but I thinks that sense is difficult to finish.",
  "Solution_4": "Guys, don't kid yourself, the problem is still open as far as I know.",
  "Solution_5": "[quote=\"harazi\"]Guys, don't kid yourself, the problem is still open as far as I know.[/quote]\r\nI want to discuss this problem, and if any has solutions for some n, please write them here.\r\nfor n<20000, it has been done by computer aid.",
  "Solution_6": "for $n$ prime it's not hard... Use the lemma posted here\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=440913#p440913"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "A particel is moved along the x-axis by a force that measures 10/(x+1)^2 pounds at a point x feet from the origin. Find the work done in moving the particle from the origin to a distance of 9 feet.\r\n\r\ncan you give me the example and answer, here's my work\r\n\r\ni antidifferentiated 10/(1+x)^2 into U^-1 or 10U^-1 if you want to leave the 10 their, then i put in 1+x into the u,\r\n\r\n1/2 integral x=0, x=9, 10(1+9)^-1= 1  1 times 1/2 equals 1/2\r\n\r\nthen i plugged in 0\r\n\r\nand i got 5, then i subtracted 1/2 - 5=and this equals -9/2 \r\nand the answer is 9 ft-lb\r\n\r\nso what did i do wrong, and can u show me how u got the answer",
  "Solution_1": "[quote=\"afcweswarrior\"]i antidifferentiated 10/(1+x)^2 into U^-1 or 10U^-1[/quote] Which should have been $-10U^{-1}$ or $-10(1+x)^{-1}$. \n\n[quote=\"afcweswarrior\"]1/2 integral[/quote] Why did you put $1/2$ here?",
  "Solution_2": "when i substituted (\r\n1+x)\r\n\r\noh man, that's what i did wrong i substituted 1+x^2 \r\ni was suppost to let u=1+x"
}
{
  "Tag": [
    "calculus",
    "Olimpiada de matematicas"
  ],
  "Problem": "Este problema esta sencillo, pero bueno\r\n\r\nHalle todas las soluciones enteras al sistema de ecuaciones\r\n$a+b+c=1$\r\n$a^3+b^3+c^2=1$",
  "Solution_1": "Bonito problema. Tengo dos soluciones, pero prefiero no publicarlas todavia. Ahora, doy una \"pista\" bastante evidente que puede ayudar:\r\n\r\n[hide=\"Pista\"]\n1. Sirve aplicar productos especiales, reacomodando los terminos y dividiendo en casos. No son tantos.\n[/hide]",
  "Solution_2": "Quien quiere colocar su solucion?",
  "Solution_3": "Ok, coloco mi solucion \r\n\r\n[hide=\"\"Solucion\"\"] Primero, si c=1, obtenemos que \n$a+b=0\\Rightarrow a^3+b^3=(a+b)(a^2-ab+b^2=0=1-c^2$\nEntonces todas las triplas (a,b,c)=(x,-x, 1) cumplen. Ahora si $c\\not=1$ podemos escribir la segunda ecuacion como como\n$(a+b)(a^2-ab+b^2)=(1-c)(1+c)$\nComo $c-1\\not=0$ podemos dividir a ambos lados por $c-1=a+b$ y nos queda\n$a^2-ab+b^2=1+c$\nAdemas, sabemos que \n$a^2+2ab+b^2=(1-c)^2$\nRestabdi ka segunda menos la primera, obtenemos que\n$3ab=c(c-3)$\nEntonces vemos que $3|c$, de donde $c=3k$ y queda que\n$ab=3k(k-1)$\n$a+b=1-3k$\nEntonces se tiene que $a,b$ son las raices de la ecuacion\n$X^2-(1-3k)X+3k(k-1)=0$\nDonde $\\Delta=-3k^2+6k+1$. Queremos que $\\Delta>0$, y con simples calculos obtenemos que $0\\leq k\\leq 2$. Por ultimo vemos que en cada uno de estos casos, $\\Delta$ es un cuadrado perfecto, entonces resolvemos y queda\nSi $k=0$, entonces $\\{a,b\\}=\\{0,1\\}$ y obtenemos las triplas $(a,b,c)=(0,1,0);(1,0,0)$. Si $k=1$, $\\{a,b\\}=\\{-2,0\\}$ de donde se tienen las triplas $(a,b,c)=(-2,0,3);(0,-2,3)$. Por ultimo, cuando $k=2$, obtenemos $\\{a,b\\}=\\{-3,-2\\}$, dandonos las triplas $(a,b,c)=(-3,-2,6);(-2,-3,6)$. Con esto se tienen todas las triplas y se sigue lo pedido.[/hide]"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "let a,b,c are the sides of a triangle.prove that:\r\n\\[ \\sum_{a,b,c}(\\frac{\\sqrt{b}\\plus{}\\sqrt{c}\\minus{}\\sqrt{c}}{\\sqrt{b\\plus{}c\\minus{}a}})\\leq3\\]",
  "Solution_1": "[quote=\"stvs_f\"]let a,b,c are the sides of a triangle.prove that:\n\\[ \\sum_{a,b,c}(\\frac {\\sqrt {b} \\plus{} \\sqrt {c} \\minus{} \\sqrt {c}}{\\sqrt {b \\plus{} c \\minus{} a}})\\leq3\n\\]\n[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=130812"
}
{
  "Tag": [
    "integration"
  ],
  "Problem": "Show that \\[\\sum_{i = 1}^{n + 1} \\frac{2^i}{i} \\cdot {n \\choose {i - 1}} = \\frac{3^{n + 1} - 1}{n + 1}.\\]",
  "Solution_1": "Hi,\r\n\r\n\r\nMay be proved by noting that   \\[{{n+1} \\choose {i}} = \\frac{n+1}{i} {{n} \\choose {i-1}}\\]\r\n\r\n(or by integrating binomial sum)",
  "Solution_2": "$\\sum_{i=1}^{n+1} \\frac{2^i}{i}\\ _nC_{i-1}=\\sum_{r=0}^n \\frac{2^{r+1}}{r+1}\\ _nC_r$ where $r=i-1$\r\n\r\n$=\\sum_{r=0}^n 2^{r+1} \\int_0^1 x^r dx\\cdot\\ _nC_r $\r\n\r\n$=2\\int_0^1 \\sum_{r=0}^n \\ _nC_r (2x)^r dx$\r\n\r\n$=2\\int_0^1 (1+2x)^n dx$\r\n\r\n$=2\\left[\\frac{(1+2x)^{n+1}}{2(n+1)}\\right]_0^1 =\\frac{3^{n+1}-1}{n+1}$",
  "Solution_3": "Here is a standard solution. \r\n\r\n$\\sum_{i=1}^{n+1} \\frac{2^i}{i} \\ _nC_{i-1}$\r\n\r\n$=\\sum_{i=1}^{n+1} \\frac{2^i}{i} \\frac{n!}{(i-1)!(n-i+1)!}$\r\n\r\n$=\\sum_{i=1}^{n+1} 2^i\\cdot \\frac{(n+1)!}{i!(n-i+1)!}\\cdot\\frac{1}{n+1}$\r\n\r\n$=\\frac{1}{n+1} \\sum_{i=1}^{n+1}\\ _{n+1}C_i\\ 2^i$\r\n\r\n$=\\frac{1}{n+1}\\left(\\sum_{i=0}^{n+1}\\ _{n+1}C_i\\ 2^i - \\ _{n+1}C_0\\ 2^0\\right)$\r\n\r\n$=\\frac{1}{n+1}\\{(1+2)^{n+1}-1\\}=\\frac{3^{n+1}-1}{n+1}$"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "LaTeX"
  ],
  "Problem": "For each of following systems of lin.equations Ax = B, factor A as LU, solve LC = B for c using forward substitution, solve UX = C be back substitution.  \r\n\r\n\r\n1)  2x + y = -3\r\n     6x + 4y = -4\r\n\r\n\r\n2) x + 2y + 0z = 3\r\n    2x + 6y -z =7\r\n    5x - 4y + 10z =4\r\n\r\n\r\nThank you",
  "Solution_1": "Can someone show me how to forward and backward substitute?",
  "Solution_2": "So do you mean elimination and back substitution? I am just learning so even if that is not so, may I try to solve?\r\n\r\nFOR #1 I think:\r\n\r\n1) 2x + y = -3 \r\n6x + 4y = -4 \r\n\r\nthe matrix is(sorry i don't have latex)\r\n\r\n2 1 -3\r\n6 4 -4\r\n\r\nso we mult. by -3 and add to row 2(note that 2 & 4 are pivots..):\r\n\r\n2 1 -3\r\n0 1  5\r\n\r\nand thus we have\r\n\r\n2x+y=-3\r\ny=5\r\nso \r\nx=-4\r\n\r\nIf you are still on this site please tell me if this is what you were looking for and I can show you #2.",
  "Solution_3": "First you need to perform the LU factorization.  For small systems with nice numbers you can usually do this in your head:\r\n\r\nFactorization:\r\n$A = LU \\implies \\begin{bmatrix}2 & 1 \\\\ 6 & 4 \\end{bmatrix}= \\begin{bmatrix}1 & 0 \\\\ 3 & 1 \\end{bmatrix}\\begin{bmatrix}2 & 1 \\\\ 0 & 1 \\end{bmatrix}$\r\n\r\nForward substitution goes from the top row to the bottom row.  Each row you solve gives you just enough values to solve for the next row:\r\n$L\\widehat{c}= B \\implies \\begin{bmatrix}1 & 0 \\\\ 3 & 1 \\end{bmatrix}\\begin{bmatrix}c_{x}\\\\ c_{y}\\end{bmatrix}= \\begin{bmatrix}-3 \\\\-4 \\end{bmatrix}\\implies \\begin{bmatrix}c_{x}\\\\ c_{y}\\end{bmatrix}= \\begin{bmatrix}-3 \\\\ 5 \\end{bmatrix}$\r\n\r\nBackward substitution is like forward substitution, except you start from the bottom row and work your way up:\r\n$U\\widehat{x}= C \\implies \\begin{bmatrix}2 & 1 \\\\ 0 & 1 \\end{bmatrix}\\begin{bmatrix}x \\\\ y \\end{bmatrix}= \\begin{bmatrix}-3 \\\\ 5 \\end{bmatrix}\\implies \\begin{bmatrix}x \\\\ y \\end{bmatrix}= \\begin{bmatrix}-4 \\\\ 5 \\end{bmatrix}$\r\n\r\nYou solve the second problem the same way you solve the first:\r\n$A = LU \\implies \\begin{bmatrix}1 & 2 & 0 \\\\ 2 & 6 &-1 \\\\ 5 &-4 & 10 \\end{bmatrix}= \\begin{bmatrix}1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 5 &-7 & 1 \\end{bmatrix}\\begin{bmatrix}1 & 2 & 0 \\\\ 0 & 2 &-1 \\\\ 0 & 0 & 3 \\end{bmatrix}$\r\n\r\nI've leave the rest up to you.",
  "Solution_4": "I like the way you do it! I'm going to remember it..."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Determine all natural numbers $ n$ and $ k\\geq 1$ such that $ k$ divides each of the numbers\r\n\r\n$ \\binom{n}{1},\\binom{n}{2},...,\\binom{n}{n\\minus{}1}$",
  "Solution_1": "Let $ p$ is prime divisor of k. Then $ p|\\binom{n}{1}\\equal{}n$. Let $ p^m|n, p^{m\\plus{}1}\\not |n$.\r\nIf $ n\\not \\equal{}p^m$ we consider $ p\\not |\\binom{n}{p^m}$. Therefore $ n\\equal{}p^m$. It give , that k had only one prime divisor p.\r\nIf $ k\\equal{}p^l,l>1$ we consider $ p^l\\not |\\binom{p^m}{p^{m\\minus{}1}}$. It give $ k\\equal{}p,n\\equal{}p^m$."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "[color=purple] Let a,b,c are three positive real numbers satisfy $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that :\n    $ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\plus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} x)(1 \\plus{} y)(1 \\plus{} z)}{\\sqrt {(1 \\minus{} x)(1 \\minus{} y)(1 \\minus{} z)}}$\n  It's of Walther Janous welcome hungkhtn, arqady, can_hang2007, Vacs, hope you can help me to prove this problem  :) [/color]",
  "Solution_1": "The left expression has 3 variables a, b, c but in the right expression has 3 variables x, y, z.\r\nCan you write exactly the problem?",
  "Solution_2": "[quote=\"evarist\"][color=purple] Let a,b,c are three positive real numbers satisfy $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that :\n    $ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\plus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} x)(1 \\plus{} y)(1 \\plus{} z)}{\\sqrt {(1 \\minus{} x)(1 \\minus{} y)(1 \\minus{} z)}}$\n  It's of Walther Janous welcome hungkhtn, arqady, can_hang2007, Vacs, hope you can help me to prove this problem  :) [/color][/quote]\r\n\r\nI think correct problem is:\r\n$ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\plus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} a)(1 \\plus{} b)(1 \\plus{} c)}{\\sqrt {(1 \\minus{} a)(1 \\minus{} b)(1 \\minus{} c)}}$",
  "Solution_3": "[quote=\"NguyenDungTN\"][quote=\"evarist\"][color=purple] Let a,b,c are three positive real numbers satisfy $ a \\plus{} b \\plus{} c \\equal{} 1$. Prove that :\n    $ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\plus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} x)(1 \\plus{} y)(1 \\plus{} z)}{\\sqrt {(1 \\minus{} x)(1 \\minus{} y)(1 \\minus{} z)}}$\n  It's of Walther Janous welcome hungkhtn, arqady, can_hang2007, Vacs, hope you can help me to prove this problem  :) [/color][/quote]\n\nI think correct problem is:\n$ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\plus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} a)(1 \\plus{} b)(1 \\plus{} c)}{\\sqrt {(1 \\minus{} a)(1 \\minus{} b)(1 \\minus{} c)}}$[/quote]\r\n\r\nIt's   $ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\plus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} a)(1 \\plus{} b)(1 \\plus{} c)}{\\sqrt {(1 \\minus{} a)(1 \\minus{} b)(1 \\minus{} c)}}$\r\n\r\nor   $ (1 \\plus{} a)\\sqrt {\\frac {1 \\minus{} a}{a}} \\plus{} (1 \\plus{} b)\\sqrt {\\frac {1 \\minus{} b}{b}} \\plus{} (1 \\plus{} c)\\sqrt {\\frac {1 \\minus{} c}{c}}\\ge\\frac {3\\sqrt {3}}{4}.\\frac {(1 \\plus{} a)(1 \\plus{} b)(1 \\plus{} c)}{\\sqrt {(1 \\minus{} a)(1 \\minus{} b)(1 \\minus{} c)}}$ ?\r\n\r\nI think the second is problem which evarist wants to post  :wink:",
  "Solution_4": "[color=purple] thank you Dung and doduclam the corriction problem must be as you posted. Sorry ervery one  :) [/color]"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c,d$ be reals such that $ a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2\\equal{}1$.\r\n\r\nProve that\r\n\r\n$ \\frac{1}{1\\minus{}ab}\\plus{}\\frac{1}{1\\minus{}ac}\\plus{}\\frac{1}{1\\minus{}ad}\\plus{}\\frac{1}{1\\minus{}bc}\\plus{}\\frac{1}{1\\minus{}bd}\\plus{}\\frac{1}{1\\minus{}cd}\\leq 8$",
  "Solution_1": "[u]Solution[/u] \r\n\\[ S \\equal{} \\frac {1}{1 \\minus{} ab} \\plus{} \\frac {1}{1 \\minus{} bc} \\plus{} \\frac {1}{1 \\minus{} cd} \\plus{} \\frac {1}{1 \\minus{} ca} \\plus{} \\frac {1}{1 \\minus{} bd} \\plus{} \\frac {1}{1 \\minus{} da}\r\n\\]\r\n\r\n\\[ \\therefore S \\minus{} 6 \\equal{} \\frac {ab}{1 \\minus{} ab} \\plus{} \\frac {bc}{1 \\minus{} bc} \\plus{} \\frac {cd}{1 \\minus{} cd} \\plus{} \\frac {ca}{1 \\minus{} ca} \\plus{} \\frac {bd}{1 \\minus{} bd} \\plus{} \\frac {da}{1 \\minus{} da}\r\n\\]\r\n\r\n\\[ \\equal{} \\sum_{cyc}\\frac {2ab}{2 \\minus{} 2ab}\r\n\\]\r\n\r\n\\[ \\equal{} \\sum_{cyc}\\frac {2ab}{2a^2 \\plus{} 2b^2 \\plus{} 2c^2 \\plus{} 2d^2 \\minus{} 2ab}\r\n\\]\r\n\r\n\\[ \\leq \\sum_{cyc}\\frac {2ab}{a^2 \\plus{} b^2 \\plus{} 2c^2 \\plus{} 2d^2}\r\n\\]\r\n\r\n\\[ \\equal{} \\sum_{cyc}\\frac {2ab}{(a^2 \\plus{} c^2 \\plus{} d^2) \\plus{} (b^2 \\plus{} c^2 \\plus{} d^2)}\r\n\\]\r\n\r\n\\[ \\leq \\frac {1}{2}\\sum_{cyc}(\\frac {a^2}{a^2 \\plus{} c^2 \\plus{} d^2} \\plus{} \\frac {b^2}{b^2 \\plus{} c^2 \\plus{} d^2}) \\equal{} 2\r\n\\]\r\nQED."
}
{
  "Tag": [],
  "Problem": "There is a secret message in this post.  First one to guess the secret message gets a prize. \r\n\r\n :D \r\n[hide=\" :D \"] The secret message is: The elephant ate my balloon. [/hide]  \r\n :D",
  "Solution_1": "The elephant ate my balloon.",
  "Solution_2": "There is a secret message in this post. First one to decode it gets absolutely nothing.\r\n\r\n[color=white]zngu92, gung jnf ernyyl rnfl.[/color]",
  "Solution_3": "math92, that was really easy.",
  "Solution_4": "The secret message is \r\nzngu92, gung jnf ernyyl rnfl.\r\n\r\nYay, I get absolutely nothing!  :lol:  Hey, wait a sec.",
  "Solution_5": "Haha, it is halfway almost very entertaining. A secret encoded crytopuzzle resides elegantly throughout my e-missive, so solve :P. A great encryption truly, only occasionally. Doesn't really always turn just while done... \r\n\r\n:D",
  "Solution_6": "to mathnerd314 or watever:\r\nzngu92, gung jnf ernyyl rnfl.",
  "Solution_7": "alan and ra, you guys are wrong\r\nyour things can be deciphered more  :)",
  "Solution_8": "[quote=\"ra5249\"]to mathnerd314 or watever:\nzngu92, gung jnf ernyyl rnfl.[/quote]\r\n\r\nMath92, that was really easy...\r\n\r\nNobody's solved mine yet :P",
  "Solution_9": "[hide=\"Solution\"]\nh i i h a v e a s e c r e t m e s s a g e t o o d r a t j w d\nThe conclusion follow immediately.  :P \n[/hide]",
  "Solution_10": "\"Hi, I have a secret message too. Dratjwd.\" What's with the last part?",
  "Solution_11": "[quote=\"bookaholic\"]\"Hi, I have a secret message too. Dratjwd.\" What's with the last part?[/quote]\r\n\r\nHi, I have a secret message too. Drat. JWD",
  "Solution_12": "I used to have this one for a signature, though as far as I know, no one has decrypted it. It's easy though if you just see the little trick... two words and you'll find it trivial.\r\n\r\n**N**0**1**7**P**Y**R**C**N**3**Y**7**5**4**N**5**1**H**7**3**3**5**U**N**4**C**",
  "Solution_13": "[quote=\"Peter VDD\"]**N**0**1**7**P**Y**R**C**N**3**Y**7**5**4**N**5**1**H**7**3**3**5**U**N**4**C**[/quote]\r\n[hide=\"My response to that, which leads to an obvious answer\"]Yes, I can see that nasty encryption.[/hide]",
  "Solution_14": "Huh, I used to wonder about that, and then I looked at again today, and got it without even looking at hint...",
  "Solution_15": "Okay, there's a secret message hidden in this post too.  Though I kinda stole it from somebody else.\r\n\r\nEDIT: Wait, I forgot to put the actual message in.[hide=\" :blush: \"][color=white][size=9]This is the message.[/size][/color][/hide]\r\n\r\n[color=white]This is not the message.  Keep looking.[/color]",
  "Solution_16": "OK, to miyomiyo, it is \"This is the message\".\r\nAnd to PeterVDD, I still can't figure out ur code...  :blush:",
  "Solution_17": "@ra5429\r\n\r\nthink 1337speak",
  "Solution_18": "Aargh, I finally get it!\r\n\r\n(For the record: I didn't see the post by sapphyre571)",
  "Solution_19": "[quote=\"math92\"]There is a secret message in this post.  First one to guess the secret message gets a prize. \n\n :D \n[hide=\" :D \"] The secret message is: The elephant ate my balloon. [/hide]  \n :D[/quote]\n\nLol, math92. I get it...It took me a whole day, but I get it... :rotfl: \n\nIn case someone wants one: [hide=\" A hint \"]Try to reply with a quote...[/hide]\n\n[hide=\" The answer \"] The elephant ate my balloon. ra5249 wasn't kidding. :wink: [/hide]",
  "Solution_20": "Uh...good job guys on decoding the post! You guys should all be cryptologists for the CIA.",
  "Solution_21": "[quote=\"math92\"]Uh...good job guys on decoding the post! You guys should all be cryptologists for the CIA.[/quote]\r\n\r\n:roll:\r\n\r\nFgiweoru bjerm qwivk iowp jmvegf.",
  "Solution_22": "[quote=\"KrazyNerd\"]ra5249 wasn't kidding.[/quote]Who claimed he was, huh? :huh: \r\n\r\nAs for peter's quote, I had tried leetspeek for some time now, even before seeing the hints, though why I had always started at the front will now remain a mystery. :)",
  "Solution_23": "[hide=\"decode this\"]\npublic void paint (Graphics g)\n\t{\n\t\tg.setColor(Color.YELLOW);\n\t\tPolygon ears=new Polygon();\n\t\tears.addPoint(33,7);\n\t\tears.addPoint(77,33);\n\t\tears.addPoint(118,32);\n\t\tears.addPoint(174,26);\n\t\tears.addPoint(133,47);\n\t\tears.addPoint(63,45);\n\t\tg.fillPolygon(ears);\n\t\t\n\t\tg.setColor(Color.BLACK);\n\t\tPolygon blackEars=new Polygon();\n\t\tblackEars.addPoint(43,14);\n\t\tblackEars.addPoint(48,28);\n\t\tblackEars.addPoint(34,8);\n\t\tg.fillPolygon(blackEars);\n\t\t\n\t\tPolygon blackEars2=new Polygon();\n\t\tblackEars2.addPoint(161,28);\n\t\tblackEars2.addPoint(174,26);\n\t\tblackEars2.addPoint(155,37);\n\t\tg.fillPolygon(blackEars2);\n\t\t\n\t\t//face\n\t\tg.setColor(Color.YELLOW);\n\t\tg.fillOval(119,41,20,49);\n\t\tg.fillOval(59,40,20,49);\n\t\tg.fillRect(67,40,61,49);\n\t\t\n\t\t//eyes\n\t\tg.setColor(Color.BLACK);\n\t\tg.fillOval(71,56,12,12);\n\t\tg.fillOval(117,55,12,12);\n\t\tg.setColor(Color.WHITE);\n\t\tg.fillOval(73,57,6,6);\n\t\tg.fillOval(119,56,6,6);\n\t\t\n\t\t//cheeks\n\t\tg.setColor(Color.RED);\n\t\tg.fillOval(63,70,12,12);\n\t\tg.fillOval(125,70,12,12);\n\t\t\n\t\tg.setColor(Color.BLACK);\n\t\tPolygon nose=new Polygon();\n\t\tnose.addPoint(100,73);\n\t\tnose.addPoint(97,70);\n\t\tnose.addPoint(103,70);\n\t\tg.fillPolygon(nose);\n\t\t\n\t\t//mouth\n\t\tg.drawArc(82,62,20,20,250,70);\n\t\tg.drawArc(98,62,20,20,220,70);\n\t\t\n\t\t//body\n\t\tg.setColor(Color.YELLOW);\n\t\tg.fillOval(58,87,32,68);\n\t\tg.fillOval(111,87,32,68);\n\t\tg.fillRect(74,87,53,68);\n\t\t\n\t\t//arms\n\t\tg.setColor(Color.BLACK);\n\t\tg.drawLine(75,102,87,129);\n\t\tg.drawLine(87,129,83,128);\n\t\tg.drawLine(83,129,83,131);\n\t\tg.drawLine(83,131,80,129);\n\t\tg.drawLine(80,129,80,134);\n\t\tg.drawLine(80,134,64,122);\n\t\tg.drawLine(121,106,114,127);\n\t\tg.drawLine(114,127,118,128);\n\t\tg.drawLine(118,127,119,130);\n\t\tg.drawLine(119,130,121,129);\n\t\tg.drawLine(121,129,123,131);\n\t\tg.drawLine(123,131,137,113);\n\t\t\n\t\t//tail\n\t\tColor brown=new Color(102,0,0);\n\t\tg.setColor(Color.YELLOW);\n\t\tPolygon tail=new Polygon();\n\t\ttail.addPoint(143,120);\n\t\ttail.addPoint(148,115);\n\t\ttail.addPoint(143,107);\n\t\ttail.addPoint(154,95);\n\t\ttail.addPoint(142,82);\n\t\ttail.addPoint(186,36);\n\t\ttail.addPoint(191,68);\n\t\ttail.addPoint(167,88);\n\t\ttail.addPoint(173,96);\n\t\ttail.addPoint(155,109);\n\t\ttail.addPoint(159,117);\n\t\ttail.addPoint(149,125);\n\t\ttail.addPoint(150,130);\n\t\ttail.addPoint(143,134);\n\t\tg.fillPolygon(tail);\n\t\t\n\t\tg.setColor(brown);\n\t\tPolygon brownTail=new Polygon();\n\t\tbrownTail.addPoint(143,120);\n\t\tbrownTail.addPoint(148,115);\n\t\tbrownTail.addPoint(156,112);\n\t\tbrownTail.addPoint(159,117);\n\t\tbrownTail.addPoint(149,125);\n\t\tbrownTail.addPoint(150,130);\n\t\tbrownTail.addPoint(143,134);\n\t\tg.fillPolygon(brownTail);\n\t\t\n\t\t//feet\n\t\tg.setColor(Color.YELLOW);\n\t\tPolygon leftFoot=new Polygon();\n\t\tleftFoot.addPoint(77,155);\n\t\tleftFoot.addPoint(72,160);\n\t\tleftFoot.addPoint(76,160);\n\t\tleftFoot.addPoint(77,160);\n\t\tleftFoot.addPoint(78,161);\n\t\tleftFoot.addPoint(81,163);\n\t\tleftFoot.addPoint(87,155);\n\t\tg.fillPolygon(leftFoot);\n\t\t\n\t\tPolygon rightFoot=new Polygon();\n\t\trightFoot.addPoint(127,155);\n\t\trightFoot.addPoint(135,159);\n\t\trightFoot.addPoint(131,159);\n\t\trightFoot.addPoint(132,160);\n\t\trightFoot.addPoint(129,160);\n\t\trightFoot.addPoint(128,162);\n\t\trightFoot.addPoint(118,155);\n\t\tg.fillPolygon(rightFoot);\n[/hide]",
  "Solution_24": "Oh, no...I don't have the program to run that, but I have a bad feeling...\r\n\r\nIt's not a pikachu, is it?",
  "Solution_25": "[quote=\"bubka\"][quote=\"KrazyNerd\"]ra5249 wasn't kidding.[/quote]Who claimed he was, huh? :huh: \n\nAs for peter's quote, I had tried leetspeek for some time now, even before seeing the hints, though why I had always started at the front will now remain a mystery. :)[/quote]\r\ni don't get it \r\nwhat is the message?",
  "Solution_26": "[quote=\"KrazyNerd\"]Oh, no...I don't have the program to run that, but I have a bad feeling...\n\nIt's not a pikachu, is it?[/quote]you win",
  "Solution_27": "@iamagenius: read backwards (obviously).\r\n\r\nI haven't got mathnerd's yet. It is not a Caesar shift, it may be a transposition.",
  "Solution_28": "[quote=\"bubka\"]I haven't got mathnerd's yet. It is not a Caesar shift, it may be a transposition.[/quote]\r\n\r\nIt's unbreakable.\r\n\r\n[hide]I just banged on the keyboard randomly and made it look convincingly like a code. Sorry about that.[/hide]",
  "Solution_29": "I have one! No one will find this one!! :D  :D",
  "Solution_30": "[quote=\"laughinghead505\"]I have one! No one will find this one!! :D  :D[/quote]Because there is none...",
  "Solution_31": "[quote=\"laughinghead505\"]I have one! No one will find this one!! :D  :D[/quote]Hmm, by trial and error, it's not by clicking the sixth box....",
  "Solution_32": "bnel aerl aq[]  2l3lkaso?",
  "Solution_33": "I have a code. \r\nit is:\r\n010000110110000101101110001000000111100101101111011101010010000001110010011001010110000101100100001000000111010001101000011010010111001100111111",
  "Solution_34": "C 4 N U 5 3 3 7 H 1 5 N 4 5 7 Y 3 N C R Y P 7 1 0 N\r\n\r\ncan you decipher ......... cryption...?\r\n\r\ni don't really get it?",
  "Solution_35": "Key: 13375P34K.",
  "Solution_36": "no one can de[code] this message[/code]",
  "Solution_37": "[quote=\"iamagenius\"]C 4 N U 5 3 3 7 H 1 5 N 4 5 7 Y 3 N C R Y P 7 1 0 N\n\ncan you decipher ......... cryption...?\n\ni don't really get it?[/quote]C4N = can\r\nU = you\r\n533 = see\r\n7H15 = this\r\nN457Y = nasty\r\n3NCRYP710N = encryption\r\n\r\n--> Can you see this nasty encryption?",
  "Solution_38": "[quote=\"sapphyre571\"]@ra5429\n\nthink 1337speak[/quote]\r\nlol\r\ni just figured it out  :blush:",
  "Solution_39": "[quote=\"hunter34\"]I have a code. \nit is:\n010000110110000101101110001000000111100101101111011101010010000001110010011001010110000101100100001000000111010001101000011010010111001100111111[/quote]\r\n\r\nDid you use [url=http://nickciske.com/tools/binary.php]this site[/url] to get binary from ASCII?\r\n\r\n[hide=\"Answer\"]Can you read this??[/hide]"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Hello\r\n\r\nHow can you prove mathematically that a complete bipatite graph\r\ncan only have a even number of edges?\r\n\r\nI can explain this by words or intuition but Im having trouble formulating\r\nthis is methematical forms.\r\n\r\nThanks!\r\nX",
  "Solution_1": "The complete bipartite graph $B_{1,1}$ has only one edge...",
  "Solution_2": "great I've found the answer:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=15325",
  "Solution_3": "[quote=\"eXKoR\"]great I've found the answer:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=15325[/quote]\r\n\r\nanswer to what?",
  "Solution_4": "To the problem given there, what he wrote is just nonsense."
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "calculus",
    "integration",
    "real analysis",
    "invariant",
    "geometric transformation"
  ],
  "Problem": "I have no idea where to put this, so I will put this here:\r\n\r\nIf a finite collection of points has an area of 0, and sometimes even an infinite collection of points (i.e. a line) has an area of 0, what justifies that a different infinite collection of points (like a rectangle) to have an area that is not 0?",
  "Solution_1": "Definition of area.  If a rectangle has area=0, then what is area?",
  "Solution_2": "Infinity is not always the same.  It's sort of weird but it's true.  Consider this example: The set of rational numbers, $ \\mathbb{Q}$ is obvioulsy infinte, as is the set of irrational numbers.  However, the set of irrational numbers is much, much larger.  Check this link out.  Interesting stuff.\r\n\r\nhttp://en.wikipedia.org/wiki/Countable_sets",
  "Solution_3": "Good question.\r\n\r\nNotions such as length, area, and volume (known as [u]measure[/u]) are not the same as the notion of [u]cardinality[/u]. They're both notions of \"amount\", but measure typically tells you more about the relative stucture/geometry of a set than it's cardinality.\r\n\r\nSee [url=http://en.wikipedia.org/wiki/Measure_Theory]Measure Theory[/url]. (Note: this is an abstract presentation of the idea of measure - something you wouldn't normally learn about until well into College.)",
  "Solution_4": "Actually, I was reading this in Apostol (the single variable calculus volume) where he defines area (for defining integration) firstly if a set is \"measurable.\" Perhaps my question would be better satisfied if someone also explained what the important of this is.",
  "Solution_5": "Here's the importance: You're reading about integration, and integrals (at their basic level) are based on limits of sums of areas. Well... what is the definition of area? This is a complicated question and a very important question.\r\n\r\nThe full blown answer to the question begins with the definition of a measurable set with respect to a measure (usually the Lebesgue measure). It turns out that there are some sets which are not (Lebesgue) measurable - basically meaning that they don't have a well-defined area in any usual sense.\r\n\r\nIf you think that area is an obvious concept that doesn't need definition, ask yourself the question, what exactly does it mean for the area of a circle to be $ \\pi r^2$ [i]square units[/i]?\r\nWhat is the area of the set $ S \\equal{} \\{(x,y) | x$ has no 2's in its decimal representation and $ y$ has at least one 2 in it's decimal representation$ \\} \\cap [0,1]X[0,1]$?",
  "Solution_6": "I know area is not an obvious concept and hence needs axioms to describe it, which is why I did not have an obvious answer to my own question. :) When I said importance, I meant the importance of measure and being measurable in this definition of area that he is describing, not the importance of defining area.\r\n\r\nAfter reading through those Wikipedia entries, it seems that measure is a $ \\sigma$-algebra, and hence its unions must countable. Thus the union of infinite points with area zero is not zero. Am I right? Thus area must be defined in a different manner...\r\n\r\nActually, someone can explain it to me anyway...",
  "Solution_7": "If you try to find a measure (volume) with \"expected\" properties (like: a rectangle/cuboid/etc has the canonical volume, it's invariant under rotation and translation, the volume of a disjoint union of the sum of the individual measures) for all sets of $ n$-dimensional space, then you get problems. The probably best known example is the Banach-Tarski-Pardoxon ( http://en.wikipedia.org/wiki/Banach-Tarski ).\r\nThe solution is simple: just don't try to measure everything.\r\nOne still want's to measure \"most\" sets, e.g. all open or closed sets. Then one can have the mentioned properties, finally leading to the Borel- and the Lebesgue-measure (or in more general manner to general measure theory). See http://en.wikipedia.org/wiki/Measure_(mathematics) .\r\nThen this has lots of good properties, and one can build integration theory solely on this (without much more analysis). This kind of integration has much more properties than \"standard\" integration.",
  "Solution_8": "I will say that right now you shouldn't worry about measurable sets and all that. That will come later when you study Real Analysis. Just be aware for now that there are some crazy subsets of $ \\mathbb{R}$ and $ \\mathbb{R}^2$ that don't have well-defined \"length\" or \"area\".\r\n\r\nSince you're studying integration, here are some not-too-theoretical, but fun, integration questions:\r\n\r\n1. What is the average distance of all the points in a circle from the center?\r\n2. What is the average distance of all the points in a sphere from the center?\r\n3. Prove that the gravity inside a hollow sphere of mass M equals 0 (physics knowledge required).",
  "Solution_9": "Question 3 is true by the shell theorem and can be shown for electrostatics as well.  Another interesting piece on area is the Dirac Delta function.  This is a function such that it is infinitesimally thin, yet has an area of 1.\r\n\r\nhttp://en.wikipedia.org/wiki/Dirac_delta_function",
  "Solution_10": "It isn't a function at all (and thus has no \"area\"). It's more a set of functions, the real limit function has area $ 0$ (or better: should not be considered).",
  "Solution_11": "[i]\n1. What is the average distance of all the points in a circle from the center?\n2. What is the average distance of all the points in a sphere from the center? [/i]\r\n\r\n[hide]I got 2/3 the radius for the first question and 3/4 the radius for the second question. What I did was look at all the concentric circles or spheres inside the given circle or sphere, multiply each concentric circle or sphere by its radius, add up all these values, and then divide by the area for #1 and the volume for #2.[/hide]",
  "Solution_12": "The dirac delta is only treated as a function for convenience in physics, it fits in better as an elementary example of a generalised function (or as they are called, a distribution).\r\n\r\nArea is notoriously difficult to define in elementary terms (the best we usually get away with is by defining axiomatically that a unit area exists and then describing the notion of area by a set of its properties), and better left till you're exposed to a rigorous theory of measures.",
  "Solution_13": "i know this may be somewhat off-topic, but you may be interested in looking up the Koch star. Similar to Gabriel's trumpet, it has the property of finite area with infinite perimeter. \r\n\r\nGabriels trumpet is a solid of revolution generated by rotating the graph of y=1/x about the x axis, and considering it from x=1-infinity. it has finite volume,(of pi if i remember correctly?), while it has infinite surface area.\r\n\r\nUnlike these two objects, Can any object have infinite volume(area), while still having finite surface area(perimeter)?",
  "Solution_14": "What is the definition of infinite volume/area? It means that at least one of its dimensions must stretch indefinitely (in informal terms). Thus its surface area/perimeter is always...",
  "Solution_15": "[quote=\"mathking123\"]i know this may be somewhat off-topic, but you may be interested in looking up the Koch star. Similar to Gabriel's trumpet, it has the property of finite area with infinite perimeter. \n\nGabriels trumpet is a solid of revolution generated by rotating the graph of y=1/x about the x axis, and considering it from x=1-infinity. it has finite volume,(of pi if i remember correctly?), while it has infinite surface area.\n\nUnlike these two objects, Can any object have infinite volume(area), while still having finite surface area(perimeter)?[/quote]\r\n\r\n\r\nHow about (and I'm speaking loosely here) rotating the Dirac Delta 'function' about the x-axis? The resulting surface area would be infinite (basically a disc of infinite radius) but the volume would be finite I think.\r\n\r\nBut I'm having trouble thinking of a standard function to fit this description."
}
{
  "Tag": [
    "blogs",
    "\\/closed"
  ],
  "Problem": "How can you create a second blog?\r\nI believe you could...",
  "Solution_1": "I am fairly certain that each user can only create one blog. You may, however, change the name of your original blog and then delete all the entries and shouts, so it would be a new blog, sort of.",
  "Solution_2": "See here:\r\n[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=967[/url]\r\n[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=893[/url]",
  "Solution_3": "I'm confused - Nerd_of_the_Ages apparently has two avatars.",
  "Solution_4": "[quote=\"isabella2296\"]I'm confused - Nerd_of_the_Ages apparently has two avatars.[/quote]\r\n\r\nThis was an issue that was dealt with before your time.",
  "Solution_5": "But this blog seems to have 'orginated' relatively recently...",
  "Solution_6": "Its a clone",
  "Solution_7": "[quote=\"#H34N1\"]Its a clone[/quote]\r\n\r\nThanks, I was trying not to give them any clever ideas.  :dry:"
}
{
  "Tag": [],
  "Problem": "When two lines are drawn in a plane, they can create four non-overlapping regions.  If five straight lines are drawn in the plane, what is the largest number of non-overlapping regions they can create?",
  "Solution_1": "The formula for maximum intersections of $ n$ lines in a plane is $ 1$ more than the $ n$th triangular number. \r\n\r\nThis formula can be written mathematically as $ \\frac {n(n \\plus{} 1)}{2} \\plus{} 1$.\r\n\r\nSo we plug in for $ n \\equal{} 5$.\r\n\r\n$ \\frac {5(5 \\plus{} 1)}{2} \\plus{} 1$\r\n\r\n$ \\frac {30}{2} \\plus{} 1$\r\n\r\n$ 15 \\plus{} 1$\r\n\r\nThe answer is $ \\boxed{16}$ intersections.",
  "Solution_2": "[quote=\"Textangle\"]The formula for maximum intersections of $ n$ lines in a plane is $ 1$ more than the $ n$th triangular number. \n...\nThe answer is $ \\boxed{16}$ intersections.[/quote]\r\n\r\nErr... this question isn't asking anything about intersections of lines, it's talking about regions of the plane.\r\n\r\nI'm also curious where you found this formula... please let me know, thanks!\r\n\r\nThe most regions I've been able to cook up is 15 :/  EDIT: and then like 30 seconds later, 16",
  "Solution_3": "i'm guessing text was solving for regions, but put intersections by mistake (because that's the right answer)",
  "Solution_4": "Wouldn't the maximum intersection of lines be the same as the number of regions?",
  "Solution_5": "You revived!  :mad: \n\nAnyway, no it isn't. The $n$th triangular number is the maximum number of intersections created by $n$ lines, however, the number of regions created is the $n$th triangular number plus $1$ (as said by Textangle).\n\nEdit: Facepalm.",
  "Solution_6": "[quote=\"$LaTeX$\"]You revived!  :mad:  [/quote]\n\nwhat do you mean?",
  "Solution_7": "Usually, you don't want to post in a thread that hasn't been posted on in three years. Or four months, for that matter..."
}
{
  "Tag": [],
  "Problem": "A huge-room has $ 1000$ doors, each with a lock provided and numbered $ 1$ to $ 1000$. A mad-man enters and at the first round, he unlocks all doors. At the second round, he locks only doors that have numbers the multiples of $ 2$. At the third round, he operates locks of doors which have a multiple of three as their number-he unlocks locked doors and locks doors which are unlocked. And this goes on for long. At his $ 1000^{th}$ round, how many doors are locked?",
  "Solution_1": "it does not really belong this sub-forum, does it ?",
  "Solution_2": "Every door changes it's state as many times, as many divisors it has. So, the unlocked doors have odd number of divisors, so they are perfect squares.",
  "Solution_3": "i'm confused...what happens in round 1000?"
}
{
  "Tag": [],
  "Problem": "Is it bad to take, say, SAT II Physics after only taking physics honors? is it necessary to learn teh AP material? Thanks.",
  "Solution_1": "i have no idea what [i]your[/i] school's honors physics class's curriculum is, but the SAT II tests are made for testing high school learning, not freshman college learning, which is what the AP tests (are supposed to) do. So no matter what physics class you took, as long as you took a high school course, you should be fine with some reviewing on the SAT IIs.",
  "Solution_2": "If I were you I would try to get a SAT II Physics review book, either from the library or purchase one, and see how much you know of it.  I agree with khan-it depends on the quality of your schools honors program.",
  "Solution_3": "I cant speak for Physics, but I took first year chemistry at my high school, got an A for the final grade, and I took the SAT IIs in June and got a 760.",
  "Solution_4": "don't take it unless you are going to study very HARD by yourself, very hard"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "If $f(x+1)=x^2+3x+5$, then find $f(x)$",
  "Solution_1": "[quote=\"4everwise\"]If $f(x+1)=x^2+3x+5$, then find $f(x)$[/quote]\r\n[hide]\n$f(x)=(x-1)^2+3(x-1)+5$\n\n$=x^2-2x+1+3x-3+5$\n\n$=x^2-x-3$[/hide]\r\n\r\nI hope that's right.  :(",
  "Solution_2": "[quote=\"Drunken_Math\"][hide]\n$f(x)=(x-1)^2+3(x-1)+5$[/hide]\n[/quote]\n\n[hide=\"clarification\"]\nHe just substituted $(x-1)$ for $(x+1)$ to get $f((x-1)+1)=f(x)$\n[/hide]"
}
{
  "Tag": [
    "articles",
    "geometry",
    "geometric transformation",
    "reflection"
  ],
  "Problem": "The cover edition of the latest edition of business week is found [url=http://www.businessweek.com/magazine/content/06_04/b3968001.htm]here[/url]. \r\n\r\nIt's an interesting article talking about more opportunities for math in the world, particularly in business and analyzing data.\r\n\r\nAny comments?",
  "Solution_1": "I notice the small numbers and the fact that we are finally having sufficient computer power and information available to make this a potential future area of great interest.  No market driven company like Ford or P&G can avoid the potential to be on the ground floor for the very small cost it takes to form such a group.\r\n\r\nIt may well be like those modeling the planet's climate -- the model is at least several orders of magnitude too simple to have meaningful results.  I'm hopefull, but expect this to be a dead end for at least a while.  Some areas, limited in scope and time, will probably work, but there will be large mistakes (model doesn't reflect real market).  For example with the test markets the army of MBAs unleached on businesses -- Polaroid almost lost the company when its test market for SX-70 failed to note that some folks who did not live in the test markets bought cameras anyway and the factory was built ten times too large.\r\n\r\nIt'll be great to look back 30 years from now and see how mislead we were in discussing this potential.  I'm sure we will be wrong -- I just don't know which way."
}
{
  "Tag": [
    "floor function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "The sum of $ k$ different even and $ l$ different odd natural numbers equals\r\n1997. Find the maximum possible value of $ 3k + 4l$ .",
  "Solution_1": "The sum of the odd numbers is at least $k^2$ (the sum of the first $k$ odd numbers), and the sum of the even numbers is at least $l^2+l$ (the sum of the first $l$ even numbers), so $3l+4k\\le3l+4\\lfloor\\sqrt{1997-l^2-l}\\rfloor\\le221$ for $0\\le l\\le44$ ($l<45$ since $45^2+45=2060>1997$).\r\n\r\nThis bound is achievable, taking all the even numbers less than $56$, all the odd numbers less than $67$, and the number $85$ (for instance).",
  "Solution_2": "[quote=\"Diarmuid\"]  $3l+4k\\le3l+4\\lfloor\\sqrt{1997-l^2-l}\\rfloor\\le221$ for $0\\le l\\le44$ ($l<45$ since $45^2+45=2060>1997$).\n\nThis bound is achievable, taking all the even numbers less than $56$, all the odd numbers less than $67$, and the number $85$ (for instance).[/quote]\r\nSo the maximum value is 436?And an another how do you find the number 221?",
  "Solution_3": "$436$? No, the maximum value is $221$.\r\n\r\nI found it by making a table of the relevant values, and looking (I did it in Excel, but it's small enough that you could make it by hand in an exam).  I don't know of any smarter way of doing it.",
  "Solution_4": "[quote=\"Diarmuid\"]\nThis bound is achievable, taking all the even numbers less than $56$, all the odd numbers less than $67$, and the number $85$ (for instance).[/quote]\r\nHow do you find this?",
  "Solution_5": "Oops - that should be $54$, not $56$.  The set chosen is the first $27$ even numbers, the first $34$ odd numbers, and $85$ as the remainder necessary to make up the sum.\r\n\r\nThe bound I found was that the maximum value of $3k+4l$ was $221$.  This can be achieved with $k=27$, $l=35$ (or with $k=23$, $l=38$, but we need an odd $l$ to have an odd sum).  So we take the first $27$ even numbers, and the first $35$ odd numbers, and add them up - but this is only $1981$, so we need another $16$ -  I got this by adding $16$ to one of the numbers (in particular, $69$)."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "trigonometry",
    "area of a triangle"
  ],
  "Problem": "Triangle ABC is equilateral  Extend $ AC$ through $ C$ to a point $ D$ such that $ AC \\equal{} DC$.  Let $ E$ be the midpoint of $ BC$.  Draw the segment from $ D$ to $ E$.  Extend $ DE$ to hit side $ AB$ at $ F$. Find the ratio of $ BF$ to $ FA$.",
  "Solution_1": "[hide]\nFirst, just set a side of $ ABC$ to be 2 (for the sake of ease) and $ BF$ to be $ x$.\nNow, we will find the ratio of the area of $ AFD$ to $ AED$ in two different ways.\n\nFirst, the altitude from F to AD has length $ \\frac{(2\\minus{}x)\\sqrt{3}}{2}$ since it is the longer leg of a 30-60-90 triangle.\nThe altitude from E to AD is $ \\frac{\\sqrt{3}}{2}$ since it is the shorter leg of a 30-60-90 triangle.\nNow, since $ AFD$ and $ AED$ have the same base of 4, the ratio of the areas is just $ 2\\minus{}x$.\n\nNext, a different way to find this ratio:\n$ [AFE] \\equal{} \\frac{2\\minus{}x}{4}[AED]$ so $ \\frac{[AFD]}{[AED]} \\equal{} \\frac{6\\minus{}x}{4}$.\n\nEquating the 2 ratios and solving, we have $ x \\equal{} \\frac{2}{3}$ so $ \\frac{BF}{FA} \\equal{} \\frac{1}{2}$.[/hide]\r\n\r\nPretty complicated for a MC problem...",
  "Solution_2": "[quote=\"alkjash\"]\n\nNext, a different way to find this ratio:\n$ [AFE] \\equal{} \\frac {2 \\minus{} x}{4}[AED]$ so $ \\frac {[AFD]}{[AED]} \\equal{} \\frac {6 \\minus{} x}{4}$.\n\nPretty complicated for a MC problem...[/quote]\r\n\r\nI think you should probably explain this.  Or I will =)... So $ \\angle EAD\\equal{}\\angle EAB$.  Since the area of a triangle can be computed by $ \\frac{1}{2}ab\\sin\\alpha$, the ratio of the two areas is simply the ratio of the two corresponding sides.  Since $ AE$ is a common side, we simply need $ \\frac{AB}{AD}\\equal{}\\frac{2\\minus{}x}{4}$.\r\n\r\nNice solution by the way.  I made the problem up xD. Use menelaus if you know it.  Gives you the answer in one step.",
  "Solution_3": "Ahh... right looked something like that..\r\nHaven't gotten around to memorizing Menelaus.  :blush: \r\nDoesn't have as nice a pattern as Ceva, you know?",
  "Solution_4": "Ceva's is found using menelaus :wink:",
  "Solution_5": "Ummm...\r\nNot necessarily...\r\nAt least the way I saw it proved..."
}
{
  "Tag": [
    "function",
    "inequalities",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $f$ be a function from $R$ to $R$ sastisfying $\\exists f''(x)\\ \\ \\ \\forall x\\in R$\r\nAssume that : $|f(x)|\\le a$ \r\n$|f'(x)|\\le b$\r\n$|f''(x)|\\le c$\r\nWhere $a,b,c$ are constant \r\nProve the inequality : $b\\le \\sqrt{2ac}$",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=44225"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "prove that in a triangle we have\r\nsinA*sinB+sinB*sinC+sinC*sinA<=cos(A/2)*cos(B/2)+cos(B/2)*cos(C/2)+cos(C/2)*cos(A/2);\r\nDo you have a strongert ineq.?",
  "Solution_1": "$2\\sum{sinAsinB}=\\sum(cosC+cos(A-B))$\r\n$2\\sum{cos(A/2)cos(B/2)}=\\sum(sin(C/2)+cos\\frac{A-B}{2})$\r\nBut $\\sum cosC \\le \\sum sin(C/2)$\r\n$\\sum cos(A-B) \\le \\sum cos\\frac{A-B}{2}$\r\n\r\nAm I wrong?"
}
{
  "Tag": [
    "limit",
    "calculus",
    "calculus computations"
  ],
  "Problem": "if $ A_n$ is not = 0 and $ {[A_n]}^{\\infty}_1$ Converges $ \\lim_{n \\rightarrow \\infty} \\frac{A_{n\\plus{}1}}{A_n} \\equal{} L \\Rightarrow |L| \\leq 1$\r\nhow can i prove it",
  "Solution_1": "Compare it with some geometric series."
}
{
  "Tag": [
    "function",
    "trigonometry",
    "limit",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Let $f,g\\;: \\mathbb R\\rightarrow\\mathbb R$ with $g(x)=\\sqrt{\\{x\\}-\\{x\\}^2}\\qquad$   and    $\\qquad f(x)=\\arccos(\\cos x)$; $\\{x\\}$ is the usual fractional part of x.\r\nI) Prove that the two functions are integrable in every conpact set of $\\mathbb R$\r\nII) Compute\r\n\\[ \\lim_{x\\to \\infty}\\frac{\\int_0^xf(t)\\,dt}{\\int_0^xg(t)\\,dt}  \\]",
  "Solution_1": "You can easily compute the upper integral for $x=k\\pi$, and the lower one for $x$ positive integer. Since the numerator goes to infinity as $x \\rightarrow + \\infty$, you can compute the quotient and you'll find $4$."
}
{
  "Tag": [
    "function",
    "topology"
  ],
  "Problem": "Define a function f from the interval I = [0, 2\u03c0) of R to R2 by setting f(t) = (cos t, sin t).\r\nShow that the inverse function  from f(I) to I is not continuous.",
  "Solution_1": "f : I -> R^2 has image f(I) = S^1 which is a compact subset of R^2; but the inverse map from f(I) to R maps S^1 to [0, 2 \\pi ) which is not compact  in R. therefore as the inverse image of a compact set is not compact, f inverse is not continuous.  :maybe:",
  "Solution_2": "[quote=\"buddhu\"]f : I -> R^2 has image f(I) = S^1 which is a compact subset of R^2; but the inverse map from f(I) to R maps S^1 to [0, 2 \\pi ) which is not compact  in R. therefore as the inverse image of a compact set is not compact, f inverse is not continuous.  :maybe:[/quote]\r\n\r\nThis works in this case since we are showing that the [i]inverse[/i] map is not continuous, but I just want to note the following: the continuous image of a compact set is compact (this is the result we used to show $ f^{\\minus{}1}$ is not continuous), but the preimage of a compact set under a continuous map is not necessarily compact. If the preimage of every compact set under a continuous map is always compact, then the map is called a proper map.\r\n\r\nIn other words, this exercise asks to show that $ f$ is not a proper map. I think buddhu understood this already, but I just wanted to make sure diophantine interprets the solution correctly."
}
{
  "Tag": [
    "function",
    "floor function",
    "trigonometry",
    "ceiling function",
    "logarithms",
    "calculus",
    "integration"
  ],
  "Problem": "Alright, so I'm looking for a way to rewrite the floor function as $x-$ (some composition with trigonometric functions)\r\n\r\nOne idea I have is using Arcsin(sin(x)) or something of that nature somehow. Any help in constructing an equivilent to [x] would be appreciated  :)",
  "Solution_1": "What's the problem with the current definition?",
  "Solution_2": "Nothing, I just want to see if floor can be written using trigonometric functions.",
  "Solution_3": "Well you're not going to be able to write it with continuous functions or any composition of such, because it's discontinuous at infinitely many points.",
  "Solution_4": "Alright, so I've made a bit of progress, but now I'm stuck:\r\n\r\nSo far I have:\r\n\r\n$\\lfloor x\\rfloor=x-\\sqrt{\\left(\\frac{\\sin^{-1}(\\cos (\\pi x))}{\\pi}-\\frac{\\sin^{-1}(\\sin (\\pi x))}{2\\sqrt{(\\sin^{-1}(\\sin (\\pi x)))^{2}}}\\right)^{2}}$\r\n\r\nHere's the thing: it works for all non-integers, but whenever I put in an integer, it encounters an indeterminate from the $\\sin (\\pi x)$.\r\n\r\nThe only reason that whole thing is there is to take care of the fact that the $\\frac12$ is either positive or negative depending on the value of $\\sin (\\pi x)$...",
  "Solution_5": "I was wondering if it was possible to define the ceiling function in terms of $\\lfloor x\\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.",
  "Solution_6": "[quote=\"lotrgreengrapes7926\"]I was wondering if it was possible to define the ceiling function in terms of $\\lfloor x\\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.[/quote]\r\n$\\lceil x\\rceil =-\\lfloor-x \\rfloor$\r\nI like to imagine it as a composition of reflections.",
  "Solution_7": "I'm not sure I understand exactly what you want. Can you tell me what's wrong with simply $\\lfloor x \\rfloor=x-\\frac{sin^{-1}(sin(2\\pi x))}{2\\pi}$",
  "Solution_8": "Because its not, since Arcsin(sin(x)) still increases and decreases.\r\n\r\nSo your suggestion would resemble the attached:",
  "Solution_9": "$\\frac{\\tan \\pi x/2 }{|\\tan \\pi x/2 |}= (-1)^{\\lfloor x \\rfloor}$, and then you just have to take a logarithm base $-1$ ...  :P \r\n\r\n[quote]...more \"accurately\" defined functions. [/quote]\r\n\r\nThere is no sense in which $\\sin$, $\\sin^{-1}$ or any other trigonometric function is \"more accurately defined\" then the plain old floor function.  (There is still, of course, the interest in doing something for the sake of doing it.)",
  "Solution_10": "I guess its more of that, just trying to do it for the sake of doing it.\r\n\r\nSee, the thing about your tan thing is, tangent function becomes zero at some integral values, so you'd get an indeterminate (which is what's wrong with\r\n$\\lfloor x\\rfloor=x-\\sqrt{\\left(\\frac{\\sin^{-1}(\\cos (\\pi x))}{\\pi}-\\frac{\\sin^{-1}(\\sin (\\pi x))}{2\\sqrt{(\\sin^{-1}(\\sin (\\pi x)))^{2}}}\\right)^{2}}$\r\nas well...",
  "Solution_11": "Do you just want a [i]finite[/i] function?  Because $x-\\lfloor x \\rfloor$ has a very simple Fourier series.",
  "Solution_12": "[quote=\"lotrgreengrapes7926\"]I was wondering if it was possible to define the ceiling function in terms of $\\lfloor x\\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.[/quote]\r\n$\\lceil x\\rceil=\\lfloor x\\rfloor+1-\\left\\lfloor 1-x+\\lfloor x\\rfloor\\right\\rfloor$.",
  "Solution_13": "[quote=\"rnwang2\"][quote=\"lotrgreengrapes7926\"]I was wondering if it was possible to define the ceiling function in terms of $\\lfloor x\\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.[/quote]\n$\\lceil x\\rceil=\\lfloor x\\rfloor+1-\\left\\lfloor 1-x+\\lfloor x\\rfloor\\right\\rfloor$.[/quote]\n\nBictor's answer\n\n[quote=\"Bictor717\"] $\\lceil x \\rceil =-\\lfloor-x \\rfloor$ [/quote]\r\n\r\nseems a little nicer to me.   :P",
  "Solution_14": "[quote=\"t0rajir0u\"][quote=\"rnwang2\"][quote=\"lotrgreengrapes7926\"]I was wondering if it was possible to define the ceiling function in terms of $\\lfloor x\\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.[/quote]\n$\\lceil x\\rceil=\\lfloor x\\rfloor+1-\\left\\lfloor 1-x+\\lfloor x\\rfloor\\right\\rfloor$.[/quote]\n\nBictor's answer\n\n[quote=\"Bictor717\"] $\\lceil x \\rceil =-\\lfloor-x \\rfloor$ [/quote]\n\nseems a little nicer to me.   :P[/quote]\r\n :rotfl:"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "incenter",
    "Euler",
    "power of a point",
    "radical axis",
    "geometry unsolved"
  ],
  "Problem": "Let P be the intersection of the diagonals AC and BD of the convex quadrilateral ABCD for which AB=CA=BD.\r\nLet O and I be the circumcenter and incenter of the triangle ABP.\r\nProve that if $ O \\neq I$, then $ OI \\perp CD$.\r\n :help: please~",
  "Solution_1": "If $ a \\equal{} BP, b \\equal{} PA, p \\equal{} AB$ are side lengths of the $ \\triangle ABP,$ $ \\overline{CP} \\equal{} p \\minus{} b, \\overline{DP} \\equal{} p \\minus{} a.$ External bisectors of the angles $ \\angle PAB, \\angle ABP$ cut $ BP, AP$ at $ X, Y$ and $ \\frac {\\overline{XP}}{\\overline{XB}} \\equal{} \\frac {b}{p},$ $ \\frac {\\overline{YP}}{\\overline{YA}} \\equal{} \\frac {a}{p}$ $ \\Longrightarrow$ $ \\overline{XP} \\equal{} \\frac {ab}{p \\minus{} b},$ $ \\overline{YP} \\equal{} \\frac {ab}{p \\minus{} a}$ $ \\Longrightarrow$ $ \\frac {\\overline{DP}}{\\overline{CP}} \\equal{} \\frac {p \\minus{} a}{p \\minus{} b} \\equal{} \\frac {\\overline{XP}}{\\overline{YP}}$ $ \\Longrightarrow$ $ YX \\parallel CD.$\r\n\r\nIf $ I_a, I_b, I_p$ are excenters of the $ \\triangle ABP,$ $ A, B, P$ are the altitude feet of the excentral $ \\triangle I_aI_bI_p,$ $ I$ is its orthocenter, $ (O)$ its 9-point circle, $ OI$ its Euler line. The circumcenter $ O'$ of the $ \\triangle I_aI_bI_p$ is on $ OI.$ $ API_aI_p$ and $ BPI_bI_p$ are cyclic because of the right angles $ \\angle I_pAI_a, \\angle I_pPI_a$ and $ \\angle I_pBI_b, \\angle I_pPI_b$ $ \\Longrightarrow$ $ \\overline {YP} \\cdot \\overline {YA} \\equal{} \\overline {YI_a} \\cdot \\overline {YI_p}$ and $ \\overline {XP} \\cdot \\overline {XB} \\equal{} \\overline {XI_b} \\cdot \\overline {XI_p}$ $ \\Longrightarrow$ $ YX$ is the radical axis of $ (O), (O')$ perpendicular to the center line $ O'O$ identical with $ OI$ $ \\Longrightarrow$ $ CD \\perp OI.$",
  "Solution_2": "Thanks a lot ~ I am happy for your solution.\r\n:coolspeak:"
}
{
  "Tag": [
    "MATHCOUNTS",
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "This is the thread that mock Mathcounts that I'll make would be posted.\r\n\r\nI'm not quite confident whether I'll be able to hold any timed contest (since I have other stuff to do), here is what I'll do.\r\n\r\nI'll post the questions on PDF.\r\n\r\nAfter a day or two, I'll post the solution packet on another post (PDF also).\r\n\r\nI'll probably include all sprint, target, team, countdown (I'm thinking of around 40 questions? Is that sound okay? :) ), and if I have good idea, maybe a master round (I have All-Time Greatest Mathcounts Problems so I might be able to know what they require in there) and like always, those solutions are what I think, not what you should have. :lol: There are always more than one way to solve the problems.\r\n\r\nBefore the answer packet is posted, you can discuss the questions or whatever you prefer to do.\r\n\r\nIf you're okay or happy with this idea, please post here. \r\n\r\nThanks and Good luck on your National MC.\r\n\r\nSilverfalcon",
  "Solution_1": "Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.",
  "Solution_2": "cool!",
  "Solution_3": "Should we follow regular time rules (40 min sprint 24 target 20 team)?",
  "Solution_4": "It depends on you.\r\n\r\nIf you want to be real, do it that way.\r\n\r\nIf you just want to see how it goes, do whatever you prefer.",
  "Solution_5": "I think this would be great! Old national problems are so hard to access  . . . I need more problems to practice on  . . ..\r\n\r\nAround what difficulty will this test be?",
  "Solution_6": "sounds good!",
  "Solution_7": "i'm in",
  "Solution_8": "I'll do it",
  "Solution_9": "I'm in.",
  "Solution_10": "Okay.",
  "Solution_11": "[quote=\"DarkKnight\"]Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.[/quote]\r\n\r\nAren't you in high school?",
  "Solution_12": "I thought he was in 6th grade--I could be wrong.",
  "Solution_13": "I'll be in it\r\n\r\nKaran Batra",
  "Solution_14": "[quote=\"Max300\"][quote=\"DarkKnight\"]Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.[/quote]\n\nAren't you in high school?[/quote]\r\n\r\nUhh? What? I'm in 6th grade. How'd you get the idea I was in high school?",
  "Solution_15": "[quote=\"Max300\"][quote=\"DarkKnight\"]Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.[/quote]\n\nAren't you in high school?[/quote]\r\n\r\nlol, akash, i think you have the wrong andrew...",
  "Solution_16": "[quote=\"DarkKnight\"][quote=\"Max300\"][quote=\"DarkKnight\"]Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.[/quote]\n\nAren't you in high school?[/quote]\n\nUhh? What? I'm in 6th grade. How'd you get the idea I was in high school?[/quote]\r\n\r\nthere was a guy who got 5th in state last year named Andrew Wang (at least in texas)",
  "Solution_17": "Yeah Andrew qualified for the USAMO.",
  "Solution_18": "[quote=\"tarquin\"]Yeah Andrew qualified for the USAMO.[/quote]\r\n\r\nYeah. I was looking at the USAMO qualifiers and that guy had the same name as me. :P I'm only in 6th grade.",
  "Solution_19": "i guess it's an easy mix up, but last year, andrew and mark were the two to beat all year in mathcounts, they both went to sartartia and always got like 1 and 2 with a 46 and 45, now mark moved to fort settlement and has dennis on his team instead...lol, hard to imagine mark zhang getting a non-perfect in mathcounts"
}
{
  "Tag": [],
  "Problem": "Posts related to Inorganic and Analytical Chemistry.",
  "Solution_1": "[url=http://www.mathlinks.ro/viewtopic.php?p=999973#999973]identify A,B...E[/url]"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "FTW"
  ],
  "Problem": "The surface of a $ 4'' \\times 5'' \\times 6''$ block is painted yellow. If the block is separated into one-inch cubes, what is the number of cubes with exactly one yellow face?",
  "Solution_1": "Uhh.... isnt this one just (2*3+3*4+2*4)*2=52????\r\n\r\nThe answer on FTW was 26, and i don't see what i did wrong.",
  "Solution_2": "I don't see what you did wrong either.  Maybe the answer-typer forgot to multiply by 2?",
  "Solution_3": "That's probably it.",
  "Solution_4": "I just reported it.",
  "Solution_5": "It is still not fixed.",
  "Solution_6": "It has been fixed. "
}
{
  "Tag": [
    "inequalities",
    "puzzles"
  ],
  "Problem": "Determine the maximum value that is obtained by multiplying together a set of positive integers which are all different and whose sum is 70.",
  "Solution_1": "[hide=\" I think it helps..\"]AM - GM gives us for the set of positive integers $ \\{a_1, a_2, ...,a_n\\}$\n\n$ \\frac{a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n}{n} \\geq \\sqrt[n]{a_1a_2...a_n}$\n\n$ \\frac{70}{n} \\geq \\sqrt[n] {a_1a_2...a_n}$\n\n$ \\left( \\frac{70}{n} \\right) ^n \\geq a_1a_2...a_n$\n\nMaximum occurs when there is equality. So now the problem is just how to optimize $ \\left( \\frac{70}{n} \\right) ^n$.[/hide]",
  "Solution_2": "[hide=\"comment\"]Nice methodology. However in terms of the given conditions the set of positive integers must be all different, and thus a(1), a(2),...., a(n) cannot be equal.[/hide]",
  "Solution_3": "Solution:\r\n[hide]Mark the numbers we choose for the set like this:\n_, 2, 3, 4, 5, _, _, _, 9, 10, 11, 12, _, 14\n\nThen if we ever have a blank after a number (like the 5) and then a different blank before a bigger number (like the 9), we can move those two numbers closer together (in this example changing them to 6 and 8). This will increase the product by AM/GM or Jensen's or whatever inequality you want to use.\n\nWe can perform this operation whenever there are two different blanks somewhere inside our set of numbers.\n\nThe set with maximal product, therefore, can have at most one blank inside it, so it must be a sequence of consecutive numbers with one of the numbers missing.\n\nAlso our set must have at least one of 2 and 3; otherwise we could just put 2 into the set and decrease the lowest odd number by 2, which increases the product. So, the string of consecutive numbers starts with either a 2 or a 3.\n\nFrom there getting the solution is easy:\n2, 3, 4, 5, 6, _, 8, 9, 10, 11, 12[/hide]",
  "Solution_4": "The AM-GM approach would still work if done in cases.  If we want $ k$ numbers that multiply to a maximal value, we can make them as close as possible to $ \\frac{70}{k}$ while still being distinct.  I'll have to check the cases on that though...",
  "Solution_5": "[quote=\"cyberspace\"]If we want $ k$ numbers that multiply to a maximal value, we can make them as close as possible to $ \\frac {70}{k}$ while still being distinct.[/quote]\r\nIn that case, the obvious thing to do is to assume that the numbers are consecutive. Suppose the numbers are $ n,\\,n \\plus{} 1,\\,n \\plus{} 2,\\,\\ldots,\\,n \\plus{} k \\minus{} 1.$ The sum of these numbers is\r\n\r\n[list]$ \\frac {k(n \\plus{} n \\plus{} k \\minus{} 1)}2\\ \\equal{} \\ 70$[/list]\ngiving $ k(2n \\plus{} k \\minus{} 1) \\equal{} 140.$ So $ k$ is a divisor of 140. Combining this with the given conditions, we find that the following are the only possibilities:\n\n[list]$ k \\equal{} 4,\\ n \\equal{} 16 \\\\\n[2mm] k \\equal{} 5,\\ n \\equal{} 12 \\\\\n[2mm] k \\equal{} 7,\\ n \\equal{} 7$[/list]\nHence the maximum product for a selection of consecutive integers is\n\n[list]$ 7\\times8\\times9\\times10\\times11\\times12\\times13 \\equal{} 8\\,648\\,640.$[/list]\r\n\r\nSo, are there non-consecutive integers whose product is higher than this (or can we prove that there aren\u2019t)? ;)"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Where x,y,z are positive reals, show that $x^{2}(x-y)(x-2y)+y^{2}(y-z)(y-2z)+z^{2}(z-x)(z-2x)\\geq 0$\r\n\r\nCan this be generalised Schur-style to $x^{r}(x-y)(x-2y)+y^{r}(y-z)(y-2z)+z^{r}(z-x)(z-2x)\\geq 0$?\r\n\r\nSorry if either of these questions is obvious, I have only just started learning inequalities properly.",
  "Solution_1": "the first ineq is a very strong ineq proposed by vasc.",
  "Solution_2": "Yeah sorry, I should have said the source of the first is Vasc (via the Hojoo Lee sheet). The second is my own hypothesis based on the first.\r\n\r\nAnyone got a proof?",
  "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?p=251877#p251877\r\nThere are three proofs of the first ineq there, one by Namdung, one by Vasc, oen by Stefan V. (posted by darij)",
  "Solution_4": "For the record, the generalisation is false. :("
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "1.  Find positive integers k, n such that kn! = (((3!)!)!/3! and n is as large as possible.  \n\n\n\n\n\nI got this off Kalva, but I think this question is bugged....\n\n\n\nAnswer and my solution in spoiler:\n\n\n\n[hide]\n\nI don't think Kalva had this problem correct.\n\n\n\n(((3!)!)!)/3!=720!/3!... So the largest n would be 720 while k would be 6.... I dunno where I went wrong.\n\n[/hide]",
  "Solution_1": "[quote=\"Ragingg\"]1.  Find positive integers k, n such that kn! = (((3!)!)!/3! and n is as large as possible.  \n[/quote]\r\n\r\nk*n!=720!/6\r\nk*n!=120*719!\r\n\r\nSo:\r\nk=120\r\nn=719",
  "Solution_2": "Erm.\r\n\r\n720!*6 is NOT equal to 720!/6.",
  "Solution_3": "Oh... Hmm I read the problem wrong. Woops! Wow... That was a stupid mistake.  :oops:",
  "Solution_4": "The question is indeed bugged. The original question (I hope AMC doesn't get mad at me for copyright issues) was\r\n1. Given that ((3!)!)!/(3!)=k*n!, where k and n are positive integers and n is as large as possible, find k+n.",
  "Solution_5": "[quote=\"captcha000\"]The question is indeed bugged. The original question (I hope AMC doesn't get mad at me for copyright issues) was\n1. Given that ((3!)!)!/(3!)=k*n!, where k and n are positive integers and n is as large as possible, find k+n.[/quote]\r\n\r\nI don't think the question was bugged; he just misinterpreted it for a second.",
  "Solution_6": "Yeah theres nothing wrong with the question.. thats exactly the same thing that you posted captcha :)"
}
{
  "Tag": [],
  "Problem": "What is the largest possible product of a set of positive integers whose sum is 20?",
  "Solution_1": "[quote=\"happyme\"]What is the largest possible product of a set of positive integers whose sum is 20?[/quote]\r\n\r\nEdit: hmm... \r\nMoldytape pointed that out...\r\nI should think about it more ...",
  "Solution_2": "[hide]Actually, 2^7x3^2=1152 is bigger and 2x7+3x2=20[/hide]",
  "Solution_3": "[hide]\n$3^{6}\\cdot2=1458$[/hide]",
  "Solution_4": "[hide]3s optimize, so put in as many 3s as possible (note that 2 2s is preferable to a 3 and a 1).[/hide]",
  "Solution_5": "bpms and DarkKnight are both right, while you can get 2^10 and 4^5, they both equal 1024.  So 3^6*2=1458. :D",
  "Solution_6": "[quote=\"happyme\"]What is the largest possible product of a set of positive integers whose sum is 20?[/quote]\r\n\r\nCan you please be a bit more specific on 'set'?"
}
{
  "Tag": [],
  "Problem": "How many integer solutions does $ xy \\plus{} 9(x \\plus{} y) \\equal{} 2006$ have?",
  "Solution_1": "Rewrite as $ (x \\plus{} 9)(y \\plus{} 9) \\equal{} 1925 \\equal{} 7*11*25$.\r\n\r\n$ x \\plus{} 9$ has to be one of the positive or negative factors of this number. Since there are 8 positive factors, there are 8 negative factors, making 16 solutions.",
  "Solution_2": "[quote=\"rofler\"]Rewrite as $ (x \\plus{} 9)(y \\plus{} 9) \\equal{} 1925 \\equal{} 7*11*25$.\n\n$ x \\plus{} 9$ has to be one of the positive or negative factors of this number. Since there are 8 positive factors, there are 8 negative factors, making 16 solutions.[/quote] You have to add $ 81$ not subtract it. So,$ (x \\plus{} 9)(y \\plus{} 9) \\equal{} 2087$\r\nAnd $ 2087$ is a prime. Which implies that $ 2$ positive and $ 2$ negative solutions and $ 4$ solutions in total.\r\nMoreover,I think that $ 1925 \\equal{} 7.11.5^2$ has $ 12$ positive divisors. BTW nice try. :)"
}
{
  "Tag": [
    "factorial"
  ],
  "Problem": "Find the smallest integer between 100 and 999 which is equal to the sum of the factorials of its digits.",
  "Solution_1": "[hide]I used guess and check and I got 145.[/hide]",
  "Solution_2": "If only I can simplify this: $100a+10b+c=a!+b!+c!$....., which is kind of hard...\r\n\r\nP.S. Robinhe, how did you just guess and check?",
  "Solution_3": "Well, I knew 5! was 120, so I thought that the number might be close to 120. So, I tried 125, 135, and I came to 145, which is my answer.  I just got lucky.",
  "Solution_4": "I think what Robinhe did is what I did. You want the lowest three digit one, and to get into the hundreds you need a 5, which would best go in the ones place to make it smallest. Since 5! is 120, and you need two more digits before the five, you try 1_5 and it works once you get to 145. Does this make sense? It's not the most elegant solution. Any better ideas anyone?\r\n[b]EDIT[/b] Beaten to it yet again.",
  "Solution_5": "Okay I understand, but still there should be a way to simplify it enough to deduce an answer....",
  "Solution_6": "If you start with 5 as your highest digit, you can make a table of factorial values to help you find an answer. \r\n\r\n5!+1!+2!=123\r\n5!+1!+3!=127\r\n...\r\n\r\nTHere's only like 10 possibilities."
}
{
  "Tag": [
    "real analysis",
    "function",
    "integration",
    "limit",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "I am just learning about distributions on my own for an interesting problem of how generally I may apply some idea. Anyway, my personal story is not of much importance to this question. \r\n\r\nIn all that follows I am thinking of tempered distributions in particular. \r\n\r\nI have just read that we cannot in general define pointwise product or convolution for distributions. In fact we can define it when one of the distributions has compact support. For example, the Dirac distribution is the identity operator for convolution, so why shouldn't that hold true when we convolve it with a distribution instead of a function? From this it follows we can only define pointwise product for distributions if the Fourier transform of one of them has compact support. \r\n\r\nI am told that Schwartz proved this ages ago. But I don't really see why it is so. What problem do we encounter? Thanks...",
  "Solution_1": "[quote=\"Xevarion\"]What problem do we encounter? Thanks...[/quote]\r\nI give you two versions of the same problem.\r\n\r\n1. Consider the Dirac delta, $\\delta.$ What is the pointwise product of $\\delta$ with itself?\r\n\r\nIf $f$ is a smooth function, then the pointwise product $f\\cdot \\delta$ is simply $f(0)\\delta.$ But since $\\delta(0)$ makes no sense, we clearly can't do that.\r\n\r\n2. Let $1$ be the function on $\\mathbb{R}$ whose constant value is $1.$ How can you possibly define $1*1?$\r\n\r\nYou can define convolution when one of the distributions is compactly supported - and that makes $\\delta$ the identity element for convolution even with distributions. Actually, with tempered distributions, you can go a little weaker than compactly supported, if you can make some sense out of what \"vanishes rapidly at infinity\" should mean for a distribution.\r\n\r\nYou can define pointwise multiplications when one of the distributions is repesented by a (tempered) smooth function - and that makes $1$ the identity element for pointwise multiplication.",
  "Solution_2": "Thanks, Professor Merryfield. \r\n\r\nOkay, so I guess the idea is that if I apply a distribution to a test function, it is no longer necessarily a test function. So then if we try to apply the second distribution, it doesn't make sense. \r\n\r\nOk, but how about this? Suppose we make a sequence of functions (nice functions, maybe $L^{2}$) that converges to the distribution. Then we can do stuff like pointwise products or convolutions with impunity. So what happens when we look at the limit of the pointwise product, inner product-ed with a test function? If the pointwise product makes sense for the distribution, will this converge to what we want, but when it doesn't, there is no limit?",
  "Solution_3": "Sure, let's do that. The only kind of limits that make sense are weak limits - that is, limits in the sense of distributions.\r\n\r\nLet $\\varphi$ be a test function such that $\\int\\varphi=1$ and $\\varphi(0)=1.$ We should be able to manufacture something like that. $\\varphi*\\varphi=\\psi$ is also a test function with $\\int\\psi=1.$\r\n\r\nNow let $f_{n}(x)=n\\varphi(nx)$ and $g_{n}(x)=\\varphi\\left(\\frac{x}{n}\\right).$ Take the limits: $f_{n}\\to\\delta$ and $g_{n}\\to1.$\r\n\r\nThe pointwise product $f_{n}\\cdot f_{n}$ is the function $n^{2}\\varphi^{2}(nx).$ If $\\eta$ is any test function such that $\\eta(0)\\ne0,$ then \r\n\r\n$\\lim\\int n^{2}\\varphi^{2}(nx)\\eta(x)\\,dx=\\lim n\\int\\varphi^{2}(u) \\eta(u/n)\\,du$ does not exist.\r\n\r\nThus collapses our effort to use such limits to find $\\delta\\cdot\\delta.$\r\n\r\nWe can compute that $g_{n}*g_{n}(x)=n\\psi(x/n).$ That is not going to have a weak limit, which collapses our effort to use such limits to find $1*1.$\r\n\r\nThat's the easy part. The question you're really asking is this: had those limits existed, would they be well-defined? Let me think about that."
}
{
  "Tag": [
    "function",
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find the maximum value of the function:\r\n\r\n$ f(x)\\equal{}\\frac{x}{x^2\\plus{}9}\\plus{}\\frac{1}{x^2\\minus{}6x\\plus{}21}\\plus{}\\cos {2 \\pi x}, x>0.$",
  "Solution_1": "$ f(x) \\equal{} \\frac {x}{x^2 \\plus{} 9} \\plus{} \\frac {1}{x^2 \\minus{} 6x \\plus{} 21} \\plus{} \\cos {2 \\pi x}\\ (x > 0).$\r\n\r\n$ \\leq \\frac {x}{(x \\minus{} 3)^2 \\plus{} 6x} \\plus{} \\frac {1}{(x \\minus{} 3)^2 \\plus{} 12} \\plus{} \\cos 2\\pi x\\leq \\boxed{\\frac {5}{4}}$ when $ x \\equal{} 3.$\r\n\r\nYou can also use AM-GM."
}
{
  "Tag": [],
  "Problem": "hi everyone, i just joined to your forum  :)",
  "Solution_1": "Welcome. \r\n\r\nWie geht's?",
  "Solution_2": "dankeschn, gut, und selbst ?  :D"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Find the point P in triangle ABC,to let the triangle PBC \u3001PAC and PAB have the same radius.\r\n :(  My English is so poor!!!",
  "Solution_1": "No one to help me... :(",
  "Solution_2": "I think that P is not constructible with ruler-compass since it must satisfy\n\n$ \\frac {[\\triangle APC]}{PC \\plus{} PA \\plus{} AC} \\equal{} \\frac {[\\triangle BPC]}{PB \\plus{} PC \\plus{} BC} \\ , \\   \\frac {[\\triangle APB]}{PA \\plus{} PB \\plus{} AB} \\equal{} \\frac {[\\triangle BPC]}{PB \\plus{} PC \\plus{} BC}$\n\nBoth loci are  not constructible by R & C.",
  "Solution_3": "I got this problem from a math magazine....I guess the point maybe the internal point...\r\nThank you all the same..."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Inegalitatea Medilor \r\n\r\n\\[(\\sum{x_i}\\frac{1}{n})^n\\geq{\\prod{x_i}}\\]\r\n\r\nInegalitatea Cauchy\r\n\r\n\\[\\sum{x^2_i}\\sum{y^2_i}\\geq(\\sum{x_iy_i})^2\\]\r\n\r\nInegalitatea Cebasev\r\n\r\n\\[n\\sum{x_iy_i}\\geq\\sum{x_i}\\sum{y_i} \\\\ unde \\ x = \\ ( \\ x_1 \\ , \\ x_2 \\ , \\ ..... , \\ x_n ) \\ \\ si \\ \\  y = \\ (  \\ y_1  \\ , \\ y_2 \\ , \\ ... \\ y_n ) \\ sunt \\\\ n-uple \\ la \\ fel \\ ordonate \\ \\\\ daca \\ \\ \\ \\ n-uplele \\ sunt \\ invers \\ ordonate \\ atunci \\ avem : \\\\n\\sum{x_iy_i}\\leq\\sum{x_i}\\sum{y_i}\\]\r\n\r\nInegalitatea Minkovsky\r\n\r\n\\[ \\sqrt{\\sum( x_i \\ + \\ y_i \\ )^2}\\leq\\sqrt\\ \\sum(x_i)^2} +\\sqrt\\ \\sum(y_i)^2 \\]\r\n\r\nInegalitatea Schur\r\n\r\n\\[ a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\\ge 0 \\]\r\n\r\nInegalitatea Holder\r\n\r\n\\[ \\sum(p_ix_iy_i)\\leq(\\sum(p_ix_i^\\alpha)^\\frac{1}{\\alpha})(\\sum(p_iy_i^\\beta)^\\frac{1}{\\beta})\\\\\\ unde \\frac{1}{\\alpha}+\\frac{1}{\\beta}=1\\]\r\n\r\nInegalitatea Jensen\r\n\r\n\\[ f( \\sum(p_ix_i))\\leq\\ \\sum(p_if(x_i)) \\ \\ unde \\ f \\ este \\ convexa \\ pe \\ intervalul \\ I \\\\ \\sum(p_i)=1 , p_1,p_2,....,p_n\\geq0 \\ \\ x_1,x_2,....,x_n\\ \\in \\ I \\]",
  "Solution_1": "Daca tot vrei sa faci o lista, vezi si [url=http://www.animath.fr/cours/inegalites.pdf]materialul asta.[/url]",
  "Solution_2": "[quote=\"rockymarcianocosmin99\"]Inegalitatea Medilor \n\n\\[ (\\sum{x_i}\\frac{1}{n})^n\\geq{\\prod{x_i} }\\]\n\nInegalitatea Cauchy\n\n\\[ \\sum{x^2_i}\\sum{y^2_i}\\geq(\\sum{x_iy_i})^2 \\]\n\nInegalitatea Cebasev\n\n\\[ n\\sum{x_iy_i}\\geq\\sum{x_i}\\sum{y_i} \\\\ unde \\ x = \\ ( \\ x_1 \\ , \\ x_2 \\ , \\ ..... , \\ x_n ) \\ \\ si \\ \\ y = \\ ( \\ y_1 \\ , \\ y_2 \\ , \\ ... \\ y_n ) \\ sunt \\\\ n-uple \\ la \\ fel \\ ordonate \\ \\\\ daca \\ \\ \\ \\ n-uplele \\ sunt \\ invers \\ ordonate \\ atunci \\ avem : \\\\n\\sum{x_iy_i}\\leq\\sum{x_i}\\sum{y_i} \\]\n\nInegalitatea Minkovsky\n\n\\[ \\sqrt{\\sum( x_i \\ + \\ y_i \\ )^2}\\leq\\sqrt\\ \\sum(x_i)^2} +\\sqrt\\ \\sum(y_i)^2  \\]\n\nInegalitatea Schur\n\n\\[ a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\\ge 0  \\]\n\nInegalitatea Holder\n\n\\[ \\sum(p_ix_iy_i)\\leq(\\sum(p_ix_i^\\alpha)^\\frac{1}{\\alpha})(\\sum(p_iy_i^\\beta)^\\frac{1}{\\beta})\\\\\\ unde \\frac{1}{\\alpha}+\\frac{1}{\\beta}=1 \\]\n\nInegalitatea Jensen\n\n\\[ f( \\sum(p_ix_i))\\leq\\ \\sum(p_if(x_i)) \\ \\ unde \\ f \\ este \\ convexa \\ pe \\ intervalul \\ I \\\\ \\sum(p_i)=1 , p_1,p_2,....,p_n\\geq0 \\ \\ x_1,x_2,....,x_n\\ \\in \\ I  \\][/quote]\r\n\r\n     Ai uitat Inegalitatea lui Bergstrom! :mad: Este des utilizata!"
}
{
  "Tag": [
    "Support"
  ],
  "Problem": "The war has to be brought speedily to a successful conclusion and the destruction of Japanese cities by means of atomic bombs may very well be an effective method of warfare. We feel, however, that such an attack on Japan could not be justified in the present circumstances. We believe that the United States ought not to resort to the use of atomic bombs in the present phase of the war, at least not unless the terms which will be imposed upon Japan after the war are publicly announced and subsequently Japan is given an opportunity to surrender.\r\nIf such public announcement gave assurance to the Japanese that they could look forward to a life devoted to peaceful pursuits in their homeland and if Japan still refused to surrender, our nation would then be faced with a situation which might require a re-examination of her position with respect to the use of atomic bombs in the war.\r\nAtomic bombs are primarily a means for the ruthless annihilation of cities. Once they were introduced as an instrument of war it would be difficult to resist for long the temptation of putting them to such use.  Both sides possess certain weapon capabilities (for example, poison gas), which, though not expressly outlawed, are understandably forbidden.  Every nation is held responsible for resisting the use of these tempting, yet inhumane, weapons.  If one nation was to commit a wartime atrocity, then all other nations would be justified in the\r\nThere are several alternates to the atomic bomb, which would be just as effective or perhaps more effective.  \r\nThe last few years show a marked tendency toward increasing ruthlessness. At present our Air Forces, striking at the Japanese cities, are using the same methods of warfare which were condemned by American public opinion only a few years ago when applied by the Germans to the cities of England. Our use of atomic bombs in this war would carry the world a long way further on this path of ruthlessness.\r\nAtomic power will provide the nations with new means of destruction. The atomic bombs at our disposal represent only the first step in this direction and there is almost no limit to the destructive power which will become available in the course of this development. Thus a nation which sets the precedent of using these newly liberated forces of nature for purposes of destruction may have to bear the responsibility of opening the door to an era of devastation on an unimaginable scale.\r\n\r\nWhat do you guys think?",
  "Solution_1": "I think that you should read the book of Revelation in the bible which will tell you what will happen in the (possibly near) future.",
  "Solution_2": "Do not think that anybody actually supports the use of atomic bombs or similar WMD. But the way things are moving, a nuclear war may become inevitable - it may even come in our lifetime!\r\nSo enjoy life and do your good deeds while you have time. If anything can persuade the beasts in the US Administration to stay away from the inhumane deeds, it won't be our innocent opinions. It will be more advisable to concentreate on games and maths!",
  "Solution_3": "I think this belongs in Round Table.",
  "Solution_4": "If a nuclear war breakes out, I hope you all are saved through Jesus!  Otherwise it would be a dark day in your life.  Eventually someone is going to be completely defeated through normal warfare, and in desperation destroy the world with bombs, bialogical diseases, and the like.  I think it was the right descision for us to bomb Japan, because if we had not done so, they would have continued fighting until every last one of them had been killed off (Examples at the many islands in the Pacific Ocean, the Japanese totally fought until they were all gone)  That is my opinion on the matter.",
  "Solution_5": "filletwho, I have an interest in history too, but I really hope a moderator moves this to other topics->round table, this is more the place to chill and talk about games, and ask random questions\r\n\r\nUS officials sometimes argued that if the war had continued without the use of the bombs, more japanse would have died\r\n\r\nI can't estimate whether or not this is true.\r\n\r\nPoint is, the bomb was developed because of fear the germans were developing one too.  Later inquiries after the liberation of western europe apparently pointed out the germans were never that close.  On the other hand, some historians claim that germany actually tested a small nuclear device during the war.\r\nAnyway, after huge spending in the manhattan project, I think they needed to show that it had paid off actually.  Also it was a good way of demonstrating superiority towards the russians.\r\n\r\n\r\nOne question, maybe a bit morbid, don't you think the russians actually saved thousands, perhaps millions of lives, by developing nuclear weapons too.  During the Korean war, general McArthur proposed bombing several chinese cities (he was later replaced because he appeared out of control).  The plan was rejected, partially because of fear of intervention of the Soviet Union, at that time already a nuclear power.\r\n\r\nCome to think of it, it must have been pretty scary for the soviet union to see the usa becoming the ultimate superpower, as the only nations with such weapons, even though it was only for a couple of years.  \r\n\r\nOne thing bothers me, the USA is the first to point fingers at nations developing nuclear weapons, and at the same time it is the only nation to have ever bombed another country. And as despicable as you may find nuclear weapons, it is a fact that nations like Pakistan or North Korea are less likely to be attacked, because they DO have nukes.  (On the other hand, the conflict at the end of sixties between Soviet Union and China shows that two nuclear nations clashing does not immediately imply a nuclear war)\r\n\r\nI am interested in your thoughts about this too."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "i send you the second problem.\r\ni wish you'll help me\r\nHere the problem,\r\nSolve equation:\r\n$ \\sqrt{1\\plus{}\\sqrt{1\\minus{}x^2}}\\equal{}x(1\\plus{}2\\sqrt{1\\minus{}x})$\r\n\r\nMany thanks to you!\r\nBest, Luan.",
  "Solution_1": "[hide]\nWell $ x\\equal{}0$ $ x\\equal{}1$ are solutions.\nI believe there are two others.[/hide]",
  "Solution_2": "Fraenkel, $ x\\equal{}0$ is an extraneous root.",
  "Solution_3": "[quote=\"10000th User\"]Fraenkel, $ x \\equal{} 0$ is an extraneous root.[/quote]\r\n\r\nlol my bad.\r\ni was seeing the wrong equation when i said that.",
  "Solution_4": "Thanks to interest the problem!\r\nBut i saw that, this equation has a root $ x\\equal{}1$. i can't find any other roots.can you see again and help me solve it.\r\nBest, Luan.",
  "Solution_5": "do you mind doing the algebra for your solution in latex for us?",
  "Solution_6": "Wait... did you just \"guess and check\" or actually do algebra to get $ x\\equal{}1$ ?\r\n\r\nBecause I saw that $ 1$ works, but it's too tedious to do it the long way... :D",
  "Solution_7": "lol i just looked at the equation and said x=1 is a solution.\r\n\r\nI think there is another real root. Having trouble finding it."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Is it true that if $a_{i}\\in[-1;1]$ and $\\sum_{i=1}^{2n}a_{i}^{3}=0$ then $\\sum_{i=1}^{2n}a_{i}\\leq 0?$",
  "Solution_1": "[quote=\"bilarev\"]Is it true that if $a_{i}\\in[-1;1]$ and $\\sum_{i=1}^{2n}a_{i}^{3}=0$ then $\\sum_{i=1}^{2n}a_{i}\\leq 0?$[/quote]\r\n\r\nIsn't this absurd? We can replace all $a_{i}$ by $-a_{i}$, the first condition remains true but the second changes its sign (and it's easy to see that equality can't always hold)?\r\n\r\n  darij",
  "Solution_2": "Yeah, I saw it after I posted the problem. It was stupid.\r\nThanks Darij :roll:",
  "Solution_3": "Let $a_{i}$=sin $x_{i}$\r\nWith $\\sum_{i=1}^{n}a_{i}^{3}$=0 \r\nWe can prove |$\\sum_{i=1}^{n}a_{i}$| $\\leq$ $\\frac{1}{3}$\r\n\r\nIt's simple,use sin3x=3sinx-4sin$x^{3}$ only",
  "Solution_4": "you should replace $\\frac{1}{3}$ with $\\frac{n}{3}$ :wink:"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "$a, b, c \\ge 0$.\r\n$x=a^{2}+bc$\r\n$y=b^{2}+ca$\r\n$z=c^{2}+ab$\r\nprove that\r\n\\[x^{3}+y^{3}+z^{3}+(6\\sqrt{2}-3)xyz\\ge \\sqrt{2}\\sum_{sym}x^{2}y \\]\r\n.",
  "Solution_1": "The inequalit is equivalent to\r\n\\[\\sum (x+y+z)(x-y)^{2}\\ge 2\\sqrt{2}\\sum z(x-y)^{2}\\]\r\nOr\r\n\\[\\sum (x-y)^{2}(x+y+(1-2\\sqrt{2})z) \\ge 0 \\]\r\nNote that $x-y=(a-b)(a+b+c)$, then it is equivalent to\r\n\\[(a+b+c)^{2}\\sum (a-b)^{2}(a^{2}+b^{2}+(1-2\\sqrt{2})c^{2}+ac+bc+(1-2\\sqrt{2})ab) \\ge 0 \\]\r\nor\r\n\\[\\sum (a-b)^{2}(a^{2}+b^{2}+(1-2\\sqrt{2})c^{2}+ac+bc+(1-2\\sqrt{2})ab) \\ge 0 \\]\r\n\\[\\sum S_{c}(a-b)^{2}\\ge 0 \\]\r\nwhere\r\n\\[S_{a}=(1-2\\sqrt{2})a^{2}+b^{2}+c^{2}+(1-2\\sqrt{2})bc+ab+ac \\]\r\n\\[S_{b}=a^{2}+(1-2\\sqrt{2})b^{2}+c^{2}+ab+(1-2\\sqrt{2})ac+bc \\]\r\n\\[S_{c}=a^{2}+b^{2}+(1-2\\sqrt{2})c^{2}+ac+bc+(1-2\\sqrt{2})ab \\]\r\nWLOG, assume $a \\ge b \\ge c \\ge 0$, we have\r\n\\[S_{b}\\ge (3-2\\sqrt{2})b^{2}+2(1-\\sqrt{2})bc+c^{2}=((1-\\sqrt{2})b+c)^{2}\\ge 0 \\]\r\nSimilarly $S_{c}\\ge 0$. Furthermore,\r\n\\[a^{2}S_{b}+b^{2}S_{a}\\ge 0 \\]\r\nThen we are done. :)",
  "Solution_2": "$x-y=(a-b)(a+b-c)$, not $(a-b)(a+b+c)$.\r\nSo your solution is wrong.  :(",
  "Solution_3": "Sorry, that's a big mistake.  :(  I will try it again.",
  "Solution_4": "In case no one's figured out yet, its false.\r\n[img]8631[/img]\r\nand that would have been an amazing proof, by the way :P",
  "Solution_5": "I think you are wrong.\r\nWhen $a=0.08, b=1, c=1$ -> $LHS-RHS=0.000605067 >0$.\r\nPlease check your calculation.\r\nIf $a=b$ -> $LHS-RHS = (3-2\\sqrt{2})(a-c)^{2}c^{2}((1+\\sqrt{2})c-a)^{2}>0$.\r\nI checked my solution, But I can't find any mistakes.",
  "Solution_6": "I'm not sure how it could be wrong, I'll double check the file but I'm pretty sure there is no mistake...",
  "Solution_7": "I think your program is not precise.\r\nI used Mathematica to calculate it.\r\nWhat program did you use?",
  "Solution_8": "[quote=\"me@home\"]In case no one's figured out yet, its false.\n[img]8631[/img]\nand that would have been an amazing proof, by the way :P[/quote]\r\n\r\nif $a=0.08$ $b=1$ $c=1$ then $x=1.0064$ and not $x=1.01$\r\n\r\nand moreover $LHS = 9.977726743....$ and $RHS = 9.977121676....$\r\n\r\nso $LHS-RHS=6.0506742*10{}^{-}{}^{4}$",
  "Solution_9": "Sorry for posting an irrelevant subject.\r\nI am planning to buy computer software for math.\r\nIs Mathematica better or Maple better?\r\nThey seem to be very expensive, so I'm thinking about it."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "geometric transformation",
    "reflection",
    "power of a point",
    "Pythagorean Theorem"
  ],
  "Problem": "$AB$ is a diameter of circle $O$. $AB$ is extended to $C$ so that $BC=2$. ($A$, $B$ and $C$ are collinear in this order.) $D$ is a point on circle $O$ so that $CD$ is tangent to the circle. If $AB=6$, find the length of $BD$.",
  "Solution_1": "[hide=\"Hint 1\"] Power-of-a-Point Theorem  :) [hide=\"Then, a second hint...\"] [hide=\"Only if you really cannot figure out what to do next  :P \"] Stewart's Theorem.  :wink:  [/hide] [/hide] [/hide]",
  "Solution_2": "Yeah, Stewart's Theorem works, but you don't even need to use it. There's an easier way :wink:",
  "Solution_3": "Well you could be lazy and use [hide=\"a piece of pie\"] Law of Cosines[/hide] also.",
  "Solution_4": "[hide]\nCD=4\nLet BD=a\nLet AD=b\n$a^{2}+b^{2}=36$\nFrom stewart's, we see that \n$96+2b^{2}=96+8a^{2}$\n$b^{2}=4a^{2}$\n$b=2a$\n$a^{2}+4a^{2}=36$\n$a=\\frac{6\\sqrt{5}}{5}$\n$b=\\frac{12\\sqrt{5}}{5}$[/hide]",
  "Solution_5": "[quote=\"cincodemayo5590\"]Well you could be lazy and use [hide=\"a piece of pie\"] Law of Cosines[/hide] also.[/quote]\r\n\r\nlol sam...you cannot use that if your solution sucks...random words are reserved for arnav/alex zhai who totally pwn\r\n\r\ni guess the fastest i see is using pythagorean theorem to get $AD^{2}+BD^{2}=36$ and $\\triangle ADC\\sim\\triangle DBC$ to get $DC=4$ and $\\frac{AD}{DB}=2$...then solve from there\r\n\r\ni suppose since i feel like showing off random geometry...here are two more solutions...\r\n\r\nreflect $B$ and $C$ over $AD$ to $B'$ and $C'$. Let $C'D$ intersect $AC$ at $E$. By menelaus' theorem, $\\frac{AE}{EB}*\\frac{BD}{DB'}*\\frac{B'C'}{C'A}=1$, $\\frac{AE}{EB}=4$...and $AE+EB=6$, so $AE=\\frac{24}{5}$.\r\n\r\nwe also have $\\angle BDC=\\angle B'DC'=\\angle EDB=\\angle DAB$, so $\\triangle DAB\\sim\\triangle EDB$, so $DB^{2}=EB*BA=6*6*4*\\frac{1}{5}$ so $DB=\\frac{12\\sqrt{5}}{5}$\r\n\r\n--\r\n\r\nor drop a perpendicular from D to AC, say D'...then AD'BC are in a harmonic range (consider other tangent from C, etc) $\\frac{AD'}{D'B}*\\frac{BC}{CA}=1$, $\\frac{AD'+D'B}{D'B}=5$, and $AD'+D'B=6$, so $AD'=\\frac{24}{5}$...using similarity, $\\triangle ADB\\sim\\triangle DD'B$, so $DB^{2}=D'B*AB$...same answer...",
  "Solution_6": "Um i dont know where i am going wrong but here\r\n\r\n[hide]you know that $OD=3$ and that $CD=4$. Now drop an altitude to $OC$ which has length $\\frac{12}{5}$. Call the foot of this altitude $N$. Now you know that $NC=5-\\frac{9}{5}=\\frac{16}{5}$ Now you know that since $BC=2$ that $NB=\\frac{16}{5}-2=\\frac{6}{5}$ now you have a right triangle $DNB$ with legs $\\frac{6}{5}$ and $\\frac{12}{5}$ so then $DB=\\frac{6\\sqrt5}{5}$[/hide]",
  "Solution_7": "[hide=\"I am just going to use Stewart for this...\"]$c=3$\n\n$b=4$\n\n$m=3$\n\n$n=2$\n\n$cnc+bmb=dad+man$\n\n$18+48 = d^{2}5+30$\n\n$d^{2}= \\frac{36}{5}$\n\n$d=\\frac{6\\sqrt{5}}{5}$\n[/hide]\r\nWHAT??? I'm getting the same answer as sonny.",
  "Solution_8": "i probably made a mistake...w/e"
}
{
  "Tag": [
    "limit",
    "function"
  ],
  "Problem": "If x is greater than 0, how large  must x be so that sqrt(x^2+x)-x shall differ from 1/2 by less than 0.02?",
  "Solution_1": "We solve the equation $ \\sqrt{x^{2}+x}-x=.48$, as it is apparent that $ \\sqrt{x^{2}+x}$ is always greater than $ x=\\sqrt{x^{2}}$.\r\n\r\n$ \\sqrt{x^{2}+x}-x=.48$\r\n$ x^{2}+x=(x+.48)^{2}$\r\n$ .04x=.2304$\r\n$ x=5.76$\r\n\r\nSo $ x$ must be greater than $ 5.76$ for the expression above to differ from $ \\frac{1}{2}$ by less than $ 0.02$.",
  "Solution_2": "<deleted>",
  "Solution_3": "Firstly, we need to show that $ \\lim_{x \\to+\\infty}({\\sqrt{x^{2}+x}-x}) = \\frac{1}{2}$.\r\n\r\nLet ${ y = \\sqrt{x^{2}+x}-x}$\r\n\r\n$ \\Longrightarrow y+x ={\\sqrt{x^{2}+x}}$\r\n$ \\Longrightarrow y^{2}+2xy+x^{2}= x^{2}+x$\r\n$ \\Longrightarrow y^{2}+2xy = x$\r\n$ \\Longrightarrow x(1-2y) = y^{2}$\r\n$ \\Longrightarrow x = \\frac{y^{2}}{1-2y}$\r\n\r\nAs $ y \\to \\frac{1}{2}$ from below, $ x \\to+\\infty \\Longleftrightarrow$ as $ x \\to+\\infty$,  $ y \\to \\frac{1}{2}$ from below.\r\n\r\nHaving shown that $ \\lim_{x \\to+\\infty}(\\sqrt{x^{2}+x}-x) = \\frac{1}{2}$, we now require the solution to $ {\\sqrt{x^{2}+x}-x}> 0.5-0.02 = 0.48$\r\n\r\n$ \\Longrightarrow x > \\frac{0.48^{2}}{1-2{\\cdot}0.48}$\r\n$ \\Longrightarrow x > 5.76$",
  "Solution_4": "[quote=\"sludgethrower\"]Firstly, we need to show that $ \\lim_{x \\to+\\infty}({\\sqrt{x^{2}+x}-x}) = \\frac{1}{2}$.[/quote]\n\nThis isn't actually necessary to address the stated problem, although it is the result that the problem suggests.  \n\n[hide=\"One line\"] $ \\sqrt{x^{2}+x}-x = \\frac{x}{ \\sqrt{x^{2}+x}+x}= \\frac{1}{ \\sqrt{1+\\frac{1}{x}}+1}$ [/hide]\n\n[quote=\"davidyko\"]We solve the equation $ \\sqrt{x^{2}+x}-x=.48$, as it is apparent that $ \\sqrt{x^{2}+x}$ is always greater than $ x=\\sqrt{x^{2}}$.\n\n$ \\sqrt{x^{2}+x}-x=.48$\n$ x^{2}+x=(x+.48)^{2}$\n$ .04x=.2304$\n$ x=5.76$\n\nSo $ x$ must be greater than $ 5.76$ for the expression above to differ from $ \\frac{1}{2}$ by less than $ 0.02$.[/quote]\r\n\r\nThe problem with this is that you assumed $ f(x) = \\sqrt{x^{2}+x}-x$ was an increasing function; without this knowledge, you also need to solve $ \\sqrt{x^{2}+x}-x = 0.52$, and even then you're assuming that $ f(x)$ is monotonic.",
  "Solution_5": "[quote=\"t0rajir0u\"][hide=\"One line\"] $ \\sqrt{x^{2}+x}-x = \\frac{x}{ \\sqrt{x^{2}+x}+x}= \\frac{1}{ \\sqrt{1+\\frac{1}{x}}+1}$ [/hide][/quote]\r\n\r\nDammit, it's solutions like these that make me want to learn more.\r\n\r\n\r\nAnd I suppose you could calc bash it into submission, as you can easily get the fact that it's monotonically increasing by taking the first derivative.",
  "Solution_6": "Fortunately, the fact that it's monotonic follows pretty easily from the rewrite I gave, which is perfectly elementary.   :) \r\n\r\nAnd if you're wondering about the motivation, I've seen many problems use that conjugation concept; for example, to telescope a summation.",
  "Solution_7": "\"Take the conjugate\" is a title of one of the chapters of [i]Math Olympiad Treasures[/i] by Titu Andreescu. Consequently, it is a technique that you should be able to recognize quickly."
}
{
  "Tag": [
    "function",
    "AMC"
  ],
  "Problem": "Which of the following is equivalent to $ \\displaystyle \\sqrt {\\frac {x}{1 \\minus{} \\frac {x \\minus{} 1}{x}}}$ when $ x < 0$?\r\n\r\n$ \\textbf{(A) } \\minus{} x \\qquad \\textbf{(B) } x \\qquad \\textbf{(C) } 1 \\qquad \\textbf{(D) } \\sqrt {\\frac x2} \\qquad \\textbf{(E) } x\\sqrt { \\minus{} 1}$",
  "Solution_1": "[hide=\"Answer\"]Algebraically simplifying, we have $\\sqrt{x^2}=x$. However, this is for $x<0$, and the square root function gives the positive root, so our answer is $\\boxed{A}$.[/hide]",
  "Solution_2": "i got A, but some of my freinds messed up with the sign and got B. which is really suprising considering it's only number 7.",
  "Solution_3": "I don't mean to revive, but I have a question:\n\nFor $ \\sqrt{x^{2}}=x $, doesn't squaring a negative make it positive? Then the square root of a positive is still positive, so I got $B$.... what's wrong with that?",
  "Solution_4": "Yes, but if $\\sqrt{(-x)^2}=x$, the answer is not $x$, it's $-x$.",
  "Solution_5": "But x is already a negative number, so -(-x) = positive x",
  "Solution_6": "But $x$ is negative. Think about it, $\\sqrt{ (-4)^2}=4=-(-4)$. It doesn't make sense for $\\sqrt{x^2}=x$ when $x$ is negative because square roots are always greater than or equal to $0$.",
  "Solution_7": "So if it was to be negative, it would actually be $E$? I'm confused right now....",
  "Solution_8": "Can you rephrase on what you are confused with?\nThe definition of square root is\n$\\sqrt{x^2}=| x |$\nIf $x\\ge 0$, then $\\sqrt{x^2}=x$\nIf $x<0$, then $\\sqrt{x^2}=-x$\n\n$\\sqrt{-x^2}=x\\sqrt{-1}$",
  "Solution_9": "Right so the answer isn't $A$, it's $E$?",
  "Solution_10": "If you square (E) you get $16*-1=-16$, not 16. \n\nJust think about it: a positive is the negative of a negative number. \n1=-(-1), doesn't it?\n\nLet $x=-1$\nSo $\\sqrt{x^2}=\\sqrt{(-1)^2}=\\sqrt{1}$\nWe always take the positive root (it's a definition that when we see the square root of something it is positive), so $\\sqrt{1}$ is equal to 1. \n1 does equal $-x$, as we said before.",
  "Solution_11": "$ \\sqrt{\\frac{x}{1-\\frac{x-1}{x}}}=\\sqrt{x^2} $ NOT $\\sqrt{-x^2}$\n\nLook at the diference\n$ \\sqrt{\\frac{x}{1-\\frac{x-1}{x}}}=\\sqrt{x^2}=| x |$ This would be $-x$ if $x<0$\n$ \\sqrt{\\frac{x}{\\frac{x-1}{x}-1}} =\\sqrt{-x^2}=|x|\\sqrt{-1}$ This would be $-x\\sqrt{-1}$ if $x<0$",
  "Solution_12": "No, the answer is A. The expression simplifies to $\\sqrt{x^2}$. Since $x=-k$ is negative, we have $\\sqrt{(-k)^2}=k=-x$.",
  "Solution_13": "Technically, couldn't it be both A and B?",
  "Solution_14": "No since $x<0$",
  "Solution_15": "Remember, it's the principle square root, so it's positive.  And since $x$ is negative, we want $-x$ which is positive.",
  "Solution_16": "\\begin{align*}\n\\sqrt{\\frac{x}{1-\\frac{x-1}{x}}} &= \\sqrt{\\frac{x}{\\frac{1}{x}}} \\\\\n&= \\sqrt{x^2} \\\\\n&= |x| \\\\\n&= \\boxed{-x}\n\\end{align*}"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "If $x,y,z > \\frac{1}{2}$ and $xyz \\geq 8$ prove that:\r\n\r\n$xy+yz+zx + \\frac{1}{\\sqrt{2}}(yz\\sqrt{x}+zx\\sqrt{y}+xy\\sqrt{z}) \\geq 3xyz$\r\n\r\n :)  :)  :)",
  "Solution_1": "[quote=\"gemath\"]If $x,y,z > \\frac{1}{2}$ and $xyz \\geq \\frac{1}{8}$[/quote]\r\n\r\nIsn't there any redundancy? :?",
  "Solution_2": "[quote=\"spider_boy\"][quote=\"gemath\"]If $x,y,z > \\frac{1}{2}$ and $xyz \\geq \\frac{1}{8}$[/quote]\n\nIsn't there any redundancy? :?[/quote]\r\n\r\nOK OK!! I have fixed it Thank you!!  :)",
  "Solution_3": "Why noboby help me??  :? or can you show me is it wrong ??"
}
{
  "Tag": [
    "inequalities solved",
    "inequalities"
  ],
  "Problem": "Let $a,b,c \\in (0,1)$ and $a+b+c=2$. Prove that  $4(a^{2}+b^{2}+c^{2}) \\geq 8-9abc$.\r\nWhen do we have equality?",
  "Solution_1": "homogenize and it's equivalent to schur's\r\n :D",
  "Solution_2": "What's about this problem ?\r\n\r\nLet 0< a, b ,c < 1 . a + b + c = 2 .\r\n  Prove that :\r\n  a  <sup>2</sup> + b <sup>2</sup>  + c <sup>2</sup>    \\leq  2 - 2abc",
  "Solution_3": "Homogenize, then put  x = m+n, y = n+p, z = p+m, m, n, p>0 (why we can?) and expand both sides, we come to an trivial inequality.\r\n\r\nNamdung"
}
{
  "Tag": [
    "invariant",
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Consider $K$ an arbitrary field and let $\\sigma_1, \\sigma_2$ two permutations in $S_n$ and $P(\\sigma_1), P(\\sigma_2)$ the permutation matrices associated to these permutations. Prove that $\\sigma_1, \\sigma_2$ are conjugated in $S_n$ if and only if the two matrices are similar. It is not that easy problem in any field.",
  "Solution_1": "The non-trivial implication is showing that $\\sigma_1,\\sigma_2$ are conjugates of one another if $P(\\sigma_i)$ are similar. \r\n\r\nWe can work in the algebraic closure of $K$ to conclude that the matrices have the same Jordan form. However, I think it's not that hard to see that for each $i\\in\\overline{1,n}$ the number of Jordan blocks of dimension $i$ in the Jordan form of $P(\\pi)$ is precisely the number of cycles of length $i$ in $\\pi$, so the conclusion follows, since two permutations are conjugate iff they have the same cycle structure.",
  "Solution_2": "Grobber, could you please be a little bit more explicit? I'm not familiar with Jordan form and I cannot prove your statement about the number of cycles. The solution I have is more difficult, but elementary. Your approach looks very nice, but I don't understand very well. Sorry for being dumb. :(",
  "Solution_3": "I've erased my previous response.\r\n\r\nThe whole space can be decomposed in subspaces invariant to the transformation represented by the matrix in a unique way s.t. the subspaces involved are indecomposable, i.e. cannot be further decomposed in invariant subspaces. The dimensions of these spaces are precisely the dimensions of the Jordan blocks of a matrix in Jordan form. \r\n\r\nIt's pretty clear now that for each cycle of length $i$ in the permutation, there is an indecomposable invariant subspace of dimension $i$, and that these spaces can be joined as a direct sum to form the large space, so the cycles correspond to the Jordan blocks.",
  "Solution_4": "Ok, thanks. :)"
}
{
  "Tag": [
    "vector",
    "linear algebra",
    "matrix",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Prove that the vectors {a_1,a_2,....,a_k} satisfying the $\\xi_1 a_1+\\xi_2 a_2+...+\\xi_n a_n=0 \\text{ where } \\xi_i\\ne 0 \\forall i$ make the linear (n-1)th-space  :blush: \r\nsorry i dont know really how this problem should be writen in english...",
  "Solution_1": "Did you mean by this that $\\textrm{rank}(a_1,a_2,\\dots,a_n)=n-1$?",
  "Solution_2": "i have found a solution involving the $rank(\\xi_1,...,\\xi_n)=1$so  the fundamental system of solution have the rank=n-1 as a corrallary from the theorem that if the matrix of the coefficients $a_{ij}, i,j=1 \\text{ to } n$ has the rank=k, then the rank of the vectors in the fundamental system of solutions( the system of  solutions,such that any other solution is the linear combination of this solution) equals to n-r .\r\n but i want to know is there another solution that doesn't  ******* to this theorem"
}
{
  "Tag": [
    "real analysis",
    "topology",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Is this result true?\r\n\r\nLet $ (\\Omega,\\mathcal{B})$ be a Borel-measure space and let $ \\mu$ be a $ \\sigma$-finite measure defined on it. If $ B\\in\\mathcal{B}$ such that $ \\mu(B)> 0$ then there exists a non-empty open set $ V$ that is contained in $ B$.",
  "Solution_1": "It is false. Just take $ \\mathbb R\\setminus\\mathbb Q$ and the Lebesgue measure on the line. It is even more apparently false if you take the $ \\delta$-measure for $ \\mu$ and the one point set it is supported by for $ B$.",
  "Solution_2": "A result which is true: $ B\\minus{}B$ contains an open set.",
  "Solution_3": "@ZetaX: could you explain that a bit? I didn't get what is $ B\\minus{}B$\r\n\r\nand what if $ \\Omega$ is locally compact and Hausdorff?",
  "Solution_4": "$ \\Omega \\equal{} \\mathbb R \\backslash \\mathbb Q$ is still a counterexample. You always get counterexamples if there is a countable basis for the topology.\r\n\r\nWith $ B \\minus{} B$ I meant the Minkowski-difference, meaning: $ B \\minus{} B \\equal{} \\{b_1 \\minus{} b_2 | b_i \\in B\\}$."
}
{
  "Tag": [
    "function",
    "algebra",
    "functional equation"
  ],
  "Problem": "Solve the functional equation $ f(x\\plus{}t)\\minus{}f(x\\minus{}t)\\equal{}4xt$\r\n\r\nhow do you [i]solve[/i] something like this?",
  "Solution_1": "Consider in particular the case $ x \\equal{} t$, then $ f(2x) \\equal{} 4x^2 \\plus{} f(0)$, which is satisfied only if $ f(x) \\equal{} x^2 \\plus{} c$ for some constant $ c$.",
  "Solution_2": "and what if t is something else? and hwo do u knwo its asking for f(x)?",
  "Solution_3": "The 'solution' to a functional equation is, by definition, an expression for the function in terms of its parameters.  So in this case, we are looking for $ f(x)$ in terms of $ x$.  Since the functional equation satisfies the equality for [i]any[/i] $ x$ and $ t$ it must satisfy it, in particular, for the case when $ x\\equal{}t$.  Thus, any $ f$ that satisfies your functional equation must necessarily satisfy $ f(x)\\equal{}x^2\\plus{}c$ for some constant $ c$.  We say, then, that $ f(x)\\equal{}x^2\\plus{}c$ is [i]the[/i] family of solutions to the functional equation.",
  "Solution_4": "[quote=\"JoeBlow\"]The 'solution' to a functional equation is, by definition, an expression for the function in terms of its parameters.[/quote]\r\n\r\noh...that helps a lot, wish i knew that earlier\r\n\r\n(no sarcasm)\r\n\r\nthanks\r\n\r\n :lol:",
  "Solution_5": "JoeBlow, can you show why that is the only solution?",
  "Solution_6": "He just did, in the post above.  :wink: (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191785]here[/url] also)."
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "What is next in the series:\r\n13,17,31,37,71,73,79,107,____",
  "Solution_1": "[hide]113,149,157,167[/hide]",
  "Solution_2": "jmadsen, what is the formation of the sequence?????",
  "Solution_3": "Primes everywhere, but I don't know how he got them.",
  "Solution_4": "[hide]Try reversing the digits.[/hide]",
  "Solution_5": "i dont understand... maybe it is like if you reverse it is still prime????",
  "Solution_6": "[quote] [b]Jmadsen wrote:[/b]\n113,149,157,167\nand\nTry reversing the digits\n[/quote]\r\n\r\n\r\n\r\nThat's right. The series contain primes whose reversal of digits is a different prime in ascending order.",
  "Solution_7": "You missed 97.",
  "Solution_8": "Well, that is a good one!\r\n:first:",
  "Solution_9": "[quote]\n[b]JBL wrote:[/b]\nYou missed 97[/quote]\r\n\r\nOops! Sorry for that."
}
{
  "Tag": [
    "calculus",
    "geometry"
  ],
  "Problem": "EVERYONE\r\n\r\nconway may seem pretty boring at first\r\nthis is because it's the first lecture and you aren't really used to getting lectured so much\r\nbut he really knows what he's talking about\r\nit's also pretty confusing at times, he goes on tangents often, but they're always related and relevant\r\n\r\nMAKE SURE YOU LISTEN IN HIS LECTURES",
  "Solution_1": "hey wait a minute are you allowed to record lectures?",
  "Solution_2": "no one can really stop you...",
  "Solution_3": "ahem\r\n\r\nalways relevant?\r\n\r\nwhy was he talking about like the earth or whatever when he was supposed to be explaining rainbows?\r\n\r\nAND SINCE WHEN WAS RAINBOWS THAT IMPORTANT TO MATH?!",
  "Solution_4": "well, rainbows is pretty physics-y\r\n\r\nlike I'm pretty sure on one of my physics tests two years ago, one of the questions asked us to explain rainbows\r\n\r\nand physics is kinda math?\r\n\r\nlike it's based on snell's law, I suppose\r\n\r\nand you could derive snell's law based on fermat's principle\r\n\r\nusing calculus of variations\r\n\r\nAND HA THAT'S MATH",
  "Solution_5": "[quote=\"Ubemaya\"]ahem\n\nalways relevant?\n\nwhy was he talking about like the earth or whatever when he was supposed to be explaining rainbows?\n\nAND SINCE WHEN WAS RAINBOWS THAT IMPORTANT TO MATH?![/quote]\r\n\r\n\r\nwell yeah lol i believe i fell asleep during that lecture\r\nBUT THAT WAS THE ONLY TIME DURING CAMP I SWEAR\r\nI SAT THROUGH ALL OF BERNDT'S LECUTRES (\"cue (q) cue cubed sub infinity, cue cubed cue cubed sub zero, cue squared cue cubed sub n cubed...\")\r\n\r\nbut yeah he was pretty funny in that lecture\r\n\"i'm one of only two people who know how rainbows work\"\r\n\r\nand yeah watch out for the \"i have one fault, i'm too modest\" thing, he'll say that at least once\r\n\r\nhis other lectures were pretty good though (bernoulli numbers, n-D geometry, fibonacci/lucas/similar sequences, infinite numbers), it would have been pretty cool if he had done cellular automata but w/e",
  "Solution_6": "In 07 he talked about pinecones :D",
  "Solution_7": "pinecone bombs? those are pretty awesome. my friends made one.",
  "Solution_8": "no, just relating them to Fibonacci Numbers",
  "Solution_9": "pft. ok  :rotfl:"
}
{
  "Tag": [],
  "Problem": "\u03a0\u03a1\u039f\u03a4\u0391\u03a3\u0397  3. \u0394\u03af\u03b4\u03b5\u03c4\u03b1\u03b9 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03bf \u0391\u0392\u0393 \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 \u0394 \u03c4\u03c5\u03c7\u03cc\u03bd \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03c0\u03af \u03c4\u03b7\u03c2 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ac\u03c2 \u0392\u0393. \u03a0\u03c1\u03bf\u03c3\u03b4\u03b9\u03bf\u03c1\u03af\u03c3\u03c4\u03b5 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u039c \u03b5\u03c0\u03af \u03c4\u03b7\u03c2 \u03b5\u03c5\u03b8\u03b5\u03af\u03b1\u03c2 \u0391\u0394, \u03ad\u03c4\u03c3\u03b9 \u03ce\u03c3\u03c4\u03b5 \u03c4\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 \u0391\u039c\u0392, \u0391\u039c\u0393, \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bd \u03af\u03c3\u03bf\u03c5\u03c2 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5\u03c2.",
  "Solution_1": "\u039f\u03c5\u03c3\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 (1)  :wink:",
  "Solution_2": "\u03a3\u03c4\u03b9\u03c2 \u03c3\u03b7\u03bc\u03b5\u03b9\u03ce\u03c3\u03b5\u03b9\u03c2 \u03bc\u03bf\u03c5 \u03bf\u03b9 \u03c0\u03c1\u03bf\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 (1), (3) \u03b5\u03af\u03bd\u03b1\u03b9 \u03b5\u03bd\u03bf\u03c0\u03bf\u03b9\u03b7\u03bc\u03ad\u03bd\u03b5\u03c2 \u03c3\u03b5 \u03bc\u03af\u03b1 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7, \u03bb\u03cc\u03b3\u03c9 \u03b1\u03c5\u03c4\u03ae\u03c2 \u03b1\u03ba\u03c1\u03b9\u03b2\u03ce\u03c2 \u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7\u03c2. \u03a4\u03b9\u03c2 \u03c0\u03b1\u03c1\u03bf\u03c5\u03c3\u03af\u03b1\u03c3\u03b1 \u03b5\u03b4\u03ce \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03bc\u03ad\u03bd\u03b5\u03c2, \u03b3\u03b9\u03b1\u03c4\u03af \u03ad\u03c4\u03c5\u03c7\u03b5 \u03ba\u03b1\u03b9 \u03b5\u03b3\u03ce \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03ce \u03bc\u03b5 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03be\u03b5\u03c7\u03c9\u03c1\u03b9\u03c3\u03c4\u03ac. \u0397 (3) \u03bc\u03c0\u03bf\u03c1\u03b5\u03af \u03bd\u03b1 \u03c3\u03c4\u03b1\u03b8\u03b5\u03af \u03bc\u03cc\u03bd\u03b7 \u03c4\u03b7\u03c2 \u03c3\u03b1\u03bd \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 \u03ba\u03b1\u03b9 \u03b7 (1) \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03c5\u03c3\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03ac \u03c0\u03cc\u03c1\u03b9\u03c3\u03bc\u03b1 \u03c4\u03b7\u03c2 (3).",
  "Solution_3": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd \u03b1\u03c6\u03bf\u03cd \u03bc\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ae\u03b8\u03b7\u03ba\u03b5 \u03bd\u03b1 \u03b2\u03ac\u03bb\u03bb\u03c9 \u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 here it is  :wink: \r\n\r\n\u0388\u03c3\u03c4\u03c9 \u0393' \u03c4\u03bf \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03cc \u03c4\u03bf\u03c5 \u0393 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7\u03bd \u0391\u0394 . \u03a4\u03cc\u03c4\u03b5 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 <\u0391\u0393\u039c = <\u0391\u0393'\u039c = <\u0391\u0392\u039c \r\n\u039a\u03b1\u03b9 \u03c4\u03cc\u03c4\u03b5 \u03c4\u03b1 \u0393', \u0392 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03af\u03b4\u03b9\u03bf \u03bc\u03ad\u03c1\u03bf\u03c2 \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf\u03c5 \u03c0\u03bf\u03c5 \u03c7\u03c9\u03c1\u03af\u03b6\u03b5\u03b9 \u03b7 \u0391\u0394 \u03ac\u03c1\u03b1 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03bf \u03c3\u03b7\u03bc\u03b5\u03af\u03bf \u03c4\u03bf\u03bc\u03ae\u03c2 \u03c4\u03bf\u03c5 \u03c0\u03b5\u03c1\u03b9\u03b3\u03b3\u03b5\u03b3\u03c1\u03b1\u03bc\u03bc\u03ad\u03bd\u03bf\u03c5 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5 \u03c4\u03bf\u03c5 \u0391\u0392\u0393' \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03bf\u03c5 \u03bc\u03b5 \u03c4\u03b7\u03bd \u0391\u0394 .  :)",
  "Solution_4": "\u0388\u03c4\u03c3\u03b9 \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b1 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b1, \u03c4\u03b1 \u03c3\u03c5\u03bc\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03ac \u03c4\u03c9\u03bd \u0392, \u0393, \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b7\u03bd \u0391\u0394, \u03ba\u03b5\u03af\u03c4\u03b1\u03b9 \u03b5\u03c0\u03af \u03c4\u03c9\u03bd \u03af\u03c3\u03c9\u03bd \u03ba\u03cd\u03ba\u03bb\u03c9\u03bd ( \u03bc\u03af\u03b1 \u03bc\u03b9\u03ba\u03c1\u03ae \u03bc\u03cc\u03bd\u03bf \u03c4\u03c5\u03c0\u03bf\u03b3\u03c1\u03b1\u03c6\u03b9\u03ba\u03ae \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b7: \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 <\u0391\u0393\u039c = <\u0391\u0393'\u039c = <\u0391\u0392\u039c ). \r\n\r\n     \u0391\u03bd \u03ba\u03b1\u03b9 \u03b3\u03b5\u03bd\u03b9\u03ba\u03ac \u03bf\u03b9 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ad\u03c2 ( \u03c0\u03b9\u03b8\u03b1\u03bd\u03cc\u03bd ) \u03b4\u03b5\u03bd \u03ad\u03c7\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03b1\u03af\u03b3\u03bb\u03b7 \u03c4\u03c9\u03bd \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03b9\u03ba\u03ce\u03bd \u03c0\u03c1\u03bf\u03c4\u03ac\u03c3\u03b5\u03c9\u03bd, \u03cc\u03c4\u03b1\u03bd \u03b6\u03b7\u03c4\u03b5\u03af\u03c4\u03b1\u03b9 \u03bf \u03c0\u03c1\u03bf\u03c3\u03b4\u03b9\u03bf\u03c1\u03b9\u03c3\u03bc\u03cc\u03c2 \u03c4\u03b7\u03c2 \u03b8\u03ad\u03c3\u03b7\u03c2 \u03b5\u03bd\u03cc\u03c2 \u03c3\u03b7\u03bc\u03b5\u03af\u03bf\u03c5, \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03b5\u03af\u03c4\u03b1\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03c3\u03c5\u03bd\u03b8\u03ae\u03ba\u03b7, \u03c3\u03c4\u03b7\u03bd \u03bf\u03c5\u03c3\u03af\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c0\u03ac\u03bb\u03b9 \u03bc\u03af\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03ba\u03c4\u03b9\u03ba\u03ae \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7.\r\n\r\n     \u03a3\u03b5 \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03b3\u03b9\u03b1 \u03c4\u03bf \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd \u03c3\u03bf\u03c5.\r\n\r\n     \u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.",
  "Solution_5": "\u0395\u03b3\u03ce \u03b5\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce \u03b3\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03c9\u03c1\u03b1\u03af\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 . \r\n\u039d\u03b1 \u03c3\u03b7\u03bc\u03b5\u03b9\u03ce\u03c3\u03c9 \u03b5\u03c0\u03af\u03c3\u03b7\u03c2 \u03cc\u03c4\u03b9 \u03b1\u03c6\u03bf\u03cd \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ac\u03c3\u03c4\u03b7\u03ba\u03b5 \u03b1\u03c5\u03c4\u03ae \u03ac\u03bc\u03bc\u03b5\u03c3\u03b1 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03ac\u03c3\u03c4\u03b7\u03ba\u03b5 \u03ba\u03b1\u03b9 \u03b7 \u03c0\u03c1\u03cc\u03c4\u03b1\u03c3\u03b7 (1) \u03cc\u03c0\u03c9\u03c2 \u03b5\u03af\u03c7\u03b1\u03bc\u03b5 \u03c0\u03c1\u03bf\u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03b5\u03b9  :wink:"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "search"
  ],
  "Problem": "Let $A, B \\in Mat_{n}(K)$. Prove that:\r\n1) $|rankA-rankB| \\leq\\ rank(A+B) \\leq\\ rank(A)+rank(B)$\r\n2) $rankA+rankB-n \\leq\\ rank(AB) \\leq\\ min (rank(A), rank(B))$",
  "Solution_1": "[quote=\"kingkong\"]Let $A, B \\in Mat_{n}(K)$. Prove that:\n1) $|rankA-rankB| \\leq\\ rank(A+B) \\leq\\ rank(A)+rank(B)$\n2) $rankA+rankB-n \\leq\\ rank(AB) \\leq\\ min (rank(A), rank(B))$[/quote]\r\n\r\nNo one can prove!!??!!",
  "Solution_2": "You should consider using the Search function.\r\nRelated to the second part: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=252804477&t=43691"
}
{
  "Tag": [
    "Support"
  ],
  "Problem": "what exactly is the spam policy?",
  "Solution_1": "Right, I think some serious action needs to be taken over something as stupid as this.. perhaps a banning for a couple of weeks? And then a permanent one if it continues? Thats usually what happens in other forums..",
  "Solution_2": "yes...he's just trying to annoy me...he kept doing the same in Fun & Games, but I deleted them all. He started here because I can't get to them...",
  "Solution_3": "i must be missing something...",
  "Solution_4": "apparently... me too...\r\n\r\nwell...\r\n\r\ndo some contests, like the Getting Started Contest.\r\n\r\n(yes, that was an ad, lol)",
  "Solution_5": "Ah you must have only seen this after it was edited.. mystic just came through posting about 3 separate messages with their subjects and contents just the word \"spam\".\r\n\r\nOh by the way.. as for the edited topic.. I don't think this should be in fun and games, either \"support\" or \"advanced problems\" would be very suitable for your spam problem..",
  "Solution_6": "lol!",
  "Solution_7": "If you spam in Advanced Problems or Other Problems Solving Topics, your spams will quickly be removed. I shall be sure to enforce that. I think TripleM will enforce it in Advanced Problems as well if he gets to it before I do.",
  "Solution_8": "I pledge my support.  I will try to remove any spam from AMC if it appears there as soon as I see it.  \r\n\r\nMaybe we should have a treaty of moderators or something."
}
{
  "Tag": [
    "geometry",
    "geometry theorems"
  ],
  "Problem": "Brahmagupta's formula gives the area of cyclic quadrilaterals. Is there a general formula which gives the area of cyclic polygons?",
  "Solution_1": ":?  this is the same as me... I've asked my self this and asked my self about what about the product of diagonals of a incribed 2n-gon (you know if n=4 ptolomey, if n=6 furhmann)",
  "Solution_2": "thx for the response... I would like to know the furhmann formula...."
}
{
  "Tag": [
    "inequalities",
    "inequalities open"
  ],
  "Problem": "If $ a,b,c,d,k \\ge 0$ prove that\r\n\\[ \\sum_{cyc} \\frac {a^k}{(a \\plus{} b \\plus{} c)^k} \\le \\max \\{ 2, \\frac {4}{3^k}\\}\r\n\\]\r\nI have solution only for $ k \\ge 1$ and $ k \\le \\frac {1}{2}$. :(",
  "Solution_1": "I think there is an inequality \r\n\r\nFor $ a,b,c \\geq 0 ,  s>r>0$\r\n$ (\\sum_{cyc} a^s)^ {\\frac {1}{s}} < (\\sum_{cyc} a^r)^ {\\frac {1}{r}}$  (Is it Jensen?)\r\n\r\nFor $ k \\geq 1$ we have  $ \\frac {\\sum_{cyc} a^k}{(a+b+c)^k} \\leq 1 \\leq RS$ \r\n\r\nFor $ k<1$ use power-mean\r\n\r\n${ \\frac {\\sum_{cyc} a^k}{(a+b+c)^k} \\leq 3^{(1-k)}}$\r\nObviosly $ 3^{(1-k)} \\leq \\frac {4}{3^k} \\leq RS$",
  "Solution_2": "i will try to come up with another solution if i can"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "$\\begin{gathered} f: N \\to N,p,k \\in N. \\hfill \\\\ f^{(1)} (n) = f(n),f^{(m + 1)} (n) = f^{(m)} (f(n)). \\hfill \\\\ \\end{gathered}$\r\nIf $f^{(p)} (n) = n + k$.Prove $p|k$.",
  "Solution_1": "It'a cute problem :) Let us define also $f^{(0)}(n)=n$. Now we have $f^{(p)}(n)=n+k$ Thus: \\[ f(n+k)=f^{(p+1)}(n)=f^{(p)}(f(n))=f(n)+k \\] so we've obtained $f(n+k)=f(n)+k$. Our first equation means that $f^{(p)}\\equiv f^{(0)}mod (k)$ It can be shown ( I can show it later if you want) that in that case all numbers $f^{(0)}(n),f^{(1)}(n),...,f^{(p-1)}(n),$ have different residues modulo $k$. One can also show that if for some $o,l,m,n$ $f^{(l)}(n) \\not\\equiv f^{(o)}(m)$ then for any $x,y$ $f^{(x)}(n) \\not\\equiv f^{(y)}(m)$. Let's take numbers $f^{(0)}(1),f^{(1)}(1),...,f^{(p-1)}(1)$ modulo $k$ these are $p$ diferent numbers. Now let's take the smallest residue  modulo $k$ which wasn't  chosen earlier. Let it be residue $r$. Then numbers $f^{(0)}(r),f^{(1)}(r),...,f^{(p-1)}(r)$ are different and so we have now chosen  $2p$  distinct residues modulo $k$....and so on ...I'm sure  you can see this :) we are always able to chose $p$ different residues, different to all residues we've chosen earlier. And as we have in total $k$ residues then $p|k$. Iknow this proof was not clear so I if you want me to write sth. more precisely just ask :)",
  "Solution_2": "[quote=\"jastrzab\"]It can be shown ( I can show it later if you want) that in that case all numbers $f^{(0)}(n),f^{(1)}(n),...,f^{(p-1)}(n),$ have different residues modulo $k$. One can also show that if for some $o,l,m,n$ $f^{(l)}(n) \\not\\equiv f^{(o)}(m)$ then for any $x,y$ $f^{(x)}(n) \\not\\equiv f^{(y)}(m)$. [/quote]\r\nWould you please show it more particularly?\r\nIt is not so easy to catch for me. ;)  ;)",
  "Solution_3": "Sorry....I've totally forgotten about this post  :)  Well first of all if for some $x,y$ we have $f^{(x)}(n )\\equiv f^{(y)}(m) \\; mod \\; k$ then for every $w$ we have  $f^{(x+w)}(n )\\equiv f^{(y+w)}(m) \\; mod \\; k \\; (*)$ Suppose that in the set $f^{(0)}(n), f^{(1)}(n),..., f^{(p-1)}(n)$ there are two numbers with the same residue $mod \\; k$ From this fact, there exists such indice $r<p-1$ for which $f^{(0)}(n) \\equiv f^{(r)}(n)$ But this means that $r|p$ Now consider  $f^{(r)}(n)$ if it's $f^{(r)}(n)=n$ we would also have  $f^{(p)}(n)=n$ which is not good for us... And if $f^{(r)}(n)=n+ak$ where $k\\geq1$ then we would have $f^{(p)}(n)=n+ak(\\frac{p}{r})>n+k$ so we have a contradiction which shows that in set \r\n$f^{(0)}(n), f^{(1)}(n),..., f^{(p-1)}(n)$ there are exactly $p$ different residues $mod \\; k$. Now the second part of this what was unclear...well actually I'think I haven't written precisely what I wanted to write...Maybe now I'll do it more clear :)  I want to show that if for some $l,o,m,n$ we have $f^{(l)}(n)\\equiv f^{(o)}(m)$ then thet ser of residues $f^{(0)}(n), f^{(1)}(n),..., f^{(p-1)}(n)$ is the same as set $f^{(0)}(m), f^{(1)}(m),..., f^{(p-1)}(m)$of course terms may be in different order, but sets are the same. This result comes straightforward from $(*)$ Now we are able to choose this disjoints sets of residues as we made in final part of this unclear proof",
  "Solution_4": "thank you very much!!\r\ntwo more questions:\r\n(1)why $f^{(0)}(n )\\equiv f^{(r)}(n)$\r\n(2)why $f^{(p)}(n)=n+ak(\\frac{p}{r})>n+k$\r\nforgive my foolishness :P",
  "Solution_5": "Well when we assume that there exist two different indices $i,j$ and $i<j$ such that $f^{(i)}(n) \\equiv f^{(j)}(n) \\; mod \\; k$ then from the fact we've mentioned earlier $f^{(i+(p-i))}(n) \\equiv f^{(j+(p-i))}(n) \\; mod \\; k$ but as $f^{(p)}(n) \\equiv f^{(0)}(n) \\; mod \\; k$  we obtain $f^{(0)}(n) \\equiv f^{(p+(j-i))}(n) \\equiv f^{(j-i)}(n) \\; mod \\; k$  so we have our first statement.... :)  now when we got that $r|p$ and $r<p$ we have also $p\\geq 2r$ ,so $n+(\\frac{p}{r})k \\geq n+2k$",
  "Solution_6": "oh ,I see~~\r\nthank you again,jastrzab"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "G is a group of order n and p is an odd prime. We know that there exist distinct elements of G, x<sub>i</sub> with i from 1 to p, s.t. x<sub>i</sub>x<sub>i+1</sub> is the same no matter what i we choose (the indices are considered modulo p). Show that 2p | n.",
  "Solution_1": "Looks the same like this one:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=35507"
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "You're given 9 empty boxes and the digits from 1 to 9. Each digit may only be used once and only one digit can be used in one box. Find some permutation of the digits such that \\[\\frac{\\square}{\\square}\\times \\square+\\frac{\\square \\times \\square \\times \\square}{\\square}+\\square \\times \\square =100.\\] Have fun. :D\r\n\r\n[hide=\"Answer\"]$\\frac{6*4}{8}+\\frac{2*5*7}{1}+3*9$[/hide]",
  "Solution_1": "Is that the only way to do it?\r\n\r\nHow would you arrive to that answer other than doing guess and check or maybe brute force?\r\n\r\n :ninja:  :ninja:",
  "Solution_2": "Nah, I've found 1 more. \r\n\r\n[hide=\"Approach\"]I considered the parity in two cases. You have\n\neven+even+even=even or\neven+odd+odd=even \n\n :) \n[/hide]"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "absolute value",
    "algebra unsolved"
  ],
  "Problem": "Let $ a,b_1,b_2,\\cdots,b_n,c_1,c_2,\\cdots,c_n$ be real numbers such that\r\n\r\n$ x^{2n}\\plus{}ax^{2n\\minus{}1}\\plus{}ax^{2n\\minus{}2}\\plus{}\\cdots\\plus{}ax\\plus{}1\\equal{}(x^2\\plus{}b_1x\\plus{}c_1)(x^2\\plus{}b_2x\\plus{}c_2)\\cdots(x^2\\plus{}b_nx\\plus{}c_n)$\r\n\r\nfor all real numbers $ x$. Prove that $ c_1\\equal{}c_2\\equal{}\\cdots\\equal{}c_n\\equal{}1$.",
  "Solution_1": "A partial proof:\r\nLet $ p(x)$ be the studied polynomial. $ f(x)\\equal{}p(x)(1\\minus{}x)\\equal{}\\minus{}x^{2n\\plus{}1}\\plus{}x^{2n}(1\\minus{}a)\\plus{}x(a\\minus{}1)\\plus{}1$. \r\n$ f''(x)\\equal{}x^{2n\\minus{}2}(\\minus{}\\alpha{x}\\plus{}\\beta)$ with $ \\alpha>0$. Thus $ f'$ is increasing then decreasing. Therefore $ f$ has at most 3 (with multiplicity) real roots and $ p$ has at most 2 (with multiplicity) real roots in the form $ u,1/u$. The $ p$- decomposition in irreducible over $ \\mathbb{R}$ is in one of the following forms: $ (x^2\\plus{}b_1x\\plus{}c_1)\\cdots(x^2\\plus{}b_{n\\minus{}1}x\\plus{}c_{n\\minus{}1})(x\\minus{}u)(x\\minus{}1/u)$ or $ (x^2\\plus{}b_1x\\plus{}c_1)\\cdots(x^2\\plus{}b_{n}x\\plus{}c_{n})$. It remains to prove that if $ z$ is a non real root of $ p$ then $ |z|\\equal{}1$.",
  "Solution_2": "Let $ z\\equal{}e^{2i\\alpha}.$ Then $ \\frac{z^{2n}\\plus{}1}{z^{2n\\minus{}1}\\plus{}...\\plus{}z}\\equal{}\\frac{\\sin(2n\\plus{}1)\\alpha}{\\sin(2n\\minus{}1)\\alpha}\\minus{}1.$ (Prove)\r\nNow consider $ f(\\alpha)\\equal{}\\sin(2n\\plus{}1)\\alpha \\plus{} (a\\minus{}1)\\sin(2n\\minus{}1)\\alpha\\equal{}0.$ Prove that it has $ n\\minus{}1$ roots in $ (0,\\pi /2).$ Thus the given polynomial has at least $ n\\minus{}1$ roots on the unit circle with arguments in $ (0, \\pi /2).$ Since $ p(x)\\in\\mathbb{R}[x]$ it has at least $ 2n\\minus{}2$ roots on the unit circle which yields to the statement from the problem.",
  "Solution_3": "[quote=\"loup blanc\"]and $ p$ has at most 2 (with multiplicity) real roots in the form $ u,1/u$[/quote]\r\nOnly if $ p(x)$ has a real root $ u\\not \\equal{}{1\\over u}$. \r\nOtherwise you need to show, that $ 1$ and $ \\minus{}1$ can't be both roots pf $ p(x)$.\r\nAlso if $ z$ is a non-real root of $ p(x)$ of absolute value $ \\not \\equal{}1$,\r\nthen $ z,\\bar{z},{1\\over z}$ are $ 3$ different such roots, which contradicts bilarev's findings.\r\n\r\nIncredible work, loup blanc and bilarev!",
  "Solution_4": "Hi olorin,\r\nyou are right; we must prove that $ \\pm1$ can't be both roots of $ p$; it's easy because else $ a\\equal{}2$ and $ n\\equal{}0$."
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "LaTeX",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Here are two problems for those who are interrested:\r\n \r\n $\\bb First Problem $\r\n\r\n   Let $f:[0,1]\\to\\mathbb{R}$ a continuous function such that $\\int_{0}^{1} f(x)dx=0$.Prove that there is $c\\in (0,1)$ such that $\\int_{0}^{c} f(x)dx=f(c)$.\r\n\r\n  $\\bb Second Problem$\r\n\r\n    Let $f:[a,b]\\to\\mathbb{R}$ a continuous function such that $\\int_{a}^{b} f(x)dx=0$.Prove that  there is $c\\in (a,b)$ such that $\\int_{a}^{c} f(x)dx=c\\cdot f(c)$",
  "Solution_1": "Still nobody? :(",
  "Solution_2": "Be patient. You only posted them 3 hours ago. :)",
  "Solution_3": "For the first problem consider the function $g(x)=e^{-x} \\cdot \\int_{0}^{x} f(t)dt$ \r\nObserve that $g(0)=g(1)=0$ and apply the Rolle's theorem.\r\n\r\nIs this ok?? :)  :)\r\n\r\nThe second is solved by the same way: taking the auxiliary function $h(x)=\\frac{\\int_{a}^{x} f(t)dt}{x}$\r\n\r\nBy the way, could you give me a latex form for the power that has base $e$ and exponent $-x$",
  "Solution_4": "At first glance,Socrates,I think it is correct. ;)",
  "Solution_5": "Ok.\r\nThey are quite clever problems from analysis!!! :D  ;)  :)"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $ x,y,z\\in R, x, y, z \\neq 0$ and $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} \\frac {1}{x \\plus{} y \\plus{} z}$\r\nProve that: $ \\frac {1}{x^{2009}} \\plus{} \\frac {1}{y^{2009}} \\plus{} \\frac {1}{z^{2009}} \\equal{} \\frac {1}{x^{2009} \\plus{} y^{2009} \\plus{} z^{2009}}$\r\n\r\n\r\n--------------------------------------------------------------\r\nMy school site: http://www.chuyenquangtrung.com.vn",
  "Solution_1": "If $ x\\plus{}y\\ge 0, y\\plus{}z\\ge 0, z\\plus{}x\\ge 0$, then only one of $ x,y,z$ may be negative. From $ xyz>0$ we get, that $ x>0,y>0,z>0$, therefore $ \\frac 1x\\plus{}\\frac 1y \\plus{}\\frac 1z >\\frac{1}{x\\plus{}y\\plus{}z}$.  :lol:",
  "Solution_2": "Thank Rust. I changed it, so.",
  "Solution_3": "[quote=\"thanhnam2902\"]Let $ x,y,z\\in R, x, y, z \\neq 0$ and $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} \\frac {1}{x \\plus{} y \\plus{} z}$\n[/quote]\r\nThis is equivalent to $ (x\\plus{}y)(y\\plus{}z)(z\\plus{}x)\\equal{}0$ . Done!  :P",
  "Solution_4": "Yes. Thank N.T.Tuan very muck. It's easy."
}
{
  "Tag": [
    "number theory open",
    "number theory"
  ],
  "Problem": "i read the following problem's solution and did not\r\nunderstand at some point allong the way:\r\n\r\nstatement:\r\nlet $p>0$ be prime and $p\\equiv 3 (mod 4)$\r\nproove that $p$ is prime also in $\\mathbb{Z}[i]$\r\n\r\nproof(that i partially understand):\r\nif p would not be prime in $\\mathbb{Z}[i]$ there would\r\nexist $\\alpha,\\beta \\in\\mathbb{Z}[i]$ so that\r\n$p^{2}=N(p)=N(\\alpha)N(\\beta)$ (N beeing the norm $N(a+bi)=a^{2}+b^{2}$)\r\nand through hypothesis( ?? ),we conclude that \r\n$N(\\alpha)=p$ but that would not be possible since the norm $N$ is\r\na sum of 2 squares wich is never congruent to 3 modulo 4.\r\n\r\n\r\n\r\nwhere ive posed the question marks is where i do not understand,\r\nwhat from hypothesis is used to state that $N(\\alpha)=p$,in my\r\noppinion $N(\\alpha)$ could also be $p^{2}$.\r\nwhere am i wrong ?",
  "Solution_1": "If $N(\\alpha)=p^{2}$ then $N(\\beta)$ is $1$ so $\\beta$ is a divisor of $1$ and the factorization $p=\\alpha\\beta$ is trivial..."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $G$ be a finite group, and $p$ be a prime dividing $\\left|G\\right|$. Let $P$ be a $p$-Sylow subgroup of $G$, and let $H$ be a subgroup of $G$ such that $N_{G}\\left(P\\right)\\subseteq H\\subseteq G$. Show that $H=N_{G}\\left(H\\right)$.\n\nHere, for any subset $T$ of $G$, we denote by $N_{G}\\left(T\\right)$ the normalizer of $T$ in $G$, defined by $N_{G}\\left(T\\right)=\\left\\{g\\in G\\mid gTg^{-1}=T\\right\\}$.\n\nEven a combinatorics \"genius\" like me has solved it, so don't expect it to be particularly challenging, but it is one of our nicer homework problems.\n\n  Darij",
  "Solution_1": "I remember mathmanman telling me this problem and having came with a nice solution. Here it is: take $g\\in N_{G}(H)$, then $gPg^{-1}$ is a $p$-Sylow of $H$ (it is already one of $G$!), as well as $P$. By Sylow theorems, they are conjugated IN $H$. Then $h^{-1}g\\in N_{G}(P)$ and thus in $H$, thus $g$ is also in $H$. Done. Even if a \"genius of combinatorics\" like me solved it quickly, I really found it interesting and nice (otherwise, I would have already forgot about it...)."
}
{
  "Tag": [
    "search",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "g is  a simple graph with n vertices none of the vertices have n-1 degrees\u3002and suppose any two of the vertices have and only have an adjacent vertex.prove \r\nthat every vertex has the same degree",
  "Solution_1": "If I understand the question correctly then the conclusion can in fact be much stronger: there is no such graph. My interpretation of the condition: $ G$ is a simple graph such that for every two distinct vertices $ u,v$ there is precisely one vertex which is adjacent to both $ u$ and $ v$ (this condition holds regardless of whether or not $ u$ and $ v$ are themselves adjacent. Since $ G$ is simple, $ z$ must be distinct from $ u$ and $ v$).\r\n\r\nThe only graphs which satisfy this condition are \"windmills\": graphs which are made up of triangles all fused at a single vertex. In other words, those are graphs where one vertex $ z$ is connected to all other vertices in the graph, which in turn are hooked up in pairs. \r\n\r\nIn particular, these graphs do have a vertex of degree $ n\\minus{}1$ where $ n$ is the number of vertices. If you exclude those, as this question seems to do, there's nothing left.\r\n\r\nI don't know any really \"elementary\" proof of the fact I quoted above. It's a nice example of a problem that can be effectively handled by algebraic methods. I'm willing to bet it was discussed here before, but if needed I can present a complete proof.",
  "Solution_2": "When you say \"handled by algebraic methods,\" do you mean by an eigenvalue computation?  I can sort of see how the proof would go, but I'm curious about the details.",
  "Solution_3": "The friendship theorem: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=132046071&t=48500\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=132046071&t=19197\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=55513"
}
{
  "Tag": [
    "number theory",
    "Divisibility"
  ],
  "Problem": "Prove that for any set of distinct positive integers $n_1, n_2, \\dots, n_k$ there exists a set of integer coefficients $a_1, a_2, \\dots, a_k$ such that\n$\\sum_{i=1}^k a_i {n_i p\\choose p}$ is divisible by $p^{2k-1}$ for all large enough primes $p$.\n\nExamples:\n$2 {p\\choose p} - {2p\\choose p}$ is divisible by $p^3$ for all primes $p\\geq 3$.\n$12{p\\choose p}  - 9{2p\\choose p} + 2{3p\\choose p}$ is divisible by $p^5$ for all primes $p\\geq 7$.\n$60{p\\choose p}  - 54{2p\\choose p} + 20{3p\\choose p} - 3{4p\\choose p}$ is divisible by $p^7$ for all primes $p\\geq 11$.\n$840{p\\choose p}  - 840{2p\\choose p} + 400{3p\\choose p} - 105{4p\\choose p} + 12{5p\\choose p}$ is divisible by $p^9$ for all primes $p\\geq 3$.",
  "Solution_1": "[url=https://www.artofproblemsolving.com/community/user/16261]Rust[/url] and I have proved this type of divisibilities in general in the preprint [url=http://arxiv.org/abs/1602.02632]On p-adic approximation of sums of binomial coefficients[/url]."
}
{
  "Tag": [
    "trigonometry",
    "integration",
    "function",
    "calculus",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove that the series $\\sum_{n\\geq1}\\frac{e^{i \\sqrt n}}{n}$ converges",
  "Solution_1": "Aplying the condensation test we get that $\\sum_{n\\ge 1}\\frac{e^{i\\sqrt n}}{n}$convrges iff $\\sum_{k\\ge 1}2^k \\frac{e^{i(\\sqrt 2)^k}}{2^k}=\\sum_{k\\ge 1}e^{i(\\sqrt 2)^k}$onverges. The last sum can be splited into $\\sum \\sin \\sqrt n$ and $\\sum \\cos \\sqrt n$ which diverges since $(\\sin{\\sqrt{n}})_n$ is dense in $[0,1]$ and , hence, do not tend to $0$. Am I wrong? :?",
  "Solution_2": "Are you sure that condensation test can be applied  :?: I think all summands need to be positive",
  "Solution_3": "For the condensation test, the terms have to be positive and decreasing. It's a test for absolute convergence, and the convergence here is not absolute.\r\n\r\nNote that $\\int_1^\\infty \\frac{e^{i\\sqrt{x}}}{x}\\,dx$ converges (Integrate by parts to get an equal absolutely convergent integral). We can show that $\\left|\\frac{e^{i\\sqrt{n}}}{n}-\\int_n^{n+1}\\frac{e^{i\\sqrt{x}}}{x}\\,dx\\right| =O(n^{-\\frac32})$, so the series differs from a known convergent series by an absolutely convergent series, and the original series converges.\r\n\r\nThere are some details to fill in- I might get back to this later.",
  "Solution_4": "The [hide=\"full version\"]\n\nLemma 1: $\\int_1^\\infty \\frac{e^{i\\sqrt{x}}}{x}\\,dx$ converges.\n[hide=\"Proof\"]$\\int_1^M \\frac{e^{i\\sqrt{x}}}{x}\\,dx=\\int_1^M \\frac{e^{i\\sqrt{x}}}{2\\sqrt{x}}\\cdot\\frac2{\\sqrt{x}}\\,dx$\nIntegrating by parts (integrate the first term, differentiate the second),\n$\\int_1^M \\frac{e^{i\\sqrt{x}}}{x}\\,dx= \\left[-ie^{i\\sqrt{x}}\\cdot \\frac2{\\sqrt{x}}\\right]_{x=1}^{x=M} -\\int_1^M-ie^{i\\sqrt{x}}\\cdot-x^{-\\frac32}\\,dx$\nBoth terms converge as $M\\to\\infty$; the first converges because the function we are evaluating goes to zero, and the second converges by comparison to the absolutely convergent integral $\\int_1^M x^{-\\frac32}\\,dx$.[/hide]\n\nCorollary: The series $\\sum_{n=1}^\\infty \\int_n^{n+1} \\frac{e^{i\\sqrt{x}}}{x}\\,dx$ converges.\n\nLemma 2 (Error estimate for a Riemann sum): If $f$ is differentiable on an interval $I=[a,b]$ and $|f'(x)|\\le M$ on $I$, $\\left|\\int_a^b f(x)\\,dx-(b-a)f(a)\\right|\\le \\frac M2(b-a)^2$. For this problem, allow $f$ to be complex-valued.\n[hide=\"Proof\"]Let $g(x)=f(x)-f(a)$, so $\\int_a^b g(x)\\,dx=\\int_a^b f(x)\\,dx-(b-a)f(a)$. If $\\int_a^b g(x)\\,dx=e^{i\\alpha}K$ for some positive real number $K$, let $h(x)=e^{-i\\alpha}g(x)$, so $\\int_a^b h(x)\\,dx=K$ is a positive real number.\nDefine $H(x)=\\int_a^x \\text{Re}(h(t))\\,dt$. Taylor's theorem applied to $H$ says that $K=H(b)=H(a)+(b-a)(\\text{Re}(h(a)))+\\frac{(b-a)^2}{2}\\cdot \\text{Re}(h'(x))$\n$=\\frac{(b-a)^2}{2}\\cdot \\text{Re}(h'(x))$ for some $x\\in I$. Since $\\text{Re}(h'(x))\\le |h(x)|\\le M$, we have $K\\le \\frac M2(b-a)^2$ as desired.[/hide]\n\nBy this estimate,\n$\\left|\\int_n^{n+1}\\frac{e^{i\\sqrt{x}}}{x}\\,dx-\\frac{e^{i\\sqrt{n}}}{n} \\right|\\le \\max_{n\\le x\\le n+1}\\left|\\frac{\\frac{d}{dx}\\frac{e^{i\\sqrt{x}}}{x}}{2}\\right| \\le \\max_{n\\le x\\le n+1}\\left|\\frac{ie^{i\\sqrt{x}}}{4x^{\\frac32}} -\\frac{e^{i\\sqrt{x}}}{2x^2}\\right|$\n$\\left|\\int_n^{n+1}\\frac{e^{i\\sqrt{x}}}{x}\\,dx-\\frac{e^{i\\sqrt{n}}}{n} \\right|\\le \\frac1{4n^{\\frac32}}+\\frac1{2n^2}$\nDefine $e_n=\\int_n^{n+1}\\frac{e^{i\\sqrt{x}}}{x}\\,dx-\\frac{e^{i\\sqrt{n}}}{n}$. By comparison to the absolutely convergent series $\\sum_n \\frac1{4n^{\\frac32}}+\\frac1{2n^2}$, $\\sum_n e_n$ converges.\n\nLemma 3: If $\\sum_n a_n$ converges to $A$ and $\\sum_n b_n$ converges to $B$, $\\sum_n a_n+b_n$ converges to $A+B$.\n[hide=\" Easy Proof\"]\nLet $A_n=\\sum_{i=1}^n a_i$ and $B_n=\\sum_{i=1}^n b_i$. By hypothesis, $A_n\\to A$ and $B_n\\to B$. It is clear that $\\sum_{i=1}^n a_i+b_i=A_n+B_n$. By the standard arithmetic theorem on limits, this converges to $A+B$ and we are done.[/hide]\n\nLet $I_n=\\int_n^{n+1} \\frac{e^{i\\sqrt{x}}}{x}\\,dx$.\nPutting the pieces together,\n$\\sum_{n=1}^\\infty \\frac{e^{i\\sqrt{n}}}{n}=\\sum_{n=1}^\\infty I_n-e_n$. Since $\\sum_n I_n$ converges by the corollary to Lemma 1 and $\\sum_n -e_n$ converges by the remarks following Lemma 2, their sum $\\sum_n I_n-e_n$ converges by Lemma 3 and we are done.[/hide]"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "ratio",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "The question:\r\n\r\nIf G is the additive group Q/Z, what are the elements of the subgroup G(2)? Of G(P) for any positive prime P?\r\n\r\nWhere G(n)={a e G| |a| = n^(k) for some k is greater than or equal to 0}...That is the set of all a in G, s.t. the order of a is some power of n. (But since it is the additive group, I suppose it would just a be a multiple of n)\r\n\r\nHow do I even begin with this? Aren't the elements of Q/Z sets? The collections of right cosets? and don't they have infinite order?....\r\n\r\nThanks..^^",
  "Solution_1": "$ \\mathbb{Q}/\\mathbb{Z}$ can be thought of as the set of rational numbers in $ [ 0, 1 )$ where the group operation is addition $ \\bmod 1$.  It is true that the quotient is constructed abstractly as a collection of cosets, but the above is also perfectly valid.\r\n\r\nEvery element of this group does in fact have finite order (and in fact this group is the torsion subgroup of the [url=http://en.wikipedia.org/wiki/Circle_group]circle group[/url]).  For example, $ \\frac {1}{2}$ has order $ 2$.  (This example should help you figure out the question pretty well now.)",
  "Solution_2": "The operation is addition mod 1? Wouldn't that mean that 1/2 is the only order 2 element in G(2)? 1/2 + 1/2 = 1 ... since 1 is the only number to divide 1...",
  "Solution_3": "Did you mean ration number in (0,1]? or really [0,1) ...anyway...I think I understand...actually, any element n/2 when n is an integer will have order 2... But what is so special about order p, where p is a prime?",
  "Solution_4": "[quote=\"Frank1987\"]Did you mean ration number in (0,1]? or really [0,1)[/quote]\nIt doesn't matter; we just change the name of the identity, which is $ 1$ in the first range and $ 0$ in the second.\n\n[quote=\"Frank1987\"]any element n/2 when n is an integer will have order 2... [/quote]\nYes, but these elements are all identified with each other under the quotient.  There are only two distinct such elements; the identity and $ \\frac {1}{2}$, and the identity has order $ 1$.  Now:  what are the elements of order $ 3$?\n\n[hide=\"Answer\"] $ \\frac{1}{3}$ and $ \\frac{2}{3}$. [/hide]\n[quote=\"Frank1987\"] But what is so special about order p, where p is a prime?[/quote]\r\nNothing here; elements of every order exist.  But usually it is easier to understand the elements of prime order, and in some sense one can build up elements of other order from them."
}
{
  "Tag": [
    "limit",
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "compute:\r\n\r\n$ \\lim_{n \\to \\plus{} \\infty} \\sum_{k \\equal{} 1}^{n} coh(\\frac {1}{\\sqrt {n \\plus{} k}}) \\minus{} n$\r\n\r\n\r\nwith $ coh(x) \\equal{} \\frac {e^x \\plus{} e^{ \\minus{} x}}{2}$",
  "Solution_1": "This doesn't look like an infinite series.",
  "Solution_2": "It's corrected.",
  "Solution_3": "$ \\sum_{k \\equal{} 1}^n Cosh(\\frac {1}{\\sqrt {k \\plus{} n}}) \\minus{} n \\equal{} \\sum_{k \\equal{} 1}^n \\sum_{j \\equal{} 0}^{\\infty}\\frac {1}{(2j!)(k \\plus{} n)^j} \\minus{} n$\r\n\r\n$ \\equal{} \\sum_{k \\equal{} 1}^n \\frac {1}{2(k \\plus{} n)} \\plus{} \\frac {1}{24(k \\plus{} n)^2} \\plus{} ...$\r\n$ \\equal{} \\frac {1}{2n}\\sum_{k \\equal{} 1}^n \\frac {1}{\\frac {k}{n} \\plus{} 1} \\plus{} \\frac {1}{24n}\\sum_{k \\equal{} 1}^n \\frac {1}{n(\\frac {k}{n} \\plus{} 1)^2} \\plus{} ...$\r\n\r\nAs $ n \\rightarrow \\infty$, we see we have alternate representations of riemann integrals\r\n\r\nAll of the terms except the first will be dominated by the factors of n in the denominator, making them all goto zero. So all we are concerned with is the first term which is\r\n\r\n$ \\lim_{n\\to \\plus{}\\infty} \\frac {1}{2n}\\sum_{k \\equal{} 1}^n \\frac {1}{\\frac {k}{n} \\plus{} 1} \\equal{} \\frac {1}{2}\\int_0^1\\frac {1}{1 \\plus{} x}dx \\equal{} ln{\\sqrt {2}}$\r\n$ \\therefore \\lim_{n \\to \\plus{} \\infty} \\sum_{k \\equal{} 1}^{n} coh(\\frac {1}{\\sqrt {n \\plus{} k}}) \\minus{} n \\equal{} ln{\\sqrt {2}}$"
}
{
  "Tag": [
    "inequalities",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let there $ S(N)$ be the sum of digits , for every $ N \\in \\mathbb{N}$\r\n\r\nExample : $ S(\\overline{a_1a_2 \\dots a_n}) \\equal{} a_1 \\plus{} a_2 \\plus{} \\dots \\plus{} a_n$ .\r\n\r\n\r\nProve that : $ \\boxed{\\frac {S(8N)}{S(N)} \\geq \\frac {1}{8}}$ .",
  "Solution_1": "obviosly, $ S(10^k N)\\equal{}S(N)$. So i get\r\n$ S(1000N)\\equal{}S(800N\\plus{}80N\\plus{}80N\\plus{}8N\\plus{}8N\\plus{}8N\\plus{}8N\\plus{}8N)$\r\nalso, $ S(a\\plus{}b) \\leq S(a)\\plus{}S(b)$.\r\nso i get $ S(N)\\leq 8S(8N)$ and hence the inequality holds.\r\n\r\nNaphthalin"
}
{
  "Tag": [
    "inequalities",
    "inequalities solved"
  ],
  "Problem": "Let A,B,C be three angles of triangle ABC with A,B,C \\in [0, \\pi /2] \r\nProve that :  \\sum 1/cos(A/2) - \\sum tg(A/2)  \\geq  \\sqrt 3",
  "Solution_1": "just the kinda problems i need to solve these days (with no time for maths :D  :D )\r\nput X=90 - A/2 and similarly for the others.we see that X,Y,Z are also the angles of a triangle (with each less than 90)\r\ninequality equivalent to proving\r\n \\sum tan (X/2) \\geq   \\sqrt 3\r\nisnt this known :D  :D"
}
{
  "Tag": [
    "geometry",
    "trapezoid",
    "parallelogram"
  ],
  "Problem": "Given trapezoid $ABCD$, $E$ and $F$ are points on $BC$ such that $DE\\parallel AB$ and $AF\\parallel DC$. Let $G$ be the intersection of $DE$ and $AF$. The area of parallelogram $ABED$ is $36$. If $AD=5$, $AG=3$ and $DG=4$, find the area of trapezoid $ABCD$.",
  "Solution_1": "Is $BC\\parallel{AD}$? :?",
  "Solution_2": "[quote=\"riddler\"]Is $BC\\parallel{AD}$? :?[/quote]\r\nThat was the only way I was able to draw the diagram.",
  "Solution_3": "now I realise. :blush:",
  "Solution_4": "[hide=\"I think...\"]Using the given information, we find that $\\triangle AGD$ is a $3-4-5$ right triangle. Because of congruent angles, $\\triangle EFG$ is similar to $\\triangle AGD$.\n\nThe height of $ABED$ is $7.2$. By drawing line segment $AE$, we can divide $ABED$ in half. So the area of $\\triangle AED$ is $18$, and we can find that $EG$ is $8$. Thus, the length of $EF$ is $10$.\n\nThe area of the trapezoid is $\\frac{1}{2}h(b_1+b_2)$. So it is $\\frac{1}{2}\\cdot7.2(5+10+5+5)=\\boxed{90}$.[/hide]",
  "Solution_5": "[quote=\"catcurio\"][quote=\"riddler\"]Is $BC\\parallel{AD}$? :?[/quote]\nThat was the only way I was able to draw the diagram.[/quote]Yeah, I forgot to mention that :blush: \n\n[quote=\"catcurio\"][hide=\"I think...\"]Using the given information, we find that $\\triangle AGD$ is a $3-4-5$ right triangle. Because of congruent angles, $\\triangle EFG$ is similar to $\\triangle AGD$.\n\nThe height of $ABED$ is $7.2$. By drawing line segment $AE$, we can divide $ABED$ in half. So the area of $\\triangle AED$ is $18$, and we can find that $EG$ is $8$. Thus, the length of $EF$ is $10$.\n\nThe area of the trapezoid is $\\frac{1}{2}h(b_1+b_2)$. So it is $\\frac{1}{2}\\cdot7.2(5+10+5+5)=\\boxed{90}$.[/hide][/quote]That's correct! ;)"
}
{
  "Tag": [
    "number theory open",
    "number theory"
  ],
  "Problem": "Suppose that $ a$ is a positive integer. Is it always true that there is a prime $ p$ such that $ p^2\\mid a^{p\\minus{}1}\\minus{}1$?",
  "Solution_1": "Heuristic arguments suggest there this statement is likely to be true.\nBut it may be very hard to prove it rigorously.\n\nSee also\nhttp://oeis.org/A039951\nhttp://oeis.org/A096082"
}
{
  "Tag": [],
  "Problem": "where were u born?",
  "Solution_1": "CHINA!!!!!!!!!!!!!!!",
  "Solution_2": "i am glad with the added options!\r\nCHINA!",
  "Solution_3": "INDIA!!!! \r\n[hide=\"Indian Song\"]Welcome to India where the cows eat hey,\nand we drive auto-richshaws everyday,\nGoat Meats, Yummy Sweets, Wild Monkeys roaming\nThe Roosters Dont Crow till 5 in the morning[/hide]",
  "Solution_4": "hmm only one from west europe (belgium)",
  "Solution_5": "i'm from england!",
  "Solution_6": "US! Illinois.",
  "Solution_7": "CCCCCCCCCCCCCCHINA!!!!!!!!!",
  "Solution_8": "Hmm, I'm the first Russian! (AKA Soviet)",
  "Solution_9": "[quote=\"DarkKnight\"]US! Illinois.[/quote]\r\n\r\nWoot! Go Illinois!",
  "Solution_10": "Hi, Sponge. I am also Russian. Where you from? Myth, who posts on AoPS is russian and imo gold medalist.",
  "Solution_11": "from Seoul, Korea.",
  "Solution_12": "I posted this on the other thread, but....ARIZONA!!!!!! Parents born in Taiwan, though.",
  "Solution_13": "[quote=\"xxreddevilzxx\"]from Seoul, Korea.[/quote]\r\nWhee~ Seoul! Me, too..",
  "Solution_14": "cool. when did you come to america?",
  "Solution_15": "China!  But I live in Illinois right now.",
  "Solution_16": "holdin it down in South Carolina, USA.",
  "Solution_17": "US of A...more specifically, good old INDIANA!!!",
  "Solution_18": "South Cackalackey",
  "Solution_19": "I was born in CHINA! :cool:",
  "Solution_20": "[quote=\"hello\"]Hi, Sponge. I am also Russian. Where you from? Myth, who posts on AoPS is russian and imo gold medalist.[/quote]\r\nI was born in Kiev in the Soviet days. Now, I live in Massachusetts. Meh, I'm Russian by ancestry and first language (first as in first learned, not better volcabulary), Ukranian by former/partial citizenship, and Soviet/Ukranian by birth. Now, I'm living in America. I'm reminded of a Chinese immigrant from Australia friend of mine.",
  "Solution_21": "turkey-istanbul\r\nthis is better than my post and poll.",
  "Solution_22": "summerville, SC",
  "Solution_23": "i'm from japan, moved to the us when i was 8",
  "Solution_24": "Wow, not many people in the \"rest of north of america\"  I'm sure there's gotta be some others!",
  "Solution_25": "[quote=\"MathFiend\"]US of A...more specifically, good old INDIANA!!![/quote]\r\nDo you still live in Indiana and if you do what school do you go to?\r\nI was wondering if I might know you from one of these competitions."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "(X with the power of 1 )(X with the power of 2)(X with the power of 3).......................(X with the power of 2n) what does this question wants ? I dun understand what is the power of 2n. And X with the power of 1 multiply X with the power of 2 isnt it X with the power of 1+2 that is X with the power of 3?",
  "Solution_1": "could you rephrase that and use the \"^\" sign? \r\ni have no idea what u r talking about and it would be helpful if u replaced \"with the power of\" with \"^\" By the way, \"^\" basically translates into \"to the power of\"\r\n\r\n3^4 means\r\n3 to the power of 4\r\n3*3*3*3 \r\n81",
  "Solution_2": "The first one: $ (x^1)(x^2)(x^3)(x^4)...(x^{2n})$.  This is equal to $ x^{1\\plus{}2\\plus{}3\\plus{}...\\plus{}2n}\\equal{}x^{n(2n\\plus{}1)}$.\r\nThe second one: $ x^1\\times x^2$, which does equal $ x^3$.\r\n\r\nIn the meantime, I'd suggest going to the LaTeX guide by [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]clicking here[/url]",
  "Solution_3": "What is latex? I dun understand the information stated  there",
  "Solution_4": "[quote=\"shaun ong\"]What is latex? I dun understand the information stated  there[/quote]\r\n\r\nLaTeX is a programming language that is used by mathematicians to express math symbols in an easier way. \r\n\r\nYou can say,\r\n\r\n(x+y^5-2x-y)/x+2z) and we'll have no idea what you're saying(or if we do, then we'll have to spend 5 painful minutes trying to unravel the mess).  \r\n\r\nIn LaTeX, it can be expressed as \r\n\r\n$ \\frac{(x\\plus{}y^{5\\minus{}2x}\\minus{}y)}{x\\plus{}2z}$.\r\n\r\nClarity and easy to understand.",
  "Solution_5": "there is a LATEX program where you can learn it"
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "polynomial",
    "inequalities proposed"
  ],
  "Problem": "i)$a,b,c >0$.Prove that:\r\n               $\\frac{ab(a+b)}{a^3+b^3+2c^3}+\\frac{bc(b+c)}{b^3+c^3+2a^3}+\\frac{ac(a+c)}{a^3+c^3+2b^3} \\leq \\frac{3}{2}$\r\nii)Find the best contanst $k$ such that the inequality hold:\r\n               $\\sum \\frac{ab(a+b)}{a^3+b^3+kc^3} \\leq \\frac{6}{k+2}$ $(a,b,c >0)$",
  "Solution_1": "SOS again... Come on...",
  "Solution_2": "How to solve this one by SOS?I have some trouble.",
  "Solution_3": ":rotfl:  ii) The maximum of $k\\,$ is the smaller positive root of the following polynomial:\r\n\r\n$23\\,k^6+374\\,k^5-3733\\,k^4+7456\\,k^3+1156\\,k^2-8192\\,k+3888,$\r\n\r\nthat is,  $2.77948796...$",
  "Solution_4": "How to do the first one by SOS?"
}
{
  "Tag": [],
  "Problem": "\u725b\u987f\u5b9a\u7406\uff1aABCD\u662f\u5706\u5185\u63a5\u56db\u8fb9\u5f62\uff0c\u76f4\u7ebfAB\u4e0e\u76f4\u7ebfCD\u4ea4\u4e8eE\uff0c\u76f4\u7ebfBC\u4e0e\u76f4\u7ebfAD\u4ea4\u4e8eF\uff0cM,N,Q\u5206\u522b\u4e3aAC,BD,EF\u7684\u4e2d\u70b9\uff0c\u6c42\u8bc1\uff1aM,N,O\u5171\u7ebf.",
  "Solution_1": "See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20264]http://www.mathlinks.ro/Forum/viewtopic.php?t=20264[/url].",
  "Solution_2": "\u5f88\u5947\u602a. \u6211\u7684\u4e00\u672c\u4e66\u4e0a\u8bf4: \u8fd9\u4e09\u70b9\u7684\u8fde\u7ebf\u53eb\u505a\"\u9ad8\u65af\u7ebf\", \u53ef\u4e3a\u4ec0\u4e48\u9898\u76ee\u53eb\u725b\u987f\u5b9a\u7406? \r\n\r\nBTW: \u5173\u4e8e\u5706\u7684\u6761\u4ef6\u53ef\u4ee5\u53bb\u6389. \u9898\u76ee\u7684\u7ed3\u8bba\u5bf9\u4e8e\u4efb\u610f\u51f8\u56db\u8fb9\u5f62\u90fd\u662f\u6210\u7acb\u7684.",
  "Solution_3": "\u5bf9\u554a,\u6211\u591a\u7ed9\u4e86\u4e00\u4e2a\u6761\u4ef6 :blush:  :blush:  :blush:",
  "Solution_4": "\u4f3c\u4e4e\u5927\u5bb6\u73fe\u5728\u90fd\u5f88\u5fd9. \u9019\u88e1\u53c8\u518d\u51b7\u6e05\u8d77\u4f86.",
  "Solution_5": "\u6d88\u70b9\u5427. \u628a\u7ebf\u6bb5\u6bd4\u6362\u6210\u9762\u79ef\u6bd4, \u518d\u7528\u4e00\u6b21\u5b9a\u6bd4\u5206\u70b9\u516c\u5f0f.",
  "Solution_6": "\u4e8b\u5b9e\u4e0a\u53ef\u4ee5\u7528Menelaus\u5b9a\u7406\u89e3\u51b3\u3002",
  "Solution_7": "[quote=\"dingdongdog\"]\u4e8b\u5b9e\u4e0a\u53ef\u4ee5\u7528Menelaus\u5b9a\u7406\u89e3\u51b3\u3002[/quote]\r\nHow?",
  "Solution_8": "\u8fd9\u9898\u53ef\u4ee5\u7ed3\u5408\u9762\u79ef\u6cd5\u89e3\u51b3\u3002\r\n\u6b32\u8bc1O\\M\\N\u5171\u7ebf\uff0c\u53ea\u9700\u8bc1S(FMN)=S(EMN)\r\n\u53d6AB\u4e2d\u70b9P,S(FMN)=S(FPM)+S(FPN)+S(PMN)=S(BPM)+S(APN)+S(PMN)=S(ANMB)\r\n=S(ADMB)/2=S(ABCD)/4\r\n\u540c\u7406S(EMN)=S(ABCD)/4\r\n\u4ece\u800cS(FMN)=S(EMN),\u547d\u9898\u5f97\u8bc1\u3002"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "logarithms",
    "function",
    "rational function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "prove that there aren't polynominal $P ,Q$ such that:$\\pi(n)=\\frac {P(x)}{Q(x)}$",
  "Solution_1": "Are you referring to polynomials $P, Q$ with integer coefficients? Or the coefficients can take on any real values?",
  "Solution_2": "It doesn't really matter. The Prime Number Theorem says that $\\pi(n)$ increases like $\\frac n{\\ln n}$, so it couldn't be represented as a rational function in $n$, since such functions behave like polynomials as $n\\to\\infty$."
}
{
  "Tag": [
    "probability",
    "floor function",
    "probability and stats"
  ],
  "Problem": "If m balls are distributed to a boys and m girls. Find the probability that even no. of balls are distributed to boys :(",
  "Solution_1": "[hide=\"hint\"]The first ball is given to one of $ m \\plus{} a$ people, what's the probability that it will be given to one of the $ m$ girls?  The second ball is given to one of the remaining $ m \\plus{} a \\minus{} 1$ people, what's the probability that it will be given to one of the remaining $ m \\minus{} 1$ girls?  Keep going until there are no balls left.  These $ m$ events are pairwise independent; how does that help?[/hide]",
  "Solution_2": "# ways to distribute balls:\r\n\r\n$ _{a\\plus{}m}C_m \\equal{} \\frac{(a\\plus{}m)!}{a!m!}$\r\n\r\n# ways to give boys 2k balls:\r\n\r\n$ _a C_{2k} \\cdot  _m C_{m\\minus{}2k} \\equal{} \\frac {a!m!}{2k! (a\\minus{}2k)! 2k! (m\\minus{}2k)!}$\r\n\r\nprob of giving boys 2k balls:\r\n\r\n$ P(X\\equal{}2k) \\equal{} \\left(\\frac{a!m!}{k!}\\right)^2 \\frac{1}{(a\\minus{}2k)!(m\\minus{}2k)!(a\\plus{}m)!}$\r\n\r\n$ P(X even) \\equal{} \\sum_{k\\equal{}0}^{\\lfloor \\frac{a}{2} \\rfloor} P(X\\equal{}2k)$\r\n\r\ni dont see this cleaning up nicely though...",
  "Solution_3": "ans.\r\n\r\n1/2{(a+b)^m-(a+b)^n}/a+b)^m"
}
{
  "Tag": [
    "inequalities",
    "limit",
    "inequalities unsolved"
  ],
  "Problem": "Let $ a,b,c$ are positive real numbers and $ n \\geq1$ prove that\r\n\r\n\r\n$ \\frac{a^{2}}{b}\\plus{}\\frac{b^{2}}{c}\\plus{}\\frac{c^{2}}{a}\\geq3\\frac{a^{n}\\plus{}b^{n}\\plus{}c^{n}}{a^{n\\minus{}1}\\plus{}b^{n\\minus{}1}\\plus{}c^{n\\minus{}1}}$",
  "Solution_1": "[quote=\"mestav\"]Let $ a,b,c$ are positive real numbers and $ n \\geq1$ prove that\n\n\n$ \\frac {a^{2}}{b} \\plus{} \\frac {b^{2}}{c} \\plus{} \\frac {c^{2}}{a}\\geq3\\frac {a^{n} \\plus{} b^{n} \\plus{} c^{n}}{a^{n \\minus{} 1} \\plus{} b^{n \\minus{} 1} \\plus{} c^{n \\minus{} 1}}$[/quote]\r\n\r\n(1), we can prove that\r\n\r\n\\[ \\frac {{a^{n \\plus{} 1} \\plus{} b^{n \\plus{} 1} \\plus{} c^{n \\plus{} 1} }}{{a^n \\plus{} b^n \\plus{} c^n }} \\ge \\frac {{a^n \\plus{} b^n \\plus{} c^n }}{{a^{n \\minus{} 1} \\plus{} b^{n \\minus{} 1} \\plus{} c^{n \\minus{} 1} }}\\]\r\nholds for any $ a,b,c > 0,n\\ge1$.\r\n\r\n(2), we can prove that\r\n\r\n\\[ \\mathop {\\lim }\\limits_{n \\to \\plus{} \\infty } \\frac {{a^n \\plus{} b^n \\plus{} c^n }}{{a^{n \\minus{} 1} \\plus{} b^{n \\minus{} 1} \\plus{} c^{n \\minus{} 1} }} \\equal{} \\max \\left\\{ {a,b,c} \\right\\}\\]\r\nholds for any $ a,b,c > 0$\r\n\r\n\r\nso, it's equivalent to prove that\r\n\\[ \\frac {a^{2}}{b} \\plus{} \\frac {b^{2}}{c} \\plus{} \\frac {c^{2}}{a}\\geq 3 \\max \\left\\{ {a,b,c} \\right\\}\\]\r\nbut this is not hold! :mad: \r\n\r\nso, the original inequality must have a counter example!\r\n\r\nI will try to find it...",
  "Solution_2": "OK, found, counter example:\r\n\r\n$ a \\equal{} 9, b \\equal{} 15, c \\equal{} 10, n \\equal{} 4$\r\n\r\n$ \\frac {a^{2}}{b} \\plus{} \\frac {b^{2}}{c} \\plus{} \\frac {c^{2}}{a} \\minus{} 3 \\cdot \\frac {a^4 \\plus{} b^4 \\plus{} c^4}{a^3 \\plus{} b^3 \\plus{} c^3} \\equal{} \\minus{} \\frac {55019}{114840} < 0$",
  "Solution_3": "by the way, when $ n \\equal{} 1,2,3$, inequality holds. I have proved that $ n \\equal{} 1,2$, who can prove that $ n \\equal{} 3$?\r\n\r\n$ \\frac {a^{2}}{b} \\plus{} \\frac {b^{2}}{c} \\plus{} \\frac {c^{2}}{a} \\minus{} (a \\plus{} b \\plus{} c) \\equal{} \\sum{\\frac {(a \\minus{} b)^2}{b}} \\ge 0$\r\n\r\n$ \\frac {a^{2}}{b} \\plus{} \\frac {b^{2}}{c} \\plus{} \\frac {c^{2}}{a} \\minus{} 3\\cdot \\frac {a^2 \\plus{} b^2 \\plus{} c^2}{a \\plus{} b \\plus{} c} \\equal{} \\sum{\\left(\\frac {1}{b} \\minus{} \\frac {1}{a \\plus{} b \\plus{} c}\\right)(a \\minus{} b)^2} \\ge 0$\r\n\r\nand ... $ n_{\\max}$ is ... ?   :huh:"
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "This year I didn't take the AMC because basically nobody at my school could have qualified for AIME except possibly me with my special skill at guessing correctly without a clue.  But, I want to know how I would have done so does anyone know of a link to a site with the questions?  Thanks in advance!",
  "Solution_1": "No, but you could always buy the questions for $1 soon - [url]http://www.unl.edu/amc/d-publication/pubform.html[/url]"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "<= in the code is parsed to \r\n<\r\n=\r\neffectively disabling the code in the posts containing it. (>=, probably too, != no)\r\nCan we fix it?",
  "Solution_1": "I don't understand what's going on. Could you be more specific? What code? Asy code?",
  "Solution_2": "He's saying if we had something like\r\n\r\n[asy]real i; path p = (0,0)--(1,1); for(i=1; i<=10; ++i){ draw((p=shift((1,0))*p)); }[/asy]\r\n\r\nIt won't work because <= gets broken up on new lines. It seems to work for me though.\r\n\r\n\r\n[asy]real i; path p = (0,0)--(1,1); for(i=0; i<10; ++i){ draw((p=shift((1,0))*p)); }[/asy]",
  "Solution_3": "Oh, well that's just the display part (for the human eye :) ). It shouldn't break after < though, it's a good catch. I'll try to fix it.",
  "Solution_4": "Try to edit or quote Hamster1800 post without touching the code and resubmit. You'll get an error. The first time you post it works OK though and I accidentally discovered it when I tried to add a smile to a perfectly functioning post and got an error message. It took me half an hour to figure out what was going on  :mad: :lol:",
  "Solution_5": "Ok, I've fixed that error, and also fixed the presentation for the for. Happy now? :P",
  "Solution_6": "[quote=\"Valentin Vornicu\"]Ok, I've fixed that error, and also fixed the presentation for the for.[/quote]\n:coolspeak: \n[quote] Happy now? :P[/quote]\r\nWay up, but I still have to multiply by 50  :heli:"
}
{
  "Tag": [
    "geometry",
    "Euler",
    "invariant"
  ],
  "Problem": "Feuerbach's theorem:\r\nProve that in any triangle, the inscribed circle and the 3 exinscribed circles are tangent to Euler's circle.",
  "Solution_1": "i think i have a proof for the incircle touching euler's circle part,two of them to be precise.they are long ,i'll post em later.btw,i believe u do have the proofs ,mate :? ,and u just wanna know sum different solutions,dont ya??\r\ncheers",
  "Solution_2": "Well, the problem is that I remember having proved this, but I have no idea how.. :? So I have to say no, I don't currently have a proof. I could look in some books of course, but I hate reading proofs in books. :D",
  "Solution_3": "Okay .. let's see now... \r\n\r\nwe need two lemmas first:\r\n\r\n [b] Lemma 1 [/b]: Let A' be the middle of [BC] of triangle ABC. Using an inversion(grobber's favourite method :D ) T(A',k) (k is arbitrary) Euler circle of triangle ABC ( from which we take out point A') becomes a antiparalel with BC from A.\r\n\r\n[b] Proof of lemma 1 [/b]: It is well known that A' is on Euler circle. So using the inversion T(A',k) Euler circle( from which like I said we take out A') transforms itsself into a line d which is perpendicular on A'w, where w is the center of Euler circle. We know that w is the middle of HO (H- ortocenter and O - the center of the circumscribed circle). Let H' be the simetrical of H from A' . Then H' is diametrally opposite of A. \r\n\r\nIn the triangle HOH' the segment A'w is the middle line, so A'w || AH'. Because d is perpendicular on A'w is perpendicular on AH', which means that d is paralel with the tangent in point H' of the circumscriebed circle of triangle ABC. Let E be the point in with the tangent in H' intersects AB.\r\nThen m(AEH')=90-m(EAH')=m(BH'A)=n(ACB).\r\n\r\nSo we have that BC and EH' are antiparalel. Because d is paralel to EH' we get that d is antiparalel with BC.\r\n\r\n[b] Lemma 2 [/b]: Let C(I,r) be the incircle of triangle ABC and let C(I',r') be the exincribed circle corespunding to A. Let A' be the middle of BC and let A1 be the point in which the bisector of angle BAC intersects BC. If A\" is the projection of A on BC and F the projection of I on F and F' the projection of I' on BC then:\r\ni) BF=CF'\r\nii) A'F^2=A'A\"*A'A1\r\n\r\n[b] Proof of lemma 2 [/b]: For more clarity on your drawings lets consider that the triangle ABC has angle m(ABC)>90.\r\n\r\n[b] i) [/b]: If we denote F1=the projection of I on ABC, x=AF1, y=BF, z=CF, a=BC, b=CA, c=AB then:\r\nc=x+y\r\na=y+z\r\nb=z+x\r\nwe get that x+y+z=p which gives us that y=BF=p-b where p is the semiperimeter of ABC.\r\nusing the following notations: u=AF\", v=BF', alpha=CF' we get that\r\na=v+alpha\r\nb=u-alpha\r\nc=u-v\r\nwhich gives us that alpha=CF'=p-b so BF=CF'.\r\n\r\n[b] ii) [/b] from the bisector theorem we get that BA1/CA1=c/b from which we get that: BA1/(BA1+A1C)=c/(b+c) so BA1=ac/(b+c)\r\nso we get A'A1=a/2-ac/(b+c) or A'A1=a(b-c)/(2(b+c)) (*)\r\nfrom the right angled triangles AA\"B and AA\"C we get that :\r\nA\"A^2=AB^2-A\"B^2 and A\"A^2=AC^2-A\"C^2\r\nfrom the last two equalities we get that:\r\nb^2-c^2=(A\"C+A\"B)(A\"C-A\"B) or b^2-c^2=a(a+2A\"B) so we get that \r\nA\"B=(b^2-c^2-a^2)/(2a); because A'A\"=A'B+A\"b we get that A'A\"=(b^2-c^2)/(2a) (**)\r\nBecause BF=p-b we get that A'F=A'B-BF=a/2-(p-b) so A'F=(b-c)/2 (***)\r\nso now if we look at (*), (**) and (***) we get that:\r\nA'A1*A'A\"=a*(b-c)/(2(b+c))* (b^2-c^2)/(2a)=(b-c)^2/4=A'F^2\r\n\r\nNow we are done with the lemmas, back to the problem :D.\r\n\r\n[b] Proof of the problem [/b]We keep the denotes we made in lemma 2 and now we transfor that drawing using an inversion of pole A' and we use the power k=p ( the power of the point A' from the inscribed circle) so k=A'F^2.\r\n\r\nUsing this inversion the circle C(I,r) is transformed into itself. Because A'F=A'F' we also get that the circle C(I',r') is transformed into itself.\r\n\r\nThe equality A'A1*A'A\"=A'F^2 (lemma 2) shows that the point A\" ( which is on Euler's circle) is transformed into A1. Using lemma 1 we get that EUler circle (from which we take out A') is transformed in the straight line d which is antiparalel with BC which goes through A1. Because d pases through A1 and because it is antiparalel with BC we get that this is the 2nd exterior common tangent  of circles C(I,r) and  C(I',r'). d the inverse of Euler circle (from which we take out A') because it is tangent to C(I,r) and C(I',r')  we get that also Euler circle will be tangent to C(I,r) and C(I',r') ( they are the invariants in the considered inversion)\r\n\r\nLet E (E' respectively) be the point in which d is tangent to circle C(I,r) (C(I',r') respectively). Then we have that Euler's circle is tangent to circle C(I,r) in the transformation of point E, and to the circle C(I',r')  in the transformation of point E' with the considered inversion. We do the same to prove that Euler's circle is tangent to the exinscriebed circles coresponding to B and C.\r\n\r\nProof done! Hope you will enjoy it!  :D",
  "Solution_4": "Holy Mother of Christ! That is enormous! :D But it works anyway. I'll try to remember what I came up with.",
  "Solution_5": "hmm.. grobber that was the exact thing that I said too :D\r\n\r\nguess that this is indeed(hopefully) the first and last solution I will give with inversions :D\r\n\r\nI really hate geometrical transformations! they are ingenious sometimes and also long sometimes :D",
  "Solution_6": "there is geometry book which tells a lot of Euclidean geometrical knowledge including Feuerbach's theorem.\r\n\r\n\"Modern Geometry\"\r\nby Roger A. Johnson\r\n(C) Houghton Mifflin 1929",
  "Solution_7": "well refer brindon's geometry series , though I don't have it",
  "Solution_8": "View: \n[i]Nathan Altshiller-Court[/i]: [b]COLLEGE GEOMETRY [/b](second Edition); p.105; Theorem 215.",
  "Solution_9": "Dear Mathlinkers, \nreading some old message, I found this one... \nIn 2003, I found an enterely synthetic proof that I put on my website in 2007. \nhttp://perso.orange.fr/jl.ayme (vol. 1 le th\u00e9or\u00e8me de Feuerbach) in French \nor : revistaoim directed by professor Francisco Bellot Rosado (in spanish). Revista Escolar de la Olimpiada \nIberoamericana de Matematica (Espagne) 26 (2006): http://www.oei.es/oim/revistaoim/index.html\nSincerely \nJean-Louis"
}
{
  "Tag": [
    "college contests"
  ],
  "Problem": "I was wondering how many people here are going to participate in this year's competition... so here I am, the first one :). Anybody else?",
  "Solution_1": "Oh well, never mind :(. If there's anybody interested, here are the problems:\r\n\r\nhttp://www.osu.cz/fpr/kma/dokumenty/vjaimc/year2007.html\r\n\r\nCategory 1 wasn't too hard, five people managed to score full points. I'm not quite competent to judge the difficulty of category 2 problems though.",
  "Solution_2": "http://jarnik.osu.cz/index.php?page=home",
  "Solution_3": ":maybe: I think there will be none from Indonesia. (Sorrry if I do a junk, can the moderators make a forum for the junkers, about any news besides the topic that has been in the forum). But, Votech jarnik 2007 has ended. I hope I can join the competition in 2008, if I had a money  :rotfl:  :(",
  "Solution_4": "What was the turnout and # of teams for this competition?",
  "Solution_5": "[quote=\"jytang\"]What was the turnout and # of teams for this competition?[/quote]\r\n\r\nThis can be seen from the page with (non-zero) results:\r\nhttp://jarnik.osu.cz/index.php?page=year2007"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "How do you graph the inequality x+y+|z| <= 3?",
  "Solution_1": "First, I would like to know, how do you draw a 3-d graph on 2-d paper?\r\n\r\nIf you have a way to do that, the best way would probably be to graph it for positive y, then take that graph, plus the graph resulting from it being reflected across the x-y plane, and you're done.",
  "Solution_2": "Do you have a graphing program on your computer?\r\nIf so you can plug it in, get the intercepts, and transfer it to paper.\r\nI'd like to know how to gragh 3D on 2D paper too. :D"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "calculus",
    "integration",
    "search",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Prove or disprove that: if $P(x)\\in Z[x]$ and for each $n\\in N*$  there exist an integer $k\\in N*$ such that $n|P(k)$ then $P(x)$ has at least one integeral root.  :) \r\n\r\n\r\nit's just my idea!",
  "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=110231 .",
  "Solution_2": "I posted it some time ago, with additional question: find the minimal degree of a counterexample. Try to search for this topic.",
  "Solution_3": "$P(x)=(x^{2}-17)(x^{2}+1)(x^{2}-2)(x^{2}+2)$.",
  "Solution_4": "@Harazi: you are too slow  :P  :D \r\n\r\n@Rust: no, try $\\mod 16$.",
  "Solution_5": "I edit, before I forget even prime 2."
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "1.  Given the polynomial identity \r\n\r\n$ (x^6 \\plus{} 1) \\equal{} (x^2 \\plus{} 1)(x^2 \\plus{} ax \\plus{} 1)(x^2 \\plus{} bx \\plus{} 1)$\r\n\r\nWhat is the value of $ ab$?\r\n\r\nEDIT: I see what I did  :oops:",
  "Solution_1": "[quote=\"Stokes93\"]1.  Given the polynomial identity \n\n$ (x^6 \\plus{} 1) \\equal{} (x^2 \\plus{} 1)(x^2 \\plus{} ax \\plus{} 1)(x^2 \\plus{} bx \\plus{} 1)$\n\nWhat is the value of $ ab$?\n\n2.  What is the number of distinct real solutions to the equation\n\n$ x^4 \\plus{} 6x^2 \\plus{} 9 \\equal{} 36x^2 \\minus{} 72x \\plus{} 36$ \n\nI know you can solve it by factoring... but I don't see what is wrong with my solution (below)\n\n[code]We have $x^4 - 30x^2 - 25 = 0$ [/code]\n\n let $ p \\equal{} x^2$ $ \\Rightarrow$ $ p^2 \\minus{} 30p \\minus{} 25 \\equal{} 0$ \n\n$ \\Rightarrow$ $ p \\equal{} 15 \\pm 5 \\sqrt {10}$ $ \\equal{}$ $ x^2$ \n\n$ \\Rightarrow$ $ x \\equal{}$ $ \\pm \\sqrt { 15 \\pm 5 \\sqrt {10}}$\n\nNow we can have $ \\plus{}$ or $ \\minus{}$ on the outside, but for a real solution the $ \\pm$ on the inside must be $ \\plus{}$ so we have two real distinct solutions.[/quote]\r\n\r\nHow does this come?",
  "Solution_2": "Oh!   :blush: I totally left out the $ \\minus{}72x$ when I did it...  \r\n\r\nThus substitution will not work well.... so never mind  :oops:",
  "Solution_3": "[quote=\"Stokes93\"]\n\n2.  What is the number of distinct real solutions to the equation\n\n$ x^4 \\plus{} 6x^2 \\plus{} 9 \\equal{} 36x^2 \\minus{} 72x \\plus{} 36$ \n\nI know you can solve it by factoring... but I don't see what is wrong with my solution (below)\n\nWe have $ x^4 \\minus{} 30x^2 \\minus{} 25 \\equal{} 0$ \n\n let $ p \\equal{} x^2$ $ \\Rightarrow$ $ p^2 \\minus{} 30p \\minus{} 25 \\equal{} 0$ \n\n$ \\Rightarrow$ $ p \\equal{} 15 \\pm 5 \\sqrt {10}$ $ \\equal{}$ $ x^2$ \n\n$ \\Rightarrow$ $ x \\equal{}$ $ \\pm \\sqrt { 15 \\pm 5 \\sqrt {10}}$\n\nNow we can have $ \\plus{}$ or $ \\minus{}$ on the outside, but for a real solution the $ \\pm$ on the inside must be $ \\plus{}$ so we have two real distinct solutions.[/quote]\r\n\r\nYou forgot about the 72x.\r\n\r\n\r\n\r\n[hide=\"#1\"]\nMultiplying out we have\n\n$ x^6\\plus{}1\\equal{}x^6\\plus{}(a\\plus{}b)x^5\\plus{}(ab\\plus{}3)x^4\\plus{}(2a\\plus{}2b)x^3\\plus{}(ab\\plus{}3)x^2\\plus{}(a\\plus{}b)x\\plus{}1$\n\nMatching we have\n\n$ a\\plus{}b\\equal{}0$ and $ ab\\plus{}3\\equal{}0$\n\nWe see $ a\\equal{}\\sqrt3$ and $ b\\equal{}\\minus{}\\sqrt3$ (or vice versa)[/hide]",
  "Solution_4": "[quote=\"Stokes93\"]1.  Given the polynomial identity \n\n$ (x^6 \\plus{} 1) \\equal{} (x^2 \\plus{} 1)(x^2 \\plus{} ax \\plus{} 1)(x^2 \\plus{} bx \\plus{} 1)$\n\nWhat is the value of $ ab$?\n\n2.  What is the number of distinct real solutions to the equation\n\n$ x^4 \\plus{} 6x^2 \\plus{} 9 \\equal{} 36x^2 \\minus{} 72x \\plus{} 36$ \n\nI know you can solve it by factoring... but I don't see what is wrong with my solution (below)\n\nWe have $ x^4 \\minus{} 30x^2 \\minus{} 25 \\equal{} 0$ \n\n let $ p \\equal{} x^2$ $ \\Rightarrow$ $ p^2 \\minus{} 30p \\minus{} 25 \\equal{} 0$ \n\n$ \\Rightarrow$ $ p \\equal{} 15 \\pm 5 \\sqrt {10}$ $ \\equal{}$ $ x^2$ \n\n$ \\Rightarrow$ $ x \\equal{}$ $ \\pm \\sqrt { 15 \\pm 5 \\sqrt {10}}$\n\nNow we can have $ \\plus{}$ or $ \\minus{}$ on the outside, but for a real solution the $ \\pm$ on the inside must be $ \\plus{}$ so we have two real distinct solutions.[/quote]\r\n\r\n[hide=\"#1\"]\n$ (x^2 \\plus{} 1)(x^2 \\plus{} ax \\plus{} 1)(x^2 \\plus{} bx \\plus{} 1)$\n$ \\equal{}x^2(x^2 \\plus{} ax \\plus{} 1)(x^2 \\plus{} bx \\plus{} 1)\\plus{}(x^2 \\plus{} ax \\plus{} 1)(x^2 \\plus{} bx \\plus{} 1)$\n$ \\equal{}x^2(x^4\\plus{}(a\\plus{}b)x^3\\plus{}(ab\\plus{}2)x^2\\plus{}(a\\plus{}b)x\\plus{}1)\\plus{}(x^4\\plus{}(a\\plus{}b)x^3\\plus{}(ab\\plus{}2)x^2\\plus{}(a\\plus{}b)x\\plus{}1)$\n$ \\equal{}x^6\\plus{}(a\\plus{}b)x^5\\plus{}(ab\\plus{}3)x^4$+2(a+b)x^3+(ab+3)x^2+(a+b)x+1$ Then we have$=x^6+(a+b)x^5+(ab+3)x^4$ \\plus{}2(a\\plus{}b)x^3\\plus{}(ab\\plus{}3)x^2\\plus{}(a\\plus{}b)x\\plus{}1\\equal{}x^6\\plus{}1$\n               $ ab\\plus{}3\\equal{}0$\n                $ ab\\equal{}\\minus{}3$\n\n[/hide]"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Well can anyone please provide me the link to the proof , or the proof itself, for the Van Aubels;s Theorem (For Triangles) . I have been trying to prove it for past several Days but have failed to devise any proof\r\n  \r\n :maybe:",
  "Solution_1": "hello, see here\r\nhttp://planetmath.org/encyclopedia/ProofOfVanAubelsTheorem.html\r\nSonnhard.",
  "Solution_2": "Thank you very much !! Finally I got the proof !!! What a relief\r\n\\ :roll:  :oops:"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "$a,b\\ge1 , c\\ge0$and  $a,b$are real numbers and $n$ is natural number then prove it:\r\n$(ab+c)^{n}-c\\le((b+c)^{n}-c)a^{n}$",
  "Solution_1": "who solved it???"
}
{
  "Tag": [
    "geometry",
    "inradius",
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "let $ a$ , $ b$ , $ c$ be the lenghts of the sides of a triangle $ r$ inradius and $ R$ outradius , prove that :\r\n\r\n$ \\frac{(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)}{4abc} \\le \\frac{R}{r}$",
  "Solution_1": "this is from Iran 98...\r\n\r\nfirst of all let $ p = \\frac {a + b + c}2$ now note that:\r\n\r\n\\begin{eqnarray*}(a + b)(b + c)(c + a) = a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc = 2p(a^2 + b^2 + c^2) - (a^3 + b^3 + c^3) + 2abc\\end{eqnarray*}\r\n\r\nnow note that:\r\n\r\n$ a^2 + b^2 + c^2 = 2(p^2 - r^2 - 4Rr)$\r\n\r\n$ a^3 + b^3 + c^3 = 2p(p^2 - 3r^2 - 6Rr)$\r\n\r\nso we have to show that:\r\n\r\n$ \\frac {r}{4R}\\leq\\frac {4Rpr}{4p(p^2 - r^2 - 4Rr) - 2p(p^2 - 3r^2 - 6Rr) + 8Rpr}$\r\n\r\n$ \\iff 8R^2 - 2Rr - r^2\\geq p^2$\r\n\r\nbut we have:\r\n\r\n$ p^2\\leq 4R^2 + 4Rr + 3r^2$\r\n\r\nso it's sufficient to show that:\r\n\r\n$ 8R^2 - 2Rr - r^2\\geq 4R^2 + 4Rr + 3r^2$\r\n\r\n$ \\iff 4R^2 - 6Rr - 4r^2\\geq 0$\r\n\r\n$ \\iff 2(R - 2r)(2R + r)\\geq 0$\r\n\r\n$ \\iff R\\geq 2r$\r\n\r\nwhich is true...",
  "Solution_2": "[quote=\"y-a-s-s-i-n-e\"]let $ a$ , $ b$ , $ c$ be the lenghts of the sides of a triangle $ r$ inradius and $ R$ outradius , prove that :\n\n$ \\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{4abc} \\le \\frac {R}{r}$[/quote]\r\n$ \\frac{a\\plus{}b}{c}\\equal{}\\frac{sin(A)\\plus{}sin(B)}{sin(C)}\\equal{}\\frac{2sin{((A\\plus{}B)/2)cos((A\\minus{}B)/2)}}{2sin(C/2)cos(C/2)}\\equal{}$ $ \\frac{cos((A\\minus{}B)/2)}{sin(C/2)} \\le \\frac{1}{sin(C/2)}$\r\nso : $ \\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{4abc} \\le \\frac{1}{4sin(A/2)sin(B/2)sin(C/2)}\\equal{} \\frac{R}{r}$; :)",
  "Solution_3": "thank's a lot\r\n nice solutions  :)",
  "Solution_4": "I need some help regarding my solution approach.:help: \r\nWe know that A=area of triangle=abc/4R\r\nthus, R=abc/4A\r\nagain let S=(a+b+c)/2\r\nthus, r=A/S\r\nthus, R/r=abcS/4A^2=abc(a+b+c)/8[(a+b+c)(a+b-c)(b+c-a)(c+a-b)/16]\r\nor, R/r=2abc/(a+b-c)(b+c-a)(c+a-b)\r\nthus, the inequality becomes,\r\n          (a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b) <= 8a^2b^2c^2\r\nNow, [(a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b)]^1/6 <= [a+b+b+c+c+a+a+b-c+b+c-a+c+a-b]/6=3[a+b+c]/6=S\r\nthus enough to show, S^3 <= 8^1/2*abc...............Hereafter I failed to continue.........:help:"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find out the first nonzero rightmost digit when $ 125!$ is represented in base $ 5$.",
  "Solution_1": "Let $ f(k)\\equal{}5k\\plus{}1)(5k\\plus{}2)(5k\\plus{}3)(5k\\plus{}4)$.\r\n\\[ 125!\\equal{}\\prod_{k\\equal{}0}^{24}f(k) *5^{25} \\prod_{k\\equal{}0}^4f(k)*5^{5}f(0).\\]\r\n$ 1*2*3*4 \\equal{} 4\\mod 5$, therefore $ f(k)\\equal{}4\\mod 5$ and $ 125!\\equal{}5^{25\\plus{}5\\plus{}1}*a, a\\equal{}4^{25\\plus{}5\\plus{}1} \\equal{} 4\\mod 5$."
}
{
  "Tag": [
    "limit",
    "algebra",
    "polynomial"
  ],
  "Problem": "(a) Show that for every $n\\in\\mathbb{N}$ there is exactly one $x\\in\\mathbb{R}^+$ so that $x^n+x^{n+1}=1$. Call this $x_n$.\r\n(b) Find $\\lim\\limits_{n\\rightarrow+\\infty}x_n$.",
  "Solution_1": "Equation can be rewritten $1+ x = 1/x^n$\r\n\r\nLHS is strictly increasing from 1 to $+\\infty$, RHS is strictly decreasing from $\\infty$ to $0^+$ this proves existence and uniqueness of solution\r\n\r\n$x^n+x^{n+1}$ is strictly increasong and $1^n+1^{n+1} > 1$ so $1 > x_n > 0$ and $(1+x)^{1/n} \\rightarrow 1$ (typo corrected).\r\n\r\nEquation can be rewritten $x(1+ x)^{1/n} = 1$ proving that $x_n \\to 1$",
  "Solution_2": "i disagree since if you have n being odd, then the polynomial has an even lead coefficient, and since the roots pair up, there cannot be just 1 real root, there must be a multiple of 2 of real roots",
  "Solution_3": "[quote=\"Altheman\"]i disagree since if you have n being odd, then the polynomial has an even lead coefficient, and since the roots pair up, there cannot be just 1 real root, there must be a multiple of 2 of real roots[/quote]yes, I forgot a detail: x should be positive. The second solution will be negative in your case.\r\n\r\nT\u00b5t\u00b5 is right anyway. :)",
  "Solution_4": "[quote=\"t\u00b5t\u00b5\"] and $(1+x)^{1/n} \\rightarrow 0$.[/quote]\r\nbut $(1+x)^{1/n} \\rightarrow 1$ no?!",
  "Solution_5": "yes. he meant 1 I guess, since the rest of his solution uses 1. :)",
  "Solution_6": "(a)\n$x^n+x^{n+1}$ is strictly increasing and continuous on $\\mathbb R^{+}$.And $x^n+x^{n+1}$ goes from $0$ to $+\\infty$ on $\\mathbb R^{+}$.Then by [b]intermediate value theorem,[/b]\n$\\exists ! x_n>0$ s.t. $x_n^n+x_n^{n+1}=1\\blacksquare$ :coool:\n\n(b)\nObviously $0<x_n<1$.If $x_{n+1}<x_n,$then \n$1=x_n^n+x_n^{n+1}>x_{n+1}^{n+1}+x_{n+1}^{n+2}$ which is absurd.Thus $x_{n+1}\\ge x_n$.Then $\\{x_n\\}$ is increasing and upper-bounded.Therefore $\\exists \\alpha$ s.t. $\\lim_{n\\to +\\infty} x_n=\\alpha$.$0<\\alpha \\le 1$.If $0<\\alpha <1,$then $\\lim_{n\\to +\\infty} \\ln x_n^n=\\lim_{n\\to +\\infty} n\\ln x_n=-\\infty$.Hence\n$\\lim_{n\\to +\\infty} x_n^n=0$.And $0<x_n^{n+1}<x_n^n,$by [b]squeeze theorem,[/b]\n$\\lim_{n\\to +\\infty} x_n^{n+1}=0$.Thus\n$\\lim_{n\\to +\\infty} x_n^n+x_n^{n+1}=0$ which is absurd.Therefore $\\alpha =1\\blacksquare$ :coool:"
}
{
  "Tag": [],
  "Problem": "Find all positive integers $ m$ and $ n$ such that $ m^{2}\\plus{} n^{5} \\equal{} 252$.",
  "Solution_1": "$ n^5\\le 252$\r\n\r\nso $ n\\le 3$\r\n\r\nFrom the above we can easily say that \"n\" and \"m\" must be equal to 3"
}
{
  "Tag": [
    "quadratics",
    "calculus",
    "derivative",
    "function",
    "geometry solved",
    "geometry"
  ],
  "Problem": "Let a;b;c;d are real numbers satisfying a;b;c;d<1.Prove that:\r\n$x(1-t)+t(1-z)+z(1-y)+y(1-x) \\geq 1$",
  "Solution_1": "[quote=\"chien than\"]Let a;b;c;d are real numbers satisfying a;b;c;d<1.Prove that:\n$x(1-t)+t(1-z)+z(1-y)+y(1-x) \\geq 1$[/quote]\r\n\r\nWhat are $x,y,z,t$ ?",
  "Solution_2": "Sorry!\r\na;b;c;d are x;y;z;t",
  "Solution_3": "Why can't you order a,b,c,d and use rearrangement then minimize the quadratics?\r\n\r\nOr, if that fails, use derivatives? The function itself seems very tame",
  "Solution_4": "[quote=\"chien than\"]Let a;b;c;d are real numbers satisfying a;b;c;d<1.Prove that:\n$x(1-t)+t(1-z)+z(1-y)+y(1-x) \\geq 1$[/quote]\r\nWhat about $x=y=z=t=0$?",
  "Solution_5": "[quote=\"bilarev\"][quote=\"chien than\"]Let a;b;c;d are real numbers satisfying a;b;c;d<1.Prove that:\n$x(1-t)+t(1-z)+z(1-y)+y(1-x) \\geq 1$[/quote]\nWhat about $x=y=z=t=0$?[/quote]\r\nMoreover, taking $x=y=z=t$ shows that even if we assume $x,y,z,t>0$, the $\\mathrm{LHS}$ can have any value from the interval $\\langle 0;2\\rangle$ (with the additional assumption [b]me@home[/b]'s method is an easy way to prove that $2$ is the maximum).",
  "Solution_6": "${(1-t) x+(1-x) y+t (1-z)+(1-y) z\\geq 1}$\r\n\r\n${(1-t) x+(1-x) y+t (1-z)+(1-y) z\\geq 1}$\r\n${-x t-z t+t+x-x y+y-y z+z\\geq 1}$\r\n${(t+y-1) (x+z-1)\\leq 0}$\r\n\r\n${\\{x,y,z,t\\}<\\{0,0,0,0\\}}$\r\n\r\nSo it is true",
  "Solution_7": "[quote=\"Thales418\"]${(1-t) x+(1-x) y+t (1-z)+(1-y) z\\geq 1}$\n\n${(1-t) x+(1-x) y+t (1-z)+(1-y) z\\geq 1}$\n${-x t-z t+t+x-x y+y-y z+z\\geq 1}$\n${(t+y-1) (x+z-1)\\leq 0}$\n\n${\\{x,y,z,t\\}<\\{0,0,0,0\\}}$\n\nSo it is true[/quote]\r\nBut what if $x+z>1$ and $y+t>1$?",
  "Solution_8": "${x<0\\land y<0\\land z<0\\land t<0}$\r\n\r\n${x=-a\\land y=-b\\land z=-c\\land t=-d}$\r\n\r\n${a>0\\land b>0\\land c>0\\land d>0}$\r\n\r\n${(t+y-1) (x+z-1)\\leq 0}$\r\n\r\n${(-a-c-1) (-b-d-1)\\leq 0}$\r\n${-(a+c+1) (-1) (b+d+1)\\leq 0}$\r\n${(a+c+1) (b+d+1)\\leq 0}$\r\n\r\nSince   ${a>0\\land b>0\\land c>0\\land d>0}$\r\n\r\nIt' s false\r\n\r\nMy error ... Sorry"
}
{
  "Tag": [
    "geometry",
    "calculus",
    "integration",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "How do I find the area for the intersection of 2 polar curves without converting to Cartesian? \r\nFor example how do I get the integral for the area between $ r\\equal{} \\sin \\theta$ and $ r\\equal{}\\cos\\theta?$",
  "Solution_1": "The area of a \"slice\" of a polar curve $ r(\\theta)$ is given by $ \\frac{1}{2} \\int_{\\theta_1}^{\\theta_2} r^2 \\, d \\theta$.  You need to find intersections $ \\theta_1, \\theta_2$ and then calculate the difference in the areas."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "probability",
    "videos",
    "AMC 10"
  ],
  "Problem": "Who's the youngest person ever to make the US IMO team? What was his/her name?\r\n\r\nAnd do you think that Evan O'Dorney dude will make it onto the team this year?",
  "Solution_1": "I probably could get 22 or higher but in a testing situation I'll probably end up 15-21.",
  "Solution_2": "whos evan o dorney?",
  "Solution_3": "He's a middle school math beast.\r\n\r\nHe also recently won the Scripps Spelling Bee, which earned him a lot of attention. A TV interview with some news woman earned him even more publicity (you can see it on youtube).",
  "Solution_4": "how beast imo?",
  "Solution_5": "There was this guy who made it to the IMO when he was 10. He lived in Australia at the time, I think.",
  "Solution_6": "[quote=\"1=2\"]There was this guy who made it to the IMO when he was 10. He lived in Australia at the time, I think.[/quote]\r\nYeah, I think you thought about Terence Tao :wink: Real genius  :)",
  "Solution_7": "Yeah, Terence Tao won the Fields medal.",
  "Solution_8": "[quote=\"The Zuton Force\"]He's a middle school math beast.\n\n[b]He also recently won the Scripps Spelling Bee, which earned him a lot of attention. A TV interview with some news woman earned him even more publicity (you can see it on youtube).[/b][/quote]\r\n\r\nyou're kidding right?\r\n\r\ni remember all that publicity was very negative if you read all the comments...\r\npeople dont appreciate his talents.",
  "Solution_9": "Theres no such thing as bad publicity?",
  "Solution_10": "i watched an interview on youtube of Evan O. He seems like he has some issues is he autistic? some times academic achievement is not the most important thing. did he make IMO?",
  "Solution_11": "IMO, Evan O'Dorney won't make IMO. O.K. he made MOP last year. That was great. Making MOP in eighth grade is great. Yet, I think that others (Krishnau, Alex Anderson, etc) are more pro. Evan O'Dorney may have a chance next year.",
  "Solution_12": "[quote=\"blackbelt14253\"]Who's the youngest person ever to make the US IMO team? What was his/her name?\n\nAnd do you think that Evan O'Dorney dude will make it onto the team this year?[/quote]\r\n\r\nI'm not sure, but I'd think it would be either Reid Barton or Tiankai Liu? I'm guessing they were about 14 at the time. Actually, come to think of it, Ricky Liu and Po-Ru Loh are also possible.",
  "Solution_13": "Well it's impressive and cool that Evan is so successful, but I do agree that it is not good publicity. I feel like those interviews reinforce the negative stereotypes that people have about people who like math. Those interviews were the most awkward thing that I've ever seen. It portrays math students as lacking any social skills whatsoever.",
  "Solution_14": "USAMO top twelve last year who have not graduated:\r\n\r\n- Sergei Bernstein (11)\r\n- Eric Larson (11)\r\n- Haitao Mao (12)\r\n- Delong Meng (11)\r\n- Krishanu Shankar (12)\r\n- Jacob Steinhardt (12)\r\n- Alex Zhai (12)\r\n\r\nHonorable Mentions last year who have not graduated:\r\n\r\n- David Benjamin (11)\r\n- Yingyu (Dan) Gao (12)\r\n- Yan Li (12)\r\n- Gaku Liu (12)\r\n- Jeffrey Manning (12)\r\n- Palmer Mebane (12)\r\n- [b]Evan O'Dorney (9)[/b]\r\n- Alexander Remorov (12)\r\n- Max Rosett (12)\r\n- Qiaochu Yuan (that's me! :) ) (12)\r\n\r\nwhich puts Evan O'Dorney in some pretty respectable company.  In any case, social skills (or lack thereof) notwithstanding, he's got three more years to make it onto the team and a pretty good head start.  :)",
  "Solution_15": "[quote=\"PI-Dimension\"][quote=\"n^4+4\"]Haha is Jeremy Kahn approximately equal to Jeremy Hahn?[/quote]\n\n\nLol sure.\n\nI was like wow that looks familiar.[/quote]\r\n\r\nThis is on topic?\r\n\r\nBesides, if you watched the videos of Evan being interviewed, you would get it.",
  "Solution_16": "[quote=\"mathcrazed\"]Evan O'Dorney has Asperger's... [/quote]\r\n\r\nWhat is Asperger's?",
  "Solution_17": "aspergers is a type of autism",
  "Solution_18": "As a rule of thumb, you can't have autism and make MOP in eighth grade",
  "Solution_19": "A human with Asperger syndrome displays intense interest in restricted interests or activities. This may lead to that human being extremely knowledgeable or good at that activity/interest. However, because this is still a brain disorder, it is labeled as autism.\r\nI don't know whether Evan O'Dorney has Aspergers or not, but someone with autism [i]can[/i] make MOP. Actually, with some types of autism, you're more likely to.",
  "Solution_20": "Asperger's syndrome is a syndrome on the autism [i]spectrum.[/i]  This is not quite the same thing as being autistic.  In any case, the autism spectrum is quite wide and there are many gradations of autistic behavior.",
  "Solution_21": "[quote=\"#H34N1\"]As a rule of thumb, you can't have autism and make MOP in eighth grade[/quote]\r\n\r\nI would take offense to that if I was close to someone with autism. I don't know what your intention was with that comment, but one could interpret it as you completely berating autistic people. As far as I know, some autistic people have extraordinary abilities, in things such as memorizing pi and music. Who knows, maybe there is an autistic person out there, if Evan O'Dorney doesn't have Asperger's as some people are oddly suggesting, that excels at olympiad math. While I still may be grossly wrong, unless you really know your stuff about autism, you probably shouldn't make a generalization like that.",
  "Solution_22": "Evan shows signs of Asperger's... Although he does attempt to engage in social activities, he is unable to understand some types of irony, sarcasm, etc.  He does have an intense interest in math, as uldivad9 implied, which is also a symptom of Asperger's.",
  "Solution_23": "I think we should all stop speculating, it's Evan's business, and by no means ours.\r\n\r\nAlso, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159137]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162184]this[/url]",
  "Solution_24": "Yeah these threads seem to get really popular about him. Guys it shouldn't be our business to speculate his mental condition, and he probably gets enough beef if a lot of people perceive a problem anyway. At least not on an online forum  :wink:. It isn't on topic anyway...\r\n\r\nAnyway I think this year has a lot of really good seniors and that USA will do really well at IMO, in my opinion. We might even win!  :D",
  "Solution_25": "[quote=\"CatalystOfNostalgia\"][quote=\"#H34N1\"]As a rule of thumb, you can't have autism and make MOP in eighth grade[/quote]\n\nI would take offense to that if I was close to someone with autism. I don't know what your intention was with that comment, but one could interpret it as you completely berating autistic people. As far as I know, some autistic people have extraordinary abilities, in things such as memorizing pi and music. Who knows, maybe there is an autistic person out there, if Evan O'Dorney doesn't have Asperger's as some people are oddly suggesting, that excels at olympiad math. While I still may be grossly wrong, unless you really know your stuff about autism, you probably shouldn't make a generalization like that.[/quote]\r\n\r\nI'm not berating autistic people. I'm just saying that it doesn't make sense to claim that someone is autistic when they are pwnage at math skillz :|",
  "Solution_26": "h341\r\nfor the last time many people with autism are very good at math",
  "Solution_27": "[quote=\"#H34N1\"][quote=\"CatalystOfNostalgia\"][quote=\"#H34N1\"]As a rule of thumb, you can't have autism and make MOP in eighth grade[/quote]\n\nI would take offense to that if I was close to someone with autism. I don't know what your intention was with that comment, but one could interpret it as you completely berating autistic people. As far as I know, some autistic people have extraordinary abilities, in things such as memorizing pi and music. Who knows, maybe there is an autistic person out there, if Evan O'Dorney doesn't have Asperger's as some people are oddly suggesting, that excels at olympiad math. While I still may be grossly wrong, unless you really know your stuff about autism, you probably shouldn't make a generalization like that.[/quote]\n\nI'm not berating autistic people. I'm just saying that it doesn't make sense to claim that someone is autistic when they are pwnage at math skillz :|[/quote]\r\n\r\nPeople weren't claiming that he had Asperger's, or he was autistic, just because he was smart, they were claiming that because of his personality as seen on television. And I don't exactly see how what you are/were \"just saying\" can ever be interpreted from your comment. So long as we're clear...",
  "Solution_28": "Alright guys let's get back on topic? Evan O'Dorney may or may not have a mental condition, but I think that is a really private thing.\r\n\r\nAny subsequent posts off topic will be deleted. Feel free, of course, to discuss what you like over PM.",
  "Solution_29": "Agreed. I will be watching this topic as well."
}
{
  "Tag": [
    "number theory"
  ],
  "Problem": "it is well known that there is no known closed-form of the formula for partitions.\r\nHas it been proven that there is no formula? If so, could someone prove it here if it isn't too complicated?",
  "Solution_1": "That's not entirely true.\r\nThere is the Rademacher Series, which is asymptotically correct.\r\nHere's the wikipedia link, but beware, the Rademacher series is pretty complex and involves Dedekind sums.\r\n[url=http://en.wikipedia.org/wiki/Partition_(number_theory)#Rademacher.27s__series]Here[/url]",
  "Solution_2": "[quote=\"DiscreetFourierTransform\"]That's not entirely true.\nThere is the Rademacher Series, which is asymptotically correct.\nHere's the wikipedia link, but beware, the Rademacher series is pretty complex and involves Dedekind sums.\n[url=http://en.wikipedia.org/wiki/Partition_(number_theory)#Rademacher.27s__series]Here[/url][/quote]\r\nI meant a closed form.",
  "Solution_3": "Well, I suppose that's as close to a closed form as we have. \r\nHowever, you're right that there is no exact closed form.\r\nThat link has all the information in it.\r\n\r\nEDIT: The link doesn't like the URL tags.\r\nJust copy+paste it into the browser."
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "prove that for every n\\in N* the following ecuation\r\n\\sum {k=1,n}1/ \\sqrt (nx+k) = \\sqrt(n)\r\nhas a unique real root x=x_n and find lim{n->oo}x_n.",
  "Solution_1": "Let f(x) be the function f(x) =  \\sum(k=1,..,n){1/\\sqrt (nx+k)} ,  defined for x > -1/n .\r\n\r\nIt's easy to see that lim(x-->+\\infty){f(x)} =  0 <  \\sqrt(n) and lim(x-->-1/n){f(x)} = +\\infty > \\sqrt(n) , so there is at least a real root x(n) , -1/n < x(n) < 0\r\n\r\nManifestly the derivative f'(x) is ngative, so f(x) is decreasing , so the root x(n) is  unique.\r\n\r\n-1/n < x(n) < 0  ===> lim(n-->\\infty){x(n)} = 0\r\n........",
  "Solution_2": "not very correct\r\nx_n is in (-1/n,oo)\r\nso its limit is not 0.\r\ncheck again your proof\r\nu can treat the more genrealy problem\r\n\r\n[i]f(x)=a \\sqrt n has a unique root x=x_n and find it's limit, where a>0[/i][/i]",
  "Solution_3": "Oh god, Diogene is really too  bad, he mistook 0 for  \\infty    !  So I , his \" alter-ego\", decided to take his place . :D  :D \r\nAnd now, come  back to the problem ..\r\n\r\nS_n(x) = (1/n) \\sum(k=1,..,n){1/\\sqrt(x+k/n)} = A , A > 0 ; We know that there is an unique solution x(n), x(n) > -1/n .\r\n\r\nBriefly, we can relate S_n(x) to the Reamann sums for I(x) = \\int(0 < t <1){1/sqrt(1/(x+t)).dt} = 2*(\\sqrt (x+1) + \\sqrt(x)) = I(x) , for x>0.\r\n\r\nFinally the result is : \r\n0 < A =< 2 ===> I(x)=A  has an unique solution  ===> lim(n-->\\infty){x(n)} = (4-A<sup>2</sup> )<sup>2</sup>/(16*A<sup>2</sup>)\r\n2 < A          ===> I(x)=A  has not  solution ===> lim(n-->+\\infty ){x(n)} = 0\r\n\r\nIn general, if the  (1/n)* \\sum (k=1,..,n){f(x,k/n)} = g(x) has a solution x(n) ,\r\n\r\nthen the study of the behavior of the x(n) sequence is relating to the equation   \\int(0< t <1){f(x,t).dt} = g(x)\r\n...",
  "Solution_4": "very good!! :D:D"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "The farmer\u2019s pivot has an effective irrigation radius of \u00bc mile (640 acres in a square mile) and is positioned alongside a straight section of highway. He wants to apply a special fertilizer at the rate of 0.7 gallons per acre. How much does he need to get for the irrigation system?\r\n\r\nA) 45 gallons B) 88 gallons C) 32 gallons D) 50 gallons",
  "Solution_1": "His lot is a semi-circle with radius $ \\frac14$. The area is $ \\frac\\pi{32}$ square miles or $ 20\\pi$ acres.\\[ 20\\pi\\cdot0.7\\approx43.98\\]This is close to A."
}
{
  "Tag": [
    "floor function",
    "induction",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "A word is a sequence of n letters of the alphabet {a, b, c, d}. A word is said to be complicated if it contains two consecutive groups of identic letters. The words caab, baba and cababdc, for example, are complicated words, while bacba and dcbdc are not. A word that is not complicated is a simple word. Prove that the numbers of simple words with n letters is greater than $2^n$, if n is a positive integer.",
  "Solution_1": "It is also Romanian TST 2003. And as I remember, this problem were discussed before on forum.",
  "Solution_2": "This problems is still unsolved.  :(",
  "Solution_3": "Let us denote by $S(n)$ the set of simple words with $n$ letters and by $s_n$ the number of elements in $S(n)$. If we put a letter at the end of each of the simple words from $S(n)$ we obtain a set $T(n+1)$ of $t_{n+1}$ words of length $n+1$, their number being $t_{n+1}=4s_n$. Obviously $S(n+1) \\subset T(n+1)$, $S(n+1)\\neq T(n+1)$.\r\n\r\nLet $T_1(n+1)$ be the set of those words from $T(n+1)$ which have the last two letters the same, $T_k(n+1)$ the set of $T(n+1)$ which end in two consecutive identical groups of $k$ letters, for each $k\\in\\{1,2,\\ldots,m\\}$, where $m=\\displaystyle \\left\\lfloor \\frac { n+1 } 2 \\right\\rfloor$. Obviously\r\n\\[ f(n+1)\\geq t_{n+1}-|T_1(n+1)|-|T_2(n+1)|-\\cdots - |T_m(n+1)|\\]\r\nWe have $t_{n+1}=4f(n)$, $|T_1(n+1)|=f(n)$, and furthermore $|T_k(n+1)|\\leq f(n+1-k)$, because of the fact that the mapping of $S(n+1-k)$ into $T_k(n+1)$ given by adding to a simple word of $n+1-k$ letters its own last $k$ letters is obviously surjective.\r\n\r\nWe have $f(1)=4$, $f(2)=12>4$. By induction we want to prove $f(k+1)>2f(k)$. We have\r\n\\[ f(n+1)\\geq 4f(n)-f(n)-\\frac 12 f(n) - \\frac 14 f(n)- \\cdots > 2 f(n) \\] from which the conclusion follows.",
  "Solution_4": "But still the obvious next questions are :\r\nQ1: Is 2^n  a tight bound ? \r\nQ2: Is the bound dependent on the alphabet size ? \r\n\r\n A precision : when I say obvious I mean the esayness to formulate such questions\r\n not that the answers are necessarily easy.\r\n\r\n REMARK : \r\n It seems that too often that thes olympiad/IMO problems concentrated on one fact. \r\n I believe it is nice to follow up , to give perspective to the result by showing how it\r\n hinges on the hypothesis, otherwise the solution as little meaning or constitutes merrely\r\n a useful technique. See and read Polya about problem solving in general and making \r\nVARIATION on problems.",
  "Solution_5": "Valentin Vornicu wrote:\r\n\r\n\\box{\r\nWe have $t_{n+1}= 4f(n)$, $|T_{1}(n+1)| = f(n)$, and furthermore $|T_{k}(n+1)| \\leq f(n+1-k)$, because of the fact that the mapping of $S(n+1-k)$ into $T_{k}(n+1)$ given by adding to a simple word of $n+1-k$ letters its own last $k$ letters is obviously surjective.}\r\n\r\nThe mapping of $S(n+1-k)$ into $T_{k}(n+1)$  defined latter is wrong defined. The mapping of $T_{k}(n+1)$ into $S(n+1-k)$ given by removing to a word of $T_{k}(n+1)$ its own last $k$ letters is obviously injective (and this mapping is well defined).",
  "Solution_6": "another solution I came across by [b]math154[/b]  from [url=https://artofproblemsolving.com/community/c285h400665p2237686]here[/url]\n\n[hide=\"Solution\"]Let $f(n)$ denote the number of simple words with $n$ letters. We'll show by strong induction that $f(n+1)>2f(n)$ for all $n\\ge0$. Since $f(0)=1$ and $f(1)=4$, the base case is clear.\n\nNow consider a simple word $W$ with $n\\ge1$ letters, and prepend a letter $x\\in\\{a,b,c,d\\}$ to it. There exist strings $S,T$ (possibly empty) such that\n\\[x+W=x+S+x+S+T \\Longleftrightarrow W=S+x+S+T\\]iff $x+W$ is complicated (note that $x+S+T\\in W$ and is thus simple). Hence there is a surjection from simple words with $1\\le k=|x+S+T|\\le n$ letters to complicated words of the form $x+W$ (i.e. prepending the first $n+1-k$ letters of $x+S+T$ to get $x+W$), so by complementary counting,\n\\[f(n+1) \\ge 4f(n) - \\sum_{k=1}^{n}f(k) > 4f(n) - \\sum_{k=1}^{n}\\frac{f(n)}{2^{n-k}} > 2f(n),\\]as desired.[/hide]\n\nI hope it is not exactly the same as above"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $x,y,z\\geq 2$. Find the least $k$ such that \r\n$(x^3+y)(y^3+z)(z^3+x)\\leq kxyz$",
  "Solution_1": "For every $k$ we can add $y=z=2$ and $x$ can be a large number; so the left side will be greater than the right side.\r\nI think you mean this one : $x,y,z\\geq2$\r\n$(x^3+y)(y^3+z)(z^3+x)\\geq125xyz$",
  "Solution_2": "[quote=\"Ramtin Madani\"]\nI think you mean this one : $x,y,z\\geq2$\n$(x^3+y)(y^3+z)(z^3+x)\\geq125xyz$[/quote]\r\n\r\nRight, can you prove it?",
  "Solution_3": "$(x^3+y)(y^3+z)(z^3+x)\\geq(4x+y)(4y+z)(4z+x)\\geq 5\\sqrt[5]{x^4y}\\cdot 5\\sqrt[5]{y^4z}\\cdot 5\\sqrt[5]{z^4x}=125xyz$\r\nRamtin, sorry.  :)"
}
{
  "Tag": [
    "inequalities",
    "inequalities theorems"
  ],
  "Problem": "Hello \r\n\r\nPlz help me to find some free books about inequalities.\r\n\r\nThanks",
  "Solution_1": "hello, see this here\r\nhttp://www.eleves.ens.fr/home/kortchem/olympiades/Cours/Inegalites/tin2006.pdf\r\nSonnhard.",
  "Solution_2": "Thank you a lot for the book\r\nit's great\r\n\r\nare there any other books ? not about inequalities only\r\n\r\n^^\r\n\r\n10x"
}
{
  "Tag": [
    "number theory solved",
    "number theory"
  ],
  "Problem": "solve in positive integers the following equation\r\n$x^2y^2=z^2(z^2-x^2-y^2)$",
  "Solution_1": "hmm.. discriminant of this equation in z^{2} is x^{4}+y^{4}+6x^{2}y^{2} which is never a perfect square .i proved this yesterday.and btw wot is the source of this problem or u made it up?\r\nnice one ,neway",
  "Solution_2": "This is a Bulgarian problem I think, i'm not sure :?",
  "Solution_3": "Bulgaria 1998 :D \r\nOn the site of the Bulgarian MO, another solution is given"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all functions $ f: [0,1] \\to  [0,1]$ and \\[ f(2x - f(x)) = x\\] for all $x \\in [0,1]$.",
  "Solution_1": "[quote=\"MJ GEO\"]$ f$ form $ [0,1]$ to $ [0,1]$ and $ f(2x \\minus{} f(x)) \\equal{} x$[/quote]\r\n\r\nThis immediately implies $ f((n\\plus{}1)x\\minus{}nf(x))\\equal{}nx\\minus{}(n\\minus{}1)f(x)$ and so $ 1\\ge nx\\minus{}(n\\minus{}1)f(x)\\ge 0$ and so $ \\frac n{n\\minus{}1}x\\ge f(x)\\ge \\frac{nx\\minus{}1}{n\\minus{}1}$ and so $ \\boxed{f(x)\\equal{}x}$ which, indeed, is a solution.",
  "Solution_2": "i cant understand your solution.coud you write more about it? its mine\r\n$ g(x)\\equal{}2x\\minus{}f(x)$ so $ f(g(x))\\equal{}x$. $ g(g(x))\\equal{}2g(x)\\minus{}f(g(x))\\equal{}(2g(x)\\minus{}2x)\\plus{}x$ and similiary $ g^n(x)\\equal{}(ng(x)\\minus{}nx)\\plus{}x$ so $ g(x)\\equal{}x$ so $ f(x)\\equal{}x$"
}
{
  "Tag": [
    "floor function",
    "logarithms",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Please help me with this one:\r\n$A=3^{2004}+25$\r\n1. How many digits A have(in base $10$)? (easy, I found that $957$ digits)\r\n2.  $S=$sum of all this digits. How much is S? \r\n3. How many do zeroes digits appear in A?",
  "Solution_1": "I doubt whether this can be done without using a computer.",
  "Solution_2": "Yes, Arnel. But using caculor is not a good way and it is meanless.\r\n   I hope some could post a good solution for it! I know that it seems difficult but let try it ! :D",
  "Solution_3": "A \"good\" (elegant) solution simply cannot exist...",
  "Solution_4": "I know the answer to 1, however, I'm still thinking about 2 and 3.  Here\r\ns 1's solution that can somehow be done without a calculator.  I'm still thinking about doing it without a calculator...hopefully, it is possible.\r\n[hide]\n$\\\\\\lfloor\\log 3^2004+1\\rfloor\\\\=\\lfloor2004\\log 3+1\\rfloor\\\\=\\lfloor 956.15+1\\rfloor\\\\=957$\n[/hide]\r\n\r\nMasoud Zargar"
}
{
  "Tag": [],
  "Problem": "Do you know of any free (preferably free as in freedom) non-demo non-trial non-shareware music notation programs for Windows that can save/export as MIDI?\r\n\r\nI'm currently using Finale Notepad 2006, but it doesn't save/export as a MIDI file.",
  "Solution_1": "NoteWorthy Composer, I'm not sure you can save as a MIDI file, but I know it works.  You could try.",
  "Solution_2": "The website says that it's a 30-day evaluation trial, and according to download.com and other sites, it's shareware. :(",
  "Solution_3": "-_- I've been on 30-day trial for a few years then...\r\n\r\nIs Finale Notepad 2006 good for just writing though?",
  "Solution_4": "Yes.  Mr. Muer recommends it.\r\n\r\nDoes the demo have nag screens, etc.?",
  "Solution_5": "[quote=\"Eli\"]Yes.  Mr. Muer recommends it.\n\nDoes the demo have nag screens, etc.?[/quote]\r\n\r\nnag screens?  \r\n\r\nDo we have to use notepad for the assignment (and it's optional, correct?)?",
  "Solution_6": "What I mean is if there are major annoyances, etc.  Are there any important features you can't use?\r\n\r\nMr. Muer wants us to use Notepad (probably because he doesn't know of anything else), but if it's a music file, he'll probably be fine with it.",
  "Solution_7": "[quote=\"Eli\"]What I mean is if there are major annoyances, etc.  Are there any important features you can't use?\n\nMr. Muer wants us to use Notepad (probably because he doesn't know of anything else), but if it's a music file, he'll probably be fine with it.[/quote]\r\n\r\nNo, I believe it does not have \"nag screens.\"  But don't take my word for it.",
  "Solution_8": "what is a MIDI?\r\n\r\nFor future purposes, if you don't need a WYSIWYG music editor then use musixtex package in MiKTeX.  Ask on the TeX forum.  I don't know how to use it myself :P",
  "Solution_9": "It's a type of music file (see http://en.wikipedia.org/wiki/Musical_Instrument_Digital_Interface), like mp3, wav, etc.\r\n\r\nI know.  I like to have a WYSIWYG editor for music.  I don't have to worry about the syntax with them.",
  "Solution_10": "Currently I use GenieSoft Score Writer v2.6. It costs money, but it's only about 30 dollars. It does all the MIDI import/export stuff with ease and multiple tracks. It has a WYSIWYG interface with easy-to-use mouse and keyboard controls. With the given price, I'd say it's one of the more economical software packages out there.\r\n\r\nHowever, changing the scores to PDF give some problems (yes, I use PDF pro v7).",
  "Solution_11": "sibelius :lol:"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "$ S$ is a set of functions such that $ f: N \\rightarrow R$ and \r\n(1)$ f(1) \\equal{} 2$ \r\n(2) $ f(n \\plus{} 1) \\ge f (n) \\ge \\frac {n}{n \\plus{} 1} f(2n)$ .\r\n Determine the smallest $ M$ such that  $ f(n) < M$  (for any $ f \\in S$ and for any $ n \\in N$)",
  "Solution_1": "Obviosly $ f(n)\\le f(2n)\\le (1\\equal{}\\frac 1n )f(n)$, therefore $ f(\\infty )\\le \\prod_{k\\equal{}0}^{\\infty }(1\\plus{}\\frac{1}{2^k})\\equal{}g(\\frac 12$, were $ g(x)\\equal{}\\prod_{k\\equal{}0}^{\\infty }(1\\plus{}x^k)$ (|x|<1).",
  "Solution_2": "I also managed to prove that $ f(\\infty) \\leq g(1/2)$ , but why $ g(1/2)$ would be the minimum value ?",
  "Solution_3": "$ g(\\frac {1}{2})$=? It's not finally answer.",
  "Solution_4": "[quote=\"Svejk\"]I also managed to prove that $ f(\\infty) \\leq g(1/2)$ , but why $ g(1/2)$ would be the minimum value ?[/quote]\r\nWe can define (maximal possible mean) $ f(n)\\equal{}\\prod_{i\\equal{}0}^{k}(1\\plus{}\\frac{1}{2^i})$, were $ k\\equal{}[log_2(n\\minus{}1)],n>1$.\r\n$ g(x)$ is special function considered by Euler."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "Six people arrive to purchase $10 tickets for a movie. Three of them have only $10 dollar bills and 3 have only $20 dollar bills. The person selling the tickets initially has no change. What is the probability that the people line up in such a way that the person selling the tickets always has enough change?",
  "Solution_1": "[hide]THe probatilty is 1/4.\nObviously, the last person to enter the movie theater must have been a person with a 20 dollar bill.  THerefore, all three people with 10 dollar bills and 2 people with 20 dollar bills must have entered before he did.  3/6*2/5*1/4*3/3*2/2=1/20.  Of course, the people with 10 dollar bills didn't have to go first, but it makes no difference becuase the numerators and denominators are just switched around accordingly.  Now, we just need to figure out how many ways this can be done.  Obviously, the first person has a 10 dollar bill and the last has a 20 dollar bill.  $4 \\choose 2$ is 6 so there's 6 ways to choose the remaining 4 people, except that 10,20,20 is not allowed.  THerefore, there's 6-1=5 middle possibilities and 5*1/20=1/4.[/hide]"
}
{
  "Tag": [],
  "Problem": "Whats me to solve for this system\r\n\r\n(  2x-4y=32\r\n{3x+5y=4\r\n(",
  "Solution_1": "this system can be turned into 6x-12y=96 and 6x+10y=8, leaving us with y = -4 and x = 8.",
  "Solution_2": "Multiply $2x-4y=32$ by $3$ to get $6x-12y=96$. Multiply $3x+5y=4$ by $2$ to get $6x+10y=8$. Now, subtract the first equaiton from the second to get $22y=-88$. Divide both sides by $4$ to get $y=-4$. Now, plug in y back into the first equation to get $2x-4(-4)=32$ or $2x=16$. Divide both sides by $2$ to get $x=8$.\r\n\r\nProblems like these are more middle school level and not high school level and can be posted here: http://www.artofproblemsolving.com/Forum/index.php?f=299 instead.",
  "Solution_3": "[hide]I would just do substitution myself, because i'm wierd.  Add 4y to both sides to get $\\displaystyle{2x=4y+32}$  Then divide by 2...\n$\\displaystyle{x=2y+16}$  Substitute that for x and solve for y... then sub y in for x in this equation and you get x=8 and y=-4...[/hide]",
  "Solution_4": "[quote]because i'm wierd[/quote]\r\n\r\nur also weird cuz u spelled weird wrong. :rotfl: \r\njust kidding.\r\nno offense."
}
{
  "Tag": [
    "trigonometry",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "$ \\sqrt[3]{x^3\\minus{}12x^2\\plus{}27x\\minus{}8}\\equal{}2x\\plus{}\\sqrt[3]{x^3\\minus{}3x}$",
  "Solution_1": "hello, this equation has no real solution.\r\nSonnhard.",
  "Solution_2": "i think there is some mistakes because this equation has more than one real root",
  "Solution_3": "[quote=\"hoangvn.fix\"]$ \\sqrt [3]{x^3 \\minus{} 12x^2 \\plus{} 27x \\minus{} 8} \\equal{} 2x \\plus{} \\sqrt [3]{x^3 \\minus{} 3x}$[/quote]\r\n\r\n$ x_1 \\equal{} 2\\cos \\frac {4\\pi}9$\r\n\r\n$ x_2 \\equal{} 2\\cos \\frac {10\\pi}9$\r\n\r\n$ x_3 \\equal{} 2\\cos \\frac {16\\pi}9$\r\n\r\n[hide=\"Why?\"]Let $ u\\equal{}\\sqrt[3]{x^3\\minus{}3x}$.\n\nThe equation is $ \\sqrt[3]{u^3\\minus{}12x^2\\plus{}30x\\minus{}8}\\equal{}2x\\plus{}u$\n\nRaising to power $ 3$ gives : $ u^3\\minus{}12x^2\\plus{}30x\\minus{}8\\equal{}8x^3\\plus{}12x^2u\\plus{}6xu^2\\plus{}u^3$\n\n$ \\iff$ $ u^3\\minus{}12x^2\\plus{}30x\\minus{}8\\equal{}8u^3\\plus{}24x\\plus{}12x^2u\\plus{}6xu^2\\plus{}u^3$\n\n$ \\iff$ $ 4(u^3\\plus{}1)\\plus{}6x^2(u\\plus{}1)\\plus{}3x(u^2\\minus{}1)\\equal{}0$\n\n$ \\iff$ $ (u\\plus{}1)(4u^2\\minus{}4u\\plus{}4\\plus{}6x^2\\plus{}3ux\\minus{}3x)\\equal{}0$\n\n$ \\iff$ $ (u\\plus{}1)\\frac{87(8u\\plus{}3x\\minus{}4)^2 \\plus{}(87x\\minus{}12)^2 \\plus{}4032}{87\\cdot 16}\\equal{}0$\n\n$ \\iff$ $ u\\equal{}\\minus{}1$ (since the other factor is always $ >0$\n\n$ \\iff$ $ x^3\\minus{}3x\\plus{}1\\equal{}0$.\n\nIt's easy to see that this equation has 3 real roots all in $ (\\minus{}2,\\plus{}2)$ and so let $ x\\equal{}2\\cos v$ and so $ 8\\cos^3v\\minus{}6\\cos v\\equal{}\\minus{}1$\n\n$ \\iff$ $ \\cos 3v\\equal{}\\minus{}\\frac 12 \\equal{}\\cos \\frac{4\\pi}3$\n\nHence the result[/hide]"
}
{
  "Tag": [
    "MIT",
    "college",
    "Support"
  ],
  "Problem": "Can anyone tell me how much money cost one year at MIT and is there some scholarships?",
  "Solution_1": "[url]http://web.mit.edu/sfs/[/url]\r\nHope this helps. :)",
  "Solution_2": "Yeah,thanks.",
  "Solution_3": "MIT sure is an 'expensive' destination for an international student.",
  "Solution_4": "[quote=\"Gen8\"]MIT sure is an 'expensive' destination for an international student.[/quote]\r\nWell,there're scholarschips,right.",
  "Solution_5": "I am  sure that If you get  outstanding academic achivements,your government will support you with a large fund,quite enough for you to follow some higher degree in foreign countries.",
  "Solution_6": "I hope to have outstanding results,but I'm still at the beginning.What exams should I take to enter an American university?I know SAT 1 is obligatory, but how many  SAT subjects are reqiured?My major is maths but what should the other subjects be? Are they optional or obligatory?\r\nWhat scores do I have to get to have to have chances for some grant or scholarships. I would appreciate any information.",
  "Solution_7": "[quote=\"barcelona\"]I hope to have outstanding results,but I'm still at the beginning.What exams should I take to enter an American university?I know SAT 1 is obligatory, but how many  SAT subjects are reqiured?My major is maths but what should the other subjects be? Are they optional or obligatory?\nWhat scores do I have to get to have to have chances for some grant or scholarships. I would appreciate any information.[/quote]\r\n\r\nmost schools require 2-3 subject tests and some have a requirement that you take Math II as one of them. What score should you get? As high as you can if you're applying to MIT. But an 800  on everything wont guarantee a scholarship. Scholarships are almost never based on standardized tests."
}
{
  "Tag": [
    "analytic geometry",
    "function"
  ],
  "Problem": "The point $ (a,b)$ is located in the third quadrant in the Cartesian plane.  The coordinates are changed according to the functions given: $ f(a)\\equal{}a^2 \\plus{} 3$ and $ g(b)\\equal{}b\\minus{}4$.  In which quadrant (1, 2, 3, or 4) is $ (f(a),g(b))$ located?",
  "Solution_1": "Third quadrant means (neg,neg). Say that it is (-1,-1). Using the functions, we would get (4,-5), which is in the fourth quadrant. Our answer is 4.",
  "Solution_2": "As Kiseki said, the third quadrant consists of the points that are (neg,neg).\r\n\r\nThe square of all negative numbers are positive numbers, and adding a positive number to a positive number always yields a positive number. \r\n\r\nAnd negative number subtracted by a positive numbers is a negative number.\r\n\r\nSo, we have the answer (pos,neg). This is represented by the $ \\boxed{4}$ the quadrant."
}
{
  "Tag": [
    "trigonometry",
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "u''(x)+c*u(x)=0 -differential equation.\r\n\r\nI was studying differential equations and when i reached the equations of grade 2 i encountered this equation.But it was saying''it is verified that'' and it just gave me the form of u(x).\r\nCould somebody please demonstrate the form of u(x)?",
  "Solution_1": "A solution of characteristic equation $\\lambda^{2}+c=0$ is $\\lambda=\\pm i\\sqrt{c}$.\r\nSo, a general solution to find is $y=C_{1}\\cos (\\sqrt{c}x)+C_{2}\\sin (\\sqrt{c}x)$.\r\n($C_{1}, C_{2}$ : Constant of Integration)",
  "Solution_2": "This is what i am trying to understand:why is y=C1cos(sqrt(c)*x)+....  ?...and i don't want to begin with the y and reach that y''+cy=0 .\r\nThere must be a way of reaching the general solution!",
  "Solution_3": "[quote=\"radac_mail\"]This is what i am trying to understand:why is y=C1cos(sqrt(c)*x)+....  ?...and i don't want to begin with the y and reach that y''+cy=0 .\nThere must be a way of reaching the general solution![/quote]\r\nIf it defines it as $z_{1}=\\cos \\sqrt{c}x+i \\sin \\sqrt{c}x, \\ z_{2}=\\cos \\sqrt{c}x-i \\sin \\sqrt{c}x$, \r\n$z_{1}'=-\\sqrt{c}\\sin \\sqrt{c}x+i \\sqrt{c}\\cos \\sqrt{c}x=i \\sqrt{c}(\\cos \\sqrt{c}x+i \\sin \\sqrt{c}x)=\\lambda_{1}z_{1}$\r\n$z_{1}''=\\lambda_{1}z_{1}'=\\lambda_{1}^{2}z_{1}$\r\nSo, $z_{1}''+c\\cdot z_{1}=(\\lambda_{1}^{2}+c)z_{1}=0$.\r\nLikewise, $z_{2}''+c\\cdot z_{2}=(\\lambda_{2}^{2}+c)z_{2}=0$\r\n$z_{1}, z_{2}$is a complex number value solution of $U''(x)+c\\cdot U(x)=0$.\r\nThese linear combination\r\n$U_{1}=\\frac{z_{1}+z_{2}}{2}=\\cos \\sqrt{c}x, \\ U_{2}=\\frac{z_{1}-z_{2}}{2i}=\\sin \\sqrt{c}x$\r\nis solutions of $U''(x)+c\\cdot U(x)=0$ again.\r\nTherefore, $U(x)=C_{1}\\cos (\\sqrt{c}x)+C_{2}\\sin (\\sqrt{c}x)$. \r\n($C_{1}, C_{2}$ : Constant of Integration)"
}
{
  "Tag": [],
  "Problem": "Se considera numerele reale pozitive $\\ a,b,c,x,y,z$, cu $\\ a>x,b>y,c>z$ si $\\ xyz=\\frac{1}{256^{2}}$.Aratati ca e adevarata inegalitatea $\\ (a+x^{2})(b+y^{2})(c+z^{2})\\geq\\frac{(a-x)(b-y)(c-z)}{(5x^{2}+a^{2}+9)^{4}(5y^{2}+b^{2}+9)^{4}(5z^{2}+c^{2}+9)^{4}}$",
  "Solution_1": "Nimeni nu se baga la inegalitatea asta? :(",
  "Solution_2": "ma io nu pricep o chestie...\r\n$a+x^{2}>a$.\r\nnumitorul membrului drept este enorm (mai mare decat $9^{12}$).\r\ndeci\r\nfacand cele mai directe aproximari, obtin\r\n$LHS>abc>(a-x)(b-y)(c-z)>9^{-12}(a-x)(b-y)(c-z)>RHS$.\r\nori am baut eu gaz, ori ceva nu`i bine p`acolo...",
  "Solution_3": "porneste de la faptul ca $\\frac{a+x^{2}}{a-x}=\\frac{x^{2}+x}{a-x}+1$...",
  "Solution_4": "io intrebam altceva\r\nce e gresit in rezolvarea mea de mai sus ?\r\ndaca rezolvarea mea e corecta, atunci inegalitatea e al dracului de slaba, si conditia cu xyz e in plus",
  "Solution_5": "Nu te-nfuria mai...cred ca e slaba.. :roll:",
  "Solution_6": "nu ma infurii\r\ndar am mai patit`o cu inegalitati sa dau o solutie \"corecta\" si sa ma prind dupaia ca de fapt nu e bine. \r\nd`asta cand vad semnu $\\geq$ ma gandesc ca trebuie sa fie un caz de egalitate, si daca mie imi da cu $>$ atunci ceva nu e bine..",
  "Solution_7": "Stii, unii propunatori mai pun si semnul asta $\\geq$ ca sa mai deruteze pe elevi..chiar daca normal ar trebui sa puna $\\ >$.Asta am vrut eu aici..pacalici. :)",
  "Solution_8": "Nu.O problema[sau parte din anuntul ei] nu  trebuie sa deruteze elevul..Si semnul $\\geq$ nu are niciun sens dak egalitatea nu are loc..DEsi logic vorbind inseamna\"mai mare[b] sau[/b] egal..parerea mea",
  "Solution_9": "Mai, eu asa am mai vazut.Inegalitati menite sa mai deruteze concurentul..intr-un fel, asta e lipsa de fair-play.. :)"
}
{
  "Tag": [
    "modular arithmetic",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "if   $\\frac{p^a-1}{p-1}=2^n$  that $p$ is a prime number and $n,a$ are positive integers.\r\nfind number of divisor of $n.a$",
  "Solution_1": "If a=2 then 2^n-1 is prime so the number of divisors of n is 2 since n must be prime. For a=2,3 its fairly easy to see there are no solutions. I don't know about the number of divisors in general.",
  "Solution_2": "1) Obviously $p>2$ else LHS would be odd\r\n2) We note that $a$ must be a power of $2$ in fact if it's not we would have let's say $d|a$ with $d$ an odd number $>1$. Then we will have $\\frac{p^d-1}{p-1} | \\frac{ p^a -1 } {p-1} = 2^n$ and so $p^{d-1} + p^{d-2} + ... + 1|2^n$. Thus $p^{d-1} + p^{d-2} + ... + 1$ must be a power of $2$ but it's a sum of an odd number of odd number and thus is odd and it's obviously $>1$.\r\n\r\n3) Thus $a=2^k$. Rewriting we get:\r\n\r\n$\\frac{p^{2^k} -1}{p-1} = (p^{2^{k-1}}+1)(p^{2^{k-2}} +1 )... (p+1) = 2^n$\r\n\r\nThus all the factors in LHS must be powers of $2$.\r\nBut $p^2+1 \\equiv 2 \\pmod{4}$ and so it can't be a power of $2$.\r\nThus $k=1$\r\n\r\n4) For $k=1$ and so $a=2$ we get $p= 2^n-1$ and so $n$ must be prime. So $an$ has always $4$ divisors"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AIME",
    "AMC 12",
    "AMC 10",
    "AMC 10 A"
  ],
  "Problem": "Will I get 1.5 or 2.5 points for not answering a question? \r\n\r\nBilly",
  "Solution_1": "2.5?\r\n\r\nIs this a trick question?",
  "Solution_2": "I think it's changing to 1.5 in 2007. Forgot where I read that, though, so I may be wrong.",
  "Solution_3": "[quote=\"bookaholic\"]I think it's changing to 1.5 in 2007. Forgot where I read that, though, so I may be wrong.[/quote]\r\n\r\nThat's correct. It was in the USAMO qualifier book as well as on the AMC site (I think).",
  "Solution_4": "[quote=\"bookaholic\"]I think it's changing to 1.5 in 2007. Forgot where I read that, though, so I may be wrong.[/quote]\r\n\r\nif they change it to 1.5, are they changing the minimum score?  :|",
  "Solution_5": "er, the minimum score is always going to be 0, pretty much?",
  "Solution_6": "I think he means minimum score to qualify, and no, as far as I know, they are not planning to. Why should they?",
  "Solution_7": "[quote=\"JesusFreak197\"]I think he means minimum score to qualify, and no, as far as I know, they are not planning to. Why should they?[/quote]\r\n\r\nErr.. maybe because most people will have a significantly lower score than before?",
  "Solution_8": "According to the 2005 Summary of AMC Results/Awards,\r\n\r\n[quote=\"AMC 12 Chair\"]Beginning in 2002 the number of points awarded for a blank response on the AMC12 was increased from 2 to 2.5. As a result, students could qualify for the AIME by solving as few as 11 problems. Analysis of student performances since then reveals that the average number of blank responses by AIME qualifiers has increased from about 8 to 11. This trend, suggesting that many bright students are simply igoring problems of even moderate difficulty, runs counter to our goals. For that reason, beginning in 2007, the reward for a blank response will decrease from 2.5 to 1.5 points. It should be emphasized that in making this change, we are not attempting to reduce the number of AIME qualifiers. Because we are simultaneously working to improve the overall success rate of the first ten problems, we expect that the number of students achieving the minimum qualifying score of 100 will remain roughly constant.[/quote]\r\n\r\nBTW I like this change, because 2 blank responses is almost 1 correct answer is... just not right.",
  "Solution_9": "Scoring on the 2006 AMC 10 A, 2006 AMC 10 B, 2006 AMC 12 A and the 2006 AMC 12 B will be 6 points for each correctly answered question, 2.5 points for each blank, or unanswered, question and 0 points for an incorrect answer.\r\n\r\nSee for instance item 4 on the facisimile of the 2006 AMC 10 A cover page which is on our website at:\r\nhttp://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/html/06tm-26-10Acover.html\r\n(or go to the AMC webstie, choose 2006 AMC 10A/ AMC 12 A Teacher's Manual from the What's New column, then using the table of contents, go to page 26).\r\n\r\nIn 2007, the scoring will change to 6 points for each correctly answered question, 1.5 points for each blank, or unanswered, question and 0 points for an incorrect answer.  This scoring change was announced in the 2005 AMC 101/12 Summary of Results and Awards, will be announced again in the 2006 AMC 10/12 Summary of Results and Awards and will be noted on the advertising brochures for the 2007 contests.  The scoring system will be noted on the cover of the 2007 AMC contests.  The implications of the scoring change will be noted in the 2007 AMC 10/AMC 12 Teachers' Manual.\r\n\r\nQualification for the 2007 AIME from the 2007 AMC 10 will continue to be a total of 120 points, or in the top 1% on the AMC 10.\r\nQualification for the 2007 AIME from the 2007 AMC 12 will continue to be a total of 100 points or in the top 5% on the AMC 12.\r\n\r\nSteve Dunbar\r\nAMC Director",
  "Solution_10": "[quote=\"probability1.01\"][quote=\"JesusFreak197\"]I think he means minimum score to qualify, and no, as far as I know, they are not planning to. Why should they?[/quote]\n\nErr.. maybe because most people will have a significantly lower score than before?[/quote]\r\n\r\nNot all that much. I'm pretty sure that part of the point is to eliminate those people who qualify for AIME by answering maybe 13 questions on the AMC12 (in case of one wrong) and leaving the rest blank. That shouldn't allow you to qualify. Now, you'll have to answer at least 14 correctly, probably more, and 19 on the AMC10.",
  "Solution_11": "Umm... well even if the average borderline qualifying person leaves 8 questions blank, that's still 8 whole points, which disqualifies a lot of people, at least 1000 from a cursory glance at last year's statistics. Obviously I'm ignoring some factors here, but it still strikes me as quite a lot.",
  "Solution_12": "But it seems that the AMC people will make the first 10 problems easier.",
  "Solution_13": "It gives 2.5 for unanswered Q is becoz it doesn't want us to GUESS\r\nbut....now has been changed to 1.5\r\nwut's that mean?",
  "Solution_14": "But i figured that it would be a little pointless to change the minimum qualification score because then the qualification would be about the same difficulty, and there's a smaller penalty for guessing.",
  "Solution_15": "Maybe this will decrease the inflating USAMO qualifying indexes, and place some more emphasis on AMC over AIME.  \r\n\r\n...on second thought, it probably will only have a small effect on that.",
  "Solution_16": "It shouldn't have much effect on the USAMO qualifying indices (although the USAMO expansion should).  This change is targeted at the borderline AIME qualifiers, who generally do not have much effect on USAMO cutoffs.  People on the borderline for USAMO probably aren't leaving very many questions blank, so they won't lose many points.  I suppose it could still lower the cutoff by a few points, though.",
  "Solution_17": "AMC over AIME? Not in this lifetime...  :P \r\n\r\nThe AIME is the coolest part, no doubt. Plus, if more people demonstrate that they can do better on the AMCs, I'm thinking they will have to make AIMEs harder. Uh-oh. There goes my hopes...",
  "Solution_18": "Doesn't matter much to me.\r\nIn the past, I've answered every question anyways.  Well, that was on the AMC10.  And, of course, I would make 2 or so minor errors or mistakes.",
  "Solution_19": "It would ruin the point of changing it from 2.5 to 1.5 if they also lowered the cutoff for AIME.",
  "Solution_20": "[quote=\"eryaman\"]Maybe this will decrease the inflating USAMO qualifying indexes, and place some more emphasis on AMC over AIME.  \n\n...on second thought, it probably will only have a small effect on that.[/quote]\r\n\r\n\r\nWell they will also try to make the AMC's easier(first 10 problems easier, which hmmm does not really help too many people) so yeah your score will stay roughly the same.\r\n\r\nBut this does hurt a lot if you are leaving like 5-10 blanks (so yeah try not to leave too many blanks that's the intention of the AMC people)",
  "Solution_21": "Really good change ... about time. It was ridiculous that 21 right and 4 blank got you almost the same score as 23 right. Generally there isn't a single challenging problem until #21 at the absolute earliest.",
  "Solution_22": "Yeah, it was quite ridiculous to qualify for AIME without getting a good number right.",
  "Solution_23": "Yeah 20 right, 5 blank gets you a higher score than 22 right... that's not fair also. Basically I don't see how the current AMC scores mean anything, like a difference of 0.5 point is a much bigger difference than a difference of 6 points.\r\n\r\nWhy can't they just make AMC short-answer format(like AIME) instead of multiple choice?",
  "Solution_24": "[quote=\"beta\"]Yeah 20 right, 5 blank gets you a higher score than 22 right... that's not fair also. Basically I don't see how the current AMC scores mean anything, like a difference of 0.5 point is a much bigger difference than a difference of 6 points.\n\nWhy can't they just make AMC short-answer format(like AIME) instead of multiple choice?[/quote]\r\n\r\nIt would scare away newbies.",
  "Solution_25": "I've noticed that on the AMC 10 booklet, it does not actually say that the cutoff is 120 points, just top 1%. So is 120 a generally accepted cutoff score to represent the top 1%, or is it actually part of the competition policy?",
  "Solution_26": "[quote=\"log0\"]I've noticed that on the AMC 10 booklet, it does not actually say that the cutoff is 120 points, just top 1%. So is 120 a generally accepted cutoff score to represent the top 1%, or is it actually part of the competition policy?[/quote]\r\n\r\nYou might have an old booklet. I believe the policy changed in 2004 to include any 120+ AMC10 scores as qualifying.",
  "Solution_27": "[quote=\"paladin8\"][quote=\"log0\"]I've noticed that on the AMC 10 booklet, it does not actually say that the cutoff is 120 points, just top 1%. So is 120 a generally accepted cutoff score to represent the top 1%, or is it actually part of the competition policy?[/quote]\n\nYou might have an old booklet. I believe the policy changed in 2004 to include any 120+ AMC10 scores as qualifying.[/quote]\r\n\r\nI noticed that when reading the instructions last year. So basically the baseline is either 120 or the top 1%, whichever is lower.",
  "Solution_28": "[quote=\"hockeyadi23\"][quote=\"paladin8\"][quote=\"log0\"]I've noticed that on the AMC 10 booklet, it does not actually say that the cutoff is 120 points, just top 1%. So is 120 a generally accepted cutoff score to represent the top 1%, or is it actually part of the competition policy?[/quote]\n\nYou might have an old booklet. I believe the policy changed in 2004 to include any 120+ AMC10 scores as qualifying.[/quote]\n\nI noticed that when reading the instructions last year. So basically the baseline is either 120 or the top 1%, whichever is lower.[/quote]\r\n\r\nYeah. It's more like the AMC12 qualifying now, only 20 points higher and 1% instead of 5%.",
  "Solution_29": "A kid from our school actually did answer only 12 questions. Checked them over several times and slept for the remaining time. Poor kid bubbled one in wrong...\r\n\r\nI think this change is really necessary in order to avoid people from spending time strategizing over the test. In that same time, the person, if preparing, could practice more math, or if it's during the actual test, could be working on math. I personally would prefer an SAT-type system (say, base score 25, +5 for correct answer, -1 for wrong answer)"
}
{
  "Tag": [
    "number theory",
    "relatively prime"
  ],
  "Problem": "1.) Prove that for any even $n>1$, $n^{2}-1$ divides $2^{n!}-1$. \r\n\r\n2.) Prove that for any $n>2$ the number $1989$ divides $n^{n^{n^{n}}}-n^{n^{n}}$.",
  "Solution_1": "What, how can 2) be true? 1989=9*13*17, am I missing something?",
  "Solution_2": "Hm these problems look familiar.\r\n\r\nOh wait I've seen these before.\r\n\r\nOh wait that's right I went to AMP :blush: \r\n\r\nYes the problem is right. Amazing isn't it :lol:",
  "Solution_3": "[hide=\"1\"] \\[n^{2}-1|2^{n!}-1 \\iff 2^{n!}\\equiv 1\\mod{(n^{2}-1)}\\]  \\[\\iff 2^{n!}\\equiv 1\\mod{n-1}, 1\\mod{n+1}\\] because n-1 and n+1 are relatively prime, being both odd.\n\nBut 2 is also relatively prime to both $n-1$ and $n+1$, and therefore is a unit in both mods.\nThus, $2^{n!}\\equiv 1\\mod{n-1}, \\mod{n+1}\\iff \\phi(n-1)|n!, \\phi(n+1)|n!$\n\nBut $\\phi(k)<k$ for any natural $k$.  Thus, $\\phi(n-1) \\le n-2, \\phi(n+1) \\le n$, and any number less than or equal to $n$ must divide $n!$.  Thus, \\[\\phi(n-1)|n! \\implies 2^{n!}\\equiv 1\\mod{n-1}\\]  \\[\\phi(n+1)|n! \\implies 2^{n!}\\equiv 1\\mod{n+1}\\]  \\[\\implies 2^{n!}\\equiv 1\\mod{n^{2}-1}\\implies n^{2}-1|2^{n!}+1\\] [/hide]",
  "Solution_4": "No seriously, can somebody please explain how 2) can possibly be true?",
  "Solution_5": "Why not?  Are you sure you aren't thinking the other way?\r\n\r\nYou have to show \\[n^{(n^{[n^{n}]})}\\equiv n^{(n^{n})}\\mod 9, \\mod 13, \\mod 17\\] Well, if you're looking at that congruence $\\mod{x}$, then you look at the first level of exponents $\\mod{\\phi(x)}$, and the second level $\\mod{\\phi(\\phi(x))}$.  And $\\phi(\\phi(9))=\\phi(6)=2$, $\\phi(\\phi(13))=\\phi(12)=4$, $\\phi(\\phi(17))=\\phi(16)=8$.\r\n\r\nThere's gotta be a better way then just checking all three mods with exponent reduction...",
  "Solution_6": "[quote=\"K81o7\"]\nThus, $2^{n!}\\equiv 1\\mod{n-1}, \\mod{n+1}\\iff \\phi(n-1)|n!, \\phi(n+1)|n!$\n[/quote]\r\n\r\nThat's not true! It's only an implication, not a equivalence! (but of course the implication has a good direction so the solution is correct)",
  "Solution_7": "[quote=\"TomciO\"][quote=\"K81o7\"]\nThus, $2^{n!}\\equiv 1\\mod{n-1}, \\mod{n+1}\\iff \\phi(n-1)|n!, \\phi(n+1)|n!$\n[/quote]\n\nThat's not true! It's only an implication, not a equivalence! (but of course the implication has a good direction so the solution is correct)[/quote]\r\n\r\nArgh...I made the same mistake the same day on a different problem...it should be $\\Leftarrow$..."
}
{
  "Tag": [
    "analytic geometry",
    "algorithm",
    "ceiling function",
    "modular arithmetic",
    "number theory",
    "Euclidean algorithm"
  ],
  "Problem": "Source: AMC-12 2000, problem 16\r\n\r\nA checkerboard of $13$ rows and $17$ columns has a number written in each squares, beginning in the upper left corner, so that the first row is numbered $1, 2, ..., 17$, the second row $18, 19, ..., 34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1, 2, ..., 13$, the second column, $14, 15, ..., 26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).\r\n\r\n$(A)\\text{ 222}$\r\n$(B)\\text{ 333}$\r\n$(C)\\text{ 444}$\r\n$(D)\\text{ 555}$\r\n$(E)\\text{ 666}$\r\n\r\n[I noticed that the AMC-12 2000 only has problems 1-11; I currently have problems 12-25, but is there a reason for the problem absences?]",
  "Solution_1": "[hide=\"my solution... if anyone cares\"]\n\nImagine the numbers on a coordinate plane.\n\n$17(13-y)+x=13x-(y-1)$\n\nNice diophantine.\n\n$-12x-16y+220=0$ (Using generalized Euclidean Algorithm)\n$x=5-4n$\n$y=10+3n$\n\nTest all $n$ such that $x,y \\ge 0$\n\n$111+56+166+221+1=\\boxed{555}$\n\nThis problem took me longer than I expected and I'm unhappy with my solution. I know there's some clever way (something to do with 222) to do this... anyone?\n[/hide]",
  "Solution_2": "[hide]We see that $111$ is in the center of the board under either system.  If $n$ is a solution, then $222-n$ is also a solution.  Then the sum of solutions is $111+222(\\text{number of solutions from 1 to 110})$.\n\nUnder the first system, $n$ is in row $\\left\\lceil\\frac{n}{17}\\right\\rceil$ and column $n\\pmod{13}$.  Under the second system, $n$ is in row $\\left\\lceil\\frac{n}{13}\\right\\rceil$ and column $n\\pmod{17}$.  Therefore, $\\left\\lceil\\frac{n}{17}\\right\\rceil\\equiv n\\pmod{13}$ and $\\left\\lceil\\frac{n}{13}\\right\\rceil\\equiv n\\pmod{17}$.\n\nWe immediately see that $1$ is a solution.  Once we find that $56$ is another solution, we stop.  The answer is D.[/hide]"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let a be real parameter. Find all a for which the equality $x^{2}+(1-2a)x+2a^{2}+lg(4x^{2}-(8a-1)x+5a^{2})=lg(x^{2}-2(a+1)x-a^{2})$ has only one solution.",
  "Solution_1": "[hide=\"hint\"]$[4x^{2}-(8a-1)x+5a^{2}]-[x^{2}-2(a+1)x-a^{2}]=3[x^{2}+(1-2a)x+2a^{2}].$[/hide]",
  "Solution_2": "that don`tt help me",
  "Solution_3": "[hide=\" 2nd Hint\"]That equation is equivalent to $f(4x^{2}-(8a-1)x+5a^{2})=f(x^{2}-2(a+1)x-a^{2})$, here $f(x)=\\lg{x}+\\frac{x}{3}$ is monotone on $(0,\\infty)$.[/hide]",
  "Solution_4": "[quote=\"Beat\"]Let a be real parameter. Find all a for which the equality $x^{2}+(1-2a)x+2a^{2}+lg(4x^{2}-(8a-1)x+5a^{2})=lg(x^{2}-2(a+1)x-a^{2})$ has only one solution.[/quote]\r\n$f(x)=lgx+3x$\r\n(V\u1eddi$x >0$)\r\nV\u00e0  $f(4x^{2}-(8a-1)x+5a^{2})=f(x^{2}-2(a+1)x-a^{2})$................\r\n....$\\to 3x^{2}-6ax+6a^{2}=0$\r\nx^2-2ax+2a^2=0\r\n........... a=0?????",
  "Solution_5": "[quote=\"VANCHANH123\"][quote=\"Beat\"]Let a be real parameter. Find all a for which the equality $x^{2}+(1-2a)x+2a^{2}+lg(4x^{2}-(8a-1)x+5a^{2})=lg(x^{2}-2(a+1)x-a^{2})$ has only one solution.[/quote]\n$f(x)=%Error. \"logx\" is a bad command.\n+3x$\n(V\u1eddi$x >0$)\nV\u00e0  $f(4x^{2}-(8a-1)x+5a^{2})=f(x^{2}-2(a+1)x-a^{2})$................\n....$\\to 3x^{2}-6ax+6a^{2}=0$\nx^2-2ax+2a^2=0\n........... a=0?????[/quote]\r\nnot ....\r\nno value a"
}
{
  "Tag": [
    "inequalities",
    "complex numbers",
    "inequalities proposed"
  ],
  "Problem": "The  triangle  ABC  has  A=30  and  AB=3/4 AC. Find  the  point  P  inside  the triangle  which  minimises  5PA+4PB+3PC\r\n\r\n(16 th  vietman  mo , 1978)  thanks!",
  "Solution_1": "Here's the best I could do up until now:\r\n\r\nWe take A to be the origin of the plane, B(3i), C(4w) and D(4) (the things in the brackets are complex numbers; w=cos60+i*sin60). We need to find a z s.t. the sum S=5|z|+4|z-3i|+3|z-4w| is minimised.\r\n\r\n5|z|=|4z-3iz|; 4|z-3i|=|12i-4z|; 3|z-4w|=|3iz-12iw|, so \r\nS=|4z-3iz|+|12i-4z|+|3iz-12iw|>=|4z-3iz+12i-4z+3iz-12iw|=|12i-12iw|=12, but I have no idea if this value is reached, or if it's reached for a z inside the triangle ABC. The condition is that the complex numbers z_1=4z-3iz, z_2=12i-4z and z_3=3iz-12iw are collinear and have the same direction: z_3/z_2 and z_1/z_2 are both positive reals. However, I'm a bit scared of the calculations. Can anyone check this? It could work.\r\n\r\nI was especially encouraged by the nice-looking result (12) :D It doesn't mean it's correct, though.."
}
{
  "Tag": [
    "Euler",
    "number theory",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Calculate: $ \\sum_{n\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^{n\\plus{}1}}{2n\\plus{}1}\\cdot\\frac{\\zeta (2n)}{2^{2n}}.$",
  "Solution_1": "There's an obvious thing to try, which is to insert the definition of $ \\zeta$ and interchange the order of summation. Doing that, we discover that is it the exact same problem as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218877]this.[/url]",
  "Solution_2": "$ \\sum_{n \\equal{} 1}^{\\infty}\\frac {( \\minus{} 1)^{n \\plus{} 1}}{2n \\plus{} 1}\\cdot\\frac {\\zeta (2n)}{2^{2n}} \\equal{} ?$\r\n\r\nThis is all I got so far.\r\n\r\n$ \\sum_{n \\equal{} 1}^{\\infty}\\frac {( \\minus{} 1)^{n \\plus{} 1}}{2n \\plus{} 1} \\equal{} 1 \\minus{} \\frac {\\pi}{4}$\r\n\r\n$ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{2^{2n}} \\equal{} \\frac {1}{3}$\r\n\r\n$ \\zeta (2n) \\equal{} \\frac {2^{2n \\minus{} 2}\\pi^{2n}|B_{2n}|}{2n!}$ This is some sort of Euler identity from my Intro to Number Theory book. I just can't figure out how to incorporate it into the summation. Maybe someone will be able to telescope some part of it? ^_^ Or not use the derived values for the other parts and be able to cancel stuff. (Just trying to add input >_<) Also $ B$ is a Bernoulli number.\r\n\r\n\r\nEdited post with regards to Carcul.\r\nP.S -> My apologies :)",
  "Solution_3": "Exactly how did you found that $ \\sum_{n \\equal{} 1}^{\\infty}\\frac {1}{2^{2^{n}}} \\equal{} \\frac {1}{3}$?",
  "Solution_4": "It's not, of course -- it was a typo for $ \\sum_{n \\geq 1} \\frac{1}{2^{2n}}$."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "analytic geometry",
    "trigonometry",
    "inequalities",
    "calculus",
    "function"
  ],
  "Problem": "Let  $ABC$ is a triangle with the circumcircle $(O)$ and $G$ is the centroid. $AG,BG, CG$ meet $(O)$ at $A', B', C'$ respectively. Prove that $S(A'B'C')\\geq\\ S(ABC)$ where S(ABC) means the area of the triangle ABC\r\n\r\nnice?",
  "Solution_1": "no one? I think it's very nice and not so hard",
  "Solution_2": "Notice that $3\\frac{S(GB'C')}{S(ABC)} = \\frac{S(G'B'C')}{S(GBC)} = \\frac{GB'.GC'}{GB.GC} = \\frac{(R^2-OG^2)^2}{GB^2.GC^2}$.\r\n\r\nAccording to Leibnitz's theorem, we have\r\n\r\n$3R^2-3OG^2 = GA^2+GB^2+GC^2$\r\n\r\nHence $3\\frac{S(A'B'C')}{S(ABC)} = \\sum3\\frac{S(GB'C')}{S(ABC)} = (GA^2+GB^2+GC^2)^2 \\sum \\frac{1}{GB^2.GC^2} \\geq 3$\r\n\r\nThus $S(A'B'C') \\geq S(ABC)$.",
  "Solution_3": "[quote=\"treegoner\"]Notice that $3\\frac{S(GB'C')}{S(ABC)} = \\frac{S(G'B'C')}{S(GBC)} = \\frac{GB'.GC'}{GB.GC} = \\frac{(R^2-OG^2)^2}{GB^2.GC^2}$.\n\nAccording to Leibnitz's theorem, we have\n\n$3R^2-3OG^2 = GA^2+GB^2+GC^2$\n\nHence $3\\frac{S(A'B'C')}{S(ABC)} = \\sum3\\frac{S(GB'C')}{S(ABC)} = (GA^2+GB^2+GC^2)^2 \\sum \\frac{1}{GB^2.GC^2} \\geq 3$\n\nThus $S(A'B'C') \\geq S(ABC)$.[/quote]\r\n\r\n  Your solution is very good . I have to use Kamarta ineq in my solution after proving $(A',B',C')>>(A,B,C)$",
  "Solution_4": "treegoner, can you prove that $r_{ABC}\\leq\\ r_{A'B'C'}$ (r means the inradii)",
  "Solution_5": "That is a very nice question. I did think about it after doing this problem. But up till now, I can't say anything. Are you sure of this result?",
  "Solution_6": "[quote=\"treegoner\"]That is a very nice question. I did think about it after doing this problem. But up till now, I can't say anything. Are you sure of this result?[/quote]\r\n\r\n  I have proved it for twice, but who know? maybe i have made mistakes for twice ???!!!",
  "Solution_7": "No one for the second one?",
  "Solution_8": "how it is demonstrated the Leibnitz's theorem ?  :blush:",
  "Solution_9": "I 'm glad that finally I got some free time this morning to think again about this problem. Here is my solution to the problem $r_{ABC} \\leq r_{A'B'C'}$.\r\n\r\nBut first, I need to answer js131090's question. Leibnizt's theorem says that if $(x: y: z)$ is the barycentric coordinate of a point $P$, then for every point $M$ we have the equality $xMA^2 + yMB^2 + zMC^2 = (x+y+z)MP^2 + xPA^2 + yPB^2 + zPC^2$. In particular, if $P$ is the centroid $G$ and $M$ is the circumcenter $O$, Leibnitz's theroem yields that $3R^2 = 3OG^2 + GA^2+GB^2+GC^2$.\r\n\r\nLet's get back to the problem. Let $A_1B_1C_1$ be the pedal triangle of $G$ with respect to $ABC$. Then according to sine theorem and Euler's formula for area of pedal triangle, we obtain the sides and area of $A_1B_1C_1$ are respectively $\\frac{2}{3}m_a.sinA, \\frac{2}{3}m_b.sinB, \\frac{2}{3}m_c.sinC, \\frac{4S(a^2+b^2+c^2)}{9R^2}(=\\frac{1}{4}S(1-\\frac{OG^2}{R^2}))$\r\n\r\nWe have $\\frac{B'C'}{BC} = \\frac{GC'}{GB} = \\frac{GC.GC'}{GB.GC} = \\frac{a^2+b^2+c^2}{4m_bm_c}$. Hence $B'C' = \\frac{(a^2+b^2+c^2)a}{4m_bm_c}$. Since triangles $A_1B_1C_1$ and $A'B'C'$ are similar, we obtain $\\frac{r_{A'B'C'}}{r_{A_1B_1C_1}} = \\frac{B'C'}{B_1C_1} = \\frac{3(a^2+b^2+c^2)R}{4m_am_bm_c}$. On the other hand, we know that $r_{A_1B_1C_1} = \\frac{S_{A_1B_1C_1}}{p_{A_1B_1C_1}} = \\frac{S(a^2+b^2+c^2)}{12(\\sum m_a.sinA)R^2}$. Hence $r_{A'B'C'} = \\frac{S(a^2+b^2+c^2)^2}{8m_am_bm_c(\\sum am_a)}$.\r\n\r\nWe need to show that $\\frac{S(a^2+b^2+c^2)^2}{8m_am_bm_c(\\sum am_a)} \\geq \\frac{S}{p}$, which is equivalent to $(a^2+b^2+c^2)^2p \\geq 8(\\sum am_a)m_am_bm_c$. By applying Tchebychev's inequality and AM-GM inequality, we obtain $\\sum am_a \\leq \\frac{2}{3}p(m_a+m_b+m_c)$ and $m_am_bm_c(m_a+m_b+m_c) \\leq \\frac{1}{3}(m_a^2+m_b^2+m_c)^2 = \\frac{3(a^2+b^2+c^2)^2}{16}$. Therefore, the problem is completely solved.",
  "Solution_10": "[quote=\"treegoner\"]I 'm glad that finally I got some free time this morning to think again about this problem. Here is my solution to the problem $r_{ABC} \\leq r_{A'B'C'}$.\n\nBut first, I need to answer js131090's question. Leibnizt's theorem says that if $(x: y: z)$ is the barycentric coordinate of a point $P$, then for every point $M$ we have the equality $xMA^2 + yMB^2 + zMC^2 = (x+y+z)MP^2 + xPA^2 + yPB^2 + zPC^2$. In particular, if $P$ is the centroid $G$ and $M$ is the circumcenter $O$, Leibnitz's theroem yields that $3R^2 = 3OG^2 + GA^2+GB^2+GC^2$.\n\nLet's get back to the problem. Let $A_1B_1C_1$ be the pedal triangle of $G$ with respect to $ABC$. Then according to sine theorem and Euler's formula for area of pedal triangle, we obtain the sides and area of $A_1B_1C_1$ are respectively $\\frac{2}{3}m_a.sinA, \\frac{2}{3}m_b.sinB, \\frac{2}{3}m_c.sinC, \\frac{4S(a^2+b^2+c^2)}{9R^2}(=\\frac{1}{4}S(1-\\frac{OG^2}{R^2}))$\n\nWe have $\\frac{B'C'}{BC} = \\frac{GC'}{GB} = \\frac{GC.GC'}{GB.GC} = \\frac{a^2+b^2+c^2}{4m_bm_c}$. Hence $B'C' = \\frac{(a^2+b^2+c^2)a}{4m_bm_c}$. Since triangles $A_1B_1C_1$ and $A'B'C'$ are similar, we obtain $\\frac{r_{A'B'C'}}{r_{A_1B_1C_1}} = \\frac{B'C'}{B_1C_1} = \\frac{3(a^2+b^2+c^2)R}{4m_am_bm_c}$. On the other hand, we know that $r_{A_1B_1C_1} = \\frac{S_{A_1B_1C_1}}{p_{A_1B_1C_1}} = \\frac{S(a^2+b^2+c^2)}{12(\\sum m_a.sinA)R^2}$. Hence $r_{A'B'C'} = \\frac{S(a^2+b^2+c^2)^2}{8m_am_bm_c(\\sum am_a)}$.\n\nWe need to show that $\\frac{S(a^2+b^2+c^2)^2}{8m_am_bm_c(\\sum am_a)} \\geq \\frac{S}{p}$, which is equivalent to $(a^2+b^2+c^2)^2p \\geq 8(\\sum am_a)m_am_bm_c$. By applying Tchebychev's inequality and AM-GM inequality, we obtain $\\sum am_a \\leq \\frac{2}{3}p(m_a+m_b+m_c)$ and $m_am_bm_c(m_a+m_b+m_c) \\leq \\frac{1}{3}(m_a^2+m_b^2+m_c)^2 = \\frac{3(a^2+b^2+c^2)^2}{16}$. Therefore, the problem is completely solved.[/quote]\r\n\r\nThe geometric problem is always should be solve in geometric way but i totally stupid in geometry. Here is my approach tatally calculus, which may not please treegoner:\r\n\r\n[b] lemma [b] $(A',B',C')>>(A,B,C) (rad)$.\n[b] proof [/b]. WLOG we can assume that $A\\geq\\ B\\geq\\ C$ which means $GA\\leq\\ GB\\leq\\ GC$. then by the cosin theorem we can proof that $A,B,C\\geq\\ max(A',B',C')$ and $A,B,C\\leq\\ min(A',B',C')$ . Then our lemma is proven.\r\n\r\nThis is the how  to apply our lemma:\r\nthe problem <=> $sin\\frac{A}{2}\\ sin\\frac{B}{2}\\ sin\\frac{C}{2}\\leq\\sin\\frac{A'}{2}\\ sin\\frac{B'}{2}\\ sin\\frac{C'}{2}$ <==> $\\sum\\ ln(sin\\frac{A}{2}\\ )\\leq\\sum\\ ln(sin\\frac{A'}{2}\\ )$. (2)\r\nlet consider the fucntion $f(x)=ln(sin(x))$ where $0\\leq\\ x\\leq\\pi\\ /2$. It is easy to see that $f''(x)<0$ Apply the KAMARATA ineq for the function $f(x)$ and 6 numbers $(A',B',C')>>(A,B,C)$ in (2) we have the result.\r\n\r\nhowever thank u treegoner."
}
{
  "Tag": [
    "function",
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose $ f,g : \\mathbb{R} \\to \\mathbb{C}$ are Borel. Prove that the function $ h(x,y)\\equal{}f(x\\minus{}y)g(y)$ is also Borel measurable.",
  "Solution_1": "Can you prove that $ \\tilde h(x,y)\\equal{}f(x)g(y)$ is Borel? Because $ h$ is $ \\tilde h$ composed with the homeomorphism $ (x,y)\\mapsto (x\\minus{}y,y)$."
}
{
  "Tag": [
    "logarithms",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "hallo, \r\nsolve the system:\r\n$2^{x}+3^{y}=5$ and $4^{x}+2^{y}=6$.  :)",
  "Solution_1": "Only x=y=1.\r\nLet $z=2^{x}$. We have $0<z<\\sqrt 6$ and $f(z)=\\ln{\\frac{5-z}{6-z^{2}}}=\\frac{\\ln 3}{\\ln 2}$. Because $f'(z)>0$ for $0<z<\\sqrt 6$, we have only one solution z=2."
}
{
  "Tag": [
    "induction",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be  finite group and $ \\mathcal{K}$ a conjugacy class of $ G$ that generates $ G$. Prove that the following two statements are equivalent: \r\n\r\n(1) There exists a positive integer $ m$ such that every element of $ G$ can be written as a product of $ m$ (not necessarily distinct) elements of $ \\mathcal{K}$. \r\n\r\n(2) $ G$ is equal to its own commutator subgroup. \r\n\r\n[i]J. Denes[/i]",
  "Solution_1": "From http://mathoverflow.net/questions/109590/the-powers-of-non-empty-subset-of-a-group-that-generate-a-subgroup/109652#109652 we get that $\\{e\\}\\not=N:=( \\mathcal{K} )^{|G|}\\lhd G$, so we can use Induction to group $G/N$ and it finishes solution."
}
{
  "Tag": [
    "search",
    "function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find infinitely many solutions of\r\n\r\n $(a) x^2+y^2+z^2=3xyz$\r\n\r\nDavron",
  "Solution_1": "What about search function\u00bf It's very easy to find!",
  "Solution_2": "please can you help me with that search function",
  "Solution_3": "Just search for \"x^2+y^2+z^2=3xyz\"... (isn't this the first one one tries\u00bf)"
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "I realize that this is in the teacher's manual, but somehow my school can't quite resolve this issue...\r\nSo\r\nWhen are we allowed to open the official solutions booklet for the AMC 12? Is it when the envelope with the answer sheets has been sealed, or is it when the mail gets picked up, or is it 24 hours?",
  "Solution_1": "I think it is the next day.",
  "Solution_2": "On the sealed Exan Manager's Envelope itself,  it says in large black block letters:\r\n\"Do not open until after the exam has been adminstered and all student answer forms have been mailed.\"\r\nIt also says in smaller print at the bottom:\r\n\"The exam may be discussed following its adminstration within the school, but not outside the school until the following day, since there will be schools in different time zones which have not yet given it.\"\r\nThe Teachers' Manual says basically the same thing, on page 5, second column, in Section II, 16. b.\r\n\r\n\r\nJust to forestall questions about the AIME:  The AIME Solutions pamphlet is NOT mailed with the AIME.  The AIME Solutions Pamphlet is mailed with the AIME RESULTS, not with the AIME itself.  That means your teacher/contest adminstrator does not have the official AIME answers until about 2 weeks after the AIME.\r\n\r\n\r\nSteve Dunbar\r\nAMC Director"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "Find 9 odd natural distinct numbers that satisfy\r\n1/(a1) + 1/(a2) + ... + 1/(a9) = 1\r\n\r\na1 means the first number\r\n...\r\na9 means the ninth number",
  "Solution_1": "[hide=\"Massive hint\"]Perfect numbers.[/hide]",
  "Solution_2": "sorry, but what does \"perfect number\" mean?",
  "Solution_3": "[quote=\"bibobeo\"]sorry, but what does \"perfect number\" mean?[/quote]\r\nA number who's factors (besides itself), when added, are equal to itself.\r\n\r\nThe first few are 6, 28, 496, 8128.",
  "Solution_4": "Well, then I don't think perfect numbers can be right.",
  "Solution_5": "http://en.wikipedia.org/wiki/Perfect_numbers\r\n\r\nRead the minor results section.",
  "Solution_6": "[quote=\"bibobeo\"]Find 9 odd natural distinct numbers that satisfy\n1/(a1) + 1/(a2) + ... + 1/(a9) = 1\n\na1 means the first number\n...\na9 means the ninth number[/quote]\r\n\r\n\r\n\r\nI wonder....\r\n\r\n[hide=\"idea?\"]$a_{2}a_{3}a_{4}a_{5}a_{6}a_{7}a_{8}a_{9}+a_{1}a_{3}a_{4}a_{5}a_{6}a_{7}a_{8}a_{9}+a_{1}a_{2}a_{4}a_{5}a_{6}a_{7}a_{8}a_{9}+a_{1}a_{2}a_{3}a_{5}a_{6}a_{7}a_{8}a_{9}+a_{1}a_{2}a_{3}a_{4}a_{6}a_{7}a_{8}a_{9}+a_{1}a_{2}a_{3}a_{4}a_{5}a_{7}a_{8}a_{9}+a_{1}a_{2}a_{3}a_{4}a_{5}a_{6}a_{8}a_{9}+a_{1}a_{2}a_{3}a_{4}a_{5}a_{6}a_{7}a_{9}+a_{1}a_{2}a_{3}a_{4}a_{5}a_{6}a_{7}a_{8}=a_{1}a_{2}a_{3}a_{4}a_{5}a_{6}a_{7}a_{8}a_{9}$\n\nFrom the original question, no number a_{1-9} can be 1.[/hide]",
  "Solution_7": "Oh, [b]odd.[/b]  This is a slightly different problem.  You might want to try induction instead.",
  "Solution_8": "I found one, but I'm not sure that it is unique.\r\n\r\n3   5   7   9   11   15   35   45   231",
  "Solution_9": "[hide]use the fact that $\\frac{1}{a(a+1)}+\\frac{1}{a+1}=\\frac{1}{a}$\nthen expand $\\frac{1}{1}=\\frac{1}{a}$\n[/hide]"
}
{
  "Tag": [
    "trigonometry",
    "LaTeX",
    "search",
    "function"
  ],
  "Problem": "Write each trig expression as an algebraic expression containing u and v.\r\n\r\n1) sin(sin^-1 (u) - cos^-1 (v))\r\n\r\n2) sec(tan^-1 (u) + cos^-1 (v))",
  "Solution_1": "hello, (1) is equals to $ v\\sqrt{1\\minus{}u^2}\\plus{}u\\sqrt{1\\minus{}v^2}$ and (2) $ \\frac{\\sqrt{1\\plus{}u^2}}{v\\minus{}u\\sqrt{1\\minus{}v^2}}$\r\nSonnhard.",
  "Solution_2": "I agree with the second one but, unless I've done something silly, for the first one I get:\r\n \r\nsin(a-b) = sin a cos b - cos a sin b\r\n\r\n=uv - sqrt(1 - u^2)sqrt(1 - v^2).\r\n\r\n(Sorry, I can't do LaTeX at the moment.)",
  "Solution_3": "$ \\sin (\\arcsin u \\minus{} \\arccos v)$\r\n$ \\equal{} \\sin \\left( \\frac{\\pi}{2} \\minus{} \\arccos u \\minus{} \\arccos v \\right)$\r\n$ \\equal{} \\cos (\\arccos u \\plus{} \\arccos v)$\r\n$ \\equal{} \\cos (\\arccos u) \\cos (\\arccos v) \\minus{} \\sin (\\arccos u) \\sin (\\arccos v)$\r\n$ \\equal{} \\boxed{uv \\minus{} \\sqrt {1 \\minus{} u^2} \\sqrt {1 \\minus{} v^2}}$",
  "Solution_4": "[quote=\"TZF\"]$ \\sin (\\arcsin u \\minus{} \\arccos v)$ (Equation 1)\n$ \\equal{} \\sin \\left( \\frac {\\pi}{2} \\minus{} \\arccos u \\minus{} \\arccos v \\right)$ (Equation 2)\n$ \\equal{} \\cos (\\arccos u \\plus{} \\arccos v)$ (Equation 3)\n$ \\equal{} \\cos (\\arccos u) \\cos (\\arccos v) \\minus{} \\sin (\\arccos u) \\sin (\\arccos v)$ (Equation 4)\n$ \\equal{} \\boxed{uv \\minus{} \\sqrt {1 \\minus{} u^2} \\sqrt {1 \\minus{} v^2}}$ (Equation 5)[/quote]\r\n\r\nI'm sorry, the only part I understand is when you got from Equation 3 to Equation 4. Can you explain all the other parts in more detail please. I'm still learning trig.  :) \r\n\r\nAlso, can somebody give a solution to number 2?",
  "Solution_5": "To help you, here is TZF's solution with the addition of \"Equation 1A\" and \"Equation 2A\":\r\n\r\n$ \\sin (\\arcsin u\\minus{}\\arccos v)$ (Equation 1)\r\n$ \\equal{}\\sin\\left(\\left(\\frac{\\pi}{2}\\minus{}\\arccos u \\right) \\minus{}\\arccos v\\right)$ (Equation 1A)\r\n$ \\equal{}\\sin\\left(\\frac{\\pi}{2}\\minus{}\\arccos u\\minus{}\\arccos v\\right)$ (Equation 2)\r\n$ \\equal{}\\sin\\left(\\frac{\\pi}{2}\\minus{} \\left( \\arccos u \\plus{} \\arccos v\\right) \\right)$ (Equation 2A)\r\n$ \\equal{}\\cos (\\arccos u\\plus{}\\arccos v)$ (Equation 3) \r\n$ \\equal{}\\cos (\\arccos u)\\cos (\\arccos v)\\minus{}\\sin (\\arccos u)\\sin (\\arccos v)$ (Equation 4)\r\n$ \\equal{}\\boxed{uv\\minus{}\\sqrt{1\\minus{}u^{2}}\\sqrt{1\\minus{}v^{2}}}$ (Equation 5)\r\n\r\nWe get from Equation 2A to Equation 3 by taking advantage of confunctions . The rule is $ \\sin \\left(\\frac{\\pi}{2} \\minus{} \\theta\\right) \\equal{} \\cos \\theta$ (and also $ \\cos \\left(\\frac{\\pi}{2} \\minus{} \\theta\\right) \\equal{} \\sin \\theta$). Search for \"cofunction identities\" for more information. \r\n\r\nSince sine and cosine are cofunctions, it follows that $ \\arcsin u \\plus{} \\arccos u \\equal{} \\frac{\\pi}{2}$. Therefore, $ \\arcsin u \\equal{} \\frac{\\pi}{2} \\minus{} \\arccos u$, which was used to get from Equation 1 to Equation 1A.\r\n\r\n[Note: You could solve problem (1) without using cofunctions, instead simply applying the angle difference formula for sine to the given expression. U\r\n\r\nTo get from Equation 4 to Equation 5, you can use the following idea, which I used to solve problem (2) below:\r\n\r\nLet $ \\arctan u \\equal{} \\theta$. Then $ \\tan \\theta \\equal{} u \\equal{} \\frac{u}{1}$. Draw the angle in standard position, draw the corresponding triangle, and label the sides. The opposite side is $ u$ and adjacent side is $ 1$. You can find that the hypotenuse is $ \\sqrt{1\\plus{}u^2}$ by the Pythagorean theorem. Now you can find any trig function of $ \\theta$ using the triangle you have drawn. For example, $ \\cos(\\arctan u) \\equal{} \\cos(\\theta) \\equal{} \\frac{1}{\\sqrt{1\\plus{}u^2}}$. Do the same thing for $ \\arccos v$.\r\n\r\nSolution to problem (2):\r\n\r\n$ \\sec(\\arctan u \\plus{} \\arccos v)$\r\n\r\n$ \\equal{} \\frac{1}{\\cos(\\arctan u \\plus{} \\arccos v)}$\r\n\r\n$ \\equal{} \\frac{1}{\\cos(\\arctan u)\\cos(\\arccos v) \\minus{} \\sin(\\arctan u)\\sin(\\arccos v)}$\r\n\r\n$ \\equal{} \\frac{1}{  \\left( \\frac{1}{\\sqrt{1\\plus{}u^2}} \\right) \\left( v \\right) \\minus{} \\left( \\frac{u}{\\sqrt{1\\plus{}u^2}} \\right) \\left( \\sqrt{1\\minus{}v^2} \\right) }$\r\n\r\n$ \\equal{} \\frac{1}{  \\left( \\frac{v}{\\sqrt{1\\plus{}u^2}} \\right) \\minus{} \\left( \\frac{u \\sqrt{1\\minus{}v^2}}{\\sqrt{1\\plus{}u^2}} \\right) }$\r\n\r\n$ \\equal{} \\frac{1}{ \\frac{v \\minus{} u \\sqrt{1\\minus{}v^2}}{\\sqrt{1\\plus{}u^2}} }$\r\n\r\n$ \\equal{} \\boxed{\\frac{\\sqrt{1\\plus{}u^2}}{v \\minus{} u \\sqrt{1\\minus{}v^2}}}$",
  "Solution_6": "Oh. I get it. Thanks a lot. That explains a lot.  :)"
}
{
  "Tag": [
    "vector",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "A triangle is formed by joining three points $ A,B$ and $ C$ where $ \\overrightarrow a,\\overrightarrow b$ and $ \\overrightarrow c$ are  the position vectors respectively.\r\n\r\nIf $ P(\\overrightarrow r)$ be a point inside the triangle,then can $ \\overrightarrow r$ be expressed as\r\n \r\n$ \\overrightarrow r\\equal{}x\\overrightarrow a\\plus{}y\\overrightarrow b\\plus{}z\\overrightarrow c$\r\n\r\nwhere $ x,y,z$ are scalars such that $ x\\plus{}y\\plus{}z\\equal{}1,x>0,y>0,z>0$.\r\n\r\nI am not able to generate a proof using vector methods.But first of all I would want to know whether above statement is correct or not.",
  "Solution_1": "The statement is true. $ P(\\overrightarrow r)$be inside the triangle $ ABC$ except boundary.  Use $ z\\equal{}1\\minus{}x\\minus{}y$ to prove."
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "[u][i]Problem[/i][/u]\r\n\r\n   [i] Find all real numbers  $ x$  such that the following inequality holds for all non negative \n          numbers $ a , b , c$\n\n $ [  a^{2} \\plus{} b^{2} \\plus{} (x\\minus{}1)c^{2} ] [ a^{2} \\plus{} c^{2} \\plus{} (x\\minus{}1)b^{2} ] [ b^{2} \\plus{} c^{2} \\plus{} (x\\minus{}1)a^{2} ]$\n\n $ \\leq (a^{2} \\plus{} xbc)(b^{2} \\plus{} xac)( c^{2} \\plus{} xab)$[/i]\r\n\r\n\r\n\r\n    [i]  My proposed problem[/i]\r\n\r\n[i] I hope it would be nice problem[/i]\r\n\r\n\r\n[i]  Scorpius119 , Nayel, ....Let perform [/i]\r\n\r\n\r\n\r\n\r\n [i]  Love An forever :love: \n Hu\u1ef3nh V\u00f5 Ph\u01b0\u01a1ng An[/i]",
  "Solution_1": "If we take $ c\\to 0$ we will get:\r\n$ (a^2 \\plus{} b^2)(a^2 \\plus{} (x \\minus{} 1)b^2)(b^2 \\plus{} (x \\minus{} 1)a^2)\\leq a^3 b^3 x$\r\nNow put here $ a \\equal{} b$ : \r\n$ 2a^6 x^2\\leq a^6 x$ or \r\n$ x\\left(x \\minus{} \\frac {1}{2}\\right)\\leq 0.$\r\nThis means that $ x \\in (0,\\frac {1}{2})$\r\nMaybe this helps... :)",
  "Solution_2": "for the post above x>0.\r\n$ c \\to 0 \\text{ and } b\\equal{}ka, a,k \\in R^\\plus{}$\r\nso we must x such that for all k we have:\r\n$ f(k)\\equal{}(k^2x^2)\\plus{}x(1\\minus{}k^4\\minus{}k^2\\minus{}k\\minus{}\\frac{k^3}{k^2\\plus{}1})\\plus{}(k^2\\plus{}k\\minus{}1\\minus{}k^3) \\le 0$\r\n$ \\implies f(k)\\equal{}(k^6\\minus{}2k^3\\minus{}k\\plus{}1)x\\plus{}k^2(k^2\\plus{}1)x^2\\minus{}(k^2\\plus{}1)(k\\plus{}1)(k\\minus{}1)^2 \\le 0$ but $ lim_{k \\to \\infty}\\equal{}\\plus{} \\infty$, contradiction.",
  "Solution_3": "Are you shure?\r\n$ \\frac {1}{k^2}x^2 \\plus{} x\\left(\\frac {1}{k^4} \\minus{} 1 \\minus{} \\frac {1}{k^2} \\minus{} \\frac {1}{k^3} \\minus{} \\frac {k}{k^2 \\plus{} 1}\\right) \\plus{} \\left(\\frac {1}{k^2} \\plus{} \\frac {1}{k^3} \\minus{} \\frac {1}{k^4} \\minus{} \\frac {1}{k}\\right)\\leq 0$\r\nIf we take $ k\\to \\infty$ we will have : $ \\minus{} x\\leq 0$. This is true if $ x\\geq 0$",
  "Solution_4": "[quote=\"Yuriy Solovyov\"]Are you shure?\n$ \\frac {1}{k^2}x^2 \\plus{} x\\left(\\frac {1}{k^4} \\minus{} 1 \\minus{} \\frac {1}{k^2} \\minus{} \\frac {1}{k^3} \\minus{}\\frac {1}{k(k^2 \\plus{} 1)} \\right) \\plus{} \\left(\\frac {1}{k^2} \\plus{} \\frac {1}{k^3} \\minus{} \\frac {1}{k^4} \\minus{} \\frac {1}{k}\\right)\\leq 0$\nIf we take $ k\\to \\infty$ we will have : $ \\minus{} x\\leq 0$. This is true if $ x\\geq 0$[/quote]\r\n\r\nhowever you're right, i'am wrong in calculos, in fact:\r\n$ p(x) \\equal{} x^2(k^2(k^2 \\plus{} 1)) \\plus{} x((k^2 \\minus{} 1)^2(k^2 \\plus{} 1) \\minus{} k^3) \\minus{} ((k^2 \\minus{} 1)^2(k^2 \\plus{} 1)) \\le 0$\r\nand we note $ x_1 \\le 0 < x_2 \\forall k$ and $ min(x_2) \\equal{} p(1) \\equal{} \\frac {1}{2}$\r\n\r\n(i've calcolated min{f (x) = ((x3-(((x2)-1)2)(x2+1))+sqrt((x3-(((x2)-1)2)(x2+1))2+4(x2)(x2+1)(x2+1)(x2-1)(x2-1)))/(2(x2)(x2+1))} with http://wims.unice.fr/wims/wims.cgi )\r\n\r\nin conclusion $ \\forall x \\in (0,\\frac {1}{2})$\r\n :roll:"
}
{
  "Tag": [
    "real analysis",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Prove that if $\\mathbb{R}^{k}\\subset \\bigcup_{1}^{\\infty}F_{n}$ where each $F_{n}$ is a closed subset of $\\mathbb{R}^{k}$, then at least one $F_{n}$ has a nonempty interior. \r\n\r\n(using only ideas developed up to chapter 2 in Rudin)",
  "Solution_1": "Don't double post, please. This post is nearly identical to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146107[/url]."
}
{
  "Tag": [],
  "Problem": "The more you have of it, the less you see. What is it?\r\n\r\ni have made a riddles thing. like it feel free to post riddles and ANSWERS!!!",
  "Solution_1": "it has several answers\r\nCataract,myopia,hypermetropia,................ :rotfl:  :rotfl: \r\n\r\nor\r\nKNOWLEDGE :rotfl:",
  "Solution_2": "hmmmm knowledge is a good answer, but it's actually darkness. good job though.",
  "Solution_3": "eye diseases r better option",
  "Solution_4": "riddles concerned with the eye onl;y?",
  "Solution_5": "next riddle i am posting.\r\nwat's in the middle of nowhere? common riddle.....easy",
  "Solution_6": "now[b]h[/b]ere\r\n\r\nanswer is [b]h[/b]  :rotfl:",
  "Solution_7": "very smart :rotfl:  :rotfl:  :rotfl:",
  "Solution_8": "INFINITY :rotfl:  :rotfl:",
  "Solution_9": "good job ur good\r\nwhat goes up an never comes down? this one shoud have more than one answer but it says it only has one.",
  "Solution_10": "a satellite launched with a speed greater than escape velocity :rotfl:  :rotfl:  :rotfl:",
  "Solution_11": "age :rotfl:",
  "Solution_12": "now hav a look at this one wat is such that it never degrades with respect to the human body?\r\n(ans=*****)",
  "Solution_13": "r u mad or wat.... :mad:  :o \r\nu kno i don't mind using slang :P",
  "Solution_14": "hey its not a slang word?",
  "Solution_15": "AGE! good job!!!!!!",
  "Solution_16": "and can u prove my ans wrong?? :rotfl:  :rotfl:",
  "Solution_17": "@rt and ritu,i am unable to understand you both",
  "Solution_18": "even i am not able to understand it................. :rotfl:  :rotfl:  :rotfl:",
  "Solution_19": ":rotfl: is telling a different story altogether\r\n\r\ncommon you two can tell it to me :P",
  "Solution_20": "tell this one\r\n\r\nin a boxing match, non of the men threw a punch ,yet there was a winner.How's so\r\n??????",
  "Solution_21": "what is the need of [b]throwing[/b] a punch :rotfl:  :rotfl:",
  "Solution_22": "ok\r\nnobody hit a punch.....\r\n\r\nbut throwing a punch means the same neways :lol:  :D",
  "Solution_23": "the other one gave up\r\n :rotfl:    :rotfl:  :rotfl:",
  "Solution_24": "without doing nething??????? :blush:",
  "Solution_25": "see thiss is a variation of a common ques (produced by myself) :P \r\n\r\nGeneral ans is \"it was a women's match so no men threw a punch\" :rotfl: \r\n\r\nBut i've twisted the ques  and it says \"None [b]of[/b] the men\" threw a punch that means men were there in the match!!! :P \r\n\r\nNow don't say the referee of a woman's match didn't threw the punch :P"
}
{
  "Tag": [
    "ARML",
    "probability"
  ],
  "Problem": "Here is the Weather Forecast for the Penn State site.\r\n[img]http://img98.imageshack.us/img98/7616/pennstateqb8.png[/img]\r\nSource: NOAA.gov, accessed Wed May 28, 9:30pm\r\n\r\nAssuming the Day lasts 12 hours and the Night lasts 12 hours, what is the probability it thunders for exactly 36 hours during ARML? (Showers != Thunder)",
  "Solution_1": "Well first of all, it can be any real number between 0 and 48...I don't know the techincal terms for this but wouldn't the probability be negligble? Also was your intention to put this in the ARML forum, because this isn't the ARML forum...",
  "Solution_2": "Possibly he was assuming that if it storms one night, it will storm the entire night.",
  "Solution_3": "Yeah it was, my bad. And yeah tjhance is right.",
  "Solution_4": "Are we supposed to solve that or is it just to inform us that it's likely going to rain?\r\n\r\nAnyways, answer below\r\n\r\n[hide]The probability for it raining 36 hours in a row is the probability that it rains for exactly three (in a row) of the five periods of day and night.\n\n20%*60%*60% + 60%*60%*50% + 60%*50%*40% = 37.2%[/hide]",
  "Solution_5": "That doesn't answer the question.  The question asks to find the probability that it thunders exactly 36 hours, not necessarily consecutively, during ARML.",
  "Solution_6": "So you're going to have $ \\binom{5}{2}$ terms to add up. That's just ugly.\r\n\r\nEdit: Errr... never mind.",
  "Solution_7": "The forecast for Las Vegas: Friday high 91/low 66, mainly sunny.  Saturday 94/69, sunshine.  Sunday 94/70, abundant sunshine.  What is the probability that it will ever rain in Vegas?  ;)",
  "Solution_8": "As far as I can remember, there's been a thunderstorm at Penn State every single Friday night I've gone to ARML.  (That's 7 years, I believe.)  Not that this has anything to do with the subject at hand ....",
  "Solution_9": "Not gonna happen at Las Vegas. Forecast now is for sunny both Friday and Saturday, high of 89 on Friday (a little cooler than normal), low of 70 Friday night, high of 95 on Saturday.",
  "Solution_10": "Prediction for Penn State: It rains all Friday Night and then is relatively dry and sunny Saturday. \r\nThere was absolutely no logical reasoning behind this prediction  :P but that's exactly what happened the last many years...",
  "Solution_11": "Chances have been upgraded to 70% / 70% / 60%  :("
}
{
  "Tag": [],
  "Problem": "Simplify ($ i^2 \\equal{} \\minus{} 1$):\r\n\\[ \\frac {5(3 \\plus{} i \\plus{} 2i^2)^2 \\minus{} (1 \\plus{} 2i \\plus{} 3i^2 \\plus{} 4i^3)^3 \\minus{} (i \\plus{} 2i^2)^4}{1 \\plus{} 3i \\minus{} 3i^2 \\minus{} i^3}\r\n\\]\r\n[hide]This question made me laugh when I saw it; I probably worked on it for half of the given time and still got it wrong.  :( \n\nAnd just now, I typed it into my graphing calculator and it took me half a minute to get the right answer.[/hide]",
  "Solution_1": "\\[\\frac{5(3+i+2i^{2})^{2}-(1+2i+3i^{2}+4i^{3})^{3}-(i+2i^{2})^{4}}{1+3i-3i^{2}-i^{3}}\\]\n\\[\\frac{5(i+1)^2-(-2i-2)^3-(i-2)^4}{4i+4}\\]\n\\[\\frac{5(i+1)^{\\cancel2}}{4\\cancel{(i+1)}}-\\frac{(\\cancelto{-2}{-8})(i+1)^{\\cancelto23}}{4\\cancel{(i+1)}}-\\frac{(i-2)^4}{4+4i}\\]\n\\[\\left(\\frac{5i}{4}+\\frac{5}{4}\\right)-\\left[-2(-1+2i+1)\\right]-\\frac{1+8i+24+32i+16}{4+4i}\\]\n\\[1.25i+1.25-\\left(-4i\\right)-\\frac{40i+41}{4+4i}\\]\n\\[\\&5.25i+1.25-\\left(\\frac{40i+41}{4+4i}\\right)\\left(\\frac{4-4i}{4-4i}\\right)\\]\n\\[5.25i+1.25-\\frac{160i+160+164-164i}{16+16}\\]\n\\[5.25i+1.25-\\frac{-4i+324}{32}\\]\n\\[5.25i+1.25-\\left(\\frac{-i}{8}+\\frac{81}{8}\\right)\\]\n\\[5.25i+1.25+0.125i-10.125\\]\n\\[\\boxed{5.675i-8.875}\\]\n\nI might have made a mistake.  If so, can someone point it out for me?  Thank you."
}
{
  "Tag": [
    "real analysis",
    "function",
    "real analysis unsolved"
  ],
  "Problem": "Let $\\mu$ be a measure that is defined on all Lebesgue measurable subsets of the real numbers. Suppose there is a function $u$ from $[0,\\infty)$ into $[0,\\infty)$ with 0 and 1 as fixed points, and the property that $\\mu((a,b)) = u(b-a)$ for $a < b$ in $\\mathbb{R}$. Is it true that $\\mu$ is the same as $\\lambda_{1}$?",
  "Solution_1": "it is a consequence of this: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=measure&t=114694",
  "Solution_2": "by \"measure\" i hope you mean a positive measure, don't you?\r\n\r\nif so, the function $u$ has to be monotone, and it has no atoms (since $u(1)=1<+\\infty$). so it's continuous.\r\nbut $\\mu$ is additive, so $u(c-a) = u(c-b)+u(b-a)$ whenever $0\\le a\\le b\\le c$.\r\nnow define $\\tilde{u}: \\mathbb{R}\\to \\mathbb{R}$ as follows: $\\tilde{u}(x) = u(x)$ for $x\\ge 0$, and $\\tilde{u}(x) =-u(-x)$ for $x<0$.\r\nit's easy to check that this function is still additive (this time on the whole real line, and not just on positive reals), and since it's continuous, and $\\tilde{u}(1)=1$, $\\tilde{u}$ is the identity, hence the measure is lebesgue measure."
}
{
  "Tag": [
    "function",
    "integration",
    "abstract algebra",
    "calculus",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "How can I show that if $ f \\in H^p$ then the harmonic function\r\n$ u_f(z) \\equal{} \\int_0^{2\\pi} P_z(t)|f(e^{it})|^p \\, \\frac {dt}{2\\pi}$\r\n($ P_z(t)$ is the Poisson kernel)\r\nis the least harmonic function that majorates $ |f|^p$.\r\n\r\nI can show that it does majorate (by writing $ f$ as a Blaschke product times a zero-free $ g \\in H^p$), but I have no idea about the least part.",
  "Solution_1": "Do you understand how to do it if $ f$ is smooth up to the boundary?",
  "Solution_2": "In this case I can write $ u_f$ as an integral over a circle of smaller radius (with an appropriate poisson kernel), right? So it would follow from the Poisson representation of a harmonic function that extends smoothly to the boundary of a circle.",
  "Solution_3": "Yes, that's the idea. Use smaller circles and then go to the limit pointwise.",
  "Solution_4": "Ok, thanks.  :blush:  I was convinced this problem was very hard, because it was stated without proof but also not as an exercise. Again the psychological effect of thinking something is easy/hard"
}
{
  "Tag": [],
  "Problem": "i do pop, mild rock, hip hop, maybe some rap if its good",
  "Solution_1": "Oldies through and through.",
  "Solution_2": "I voted for rock..\r\n\r\nYou missed tons of genres.. But even if there were 25 options, you would miss sth.. So I guess it`s ok..\r\n\r\nI like almost all genres.. But especially rock..\r\n\r\nMy favorite artists are: Pink Floyd, Radiohead, R.E.M., Air, Pearl Jam, A-Ha, Interpol, The Shins, Husker Du, Pixies, Suede, Vangelis, PJ Harvey, Belle & Sebastian, Bob Marley, Bad Religion, Erykah Badu...\r\n\r\nThese days I like to listen Goo Goo Dolls, Mike Oldfield, Charlatans UK, Grandaddy..",
  "Solution_3": "I listen to everything except Country.",
  "Solution_4": "I frequently find that some people miss the distinction between rock and rock'n'roll.  The latter, I prefer.",
  "Solution_5": "I love classical music - especially from the romantic period for those who care - but I'll listen to everything except rap, country, and polka. And I'll even listen to some country.\r\n\r\nBob Dylan is a literary genius, though. I tell you, there's nothing like singing Bob Dylan at the top of your lungs with the car windows down and having people stare at you as if you were a leper...\r\n\r\n\"But I was so much older then; I'm younger than that now\" has to be the most profound line in all of music.",
  "Solution_6": "Marching band music!",
  "Solution_7": "im into christian rock and punk rock\r\n\r\nswitchfoot and reliant k are two of my favorite bands",
  "Solution_8": "[quote=\"devilsbrother190\"]im into christian rock and punk rock\n\nswitchfoot and reliant k are two of my favorite bands[/quote]\r\n\r\nAh, switchfoot. Good stuff.",
  "Solution_9": "Oldies, i.e. Rock 'n Roll.  That means the Beatles, the Stones, the Bee Gees, Van Morrison, Kansas, Bob Seger, Jim Croce, Cat Stevens, Billy Joel, the Animals, BTO, the Beach Boys, Boston, CCR, Elvis, the Righteous Brothers, Queen, the Temptations, the Doobies, and so on.\r\n\r\nBut I like some modern rock (if that's the right genre) musicians - Counting Crows, DMB, Sting, and occasional songs by Matchbox 20, U2, etc.",
  "Solution_10": "[quote=\"mathfanatic\"]Oldies, i.e. Rock 'n Roll.  That means the Beatles, the Stones, the Bee Gees, Van Morrison, Kansas, Bob Seger, Jim Croce, Cat Stevens, Billy Joel, the Animals, BTO, the Beach Boys, Boston, CCR, Elvis, the Righteous Brothers, Queen, the Temptations, the Doobies, and so on.\n[/quote]\r\n\r\nNice to meet a fellow Oldies fan.",
  "Solution_11": "Everything (esp. oldies) except rap.",
  "Solution_12": "EMINEM! EMINEM!....\r\n\r\nBesides him...  Jimmy Eat World, Sum 41, Coheed and Cambria, Dashboard Confessional, Ben Folds Five, John Mayer, Funeral for a Friend, and some DMB...",
  "Solution_13": "I only listen to classical, especially Beethoven and Chopin.  I can't stand other stuff.\r\nNo angry replies, please.",
  "Solution_14": "um...shouldn't u add a poll for \"none\"? i don't listen to music.",
  "Solution_15": "I really enjoy listening to the Beatles. Their songs are really good! My favorite: \"The Long and Winding Road\"\r\n\r\nBut I like all kinds of music, even Country! Actually, now that I think about it, I do not like rap. I believe that rap is distatesful and should not be considered music. Of course, that's my opinion.....",
  "Solution_16": "so...aren't you going to add a \"none\" poll?",
  "Solution_17": "[quote=\"Danbert\"]I only listen to classical, especially Beethoven and Chopin.  I can't stand other stuff.\nNo angry replies, please.[/quote]\r\n\r\nROBINHOOD3000 ANGRY...only kidding, of course.\r\n\r\nWhat, no Lizst?",
  "Solution_18": "I like classical music and Weird Al music. I HATE rap and country. And I listen to any else. Yay classical is in second place. :)",
  "Solution_19": "[quote=\"A+MATH\"]I like classical music and Weird Al music.[/quote]\r\n\r\nThat's an interesting mix.\r\n\r\nAnd Robinhood3000, same to you.  Although I think we've already discussed this in the Oldies thread.",
  "Solution_20": "True enough, but it can't be said enough.\r\n\r\nThat, and I think my Oldies thread has all but vanished.",
  "Solution_21": "there's still no \"none\" option? grr...",
  "Solution_22": "I despise rap.\r\nJ-pop, J-rock ballads, instrumental jazz, classical, and opera.\r\nNot a real big fan of American pop music."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "incenter",
    "ratio",
    "geometric transformation",
    "reflection",
    "parallelogram"
  ],
  "Problem": "Dear Mathlinkers,\r\nlet ABC be a triangle, 1 the circumcircle of ABC, I the incentre of ABC, M a point on BC, \r\nand 2, 3 the Th\u00e9bault's circles wrt to M.\r\nProve : the radical axis of 2 and 3 passes through the midpoint of IM.\r\nAny synthetic proof for my the two last problems?\r\nSincerely\r\nJean-Louis",
  "Solution_1": "In $ AM$ cuts the circumcircle $ (O)$ again at $ M'.$ Let $ P$ be tangency point of the A-mixtilinear incircle with $ (O).$ Let $ AI$ cut $ (O)$ again at $ X$ and let $ (X)$ be a circle with center $ X$ and radius $ XB \\equal{} XC \\equal{} XI.$ Let the Thebault circles $ (2), (3)$ touch $ BC$ at $ U, V$ and $ (O)$ at $ E, F.$ Inversion in $ (X)$ takes $ BC$ to $ (O)$ and $ U, V$ to $ E, F,$ hence $ \\overline{XU} \\cdot \\overline{XE} \\equal{} XI^2 \\equal{} \\overline{XV} \\cdot \\overline{XF},$ $ UVFE$ is cyclic and moreover, $ X$ is on the radical axis of $ (2), (3)$ (well known). According to [url]http://www.mathlinks.ro/viewtopic.php?t=233986[/url], $ UVPM'$ is also cyclic. Radical axis $ UV$ of the circles $ \\odot(UVFE), \\odot(UVPM'),$ radical axis $ EF$ of the circles $ \\odot(UVFE), (O)$ and radical axis $ M'P$ of the circles $ \\odot(UVPM'), (O)$ meet at their radical center $ Q \\in BC.$ $ E, F$ are external similarity centers of the circle pairs $ (O), (2)$ and $ (O), (3),$ hence $ Q \\in BC$ is the external similarity center of $ (2), (3).$ By Thebault theorem, $ IQ$ is the common center line of the incircle $ (I)$ and the circles $ (2), (3).$ \r\n\r\nParallel to $ AM$ through $ I$ cuts $ BC$ at $ S.$ According to the proof of [url]http://www.mathlinks.ro/viewtopic.php?t=233986[/url], $ XS \\perp IQ.$ Let perpendicular to $ IQ$ through $ I$ cut $ AM$ at $ K.$ $ \\triangle AIK \\sim \\triangle IXS$ are centrally similar, having parallel sides. Let the incircle touch $ BC$ at $ D,$ let $ A'$ be midpoint of $ \\overline{BC},$ and let parallel $ k \\parallel BC$ through $ K$ and perpendicular $ p \\perp BC$ through $ I$ intersect at $ T.$ $ T, A'$ are corresponding points of the centrally similar $ \\triangle AIK \\sim \\triangle IXS$ with similarity coefficient  $ \\frac {\\overline{IT}}{\\overline{XA'}} \\equal{} \\frac {\\overline{AI}}{\\overline{IX}}.$ Let the angle bisector $ AI$ cut the side $ BC$ at $ X'$ and let $ (I_a)$ be the A-excircle of the $ \\triangle ABC$ centered at $ I_a.$ The double ratio $ \\frac {\\overline{AI}}{\\overline{AI_a}} \\equal{} \\minus{} \\frac {\\overline{X'I}}{\\overline{X'I_a}}$ is harmonic on account of $ A, X'$ being the external and internal similarity centers of $ (I), (I_a).$ In addition, $ X$ is the midpoint of $ \\overline{II_a},$ therefore $ \\frac {\\overline{AI}}{\\overline{IX}} \\equal{} \\minus{} \\frac {\\overline{IX'}}{\\overline{XX'}} \\equal{} \\minus{} \\frac {\\overline{ID}}{\\overline{XA'}}.$ As a result, $ \\overline{IT} \\equal{} \\minus{} \\overline{ID},$ so that the parallel line $ k \\equiv KT \\parallel BC$ is tangent to the incircle $ (I)$ at $ T.$ Since $ IK \\perp IQ$ by definition and $ I$ is on a midparallel $ l$ of $ k \\parallel BC,$ the line $ KQ$ is a reflection of both the incircle tangent $ BC$ in $ IQ$ and the incircle tangent $ KT$ in $ IK,$ i.e., a tangent of $ (I).$\r\n\r\nLet the midparallel $ l$ of $ k \\parallel BC$ cut $ AM$ at $ L.$ As $ IS \\parallel KL$ and $ \\overline{IS} \\equal{} \\overline{KL},$ $ KLSI$ is a parallelogram and $ SL \\parallel IK$ and $ SL \\perp IQ.$ As $ XS \\perp IQ$ as well and $ X$ lies on the radical axis of the circles $ (2), (3),$ $ XSL$ is their radical axis. But $ LMSI$ is a parallelogram, its diagonals $ SL, IM$ cutting each other in half.",
  "Solution_2": "Though it might be the same as Vladimir's, here is mine: [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=777557517&t=214426[/url] (read posts #2, #3, #5)."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "function",
    "calculus computations"
  ],
  "Problem": "Supose you have the integral $ \\int_0^\\pi\\sin(\\theta)\\,d\\theta$\r\nEveryone knows that the result of this integral is 2, but.. what if we tried a variable change like:\r\n$ y \\equal{} \\sin(\\theta)$\r\nfor $ \\theta\\equal{}0$, $ y\\equal{}0$\r\nfor $ \\theta\\equal{}\\pi$, $ y\\equal{}0$\r\nso you get an integral going from 0 to 0 which means the result is zero... where's the flaw?  :oops:  :?:  :ninja:  :huh:",
  "Solution_1": "You have to go all the way through with the substitution- what about the $ dy$ and $ d\\theta$? We have $ dy \\equal{} \\cos\\theta\\,d\\theta$, or $ d\\theta \\equal{} \\pm\\frac1{\\sqrt {1 \\minus{} y^2}}\\,dy$. That $ \\pm$ is trouble, since it changes in the middle of the integral. We could make it into $ \\int_0^1 \\frac {y}{\\sqrt {1 \\minus{} y^2}}\\,dy \\plus{} \\int_1^0 \\frac { \\minus{} y}{\\sqrt {1 \\minus{} y^2}}\\,dy \\equal{} 2\\int_0^1 \\frac {y}{\\sqrt {1 \\minus{} y^2}}\\,dy$, and that form is clearly nonzero.\r\n[Edit- corrected transformed integrals]",
  "Solution_2": "Recall what the substitution rule actually states:\r\nGiven a continuously differentiable function $ g(y)$ and a continuous function $ f(x)$ it is true that\r\n\\[ \\int_a^b{f(g(y))g'(y)dy = \\int_{g(a)}^{g(b)}{f(x)dx}\r\n}\\]\r\nNow in your case $ f(x) = \\sin (x)$, but what is your $ g$?  Well, you don't state it, but since you're formally making the substitution $ y = \\sin \\theta$, you're implicitly setting $ g(y) = \\arcsin y$.  The problem with this is that $ \\arcsin y = \\pi$ has no solution (recall the range of $ \\arcsin$ is $ \\left[-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right]$), so the substitution simply isn't valid.  Though, as jmerry suggested, there are tweaks you can do to make it work.  \r\n\r\nThe most important thing to get out of this is that the substitution rule is really a very precisely stated theorem -- and all theorems you encounter in mathematics are valid only if the stated conditions are met.  If you ignore the conditions and proceed heuristically, you'll inevitably stumble onto 'contradictions' like this one.",
  "Solution_3": "We can always force a substitution through with an inverse- if we're substituting something one-to-one. If we're not, the d_ factor doesn't work uniquely, and we have exactly this sort of problem.",
  "Solution_4": "Great answers everyone! thanks a lot.\r\nThis is why we, physicists, need mathematicians... to kick us back into doing things the proper way."
}
{
  "Tag": [],
  "Problem": "HI, \r\nI am solving \"Problem solving strategies \" by ARthur Engel.I am able to solve all the chapters , save one i.e. Extremal principle.The approaches seem to me very ingenious and hard to guess.Do you know any of the sites or material where this principle is explained in detail ? \r\nthanks \r\nkishore",
  "Solution_1": "i think i know what you are talking about.\r\nin ACoPS it is referred to as the extreme principle.\r\nbasically what you do is is arrange what you are looking at it some sort of order. then you look at the smallest or largest element. sometimes this is combined w/ proof by contridiction to prove that some statement is true or false."
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "i predict that alan from clarke will definitely make it to nats. David Fink also has a good chance, if he actually studies. Shen and Pejin also might make it. I predict that there is 1 random guy who nobody knows who gets to nats. for team, clarke 1st, diamond 2nd, acton ~7th :wink:",
  "Solution_1": "ugh, don't ask about art class today...\r\n\r\nyes, this is actually sorta relevant\r\n\r\nhint: shen (hao shen) is in my art class",
  "Solution_2": "[quote=\"illusiat\"]i predict that alan from clarke will definitely make it to nats. David Fink also has a good chance, if he actually studies. Shen and Pejin also might make it. I predict that there is 1 random guy who nobody knows who gets to nats. for team, clarke 1st, diamond 2nd, acton ~7th :wink:[/quote]\r\n\r\nNo. No predictions. They'll be wrong. They are useless. Especially a year beforehand. No. Bad.",
  "Solution_3": "Yeah, don't lower your already low self esteem.",
  "Solution_4": "He's in 8th grade, he will be in high school next year........I like how I am on the list, it makes me feel special. :lol: Art class? I am missing something?\r\n\r\nAnd you forgot about Jonathan Tidor from Clarke, that was sick this year, and all the incoming maybe beastly 5th graders.",
  "Solution_5": "Ok now you need to start lowering your ridiculously high self-esteem.",
  "Solution_6": "Can I predict 5 Clarke kids make State Countdown?  Or is that hoping?  I always get hoping and predicting mixed up :D",
  "Solution_7": "Yeah, what? Wasn't state like...9 days ago? wtf? But I'll be a hypocrite.\r\n\r\n2009 Results:\r\n\r\n1. Carl\r\n2. Carl\r\n3. Carl\r\n4. Alan Chou\r\n5. Samlmaml\r\n\r\nteam:\r\n\r\n1. Carl\r\n2. Clarke",
  "Solution_8": "Thanks cat nose. So I'm improving 2 places since last year? Well actually, since at least 2 of those carls would be DQed, I'd still make nats. But I'd probably have to go to mexico again like last year...(the reason i didn't go last year)",
  "Solution_9": "[quote]i predict that alan from clarke will definitely make it to nats. David Fink also has a good chance, if he actually studies. Shen and Pejin also might make it. I predict that there is 1 random guy who nobody knows who gets to nats. for team, clarke 1st, diamond 2nd, acton ~7th[/quote]\r\n\r\n\r\nwhy will acton get 7th next year???\r\nYou never know who will appear in RJ Grey next year......\r\n\r\n\r\nand\r\nmaybe I'll stay back in 8th grade ( I don't know why it is possible with the grades I have now... but...)\r\nand I might get first or second in state... yay!!!!   :clap: :clap:\r\n\r\nwait.... is mathcounts based on age or grade (I think it's grade)",
  "Solution_10": "Yea grade.\r\nBut it would be pointless to waste one year of your life just to participate in a Math Competition.",
  "Solution_11": "[quote=\"shentang\"]He's in 8th grade, he will be in high school next year........I like how I am on the list, it makes me feel special. :lol: Art class? I am missing something?\n\nAnd you forgot about Jonathan Tidor from Clarke, [b]who[/b] was sick this year, and all the incoming maybe beastly 5th graders.[/quote]\r\n\r\noh yeah, I forgot, it was during lunch...\r\n\r\nHao, did you just call Jonathan \"that\"?",
  "Solution_12": "Since I think that's enough of my team predicting how they'll do next year when they still haveother things to focus on(AIME, being first in the nation in SIGMA, breaking the record for most points in IMLEM, a little thing called MC nats.....) I'm going to lock this one up.  Wait until summer to start predicting your nonsense again :lol:"
}
{
  "Tag": [],
  "Problem": "what are the last three elements in the pattern A, 2, B, 0, C, 2, D, 0, E, 3, F, 3, G, 2, H, 4, I, J, 2, K, 4, L, 2, M, 2, N, 2, O, 0, P, 1, Q, 0, R, 2, S, 2, T, 3, U, 2, V, 2, W, 2, X, 4, Y, ?, ?, ?",
  "Solution_1": "Those are 3,Z,2",
  "Solution_2": "[quote=\"mr. math\"]what are the last three elements in the pattern A, 2, B, 0, C, 2, D, 0, E, 3, F, 3, G, 2, H, 4, I, J, 2, K, 4, L, 2, M, 2, N, 2, O, 0, P, 1, Q, 0, R, 2, S, 2, T, 3, U, 2, V, 2, W, 2, X, 4, Y, ?, ?, ?[/quote]is that supposed to be\r\n...H, 4, I, [b]4[/b], J, 2, K, 4...?",
  "Solution_3": "I don't get the pattern >.> ."
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "algebra unsolved"
  ],
  "Problem": "Let a,b,c,r,s be real numbers such that ar^2+br+c=0\r\n                                                        -as^2+bs+c=0\r\n  Prove that there exist u is real numbers such that (u-r)(u-s)<=0 and (a/2)u^2+bu+c=0",
  "Solution_1": "It is quite simple.  :) \r\n\r\n Note that $ar^2+br+c= 0$ and $-as^2+bs+c= 0$ implies that $a(r^2+s^2)= 0$, by substracting the two equations. Then, $a=0$ or $r^2+s^2=0$.\r\n If $a=0$, we have that $br+c=bs+c\\Rightarrow r=s$. Then, we have \r\n  \r\n $(u-r)(u-s)\\leq0\\Leftrightarrow(u-r)^2\\leq0\\Leftrightarrow (u-r)^2=0\\Leftrightarrow u=r=s$.\r\nSo, if $a=0$, we have that $bu+c=0\\Rightarrow u=r=s=-c/b$.\r\n\r\nNow, if $r^2+s^2=0$, it happens that $r=s=0$. So, we have that \r\n$(u-r)(u-s)\\leq0\\Leftrightarrow(u-0)^2\\leq0\\Leftrightarrow u^2\\leq0\\Leftrightarrow u=r=s=0$.\r\nThen, if a is a real number different form zero, u=0 satisfies all the conditions of the problem.",
  "Solution_2": "[quote=\"carlaperez_20\"] Note that $ar^2+br+c= 0$ and $-as^2+bs+c= 0$ implies that $a(r^2+s^2)= 0$, by substracting the two equations.[/quote]\r\n\r\nIf I'm subtracting correctly, we get $a(r^2+s^2)+b(r-s)=0$",
  "Solution_3": "agree with farenhajt that carlaperez_20's solution is false.Subtract 2 equation we get a(r^2+r^2)+b(r-s)=0 not \r\na(r^2+s^2)=0\r\n   carlaperez_20 can you check your solution???\r\nAnyone has the solution?? Please help me.",
  "Solution_4": "[quote=\"bui_haiha2000\"]...and (a/2)u^2+bu+c=0...[/quote]\r\nCheck it, please. I think the condition is false",
  "Solution_5": "It's easy to see that $r, s$ are the roots of the quadratic equation:     $ax^2+bx+c=0$\r\nBy Vi\u00e8te's formula:\r\n   $r+s=\\frac{-b}{a}$\r\n   $r.s=\\frac{c}{a}$\r\nWe need to prove that thre exists a real number u satisfying:\r\n   $(u-s)(u-r) \\le 0$\r\n $\\Longleftrightarrow u^2-(r+s)u+rs \\le 0$\r\n $\\Longleftrightarrow u^2+\\frac{b}{a}.u+\\frac{c}{a} \\le 0$\r\n $\\Longleftrightarrow au^2+bu+c \\le 0$\r\nThis is true because $\\Delta \\ge 0$ (see the function's graph).\r\nThen consider: \r\n $\\frac{a}{2}u^2+bu+c=0$\r\n $\\Longleftrightarrow \\frac{a}{2}u^2=au^2+bu+c$\r\n  */if $a>0$\r\n    $\\frac{a}{2}u^2=au^2+bu+c \\le 0$\r\n    $\\Longleftrightarrow u^2 \\le 0$ \r\n    $\\Longleftrightarrow u=0$\r\n  */if $a<0$\r\n    similarly like above...:$u=0$\r\n  This means $r \\le 0 \\le s$, which is equivalent to $rs=\\frac{c}{a}<0$: unsure (not appear in the problems' condition)\r\n :?:  :!:  :?:"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "Calculate \r\n\\[\\int x\\cos (1+x^2) dx\\]",
  "Solution_1": "$\\int x\\cos(1+x^2)dx = \\frac{1}{2} \\int \\cos(1+x^2)d(1+x^2) = \\frac{1}{2}\\sin (1+x^2)+c$.",
  "Solution_2": "That's right! :)",
  "Solution_3": "[quote=\"kunny\"]Calculate\n\\[\\int x\\cos (1+x^{2}) dx \\]\n[/quote]\r\nSet $u=1+x^{2}\\Longleftrightarrow du=2xdx$, which yields:\r\n\r\n\\begin{eqnarray*}\\int{x\\cos (1+x^{2})dx}&=& \\frac{1}{2}\\int{\\cos udu}\\hfill \\\\ &=& \\frac{1}{2}\\sin (1+x^{2})+c \\hfill \\\\ \\end{eqnarray*}"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "For any three positive reals $a, b, c$, we have\r\n\r\n$\\displaystyle \\frac{a^3+abc}{\\sqrt{b^2+c^2}}+\\frac{b^3+abc}{\\sqrt{c^2+a^2}}+\\frac{c^3+abc}{\\sqrt{a^2+b^2}}\\ge\\frac{\\sqrt{2}}{3}(\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca})^2$",
  "Solution_1": "WLOG assume $a \\ge b \\ge c$. By Chebysev's inequality,\r\n\\[\\displaystyle\\sum \\frac{a^3+abc}{\\sqrt{b^2+c^2}} \\ge \\frac{1}{3}(\\displaystyle\\sum a^3+abc)(\\displaystyle\\sum \\frac{1}{\\sqrt{b^2+c^2}}).\\]\r\nBy Schur's inequality,\r\n\\[\\displaystyle\\sum a^3+abc \\ge \\displaystyle\\sum a(b^2+c^2),\\]\r\nso together with Cauchy Schwarz's inequality,\r\n\\[\\frac{1}{3}(\\displaystyle\\sum a^3+abc)(\\displaystyle\\sum \\frac{1}{\\sqrt{b^2+c^2}}) \\ge \\frac{1}{3}(\\displaystyle\\sum a(b^2+c^2))(\\displaystyle\\sum \\frac{1}{\\sqrt{b^2+c^2}}) \\ge \\frac{1}{3}(\\displaystyle\\sum \\sqrt{a\\sqrt{b^2+c^2}})^2.\\]\r\nSince $\\sqrt{\\sqrt{b^2+c^2}} \\ge \\sqrt{\\sqrt{2}}\\frac{\\sqrt{b}+\\sqrt{c}}{2}$, so\r\n\\[\\displaystyle\\sum \\frac{a^3+abc}{\\sqrt{b^2+c^2}} \\ge \\frac{1}{3}(\\displaystyle\\sum \\sqrt{a\\sqrt{b^2+c^2}})^2 \\ge \\frac{1}{3}(\\displaystyle\\sum \\sqrt{a}(\\sqrt{\\sqrt{2}}\\frac{\\sqrt{b}+\\sqrt{c}}{2}))^2 =\\frac{\\sqrt{2}}{3} (\\displaystyle\\sum \\sqrt{ab})^2.\\]",
  "Solution_2": "Here is my solution:\r\n $\\frac{a^3+abc}{\\sqrt{b^2+c^2}}-\\sqrt{2}bc=\\sqrt{2} \\left(\\sqrt{\\frac{a^2}{b^2+c^2}\\frac{1}{2}}(a^2+bc)-bc\\right)\\geq \r\n\\sqrt{2}\\left( (a^2+bc) \\frac{2}{\\frac{1}{\\frac{a^2}{b^2+c^2}}+2}-bc\\right)=\\sqrt{2}\\left( \\frac{2a^4-bc(b^2+c^2)}{2a^2+b^2+c^2}\\right)\\geq \\sqrt{2}\\frac{1}{2}\\left(2a^2-b^2-c^2\\right)$\r\n\r\n \r\nSo, $\\frac{a^3+abc}{\\sqrt{b^2+c^2}}+\\frac{b^3+abc}{\\sqrt{a^2+c^2}}+\\frac{c^3+abc}{\\sqrt{a^2+b^2}}\\geq \\sqrt{2}\\left(ab+ac+bc\\right)\\geq \\frac{\\sqrt{2}}{3}\\left( \\sqrt{ab}+\\sqrt{ac}+\\sqrt{bc}\\right)^2$  :)"
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "I was having trouble sleeping last night and I tossed and turned well into the night. Our local town hall has a clock which strikes on the hour and also strikes just once on the half hour. During one of my more awake moments I heard the clock strike once, but I could not tell what time it was. Half an hour later it struck once again, but I still could not tell what time it was. Finally, half an hour later it struck once again and I knew what the time was. What time was it?",
  "Solution_1": "was it half past one ???",
  "Solution_2": "good job!\r\nyou got it right. :)"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "$ a,b,c>0$ Prove that !\r\n$ (a \\plus{} b)(b \\plus{} c)(a \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(a \\plus{} c \\minus{} b) \\le 8(abc)^2$",
  "Solution_1": "[quote=\"bigbang195\"]$ a,b,c > 0$ Prove that !\n$ (a \\plus{} b)(b \\plus{} c)(a \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(a \\plus{} c \\minus{} b) \\le 8(abc)^2$[/quote]\r\n\r\n\r\n$ \\Longleftrightarrow \\sum(a^3(b \\plus{} c)(a \\minus{} b)( a\\minus{}c)) \\plus{} (a \\minus{} b)^2(b \\minus{} c)^2(c \\minus{} a)^2 \\plus{}$ $ 4abc\\sum(a(a \\minus{} b)(a\\minus{}c))\\geq 0$\r\n\r\n\r\n :lol:"
}
{
  "Tag": [],
  "Problem": "Find the sum of even positive divisors of 10,000.",
  "Solution_1": "$ (2\\plus{}2^{2}\\plus{}2^{3}\\plus{}2^{4})(5^{0}\\plus{}5\\plus{}5^{2}\\plus{}5^{3}\\plus{}5^{4})$\r\n\r\nps: is this a well known formula or something, it looks beautiful",
  "Solution_2": "how do you get that",
  "Solution_3": "Well, $ 10^{4}\\equal{} (2^{4})(5^{4})$\r\n\r\nWe can have 0, 1, 2, 3, or 4 factors of 5 in such a divisor, and 1, 2, 3, or 4 factors of 2 (we must have at least 1 because the number should be even).\r\n\r\nSo, the product $ \\left(\\sum_{i \\equal{} 1}^{4}2^{i}\\right)\\left(\\sum_{j \\equal{} 0}^{4}5^{j}\\right) \\equal{} 23430$ gives the sum of all possible numbers $ 2^{i}5^{j},\\; i\\in\\{ 1,2,3,4\\},\\; j\\in\\{0,1,2,3,4\\}$. This is exactly the sum that Mohamed wrote."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "geometry",
    "graph theory",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Prove that bipartite graphs has no [url=http://mathworld.wolfram.com/IsospectralGraphs.html]isospectral[/url] graphs among connected graphs.",
  "Solution_1": "Let me add a few details: a graph on $n$ vertices $v_{1},\\dots,v_{n}$ can be represented by an $n\\times n$ matrix, in which the $(i,j)$-entry is the number of edges between $v_{i}$ and $v_{j}$. This [i]adjacency matrix[/i] depends on the enumeration of vertices, but its eigenvalues do not. Since the matrix is symmetric, it has $n$ eigenvalues (counting multiplicities), which form the spectrum of the graph. A natural (and very difficult) question is: which graphs are determined by the spectrum? \r\n\r\nHere we consider only [url=http://mathworld.wolfram.com/SimpleGraph.html]simple[/url] graphs (i.e. no multiple edges and no loops starting and ending at the same vertex). The adjacency matrix of such a graph has only $0$s and $1$s, with $0$s on the diagonal. For example, the complete graph on $3$ vertices (a triangle) has \r\n\\[A=\\begin{pmatrix}0& 1& 1\\\\ 1& 0 & 1 \\\\ 1& 1& 0 \\end{pmatrix}\\]\r\nLet $G$ be a [url=http://mathworld.wolfram.com/BipartiteGraph.html]bipartite graph[/url]. If we enumerate its edges so that $v_{1},\\dots,v_{m}$ are in the first part, and $v_{m+1},\\dots,v_{n}$ are in the second, then the adjacency matrix will look like this:\r\n\\[A_{G}=\\begin{pmatrix}0& C^{T}\\\\ C& 0 \\end{pmatrix}\\]\r\nwhere $C$ is a rectangular $m\\times(n-m)$ matrix. The claim is that there is no connected graph on $n$ vertices with the same spectrum as $G$. \r\n\r\nI don't have a solution, but can offer a simple observation: two $n\\times n$ matrices $A_{1}$ and $A_{2}$ have the same spectrum if and only if $A_{1}^{k}$ and $A_{2}^{k}$ have the same trace for all $k\\ge 1$ ($1\\le k\\le n$ is enough). One can try to interpret the trace of $A_{G}^{k}$ in terms of the geometry of $G$...",
  "Solution_2": "Thanks for your reply. Let me specialize the question for the moment. Prove that [i]complete[/i] bipartite graphs has no isospectral graphs among connected graphs.\r\n(PS I doubt the general case.)"
}
{
  "Tag": [
    "function",
    "algebra",
    "rational function",
    "algebra unsolved"
  ],
  "Problem": "This is a generalization of \r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=157768\r\nLet $ a,b,c$ be 3 reals s.t. $ a+b+c<0,a^{2}>bc,b^{2}>ac,c^{2}>ab$.\r\nWe seek 3 reals $ x,y,z$ s.t.: $ \\sqrt{yz}-x=a,\\sqrt{xz}-y=b,\\sqrt{xy}-z=c$.\r\n1) Find explicitly a solution which is a rational function of $ a,b,c$.\r\n2) Do there exist other solutions ?",
  "Solution_1": "Enough  to consider the equation:\r\n$ x^2\\minus{}yz\\equal{}a,y^2\\minus{}xz\\equal{}b,z^2\\minus{}xy\\equal{}c$\r\nWe has :\r\n$ (x^2\\minus{}yz)^2\\minus{}(y^2\\minus{}xz)(z^2\\minus{}xy)\\equal{}a^2\\minus{}bc$\r\nso $ x(x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz)\\equal{}a^2\\minus{}bc$\r\nCase 1:$ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz\\equal{}0$\r\nso $ x\\plus{}y\\plus{}z\\equal{}0$ or $ x\\equal{}y\\equal{}z$\r\n*)$ x\\plus{}y\\plus{}z\\equal{}0$ then\r\n$ a^2\\equal{}bc,c^2\\equal{}ba,a^2\\equal{}bc$\r\nSo we can find out $ (x,y,z)$\r\n*)$ x\\equal{}y\\equal{}z$ same method .\r\nCase 2:$ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz$ different from 0.\r\nSo $ \\frac{x}{y}\\equal{}\\frac{a^2\\minus{}bc}{b^2\\minus{}ca}$\r\n     $ \\frac{x}{z}\\equal{}\\frac{a^2\\minus{}bc}{c^2\\minus{}ab}$\r\nReplace to the first equation we can find out solution of this equations.",
  "Solution_2": "Hi TTsphn,\r\n1) you must also prove that, in my notations, necessarily $ x,y,z>0$.\r\nIn particular your case $ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz\\equal{}0$ is impossible.\r\n2) In my notations the requested solution is :\r\n$ x\\equal{}\\dfrac{(a^2\\minus{}bc)^2}{\\minus{}d},y\\equal{}\\cdots$ where $ d\\equal{}a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc\\equal{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}bc\\minus{}ca)$.\r\n3) You have proved also that there don't exist any other real solution.",
  "Solution_3": "I solve for any $ (a,b,c)\\in R$ not necessary your condition.\r\nSo can $ x^3\\plus{}y^3\\plus{}z^3\\minus{}3xyz\\equal{}0$"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "inequalities",
    "trigonometry"
  ],
  "Problem": "A convex quadrilateral $ ABCD$ with area $ 2002$ contains a point $ P$ in its interior such that $ PA \\equal{} 24$, $ PB \\equal{} 32$, $ PC \\equal{} 28$, and $ PD \\equal{} 45$.  FInd the perimeter of $ ABCD$.\r\n\r\n$ \\textbf{(A)}\\ 4\\sqrt {2002}\\qquad \\textbf{(B)}\\ 2\\sqrt {8465}\\qquad \\textbf{(C)}\\ 2\\left(48 \\plus{} \\sqrt {2002}\\right)$\r\n$ \\textbf{(D)}\\ 2\\sqrt {8633}\\qquad \\textbf{(E)}\\ 4\\left(36 \\plus{} \\sqrt {113}\\right)$",
  "Solution_1": "you have to get \"lucky\" with this question.\r\n\r\nas in noticing that 24*32+32*28+28*45+45*24=4004",
  "Solution_2": "[hide]Yea, I did it inelegantly too. I noticed that if I made the diagonals perpendicular, the area would be $ 2002$.\n\nThen, the perimeter is $ 40\\plus{}4\\sqrt{113}\\plus{}53\\plus{}51\\equal{}144\\plus{}4\\sqrt{113}\\equal{}4\\left(36\\plus{}\\sqrt{113}\\right) \\Rightarrow \\boxed{E}$.[/hide]",
  "Solution_3": "but seriously, honestly, if this is the \"best solution\", then i do not find this to be a very good problem for an AMC 12 #24 because its just pure luck that you come up with that.  Also, the solution, once you find it, is pretty trivial, just 1 step using pythagorean theorem.",
  "Solution_4": "Then again, if you actually took the 2002 AMC 12B, I would assume that you would memorize that $ 2002 \\equal{} 11 \\times 13 \\times 14$, so the solution wouldn't be so surprising (as in luck-wise). \r\n\r\nA complete solution is on the wiki: [[2002 AMC 12B Problems/Problem 24]].",
  "Solution_5": "should I add how they found the inequality by using the area formula 1/2 absin C to clear things up?",
  "Solution_6": "could someone elaborate on this? (the inequality found in the wiki solution)",
  "Solution_7": "[color=darkred][b][u]PP[/u].[/b] A convex quadrilateral $ ABCD$ with area $ 2002$ contains a point $ P$ in its interior such that \n$ PA \\equal{} 24\\ ,\\ PB \\equal{} 32\\ ,\\ PC \\equal{} 28$ and $ PD \\equal{} 45$ .  FInd the perimeter of $ ABCD$ .[/color]\n\n[b]Proof[/b] ([u][b]the future[/b][/u]). Denote the area $[ABCD]=S$ . Therefore, $2S=PA\\cdot PB\\cdot\\sin \\widehat {APB}+PB\\cdot PC\\cdot\\sin \\widehat {BPC}+$ \n\n$PC\\cdot PD\\cdot\\sin \\widehat {CPD}+PD\\cdot PA\\cdot\\sin \\widehat {DPA}\\implies$ $2S\\le PA\\cdot PB+PB\\cdot PC+PC\\cdot PD+PD\\cdot PA$ . \n\n[u]Observe that[/u] $2S=PA\\cdot PB+PB\\cdot PC+PC\\cdot PD+PD\\cdot PA$ . Thus, $\\sin \\widehat {APB}=\\sin \\widehat {BPC}=$\n\n$\\sin \\widehat {CPD}=\\sin \\widehat {DPA}=1$ , .e. $AC\\perp BD$ . In conclusion, the perimeter is $40+4\\sqrt {113}+53+51=$ $\\boxed{4\\left(36+\\sqrt {113}\\right)}$ .",
  "Solution_8": "[b]Solution:[/b] I always struggling on these $PA, PB, PC, PD$ problems but seriously what was this. So it is pretty clear that the diagonals multiply to a number greater than or equal to 4004. [hide=Explanation from AoPS Wiki] (Which is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$)[/hide] Furthermore, notice that the diagonals have to be less than 52 and 77 respectively. However, 52 * 77 = 4004. This implies that the diagonals are straight lines, perpendicular, and that $PAC$ and $PBD$ are degenerate triangles. Use Pythagorean theorem and we are done. I guess the main point of this problem was to realize that there is no way other than bounding the diagonals and then ultimately solving the question."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "What is the maximum number of non-overlapping $ 2\\times 1$ dominoes that can be placed on a $ 8\\times 9$ checkerboard if six of them are placed as shown? Each domino must be placed horizontally or vertically so as to cover two adjacent squares of the board.",
  "Solution_1": "We can place at most 34 dominoes.\r\n\r\nConsider two parts of the board: $ A$ consists of the squares above and to the left of the six dominoes and also the lower left corner, $ B$ consists of the squares below and to the right of the six dominoes and also the upper right corner.\r\n\r\nColor the squares in the usual checkerboard style, so that there are 14 black squares and 16 white squares in A, and there are 14 white squares and 16 black squares in B. Hence there are at most 14 dominoes in each part. Therefore there are at most $ 14\\plus{}14\\plus{}6\\equal{}34$. This is clearly achievable.",
  "Solution_2": "Let's prove that the answer is 34.\r\nWe color the $ 8x9$ checkerboard like a chessboard. Every domino is maked by a black square and a white square.\r\nThe square $ (a,b)$ is the intersection of the row $ a$ and column $ b$.\r\nLet divide the checkerboard into two parts as follow:\r\n\r\n\r\nWe can see that in the red part there are 17 black squares and 13 white squares.\r\nAlso we can see that in the blue part there are 13 black squares and 17 white squares.\r\nThe parts red and blue only have two intersections.\r\nThen when the maximun number of dominoes are placed, we have to place these two dominoes:\r\n1- (7,1)(8,1) and 2-(1,9)(2,9).\r\nAnd the answer is 28+6=34 because in the both parts the diference of the numbers of squares of diferents colors is at least 2.",
  "Solution_3": "Color the checkerboard in a chessboard fashion, and split it into sections, as shown to below to the left. Notice there are $14$ white squares in the orange section, and that it is not possible to place a domino that has a black square inside the orange region and white square outside the orange region. Since each domino covers a black and white square, there can be at most $14$ dominoes that cover squares in the orange region. Similarly, there are at most $14$  dominoes that cover squares in the blue region. There are $6$ dominos placed in the beginning, so we must have at most $14+14+6=34$ dominoes. A construction with $34$ dominoes is shown below to the left. $\\square$",
  "Solution_4": "We break the $8\\times 9$ grid into two sections: $A$ - including the rightmost column, and every square vertically below one of the placed dominoes and $B$ - including the leftmost column and every square which is vertically above one of the already placed dominoes. Then, we color the square in the usual chessboard style. Then, section $A$ has 14 black squares - so we can place almost 14 dominoes on it. Section $B$ has 14 white squares, so we can also place almost 14 dominoes on it. Thus, we cannot place more than $14+14+6=34$ dominoes on this board. Now, we shall give a construction which places 34 dominoes on such a $8\\times 9$ board.\n"
}
{
  "Tag": [
    "function",
    "induction",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all functions $ f: \\mathbb{Z} - \\{0\\}\\to\\mathbb{Q}$ which satisfy\n\\[f\\left(\\frac {x + y}3\\right) = \\frac {f(x) + f(y)}2,\\]\nfor all $ x,y\\in\\mathbb{Z} - \\{0\\}$ such that $ 3\\mid x + y$.",
  "Solution_1": "[quote=\"BaBaK Ghalebi\"]you are right pco,the problem is not correctly stated,the original problem is the following one:\n\n[b]find all functions $ f: \\mathbb{Z} \\minus{} \\{0\\}\\to\\mathbb{Q}$ which fulfill the following condition:\n\n$ f\\left(\\frac {x \\plus{} y}3\\right) \\equal{} \\frac {f(x) \\plus{} f(y)}2$[/b]\n\n\n\nso we must have $ 3\\mid x \\plus{} y$ and also $ x,y\\in\\mathbb{Z} \\minus{} \\{0\\}$ for example $ x \\equal{} 1,y \\equal{} 2$[/quote]\r\n\r\nLet $ P(x,y)$ be the property $ f\\left(\\frac {x \\plus{} y}3\\right) \\equal{} \\frac {f(x) \\plus{} f(y)}2$\r\n\r\n$ P(x,2x)$ implies $ f(2x)\\equal{}f(x)$ $ \\forall x\\neq 0$\r\n$ P(3x,3x)$ implies $ f(2x)\\equal{}f(3x)$ $ \\forall x\\neq 0$ and so $ f(3x)\\equal{}f(2x)\\equal{}f(x)$ $ \\forall x\\neq 0$\r\n\r\nAssume then $ f(nx)\\equal{}f(x)$ $ \\forall x\\neq 0$ for some integer $ n\\neq 0$. Then :\r\n\r\n$ P(3nx,3x)$ implies $ f((n\\plus{}1)x)\\equal{}\\frac{f(3nx)\\plus{}f(3x)}{2}$ and, since we know $ f(3x)\\equal{}f(x)$, $ f((n\\plus{}1)x)\\equal{}\\frac{f(nx)\\plus{}f(x)}{2}$, and, since $ f(nx)\\equal{}f(x)$, $ f((n\\plus{}1)x)\\equal{}f(x)$\r\nSo $ f(nx)\\equal{}f(x)$ $ \\forall x\\neq 0$ and $ \\forall n>0$\r\n\r\nThen $ P(6x,\\minus{}3x)$ gives $ f(x)\\equal{}\\frac{f(6x)\\plus{}f(\\minus{}3x)}{2}$ $ \\equal{}\\frac{f(x)\\plus{}f(\\minus{}x)}{2}$ and so $ f(\\minus{}x)\\equal{}f(x)$ $ \\forall x$\r\n\r\nAnd so $ f(nx)\\equal{}f(x)$ $ \\forall x\\neq 0$, $ \\forall n\\in\\mathbb{Z}^*$\r\n\r\nAnd so the solutions are $ f(x)\\equal{}c$ for any $ c\\in\\mathbb{Q}$ (and it is easy to check that these necessary conditions are sufficient).\r\n\r\nBut I don't understand the interest of $ \\mathbb{Z}^*$ instead of $ \\mathbb{Z}$ or the interest of $ \\mathbb{Q}$ instead of $ \\mathbb{R}$",
  "Solution_2": "[quote=\"pco\"]But I don't understand the interest $ \\mathbb{Z}^{*}$ of  instead of $ \\mathbb{Z}$ or the interest of $ \\mathbb{Q}$ instead of $ \\mathbb{R}$[/quote]\r\n\r\nthis problem was given in an iranian olympiad about 13 years ago,and the official solution uses induction to prove that $ f$ is constant,but as far as I can see,the official solution didn't use tha fact $ f: \\mathbb{Z}^{*}\\to\\mathbb{Q}$ either (instead of $ f: \\mathbb{Z}\\to\\mathbb{R}$)"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "absolute value",
    "algebra solved"
  ],
  "Problem": "Let P be a polynomial in R[x] satisfied For all x in the interval [0,1] we have |P(x)|<=1.\r\nProve that:|P(-1/n)|<=2 <sup>n+1</sup> -1.(n is degree of P)",
  "Solution_1": "Indeed, an easy, but beautiful problem. I hope Im not wrong with my computations, but here is what I think works: consider the numbers $ x_k=\\frac{k}{n} $ with $ k=0,1,,n $. Write the Lagrange interpolation formula for these points and take for x the value $ x=\\frac{-1}{n} $. Then, use the fact that $ |P(x_k)| $ is at most 1 and the triangle inequality. The absolute value of the terms that appear in the formula is $\\frac{(n+1)!}{(i+1)!(n-i)!} $ and the sum of these things is $ 2^{n+1}-1$. Is it correct?",
  "Solution_2": "Congratulation,your way is true! :D"
}
{
  "Tag": [
    "inequalities",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Consider the inequality\n\\[(a_1+a_2+\\dots+a_n)^2\\ge 4(a_1a_2+a_2a_3+\\cdots+a_na_1).\\]\na) Find all $n\\ge 3$ such that the inequality is true for positive reals.\nb) Find all $n\\ge 3$ such that the inequality is true for reals.",
  "Solution_1": "(1)$n\\geq4$,(2)$n=4$.",
  "Solution_2": "Its surely not that obvious. Does anyone no a full solution?",
  "Solution_3": "Hint: Use Abel SUmmation Formula for the RHS.",
  "Solution_4": "\\[(a_1+a_2+a_3+a_4)^2= 4(a_1a_2+a_2a_3+a_3a_4+a_4a_1)+(a_1-a_2+a_3-a_4)^2\\]\\[\\ge 4(a_1a_2+a_2a_3+a_3a_4+a_4a_1).\\]",
  "Solution_5": "(a) For $n = 3$, if we let $a_1 = a_2 = a_3 = 1$, then we get $9 \\geq 12$, which is clearly false.\n\nWe claim that the inequality is true for $n \\geq 4$. We use induction on $n$.\n\nFor $n = 4$, the inequality can be rearranged to get $(a_1 - a_2 + a_3 - a_4)^2 \\geq 0$, which is true for all reals $a_1, a_2, a_3, a_4$. \nIf the inequality is true for $n$, then we will show it is also true for $n + 1$. Note that since the inequality is cyclic, we can assume that $a_1$ is the largest among all $a_i$'s. Then $(a_1 + \\ldots + a_{n-1} + (a_n + a_{n+1}))^2 \\geq 4(a_1a_2 + \\ldots + a_{n-2}a_{n-1} + a_{n-1}(a_n + a_{n+1}) + (a_n + a_{n+1})a_1) = 4(a_1a_2 + \\ldots + a_na_{n+1} + a_{n+1}a_1) + 4(a_{n-1}a_{n+1} + a_n(a_1 - a_{n+1})) \\geq 4(a_1a_2 + \\ldots + a_na_{n+1} + a_{n+1}a_1),$ which completes the induction.\n\n(b) We have already shown that the inequality is false for $n = 3$ and true for $n = 4$. For $n \\geq 5$, let $a_1 = a_2 = 1, a_3 = 0, a_4 = -2, a_5 = \\ldots = a_n = 0$. Then we get $0 \\geq 4$, which is clearly false. So the inequality is only true for $n = 4$.\n"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "homothety",
    "geometry proposed"
  ],
  "Problem": "[color=blue]\n$ABC$ is a triangle and $B',C'$ are two points on its circumcircle so that $\\angle{C'AB}=\\angle{B'AC}$.The line trough $C'$ parallel to $AC$ intersects the line trough $B'$ parallel to $AB$ at $A'$.Prove that the points $A,A'$ and $A''$ are collinear,where $A''$ is the point of intersection of the lines $BC'$ and $CB'$.    [/color]",
  "Solution_1": "I think that the point $A'',$ must be defined as the intersection point of the segment lines $BB',$ $CC',$ instead of $BC',$ $CB'.$\r\n\r\nSo, we have that $\\angle C'CB = \\angle C'AB = \\angle B'AC = \\angle B'BC$ $\\Longrightarrow$ $B'C'\\parallel BC$ and now, we see that the sidelines of the triangles $\\bigtriangleup ABC,$ $\\bigtriangleup A'B'C',$ are parallel each other, one per one.\r\n\r\nHence, these triangles are perspective and so, we conclude that the segment lines $AA',$ $BB',$ $CC',$ are concurrent at one point. That is, the point $A''\\equiv BB'\\cap CC',$ lies on the segment line $AA'$ $($ also on the midperpendicular of the side segment $BC$ $)$ and the proof is completed.\r\n\r\nKostas Vittas.",
  "Solution_2": "Dear Vittas,actually there is a typo but it is not what you mentioned(it will be too easy that way :) ).\r\nThere should be $\\angle{C'BA}=\\angle{B'CA}$,insted of$\\angle{C'AB}=\\angle{B'AC}$    :roll: \r\n\r\nYou can also think of point $B',C'$ as two point so that $B'C'$ is perpendicular to the diametre trough the vertex $A$(probably this is a better way to define them because there again can be a confusion if both $B'$ and $C'$ lie on the midlle arc $BC$.)",
  "Solution_3": "Thank you dear Tiks for the clarification. Now is not so easy, but not difficult.\r\n\r\nWe denote as $B'',$ $C'',$ the intersection points of the circle $(O),$ from the segment lines $B'A'\\parallel AB,$ $C'A'\\parallel AC,$ respectively. Also we denote as $S,$ the intersection point of the segment lines $BB'',$ $CC''.$\r\n\r\nSo, we have that $\\angle B''CB = \\angle B'CA = \\angle C'BA = \\angle C''BC$ and hence, we have, as before, $B''C''\\parallel BC.$ \r\n\r\nBecause of the triangles $\\bigtriangleup ABC,$ $\\bigtriangleup A'B''C'',$ have their sidelines parallel each other, one per one, we conclude that these are perspective $($ also Homothetic $),$ with perpspector the point $S.$ So, the points $A,$ $A',$ $S,$ are collinear.\r\n\r\nFrom the non-convex inscribed hexagon $BC'C''CB'B'',$ by Pascal\u2019s theorem, we have that the points $A',$ $A'',$ $S,$ are collinear.\r\n\r\nHence the points $A,$ $A',$ $A'',$ are collinear and the proof is completed.\r\n\r\nKostas Vittas.",
  "Solution_4": "Congratulations with a very nice solution!!!!!!!!!!!!!!!!\r\n\r\nJust one more thing:from Dezarg's theorem we can see that this problem is equivalent to the fact that lines $BC,B'C'$ and $B_{1}C_{1}$ are concurent where $B_{1}\\in{{B'A'}\\cap{AC}}$ and $C_{1}\\in{{C'A'}\\cap{AB}}$.\r\n\r\nNow it is time to see what was mentioned in the description of this topic.\r\n\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55443[/url]\r\n\r\nTake the points $B'$ and $C'$ to be the same and you will get the problem from the link above.",
  "Solution_5": "Let $C''\\in BC'\\cap AC, B''\\in CB'\\cap AB, S\\in BC'\\cap A'B', T\\in CB'\\cap A'C'$.\r\n\r\nA little and easy angle-chasing shows that\r\n$C''AB\\sim C'A'S$\r\n$BAC\\sim SA'T$\r\n$CAB''\\sim TA'B'$\r\n$B''AC''\\sim B'A'C'$.\r\n\r\nThus, the quadrilaterals $C''BCB''$ and $C'STB'$ are also similar, having the corresponding sides parallel. Hence, there exist a homothety taking $C''BCB''$ to $C'STB'$ and thus $A$ to $A'$. But since $A''\\in C''C'\\cap B''B'$, $A''$ must be their center of similtude, hence $A'', A, A'$ are collinear.",
  "Solution_6": "I think the difficult part of this problem was to come up with the construcation of the additional points $B\"$ and $C\"$ :wink: .\r\n\r\nAnyway,thank you for your solution Yimin! :wink: \r\n\r\nAlso I am wondering,are you going to participate in the IMO 2007?",
  "Solution_7": "[quote=\"Tiks\"]Also I am wondering,are you going to participate in the IMO 2007?[/quote]\r\n\r\nProvided that I qualify...yes :)\r\nWhat about you?",
  "Solution_8": "[quote=\"Yimin Ge\"][quote=\"Tiks\"]Also I am wondering,are you going to participate in the IMO 2007?[/quote]\n\nProvided that I qualify...yes :)\nWhat about you?[/quote]\r\n\r\nMe too :wink: ,if I will qualify :) .\r\n\r\nSee you!!!!!!"
}
{
  "Tag": [
    "geometry",
    "search",
    "number theory",
    "\\/closed"
  ],
  "Problem": "Hello,\r\n\r\nMay be what I am going to ask is somewhere and I just did not look appropertly for it, but, Is there a way to post a same topic in more than one forum at once. Let me try to explain myself better trying to make an example: Pretty often you can find problems (or topics in general) that fit in more than one of the subfora (just in case, plural of subforum) on MathLinks, for example, some problems that can be solved either with Geometry, or a totally different solution using only algebra, or with number theory or combinatorics, and some of us may want to place it in all the subfora that apply.\r\n\r\nFor example, this problem \r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=26695\r\n\r\nI would like to place it both in Combinatorics and in Geometry, but if I do it placing it once on each fora, probably it turns messy, because it becomes to different discussions, that could fit nice in just one. \r\n\r\nIs there a way to do this? Was I clear enough with what I mean?\r\n\r\nBest regards,",
  "Solution_1": "In general, we prefer to discourage such things, because it takes up space in our database.  Plus it make our forum too easy to spam.  \r\n\r\nPlease don't post the same thing in more than one topic.  Try to pick the most appropriate topic, and post it there.",
  "Solution_2": "As David pointed out we don't want that. Try not to double-post a problem, although this happens a lot (given the numbers of problems here, there are roughly more than 50% chances that a problem has been already posted somewhere) and use the search function. \r\n\r\nAs for problems that fit in two categories, placing them in one category is enough, the subforums for Geometry for example are not restricted to solutions only involving Geometry (etc.)"
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Let \r\n$ A \\equal{} {\\begin{pmatrix} 1 & 1 & \\minus{}1 \\\\\r\n1 & 1 & 1 \\\\\r\n1 & 1 & 1 \\\\\r\n\\end{pmatrix}}$\r\n$ D \\equal{} {\\begin{pmatrix} 0 & 0 & 0 \\\\\r\n0 & 1 & 0 \\\\\r\n0 & 0 & 2 \\\\\r\n\\end{pmatrix}}$\r\n\r\nProve that $ \\exists P$ invertible matrix such that\r\n\r\n$ A \\equal{} P^{ \\minus{} 1}DP$",
  "Solution_1": "Hint: Why is it enough to show that both matrices have the same eigenvalues assuming that we are working, say, over the reals?",
  "Solution_2": "Observe the second and third columns of A: the sum of first elements is zero, while the sum of second, and third elements, is 2. Therefore, 2 is an eigenvalue. Also, the first two columns are equal, therefore 0 is another eigenvalue. Finally, the trace equals 3, therefore the other eigenvalue is 1. Since A has all eigenvalues distinct, it is diagonalizable, which means A is similar to the diagonal matrix of its eigenvalues, which is D. $ \\Diamond$"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "algebra",
    "polynomial",
    "invariant",
    "absolute value",
    "linear algebra unsolved"
  ],
  "Problem": "Let $A\\in M_n(Z)$ be an invertible matrix such that $|Tr(A^k)|\\leq n$ for all $k$. Prove that there exists a matrix $Q\\in GL_n(Z)$ such that $Q^{-1}AQ$ is an upper triangular matrix by blocks, with some 1 and -1 on the main diagonal and such that all the other blocks on the main diagonal have characteristic polynomials some cyclotomic polynomials.",
  "Solution_1": "It is sufficient to prove that the eigenvalues $\\alpha_i$ of $A$ are all roots of unity.  We shall prove this under the weaker assumption that $\\vert Tr A^n\\vert < P(n)$ where $P$ is a polynomial.\r\n\r\nFirstly we prove that $\\vert\\alpha_i\\vert \\le 1$ for all $\\alpha_i$.  We have $Tr A^n = \\sum_i \\alpha_i^n$.  Consider the power series $F(z) = \\sum_{n=0}^\\infty (Tr A^n) z^n = \\sum_i \\frac{1}{1 -\\alpha_i z}$.  Since the coefficient of $z^n$ in $F$ is bounded by a polynomial in $n$, we have $F$ convergent for any $z$ with $\\vert z \\vert < 1$ and so the radius of convergence of $F$ is 1.  Hence $F$ is analytic in the open unit disc.  We conclude that all the $\\vert \\alpha_i \\vert \\le 1$, otherwise there would be a pole of $F$ in the open disc.\r\n\r\nNow we show that if $f$ is a monic integer polynomial with roots non-zero algebraic integers all of modulus $\\le 1$ then the roots of $f$ are all roots of unity. This is well-known, but here's a proof I like.  Consider a recurrent sequence $(x_n)_{n=1}^\\infty$ with driving (auxiliary) polynomial $f$ and initial terms $0,0,\\ldots,0,1$ (impulse response sequence).  The general term is $x_n = \\sum_i A_i \\alpha_i^n$ for certain constants $A_i$.  Since all $\\vert \\alpha_i \\vert \\le 1$, $x_n$ is bounded and so can take on only finitely many values.  Hence the sequence must repeat from some point on and (since the constant term of  $f$ is $\\pm 1$) the sequence is reversible, so the sequence is fully cyclic.  Thus the sequence admits $X^M-1$ as a polynomial for for $M$, and so $f$ divides $x^M-1$.  Hence the roots of $f$ are all roots of unity.",
  "Solution_2": "I am sorry, but may I ask you why from the fact that the eigenvalues are roots of unity it follows what I asked? I don't think it is so straightforward... or I am blind.",
  "Solution_3": "The characteristic polynomial of $A$ is monic with integer entries, so the eigenvalues are algebraic integers.  $A$ is invertible, so they are all non-zero and indeed their product is $\\det A = \\pm 1$.  My posting aimed to show that they all have absolute value at most 1, and that algebraic integers with this property which are roots of a monic polynomial with constant term $\\pm 1$ (another way of saying that all conjugates are also bounded by 1) must be roots of unity.",
  "Solution_4": "Oh, sorry, your post again I've just read.  The quick answer is rational canonical form for the invariant subspaces.",
  "Solution_5": "Sorry, but is this theorem true in any ring? Here all matrices are supposed to have integer entries and yet the only version of the theorem you are speaking is working with fields.",
  "Solution_6": "Oh, I see your point: I'd read $Q$ for $Z$."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be a group, $ \\phi : G \\to G$ a group endomorphism.  Is the center of $ G$ necessarily stable under $ \\phi$?  Is there a finite counterexample?",
  "Solution_1": "What means stable here\u00bf\r\nIf $ \\phi$ is an automorphism, then the center is mapped to the center because it is a characteristic subgroup (it can be defined by all-quantified relations on $ G$).\r\n\r\nIf $ \\phi$ is just a general homomorphism, then this is no longer true. For example, take $ G\\equal{} S_3 \\times (\\mathbb Z / 3 \\mathbb Z)$. It's center is $ \\{ 0 \\} \\times (\\mathbb Z / 3 \\mathbb Z)$. Now take $ \\phi(a,b) \\equal{} ( (1,2,3)^b , 0)$. This is a homomorphism and maps the center not to itself."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "reflection",
    "symmetry",
    "homothety",
    "geometry proposed"
  ],
  "Problem": "Dear Mathlinkers,\r\nlet ABC be a triangle, A'B'C' the orthic triangle of ABC, A\"B\"C\" the Euler's triangle of ABC, X the meet point of B\"C' and C\"B', O the center of the circumcircle of ABC and N the center of the Euler's circle of ABC.\r\nProve : X, A and N are collinear. \r\nSincerely\r\nJean-Louis",
  "Solution_1": "Let $ K$ be the circumcenter of $ \\triangle AB'C'$ lying on $ AH,$ since rays $ AO,AH$ are isogonal WRT $\\angle BAC.$ Let $ K_0$ be the reflection of $ K$ about $ B'C'.$ Note that $ \\angle HC'B'' \\equal{} \\angle HB'C''$ $\\Longrightarrow$ Rays $C'X, C'K_0$ and $ B'X,B'K_0$ form equal and oppositely directed angles  with respect to the pairs of rays $ C'B',C'A$ and $B'C',B'A$ $\\Longrightarrow$ $X,K_0$ are isogonal conjugates WRT $ \\triangle AB'C'$ $\\Longrightarrow$ Rays $ AK_0$ and $ AX$ are isogonal WRT $ \\angle BAC.$ Since $ B'C'$ is antiparallel to $ BC,$ the axial symmetry about the internal bisector of $ \\angle BAC$  takes $ B'C'$ into $ B''C''$ parallel to $ BC,$  $ K$ into $ K''$ on the ray $ AO$ and  $ K_0$ into $ {K_0}'$ on the ray $AX.$ Then homothety with center $ A,$ that takes $ B''C''$ into $ BC,$ sends $ K''$ into $ O$ and carries ${K_0}'$ into the reflection $O_a$ of $ O$ about the sideline $ BC.$ Since $ AH \\equal{} OO_a$ $\\Longrightarrow$ $ AHO_aO$ is a parallelogram.",
  "Solution_2": "Dear Luis,\r\nwhy do you haven't thought at the Pascal's theorem?\r\nSincerely\r\nJean-Louis",
  "Solution_3": "Dear Jayme,\r\nYes Pascal's theorem is very useful if we let $ B_1, C_1$ be the midpoint of $ AC, AB$."
}
{
  "Tag": [
    "inequalities",
    "inequalities solved"
  ],
  "Problem": "2. Proove that for every x,y,z>0 we have: \r\nsqrt((x+y)/(x+z))+sqrt((x+z)/(x+y)) <=(y+z)/sqrt(yz).\r\n\r\n\r\nthis is funny... just look at M ineq 11 posted by me! :D\r\n\r\ncheers!",
  "Solution_1": "We see :\r\n            sqrt((x+y)/(x+z)) + sqrt((x+z)/(x+y))  \\leq  (y+z)/sqrt(yz)\r\n <=>    sqrt ((1+y/x)/(1+z/x) + sqrt((1+z/x)/(1+y/x))  \\leq  (y/x+z/x)/ sqrt((y/x)*(z/x))\r\n       We denote :\r\n            a = y/x\r\n            b = z/x\r\n       The inequality becomes :\r\n             sqrt((1+a)/(1+b)) + sqrt((1+b)/(1+a))  \\leq  (a+b)/sqrt(ab)\r\n  <=>     2ab + a <sup>2</sup> b + ab <sup>2</sup>  \\leq  a <sup>2</sup>  + b <sup>2</sup>  + a <sup>3</sup> +b <sup>3</sup>   \r\n        Which is obvious by Cauchy"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all natural numbers $ k$ which can be represented as the sum of two relatively prime numbers not equal to 1.",
  "Solution_1": "$ k \\notin \\{ 1,2,3,4,6 \\}$ Otherwise $ k \\in \\{ 4m,4m\\plus{}1,4m\\plus{}2,4m\\plus{}3 \\}$\r\n\r\n$ 4m\\equal{}(2m\\minus{}1)\\plus{}(2m\\plus{}1)$\r\n\r\n$ 4m\\plus{}1\\equal{}2m\\plus{}(2m\\plus{}1)$\r\n\r\n$ 4m\\plus{}2\\equal{}(2m\\minus{}1)\\plus{}(2m\\plus{}3)$\r\n\r\n$ 4m\\plus{}3\\equal{}(2m\\plus{}1)\\plus{}(2m\\plus{}2)$",
  "Solution_2": "Or even shorter :P\r\n$ 2k \\equal{} (k\\minus{}1) \\plus{} (k\\plus{}1)$ and $ 2k\\plus{}1 \\equal{} (k) \\plus{} (k\\plus{}1)$",
  "Solution_3": "[quote=\"Mathias_DK\"]Or even shorter :P\n$ 2k \\equal{} (k \\minus{} 1) \\plus{} (k \\plus{} 1)$[/quote]\r\nk=6 :?:",
  "Solution_4": "[quote=\"Akashnil\"][quote=\"Mathias_DK\"]Or even shorter :P\n$ 2k \\equal{} (k \\minus{} 1) \\plus{} (k \\plus{} 1)$[/quote]\nk=6 :?:[/quote]\r\n^^ My fault, you have to have three cases.. But it is still shorter :P"
}
{
  "Tag": [
    "geometry",
    "inequalities",
    "analytic geometry",
    "trigonometry",
    "angle bisector",
    "geometry unsolved"
  ],
  "Problem": "ABC is a right-angled triangle with the right angle at A.  Let D be the foot of the perpendicular from A to BC, and let E and F be the intersections of the bisector of B with AD and AC respectively.  Prove that DC>2EF.\r\n\r\nThanks",
  "Solution_1": "I believe a co-ordinate approach is very effective when solving problems involving right triangles and their altitudes, bisectors,etc\r\n\r\nConsider a co-ordinate system with origin at A.  Let $ B\\equiv (u,0), C\\equiv (0,v)$ and $ \\ell (BC)\\equal{}a$\r\n\r\nSo, we have as follows\r\nEquation of internal angle bisector of B:$ vx\\plus{}(u\\plus{}a)y\\minus{}uv\\equal{}0$\r\nEquation of AD:$ ux\\minus{}vy\\equal{}0$\r\n\r\nSo, $ E\\equiv (\\frac {uv^2}{a(a\\plus{}u)}, \\frac {u^2v}{a(a\\plus{}u)})$\r\n\r\n$ F\\equiv (0,\\frac {uv}{u\\plus{}a})$\r\n\r\n$ \\ell (EF)\\equal{}\\frac {uv}{a(a\\plus{}u)} \\sqrt {2a^2\\minus{}2au}$\r\n\r\n$ \\ell (CD)\\equal{}\\frac {v^2}{a}$\r\n\r\nSo,\r\n$ DC>2EF\\Leftrightarrow \\frac {v^2}{a}>2\\frac {uv}{a(a\\plus{}u)} \\sqrt {2a^2\\minus{}2au}$\r\n$ \\Leftrightarrow v^2a^2\\plus{}u^2v^2\\plus{}2v^2au\\plus{}8au^3>8u^2a^2$\r\n\r\nThat inequality should be true (it has to be if the problem is correct).\r\nCan someone prove that inequality?? as i could'nt.",
  "Solution_2": "It very easy by Coordinate - Decarter:\r\nWe chose coordinate as picture.\r\nLet $ D(0;0), B(\\minus{}1; 0), A(0; a)$. Because $ AB^2\\plus{}AC^2\\equal{}BC^2$ $ \\implies$ $ C(a^2; 0)$.\r\nWe have EA/ED = BA/BD --> coordinate of E.\r\nWe have FA/EC = BA/BC --> coordinate of F.\r\nSo we have EF and DC by a. Very easy --> CD > 2EF. \r\nOK.",
  "Solution_3": "here's a simple proof .\r\nfirst observe that in the right triangle $ \\cos^{2} \\frac {B}{2} \\equal{} \\frac {(a \\plus{} c)^{2}}{4a^{2}}$\r\nalso $ BE \\equal{} \\frac {2cx \\cos \\frac {B}{2}}{c \\plus{} x}$ and $ BF \\equal{} \\frac {2ac \\cos \\frac {B}{2}}{a \\plus{} c}$\r\nwhere $ AD \\equal{} x$\r\nhence we are supposed to prove\r\n$ a \\minus{} x > 2(\\frac {2cx \\cos \\frac {B}{2}}{c \\plus{} x} \\plus{} \\frac {2ac \\cos \\frac {B}{2}}{a \\plus{} c})$\r\non simplification this gives\r\n$ \\frac {(a \\plus{} c)(c \\plus{} x)}{4c^{2}} > \\cos \\frac {B}{2}$\r\nuse $ \\frac {x}{c} \\equal{} \\cos B$\r\nhence we need to prove $ \\cos \\frac {B}{2} > \\frac {2c}{a \\plus{} c}$\r\nwhich after putting the value of $ \\cos \\frac {B}{2}$ simplifies to proving \r\n$ (a \\plus{} c)^{4} > 16a^{2}c^{2}$ which follows from AM-GM and since $ a \\neq c$ the inequality is strict",
  "Solution_4": "[quote=\"thanhnam2902\"]It very easy by Coordinate - Decarter:\nWe chose coordinate as picture.\nLet $ D(0;0), B( \\minus{} 1; 0), A(0; a)$. Because $ AB^2 \\plus{} AC^2 \\equal{} BC^2$ $ \\implies$ $ C(a^2; 0)$.\nWe have EA/ED = BA/BD --> coordinate of E.\nWe have FA/EC = BA/BC --> coordinate of F.\nSo we have EF and DC by a. Very easy --> CD > 2EF. \nOK.[/quote]\r\n\r\nCould you explain what you mean by implies coordinate of E, F.\r\n\r\nAlso could you explain what you mean EF and DC by a.\r\n\r\nAnd then explain how that implies CD>2Ef",
  "Solution_5": "[color=darkred] [size=117][b]Nice problem ![/b][/size] \n\n[/color] [quote=\"mathgeniuse^ln(x)\"][color=darkred]$ ABC$ is a right-angled triangle with the right angle at $ A$. \nLet $ D$ be the foot of the perpendicular from $ A$ to $ BC$ and let $ E$ and \n$ F$ be the intersections of the bisector of $ B$ with $ AD$ and $ AC$ respectively. \nProve that $ DC > 2\\cdot EF$ .[/color] [/quote]\r\n[b][u]Proof.[/u][/b] $ BE^2 \\equal{} BD\\cdot BA \\minus{} ED\\cdot EA$ $ \\implies$ $ \\underline {BE^2 < BD\\cdot BA} \\equal{} \\frac {c^3}{a}$ $ \\implies$ $ BE^2 < \\frac {c^3}{a}$ .\r\n\r\nProve easily that $ EF \\equal{} \\frac {a \\minus{} c}{c}\\cdot BE$ . Thus, $ \\boxed {EF^2 < \\frac {c(a \\minus{} c)^2}{a}}$ . But\r\n\r\n$ \\boxed {\\frac {c(a \\minus{} c)^2}{a} < \\left(\\frac 12\\cdot DC\\right)^2}$ $ \\Longleftrightarrow$ $ \\frac {4c(a \\minus{} c)^2}{a} < \\left(\\frac {b^2}{a}\\right)^2$ $ \\Longleftrightarrow$\r\n\r\n$ 4ac(a \\minus{} c)^2 < (a \\minus{} c)^2(a \\plus{} c)^2$ $ \\Longleftrightarrow$ $ \\underline {4ac < (a \\plus{} c)^2}$ , what is truly. \r\n\r\nTherefore, $ EF^2 < \\frac {c(a \\minus{} c)^2}{a} < \\left(\\frac 12\\cdot DC\\right)^2$ $ \\implies$ $ \\boxed {DC > 2\\cdot EF}$ .",
  "Solution_6": "A good candidate for PWW.\r\n\r\n  To improve visibility draw bisector AX of angle DAC\r\n  and midline MN of triangle DAC naturally parallel to DC.\r\n  All 3 are concurrent.  EF is always perpendicular to AX\r\n  while MN is mostly not.\r\n  \r\n\r\n  T.Y.\r\n  M.T.",
  "Solution_7": "[quote=\"armpist\"]A good candidate for PWW.\n\n  To improve visibility draw bisector AX of angle DAC\n  and midline MN of triangle DAC naturally parallel to DC.\n  All 3 are concurrent.  EF is always perpendicular to AX\n  while MN is mostly not.\n  \n\n  T.Y.\n  M.T.[/quote]\r\n\r\nWhat is PWW?\r\n\r\nIs it like some sort of geometry method.",
  "Solution_8": "proofs without words..."
}
{
  "Tag": [
    "function",
    "search"
  ],
  "Problem": "I learned of a new nim-like game in economics, though I don't know what it's called.\r\no\r\noo\r\nooo\r\noooo\r\nooooo\r\n\r\nYou have stacks of pins (stacks of 1, 2, 3, 4, 5 pins in my class) and you can take away 1 or 2 pins at a time.  The object of the game is to force your opponent into taking the last pin.  The catch: if you take 2 pins, they must be adjacent and in the same stack.  Of course, the pins to not collapse together.\r\n\r\nI represent any position as a set of positive integers.  For instance, My class's position would be $\\{1,2,3,4,5\\}$.  Now, overload the '+' operator to combine two sets (scalar multiplication follows from addition), let $[N]=1$ if you can force a win by moving next, and $[N]=0$ if you cannot force a win.  \r\n\r\nHere are some rules I have gathered (maybe you want to verify them):\r\n1) $[N+\\{1,1\\}]=[N]$\r\n2) If $A$ is a position in which you can force a lose, and $[B]=0$, then $[A+B]=1$\r\n3) $\\{3\\}\\equiv\\{1,2\\}, \\{3,3\\}\\equiv\\{2,2\\}$\r\nwhere $A\\equiv B$ means $A$ and $B$ function the same and can replace eachother.  I couldn't find any more useful equivalences.\r\n\r\nIn the $\\{1,2,3,4,5\\}\\equiv\\{1,2,1,2,4,5\\}\\equiv\\{2,2,4,5\\}$ position, I believe it is best to take one off the end from the stack with 5 pins.  Some useful reduced losing positions to corner your opponent into are: $\\{4\\},\\{1,5\\},\\{2,2\\},\\{4,5\\},\\{1,4,4\\},\\{2,2,5\\},\\{1,2,2,4\\},\\{2,2,2,2\\},\\{2,2,4,4\\}$, and $\\{2,2,2,5\\}$\r\n\r\nBTW, does anyone know what a \"break-and-take game\" is?",
  "Solution_1": "[quote=\"juggle7\"]I learned of a new nim-like game in economics, though I don't know what it's called.[/quote]\n\nIt's probably called something like Misere k-Token Nim. The reason why I say maybe is because there are so many variations of nim out there, I can't imagine people agreeing on the same name for many variations. Misere refers to the style of play where the player that takes the last token/pin loses. k-Token Nim is the variation where each player may remove up to k-tokens each turn. In this case, k=2.\n\n\n[quote=\"juggle7\"]I represent any position as a set of positive integers.  For instance, My class's position would be $\\{1,2,3,4,5\\}$.  Now, overload the '+' operator to combine two sets (scalar multiplication follows from addition), let $[N]=1$ if you can force a win by moving next, and $[N]=0$ if you cannot force a win.[/quote]\n\nIt's great that you're coming up with your own notation; it shows that you've really been thinking about the games. In game theory, your $[N]=1$ is an N-position, which means that the next person that moves can win. $[N]=0$ is a P-position, which stands for previous player wins. Often, in game theory, we search for P-positions.\n\n\n[quote=\"juggle7\"]Here are some rules I have gathered (maybe you want to verify them):\n1) $[N+\\{1,1\\}]=[N]$\n2) If $A$ is a position in which you can force a lose, and $[B]=0$, then $[A+B]=1$\n3) $\\{3\\}\\equiv\\{1,2\\}, \\{3,3\\}\\equiv\\{2,2\\}$\nwhere $A\\equiv B$ means $A$ and $B$ function the same and can replace eachother.  I couldn't find any more useful equivalences. [/quote]\n\nHere, you are referring to Bogus-Nim Heaps, and this is a big idea in game theory. You might want to look it up. It looks like you're on the right track so far!!\n\n\n[quote=\"juggle7\"]In the $\\{1,2,3,4,5\\}\\equiv\\{1,2,1,2,4,5\\}\\equiv\\{2,2,4,5\\}$ position, I believe it is best to take one off the end from the stack with 5 pins.  Some useful reduced losing positions to corner your opponent into are: $\\{4\\},\\{1,5\\},\\{2,2\\},\\{4,5\\},\\{1,4,4\\},\\{2,2,5\\},\\{1,2,2,4\\},\\{2,2,2,2\\},\\{2,2,4,4\\}$, and $\\{2,2,2,5\\}$[\\quote]\n\nNice start. Actually, if you investigate it a bit more, nim has a VERY interesting solution for finding all P-positions. \n\n[hide] If you're still interested in exploring the game, try expressing the piles in base-2.[/hide]\n\n\n[quote=\"juggle7\"]BTW, does anyone know what a \"break-and-take game\" is?[/quote][/quote]\n\nYes, a take-and-break game is a game in which a player can remove tokens and/or (depending on the variation) \"break\" or divide the heaps into several smaller heaps. [/hide]"
}
{
  "Tag": [
    "invariant",
    "search"
  ],
  "Problem": "Suppose that an ordinary chessboard is tiled with dominos, each of which covers two adjacent squares of the chessboard.  Show that the number of horizontal dominos is even.",
  "Solution_1": "Generalizing to a $ 2k \\times 2k$ board. Let $ D$ be the number of horizontal dominoes in a tiling, and let $ a_1, a_2, \\ldots a_{2k}$ be such that $ a_i$ is the number of squares in column $ i$ covered by a horizontal domino.\r\n\r\nObserve the following facts:\r\n(1) $ a_1 \\plus{} a_2 \\plus{} a_3 \\plus{} \\ldots \\plus{} a_{2k} \\equal{} 2D$.\r\n(2) It is necessary (though not sufficient) for a full tiling to have $ 2k \\minus{} a_i$ even for all $ 1 \\leq i \\leq 2k$, since the remaining squares must be covered by vertical dominos. Equivalently, we require that all $ a_i$ are even.\r\n(3) Placing a horizontal domino between columns $ i$ and $ i\\plus{}1$ adds $ 1$ to both $ a_i$ and $ a_{i\\plus{}1}$. From this we can find the invariant $ a_1 \\plus{} a_3 \\plus{} \\ldots \\plus{} a_{2k\\minus{}1} \\equal{} a_2 \\plus{} a_4 \\plus{} \\ldots \\plus{} a_{2k}$. Using (1), we also determine that both expressions are equal to $ D$.\r\n\r\nSuppose $ D$ is odd. Then by (3), $ a_1 \\plus{} a_3 \\plus{} \\ldots \\plus{} a_{2k\\minus{}1} \\equal{} a_2 \\plus{} a_4 \\plus{} \\ldots \\plus{} a_{2k} \\equal{} D$. For those sums to be odd, the sequences $ a_1, a_3, \\ldots a_{2k\\minus{}1}$ and $ a_2, a_4, \\ldots a_{2k}$ each must contain at least one odd term. This contradicts (2). Therefore $ D$ is even.",
  "Solution_2": "Here's a wonderful link: http://www.cs.utexas.edu/users/EWD/indexEWDnums.html\r\n\r\nI discovered it by \"accident\" some time ago. Everybody should check it. Great stuff in there.\r\n\r\nThis particular problem is 1306a.",
  "Solution_3": "So, their problem is more general. For those who don't want to search the link, their problem is the number of horizontal dominoes with a white square on the left equals the number of horizontal dominoes with a white square on the right.\r\n\r\nThe approach of my solution completely fails to solve this. Time to rethink this.",
  "Solution_4": "A nice solution to the original problem is as follows:\r\n\r\nColor the columns of the chessboard alternately white and black. Then, a horizontal domino covers one of each color and a vertical domino covers two of one color. Finish with a simple parity argument.\r\n\r\nThe more general problem eludes this technique as well. (at least as far as I see)",
  "Solution_5": "[hide=\"the 1306a problem\"]\nConsider any vertical line between two columns and the set $ H$ of all horizontal dominoes crossing the line. Besides these dominoes, each domino is on one side of the line or the other. Say on the left side, there are $ a$ black squares and $ a$ white squares. Then if $ b$ is the number of dominoes in $ H$ covering a black square on the left side and $ w$ is the number of dominoes in $ H$ covering a white square on the left side, the number of uncovered white squares in the left section in $ a\\minus{}w$ and the number of uncovered black squares is $ a\\minus{}b$. Then $ a\\minus{}w\\equal{}a\\minus{}b\\implies b\\equal{}w$. We can apply this to all particular vertical lines, so the number of dominoes touching a black square on the left side is the same as the number of dominoes touching a white square on the left side.\n[/hide]",
  "Solution_6": "[b]Further Generalization:[/b] If you tile an $ nk\\times nk$ square using $ 1\\times n$ tiles, then the number of horizontal tiles is divisible by $ n$."
}
{
  "Tag": [
    "probability",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "There exists a partition of the set of positive integers in two classes such that from any three consecutive positive integers at least one is in each class and no class contains an infinite arithmetic progression.",
  "Solution_1": "Here's a nonconstructive argument.  Color each odd integer red or blue randomly.  Color each even integer differently from its predecessor.  So we have randomly partitioned the integers into two color classes that definitely meet the \"three consecutive\" condition.\r\n\r\nBut there are only countable many arithmetic progressions, and each has probability 0 of being colored one color.  So our partition has the required properties with probability 1.  In particular, there is a good partition.\r\n\r\nIf you want a constructive argument, I guess we can use a diagonalization argument.  Enumerate each of the arithmetic progressions, and kill each one as we go, maintaining the odd-even pattern.",
  "Solution_2": "Very nice construction RaviB. In fact the author indicated a concrete such partition. Put in the first class only those numbers $n$ for which $ n+[\\sqrt{n}]$ is even. But these solutions are much more natural and beautiful.",
  "Solution_3": "Thanks.  I like that concrete partition too."
}
{
  "Tag": [
    "inequalities",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let $ P$ be an interior point of triangle $ ABC$, and let $ x,y,z$ denote the distance from $ P$ to $ BC,AC,$ and $ AB$ respectively. Where should $ P$ be located to maximize the product $ xyz$?",
  "Solution_1": "It is not so complicated... :blush: \r\n\r\nPut $ a\\equal{}BC,b\\equal{}CA,c\\equal{}AB$ then we have: $ ax\\plus{}by\\plus{}cz\\equal{}2S_{ABC}$.\r\n\r\nBy AM-GM inequality: $ 2S_{ABC}\\equal{}ax\\plus{}by\\plus{}ca\\geq 3\\sqrt[3]{abcxyz}\\Rightarrow xyz\\leq\\frac{8S_{ABC}^3}{27abc}$\r\n\r\nEquality holds if and only if $ ax\\equal{}by\\equal{}cz\\Leftrightarrow S_{PAB}\\equal{}S_{PBC}\\equal{}S_{PCA}\\Leftrightarrow P$ is the centroid of $ \\triangle ABC$"
}
{
  "Tag": [],
  "Problem": "When 15 is appended to a list of integers, the mean is increased by 2.  When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1.  How many integers were in the original list?",
  "Solution_1": "[hide]let\nA: average \nX: number of integers in the original list\n\n(AX+15)/(X+1) = A+2\n(AX+15+1)/(X+1+1) = A+2-1\n\nCrossmultiply,\n\nAX+A+2X+2=AX+15\nAX+2A+X+2=AX+16\n\nSubtract second from first*2,\n3X+2=14\n\nX=4[/hide]",
  "Solution_2": "Yup.  Welcome to the NYC forum!"
}
{
  "Tag": [
    "calculus",
    "integration",
    "limit",
    "function",
    "real analysis",
    "calculus computations"
  ],
  "Problem": "Let $ f$ be defined on $ [1, \\infty]$ and suppose $ I_{n}= \\int_{1}^{n}f$ exists for all $ n \\in \\mathbb N$.  Decide whether the statement is true or false with either a (short) proof or counterexample:\r\n\r\n1.  If $ f$ monotonic decreasing and if $ \\lim_{n \\rightarrow \\infty}I_{n}$ exists, then $ \\int_{1}^{\\infty}f$ converges.\r\n\r\n2.  If $ \\lim_{x \\rightarrow \\infty}f(x) = 0$ and $ \\lim_{n}I_{n}= I$, then $ \\int_{1}^{\\infty}f$ converges to $ I$.\r\n\r\n3.  If $ I_{n}$ converges, then so does $ \\int_{1}^{\\infty}f$.\r\n\r\n4.  Suppose that $ f'$ exists and is bounded on $ [1, \\infty]$, and that $ \\lim_{n}I_{n}= I$.  Then $ \\int_{1}^{\\infty}f$ converges to $ I$.\r\n\r\n5.  If $ \\int_{1}^{\\infty}f$ converges, then $ \\lim_{x \\rightarrow \\infty}f(x) = 0$.",
  "Solution_1": "#3 is false. Consider the following counterexample:\r\n\r\nLet $ f(x)=\\sin(2\\pi x).$ Then $ I_{n}=\\int_{1}^{n}\\sin(2\\pi x)\\,dx=0$ for all $ n\\in \\mathbb{N}.$ \r\n\r\nHowever, $ \\int_{0}^{\\infty}\\sin(2\\pi x)\\,dx$ does not converge.\r\n\r\nThis shows what the issue is for #1, #2, or #4, but the extra conditions in those problems rule out this exact example. I'll let others decide whether they are true.\r\n\r\n[hide=\"Hint.\"]Of the five statements, at least two are false.[/hide]",
  "Solution_2": "[quote=\"arp\"]\n5.  If $ \\int_{1}^{\\infty}f$ converges, then $ \\lim_{x \\rightarrow \\infty}f(x) = 0$.[/quote]\r\nThis [b]can[/b] be made true with enough initial conditions on the hypothesis. \r\nBut consider a silly counterexample,\r\n\r\n$ f(x) = \\{ \\begin{array}{c}1 , \\ x\\in \\mathbb{N}\\\\ 0 , \\ x\\not \\in \\mathbb{N}\\end{array}$\r\n\r\nThis integral is clearly zero because one isolated point doest change the integral. But $ \\lim_{x\\to \\infty}f(x)$ doesnt even exist.\r\n\r\n--------\r\nFor #1 consider $ f(x) = 1$.",
  "Solution_3": "sylow_theory's example is not continuous. Very well, then, let us require continuity.\r\n\r\nDefine $ T(x)=\\max(1-|x|,0).$ This is the piecewise-linear continuous function with vertices at $ (-1,0),(0,1),$ and $ (1,0),$ taking on the value zero outside of $ [-1,1].$ (\"T\" is for \"tent\".)\r\n\r\nLet $ f(x)=\\sum_{n=1}^{\\infty}nT(n^{3}(x-n))$\r\n\r\nOnce $ n$ is large enough for the tents not to overlap, this a piecewise linear continuous function with vertices at $ (n-\\frac1{n^{3}},0),\\,(n,n)$ and $ (n+\\frac1{n^{3}},0).$\r\n\r\n$ \\int_{0}^{\\infty}f(x)\\,dx=\\sum_{n=1}^{\\infty}\\frac1{n^{2}}$ converges.\r\n\r\nBut not only does the function fail to tend to zero, it is unbounded as $ x\\to\\infty.$\r\n\r\nThe fact that this example is not smooth is unimportant; by replacing $ T$ by a $ C^{\\infty}$ \"bump\" function and repeating the idea, we could create a $ C^{\\infty}$ example with the same flaws.\r\n\r\n---\r\n\r\n[quote=\"sylow_theory\"]For #1 consider $ f(x) = 1.$[/quote]\r\nBut then the sequence $ I_{n}$ would diverge, so that's not a counterexample - at least if we read the problem as requiring $ \\lim I_{n}$ to exists as a finite number.",
  "Solution_4": "It seems like interest has been lost in this question, so I'll try to answer two more and see if someone wants to do the last.\r\n\r\nFor 1, sylow, kent is right that we are to assume $ \\lim I_{n}= I$ exists as a finite number.  In fact, this makes the statement true.  First, note that $ f \\geq 0$ everywhere because of monotocity, for if $ f(x_{0}) = l < 0$ for some $ x_{0}$, then $ f(x) \\leq l$ for all $ x \\geq x_{0}$ and so $ \\lim I_{n}$ could not exist.  There is some $ N$ s.t. for $ n \\geq N$,\r\n $ |I_{n}-I| < \\epsilon$ for every $ \\epsilon > 0$.  Let $ I(b) = \\int_{1}^{b}f$, $ b \\in \\mathbb R$.  For any $ b \\geq N$, there is an $ n \\geq N$ such that $ n \\leq b < n+1$ ($ n = [b]$).  Then $ I_{n}\\leq I(b) \\leq I_{n+1}$ since $ f$ is nonnegative everywhere, so we get $ I-\\epsilon < I_{n}\\leq I(b) \\leq I_{n+1}< I+\\epsilon$, which means $ \\int_{1}^{\\infty}f$ converges to $ I$.\r\n\r\nI'll cheat off Kent to answer 4: False. If $ f(x) = \\sin(2\\pi x)$, then $ f'$ is bounded and and $ \\lim I_{n}= 0$, but $ \\int_{1}^{\\infty}f$ doesn't even exist."
}
{
  "Tag": [
    "combinatorics solved",
    "combinatorics",
    "graph theory"
  ],
  "Problem": "Here's a question I found in a paper download from this site a long time ago (I don't know if you can find the paper here now):\r\n\r\nProve that in a group of 2n people there are 2 people which have an even number of common acquaintances, where, as usual, if A knows B then B knows A.\r\n\r\nIt looks nice and easy, but I've been thinking abt it for quite some time and I don't even know if my soln is correct. Anyway, it inspired me to think of this:\r\n\r\n1. Is it true that in every graph there is a vertex for which the number of chains of length 2 starting from that vertex is even? I have to mention that I don't know the answer for this one but it seems to be 'yes'. Could ne1 help me with it? It's closely related to this:\r\n\r\n2. Is it true that in every graph there is a vertex linked to an even number of vertices of even degree? (the answer for 1 is yes iff the answer for 2 is yes, if you think abt it for a while). Any ideas?",
  "Solution_1": "The initial problem appeared in the Tounament of towns in 1995 (with n = 25, but it doesn't matter).\r\nSince the solution is a little boring to write here, you may find it (in french) at : \r\nhttp://boumbo.salle-s.org/xavier/maths/stmalo/ \r\n\r\nChoose the paper on graph theory. This is exercise #4.\r\n\r\n[Moderator edit: Or see http://www.mathlinks.ro/Forum/viewtopic.php?t=68109 ]\r\n\r\nPierre.",
  "Solution_2": "Thanx a lot, but, like I said, I think I have a soln for that one (it's also a bit boring so i didn't exactly check everything out but...). I think the other problems I wrote there are quite interesting so maybe some1 here has a cute or interesting or useful idea abt them?",
  "Solution_3": "Hmmm...did I miss something or the answer for your pb 1 is trivially 'no' as shown by the following graph :\r\nA-B-C-D\r\n\r\nPierre.",
  "Solution_4": "Thanks a lot! It's like taking a load off my chest. Even if the answer is no, at least I know that! Thx again!"
}
{
  "Tag": [
    "induction",
    "number theory solved",
    "number theory"
  ],
  "Problem": "Prove that s_n = \\sum^{2n-1}_{k=1} 2^(k-1) * Cn(4n-2,2k) is a perfect square for all n = 1,2,3,...",
  "Solution_1": "the sum clearly equals 1/4((1+ \\sqrt 2)^(4n-2) + (1- \\sqrt 2)^(4n-2) -2)\r\ndefine an=2an-1 +an-2 with a1=6 and a2=198\r\nit remains to show that a4n-2 -2 is perfect square. \r\nwe show by induction that a4n-2 = a2n-1^2 +2 ,a4n=a2n^2 -2\r\na4n-1=1/2(a2n^2-a2n-1^2)-2 and that a4n+1=1/2(3a*2n^2+a2n-1^2)+6\r\nproof: lets assume that the hypothesises are true for all n<=4k.\r\na4k+1= 2a4k+a4k-1 = 2a2k^2-4 + 1/2(a2k^2-a2k-1^2)-2 \r\n=1/2(3a2k^2+a2k-1^2)+6\r\nsimilarly you can do it for a4k+2, a4k+3, a4k+4.",
  "Solution_2": "a type mistake .\r\nit should be a1=2, a2=6\r\nthe rest is the same."
}
{
  "Tag": [],
  "Problem": "I'm just wondering.",
  "Solution_1": "I like the ones that aren't to fake like things that you know will never happen. The type of movie that makes you afraid of the dark for like 2 weeks lol...",
  "Solution_2": "i don't like horror movies...",
  "Solution_3": "[quote=\"1=2\"]Re: Thriller/Horror movies\nWhat do you think is the best thriller/horror movie composed of?\nI'm just wondering.[/quote]\r\nWell, Have you watched any horror/thriller movies lately? (I mean, do you watch horror or thriller movies?) If you have watched some good horror movies, you wouldn't have asked a question like this.\r\n\r\n(Watch some horror movies, say like:The Exorcist, Evil Dead, The Ring etc; They're good horror movies)",
  "Solution_4": "By far the best horror movie is The exorcist 1. I mean , I couldn't sleep well for three entire weeks  :P \r\n\r\nIn second place I would place H2O , but just the first one,  because the following H2O suck  :D \r\n\r\nIn third place I would place Scream , The ring , The profecy 1 and Poltergeist 1"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "circumcircle",
    "power of a point",
    "radical axis"
  ],
  "Problem": "[color=cyan]So, here they be.  The rules are simple:  you can draw a circle centered at a given point through a given point.  You can draw a circle centered at a given point with a radius a given length.  And you can draw a straight line through two given points.  That's it.  You can use previous problems to do later problems.\n\n1.  Given 2 points, construct the midpoint of the segment joining them.\n\n2.  Given a point and a line, construct the unique perpendicular to the line passing through the point.\n\n3. Construct the incenter of a given triangle.\n\n4. Construct the circumcenter of a given triangle.\n\n5.  Construct the tangent to a given circle from a point not inside that circle.  (Note that this has two cases.)\n\n6. Given a unit length, construct a length of :sqrt:n where n is any positive integer.\n\n7. Construct a regular hexagon of given side length.\n\n8. Construct a regular pentagon of given side length at least 2 different ways.\n\n9. Construct the radical axis of two circles.  (The radical axis is the set of points whose powers with respect to the two circles are equal.  That is, given 2 circles O1 and O2 with radii r1 and r2 respectively, the radical axis consists of all points P such that PO1:^2: - r1:^2: = PO2:^2: - r2:^2:.)\n\n10. Divide a segment into n equal pieces for any positive integer n.\n\n11. Construct the common tangents of 2 circles in all 3 cases for which tangents exist (0, 1, and 2 intersection points).\n\nLook in the Advanced Forum for three more difficult problems on construction, plus whatever anyone else has added.[/color]",
  "Solution_1": "Another one I like is: given a line segment, divide it into n equal parts for any positive integer n.",
  "Solution_2": "[color=cyan]Oops, as I was writing this I was going to put that one in near the bottom but then I forgot.  So why not make that number 10?[/color]",
  "Solution_3": "Hm..that's some pretty interesting constructions.  I'm taking drafting this year, and it'll probably pretty useful for it!\r\n\r\n-interesting_move"
}
{
  "Tag": [
    "trigonometry",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Given the quadrilateral $ABCD$ with $\\angle DAB=150^{o}$, $\\angle DAC+\\angle ABD=120^{o}$ and $\\angle DBC-\\angle ABD=60^{o}$. Evaluate $\\angle BDC$.",
  "Solution_1": "Answer 45 :D\r\nPutting $\\angle{BAC}= t$ then $\\angle{CAD}=150-t$, $\\angle{ABD}=t-30$, $\\angle{DAC}=t+60$, \\angle{ADB}=60-x, \\angle{ACB}=150-3t.\r\n\r\nIf $\\angle{BDC}=x$ then \r\n$\\frac{BC}{CD}=\\frac{BC}{AC}. \\frac{AC}{CD}$ and after few sine laws we get\r\n(1): $\\frac{sin x}{sin(60-t+x)}=\\frac{sin t. sin(t+60)}{sin (2t+30). sin (t+30)}$\r\n\r\nWe can check that if $x=45$ then (1) is true for every $t$. \r\n\r\nFrom here $\\frac{sin x}{sin(60-t+x)}=\\frac{sin 45}{sin(60-t+45)}$ and therefore $x= 45$."
}
{
  "Tag": [
    "trigonometry",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Find a real number $ r\\ne1$ for which $ \\lim_{n\\to \\infty}\\sin(r^n)$ converges to a number unequal to 0,\r\n(or show that such a number does not exist).",
  "Solution_1": "Do you have a solution?",
  "Solution_2": "no  :(\r\n\r\n(The message was too small. So I made the message longer before submitting).",
  "Solution_3": "Maybe this is idea (only idea not the full solution, I really don't wanr to spend time to solving this problem  :lol:  )\r\nI will speak roughly by the approximation language).\r\nsuppose there  exist limit, then we can write \r\n$ r^{n} \\equal{} 2\\pi k_{n} \\plus{} b \\plus{} \\eta_{n}(\\pi \\minus{} 2b) \\plus{} \\varepsilon_{n}$\r\nwhere $ k_{n} \\in \\mathbb{Z}$,  $ \\eta_{n}$  takes value from the set $ \\{0,1\\}$  or  $ \\eta_{n} \\in \\{0,1\\}$  and  $ \\varepsilon_{n} \\to 0$ as $ n \\to \\infty$\r\nI also suppose that $ \\eta_{n} \\equal{} 0$ for all $ n$ (I believe that  such steps as $ r > 1$, $ \\eta_{n} \\equal{} 0$ is true and one can explain it.  :lol: )\r\nalso suppose that $ k_{0} \\equal{} 0$ then we have :\r\n$ r^{n \\plus{} 1} \\equal{} 2\\pi k_{n \\plus{} 1} \\plus{} b \\plus{} \\varepsilon_{n \\plus{} 1}$ thus we have :\r\n$ 2\\pi k_{n \\plus{} 1} \\plus{} b \\plus{} \\varepsilon_{n \\plus{} 1} \\equal{} r(2\\pi k_{n} \\plus{} b \\plus{} \\varepsilon)$ hence :\r\nfor sufficiently large $ n$ we have :\r\n$ k_{n \\plus{} 1} \\approx \\frac {b(r \\minus{} 1)}{2 \\pi} \\plus{} rk_{n}$ or \r\n$ k_{n \\plus{} 1} \\approx \\frac {b(r^{n \\plus{} 1} \\minus{} 1)}{2\\pi}$\r\non the other hand we have : $ r^{n \\plus{} 1} \\equal{} 2\\pi k_{n \\plus{} 1} \\plus{} b \\plus{} \\varepsilon_{n \\plus{} 1}$ hence,   for the sufficiently large $ n$ we have \r\n$ 2\\pi k_{n \\plus{} 1}\\approx 2\\pi k_{n \\plus{} 1}b \\plus{} b^{2}$, but it's impossible , since   such $ b$ does not exist.",
  "Solution_4": "No comment :what?:"
}
{
  "Tag": [
    "algorithm"
  ],
  "Problem": "If I want to teach myself Computer Science to the point of being able to do well on an AP Computer Science [i]course[/i] that's online, what should I learn?\r\n\r\nIt's Java based from what I hear, so is learning Java sufficient?",
  "Solution_1": "Two great books:\r\n\r\nConcrete Mathemathics by Knuth,Graham,Patashnik - contains exercises + solutions to all of them\r\nIntroductio to Algorithms by Cormen, Leiserson, Rivest, Stein - a lot of useful stuff.\r\n\r\nAnother great resource:\r\n\r\nocw.mit.edu\r\n\r\nAnd judges( in case you'd like to practice):\r\n\r\ntrain.usaco.org - worth doing\r\nacm.uva.es\r\nspoj.pl\r\n...."
}
{
  "Tag": [
    "pigeonhole principle",
    "geometry",
    "analytic geometry",
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "1) Suppose a triangle can be placed inside a unit square such that the center of the square is not inside the triangle. Show that one side of the triangle has length less than 1.\r\n\r\n2) A region M in the plane has area > 1. Prove that there are two points (a, b) and (c, d) in M such that a - c and b - d are both integers.\r\n\r\n3) In a party, n girls and n boys are paired. One observes that in each pair, the difference in height is < 10 cm. Show taht the difference in height for the kth tallest boy and the kth tallest girl is also < 10 cm for k = 1, 2, ..., n.\r\n\r\n4) Prove that from any sequence of 2003 reals, one can choose a consecutive block whose sum differs from an integer by at most 1/1000.\r\n\r\n5) In a contest with 8 problems, 8 students took part. Each problem was solved by 5 students. (a) Prove that there are two students who between them solved all eight problems. (b) Is this true if 5 is replaced by 4 ?\r\n\r\n6) Prove that among any 6 integers, there will be a pair whose sum/difference is divisible by 9.",
  "Solution_1": "(6)[hide]more generally, if we are given n numbers, there exist 2 whose sum/diff. is divisible by 2n-3. We construct (n-1) boxes, labeled 0, 1&(2n-4), 2&(2n-5), 3&(2n-6), ..., (n-2)&(n-1). We put each of the n numbers in the box that contains its residue mod (2n-3). Now, by pigeonhole, as we have n numbers and (n-1) boxes, two must be in the same box. Either their sum or difference is 0, mod 2n-3. (here n=6)[/hide]",
  "Solution_2": "(4) [hide]let the sequence of reals be given by a<sub>k</sub>, 1:le:k:le:2003. define s<sub>i</sub> as {:Sigma:(k=1 to i) a<sub>k</sub>}, where {x} represents the fractional part of x, or x-[x]. now, divide the real number line from 0 to 1 into 1000 parts. If one of the s<sub>i</sub> is in the first or last block, from 0 to 1/1000, we're done. If not, there exist two of the s<sub>i</sub> in another block. Taking the difference yields a consecutive block of that sequence, whose sum is within 1/1000 of an integer.[/hide]",
  "Solution_3": "(2) [hide]let us proceed by contradiction. superimpose (is that a real word?) this region on a cartesian plane tiled by horizontal and vertical points through the lattice line, subdividing the plane into 1 by 1 unit squares. now, if (x,y) is the ordered pair corresponding to a point in the plane, define its SPECIAL coordinates (creative, aren't i?) as ({x},{y}), where {a}=fractional part of a=a-[a]. then, assuming that there are no two points of m satisfying the given condition, there cannot be any two points in M having the same pair of SPECIAL coordinates. So take any 1 by 1 square in the plane, and note that M can be mapped into this square without any overlap by the preceding observation. however, this would mean the max area of M is 1, a contradiction.[/hide]",
  "Solution_4": "An extra question:\r\n\r\nThe bound 1/1000 for problem 4 can obviously be sharpened.\r\nWhat is the sharpest possible bound ?",
  "Solution_5": "I'm afraid I don't understand number 4 at all.\r\n\r\nMystic, I'm not entirely convinced of your number 2 -- I can design plenty of injective maps from regions of area greater than 1 to a 1x1 square -- you have to prove that just because two points aren't mapped to the same place means that the initial area is less than 1.",
  "Solution_6": "Okay, so lets do number 5.\n\n[hide]If any student solved all 8 problems, we're done.  If any student solved 7 problems, some other student must have solved the 8th, so we take those 2 students and we're done.  If any student solved 6 questions, consider the other 2 students.  Each was solved by 5 people, thus they must both have been solved by at least 1 of the same people.  Take that person and our six-question guy, and we're done.\n\n\n\nThus, the only scenario we are left with is that in which each student solved exactly 5 questions.  Now then, returning to this much-neglected problem:\n\nTake any of the students, and WLOG he has solved questions 1, 2, 3, 4 and 5.  We wish to show that some other student solved questions 6, 7, and 8.  Now, between them, questions 6, 7, and 8 have been solved a total of 15 times by the 7 other students.  By the pigeonhole principle, one of the students must have accounted for 3 of these solutions.  Since he can't solved the same question 3 times (I would have had a perfect score on USAMO last year if I could have done that ), he must have solved each of 6, 7 and 8.  Thus, we are all done.\n\n\n\nI highly suspect the same is not true for 4.  I'll get on it now.[/hide]",
  "Solution_7": "Arne wrote:An extra question:\n\nThe bound 1/1000 for problem 4 can obviously be sharpened.\nWhat is the sharpest possible bound ?\n\n\n\n[hide]If the bound is of the form 1/k, with k an integer, we construct k boxes, and define s<sub>i</sub> as before. now, we have that if any of the s<sub>i</sub> are in the first or last box, we're automatically done. now, if not, we want at least 2 of the s<sub>i</sub> in one of the remaining boxes so that we can take their difference. So i think k is 2004, and the lowest bound is 1/2004. however, i don't know why the bound should be of the form 1/k. sort of non-rigourously, we can see that if k is not an integer, we will have [k] boxes and then a mini-box in the middle, that will not sharpen the pigeonholing argument. but that's not too terribly rigorous.[/hide]\n\n\n\n\n\nugh, JBL, you're right on my #2. if i said [hide]construct the mapping from M to the unit square with vertices at (0,0), (1,1), (0,1), and (1,0), such that any point in M is mapped into its SPECIAL coordinates, and then note that this mapping must be injective by the above reasoning, and then note that this preserves area because basically splitting M based on the grid lines and then reassemble it in one square won't change area[/hide] would that fix it?",
  "Solution_8": "Yeah, I think that will do it.  The way I would phrase it is to say that it preserves area over the open lattice squares, and since the other parts (all the grid lines) don't have any area, it does in fact preserve area.  You might have to mess that around a little bit when you talk about the boundaries, but I don't think that's a really major point.",
  "Solution_9": "For 4), a hint : \r\n\r\nConsider the sequence 1/2004, 1/2004, ..., 1/2004 (2003 terms).\r\nThis sequence has k = 1/2004.",
  "Solution_10": "my bad",
  "Solution_11": "\"between them\" implies when the solutions from two students are put together, all 8 problems are solved.",
  "Solution_12": "In your particular example, students 1 and 8 (among several other pairs) between them solved all 8 questions."
}
{
  "Tag": [
    "floor function"
  ],
  "Problem": "Solve \r\n\\[ \\left\\lfloor {\\frac{{x \\minus{} p}}\r\n{p}} \\right\\rfloor  \\equal{} \\left\\lfloor {\\frac{{ \\minus{} x \\minus{} 1}}\r\n{p}} \\right\\rfloor \r\n\\]\r\n\r\nfor real $ x$ and nonzero integer $ p$",
  "Solution_1": "I think $ x$ belongs to $ [0 , \\frac {p\\minus{}1}{p}]$"
}
{
  "Tag": [
    "FTW"
  ],
  "Problem": "A salesman buys a coat at $ \\$64$ less than $ 12.5 \\%$. He then sells the coat at a gain of $ 25\\%$ of his cost after allowing a $ 20\\%$ discount on the marked price. What is the marked price, in dollars, of the coat?",
  "Solution_1": "I don't understand that first sentence.",
  "Solution_2": "i know... the first sentence doesn't make sense... maybe it's saying he bought it at a 12.5% discount, and the final price was 64? 0.o sometimes the questions and solutions for FTW are a bit shaky, like for some reason, the answers are all like $ DFRAC(7)(11)$. 0.o it would be greatly appreciated if all of those bugs could be worked out, especially in the Alpha Test version. -.-",
  "Solution_3": "The first sentence actually means \"The salesman buys a coat at \\$64 [i]minus[/i] 12.5%\". \r\n\r\n12.5% = $ \\frac{1}{8}$, so $ 64 \\cdot \\frac{7}{8} \\equal{} \\$56$\r\nHe gains 25%, so $ \\$56 \\cdot 25\\% \\equal{} \\$14$ , $ \\$56  \\plus{} 14 \\equal{} 70$.\r\nTo find the marked price, we must reverse the discount, so we divide $ \\$70$ by $ \\frac{80}{100}$. The answer and marked price is $ \\$87.5$."
}
{
  "Tag": [
    "geometry",
    "trigonometry"
  ],
  "Problem": "A surveyor is mapping a triangular plot of land. He measures two ofthe sides and the angle formed by these two sides and finds that thelengths are 400 yards and 200 yards and the included angle is 50\u00b0.What is the measure of the third side of the plot of land, to the nearest yard? What is the area of this plot of land, to the nearest square yard?",
  "Solution_1": "use the law of cosines...",
  "Solution_2": "Can you set up the equation using the law of cosines?\r\nI would be able to answer the question if I had the right equation."
}
{
  "Tag": [
    "calculus",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Find the maximum value of the expression: $ (x^3\\plus{}1)(y^3\\plus{}1)$ when $ x\\plus{}y\\equal{}1$. Without using calculus.\r\nI was trying cauchy-schwarz, but no success.",
  "Solution_1": "I solved, the answer is $ 2$.\r\nWhat do you mean by a solution without calculus? :?:\r\nEdit: don't read, I thought that $ x,y\\geq 0$.",
  "Solution_2": "[quote=\"danilorj\"]Find the maximum value of the expression: $ (x^3 \\plus{} 1)(y^3 \\plus{} 1)$ when $ x \\plus{} y \\equal{} 1$. Without using calculus.\n[/quote]\r\n$ (x^3 \\plus{} 1)(y^3 \\plus{} 1)\\leq4\\Leftrightarrow3 \\minus{} (x^3 \\plus{} y^3) \\minus{} x^3y^3\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow3(x \\plus{} y)^6 \\minus{} (x \\plus{} y)^3(x^3 \\plus{} y^3) \\minus{} x^3y^3\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow(x^2 \\plus{} 3xy \\plus{} y^2)^2(2x^2 \\plus{} 3xy \\plus{} 2y^2)\\geq0,$ which true.\r\nThe equality holds when $ x^2 \\plus{} 3xy \\plus{} y^2 \\equal{} 0$ id est, when $ x \\equal{} \\frac {1 \\plus{} \\sqrt5}{2},$ $ y \\equal{} \\frac {1 \\minus{} \\sqrt5}{2}$ or \r\n$ y \\equal{} \\frac {1 \\plus{} \\sqrt5}{2},$ $ x \\equal{} \\frac {1 \\minus{} \\sqrt5}{2}.$\r\nThus, the answer is $ 4.$",
  "Solution_3": "But how do you assure that 4 is the maximum?",
  "Solution_4": "[quote=\"danilorj\"]But how do you assure that 4 is the maximum?[/quote]\r\n\r\naquardy has proved that\r\n$ (x^3 \\plus{} 1)(y^3 \\plus{} 1) \\le 4$\r\nSo that the maximum is $ \\le 4$\r\n\r\nBut he also show example when the  equality holds.\r\nHence 4 is the answer.",
  "Solution_5": "i have   solution for  this problem:\r\n Let $ xy\\equal{}t \\leq \\frac{1}{4}$\r\nand $ P\\equal{} 1\\plus{}x^3y^3\\plus{}(x\\plus{}y)^3\\minus{}3xy(x\\plus{}y)\\equal{}t^3\\minus{}3t\\plus{}2.$\r\nWe have:$ f'(t)\\equal{}3t^2\\minus{}3\\equal{}0 \\leftrightarrow t\\equal{}\\minus{}1 ,t\\equal{}1$\r\nThus,$ MaxP\\equal{}f(\\minus{}1)\\equal{}4$\r\nLast, we solve equation $ t^2\\minus{}t\\minus{}1\\equal{}0$, to find values$ x$ and $ y$",
  "Solution_6": "Let $x,y $ be reals such that $ x + 2y = 1$. Prove that\n$$ (x^3 + 1)(y^3 + 1)\\leq \\frac{11}{2}+3\\sqrt 2$$"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "MATHCOUNTS",
    "AMC 10",
    "\\/closed"
  ],
  "Problem": "In the middle School Tournaments, there have been some tourneys that involved  USAMO and IMO problems, and I was wondering if there could be a forum such that they are separated?  (ie. High School Tournaments)",
  "Solution_1": "Well, not sure exactly which forum you are referring to. But the mods of the respective forum should rather execute their moderator rights more effectively by moving all USAMO and IMO problems to the olympiad section, or other suitable fora possibly. If you see such cases you may also want to pm the mods in charge.",
  "Solution_2": "007math wants to create a USAMO/IMO Tournaments forum.\r\n\r\nBut how well would that work? A tournament of IMO problems? I think maybe IMO problems should just be presented individually.",
  "Solution_3": "gah...i mean an overall High School Tournaments Forum (like middle school  :lol: )",
  "Solution_4": "[quote=\"007math\"]gah...[/quote]\r\n\r\noh... sorry if i frustrated you by overgeneralizing your suggestion. :)",
  "Solution_5": "I just like saying gah... :P  :D",
  "Solution_6": "Subsection of MATHCOUNTS - Middle School Tournaments\r\nThere could be a section of High School for tournaments, as tournaments in MST are very boring with only MATHCOUNTS problems, and if someone tries harder problems, mods rush in and warn that those tournaments shouldn't be there.",
  "Solution_7": "Well we could start some tournaments in the high school forums, but how much would a separate tournaments forum be used?",
  "Solution_8": "We could create a whole new forum altogether, split into Middle School Tournaments and High School Tournaments.",
  "Solution_9": "Idea collapsed?",
  "Solution_10": "Reasons against:\r\n\r\nNewbies will promptly post all their AMC problems in this forum, since \"it's for High School Competitions, and the AMC is a High School Competition\", thus making it an interminable moderator headache for whoever is unlucky enough to mod that forum\r\n\r\nThere really aren't that many fullblown AoPS competitions in high school, whereas there are quite a few active mock Mathcounts competitions and they clutter up the main Mathcounts forum. Thus, a high school forum wouldn't be helpful.",
  "Solution_11": "[quote=\"solafidefarms\"] thus making it an interminable moderator headache for whoever is unlucky enough to mod that forum\n[/quote]\n\nI would mod it :D  \nAnd also, every tournament with AMC problems, or similar could easily be locked, or set to doom.\n\n[quote=\"solafidefarms\"]\nwhereas there are quite a few active mock Mathcounts competitions and they clutter up the main Mathcounts forum. [/quote]\r\n\r\nMost users in MST are probably bored of easy MC problems (only newbies are the problem, but mostly the same people join every tournament). 007math pointed out that in his first post (MC problems are just too boring and easy to just post them in tournaments)",
  "Solution_12": "Well, I wouldn't have a problem modding it.\r\n\r\nA lot of people are past MC level, and would like a tournament with harder problems.",
  "Solution_13": "Heh, sometimes modding is more annoying than it looks. \r\n\r\nPerhaps instead tournaments with more difficult problems could just be posted in Middle School Tournaments.",
  "Solution_14": "Making harder problems that aren't computationally intensive or instantly solved by one application of something is fairly difficult. No offense to anyone, but the problems in many mock mathcount tests are quite computationally intensive. However, not all Mathcounts problems are boring. They might not be as difficult, but there are several with quite elegant solutions. \r\n\r\nI'm not sure about adding a high school tournaments forum...there are already some mock tests, although, to be fair, there aren't too many. I guess the main thing that I'm worrying about is that a lot of the tests that will come up are going to have half-baked problems. I agree with solafidefarms.\r\n\r\nBut hey, if you think that this is a good idea, go for it.",
  "Solution_15": "[quote=\"AIME15\"]We could create a whole new forum altogether, split into Middle School Tournaments and High School Tournaments.[/quote]\r\n\r\nThis is kinda a good idea.",
  "Solution_16": "I'm not sure that making a whole new forum for HST is the right way to go. I think that we should keep our Middle school tournament forum and the next time someone makes a mock test, they will just specify how hard the problems are going to be so that people who can't do USAMO problems don't sign up to some mock MATHCOUNTS test with USAMO level problems.",
  "Solution_17": "couldn't you just make a mock test in the high school forum and lable it \"mock test\" like everyone used to do in mathcounts forum before the new subforum?\r\n\r\nor is there a rule in the high school forum forbidding this?",
  "Solution_18": "xpmath, again I say that less tournaments are probably caused by lack of MC level problems.\r\nmathemagician1729, OK, but maybe renaming the forum MST to M&HST (ok, it's not good name)",
  "Solution_19": "[quote=\"vallon22\"]couldn't you just make a mock test in the high school forum and lable it \"mock test\" like everyone used to do in mathcounts forum before the new subforum?\n\nor is there a rule in the high school forum forbidding this?[/quote]\r\n\r\ni think this is a good idea. mock tests have helped me a tremendous amount for mathcounts, so im sure a mock amc 10 or 12 wouldnt hurt. or maybe, a subforum entitled High School Tournaments.",
  "Solution_20": "You could probably post Mock AMC's in the AMC forum.",
  "Solution_21": "My idea is just to move middle school tournements and put it into a forum called Tournements, which would also include High school tournements.",
  "Solution_22": "...which is EXACTLY what I said?"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "ratio",
    "trigonometry",
    "conics",
    "parallelogram",
    "geometry unsolved"
  ],
  "Problem": "There is a circle with its circumcenter as O and its radius is 1. There is a point A outside the circle. AM,AN is the tengents to the circle O with M N as the tangent points respectively. There is a point L in bad arc MN (not equal to M,N). A line passing point A parallel to line MN intesects with ML,NL at point P,Q respectively. if PQ=2, cos(angle POQ)=1/3, find the length of AM.\r\n\r\ni have no idea to use the given condition cos(angle POQ)=1/3. i dun know how to translate it.",
  "Solution_1": "Assuming that \"bad\" arc $ MN$ should be \"back\" arc $ MN$ (more distant arc from $ A.$ Let $ AL$ cut the circle $ (O)$ again at $ K$ on the front arc $ MN$ (closer to $ A$) and let $ X \\equiv MN \\cap AL$. Using directed line segments:\r\n\r\n$ \\overline{AP} \\cdot \\overline{AQ} \\equal{} \\frac{\\overline{AL}^2}{\\overline{XL}^2} \\cdot \\overline{XM} \\cdot \\overline{XN} \\equal{} \\frac{\\overline{AL}^2}{\\overline{XL}^2} \\cdot \\overline{XL} \\cdot \\overline{XK} \\equal{} \\frac{\\overline{AL}^2}{\\overline{XL}} \\cdot \\overline{XK} \\equal{}$\r\n\r\n$ \\equal{} \\frac{\\overline{AL}}{\\overline{AK}} \\cdot \\frac{\\overline{XK}}{\\overline{XL}} \\cdot (\\overline{AL} \\cdot \\overline{AK}) \\equal{} \\minus{}\\overline{AM}^2$\r\n\r\nbecause the cross ratio $ \\frac{\\overline{AL}}{\\overline{AK}} \\cdot \\frac{\\overline{XK}}{\\overline{XL}} \\equal{} \\minus{}1$ is harmonic (easily seen by projecting the circle $ (O)$ to a circle and the line $ PQ$ to infinity). Giving $ \\cos \\widehat{QOP}$ is obfuscation, we need $ \\tan \\widehat{QOP} \\equal{} ... \\equal{} 2 \\sqrt{2}.$\r\n\r\n$ \\tan \\widehat{QOP} \\equal{} \\frac{\\tan \\widehat{QOA} \\plus{} \\tan \\widehat{AOP}}{1 \\minus{} \\tan \\widehat{QOA} \\cdot \\tan \\widehat{AOP}} \\equal{} \\frac{\\frac{\\overline{QA}}{\\overline{OA}} \\plus{} \\frac{\\overline{AP}}{\\overline{OA}}}{1 \\minus{} \\frac{\\overline{QA}}{\\overline{OA}} \\cdot \\frac{\\overline{AP}}{\\overline{OA}}} \\equal{}$\r\n\r\n$ \\equal{} \\overline {OA} \\cdot \\frac{\\overline{QA} \\plus{} \\overline{AP}}{\\overline{OA}^2 \\minus{} \\overline{QA} \\cdot \\overline{AP}} \\equal{} \\frac{\\overline{OA} \\cdot \\overline{QP}}{\\overline{OA}^2 \\minus{} \\overline{AM}^2} \\equal{} \\frac{\\overline{OA} \\cdot \\overline{QP}}{\\overline{OM}^2} \\equal{} 2 \\sqrt{2}$\r\n\r\nSubstituting $ \\overline{OM}^2 \\equal{} 1,$ $ \\overline {QP} \\equal{} 2$ yields $ \\overline{OA} \\equal{} \\sqrt{2}$ and $ \\overline{AM}^2 \\equal{} \\overline{OA}^2 \\minus{} \\overline{OM}^2 \\equal{} 1.$ Done. The problem does not ask for this, but from $ \\overline{QA} \\cdot \\overline{AP} \\equal{} \\overline {AM}^2 \\equal{} 1$ and $ \\overline{QA} \\plus{} \\overline{AP} \\equal{} \\overline {QP} \\equal{} 2,$ we get $ \\overline {QA} \\equal{} \\overline {AP} \\equal{} 1,$ points $ P, Q$ are symmetrical with respect to the line $ OA$ and $ L \\in OA.$",
  "Solution_2": "[quote=\"yetti\"]\nbecause the cross ratio $ \\frac {\\overline{AL}}{\\overline{AK}} \\cdot \\frac {\\overline{XK}}{\\overline{XL}} \\equal{} \\minus{} 1$ is harmonic (easily seen by projecting the circle $ (O)$ to a circle and the line $ PQ$ to infinity).[/quote]\r\n\r\ni dun understand the reason for the harmonic division..(i am still a beginner of projective geom), can yo explain a little bit more?",
  "Solution_3": "Suppose you have 2 planes $ \\pi, \\pi',$ not parallel, and a point $ S$ not on either of them. Arbitrary line through $ S$ cuts $ \\pi, \\pi'$ at $ P, P'.$ $ P'$ is image of $ P$ in the central projection of $ \\pi$ to $ \\pi'$ from $ S.$ But lines through $ S$ parallel to $ \\pi'$ do not cut it, even though they cut $ \\pi.$ Collection of all lines parallel to $ \\pi'$ forms a plane $ \\sigma \\parallel \\pi'$ and the plane $ \\sigma$ cuts the original plane $ \\pi$ in a line $ l.$ The image $ l'$ of $ l \\in \\pi$ does not show in $ \\pi'.$ We say that the line $ l \\in \\pi$ was projected to the line at infinity $ l' \\in \\pi'.$ Each line through $ S$ parallel to $ \\pi'$ defines a direction in $ \\sigma$ or in $ \\pi'$ and collection of all these lines is a collection of all directions in $ \\sigma$ or in $ \\pi'.$ Therefore, point at infinity is the same thing as direction in a plane and line at infinity is collection of all directions in a plane. For example, 2 parallels have a common point at infinity - they have the same direction. Any 2 lines intersecting on the line $ l$ projected to infinity become parallel, because their common point is at infinity.\r\n\r\nGenerally, central projection takes a circle to a conic section and it preserves tangencies. I used a well known theorem (with long 3D proof): Given a circle and a line not intersecting this circle, a suitable central projection exists, which takes the given circle to another circle and the given line to infinity. In your problem, the circle $ (O)$ can be projected to another circle $ (O')$ and any line through $ A$ not intersecting the circle $ (O),$ for example the line $ PQ,$ to infinity. Since the image $ A'$ of $ A$ is at infinity, images $ A'M', A'N'$ of the circle tangents $ AM, AN$ and image $ A'K'L'$ of the circle secant $ AKL$ become parallel, $ M'N' \\perp K'L'$ becomes a diameter of the projected circle $ (O').$ The cross ratio of 4 points is preserved in a central projection,\r\n\r\n$ \\frac {\\overline{AL}}{\\overline{AK}} \\cdot \\frac {\\overline{XK}}{\\overline{XL}} \\equal{} \\frac {\\overline{A'L'}}{\\overline{A'K'}} \\cdot \\frac {\\overline{X'K'}}{\\overline{X'L'}} \\equal{} \\frac {\\overline{X'K'}}{\\overline{X'L'}} \\equal{} \\minus{} 1$\r\n\r\n$ \\overline{A'L'}, \\overline{A'K'}$ are infinite segments with the same direction, they cancel out, because their difference is insignificant compared to infinity. $ X'$ becomes the midpoint of $ K'L',$ $ \\overline{X'K'}, \\overline{X'L'}$ are oppositely directed segments of equal length, that's why $ \\minus{} 1.$\r\n__________________________________________\r\n\r\nYou could also use the following theorem: 2 diagonals of a complete quadrilateral cut the 3rd one harmonically. Let $ AB, BC, CD, DA$ be 4 lines forming a complete quadrilateral. $ X \\equiv AB \\cap CD,$ $ Y \\equiv BC \\cap DA,$ $ Z \\equiv AC \\cap BD.$ Lines $ AC, BD, XY$ are the 3 diagonals. How do you prove the theorem ? Project the line $ XY$ to infinity, the image $ A'B'C'D'$ of $ ABCD$ becomes a parallelogram, and use invariance of the cross ratio in a central projection. \r\n\r\nIn your problem, you also have to know that $ MN$ is a polar of the point $ A$ with respect to the circle $ (O).$ Let $ UV$ be another circle secant from $ A.$ Then $ B \\equiv KV \\cap LU,$ $ C \\equiv KU \\cap LV$ lie on the polar $ MN.$ Let $ KV, LU, KU, LV$ are the 4 lines forming a complete quadrilateral, then $ KL, UV, BC$ are the 3 diagonals. The 2 diagonals $ UV, BC \\equiv MN$ cut the 3rd diagonal $ KL$ harmonically at $ A, X,$ so that \r\n\r\n$ \\frac {\\overline{AL}}{\\overline{AK}} \\cdot \\frac {\\overline{XK}}{\\overline{XL}} \\equal{} \\minus{} 1.$\r\n\r\nThe 1st method (projecting the circle $ (O)$ to a circle and the line $ PAQ$ to infinity is much quicker and easier to imagine.",
  "Solution_4": "thank you very much,yetti~:)"
}
{
  "Tag": [],
  "Problem": "Okay firstly please be honest. And don't post srini does not sleep at all and rakesh sleeps only for 2.5 hours every year. Post proper stuff here.\r\n\r\nI used to be a typical late night sleeper. Sleeping after 2:30 and sometimes going on till 3:30. And then most of my morning was spoiled. I used to get up at 8:30 mull around for another hour. go back to sleep and then start productive work only at 1:00. However I wanted to shift my biological clock because this was not really helping me in comp exams and I was feeling sleepy and I started sleeping at 12:00 itself with disastrous results. I still feel sleepy in the morning. :D:D\r\n\r\nSo what time do you goto sleep? And how long do you sleep?",
  "Solution_1": "Used to be from 11:00 to 6:30 earlier abt a month before :D\r\nBut bcos of the influence of my friends n shre in particular, i went to late night and am not able to sleep till 12:00 am :( I try to but still, dont have the inclination!\r\nNowadays its been from 12:30 am to 8:30 am\r\n\r\nPS: Elastiboysai had written he slept from 3:50 to 9:30 in another post :)\r\nAnd ofc for the sake of saying it i shall say:\r\nsrini does not sleep at all and rakesh sleeps only for 2.5 hours every year  :P",
  "Solution_2": "ha well, my sleeping patern is shockingly similar to shreyas's :D\r\nespecially in the last few months, i've been sleeping at around 2- 2.30 and wake up at 1030- 11. Effectively, i start studying properly only at around 1 :(\r\n\r\nand yesterday, i promised myself that from now on, i'd sleep at 12 itself to \"tune my biological clock\" :D \r\nbut it was a DISASTER. i just couldn't get proper sleep and for at least an hour or 2, i was just lying down and yawning. as a result i got up only at 930 today :(\r\n*yawns*",
  "Solution_3": "Baby, why did you post the same thing so many times? :D \r\n\r\nMy sleeping pattern is probably the most erratic but still.. I usually have dinner and go to sleep at around 9 and then get up at 11-11.30, then study/do something that people would tend to mistake for studying till 2.30, watch Dawson's Creek from 2.30 -3 and then go back to sleep and then get up in the morning at 7 - 7.30...\r\n\r\nGreen tea is a very potent stimulant :D",
  "Solution_4": "The mod can edit that :D Vetti mod is always here to the rescue :D\r\nThat was a consequence of the stupid server here which was swaying between various links all by itself without me touching anything :D\r\n\r\nAnd incase u dont believe that it wasnt my fault, i couldnt have posted all that as consecutive posts wont be allowed for less that 10 mins :P",
  "Solution_5": "hey to be frank i go to sleep at around 9 and wake up at 7 am  :P and rohit dont tell i dont sleep and all for sleep is the most important aspect of my life da and one small correction rakesh sleeps for only 1:00:00:02 hrs and please dont exxagerate it to 2.5 hrs and all!!! :D",
  "Solution_6": "Dei srini, just bcos shre said u can write nething u feel is right, doesnt mean u can lie :D\r\n[quote=\"lankies\"]\nYou lie like a brute :P\n[/quote]",
  "Solution_7": "[quote=\"madness\"]Dei srini, just bcos shre said u can write nething u feel is right, doesnt mean u can lie :D\n[quote=\"lankies\"]\nYou lie like a brute :P\n[/quote][/quote]\r\nhey it shows u are offline but how did you post now???",
  "Solution_8": "dei its saying the truth da :D\r\nim offline :P :P",
  "Solution_9": "[quote=\"madness\"]dei its saying the truth da :D\nim offline [/quote]\r\nhey is there any facilities for you to change your active state!!! :?:",
  "Solution_10": "[quote=\"Albert Einstien\"][quote=\"madness\"]dei its saying the truth da \nim offline [/quote]\nhey is there any facilities for you to change your active state!!! [/quote]\r\n\r\nwell there is pathetic fallacy :P :D \r\nyou cant change the active state of a compd atleast :P",
  "Solution_11": "i don't any compromise with my sleep--i sleep at 11.30 and would have woken up at 7.30 if it weren't 4 school\r\ni've to wake at 6:15 but i do take a nice nap at school :D  :D",
  "Solution_12": "Now that u all r talkin abt it,\r\ni used 2 sleep frm 11.30 to 6.30-7.00 normally :wink: \r\nfor the past 1.5 months my sleep pattern has drastically changed.\r\nSomthin like 3.30 ,4 to 9.40 10.\r\nFunnily  I too tried to sleep early yest- 11.40, it worked bt i woke up only at 9. :rotfl:",
  "Solution_13": "I'm scared I'll sleep off during JEE paper I. :( My full momentum is usually gained only after 3. :(",
  "Solution_14": "go to sleep da now :)",
  "Solution_15": "I am scared abt paper 2 :oops: \r\nI really slept off during FT5 paper 2 (thats bcoz i had eaten 2 much n was ready 2 burst) :rotfl:",
  "Solution_16": "oh my god u all study till 3 and all ah ???how is it possible ??? u all kattans!!!!!!!!!!! :o  :o  :o",
  "Solution_17": "Look who is talking. Thats all I have to say",
  "Solution_18": "[hide=\"TO SRINI AND ROHIT:\"]\nI DON'T SLEEP AT ALL DA :P  :P \n[/hide]\r\n\r\nAnyway the truth is I sleep around 12:00,12:30 and ofc get up only at 8:30 in the morning. It has been my routine for about 2 months now. :rotfl:",
  "Solution_19": "I generally sleep at around 12- 1230 and get up between 6:30 to 7:00 .\r\n\r\nvery few times i have tried 11 - 4:45 . \r\n\r\nBoth have equal productivity",
  "Solution_20": "Shifted to my old self. Sleeping only after 2.  :rotfl:  :rotfl:",
  "Solution_21": "u r damaging ur pineal gland\r\n\r\ni think soumya is practising to prevent jetlags :rotfl:  :rotfl:"
}
{
  "Tag": [
    "USAMTS",
    "LaTeX"
  ],
  "Problem": "I was looking through the USAMTS webiste, and I noticed it said that one must send both the pdf and the latex file if using LaTeX. I didn't do this when I submitted it, does that mean my work will get disqualified?",
  "Solution_1": "No, it won't be disqualified.  If we don't have the TeX code, then we probably won't publish your solution.  But your paper will still be graded."
}
{
  "Tag": [],
  "Problem": "Show that there are an infinity of natural numbers wich cannot be written as a^3+b^3 with a,b naturals but can be written in the same form with a,b rationals(positive).",
  "Solution_1": "However any natural number gives residue 0,1,8 mod9\r\nthe number 9k+3 can't be written as a^3+b^3 with a,b naturals.\r\nNow let a=9k^3-1,b=9k^3+1,then n=(a^3+b^3)/9k^3 is natural and gives residue 3 mod9,that we had to prove."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "A fair, 6-sided die is thrown and a card is drawn from a 52-card, standard deck. Find the probability that a multiple of 3 appears on the die and a face card (Jack, Queen, King, or Ace) is drawn.",
  "Solution_1": "[hide]There are two multiples of 3 on a die, so the probability is $\\frac12.$\n\nThere are 12 face cards, so the probability is $\\frac{12}{52},$ or $\\frac{3}{13}.$\n\n$\\frac12\\cdot\\frac3{13}=\\boxed{\\frac3{26}}$[/hide]",
  "Solution_2": "[quote=\"i_like_pie\"][hide]There are two multiples of 3 on a die, so the probability is $\\frac12.$\n\nThere are 12 face cards, so the probability is $\\frac{12}{52},$ or $\\frac{3}{13}.$\n\n$\\frac12\\cdot\\frac3{13}=\\boxed{\\frac3{26}}$[/hide][/quote]\r\n\r\nwait... wouldn't the probability of the die thing be $\\frac13.$, and the face card one be $\\frac{16}{52},$ or $\\frac{4}{13}.$\r\nso then it would be $\\frac13\\cdot\\frac4{13}=\\boxed{\\frac4{39}}$\r\nam I wrong???...",
  "Solution_3": "xxxxx, It would not be 4/39 because you plus the two probabilities together, not times.",
  "Solution_4": "[quote=\"Necrowarrio0\"]xxxxx, It would not be 4/39 because you plus the two probabilities together, not times.[/quote]\r\nYou multiply, and xxxxx is correct. The correct answer is ${4\\over39}.$",
  "Solution_5": "Oh is it multiply? I thought it was plus.....  :|\r\n\r\nThe above was edited.\r\n\r\nAH SORRY!!!! IT IS MULTIPLY....................  :(",
  "Solution_6": "In probablity \"and\" usually tells you to use multiplication\r\nwhile \"or\" usually tells you to use addition\r\n\r\nAren't face cards just jack, queen, and king?  :P \r\n\r\nI guess it doesn't matter because the question tells you that it includes ace as a face card"
}
{
  "Tag": [
    "abstract algebra",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "for $S= \\{m/n \\in Q| n Is Odd  \\}$, under usual operations on real numbers. I must determine if this is a ring.  Is it sufficient to show that $S$ satisfies the ring axioms? ie: $(m_1/n_1) + (m_2/n_2) = (m_2/n_2) + (m_1/n_1)$ so $S$ is abelian under addition, then show Closure, Associativity, and distributivity",
  "Solution_1": "[quote=\"zargon\"] I must determine if this is a ring.  Is it sufficient to show that $S$ satisfies the ring axioms?[/quote]\r\n\r\nWell if an object satisfies all the properties in the definitition of a class, then it belongs to the class. Basic logic.",
  "Solution_2": "The least obvious thing to check is closure.\r\nTake $a/b,c/d\\in S$. Then $b,d$ are odd numbers. $a/b+c/d=(ad+bc)/bd$. This may not be a reduced form for $a/b+c/d$, but since $bd$ is odd, no matter by what you divide it, you still get an odd number, so $a/b+c/d\\in S$."
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "In order that $998680883748524N5070273447265625$ equal $1995^{10}$, what is N?",
  "Solution_1": "[hide]\n$1995=3\\cdot 5\\cdot 7\\cdot 19$, so $1995^{10}$ is a multiple of $9$, which means the sum of all the digits must be divisible by $9$.\n\nAdding all the digits, we get $154\\equiv 1\\pmod{9}$, so $N=8$.\n(I hope I added them correctly... :roll: )[/hide]",
  "Solution_2": "[quote=\"frt\"][hide]\n$1995=3\\cdot 5\\cdot 7\\cdot 19$, so $1995^{10}$ is a multiple of $9$, which means the sum of all the digits must be divisible by $9$.\n\nAdding all the digits, we get $154\\equiv 1\\pmod{9}$, so $N=8$.\n(I hope I added them correctly... :roll: )[/hide][/quote]\r\n\r\nI was so intimidated by the number that I didn't think to use that simple method. \r\n\r\n(and you added correctly according to my super [2 dollar] calculator)",
  "Solution_3": "[quote=\"frt\"]\n$1995=3\\cdot 5\\cdot 7\\cdot 19$, so $1995^{10}$ is a multiple of $9$.[/quote]You mean it's a multiple of 3. ;)",
  "Solution_4": "[quote=\"4everwise\"][quote=\"frt\"]\n$1995=3\\cdot 5\\cdot 7\\cdot 19$, so $1995^{10}$ is a multiple of $9$.[/quote]You mean it's a multiple of 3. ;)[/quote]\r\n\r\nWell, since 1995 has one multiple of 3, then $1995^{10}$ has a factor of $3^{10}$ and $3^2=9$.",
  "Solution_5": "Right. I'm losing it today, lol.  :blush:  :rotfl:"
}
{
  "Tag": [],
  "Problem": "$ P$ and $ Q$ are each a distinct member of the set $ \\{6, \\minus{}\\frac 12, \\minus{}3, \\frac 13\\}$. What is the least possible value of the quotient $ P\\div Q$?",
  "Solution_1": "To minimize first we try to make $ P\\div Q$ negative.  \r\n\r\nIf $ P$ is positive and $ Q$ is negative, then the minimum possible value of $ P\\div Q$ is $ \\frac {6}{ \\minus{} \\frac {1}{2}} \\equal{} \\minus{} 12$.  \r\n\r\nIf $ P$ is negative and $ Q$ is positive, then the minimum is $ \\frac { \\minus{} 3}{\\frac {1}{3}} \\equal{} \\minus{} 9$.  $ \\minus{} 12 < \\minus{} 9$, so $ \\boxed{ \\minus{} 12}$"
}
{
  "Tag": [
    "inequalities",
    "function",
    "algebra solved",
    "algebra"
  ],
  "Problem": "a,b,c,d>0 Prove that:\r\n\r\na/(b+2c+3d)+ b/(c+2d+3a)+ c/(d+2a+3b)+ d/(a+2b+3c)  \\geq 2/3",
  "Solution_1": "you can consider wlog that \r\n\r\na+b+c+d=1 (there is a post here that explains why this is a \"WLOG\" consideration)\r\nthen you use Jensen for  the convex function f : (0,+oo)-->R, f(x)=1/x with wieghts a,b,c,d.\r\n\r\ncheers!",
  "Solution_2": "I think Cauchy also works. (All sums are cyclic.) \r\nLet L denote the left - hand side. \r\nPut A = a + b + c + d and B = ab + ac + ad + bc + bd + cd.\r\nThen we get L *  \\suma(b + 2c + 3d) \\geq (a + b + c + d).\r\nNow \\suma(b + 2c + 3d) = 4B.\r\nSo we have L \\geq (a + b + c + d)/(4B).\r\nSo we want to prove that (a + b + c + d)/(4B) \\geq 2/3.\r\nThis is equivalent to 3A + 6B \\geq 8B or 3A \\geq 2B.\r\nNow we have\r\n1) A = a + b + c + d \\geq ab + bc + cd + da\r\n2) A = a + c + b + d \\geq ac + bc + bd + da\r\n3) A = b + a + c + d \\geq ab + ac + cd + bd\r\nand summing these gives indeed 3A \\geq 2B.\r\n\r\nPS. Nice solution Lagrangia !",
  "Solution_3": "This is from a shortlist isn't it? I think I've seen it before.",
  "Solution_4": "Yeah, I guess \"IS3\" means IMO 1993 Shortlist.",
  "Solution_5": "Here is my solution:\r\nwe have :\r\n\r\na/(b+2c+3d)+b/(c+2d+3a)+c/(d+2a+3b)+ d/(a+2b+3c)= a <sup>2</sup> /(ab+2ac+3ad) +b <sup>2</sup> /(bc+2bd+3ab)+ c <sup>2</sup> /(cd+2ac+3bc)+ d <sup>2</sup> /(da+2bd+3cd)  \\geq (a+b+c+d) <sup>2</sup> /4(ab+bc+cd+da+ac+bd)\r\nSo now we have:\r\n(a+b+c+d) <sup>2</sup> /4(ab+bc+cd+da+ac+bd)  \\geq 2/3\r\n3* \\sum a <sup>2</sup> + 6* \\sum ab  \\geq  8* \\sum ab\r\n3* \\sum a <sup>2</sup>  \\geq 2 *\\sum ab\r\n(a-b) <sup>2</sup> + (b-c) <sup>2</sup> + (c-d) <sup>2</sup> + (d-a) <sup>2</sup> + (a-c) <sup>2</sup> + (b-d) <sup>2</sup>   \\geq 0"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a_0, a_1, a_2,..., a_n $ be a real numbers such that:\r\n$ (i) 0=a_0 \\le a_1 \\le...\\le a_n$\r\n$(ii) a_i_+_1  - a_i \\le \\frac{1}{n}$ for  $ i=0,1,...,n-1 $\r\nProve that:\r\n      $ n(a_1^3+...+a_n^3) \\le (a_1+...+a_n)^2 $",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=33412",
  "Solution_2": "Think's Siuhochung"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "function",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ p$ be a prime. Here $ a_p$ is the number of irreducible polynomails of degree $ n$ in $ \\mathbb F_p[X]$.\r\n\r\n$ a_p(n) \\equal{} \\frac{1}{n} \\sum_{d|n}\\mu(d)p^{n/d}.$\r\nhere $ \\mu$ is the mobius function.\r\n\r\nHow can I show that $ a_p(n)$ is positive for all $ n$ and also when $ p\\geq 2$ is not a prime number.",
  "Solution_1": "It's a multiplicative function, and you can easily check that it's not zero when $ n$ is a prime power.\r\nSee [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=30473]here[/url].",
  "Solution_2": "but my question is different. I want to show that this is positive? how can I do that?",
  "Solution_3": "Well, you can just construct an irreducible polynomial in every field. \r\n[quote=\"peteryellow\"]Let $ p$ be a prime. How can I show that is positive for all and also when $ p$ is not a prime number.[/quote]\r\nDon't use same letters in different meanings. Often denotes $ \\mathbb{F}_p[X]$ the ring of polynomials over a field. In this case $ p$ is a power of prime. There are different rings with $ p$ elements if $ p$ is composite.",
  "Solution_4": "but can somebody give a hint of how to  show it.",
  "Solution_5": "One way is to show the existence of the field $ \\mathbb{F}_{p^n}$ as the splitting field of $ x^{p^n}\\minus{}x$. Then since the multiplicative group of a field is cyclic, there must be some generator $ a$, so $ a$ is a primitive element, and thus the minimal polynomial of $ a$ over $ \\mathbb{F}_p$ must be an irreducible polynomial of degree $ n$.\r\n\r\nAlternatively, you can show directly that that formula gives a positive number. I clicked jmerry's link, but I'm slightly confused why you can do that, since $ f(k) \\equal{} p^k$ is not a multiplicative function.",
  "Solution_6": "[quote=\"peteryellow\"]but my question is different. I want to show that this is positive? how can I do that?[/quote]\r\nIf you already know that it counts something, then it cannot be negative; so only 0 must be excluded,",
  "Solution_7": "[quote=\"calc rulz\"]I clicked jmerry's link, but I'm slightly confused why you can do that, since  is not a multiplicative function.[/quote]\r\nYou're right, and I made a mistake the first time around. Two quick fixes:\r\n(1): $ \\sum_{d|n}\\mu(d)q^{n/d}\\ge q^n\\minus{}q^{n\\minus{}1}\\minus{}q^{n\\minus{}2}\\minus{}\\cdots\\minus{}1\\equal{}q^n\\minus{}\\frac{q^n\\minus{}1}{q\\minus{}1}\\ge 1$\r\n(2): If $ D$ is the largest squarefree divisor, the sum is equal to $ \\mu(d)q^{n/D}$ mod $ q^{n/D\\plus{}1}$, and is not zero. Since it counts something, it's positive.",
  "Solution_8": "how do you get that:\r\n\r\n$ \\sum_{d|n}\\mu(d)q^{n/d}\\ge q^n\\minus{}q^{n\\minus{}1}\\minus{}q^{n\\minus{}2}\\minus{}\\cdots\\minus{}1$",
  "Solution_9": "$ \\mu(k)\\ge \\minus{}1$ for all $ k$, and subtracting the terms that don't match to divisors makes it smaller than not subtracting them. In fact, we never have equality, and I don't need that last $ \\minus{}1$ term. It comes closest for $ n\\equal{}2$, with a value of $ q^2\\minus{}q$ in that case.",
  "Solution_10": "can this result be related in any way to the number of irreducible polynomials with coefficients in the set $ {0,1,2,...,9}$ ?",
  "Solution_11": "No, but it can be related to something close.  $ a_p(n)$, for arbitrary $ p$, also counts the number of Lyndon words on $ p$ letters of length $ n$, which when $ p$ is prime can be put in bijection with irreducible polynomials; see the definition and proof [url=http://qchu.wordpress.com/2009/11/03/the-cyclotomic-identity-and-lyndon-words/]here[/url].\r\n\r\nTo prove that $ a_p(n)$ is always positive when $ p \\ge 2$, it suffices to show that Lyndon words of every length exist.  But this is clear; if $ a < b$ are two letters in the alphabet, then $ aaa...b$ is always a Lyndon word.  (This is actually not enough to prove that irreducible polynomials of every degree exist over $ \\mathbb{F}_p$; the bijection between the two requires the existence of finite extensions of every degree.)",
  "Solution_12": "[quote=\"t0rajir0u\"]No, but it can be related to something close.  $ a_p(n)$, for arbitrary $ p$, also counts the number of Lyndon words on $ p$ letters of length $ n$, which when $ p$ is prime can be put in bijection with irreducible polynomials; see the definition and proof [url=http://qchu.wordpress.com/2009/11/03/the-cyclotomic-identity-and-lyndon-words/]here[/url].\n\nTo prove that $ a_p(n)$ is always positive when $ p \\ge 2$, it suffices to show that Lyndon words of every length exist.  But this is clear; if $ a < b$ are two letters in the alphabet, then $ aaa...b$ is always a Lyndon word.  (This is actually not enough to prove that irreducible polynomials of every degree exist over $ \\mathbb{F}_p$; the bijection between the two requires the existence of finite extensions of every degree.)[/quote]\r\n\r\nyes but can I compute the number of irreducibles with coeffs in 0..9 using $ a_p(n)$ ?",
  "Solution_13": "No; at best, you can take $ p$ to be a prime power.  The problem is that these are the only values of $ p$ for which there exists a finite field of cardinality $ p$.  There are two other interpretations of \"irreducible polynomial with coefficients in $ \\{ 0, ... p\\minus{}1 \\}$\" when $ p$ is not a prime power: irreducible in $ \\mathbb{Z}[x]$ or irreducible in $ \\mathbb{Z}/p\\mathbb{Z}$, and neither of them are counted by this expression."
}
{
  "Tag": [
    "Gamebot"
  ],
  "Problem": "Ellen baked $ 2$ dozen cupcakes of which half contained chocolate, two-thirds contained raisins, one-fourth contained chocolate chips, and one-sixth contained nuts.  What is the largest possible number of cupcakes that had none of these ingredients?",
  "Solution_1": "Since $ \\frac23$ have raisins, $ \\frac23\\times{24}\\equal{}16$ have raisins. Each of the other ingredients can be mixed in with any of these cookies, so the maximum is $ 24\\minus{}16\\equal{}\\boxed{8}$.",
  "Solution_2": "This is one of those \"trick\" problems that love to show up on math contests. Before you plunge right in and start drawing a four-circle Venn diagram, you need to stop, take a step back, take a deep breath, and then re-read the problem. \n\nI approached this problem in the same way as gaussintraining.\n\nIn the end, we have 16 cupcakes with raisins and maybe or maybe not something else, and therefore 8 cupcakes with none of the ingredients. We don't have anywhere near enough information to fill in a four-circle Venn diagram. For one, it's entirely possible that all 16 cupcakes have raisins paired with at least one other ingredient, and it's also possible that 4 of those cupcakes are raisin only. Don't even try to wander into that quagmire; you'll never be seen again! ;).\n\nTim",
  "Solution_3": "Tim is right, bust I really think that this shouldn't be a level 19 problem...it even gives 100+ points.",
  "Solution_4": "[hide]I had a quick way, the largest venn diagram is 16, to make outside portiion the largest, we can put all other venn diagrams inside the largest one, like 4,6,12 all goes inside the 16, that way, the outside will be the largest which has none of these 4 ingredients at all, so 24-16=[b]8![/b][/hide]",
  "Solution_5": "Was it really necessary to revive a four-year old thread with a solution?",
  "Solution_6": "[quote=no_name]Was it really necessary to revive a four-year old thread with a solution?[/quote]\n\nYes, it is okay to revive alcumus threads if you are actually posting something relevant to the thread.",
  "Solution_7": "And it's not 8! It's 8",
  "Solution_8": "I think he meant that the answer was 8 with an exclamation mark behind it because he was excited or something.",
  "Solution_9": " :( could someone explain this. I don't get it to well. ",
  "Solution_10": "[hide=Solution]Because we are trying to find the maximum number of plain cupcakes Ellen could have made, we can assume that she will try and mix fillings whenever possible. Because the largest fraction of cupcakes ($2/3$) contain raisins, if she tries to mix fillings, all of the cupcakes that contain chocolate will contain raisins, too. This also applies to all of the other fillings, so we can ignore those and just focus on the raisins. $2/3$ of the cupcakes contain raisins, so $1/3$ don't. $12 * 1/3= \\boxed{4}$ [/hide]",
  "Solution_11": "Let me know if that solution is too wordy.",
  "Solution_12": "[quote=elevate]Tim is right, bust I really think that this shouldn't be a level 19 problem...it even gives 100+ points.[/quote]\n\nit is a level 23 problem",
  "Solution_13": "[quote=HWGelber]Let me know if that solution is too wordy.[/quote]\n\n(You're probably not going to see this, but..)\n\nYour solution is not wordy, but you said $1$ dozen cupcakes instead of $2$. The answer should be $8$. (But you were on the right track  :) )\n\n",
  "Solution_14": "[quote name=\"jcl-12\" url=\"/community/p15744495\"]\n[quote=elevate]Tim is right, bust I really think that this shouldn't be a level 19 problem...it even gives 100+ points.[/quote]\n\nit is a level 23 problem\n[/quote]\n\nhi, there's no point of responding when they probably haven't been on the site for 8-9 years\n\nalso alcumus was probably different then with different levels and stuff",
  "Solution_15": "Should I quit math if I got this one wrong after spending an hour on it? Seems like I'm not cut out for this.",
  "Solution_16": "[quote=arguewithplato]Should I quit math if I got this one wrong after spending an hour on it? Seems like I'm not cut out for this.[/quote]\nYou can't avoid maths. It will crop up wherever you go; when you go to the shops or become a scientists, maths will crop up [b]everywhere[/b] you go. It's in your genetics (literally). \n\ntldr: no.",
  "Solution_17": "Hi, \nI believe there\u2019s another way to do it, but it doesn\u2019t have anything to do with counting and probability.\nI might be wrong.\n\nLet the total cupcakes be the area of a square with side a, which is a^2\nSo the ones with other ingredients in it be squares or rectangles.\n(You could draw a square to start then cut that square out to get pieces that represent others, like the chocolate is half of that square, to get the nuts first cut the square a half then cut that half into three pieces. But note that all of this is to get the visual pieces of each category not taking pieces out of the square a^2 piece)\nThen laying all the pieces out \n(a x a) - all the cupcakes,\n(a x a/2) - chocolate,\n(a x (2/3)a) - raisins,\n(a/2 x a/2) -  chocolate chips, \n(a/3 x a/2) - nuts, \n\nThe question ask what is the largest possible number of cupcakes that have none of these ingredients is the same as how to crunch all the pieces together so that when fit it in the big piece, the a^2 square, it left the largest area. To get that we crunch all the smaller pieces into that biggest piece in all those smaller piece (a x (2/3)a).\nSo the largest possible space left is a^2 - (2/3)a^2 = (1/3)a^2 \nAnd a^2 = 24 so the space is 24/3 = 8\n"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $x,y,z$ be positive reals such that $4xyz = x+y+z+1$.  Prove that\r\n\r\n\\[(x+1)(y+1)(z+1) \\geq 8.\\]",
  "Solution_1": "Note that $x+y+z\\geq 3(xyz)^{1/3}$. Let define $a:=(xyz)^{1/3}$, then by assumption we have\r\n\\[\r\n(a-1)(4a^2+4a+1)=4a^3-3a-1=4xyz-3(xyz)^{1/3}-1\\geq 0\r\n\\]\r\nIt follows that $a^n\\geq a\\geq 1$ for every $n$. Now \r\n\\[\r\n(x+1)(y+1)(z+1)=xyz+(xy+xz+yz)+4xyz\\geq 5xyz+3(xyz)^{2/3}=5a^3+3a^2\\geq 8.\r\n\\]",
  "Solution_2": "Cool.  My (contrived) solution uses the substitution $a=\\frac{1}{2x+1}$, etc so our condition (after some algebra) rewrites as $a+b+c=1$.  Our inequality becomes\r\n\\[\\left(1+\\frac{1}{a}\\right)\\left(1+\\frac{1}{b}\\right)\\left(1+\\frac{1}{c}\\right) \\geq 64.\\]\r\nHowever, this follows easily by homogenizing and using AM-GM.",
  "Solution_3": "[quote=\"ThAzN1\"]Cool.  My (contrived) solution uses the substitution $a=\\frac{1}{2x+1}$, etc so our condition (after some algebra) rewrites as $a+b+c=1$.  Our inequality becomes\n\\[\\left(1+\\frac{1}{a}\\right)\\left(1+\\frac{1}{b}\\right)\\left(1+\\frac{1}{c}\\right) \\geq 64.\\]\nHowever, this follows easily by homogenizing and using AM-GM.[/quote]\r\nhuygens and amgm works as well.",
  "Solution_4": "[quote=\"ThAzN1\"]Cool.  My (contrived) solution uses the substitution $a=\\frac{1}{2x+1}$, etc so our condition (after some algebra) rewrites as $a+b+c=1$.  Our inequality becomes\n\\[\\left(1+\\frac{1}{a}\\right)\\left(1+\\frac{1}{b}\\right)\\left(1+\\frac{1}{c}\\right) \\geq 64.\\]\nHowever, this follows easily by homogenizing and using AM-GM.[/quote]\r\n\r\nHi ThazN1, this problem seem closely related to yours, check if your method also working for this problem\r\n\r\nhttp://www.mathlinks.ro/Forum/topic-41928.html\r\n\r\nLet me know it it is working. Thanks"
}
{
  "Tag": [],
  "Problem": "It is given that $x$ and $y$ are positive integers and $3x^{2}+x=4y^{4}+y$. Show that: $x-y$, $3x+3y+1$ and $4x+4y+1$ are squares of integers.",
  "Solution_1": "3[(x^2)-(y^2)]+(x+y)=y^2\r\n(x-y)(3(x-y)+6y+1)=y^2\r\nif d=(x-y,3(x-y)+6y+1) if d>1, an p is prime number.\r\np devides d and p devides (x-y) and p devides y^2 so p devides y.\r\nso p devides 3(x-y)+6y and we know p devides d so p devides 3(x-y)+6y+1 so p=1 but it is impossible so d=1 .\r\nwe know [3(x-y)+6y+1)].[(x-y)]=a^2 so (x-y)=m^2.\r\nother questions is same .",
  "Solution_2": "[quote]3[(x^2)-(y^2)]+(x+y)=y^2 \n[/quote]\r\nIt is $y^{4}$, but not $y^{2}$"
}
{
  "Tag": [
    "Princeton",
    "college",
    "search",
    "articles",
    "real analysis",
    "Duke",
    "geometric series"
  ],
  "Problem": "UP Math prof proves Princeton man wrong...\r\n\r\nhttp://www.manilatimes.net/national/2005/may/05/yehey/top_stories/20050505top4.html\r\n\r\nI'm not too sure what to believe... I know the website looks bogus, but if you search for it online, there are some other pages with the same story.\r\n\r\nWhat do you guys think?",
  "Solution_1": "*i* coulda told you that!\r\n\r\n-peter ruse\r\n-kamaldeep gandhi\r\n-ricky sharma\r\n-rathanak chuon",
  "Solution_2": "Dunno...that letter by Wiles sure sounds fake...",
  "Solution_3": "Listen ... false axioms?  Gimme a break.  This guy sounds just like Doron Shamdi (perhaps that is misspelled, but you can find his posts in Other Problem Solving Topics yourself) who has decided that mathematics is false.  I mean, fine, you don't like the axioms, but that doesn't mean that your own axioms are [i]truer[/i], just that they are an alternative choice.  This is a fairly classic example of a crank and you shouldn't worry about it.",
  "Solution_4": "If it would comfort you that modern mathematics will remain intact despite his antics, I could go through one of his articles (such as the editorial in the Manila Times found [url=http://www.manilatimes.net/national/2004/jun/18/yehey/opinion/20040618opi7.html]here[/url]) and explain why the various things he says are either nonsense or at least don't demonstrate the kind of thing he's trying to show.  Or you could just take my word for it :)",
  "Solution_5": "lol.  Everyone go to the link that JOel sent you.  There are about 10000000 ways to prove 1 = 0.9999999...\r\n\r\nyou could use the standard let x = 0.999999, then 10x = 9.9999999, etc.  \r\n\r\nor you could use a1=0.9, a2=0.09, etc. and use sum of an infinite geometric series and more\r\n\r\nWow, thank you Joel.  Thou hast saved us all.",
  "Solution_6": "[quote=\"MithsApprentice\"]Dunno...that letter by Wiles sure sounds fake...[/quote]\r\n\r\nYes it did....it sounded very \"unprofessional\"",
  "Solution_7": "eww... this guy needs some common sense!",
  "Solution_8": "a) that guy is a quack (he's a fairly regular poster on sci.math)\r\n\r\nb) that newspaper should be completely ashamed of itself, but certain ethnic groups tend to take excessive pride in the 'accomplishments' of members of said ethnic group.  A publication with higher standards, however, would have laughed that article away.",
  "Solution_9": "haha he says that average of two numbers must lie right in between them and then states that when you average 0.99999.... and 1 you get 0.99999..... which is a contradiction since it doesn't lie in between\r\n\r\nyet he just proved exactly what he tried to disprove: that 0.99.... =1 since two same numbers averaged together will give you that number itself",
  "Solution_10": "[quote]The most advanced computer today can only deal with numbers having no more that 11 digits.[/quote]\r\n\r\nSomehow, I doubt this...\r\n\r\nBut anyway, yeah, his \"proof\" of the nonsense of, well, mathematics is crazy. This guy needs to go to an asylum... for math. Why can't he just accept the beauty and logic of it all like the rest of us? :D",
  "Solution_11": "Actually, blah^3, he was the [i]editor[/i] of the math and science portion of that paper, if I read correctly.  Which explains in part why he managed to get so much space from them for his silliness.",
  "Solution_12": "Well, I guess every paper has its employees that they would like swept under the carpet (although it seems that they don't even know the magnitude of the absurdities there)\r\n\r\nThe thing that I find really ironic is that our Edgar Escultura got his Ph.D the same year, and from the same class, as Walter Rudin, the famed author of those difficult-to-the-point-of-unreadable analysis texts that every undergrad runs into sooner or later.",
  "Solution_13": "[quote=\"blahblahblah\"]The thing that I find really ironic is that our Edgar Escultura got his Ph.D the same year, and from the same class, as Walter Rudin, the famed author of those difficult-to-the-point-of-unreadable analysis texts that every undergrad runs into sooner or later.[/quote]\r\n\r\nI'm not sure what you mean, but Walter Rudin got his PhD from Duke in 1949. He was a professor at UW-Madison in the 70s, but I don't think he was Escultura's advisor. Escultura's advisor appears to have been Laurence Young - who was a very famous and well respected mathematician. I can imagine he would roll over in his grave if he could see what his former student is doing today.",
  "Solution_14": "Yeah, somehow the list of Ph.D candidates and their advisors blurred together in my head :| .  I know this because I did a google search for escultura's name and Rudin's name jumped out at me on the same webpage, and I must have just mentally associated them as getting their Ph.D's at the same time.",
  "Solution_15": "I have to admit that the complete article seems to be fake. Remember that sometimes this articles are written for jurnalists that seem not to understand what happened, even when it seems not to be the case. I remember that once I read an article about a girl, 22 in Sweden that had proven the Riemann Hypothesis, and at the end it was that she had earned a price for some research in something related (as many prices that are given to young mathematicians) and the jurnalist had listened about the Clay institute price the same day and mixed the information. It was not too long ago, three years or less. Any way, I would not trust too much such article!",
  "Solution_16": "these quacks are quack quacks.  (yes indeed!)  how?  they don't know the first thing about good quacking or *where* to start quacking.  there's nothing wrong with BS'ing if you do it skillfully, in which case it's not really so much BS anymore, but stuff actually worthy of attention.  i mean, there's Godel, who allows us to BS on certain levels.  but these superficial ones!  it's gotta be more polyphonic than that, oh hotdammit!\r\n\r\n-petee outta the blue",
  "Solution_17": "The damage has already been done:\r\n\r\n[url=http://houseonahill.net/index.php/blog/wiles-escultura-and-pierre-de-fermats-equation/]houseonahill.net/index.php/blog/wiles-escultura-and-pierre-de-fermats-equation/[/url]\r\n\r\n[quote]No wonder I couldn\u2019t understand algebra.[/quote]\r\n\r\nEdit:  The first time I read Wiles' letter, I didn't get that he was being sarcastic.  Woops.  I wouldn't have guessed him for it.  That was harsh: \"Also I\u2019d like to have the address of the guy who let you get a PhD 30 years ago. I\u2019d like to discuss few things with him. . . \"",
  "Solution_18": "I doubt wiles wrote the letter.",
  "Solution_19": "So then I was right not to have guessed him for it.",
  "Solution_20": "i highly doubt that the letter is from Wiles, it is totaly unprofessional..., it just doesnt make sense\r\non the other hand, i would be very happy to see one of the counter examples, that they have proposed... ;)",
  "Solution_21": "Well...if Wiles's proof is true, and as ameteurs, we certainly have no right to assert that it's faulty, then there are no real counterexamples.",
  "Solution_22": "My assumption is that it's phony"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "find all [b]x,y[/b] in integer number:\r\n [b]xy[/b] divisible by [b]x+y[/b].",
  "Solution_1": "$x=abc$, $y=ab(b-c)$. Right?",
  "Solution_2": "How do you think he can answer you if it is an unsolved problem for him? :D  :D  :D \r\n\r\nPlease post complete solution, otherwise this one will remain here for years before going into the solved section.\r\n\r\nPierre.",
  "Solution_3": "Let $d=(x,y)$, $x=du$, $y=dv$, then $u+v\\mid duv$, but $(u,u+v)=(v,u+v)=1$, so $u+v\\mid d$, i.e. $d=ab$, $u+v=b$. Denote $c=u$.\r\nWe see $x=abc$, $y=ab(b-c)$.\r\n\r\nThat's all."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "AwesomeMath",
    "summer program"
  ],
  "Problem": "Is a person who scored a 132 on AMC 10 and 7 on AIME (me) this year good enough to go to this camp? I just want a rough idea...",
  "Solution_1": "i think grade might matter..\r\nanyway you shouldnt have a problem getting in\r\nthey accept way worse people\r\n(except if youre a senior or something)\r\nlast year i only got slightly higher and i got in without testing",
  "Solution_2": "So AwesomeMath isn't like ridiculously high level? IF I do get in, I'm not sure if I can learn with the rest of the people there.",
  "Solution_3": "There are classes of varying difficulties. You can always find things more difficult if you ask. I think that the easier classes should be manageable for students who did well at mathcounts."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There is a n*n matrix filled with 0 and 1.We can do the following operation:\r\nIf there is more than m (m<=n) 1 in a row (or a column), then we can change all the zeros in this row (or column) into 1.  \r\nProve that:\r\nIf there is at most (m^2-1)  1s in the matrix at the very beginning then we can never change the matrix into all 1s.\r\n(Here m^2-1 is the best number)",
  "Solution_1": "Just a first observation: $ m^2\\minus{}1$ is not the\"best number\" (also, I think you mean if there are at least $ m$ ones in a row or column we can change it into all ones). For a simple example, take $ n\\equal{}10$ and $ m\\equal{}2$. Clearly $ m^2\\minus{}1 \\equal{} 3$ ones are not enough, while $ m^2 \\equal{} 4$ are enough if we position them in a $ 2 \\times 2$ subsquare. But if we take the diagonal of the square to be all ones ($ 10$ in all), we can still not make even one move. We may need to rephrase this as $ m^2\\minus{}1$ is the \"best\" in the sense that, for more, [b]there exists[/b] a configuration that allows the square to be filled with ones, but that is trivial, since a $ m \\times m$ subsquare of ones does the job.",
  "Solution_2": "Dear mavropnevma\r\nWhat you say is right.Here the best means that there exists a condition such that it can be turned into all 1s\r\nFor all number less than m^2-1,one can never turn it into all 1s.\r\nDid you find the solution to this?\r\nI think I got it. :D",
  "Solution_3": "This is one of those delightful cases when one needs to prove a stronger statement in order the get the required result as a particular case, due to the easier, unrestricted manipulation allowed by the general case. This general context therefore will be \r\n\r\n[color=blue]Given integers $ h \\geq c \\geq 1$, $ w \\geq r \\geq 1$, a rectangular $ h \\times w$ array, containing $ 0 \\leq N \\leq rc$ entries equal to $ 1$, and the rule that anytime a row contains at least $ r$ ones it may be filled with ones, while anytime a column contains at least $ c$ ones it may be filed with ones, then the only case that allows the full array to end up with all entries equal to $ 1$ is $ N \\equal{} rc$ (for some special configurations).[/color]\r\n\r\nDenote by $ x$ the number of those rows containing each at least $ r$ ones, and by $ R$ the set of these entries, while we denote by $ y$ the number of those columns containing each at least $ c$ ones, and by $ C$ the set of these entries. Then $ N \\geq | R \\cup C | \\equal{} | R | \\plus{} | C | \\minus{} | R \\cap C | \\geq xr \\plus{} yc \\minus{} xy$. It is easily seen that the problem now reduces to filling with ones a $ (h' \\equal{} h\\minus{}x) \\times (w' \\equal{} w\\minus{}y)$ array, containing $ 0 \\leq N' \\equal{} N \\minus{} | R \\cup C | \\leq rc \\minus{} (xr \\plus{} yc \\minus{} xy) \\equal{} (r\\minus{}y)(c\\minus{}x) \\equal{} r'c'$ entries equal to $ 1$. When $ x \\equal{} y \\equal{} 0$ there is status-quo, therefore no move can be made, and the array remains unfilled, while otherwise the problem reduces in steps to the base case of $ N\\equal{}0$, which is trivial. \r\n\r\nThe particular case is for $ h\\equal{}w\\equal{}n$ and $ r\\equal{}c\\equal{}m$; when $ 0\\leq N \\leq m^2\\minus{}1$ the array cannot therefore be filled up, while when $ N\\equal{} m^2$ the array may be possibly filled up, for example when the entries equal to $ 1$ make up a $ m \\times m$ subarray (but in other cases also, although not for all configurations).",
  "Solution_4": "Dear mavropnevma\r\nThat's exactly the way how I did that.\r\nIt's really interesting,isn't it?"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "I know that there are several proofs for $\\sum_{n\\in\\mathbb{N}}n^{-2}=\\pi^{2}/6$.  I was trying to prove something, but I'm not sure if I've made an error here:\r\n\r\nLet $f\\left(x\\right)=x^{2}-x+1/6$ for $0\\leq x\\leq 1$ and require that $f\\left(x\\right)=f\\left(x-n\\right)$ for $n\\leq x\\leq n+1$ where $n\\in\\mathbb{Z}$.  Part integration shows us that\r\n\r\n\\[\\int_{0}^{1}f\\left(x\\right)e^{-2\\pi inx}\\,dx=\\begin{cases}1/2\\pi^{2}n^{2}& n\\neq 0 \\\\ 0 & n=0\\end{cases}.\\]\r\nThe next step is to show that\r\n\r\n\\[f\\left(x\\right)=\\frac{1}{\\pi^{2}}\\sum_{n\\in\\mathbb{N}}\\frac{\\cos\\left(2\\pi nx\\right)}{n^{2}},\\]\r\nwhich clearly implies that $\\sum_{n\\in\\mathbb{N}}n^{-2}=\\pi^{2}/6$.  I know that since $f$ is a continuous function, we can say that\r\n\r\n\\[f\\left(x\\right)=\\sum_{n\\in\\mathbb{Z}}a_{n}e^{-2\\pi inx},\\]\r\nwhere $a_{n}=1/2\\pi^{2}n^{2}$.  However, what is the next step?",
  "Solution_1": "[quote=\"amcavoy\"]I know that there are several proofs for $\\sum_{n\\in\\mathbb{N}}n^{-2}=\\pi^{2}/6$.  I was trying to prove something, but I'm not sure if I've made an error here:\n[/quote]\r\n\r\nWhat about this?\r\n\r\nDefine $f(x)=x^{2}$ on $[-\\pi , \\pi]$.\r\nDefine $f(x+2\\pi n)=x^{2}$ for all $n\\in \\mathbb{Z}$.\r\n\r\nThis function satisfies the \"Dirichlet conditions\".\r\n\r\nExpand it as a Half-Range Cosine Series.\r\n\r\nPlay around with it until you bring it to the desired form.",
  "Solution_2": "Oh yeah:\r\n\\[f\\left(x\\right)=\\sum_{n\\in\\mathbb{Z}}a_{n}e^{2\\pi inx}=\\frac{1}{\\pi^{2}}\\sum_{n\\in\\mathbb{Z}}\\frac{e^{2\\pi inx}}{2n^{2}}=\\frac{1}{\\pi^{2}}\\sum_{n\\in\\mathbb{N}}\\frac{\\cos\\left(2\\pi nx\\right)}{n^{2}}, \\]\r\nso $f\\left(0\\right)=1/6=\\left(1/\\pi^{2}\\right)\\sum_{n\\in\\mathbb{Z}}n^{-2}$."
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "If $ f(x) \\in C(\\minus{}\\infty,\\infty)$\r\n\r\ncan we prove that there exists a interval $ (a,b) \\subset C(\\minus{}\\infty,\\infty)$ ,such that\r\n\r\n$ f(x)$ is monotony as $ x \\in (a,b)$\r\n\r\n :idea:  :roll:",
  "Solution_1": "If $ \\ f(x)$ happens to be an open map, we can say that the function is monotonic.",
  "Solution_2": "No, we can not, because this is not true (just take continuos function which is nowhere differentiable) :wink:"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c\\geq 0$ and $ ab\\plus{}bc\\plus{}ca\\equal{}1$.prove that $ \\sum \\frac{ab\\plus{}1}{a\\plus{}b}  \\geq 3$",
  "Solution_1": "[quote=\"euler_vn\"]Let $ a,b,c\\geq 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 1$.prove that $ \\sum \\frac {ab \\plus{} 1}{a \\plus{} b} \\geq 3$[/quote]\r\n\r\nVery nice  :P \r\n$ < \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\ge 3p$\r\nWith:$ p \\equal{} a \\plus{} b \\plus{} c;q \\equal{} ab \\plus{} bc \\plus{} ac \\equal{} 1;r \\equal{} abc$\r\n$ \\plus{} ) p \\ge 2$\r\n$ \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\equal{} (p \\minus{} 1)(p \\minus{} 2) \\plus{} pr \\plus{} 3r \\plus{} 3p \\ge 3p$ \r\n$ \\plus{} ) p \\le 2$\r\n$ \\equal{} > r \\ge \\frac {p(4 \\minus{} p^2)}{9}$ by shur :)\r\n$ \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\ge p^2 \\plus{} \\frac {p^2(4 \\minus{} p^2)}{9} \\plus{} 2 \\plus{} \\frac {3p(4 \\minus{} p^2)}{9} \\ge 3p$\r\nWe have prove that:\r\n$ p^2 \\plus{} \\frac {p^2(4 \\minus{} p^2)}{9} \\plus{} 2 \\plus{} \\frac {3p(4 \\minus{} p^2)}{9} \\ge 3p$\r\n$ < \\equal{} > (p \\minus{} 2)(3p \\minus{} p^3 \\minus{} 5p^2 \\minus{} 9) \\ge 0$-It's true by $ p \\le 2$\r\n  :P",
  "Solution_2": "[quote=\"caubetoanhoc94\"][quote=\"euler_vn\"]Let $ a,b,c\\geq 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 1$.prove that $ \\sum \\frac {ab \\plus{} 1}{a \\plus{} b} \\geq 3$[/quote]\n\nVery nice  :P \n$ < \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\ge 3p$\nWith:$ p \\equal{} a \\plus{} b \\plus{} c;q \\equal{} ab \\plus{} bc \\plus{} ac \\equal{} 1;r \\equal{} abc$\n$ \\plus{} ) p \\ge 2$\n$ \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\equal{} (p \\minus{} 1)(p \\minus{} 2) \\plus{} pr \\plus{} 3r \\plus{} 3p \\ge 3p$ \n$ \\plus{} ) p \\le 2$\n$ \\equal{} > r \\ge \\frac {p(4 \\minus{} p^2)}{9}$\n$ \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\ge p^2 \\plus{} \\frac {p^2(4 \\minus{} p^2)}{9} \\plus{} 2 \\plus{} \\frac {3p(4 \\minus{} p^2)}{9} \\ge 3p$\nWe have prove that:\n$ p^2 \\plus{} \\frac {p^2(4 \\minus{} p^2)}{9} \\plus{} 2 \\plus{} \\frac {3p(4 \\minus{} p^2)}{9} \\ge 3p$\n$ < \\equal{} > (p \\minus{} 2)(3p \\minus{} p^3 \\minus{} 5p^2 \\minus{} 9) \\ge 0$-It's true by $ p \\le 2$\n  :P[/quote]\r\nwhy $ p\\leq 2$  :?:",
  "Solution_3": "[quote=\"peine\"][quote=\"caubetoanhoc94\"][quote=\"euler_vn\"]Let $ a,b,c\\geq 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 1$.prove that $ \\sum \\frac {ab \\plus{} 1}{a \\plus{} b} \\geq 3$[/quote]\n\nVery nice  :P \n$ < \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\ge 3p$\nWith:$ p \\equal{} a \\plus{} b \\plus{} c;q \\equal{} ab \\plus{} bc \\plus{} ac \\equal{} 1;r \\equal{} abc$\n$ \\plus{} ) p \\ge 2$\n$ \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\equal{} (p \\minus{} 1)(p \\minus{} 2) \\plus{} pr \\plus{} 3r \\plus{} 3p \\ge 3p$ \n$ \\plus{} ) p \\le 2$\n$ \\equal{} > r \\ge \\frac {p(4 \\minus{} p^2)}{9}$\n$ \\equal{} > p^2 \\plus{} pr \\plus{} 2 \\plus{} 3r \\ge p^2 \\plus{} \\frac {p^2(4 \\minus{} p^2)}{9} \\plus{} 2 \\plus{} \\frac {3p(4 \\minus{} p^2)}{9} \\ge 3p$\nWe have prove that:\n$ p^2 \\plus{} \\frac {p^2(4 \\minus{} p^2)}{9} \\plus{} 2 \\plus{} \\frac {3p(4 \\minus{} p^2)}{9} \\ge 3p$\n$ < \\equal{} > (p \\minus{} 2)(3p \\minus{} p^3 \\minus{} 5p^2 \\minus{} 9) \\ge 0$-It's true by $ p \\le 2$\nwhy $ p\\leq 2$  :?: \n  :P[/quote][/quote]\r\n\r\nI Consider two cases $ p \\le 2;p \\ge 2$ :lol:",
  "Solution_4": "[quote=\"caubetoanhoc94\"][/quote][quote=\"peine\"][quote=\"caubetoanhoc94\"][quote=\"euler_vn\"]Let $ a,b,c\\geq 0$ and $ ab + bc + ca = 1$.prove that $ \\sum \\frac {ab + 1}{a + b} \\geq 3$[/quote]\n\nVery nice  :P \n$ < = > p^2 + pr + 2 + 3r \\ge 3p$\nWith:$ p = a + b + c;q = ab + bc + ac = 1;r = abc$\n$ + ) p \\ge 2$\n$ = > p^2 + pr + 2 + 3r = (p - 1)(p - 2) + pr + 3r + 3p \\ge 3p$ \n$ + ) p \\le 2$\n$ = > r \\ge \\frac {p(4 - p^2)}{9}$\n$ = > p^2 + pr + 2 + 3r \\ge p^2 + \\frac {p^2(4 - p^2)}{9} + 2 + \\frac {3p(4 - p^2)}{9} \\ge 3p$\nWe have prove that:\n$ p^2 + \\frac {p^2(4 - p^2)}{9} + 2 + \\frac {3p(4 - p^2)}{9} \\ge 3p$\n$ < = > (p - 2)(3p - p^3 - 5p^2 - 9) \\ge 0$-It's true by $ p \\le 2$\nwhy $ p\\leq 2$  :?: \n  :P[/quote][/quote][quote=\"caubetoanhoc94\"]\n\nI Consider two cases $ p \\le 2;p \\ge 2$ :lol:[/quote]\r\nokay, I didn't see that you considered two case, now I understand, thank's.",
  "Solution_5": "[quote=\"peine\"][/quote][quote=\"caubetoanhoc94\"][/quote][quote=\"peine\"][quote=\"caubetoanhoc94\"][quote=\"euler_vn\"]Let $ a,b,c\\geq 0$ and $ ab + bc + ca = 1$.prove that $ \\sum \\frac {ab + 1}{a + b} \\geq 3$[/quote]\n\nVery nice  :P \n$ < = > p^2 + pr + 2 + 3r \\ge 3p$\nWith:$ p = a + b + c;q = ab + bc + ac = 1;r = abc$\n$ + ) p \\ge 2$\n$ = > p^2 + pr + 2 + 3r = (p - 1)(p - 2) + pr + 3r + 3p \\ge 3p$ \n$ + ) p \\le 2$\n$ = > r \\ge \\frac {p(4 - p^2)}{9}$\n$ = > p^2 + pr + 2 + 3r \\ge p^2 + \\frac {p^2(4 - p^2)}{9} + 2 + \\frac {3p(4 - p^2)}{9} \\ge 3p$\nWe have prove that:\n$ p^2 + \\frac {p^2(4 - p^2)}{9} + 2 + \\frac {3p(4 - p^2)}{9} \\ge 3p$\n$ < = > (p - 2)(3p - p^3 - 5p^2 - 9) \\ge 0$-It's true by $ p \\le 2$\nwhy $ p\\leq 2$  :?: \n  :P[/quote][/quote][quote=\"caubetoanhoc94\"]\n\nI Consider two cases $ p \\le 2;p \\ge 2$ :lol:[/quote][quote=\"peine\"]\nokay, I didn't see that you considered two case, now I understand, thank's.[/quote]\r\n\r\nOkay.\r\nP/s:H\u1ecfi m\u00e3i m\u1edbi bik x\u00e9t 2 TH ti\u1ebfng anh l\u00e0 th\u1ebf :("
}
{
  "Tag": [
    "modular arithmetic",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $p$ be an odd prime. Assume $\\sigma$ is a permutation of $\\{1,2,\\ldots,p\\}$. Show that we can find $i\\ne j\\in\\{1,2,\\ldots,p\\}$ s.t. $i\\sigma(i)\\equiv j\\sigma(j)\\pmod p$.\r\n\r\nP.S. People will probably find this a joke :)",
  "Solution_1": "Oh, really  :lol:  :lol:  :lol: \r\nSuppose the contrary. It is clear $\\sigma(p)=p$. Then $\\sigma(1),2\\sigma(2),...,(p-1)\\sigma(p-1)$ is a permutation of $1,2,...,p-1$. Multyplying all number $i\\sigma(i)$ we obtain $(p-1)!^2\\equiv (p-1)!\\pmod{p}$. Contradiction with Wilson'n theorem.",
  "Solution_2": "First assume $\\sigma (p) =0$, because otherwise its trivial.\r\nNow assume that all $ i\\cdot \\sigma (i), i=1,...,p-1$ are different.\r\nThen you get $ \\prod  i\\cdot \\sigma (i) \\equiv -1 \\not\\equiv 1 \\equiv \\prod i \\cdot \\prod \\sigma (i) \\mod p$\r\n\r\nnice joke  :lol:"
}
{
  "Tag": [
    "blogs",
    "MATHCOUNTS",
    "function",
    "algebra",
    "polynomial",
    "AMC",
    "AIME"
  ],
  "Problem": "I am interested in collecting a small but fun list of topics for write about in my blog.  In particular, I would like to hear about topics that middle school math teamers typically say \"wow!\" about upon first discover.  What would students here suggest?",
  "Solution_1": "Maybe this is a little advanced, but a lot of people in my class were really suprised when they found out that some infinities are bigger than others and the teacher proved it.\r\n\r\n[hide=\"Also Interesting\"]There was a problem in my class: \\[x^{x^{x^{\\cdots}}}=2\\]\n\nWho's answer is: \\[\\sqrt{2}\\]\n\nThat people found interesting.\n\nHowever, they found this:\n\nWhat is x if \\[x^{x^{x^{\\cdots}}}=y\\]\n\neven more interesting because it's answer is \\[^y\\sqrt{y}\\] because then \\[\\sqrt{2}^{\\sqrt{2}^{\\cdots}}=2=4\\] which obviously isn't true because \\[2\\ne 4\\]\n\nWhat happened after that if that most people claimed that math as we know it was false, and as such, we should not learn math in school. :D \n\nAnd then the teacher explained how infinity can't be treated like a normal number, so we hadn't \"disproved\" math.[/hide]\r\n\r\nWhat I'm getting at is that proofs of things that aren't true are always interesting.",
  "Solution_2": "pi formulae are always awesome",
  "Solution_3": "I like RC-7th 's Idea....\r\n\r\nInfinity can be tough, but I think it would be fun to learn it...",
  "Solution_4": "While a perfectly good subject to blog about, I think that kind of self-similarity and infinity discussion is more advanced that what I'm asking about.\r\n\r\nPicture that MATHCOUNTS student who can score around half the points at the Chapter competition.  That's the audience I'm interested in.",
  "Solution_5": "Well, according to Treething, you have to solve infinity problems in MATHCOUNTS...\r\n\r\nI looked at some MATHCOUNTS practice problems and they definitley had infinity problems..\r\n\r\nlike I expected..\r\n\r\nand in most schools, you don't get to learn about infinities..",
  "Solution_6": "I would love to learn about infinities, i think it's hard to get around the fact that it's really going on FOREVER... most middle schools don't teach much about it because it's kinda hard except for like MATHCOUNT people... so I think it would be great. But I don't think you should start off with problems like the ones RC-7th mentioned, they are a little to hard to start off with",
  "Solution_7": "I think people would love stuff like how pi goes on forever, how to prove anything to the 0th power is 1, and how 0.999999 = 1. We've always like to ask these questions in math class, but the teachers usually don't know. I was amazed when I found out how to prove x^0 = 1... when I told my classmates, they thought that was really cool",
  "Solution_8": "[quote=\"LuCy4EvA\"]I think people would love stuff like how pi goes on forever, how to prove anything to the 0th power is 1, and how 0.999999 = 1. We've always like to ask these questions in math class, but the teachers usually don't know. I was amazed when I found out how to prove x^0 = 1... when I told my classmates, they thought that was really cool[/quote]\r\n\r\n$0^0\\ne 1$\r\n\r\nThe problem I posted was the only one the teacher had given to us that related to infinity.\r\n\r\nGame theory is pretty fun. There was a question from mathcounts that went like this : Two people are playing a game with 40 counters. On each turn, a person may take 1,2,3,4, or 5 counters. The person who takes the last counter wins. How many counters should the person who goes first take to ensure that they win?",
  "Solution_9": "is 0^0 undefined?",
  "Solution_10": "okay first of all\r\n\r\n0^0\r\n\r\nmeans nothing to the nothing power\r\n\r\ndo you understand the concept X^0 = 1 ?\r\n\r\nbut it only happens when X is like numbers not including 0\r\n\r\n0 means nothing..\r\n\r\nthere won't be nothing to power over...\r\n\r\nbut when you set a group like this, {0}\r\n\r\nyou always have to count 0 as one of the elements...\r\n\r\ngod!! this is hard to tell..",
  "Solution_11": "wait, I get ur explanation, but is it 1 or undefined?",
  "Solution_12": "$0^0$ is undefined.\r\n\r\nFor most numbers you can product chains of exponents using multiplication and go backwards in these chains using division:\r\n\\[ \\frac{2^3}{2} = \\frac{8}{2} = 4 = 2^2. \\]\r\n\\[ \\frac{2^2}{2} = \\frac{4}{2} = 2 = 2^1. \\]\r\n\\[ \\frac{2^1}{2} = \\frac{2}{2} = 1 = 2^0. \\]\r\n\\[ \\frac{2^0}{2} = \\frac{1}{2} = \\frac{1}{2} = 2^{-1}. \\]\r\n\\[ \\frac{2^{-1}}{2} = \\frac{\\frac{1}{2}}{2} = \\frac{1}{4} = 2^{-2}. \\]\r\n\r\nDivision by 0 is undefined, so this reversal of the multiplication chain is not possible.",
  "Solution_13": "I remember one of the things that first wowed me was combinatorics, especially counting arguments and (simple) generating functions. Like seeing combinatoric identities proved in three different ways: 1) just using normal algebra, 2) block-walking on Pascal's triangle, and 3) counting arguments. Finding out that those \"How many ways can you make 83 cents using pennies, nickels, and dimes\" questions could actually be solved through a real method (!) was also neat. Not sure if that stuff is too advanced, though.",
  "Solution_14": "[quote=\"kidnappedturtle\"]I remember one of the things that first wowed me was combinatorics, especially counting arguments and (simple) generating functions. Like seeing combinatoric identities proved in three different ways: 1) just using normal algebra, 2) block-walking on Pascal's triangle, and 3) counting arguments. Finding out that those \"How many ways can you make 83 cents using pennies, nickels, and dimes\" questions could actually be solved through a real method (!) was also neat. Not sure if that stuff is too advanced, though.[/quote]\r\n\r\nThis are topics I would call \"Getting Started\" topics.  A little beyond the basics.  Generating functions I would call Intermediate -- even at the simple level -- because they involve using polynomials together with counting arguments.  Students need to know two subjects fairly well and be able to go between then easily -- the kinds of skills that are required for AIME problems.  Those are certainly topics I might blog about however.  Thanks for the suggestions.",
  "Solution_15": "Things that blew my hair back:\r\n\r\n1.) Proof by contradiction.\r\n2.) Euclid's proof for infinitely many primes.\r\n3.) Fundamental theorem of arithmetic.( Every integer factors as a product of primes)\r\n4.) Parity arguments.(The checkerboard tiling one comes to mind)",
  "Solution_16": "Things that blow my hair back:\r\n1) hair dryer\r\n2) riding in convertible\r\n3) roller coaster\r\n\r\nSeriously (and on topic)\r\nI went to a middle school that didn't go very far into much of anything so I was ecstatic when we learned graphing. While the class was struggling with lines, I graphed my first *gasp* parabola! It amazes me that 4 years ago, I couldn't find slope. :what?: \r\n\r\nBut since we're talking about MathCounts, I really liked counting and geometry problems. I had a teacher who showed us how to derive the area formula for a circle, essentially doing some integral calculus for 7th graders. :wacko: He wasn't using anything we didn't understand, though. He just introduced the idea of making the pieces of the circle small enough and sticking them back together so that it would look enough like a rectangle to use base x height.\r\nSo I guess I'm really talking about limits and approximations, and proofs to some extent.",
  "Solution_17": "Proofs of irrationality are some of my favroites.",
  "Solution_18": "I did a bunch of parity argument proofs with my middle school students, and the ones who I imagine could get about 1/4 or more of the points on a MATHCOUNTS Chapter competition really seemed to enjoy them.  My favorite is one about whether or not it is possible to create a magic square with the first n<sup>2</sup> primes.\r\n\r\nThere are also things in the MATHCOUNTS handbook that could use expanding upon.  Kids are often given formulas from there without understanding why they work, like the formula for the sum of all the factors of a number.\r\n\r\nMy favorite math proof of all time was the proof that there is a countable infinity of rational numbers  (Cantor's diagonal argument).  I was much older than middle school when I first encountered it, but I think that it could be presented in a way that the MATHCOUNTS kids could understand.\r\n\r\nWhen I was in middle school myself, I was really hung up on 0<sup>0</sup>.  I would have loved to see a good explanation of why it is neither 0 nor 1, both of which seemed like perfectly reasonable values.  Similarly, 0!=1 intrigues my middle schoolers, and telling them that it just is that way because it makes all the formulas work out properly isn't very satisfactory to them :)",
  "Solution_19": "By the way, I should mention that many mathematicians, including myself, define $0^0$ to be 1.  My arguments, and counterarguments by Simon and others, appeared in a topic on a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11999]Limit Problem[/url].\r\n\r\nBasically, some mathematicians say it is undefined, and some set it to 1.  It's a matter of what is the most convenient definition.",
  "Solution_20": "but 0 means none... :(",
  "Solution_21": "Many kids are exposed to Magic Squares in elementary school.  But Middle School is when you can begin talking about nested patterns and recursion and algorithms.  Attached is a 12x12 Magic Square as an example.  Ask the students to find smaller magic squares hidden in this big magic square.  Either show the kids the pattern I used to make this square or challenge them to find the pattern for themselves.  In either case, ask [b]why[/b] does it work?  Invite them to create their own pattern that will create a 9x9, 12x12, or 15x15 square.  There are [b]many[/b] different patterns / algorithms to make these larger squares.\r\n$\\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}\r\n\\hline 113   & 1     & 81    & 127   & 15    & 95    & 126   & 14    & 94    & 116   & 4     & 84 \\\\\r\n\\hline 33    & 65    & 97    & 47    & 79    & 111   & 46    & 78    & 110   & 36    & 68    & 100 \\\\\r\n\\hline 49    & 129   & 17    & 63    & 143   & 31    & 62    & 142   & 30    & 52    & 132   & 20 \\\\\r\n\\hline 124   & 12    & 92    & 118   & 6     & 86    & 119   & 7     & 87    & 121   & 9     & 89 \\\\\r\n\\hline 44    & 76    & 108   & 38    & 70    & 102   & 39    & 71    & 103   & 41    & 73    & 105 \\\\ \r\n\\hline 60    & 140   & 28    & 54    & 134   & 22    & 55    & 135   & 23    & 57    & 137   & 25 \\\\\r\n\\hline 120   & 8     & 88    & 122   & 10    & 90    & 123   & 11    & 91    & 117   & 5     & 85 \\\\\r\n\\hline 40    & 72    & 104   & 42    & 74    & 106   & 43    & 75    & 107   & 37    & 69    & 101 \\\\\r\n\\hline 56    & 136   & 24    & 58    & 138   & 26    & 59    & 139   & 27    & 53    & 133   & 21 \\\\\r\n\\hline 125   & 13    & 93    & 115   & 3     & 83    & 114   & 2     & 82    & 128   & 16    & 96 \\\\\r\n\\hline 45    & 77    & 109   & 35    & 67    & 99    & 34    & 66    & 98    & 48    & 80    & 112 \\\\\r\n\\hline 61    & 141   & 29    & 51    & 131   & 19    & 50    & 130   & 18    & 64    & 144   & 32 \\\\\r\n\\hline\r\n\\end{tabular}$",
  "Solution_22": "what's the magic square?\r\n\r\nI don't really get it....\r\n\r\nIs it like a matrix or something? :?",
  "Solution_23": "All the rows, columns and diagonals add up to the same number.",
  "Solution_24": "to see the different solutions of the same problem was always aweful for me.particularly different solution of the same problem using different field of maths.intutive and easy solution using elemantry language r also helpful.",
  "Solution_25": "[quote=\"A+MATH\"]Proofs of irrationality are some of my favroites.[/quote]\r\n\r\nWell, whether or not that is middle-school level depends.  The proof that $\\sqrt{2}$ is irrational is very clever and uses very simple math.  Many middle schoolers could understand that.\r\n\r\nA proof that $\\pi$ is irrational would go into one of the advanced/college boards...",
  "Solution_26": "Things that surprised and intrigued me in my middle school years:\r\n\r\n1) Figuring out how to calculate the number of shortest paths from one corner to another on a grid (already mentioned)\r\n\r\n2) Finding that the sum of factors of a number like 120 is (1+5)(1+3)(1+2+4+8)\r\n\r\n3) Finding the proof that 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6\r\n\r\n4) Proof of the Pythagorean theorem\r\n\r\n5) A non-calculus proof of the volume of sphere formula (given the area of circle formula)\r\n\r\n6) Discovering the concept of instantaneous slope\r\n\r\nIn general, I usually found it most fascinating when I read about or discovered a neat way to calculate something that would normally be tedious or impossible.\r\n\r\n\r\nI also think it would be great if you could, instead of just showing a proof or idea, walk viewers through to make them discover it as much by themselves as possible.  Of course you're probably already doing that.",
  "Solution_27": "I think that you should host online 24 games.  You know that game where you make 24 with four numbers 1-10.  That would be lots of fun.",
  "Solution_28": "[quote=\"jimli\"]I think that you should host online 24 games.  You know that game where you make 24 with four numbers 1-10.  That would be lots of fun.[/quote]\r\n\r\nHosting games is somewhere down our list of priorities.",
  "Solution_29": "My FAVORITE is the decimal expansion of $\\frac{1}{n}$, where n is a prime number. Some of them are pure awesome. For example: \r\n\\[\\frac{1}{7}=0.142857. . . . . . . .\\]\r\n\\[\\frac{2}{7}=0.285714. . . . . . . .\\]\r\n\\[\\frac{3}{7}=0.428571. . . . . . . .\\]\r\n\\[. . . . . . . . . . \\]\r\nOther primes such as 13 and 17 have similar characteristics.  :)",
  "Solution_30": "lots and lots of factoring things\r\n\r\n$(x+y)^2 = x^2 + 2xy + y^$\r\n\r\n$(x+y)(x-y) = x^2 - y^2$\r\n\r\nthose sorts of things plus the fact that the coefficients of $(x+y)^n$ is the $n^{{th}}$ row of pascals triangle. I thought that that was really cool. (note that the top row, 1, is row #0.",
  "Solution_31": "Pascal's triangle and Fibonacci numbers always caught my attention.  There are some very basic but cool things you might be able to do with the golden ratio.  \r\n\r\nClever factorization is always cool.\r\n\r\nBasic number theory or rearranging of numbers to make multiplication easier (that was pretty cool, you already did that). \r\n\r\nDifferent properties of numbers, perfect numbers, or the dividing by 11 rule, etc. \r\n\r\nPerhaps go over some cool things with palindromes.  \r\n\r\nI still don't really get magic squares much  :P",
  "Solution_32": "[quote=\"LuCy4EvA\"]is 0^0 undefined?[/quote]\r\n\r\n0^0=0/0 -->undefined.",
  "Solution_33": "The most fun thing to do for me was play with a graphing calculator for the first time, after some instructions."
}
{
  "Tag": [
    "geometry",
    "algebra",
    "polynomial",
    "inequalities",
    "trigonometry",
    "vector",
    "analytic geometry"
  ],
  "Problem": "Math Zoom Year-round courses in Southern California for Fall 2008 will start in September.  The following are class schedules.  For further information, visit [url]http://www.mathzoom.org/mathzoom_yearround.html[/url].\r\n\r\n[b]- Math Challenge II[/b]\r\nIn Math Challenge II, students learn and practice in areas such as algebra and geometry on the high school level, as well as advanced number theory and combinatorics.  Topics include polynomials, inequalities, special algebraic techniques, trigonometry, triangles and polygons, collinearity and concurrency, vectors and coordinates, numbers and divisibility,  modular arithmetic, residue classes, advanced counting strategies, binomial coefficients, pigeonhole principle, sequence and series, and various other topics and problem solving techniques involved in math contests such as the American Mathematics Competition (AMC) 10, 12 and ARML, and also the beginning AIME level.\r\n[u]MC II schedules:[/u]\r\n  [b]Arcadia[/b]: Saturdays starting 9/13, 9am-11am, 41 W Santa Clara St, Arcadia, CA 91007\r\n  [b]Irvine[/b]: Fridays starting 9/5, 7:30pm-9:30pm, 4255 E Campus Drive, Suite 290, Irvine, CA 92612\r\n\r\n[b]- Math Challenge III[/b]\r\nThis course is for experienced students who can solve at least a few problems in the AIME contest, and up to the USAMO level. In this course, the students learn more in-depth math concepts and problem-solving strategies that involves more rigorous mathematical writing and a broad range of topics. The topics includes everything covered in MC II with more depth, and additional topics such as quadratic residue, diophantine equations, recurrence, proving inequalities, mathematical induction, functional equations, generating functions, inversion, projective geometry, and rigorous proof-writing practices. Through effective guidance and direction by our experienced teachers, students develop strong problem solving skills that make them perform well in AIME and USAMO.\r\n[u]MC III schedule:[/u]\r\n[b]Arcadia[/b]: Saturdays starting 9/20, 2pm-4pm, 411 E Huntington Dr, Suite 309, Arcadia, CA 91006",
  "Solution_1": "[b]- Math Challenge I[/b] (grades 6-8)\r\nStudents learn the Math skills at a deeper level with topics in beginning algebra, fundamental geometry, basic number theory concepts and counting strategies, as well as logic and probabilities. The students not only learn practical skills of challenging problem solving that are supplemental to their school curricula, but also develop skills in creative thinking, logical reasoning, oral and written presentation, and team work. This course helps 6th to 8th graders (some advanced 5th graders may attend upon approval) to participate in the American Mathematics Competition (AMC) 8, MathCounts and Math Olympiads for Elementary and Middle Schools MOEMS. Through effective guidance and direction by our experienced teachers, students develop strong problem solving skills that make them perform well in the MathCounts, AMC 8, and MOEMS Contests.\r\n[u]MC I schedules:[/u]\r\n[b]Arcadia:[/b] Saturdays starting 9/13, 11:15am-1:15pm, 41 W Santa Clara St, Arcadia, CA 91007\r\n[b]Diamond Bar:[/b] Saturdays starting 9/13, 6:30pm-8:30pm, 315 S. Diamond Bar Blvd., #C, Diamond Bar, CA 91765\r\n\r\n[b]- Young Math Olympians[/b] (grades 3-5)\r\nThis course is focused on nurturing the interests of young kids from 3rd to 5th grades in challenging Math problem solving. With the effective guidance of our experienced teachers, kids explore the fun of solving interesting and meaningful Math related problems, and develop their logical and critical thinking skills. Some basic concepts covered in the Math Olympiads for Elementary School (MOEMS) are introduced.\r\n[u]YMO schedules:[/u]\r\n[b]Arcadia:[/b] Saturdays starting 9/13, 9:30am-10:30am, 41 W Santa Clara St, Arcadia, CA 91007"
}
{
  "Tag": [],
  "Problem": "Em c\u00f3 1 b\u00e0i to\u00e1n sau : \r\ncho 2 s\u1ed1 N v\u00e0 k, \u0111i\u1ec1u ki\u1ec7n N > k v\u00e0 N kh\u00f4ng chia h\u1ebft cho k.\r\n\r\nt\u1ee9c l\u00e0 N chia k d\u01b0 Q ( hay N = Q mod k )\r\n\r\nL\u00fac \u0111\u1ea7u N s\u1ed1 \u0111ang \u1edf tr\u1ea1ng th\u00e1i A, m\u1ed7i l\u1ea7n ta \u0111\u1ed5i \u0111\u01b0\u1ee3c k s\u1ed1 sang tr\u1ea1ng th\u00e1i B. ng\u01b0\u1ee3c l\u1ea1i ta c\u0169ng th\u1ec3 \u0111\u1ed5i t\u1eeb B v\u1ec1 A, nh\u01b0ng m\u1ed7i l\u1ea7n b\u1eaft bu\u1ed9c ph\u1ea3i \u0111\u1ed5i k ng\u01b0\u1eddi. T\u1ee9c l\u00e0 n\u1ebfu \u0111\u1ed5i i ng\u01b0\u1eddi b\u00ean B v\u1ec1 A th\u00ec t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi \u0111\u1ed5i k - i b\u00ean A v\u1ec1 B. hix ! Em n\u00f3i lung tung qu\u00e1 nh\u1ec9 ! \r\nCMR : n\u1ebfu k ch\u1eb5n v\u00e0 Q l\u1ebb th\u00ec kh\u00f4ng bao h \u0111\u1ed5i h\u1ebft N s\u1ed1 sang tr\u1ea1ng th\u00e1i B \u0111\u01b0\u1ee3c !\r\nEm th\u1eed r\u1ea5t nhi\u1ec1u test v\u00e0 c\u1ea3m th\u1ea5y n\u00f3 \u0111\u00fang, nh\u01b0ng ch\u01b0a t\u00ecm ra 1 c\u00e1ch CM n\u00e0o t\u1ed5ng qu\u00e1t. C\u00e1c anh gi\u00fap em v\u1edbi nh\u00e9 ! \r\n\r\nV\u00ed d\u1ee5 : N\u1ebfu A = 333 v\u00e0 k = 50 th\u00ec kh\u00f4ng c\u00f3 c\u00e1ch.\r\n           N\u1ebfu A = 15 v\u00e0 k = 10 th\u00ec kh\u00f4ng c\u00f3 c\u00e1ch...\r\n\r\nN\u1ebfu A = 1000 v\u00e0 k = 35 th\u00ec c\u00f3 c\u00e1ch : \u0111\u1ea7u ti\u00ean ta \u0111\u1ed5i 28 l\u1ea7n \u0111\u01b0\u1ee3c 980 s\u1ed1 sang tr\u1ea1ng th\u00e1i B. C\u00f2n d\u01b0 l\u1ea1i 20 s\u1ed1. L\u1ea7n k\u1ebf ti\u1ebfp ta l\u1ea5y 10 s\u1ed1 b\u00ean B v\u00e0 25 s\u1ed1 b\u00ean A \u0111\u1ed5i l\u1ea1i. Ta \u0111\u01b0\u1ee3c 25 s\u1ed1 + 10 s\u1ed1 v\u1ec1 tr\u1ea1ng th\u00e1i A. L\u1ea7n cu\u1ed1i \u0111\u1ed5i l\u1ea1i 35 s\u1ed1 v\u1ec1 B xong. \r\nN\u1ebfu em ghi c\u00f3 ch\u1ed7 n\u00e0o kh\u00f4ng r\u00f5 r\u00e0ng th\u00ec m\u1ea5y anh c\u1ee9 n\u00f3i \u1ea1 ! Em s\u1ebd s\u1eefa l\u1ea1i. C\u00e1m \u01a1n m\u00e1y anh r\u1ea5t nhi\u1ec1u !",
  "Solution_1": "C\u00e1c b\u00e0i to\u00e1n trong box Vi\u1ec7t Nam s\u1ebd \u0111\u01b0\u1ee3c post trong box n\u00e0y http://www.mathlinks.ro/Forum/index.php?f=284 . M\u00e0 theo anh kh\u00f4ng n\u00ean th\u1ea3o lu\u1eadn v\u1ec1 To\u00e1n trong box Vi\u1ec7t Nam, c\u00e1i n\u00e0y sang ddth t\u1ed1t h\u01a1n, em n\u00ean post b\u1eb1ng ti\u1ebfng Anh ra b\u00ean ngo\u00e0i.",
  "Solution_2": "Da em hi\u1ec3u r\u1ed3i ! Do l\u1ea7n \u0111\u1ea7u em ch\u01b0a bi\u1ebft n\u00ean post nh\u1ea7m ch\u1ed7 ! Srry anh nh\u00e9  :oops:"
}
{
  "Tag": [
    "function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "What is the number of partitions of a positive integer in $ k$ integers, such that there aren't two equal numbers in the partition. \r\n\r\nExample. n=7, k=3\r\nI need only 1+2+4.",
  "Solution_1": "If we ignore the condition(to be not equal each other), it would be a well-known open problem. However, your question may have a solution. Try to use \"generating functions\".",
  "Solution_2": "I forgot how to manipulate the generating functions regarding the partitions :blush:. I think it's very difficult to find a closed form for the partitions I need.",
  "Solution_3": "Your question is equivalent to the following question \"What is the coefficient of  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a/2/9/a297bb91c9703af1975402dded3ab9b7e6431dde.gif[/img] in [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6/f/c/6fc57c3597157b964300706a9f5848b93f22e8c8.gif[/img] ?\"",
  "Solution_4": "Moreover, you can generalize the problem allowing to use i [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/1/b/a/1ba9b59bdee92f38c1698c784b67ba70f803331d.gif[/img] times. Then the question becomes \"What is the coefficient of [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/5/3/8/5381feaab83ed3656546a8cdfac424159d680155.gif[/img] in [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/9/8/d/98d7dd9bab313baf7c0edde3b6e61c4bbbd35ff1.gif[/img] ?\"",
  "Solution_5": "I've found on the internet a closed form. It's $ p_{3}(6j\\plus{}k)\\equal{}3j^{2}\\minus{}(3\\minus{}k)j$, for $ k\\in [1,5]$ and $ p_{3}(6j\\plus{}6)\\equal{}3j^{2}\\plus{}3j\\plus{}1$. Nice",
  "Solution_6": "[quote=\"matehan\"]Your question is equivalent to the following question \"What is the coefficient of $ x^{n}$ in $ (1\\plus{}x)(1\\plus{}x^{2})\\ldots(1\\plus{}x^{n})$ ?\"[/quote]\r\n\r\nThis is wrong.He asked for the number of partitions in exactly $ k$ integer.\r\n\r\nHis question is equivalent,for example,to find the coefficient of $ x^{n}y^{k}$ in \r\n\r\n$ (xy\\plus{}1)(x^{2}y\\plus{}1)\\ldots (x^{n}y\\plus{}1)$",
  "Solution_7": "Yes, you are right. I didn't see that condition.   :blush:"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $a,b,c,d$ be non-negative numbers sucht that $a^2+b^2+c^2+d^2=1$.\r\nProve that\r\n$a^3+b^3+c^3+d^3+abc+bcd+cda+dab \\geq 1$.\r\n\r\n[b]Remark1[/b]. I am interested in a solution without squaring and expanding.\r\n\r\n[b]Remark2[/b]. If $x_1^2+x_2^2+...+x_n^2=1$ and $p=\\frac 6{(n-2)(\\sqrt{n}+1)}$, then\r\n$\\sum x_1^3+p\\sum x_1x_2x_3 \\leq 1$.",
  "Solution_1": "i think we can use AC method just like in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90766[/url] what nhat did i.e let \r\n\r\n$f(a,b,c,d)=a^3+b^3+c^3+d^3+abc+abd+acd+bcd$ then\r\n\r\n$f((a+b)/2,(a+b)/2,c,d)-f(a,b,c,d)=\\frac{1}{4}(a-b)^2(c+d-3(a+b))$\r\n$f(0,a+b,c,d)-f(a,b,c,d)=ab(3(a+b)-(c+d))$\r\n\r\nSo $f(a,b,c,d)\\leq Max\\{f((a+b)/2,(a+b)/2,c,d),f(0,a+b,c,d)\\}$\r\n\r\nSo by AC method ,\r\n\r\n$f(a,b,c,d)\\leq Max\\{f(0,0,0,1),f(0,0,1/\\sqrt{2},1/\\sqrt{2}),f(0,1/\\sqrt{3},1/\\sqrt{3},1/\\sqrt{3}),f(1/2,1/2,1/2,1/2)\\}=1$\r\n\r\nequality case when $a=b=c=0,d=1$(and its permutation) or $a=b=c=d=1/2$\r\n\r\nMaybe Vasc want another solution , i guess ?",
  "Solution_2": "It is wrong, because AC-Theorem is valid if $a+b+c+d=constant$.\r\nWe must have $a^2+b^2+c^2+d^2=(a+b)^2/4+(a+b)^2/4+b^2+c^2$ and $a^2+b^2+c^2+d^2=0^2+(a+b)^2+b^2+c^2$ .\r\nSee:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=532271#p532271",
  "Solution_3": "Has anyone got a solution?"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "$a>0,b>0,c>0.$ Prove that:\r\n$\\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}\\geq\\sqrt[3]{9(a^3+b^3+c^3)}.$",
  "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=38369 and http://www.mathlinks.ro/Forum/viewtopic.php?t=43403 for an even stronger inequality which is still unsolved.\r\n\r\n  Darij",
  "Solution_2": "Thank you, Darij.",
  "Solution_3": "I hope, this is unknown:\r\n$a>0,b>0,c>0.$ Prove that $\\frac{a^3}{b^2}+\\frac{b^3}{c^2}+\\frac{c^3}{a^2}\\geq\\sqrt[4]{27(a^4+b^4+c^4)}.$",
  "Solution_4": "The next is proved.\r\nIf $a>0,b>0,c>0,a^{10}+b^{10}+c^{10}=3,$ then $\\frac{a^3}{b^2}+\\frac{b^3}{c^2}+\\frac{c^3}{a^2}\\geq3.$",
  "Solution_5": "[quote=\"arqady\"]I hope, this is unknown:\n$a>0,b>0,c>0.$ Prove that $\\frac{a^3}{b^2}+\\frac{b^3}{c^2}+\\frac{c^3}{a^2}\\geq\\sqrt[4]{27(a^4+b^4+c^4)}.$[/quote]\r\nI've found a proof of that one.  :)\r\n edit: :( No, sorry. It seems I have not."
}
{
  "Tag": [],
  "Problem": "is there any trick to do this problem:\r\n\r\nwhat is the sum of the digits in (111111111) squared?",
  "Solution_1": "All numbers of this form (edit:less than or equal to 9 digits) squared gives us a pattern of 123... up to the number of digits and then down to one, e.g. $1111^{2}=1234321$.  Thus. $111111111=12345678987654321$.  The sum of the digits is $\\frac{9(10)}{2}+\\frac{8(9)}{2}=81$.\r\n\r\nMasoud Zargar",
  "Solution_2": "Smart method. What if there are more than 9 1's?",
  "Solution_3": "Then you would just have to carry over.  You would have:\r\n\r\n1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1\r\nwhich turns into\r\n1 2 3 4 5 6 7 8 10 0 9 8 7 6 5 4 3 2 1\r\nwhich turns into\r\n1 2 3 4 5 6 7 9 0 0 9 8 7 6 5 4 3 2 1\r\n\r\nso $1111111111^{2}=1234567900987654321$.",
  "Solution_4": "Thank You Robin\r\n\r\nThat is a good thing to remember.\r\n\r\nI like that trick.\r\n\r\n111^2=12321"
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $ f: [0;1] \\rightarrow \\mathbb{R}$ be a function such that:\r\n$ (i)$ $ f(1) \\equal{} 1$,\r\n$ (ii)$ $ f(x) \\ge 0$ for all $ x \\in [0,1]$,\r\n$ (iii)$ if $ x,y$ and $ x \\plus{} y$ all lie in $ [0,1]$, then $ f(x \\plus{} y) \\ge f(x) \\plus{} f(y)$.\r\nProve that $ f(x) \\le 2x$ for all $ x \\in [0;1]$.",
  "Solution_1": "[quote=\"N.N.Trung\"]Let $ f: [0;1] \\rightarrow \\mathbb{R}$ be a function such that:\n$ (i)$ $ f(1) \\equal{} 1$,\n$ (ii)$ $ f(x) \\ge 0$ for all $ x \\in [0,1]$,\n$ (iii)$ if $ x,y$ and $ x \\plus{} y$ all lie in $ [0,1]$, then $ f(x \\plus{} y) \\ge f(x) \\plus{} f(y)$.\nProve that $ f(x) \\le 2x$ for all $ x \\in [0;1]$.[/quote]\r\n\r\nLet $ a\\in[0,1]$ such that $ f(a)>2a$\r\nIf $ a\\leq\\frac 12$, and using $ f(x \\plus{} y) \\ge f(x) \\plus{} f(y)$, we get $ f(2a)\\geq 2f(a) > 4a\\equal{}2(2a)$. Using this process as much as necessary, we get $ b\\in(\\frac 12,1]$ such that $ f(b)>2b>1$\r\nThen $ f(1)\\equal{}f(b\\plus{}(1\\minus{}b))\\ge f(b)\\plus{}f(1\\minus{}b)>1\\plus{}0$ (remember $ f(x) \\ge 0$ for all $ x \\in [0,1]$) which is a contradiction with $ f(1)\\equal{}1$\r\n\r\nSo no such $ a$ exists and $ f(x)\\leq 2x$ $ \\forall x\\in[0,1]$"
}
{
  "Tag": [
    "absolute value",
    "algebra solved",
    "algebra"
  ],
  "Problem": "The absolute value of every number in the sequence $\\{a_n\\}$ is smaller than 2005, and \\[a_{n+6}=a_{n+4}+a_{n+2}-a_n.\\] holds for all positive integers n. Prove that $\\{a_n\\}$ is periodic.\r\n\r\nIncredibly, this was probably the most difficult problem of our independent study problems in the 1st TST (excluding the final exam).",
  "Solution_1": "so just define two sequences $b, c$ the even and odd terms so that $b_{n+3} = b_{n+2} + b_{n+1} - b_n$, similar one for $c$. Then characteristic equation is $r^3 - r^2 - r + 1 = 0 \\Rightarrow (r-1)(r^2-1) = 0 \\Rightarrow r = -1, 1, 1$ so $b_n = \\alpha + n\\beta + \\gamma (-1)^n$; if $\\beta \\ne 0$, then for $n$ arbitrarily large, the modulus will be unbounded, which cannot happen, so $\\beta = 0$ and $b_n = \\alpha + \\gamma (-1)^n$ which has period 2; a similar analysis takes place for the $c_n$ so $a_n$ has period $4$."
}
{
  "Tag": [],
  "Problem": "Problem 2 (Group II). The glass of a fluorescent lamp is composed mainly by silica (SiO2), alumina (Al2O3), calcium oxide (CaO), magnesium oxide (MgO), and sodium oxide (Na2O).\r\n\r\n2.1 Regarding those compounds, select from the following statements the correct one:\r\n\r\nA - Alumina is the responsible for the formation of the net structure of glass.\r\nB - The ion Na+ is one of the responsibles for the breaking of Si-O bonds in the silica structure.\r\nC - Alumina is added in order to reduce the melting point of glass, facilitating its production.\r\nD - The less the quantity of CaO and Na2O, the smaller will be melting point of glass.\r\n\r\n2.2 Consider that the general formula of the glass used in those lamps production is $ SiO_{2}(Na_{2}O)_{m}(CaO)_{n}$. Supposing now that, in the above formula, m = 0.08 and n = 0.06, compute the minimum mass of sodium sulphate (Na2SO4) necessary for the production of 1.0 kg of that glass.\r\n\r\n$ Na_{2}SO_{4}(s)+heat \\longrightarrow Na_{2}O(s)+SO_{3}(g)$.",
  "Solution_1": "Let $ M$ be the molar mass of glass..$ r$ is molar mass of $ Na_{2}O$ .$ k$ molar mass of $ Na_{2}SO_{4}$\r\nin 1 mole of glass i.e in $ \\frac{M}{1000}$ Kg we have 0.08 moles $ Na_{2}O$ hence in 1kg we have \r\nhave $ \\frac{80m}{M}$ g of $ Na_{2}O$\r\nin $ k$ g of $ Na_{2}SO_{4}$ we have $ m$ g of $ Na_{2}O$\r\nso minimum mass of $ Na_{2}SO_{4}$ required is $ \\frac{80k}{M}$",
  "Solution_2": "Your answer is correct. How about question 2.1?"
}
{
  "Tag": [
    "email",
    "articles",
    "USAMTS",
    "function",
    "Valentin was here"
  ],
  "Problem": "Well, people were talking about me when I first joined, so let's see who will be the next person to hit 20000th user :D",
  "Solution_1": "Haven't we reached 30000 already?",
  "Solution_2": "How do we know that you were no 10000? Your profile number is 10153 (roll over username and see the status bar)",
  "Solution_3": "well here it says he was 9999th...\r\nhttp://www.artofproblemsolving.com/Forum/memberlist.php?mode=joindate&order=ASC&start=9950\r\n\r\nstrange.",
  "Solution_4": "I'm sorry 10000th user, your name is really cool but as they say, you are just 9999\r\n\r\nthe 10000th user is conway, he joined the same day, and made one post immediately to disappear again\r\n\r\nCan anyone understand mathI, he joined just one day after Creation by Valentin, waited until december [b]2004[/b] to disappear forever then??",
  "Solution_5": "I somehow think that the $9999$ comes from banning/deleting one user or somewhat like that...",
  "Solution_6": "Banning?  I thought that that was extremely rare on mathlinks?\r\n\r\nHowever one suffices, and that it is sad to take that pride away,  :o  :(  sorry 10000th user\r\n\r\n\r\nI just discovered I'm a pretty early bird, I am user 359 according to that list, I joined in july 2003. :roll:",
  "Solution_7": "How do you know that math1 has disappeared forever? He may turn up again now. Maybe he makes a post every one and a half year!\r\n\r\nAnd why does it show 78 when I roll over your name?",
  "Solution_8": "[quote=\"bubka\"]How do you know that math1 has disappeared forever? He may turn up again now. Maybe he makes a post every one and a half year!\n\nAnd why does it show 78 when I roll over your name?[/quote]\r\n\r\n :huh:  :?:   No idea, I am number 359 according to the member list\r\n\r\nyou show 17356, while you are in fact 17201",
  "Solution_9": "Possibly one counts the number of registrations, the other one the number of existing users.\r\nOr somewhat like that.\r\n\r\nPS: yes, bans are very rare, but one out of $9999$ in about one year isn't much.",
  "Solution_10": "Hmm, I heard about a member who got agressive about Iranian-Israeli relations, ultimately ending in a ban, but Valentin accepted his pleas to come back to matlinks.  I don't if I got that story completely, but CAN he do that?  When you are 'killed', is your existence wiped off of this forum?  (your private messages, number of posts, your list of posts)\r\n\r\nAnd now one other thing , if microsoft decides to prey on me, and I lose my hotmail account... are my mathlink days over then.  Or will I be able to come back with another email account, but the same username??",
  "Solution_11": "Well, there is a difference between ban and deletion, the first one makes the forum just unaccessible for you, the second one deletes everything like  e.g. PMs (but posts get just \"Guests\" or somewhat like that, not sure). At my last post, I meant more or less deletion instead of ban.\r\nAnd as much as I know, there where two users banned in these Iranian-Israeli relations, only one of them got back (but they both really only got banned I think).\r\nBut I know of at least one user that was really deleted.\r\n\r\nFor the second, you can change your mail-address mathlinks uses as long as you have the MathLinks-password. When you loose both, you get into troubles... (but there should be a way, too)",
  "Solution_12": "interesting :\r\n\r\nsome questions : \r\n\r\n1) can valentin see our passwords?  And our email accounts (that are hidden if you refuse to display them)?\r\n\r\n2) do banned/deleted people get warnings before they are thrown off?\r\n\r\n3) can moderators ban/delete as well or is that decision only up to valentin himself?\r\n\r\n4)can mathlinks do anything against double account people?\r\nI really hate that, people who get into arguments and nobody agrees with them... except one other guy that doesn't say anything else , or banned people who refuse to let the site be (you should the IMO article on wikipedia, it's pathetic on the talk page)",
  "Solution_13": "Just to clear a bit of the mystery here :P Math1 is a test account that I occasionally use to check stuff up :D",
  "Solution_14": "[quote=\"fredbel6\"]interesting :\n\nsome questions : \n\n1) can valentin see our passwords?  And our email accounts (that are hidden if you refuse to display them)?\n\n2) do banned/deleted people get warnings before they are thrown off?\n\n3) can moderators ban/delete as well or is that decision only up to valentin himself?\n\n4)can mathlinks do anything against double account people?\nI really hate that, people who get into arguments and nobody agrees with them... except one other guy that doesn't say anything else , or banned people who refuse to let the site be (you should the IMO article on wikipedia, it's pathetic on the talk page)[/quote]\r\n\r\nI think he can see your email but not your password, and while the other questions are really for moderators to answer, I would say it depends on the circumstances.",
  "Solution_15": "/bump......",
  "Solution_16": "oh my a 14 year old thread\n\nlook at how far aops has come; it now has like 500k useres i think",
  "Solution_17": "[quote=Potato2017]oh my a 14 year old thread\n\nlook at how far aops has come; it now has like 500k useres i think[/quote]\n\nI think 600k now",
  "Solution_18": "woah...\n2006",
  "Solution_19": "[quote=10000th User]Well, people were talking about me when I first joined, so let's see who will be the next person to hit 20000th user :D[/quote]\n\nRias Gremory is the OG of AoPSers! ",
  "Solution_20": "Currently the newest user is s128190 which is user #687863 (I found him)\n\nhttps://artofproblemsolving.com/community/user/687863",
  "Solution_21": "[quote=Valentin Vornicu]Haha, I know the secret identity of the user #20000 (with the username #20000) that is and it's not user #10000 :)\n\nAbout admins: no you need not 42 points in the IMO. There is no condition that can be fulfilled that guarantees us needing you in the team or anything like that. There are though conditions that, if not fulfills, automatically disqualify any attempt: being very fluent in English is one of them for example.[/quote]\n\n\nWho-",
  "Solution_22": "bruh how do u have a space in ur username",
  "Solution_23": "that's the 2nd user with there being no first",
  "Solution_24": "prob admin or dev.",
  "Solution_25": "Special users can do that\nspaces I mean",
  "Solution_26": "Who will be the 1,000,000th user? I hope they're active. [tip=nothing to see here dotted]they will probably be more orzed than dotted and cents[/tip]",
  "Solution_27": "[quote=StarshipTigerTheNext]Who will be the 1,000,000th user? I hope they're active. [tip=nothing to see here dotted]they will probably be more orzed than dotted and cents[/tip][/quote]\n\nI see what you did there  :-D ",
  "Solution_28": "How do you even check what user you are",
  "Solution_29": "Almost all of you in the top posts died anyway."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometry solved"
  ],
  "Problem": "$ABCDEF$ is a cyclic haxegon with $AB=CD=EF$. $G, H, I$ are the midpoints of $BC, DE, AF$. \r\n\r\nIs $\\triangle{GHI}$ an equalateral one? Prove your answer.",
  "Solution_1": "[quote=\"shobber\"]$ABCDEF$ is a cyclic haxegon with $AB=CD=EF$. $G, H, I$ are the midpoints of $BC, DE, AF$. \n\nIs $\\triangle{GHI}$ an equalateral one?[/quote]\r\n\r\nThis holds only when AB = CD = EF is the radius of the circumcircle of ABCDEF. This fact is known under the name \"asymmetric propeller theorem\". Check out http://www.mathlinks.ro/Forum/viewtopic.php?t=16908 and http://www.mathlinks.ro/Forum/viewtopic.php?t=45874 .\r\n\r\n  Darij",
  "Solution_2": "Thanks darij.  :)"
}
{
  "Tag": [],
  "Problem": "If the nth term of a product is $ {(4(2k\\minus{}1)^4)}\\plus{}1$ as numerator and $ {(4(2k)^4)}\\plus{}1$ as denominator find the product up to 6 terms.\r\nThis is a star question",
  "Solution_1": "i guess using sophie germain identity we will get a telescopic cancellation but with me that got too computational",
  "Solution_2": "[hide]By the Sophie Germain Identity, we have $ 4x^4\\plus{}1\\equal{}(2x^2\\minus{}2x\\plus{}1)(2x^2\\plus{}2x\\plus{}1)$. Also note that $ 2x^2\\minus{}2x\\plus{}1\\equal{}2(x\\minus{}1)^2\\plus{}2(x\\minus{}1)\\plus{}1$. Hence:\n\\[ \\frac{4\\cdot 1^4\\plus{}1}{4\\cdot 2^4\\plus{}1}\\cdot\\frac{4\\cdot 3^4\\plus{}1}{4\\cdot 4^4\\plus{}1}\\cdots\\frac{4\\cdot 11^4\\plus{}1}{4\\cdot 12^4\\plus{}1}\\equal{}\\frac{2\\cdot 1^2\\minus{}2\\cdot 1\\plus{}1}{2\\cdot 12^2\\plus{}2\\cdot 12\\plus{}1}\\equal{}\\frac 1{313}.\\][/hide]",
  "Solution_3": "Your solution is a little bit vague",
  "Solution_4": "I did not feel like typing all the computation out. Let me do this again:\r\n\\begin{align*}\\prod_{k=1}^6 \\frac{4(2k-1)^4+1}{4(2k)^4+1}&= \\frac {4\\cdot 1^4 + 1}{4\\cdot 2^4 + 1}\\cdot\\frac {4\\cdot 3^4 + 1}{4\\cdot 4^4 + 1}\\cdots\\frac {4\\cdot 11^4 + 1}{4\\cdot 12^4 + 1} \\\\\r\n&= \\frac{(2\\cdot 1^2-2\\cdot 1+1)(2\\cdot 1^2+2\\cdot 1+1)}{(2\\cdot 2^2-2\\cdot 2+1)(2\\cdot 2^2+2\\cdot 2+1)}\\cdot \\frac{(2\\cdot 1^3-2\\cdot 3+1)(2\\cdot 1^3+2\\cdot 3+1)}{(2\\cdot 4^2-2\\cdot 4+1)(2\\cdot 4^2+2\\cdot 4+1)}\\\\\r\n&\\cdots\\frac{(2\\cdot 11^2-2\\cdot 1+1)(2\\cdot 11^2+2\\cdot 11+1)}{(2\\cdot 12^2-2\\cdot 12+1)(2\\cdot 12^2+2\\cdot 12+1)}.\\end{align*}\r\nNote that since $ 2n^2-2n+1=2(n-1)^2+2(n-1)+1$ for any $ n$, this product telescopes to leave:\r\n\\begin{align*}\r\n&= \\frac {2\\cdot 1^2 - 2\\cdot 1 + 1}{2\\cdot 12^2 + 2\\cdot 12 + 1} \\\\\r\n&= \\frac 1{313}.\\end{align*}",
  "Solution_5": "i was plainly lazy to do that calculation"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation",
    "geometry proposed"
  ],
  "Problem": "The plane is covered with network of regular congruent disjoint hexagons. Prove that there cannot exist a square which has its four vertices in the vertices of the hexagons.\r\n\r\n[i]Gabriel Nagy[/i]",
  "Solution_1": "We can even assume the lattice is triangular: for each hexagon, add a point in its center and connect it to the six vertices of the hexagon. The conclusion holds for this larger lattice as well:\r\n\r\nAssume WLOG that we can find points $ A,B $ belonging to this lattice s.t. if $ O $ is the origin of the plane, $ OB $ is obtained from $ OA $ through a rotation of angle $ \\frac\\pi 2 $ around $ O $. This is equivalent to finding integers $ a,b,c,d $ s.t. $ \\frac{a+b\\varepsilon}{c+d\\varepsilon}=i $, where $ \\varepsilon $ is a solution to $ x^2+x+1=0 $ ($ a+b\\varepsilon $ represents the point $ A $, while $ c+d\\varepsilon $ - the point $ B $). In turn, this implies that we can find rationals $ u,v $ with $ u+v\\varepsilon=i $. This is claerly absurd, since eliminating $ u $ between $ u+v\\varepsilon=i,u+v\\bar\\varepsilon=-i $ gives an irrational value for $ v $."
}
{
  "Tag": [
    "inequalities",
    "trigonometry"
  ],
  "Problem": "Prove that $ (\\sin x \\plus{}\\cos x)^4 \\leq 4$",
  "Solution_1": "[quote=\"enndb0x\"]Prove that $ (\\sin x \\plus{} \\cos x)^4 \\leq 4$[/quote]\r\n\r\n$ \\Leftrightarrow (\\sin \\ x \\plus{} \\cos \\ x)^2\\le 2\\Leftrightarrow \\sin^2\\ x \\plus{} \\cos^2 \\ x \\plus{} 2\\sin\\ x\\cos \\ x\\le 2\\Leftrightarrow 2\\sin \\ x\\cos\\ x\\le$ $ 1\\Leftrightarrow \\sin \\ 2x\\le 1$\r\n\r\nbetter for intermediate forum .",
  "Solution_2": "[quote=\"enndb0x\"]Prove that $ (\\sin x \\plus{} \\cos x)^4 \\leq 4$[/quote]\r\n\r\n$ \\sin x \\plus{} \\cos x \\equal{} \\sqrt{2} \\ \\sin (x\\plus{} \\frac{\\pi}{4})$\r\n\r\n$ ( \\sin x \\plus{} \\cos x)^4 \\equal{} 4 \\ \\sin ^4 (x \\plus{} \\frac{\\pi}{4}) \\leq 4$"
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "polynomial",
    "LaTeX"
  ],
  "Problem": "Just doing some homework and I came across an equation that I can't seem to factorise. The surd has confused me  :oops: .\r\n\r\nThe equation:\r\n[b]4x^2 - 12\u221a2x + 18[/b]\r\n\r\nThis is what I have done for the next step but I'm not sure if it's right:\r\n2(2x^2 - 6\u221a2x +9)\r\n\r\nCan anyone please help?",
  "Solution_1": "[hide=\"hint\"]Let $y=x\\sqrt{2}$. The polynomial looks nicer in terms of $y$.[/hide]",
  "Solution_2": "So it would be\r\n\r\n2(2x^2 - 6yx +9)?",
  "Solution_3": "You would need to convert all of the $x$'s to $y$'s in order to make the substitution useful.\r\n\r\nThis is $2(y^2-6y+9)$, which can be factored further, of course.\r\n\r\nAfter factoring completely, rewrite in terms of $x$.",
  "Solution_4": "so it's \r\n\r\n2(y^2 -6y+9)\r\n[b]2(y-3)^2[/b]\r\n\r\nbut when you expand it again it doesn't work out back to the original equation? because the surd is in the first section too.. sorry i must seem really dumb.\r\n\r\n2(x\u221a3-3)^2\r\n2(x\u221a3-3)(x\u221a3-3)\r\n2(2x^2\u221a3 - 6\u221a3 +9)\r\n4x^2\u221a3- 12\u221a3 +18",
  "Solution_5": "2(y^2 - 6y + 9)\r\n2(y - 3)^2\r\nexpanding out again..\r\n2(sqrt[2]x - 3)^2\r\n2( (sqrt[2]^2*x^2 - 2*3*sqrt[2]*x + 3^2)\r\n2(2x^2 - 6*sqrt[2]x + 9)\r\nand you're back to the same thing =]",
  "Solution_6": "Thanks heaps. I get it now.",
  "Solution_7": "What's a surd? :?",
  "Solution_8": "square roots that cant be simplified into an integer, i think",
  "Solution_9": "[quote=\"laurenluvspink\"]Just doing some homework and I came across an equation that I can't seem to factorise. The surd has confused me  :oops: .\n\nThe equation:\n[b]4x^2 - 12\u221a2x + 18[/b]\n\nThis is what I have done for the next step but I'm not sure if it's right:\n2(2x^2 - 6\u221a2x +9)\n\nCan anyone please help?[/quote]\r\n\r\n4X^2-12\u221a2X+18=0.\r\n\r\nRoot (Only one): (3/2)\u221a2.\r\n\r\nThen, 4X^2-12\u221a2X+18=4(X-3/2\u221a2)(X-3/2\u221a2).",
  "Solution_10": "How do you type those square-root thingies without $\\LaTeX$?",
  "Solution_11": "It's a special character. You can go into a text formatter (MS Word) make one, copy/paste it, or copy/paste his into the text field in which you reply with.",
  "Solution_12": "[quote=\"JavaMan\"]square roots that cant be simplified into an integer, i think[/quote]\r\n\r\nSpecifically, square roots that are irrational--can't be simplified to rational numbers"
}
{
  "Tag": [],
  "Problem": "I understand long division but synthetic i dont like...so can someone solve this division problem using synthetic division....thanks\r\n\r\n\r\n$x^3-4x^2-19x=9/x+3$\r\n\r\nthanks",
  "Solution_1": "-3[_1__-4___-19___0___\r\n ______  -3 ___21 __-6\r\n____1 _ -7____2 __ -6\r\n\r\nX^2 - 7X + 2 Remainder -6\r\nSorry for the sloppyness I'm not sure how to use latex. Do you see what I did?\r\n\r\n:starwars:",
  "Solution_2": "[hide]\nYou use 3 for the divider, but make it -3.\nYou take the coefficents of $x^3-4x^2-19$\nThen, you bring down the one, multiple -3 by the number and put it below the next number and subtract:\n\n\n-3\n  1             -4             -19\n                 -3               3     \n_____________________\n  1             -1              -22\nSo: $x^2-x-\\frac{22}{x+3}$\nUmmm... I think that's how you do it.\nThe -3 is supposed to be under the -4, and the 3 under the -19.[/hide]"
}
{
  "Tag": [
    "geometry",
    "angle bisector",
    "area of a triangle"
  ],
  "Problem": "Prove that it is impossible to have a triangle in which the trisectors of an angle also trisect the opposite side.",
  "Solution_1": "[hide=\"Hint\"]Begin by showing that if an angle-bisector in a triangle is also the median to the opposite side, it must be perpendicular to that opposite side, and the triangle is isosceles.[/hide]",
  "Solution_2": "The area of a triangle is $ \\frac {1}{2}absin\\theta$.\r\n\r\nBy trisecting the side, you now have 3 triangles with equal area. Label the sides going from the vertex $ p,q,r,s$ from left to right.\r\n\r\n$ \\frac{1}{2}pq*sin\\theta \\equal{} \\frac{1}{2}qr*sin\\theta \\equal{} \\frac{1}{2}rs*sin\\theta$\r\n\r\nOr $ pq \\equal{} qr \\equal{} rs$ so From the first equality, $ p \\equal{} r$ from the second $ q \\equal{} s$.\r\n\r\nNote that they can't all be equal, since there can be at most 2 segments from a point to a line with equal length.\r\n\r\nNow, it is a simple argument. Ask yourself if $ p > q$ or $ p < q$ can you have the segments in that order? Hint, consider the foot of the altitude from the vertex, and the distances from the bases of the segments to the foot of the altitude."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "The sum of the first n terms of the series 1^2+2.(2^2)+3^2+2.(4^2)+5^2+2.(6^2)...... is {n(n+1)^2}/2, when n is even.What will be the sum when n is odd?",
  "Solution_1": "[quote=\"papu24\"]The sum of the first n terms of the series 1^2+2.(2^2)+3^2+2.(4^2)+5^2+2.(6^2)...... is {n(n+1)^2}/2, when n is even.What will be the sum when n is odd?[/quote]\r\n\r\nDo you mean:\r\n\r\n$ 1^2 \\plus{} 2(2^2) \\plus{} 3^2 \\plus{} 2(4^2) \\plus{} 5^2 \\plus{} 2(6^2)\\equal{}\\frac{n(n \\plus{} 1)^2}{2}$\r\n\r\n? Please use $ LaTeX$ in the future.",
  "Solution_2": "[hide]\n$ 1^2 \\plus{} 2(2^2) \\plus{} 3^2 \\plus{} 2(4^2)...\\plus{} 2(n^2) \\equal{} \\frac {n(n \\plus{} 1)^2}{2}$\n\nWe are attempting to find the value of N+1. We currently have the value of N. Set $ m \\equal{} n \\minus{} 1$, so that we can work with the odd series without being confused\n\n$ 1^2 \\plus{} 2(2^2) \\plus{} 3^2 \\plus{} 2(4^2)...\\plus{} (m^2) \\equal{} \\frac {(m\\minus{}1)(m)^2}{2}\\plus{}m^2$\n\n$ \\equal{} \\frac {(m\\minus{}1)m^2\\plus{}2m^2}{2}$\n\nFactor out m^2\n\n$ \\equal{} \\frac {(m\\minus{}1\\plus{}2)m^2}{2}$\n\n$ \\equal{} \\frac {(m\\plus{}1)m^2}{2}$[/hide]",
  "Solution_3": "Uh, in your second line, don't you mean set $ m\\equal{}n\\plus{}1$?"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "angle bisector",
    "congruent triangles",
    "geometry solved"
  ],
  "Problem": "Let ABC be a triangle. M is a point in it's plane. The bisectrix of angles BMC, AMC,AMB intersect BC,AC,AB in A1,B1,C1.\r\nShow that:\r\na) AA1 \\cap BB1 \\cap CC1=P\r\nb) PA/PA1*PB/PB1*PC/PC1=8 if and only if M is the circumcenter of ABC",
  "Solution_1": "Do you mean that M lies inside ABC ? If so, the first part is quite obvious.\r\n\r\nWe have t(a) = CA1/BA1 = CM/BM by the angle bisector theorem. Similarly, t(b) = AB1/CB1 = AM/CM and t(c) = BC1/AC1 = BM/AM. Now t(a)t(b)t(c) = (CM/BM)(AM/CM)(BM/AM) = 1 so AA1, BB1 and CC1 are concurrent at P by Ceva's theorem.\r\n\r\nNow we have to show that PA*PB*PC = 8*PA1*PB1*PC1 if and only if M is the circumcenter. Now PA/PA1 = AB/A1B, PB/PB1 = BC/B1C, PC/PC1 = AC/A1C. So we should prove that AB*BC*CA = 8*A1B*B1C*C1A if and only if M is the circumcenter. The if-part is obvious. If M is indeed the circumcenter, then triangles MAB, MBC, MCA are isosceles so A1, B1, C1 are the midpoints of BC, CA, AB. In that case AB = 2*C1A, BC = 2*A1B, CA = 2*B1C so indeed AB*BC*CA = 8*A1B*B1C*C1A. Now the \"only if\" -part remains.",
  "Solution_2": "Here's the proof of the \"only-if\" part. So we want to prove that if 8*A1B*B1C*C1A = AB*BC*CA, then M is the circumcenter.\r\nNow note that A1B*B1C*C1A = A1C*C1B*B1A (by Ceva).\r\nSo A1B*B1C*C1A = \\sqrt(A1B*A1C)*\\sqrt(B1C*B1A)*\\sqrt(C1A*C1B).\r\nThen \\sqrt(A1B*A1C) \\leq 1/2(A1B + A1C) = BC/2 with equality only if A1B = A1C, i.e. if A1 is the midpoint of BC.\r\nSimilarly, \\sqrt(B1C*B1A) \\leq AC/2 and \\sqrt(C1A*C1B) \\leq AB/2.\r\nMultiplying, we have\r\n8\\sqrt(A1B*A1C)*\\sqrt(B1C*B1A)*\\sqrt(C1A*C1B) \\leq AB*BC*CA \r\nwith equality only if A1, B1, C1 are the midpoints of BC, CA, AB.\r\nSo if 8\\sqrt(A1B*A1C)*\\sqrt(B1C*B1A)*\\sqrt(C1A*C1B) = AB*BC*CA then A1, B1, C1 are the midpoints of BC, CA, AB. This means M is the circumcenter (this is obvious, it can be proven by congruent triangles for example.)",
  "Solution_3": "It is very easy to see, that PA/PA_1=AB_1/B_1C+AC_1/C_1B= MA/MC+MA/MB.\r\n\r\n(MA/MC+MA/MB)(MC/MA+MC/MB)(MB/MC+MB/MA)\\geq 8  and equality holds iff MA=MB=MC"
}
{
  "Tag": [
    "inequalities open",
    "inequalities"
  ],
  "Problem": "If $ r\\geq 0$ and $ \\infty >f(r),g(r)\\geq 0$ then\r\n\\[ \\left(\\frac {w_1f^r(r) \\plus{} w_2g^r(r)}{w_1 \\plus{} w_2}\\right)^{\\frac {1}{r}}\\geq \\left(f^{w_1}(r)g^{w_2}(r)\\right)^{\\frac {1}{w_1 \\plus{} w_2}}\r\n\\]\r\nhere $ w_1,w_2 > 0$",
  "Solution_1": "...and $ f(r),g(r)$ is increasing functions. No one interested?",
  "Solution_2": "[quote=\"Sunjee\"]If $ r\\geq 0$ and $ \\infty > f(r),g(r)\\geq 0$ then\n\\[ \\left(\\frac {w_1f^r(r) \\plus{} w_2g^r(r)}{w_1 \\plus{} w_2}\\right)^{\\frac {1}{r}}\\geq \\left(f^{w_1}(r)g^{w_2}(r)\\right)^{\\frac {1}{w_1 \\plus{} w_2}}\n\\]\nhere $ w_1,w_2 > 0$[/quote]Is this what you mean really ? If so, simplily put $ f(r)\\equal{}: a$, $ g(r)\\equal{}: b$, $ \\frac {w_1}{w_1\\plus{}w_2}\\equal{}: \\lambda$, $ \\frac {w_2}{w_1\\plus{}w_2}\\equal{}: \\mu$,  and your ineqality becomes\r\n\r\n$ (\\lambda a^r\\plus{}\\mu b^r)^{\\frac1r}\\ge a^\\lambda b^\\mu$. Weigthed power mean.  :lol: :rotfl:",
  "Solution_3": "[quote=\"Amunaho\"][quote=\"Sunjee\"]If $ r\\geq 0$ and $ \\infty > f(r),g(r)\\geq 0$ then\n\\[ \\left(\\frac {w_1f^r(r) \\plus{} w_2g^r(r)}{w_1 \\plus{} w_2}\\right)^{\\frac {1}{r}}\\geq \\left(f^{w_1}(r)g^{w_2}(r)\\right)^{\\frac {1}{w_1 \\plus{} w_2}}\n\\]\nhere $ w_1,w_2 > 0$[/quote]Is this what you mean really ? If so, simplily put $ f(r) \\equal{} : a$, $ g(r) \\equal{} : b$, $ \\frac {w_1}{w_1 \\plus{} w_2} \\equal{} : \\lambda$, $ \\frac {w_2}{w_1 \\plus{} w_2} \\equal{} : \\mu$,  and your ineqality becomes\n\n$ (\\lambda a^r \\plus{} \\mu b^r)^{\\frac1r}\\ge a^\\lambda b^\\mu$. Weigthed power mean.  :lol: :rotfl:[/quote]\r\n\r\n $ f(r)\\equal{}a,g(r)\\equal{}b$ this is not true"
}
{
  "Tag": [
    "analytic geometry",
    "complex numbers",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "What are the Cauchy-Riemann Equations in polar coordinates FOR the modulus and the argument?",
  "Solution_1": "The Cauchy-Riemann equations state that:\r\n\r\n$ \\frac{\\partial u}{\\partial x}\\equal{}\\frac{\\partial v}{\\partial y}$ and $ \\frac{\\partial v}{\\partial x}\\equal{}\\minus{}\\frac{\\partial u}{\\partial y}$\r\n\r\nThat's if $ z\\equiv x\\plus{}iy$\r\n\r\nIf $ z\\equiv re^{i\\theta}$ we get \r\n\r\n$ \\frac{\\partial u}{\\partial r}\\equal{}\\frac{1}{r}\\frac{\\partial v}{\\partial \\theta}$ and $ \\frac{1}{r}\\frac{\\partial u}{\\partial \\theta}\\equal{}\\minus{}\\frac{\\partial v}{\\partial r}$\r\n\r\nwhich is what you were asking.",
  "Solution_2": "Thank you for your answer, let me tell you why Im confused.\r\n\r\nWe know that if  $ f(z) \\equal{} u(x,y) \\plus{} i v(x,y)$ and $ f$ is analytic then the real and the imaginary part are connected by the\r\nCauchy-Riemann equations: \r\n\r\n$ \\frac {\\partial v}{\\partial x} \\equal{} \\minus{} \\frac {\\partial u}{\\partial y},$ and $ \\frac {\\partial u}{\\partial x} \\equal{} \\frac {\\partial v}{\\partial y}.$\r\n\r\nIf $ f(z) \\equal{} u(r,\\theta) \\plus{} i v(r,\\theta)$ then we have the Cauchy-Riemann equations in polar coordinates:\r\n\r\n$ \\frac {\\partial u}{\\partial r} \\equal{} \\frac {1}{r}\\frac {\\partial v}{\\partial \\theta},$ and  $ \\frac {1}{r}\\frac {\\partial u}{\\partial \\theta} \\equal{} \\minus{} \\frac {\\partial v}{\\partial r}.$\r\n\r\nNow if we put $ f(z) \\equal{} R(x,y)e^{i \\phi(x,y)}$ then the modulus $ R$ and the argument $ \\phi$ are connected by the relations\r\n\r\n$ \\frac {\\partial R}{\\partial x} \\equal{} R \\frac {\\partial \\phi}{\\partial y}$ and $ \\frac {\\partial R}{\\partial y} \\equal{} \\minus{} R \\frac {\\partial \\phi}{\\partial x}$.\r\n\r\nSo my questions is: The latter are the Cauchy-Riemann equations FOR the modulus and the argument? If so, then I guess I can deduce them in polar coordinates, in the same way we can do it for the real and the imaginary part.\r\n\r\nI mean if $ f(z) \\equal{} R(r,\\theta)e^{i \\phi(r,\\theta)}$ what are the equations relating the modulus and the argument.\r\n\r\nThank you"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c$ be positive real numbers such that $ abc\\equal{}a\\plus{}b\\plus{}c\\plus{}2$.Prove that:\r\n$ \\frac{ab}{(a\\plus{}c)(b\\plus{}c)}\\plus{}\\frac{bc}{(b\\plus{}a)(c\\plus{}a)}\\plus{}\\frac{ca}{(c\\plus{}b)(a\\plus{}b)}\\le \\frac{3(abc\\minus{}4)}{8\\plus{}abc}$",
  "Solution_1": "[quote=\"vo thanh van\"]Let $ a,b,c$ be positive real numbers such that $ abc \\equal{} a \\plus{} b \\plus{} c \\plus{} 2$.Prove that:\n$ \\frac {ab}{(a \\plus{} c)(b \\plus{} c)} \\plus{} \\frac {bc}{(b \\plus{} a)(c \\plus{} a)} \\plus{} \\frac {ca}{(c \\plus{} b)(a \\plus{} b)}\\le \\frac {3(abc \\minus{} 4)}{8 \\plus{} abc}$[/quote]\r\nIT's easy because $ (abc)^2\\geq \\frac{9}{8}(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca) \\geq 64$! :P"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "Prove that none of the digits 2,4,7,9 can be the last digit of a number equal to the sum $1+2+3+ . . . +n$  where $n$ is an arbitrary natural number.",
  "Solution_1": "$\\sum_{i=1}^{n}{n}=\\frac{n(n+1)}{2}$ , after this just check remainders on 10.",
  "Solution_2": "[hide]It is easily verifiable for n = 1, 2, 3, 4....., 10\nSo we will prove it by induction for n+10\n\nAssume true for n=k, ie (n+1)(n)/2 not equal no 2,4,7,9 (mod 10)\nSubbing in k+10 and simplifing we get (n+1)(n)/2 + 55 + 10n\ntherefore by putting in k +10 we either add 5 or 0 (mod 10) so we end up that once again, (k+11)(k+10)/2 does not end in 2,4,7,9... and by the princable or induction, is true for all n. (This is a quick, not reall detailed proof, but not to hard to verify yourself...)[/hide]",
  "Solution_3": "Curses!I hate when someone puts in their message while your typing yours! :)",
  "Solution_4": "[hide=\"Even simpler is\"]\nmod 5 rather than 10 :) But of course, in the few seconds it takes you to think of that, you could probably have finished the mod 10 way..\n[/hide]",
  "Solution_5": "[quote=\"seamusoboyle\"]Curses!I hate when someone puts in their message while your typing yours! :)[/quote]\r\nI guess that \"someone\" don't know when you are typing your message. :)",
  "Solution_6": "Tis what i ment! :)"
}
{
  "Tag": [
    "floor function",
    "ceiling function"
  ],
  "Problem": "How many positive multiples of 13 are three-digit integers?",
  "Solution_1": "There are 900 positive 3-digit integers, from 100~999. So, our answer is 900/13, or 69.23..., and you ignore the decimal, so 69.\r\nP.S. do not make the mistake of rounding...for example, when 900/7, it is NOT 129.",
  "Solution_2": "Hmm, actually first you do $ 99 \\div 13 \\approx 69$ and then $ 100 \\div 13 \\equal{} \\approx 8$. $ 69 \\minus{} 8 \\plus{} 1 \\equal{} \\boxed{62}$.",
  "Solution_3": "[quote=\"isabella2296\"]Hmm, actually first you do $ 99 \\div 13 \\approx 69$ and then $ 100 \\div 13 \\equal{} \\approx 8$. $ 69 \\minus{} 8 \\plus{} 1 \\equal{} \\boxed{62}$.[/quote]\nI think isabella2296 meant $\\lfloor\\frac{999}{13}\\rfloor=76$ and $\\lceil\\frac{100}{13}\\rceil=8$.  Therefore, the number of 3-digit multiples of $13$ is $76-8+1=68+1=\\boxed{69}$."
}
{
  "Tag": [],
  "Problem": "Warum gelten reelle Formeln f\u00fcr $ sinx$, $ cosx$, $ sinhx$, $ coshx$, $ e^x$ automatisch auch im Komplexen? Wie ist die Situation fuer Funktionen, die nicht ueberall definiert sind?($ arctan x$, $ log x$(!) unw.)\r\n\r\n\r\n\r\nFitim",
  "Solution_1": "Diese Funktionen, wie auch alle daraus resultierenden, sind lokal holomorph/analytisch (d.h. zu jedem Punkt im Definitionsbereich gibt es eine offene Umgebung, in dem die Funktion komplex differenzierbar ist/in dem die Funktion durch eine Potenzreihe gegeben ist; beide Begriffe stimmen faktisch \u00fcberein). Nun gibt es den Eindeutigkeitssatz f\u00fcr holomorphe Funktionen:\r\n$ f,g: U \\to \\mathbb C$ seien holomorph auf einem (offenen, zusammenh\u00e4ngenden) Gebiet $ U$. Enthalte weiter $ M \\subset U$ einen seiner H\u00e4ufungspunkte. Dann gilt: Stimmen $ f,g$ auf $ M$ \u00fcberein, so auf ganz $ U$.\r\n\r\nNimmt man also einen zusammenh\u00e4ngenden Definitionsbereich in dem genannten Funktionen holomorph sind, so gilt eine \"Formel\" bereits dann, wenn sie z.B. auf einem Intervall einer positiven L\u00e4nge gilt.",
  "Solution_2": "Danke ZetaX!  :wink:"
}
{
  "Tag": [
    "inequalities",
    "function",
    "inequalities theorems"
  ],
  "Problem": "[b]Triangles relation for inequalities:[/b]\r\n\r\nHi: I just wanted to know from you, some of the relations between triangle things, specially, I would like to find expressions in terms of $R, r ,s (ABC)$, for the symetric functions of the sides $a,b,c$. If someone knows something, I will be very pleased if he could pot it here...",
  "Solution_1": "[quote=\"Pascual2005\"][b]Triangles relation for inequalities:[/b]\n\nHi: I just wanted to know from you, some of the relations between triangle things, specially, I would like to find expressions in terms of $R, r ,s (ABC)$, for the symetric functions of the sides $a,b,c$. If someone knows something, I will be very pleased if he could pot it here...[/quote]\r\n\r\nIt's well known that the following three identities hold in every triangle.\r\n$a+b+c=2s$;\r\n$ab+bc+ca=s^2+4Rr+r^2$;\r\n$abc=4Rrs$.\r\nWith them, we can ratiocinate all other symetric functions of the sides $a,b,c$.\r\nFor example,\r\n$a^3+b^3+c^3=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=2s(s^2-6Rr-3r^2)$."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "I post this not to receive a solution, because this seems to be rather impossible, but because the results is magnificient:\r\n  There is a constant $C$ such that there are no numbers of the form $ y^n$ with $y>1$ and $n>C$ in the Fibonacci sequence.",
  "Solution_1": "Harazi: I have an idea that i cant finish...\r\n\r\nSuppose no such constant exist, first we can prove we can find  a composite index $m$, such that $F_m=y^k$, now i think we can extend this lemma [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=27757[/url] to the field $a+b\\sqrt{5}$. using it we can prove that $m=ty$, and that $F_t=y^{k-1}$ but then $F_m/F_t=y$ wich fails for big enough $k$ I dont know, it is just some crazy idea.... ;)  and i have not been trainig lately ( i am on VACATION!!)"
}
{
  "Tag": [
    "trigonometry",
    "ratio"
  ],
  "Problem": "Can two triangles have two equal sides and three equal angles, and still be noncongruent?\r\n\r\nIf yes, then how? ;)",
  "Solution_1": "well we know that they are similar, but we could have it so the corresponding sides are not equal",
  "Solution_2": "No, of course they have to be congruent (unless I misunderstood what you meant  :? ). If any 2 angles and 1 side of the triangles are equal, they are congruent, and here more than 1 side and more than 2 angles are equal.",
  "Solution_3": "They must be congruent by SAS or ASA.",
  "Solution_4": "[quote=\"Altheman\"]well we know that they are similar, but we could have it so the corresponding sides are not equal[/quote]\r\n\r\nI don't think that's true, because consider that you have a $3,4,5$ triangle, and using trig we see that the angles are $36.9, 53.1, 90$ degrees, let's say we have another triangle with the same angles and sides $4$ and $3$ but let's say $4$ is the hypotenuse, but $3$ can't be another side because it goes against the ratio of the sides and law of sines."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "What is the maximum number of regions that 9 planes can divide a box?",
  "Solution_1": "I'm guessing 81 (which is 9 squared). I think there may be more, but I can't find that solution.  :blush:\r\n\r\nnever mind (edit), what the heck am I thinking?",
  "Solution_2": "[hide]\n1 plane-2\n2-4\n3-8\n4-14?\n\npattern\n\n9planes-58\n[/hide]",
  "Solution_3": "I think with four cuts, you can have 16 pieces, just cut into the shape of an 8-sided asterisk. \r\n\r\nSo now I think it's 2^9, which is 512.",
  "Solution_4": "I got fifteen regions for four cuts now...\r\n\r\n1-2\r\n2-4\r\n3-8\r\n4-15?\r\n\r\ndifferences-2,4,7...\r\n11,16,22,29,37\r\n\r\nterms-\r\n5-26\r\n6-42\r\n7-64\r\n8-93\r\n\r\n9-130\r\n\r\n\r\nEDIT:\r\nI just noticed,\r\n\r\npoints cutting a line (points, section)\r\n1-2\r\n2-3\r\n3-4\r\n4-5...difference is 1\r\n\r\nlines cutting a plane (lines, regions)\r\n\r\n1-2\r\n2-4\r\n3-7\r\n4-11...differences are the ammount of sections from \"points cutting a line\"\r\n\r\nplanes cutting a space (planes, regions)\r\n\r\n1-2\r\n2-4\r\n3-8\r\n4-15\r\n5-26...differences are the ammount of regions from \"lines cutting a plane\"",
  "Solution_5": "[quote=\"pianoforte\"]I got fifteen regions for four cuts now...\n\n[/quote]\r\n\r\nStrive for excellence: try to find one more  :D",
  "Solution_6": "[hide]\nI think each plane divides the current space into 2 parts, so 9 planes would divide this into 2^9=512 parts??!!\n[/hide]",
  "Solution_7": "[quote=\"236factorial\"][quote=\"pianoforte\"]I got fifteen regions for four cuts now...\n\n[/quote=\"pianoforte\"]\n\nStrive for excellence: try to find one more  :D[/quote]\n\nI can't find one more, and my answer fits the pattern.  :? \n\n[quote][hide]EDIT:\nI just noticed,\n\npoints cutting a line (points, section)\n1-2\n2-3\n3-4\n4-5...difference is 1\n\nlines cutting a plane (lines, regions)\n\n1-2\n2-4\n3-7\n4-11...differences are the ammount of sections from \"points cutting a line\"\n\nplanes cutting a space (planes, regions)\n\n1-2\n2-4\n3-8\n4-15\n5-26...differences are the ammount of regions from \"lines cutting a plane\"[/hide][/quote][/quote]\r\n\r\nEDIT: I don't see how the forth plane could split all of the existing 8 planes, at least one has to be left out :?",
  "Solution_8": "sorry piano forte, I was thinking of something else. How did you get 8 for three cuts?",
  "Solution_9": "One plane going side to side and up and down, \r\nanother going side to side and forward and back, \r\nthe third going up and down, and forward and back, :P\r\n\r\nlike stacking eight cubes into a bigger cube, four on top, four on bottom.",
  "Solution_10": "______              \r\n|_|_|_|_|\r\n|_|_|_|_|\r\n|_|_|_|_|\r\n|_|_|_|_| = 16*4=64\r\n\r\n64 regions"
}
{
  "Tag": [],
  "Problem": "What makes a good math teacher/tutor/professor?\r\n\r\nI believe that a good math teacher is someone who can break the material down to the evel of students.\r\n\r\nYour view?\r\n\r\nI want several replies.",
  "Solution_1": "I believe a good math teacher is fair and appreciates everyone's skill in math, whether good or bad. \r\n\r\nA good math teacher should not give a lot of homework lol",
  "Solution_2": "a good math teacher is one who only cares about kids who do math competitions and doesnt teach the other people.",
  "Solution_3": "one that lets you go fast if you are getting it but understands when you don't",
  "Solution_4": "[quote=\"funcia\"]one that lets you go fast if you are getting it but understands when you don't[/quote]\r\n\r\none who believes that mistakes do happen and should be less strict about grades and more eabout learning the material itself",
  "Solution_5": "Anyone can take something and make it more complicated. It takes true insight and understanding to take something that's complicated and make it simple. It takes a truly good teacher to do this with mathematics.\r\n\r\nBuilding on that, mathematics teachers (before college, anyway) are notorious for making math seem like a collection of formulas. This goes with taking something simple and making it more complicated. A good mathematics teacher can show the relationships between concepts and unify ideas.\r\n\r\nA good math teacher is willing to deviate from the curriculum a bit and explore different concepts. Most math teachers will not teach things like Ptolmey's, finite differences, etc.\r\n\r\nA good math teacher does not teach to the test. Most math teachers teach to their own tests, and a lot of AP teachers teach to the AP test. I find this very annoying."
}
{
  "Tag": [
    "function",
    "symmetry",
    "conics",
    "parabola",
    "calculus"
  ],
  "Problem": "Function $y=kx^2-4x-1$ is decreasing on $[3,6]$. Find the range of $k$.",
  "Solution_1": "[hide=\"Solution\"]We have $y'=2kx-4$ and it must be negative on $\\lbrack 3,6\\rbrack$. The derivation is a linear function, so its maximal value is in the left boundary for $k<0$ or in the right boundary for $k>0$:\n\n\\begin{eqnarray*}1^\\circ\\quad k<0 \\land y'(3)\\leq 0\\implies k<0 \\land 6k-4\\leq 0\\implies k<0\\land k\\leq{2\\over 3}\\implies k<0\\end{eqnarray*}\n\n\\begin{eqnarray*}2^\\circ\\quad k>0\\land y'(6)\\leq 0\\implies k>0\\land 12k-4\\leq 0\\implies k>0\\land k\\leq{1\\over 3}\\implies k\\in(0,{1\\over 3}\\rbrack\\end{eqnarray*}\n\n$3^\\circ\\quad k=0\\implies y=-4x-1$, so the function is decreasing on the whole $\\mathbb{R}$.\n\nHence, the answer is $k\\in(-\\infty,{1\\over 3}\\rbrack$[/hide]",
  "Solution_2": "[hide=\"solution\"]\n\nFor negative $k$ , we have $y=k(x-\\frac{2}{k})^2-1-\\frac{4}{k}$ . So the symmetry line of the parabola is negative value and hence over $[3,6]$ , $y$ definitely decreases .\n\nFor positive $k$ , since the symmetry line is at $x=\\frac{2}{k}$ . So in order for $y$ to decrease over $[3,6]$ , the symmetry line must be at least at $6$ which means $\\frac{2}{k}\\geq 6$ or $k\\leq \\frac{1}{3}$\n\nHence combine both cases to get the range for $k$ to be $(-\\infty,\\frac{1}{3}]$\n[/hide]",
  "Solution_3": "No calculus, Farenhajt! :)"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "what proportion  of the area of a circle lies inside an inscribed regular hexagon?",
  "Solution_1": "[hide=\"My sorution\"]\nDivide the hexagon into 3 equliteral triangles with side $s$. All of these have area $\\frac{\\sqrt{3}}{4}s^{2}$. So the area of the hexagon is\n\\[\\frac{\\sqrt{3}}{4}s^{2}\\cdot 6=\\frac{3\\sqrt{3}}{2}s^{2}\\]\nTherefore, the proportion of the area of the circle that lies inside the hexagon is\n\\[\\frac{\\frac{3\\sqrt{3}}{2}s^{2}}{\\pi s^{2}}=\\frac{3\\sqrt{3}}{2\\pi}\\]\n[/hide]"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "perimeter"
  ],
  "Problem": "The ratio of the perimeters of two similar pentagons is 3 : 4.  If the length of the shortest side of the smaller pentagon is 48, what is the length of the shortest side of the larger pentagon?\r\n\r\nThe choices are:\r\n\r\n(A) 24\r\n\r\n(B) 36\r\n\r\n(C) 64\r\n\r\n(D) 72",
  "Solution_1": "This should be in Classroom Math...\r\n\r\n[hide]$ \\frac{48}{3}=\\frac{x}{4}$\n\n$ x=64$[/hide]",
  "Solution_2": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153452[/url]\r\n\r\nPlease.",
  "Solution_3": "He has the right to post questions, but they have to be in the right forum.",
  "Solution_4": "I am sharkman not interval.  Maybe this interval person was a real pain but I am tired of being compared to someone who probably lives on the other side of the globe.\r\n\r\nI am sharkman!!  Why sharkman???\r\n\r\nI like sharks, the fish.\r\n\r\nGet it?\r\n\r\nThank you i_like_pie for your defense but I have no idea what these people are talking about.",
  "Solution_5": "He used similarity. You know, if we say:\r\n\r\n$ \\frac{1}{x}= \\frac{2}{3}$\r\n\r\nthen $ x = \\frac{3}{2}$.\r\n\r\nSimilarity is a crucial part of geometry. Good luck learning it."
}
{
  "Tag": [],
  "Problem": "What is the sum of the numbers less than 200 that have exactly 9 divisors?",
  "Solution_1": "[hide]36,100,196 so 332[/hide]",
  "Solution_2": "[quote=\"usaha\"][hide]36,100,196 so 332[/hide][/quote]\r\nHow did you get those numbers?",
  "Solution_3": "[quote=\"usaha\"][hide]36,100,196 so 332[/hide][/quote]\n\n\n[hide]A number can only have an odd number of factors if it is a perfect square. Now to find the amount of factors a number has you add 1 to each exponent and multiply the sums. So this can equal 9 only when the number equals one exponent which has to be eight or, two exponents which are 2 and 2. [/hide]"
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "[color=violet][size=134]An abelian group $ M$ is called indecomposable if it can't be writen by  a direct sum of two non-zero subgroup .\n\nIs any indecomposable abelian group  $ M$    torsion-free(no finite order elements)?  except the trival situation:$ Z[1/p]/Z,Z_{p^{n}}$ \n\nin other word ,if an indecomposable abelian  group $ M$   is not torsion-free, it must be $ Z[1/p]/Z$, or $ Z_{p^{n}}$??[/size][/color]\r\n\r\n\r\n\r\n[size=150]i don't know whether this problem has been solved ?\n\nbut  in the condition that $ M$ has only finite maximal subgroups we have got it !!\n\nthanks very much for who gives  advices ![/size] ",
  "Solution_1": "This is theorem 10, page 22 of Kaplansky's textbook on Infinite Abelian Groups: an indecomposable abelian group is either Z[1/p]/Z, Z/p^nZ, or torsion-free.",
  "Solution_2": "thanks \r\nbut i don't have read this book.\r\nbut in another  book there is this result.\r\n\r\nINFINITE ABELIAN GROUPS\r\nLd.szlo Fuchs\r\nTtclane University\r\nNew Orleans, Louisiana\r\nV O L U M E I\r\n\r\n\r\n\"Theorem 3.1. A group C is cocyclic if and only if CgZ(pk) with\r\nk = 1, 2, - - .o r 00.\r\n\r\nCorollary 27.3 (Kulikov [ l ] ) . If a group contains elements of fitiite order,\r\nthen it contains a cocyclic direct summand.\r\nCorollary 27.4 (Kulikov [l]). A directly indecomposable group is either torsion-free or cocyclic.\"\r\n\r\nso the guess is right....but it isnot proved by me.... :("
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": ":pilot:  :censored::  Let a,b,c be posive numbers such that $ a\\plus{}b>c, b\\plus{}c>a, c\\plus{}a>b$ and $ a^2\\plus{}b^2\\plus{}c^2\\equal{}3$. Prove that      $ \\sum  \\sqrt{2\\minus{}a^2}\\geq a\\plus{}b\\plus{}c$",
  "Solution_1": "[quote=\"thegod277\"]:pilot:  :censored::  Let a,b,c be posive numbers such that $ a \\plus{} b > c, b \\plus{} c > a, c \\plus{} a > b$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 3$. Prove that      $ \\sum \\sqrt {2 \\minus{} a^2}\\geq a \\plus{} b \\plus{} c$[/quote]\r\nI think, it's wrong. Try $ a\\equal{}1.4$ and $ b\\equal{}c\\equal{}\\sqrt{0.52}.$ :wink:",
  "Solution_2": "sorry. i think you right. thank you! :("
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "could someone help me factor these:\r\n -x^3+8x^2-8x-32\r\nand -x^3+3x+2",
  "Solution_1": "[quote=\"mikejones\"]could someone help me factor these:\n -x^3+8x^2-8x-32\nand -x^3+3x+2[/quote]\r\n\r\nWrong difficulty level, I think. ;) \r\n\r\nBut as for your problems, think Rational roots. ;)",
  "Solution_2": "1 . (x-4)[(4-x)(2+x)+2x]\r\n\r\n2 . (x+1)(x+1)(2-x)",
  "Solution_3": "$x^3-8x^2+8x+32\\ \\ |\\ \\cdot (-1)$\r\n$x^3+0x^2-3x-2$\r\n$============$\r\n$8x^2-11x-34\\ \\ \\ \\ \\ \\ \\ |\\ \\cdot (-x)$\r\n$x^3+0x^2-3x-2\\ \\ \\ |\\ \\ \\cdot \\ 8$\r\n$============$\r\n$11x^2+10x-16\\ \\ \\ \\ \\ \\ |\\ \\ \\ \\cdot \\ 8$\r\n$8x^2-11x-34\\ \\ \\ \\ \\ \\ \\ \\ |\\ \\ \\cdot (-11)$\r\n$============$\r\nA.S.O. These polynominals hasn't common factors. I think you make a mistake in the coefficients !"
}
{
  "Tag": [
    "calculus",
    "integration",
    "analytic geometry",
    "trigonometry",
    "Gauss",
    "calculus computations"
  ],
  "Problem": "Calculate the integral\r\n\r\n$\\int \\int_V \\int \\frac{1}{x^2 + y^2 + z^2} \\text dx \\text dy \\text dz$\r\n\r\nover the volume $V$ defined by $x^2 + y^2 + (z - 1)^2 \\le 1$.",
  "Solution_1": "I am in a hurry right now, but let's see. I suppose that the best way to do it is make a change of coordinates to spherical coordinates, and so, If I am not mistaken (As I said, right now I don't have time to review my calculations), the integral becomes,\r\n\r\n\\[\\int_{0}^{2\\pi}\\int_{-\\frac\\pi2}^{\\frac\\pi2}\\int_{0}^{2\\cos\\phi}\\frac{1}{\\rho^2}\\;d\\rho\\;d\\phi\\;d\\theta\\]\r\n\r\nWhat is easier to compute. Sorry, gotta go. Bye,",
  "Solution_2": "In hurry you forgot to multiply by the Jacobian $\\rho^2\\sin\\varphi$. ;)  If you do that the integral becomes really easy...",
  "Solution_3": "[quote=\"fedja\"]In hurry you forgot to multiply by the Jacobian $\\rho^2\\sin\\varphi$. ;)  If you do that the integral becomes really easy...[/quote]\r\n\r\nIt was not just the missing jacobian $r^2 \\sin \\theta$, but also the integration limits. If you go to spherical coordinates $r, \\phi,  \\theta$\r\n\r\n$x = r \\cos \\phi \\sin \\theta$\r\n\r\n$y = r \\sin \\phi \\sin \\theta$\r\n\r\n$z = r \\cos \\theta$\r\n\r\nthe integration limits are not constant for either of $r, \\phi$ or $\\theta$. If you go to spherical coordinates $r, \\phi,  \\theta$\r\n\r\n$x = r \\cos \\phi \\sin \\theta$\r\n\r\n$y = r \\sin \\phi \\sin \\theta$\r\n\r\n$z = r \\cos \\theta + 1$\r\n\r\nthen the integration limits are constant and the integral becomes\r\n\r\n$\\int \\int_V \\int \\frac{1}{x^2 + y^2 + z^2} \\text dx \\text dy \\text dz = \\int_0^{2 \\pi} \\int_0^{\\pi} \\int_0^1 \\frac{r^2 \\sin \\theta}{r^2 + 2r \\cos \\theta + 1} \\text dr \\text d\\theta \\text d\\phi = 2 \\pi \\int_0^{\\pi} \\int_0^1 \\frac{r^2 \\sin \\theta}{r^2 + 2r \\cos \\theta + 1} \\text dr \\text d\\theta$\r\n\r\nThis can be integrated in a rather laborious way by expanding the integrand into a power series $\\sum_{n = 0}^{\\infty} a_n(\\cos \\theta) r^n$, integrating term by term and then figuring out how to sum the results. But there is another solution, much simpler than going to spherical coordinates.",
  "Solution_4": "[quote=\"yetti\"] If you go to spherical coordinates $r, \\phi,  \\theta$\n\n$x = r \\cos \\phi \\sin \\theta$\n\n$y = r \\sin \\phi \\sin \\theta$\n\n$z = r \\cos \\theta$\n\nthe integration limits are not constant for either of $r, \\phi$ or $\\theta$. \n\n[/quote]\r\nI think djimenez used \r\n$x = \\rho \\sin \\phi \\cos \\theta$\r\n\r\n$y = \\rho \\sin \\phi \\sin \\theta$\r\n\r\n$z = \\rho \\cos \\phi$\r\n\r\nThen his limits of integration are correct ;)",
  "Solution_5": "There are too many different angle conventions for spherical coordinates here.\r\n\r\ndjimenez is using the convention (standard in calculus books) that $\\theta$ is longitude, $\\phi$ is colatitude, and the radius is $\\rho$, but his limits are slightly wrong as well as the Jacobian: it should be $\\int_0^{2\\pi}\\int_0^{\\frac\\pi2}\\int_0^{2\\cos\\phi}\\sin\\phi\\, d\\rho\\,d\\phi\\, d\\theta$\r\n(The entire ball is in the \"north\" half-space $0\\le\\phi\\le\\frac\\pi2$. The southern half-space would be $\\frac\\pi2\\le\\phi\\le\\pi$, and I don't know where $\\frac{-\\pi}2$ came from.)\r\nThis is a simple integral- easier than yetti's method.\r\n\r\nyetti is using the convention (standard in physics books) that $\\theta$ is colatitude, $\\phi$ is longitude, and the radius is $r$.\r\n\r\nSpherical coordinates centered at the origin make the integrand nice. Since the limits aren't too bad, it works better than spherical coordinates centered at the ball of integration.",
  "Solution_6": "[quote=\"jmerry\"] it should be $\\int_0^{2\\pi}\\int_0^{\\frac\\pi2}\\int_0^{2\\cos\\phi}\\sin^2\\phi\\, d\\rho\\,d\\phi\\, d\\theta$\n[/quote]\r\nOops  :blush:  jmerry is right as to the limits. As to the power of $\\sin\\phi$, I disagree:\r\nthe volume element is $d\\rho\\cdot \\rho \\,d\\phi \\cdot \\rho\\sin\\phi \\,d\\theta$, so the integral becomes $\\int_0^{2\\pi}\\int_0^{\\frac\\pi2}\\int_0^{2\\cos\\phi}\\sin\\phi\\, d\\rho\\,d\\phi\\, d\\theta$. Seems like we all are not at our best today  :P",
  "Solution_7": "Oops... I was confused about a physical interpretation when I edited that in. $\\sin^2\\phi$ comes from the $z$-component of an inverse-square force.",
  "Solution_8": "[quote=\"fedja\"]\nThen his limits of integration are correct ;)[/quote]\r\n\r\nNot entirely. It should be\r\n\r\n$\\int_0^{2 \\pi} \\int_0^{\\frac \\pi 2} \\int_0^{2 \\cos \\phi} \\frac{r^2 \\sin \\phi}{r^2} \\text dr \\text d\\phi \\text d\\theta$\r\n\r\nWhat I had in mind was to substitute $\\frac{1}{r^2} = \\text{div}\\left( \\frac{\\vec r}{r^2} \\right)$ and then use Gauss' theorem\r\n\r\n$\\int \\int_V \\int \\frac{1}{r^2} \\text dV = \\int \\int_V \\int \\text{div}\\left( \\frac{\\vec r}{r^2} \\right) \\text dV = \\int_S \\int \\frac{\\vec r}{r^2} \\cdot \\text d\\vec S$\r\n\r\nSubstituting $\\vec \\rho = \\vec r - \\vec k$, the integration surface becomes a unit ball $\\Sigma$ and\r\n\r\n$r^2 = |\\vec \\rho + \\vec k|^2 = \\rho^2 + 2\\rho \\cos \\theta + 1 = 2(1 + \\cos \\theta)$\r\n\r\n$\\vec \\rho \\cdot \\text d\\vec \\Sigma = \\text d\\Sigma$ [color=white].[/color] and [color=white].[/color] $\\vec k \\cdot \\text d\\vec \\Sigma = \\cos \\theta\\ \\text d\\Sigma$\r\n\r\n$\\int_{\\Sigma} \\int \\frac{(\\vec \\rho + \\vec k)}{|\\vec \\rho + \\vec k|^2} \\cdot \\text d\\vec \\Sigma = \\int_{\\Sigma} \\int \\frac{1 + \\cos \\theta}{2(1 + \\cos \\theta)} \\text d\\Sigma = \\frac 1 2 \\int_{\\Sigma} \\int \\text d\\Sigma = 2 \\pi$",
  "Solution_9": "[quote=\"fedja\"]In hurry you forgot to multiply by the Jacobian $\\rho^2\\sin\\varphi$. ;)  If you do that the integral becomes really easy...[/quote]\r\n\r\nUpsss :blush:  :blush: , as I told, I was in a hurry, and well, you already saw the resoult of that  :oops: .",
  "Solution_10": "[quote=\"djimenez\"]..., as I told, I was in a hurry, and well, you already saw the result of that  :oops: .[/quote]\r\n\r\nIt was very nice and quick solution. At first, I could not get it, because of the incorrect limits $-\\frac \\pi 2, \\frac \\pi 2$. This gave me the impression that the missing jacobian must be $\\rho^2 \\cos \\phi$ and than it did not add up.\r\n\r\nyetti"
}
{
  "Tag": [
    "geometry",
    "algebra",
    "polynomial",
    "quadratics",
    "absolute value"
  ],
  "Problem": "Let y = ax^2 + bx + c be a second degree polynomial with real roots, where a < 0. Find the area enclosed by the x axis and the quadratic in terms of a,b,c . \r\n\r\nAnd don't give me Archimedes' 2/3bh ! I want it in terms of a,b,c.",
  "Solution_1": "why must you know this??\r\n\r\nx1 = (-b -  :sqrt: (b^2-4ac))/2a\r\nx2 = (-b +  :sqrt: (b^2-4ac))/2a\r\n\r\n\r\ny = ax^2 + bx + c\r\n\r\nax^3 / 3 + bx^2 / 2 + cx from x1 to x2\r\nthat's assuing x2 is to the right of x1, if u get a negative answer... just absolute value it.\r\n\r\narea ( a/3*(x2)^3 +b/x*(x2)^2 +c(x2) ) -\r\n( a/3*(x1)^3 +b/x*(x1)^2 +c(x1) )\r\n\r\n\r\ni think that's right, i dunno, its a ridiculous formula, not too bad when you use actual numbers though.",
  "Solution_2": "You must go further! hehehe don't worry, keep working on it. It actually works out to a pretty nice formula",
  "Solution_3": "it's just  |integral_(x1)^(x2)(f(x))|=|[2ax+b]_x1^x2|=|2ax1+b-2ax2-b|=|2ax1-2ax2|\r\nx1=lower root of f(x)\r\nx2=upper root of f(x)\r\n|2ax1-2ax2|=|2a((-b- :sqrt: (b :^2: -4ac))/2a-(-b+ :sqrt: (b :^2: -4ac))/2a)|\r\n=|2a(-2 :sqrt: (b :^2: -4ac)/2a)|=|-2 :sqrt: (b :^2: -4ac)|=2 :sqrt: (b :^2: -4ac)"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "The area of a square is 64 square centimeters. What is the number of square centimeters in the area of a new square made by decreasing each side of the original square by $ 75\\%$?",
  "Solution_1": "Each of the sides of the square has length $ \\sqrt{64}\\equal{}8$. So, $ 8(1\\minus{}.75)\\equal{}2$. Hence, the new area is $ 2^2\\equal{}\\boxed{4}$."
}
{
  "Tag": [],
  "Problem": "Interested in what IPO is about? Here is the [url=http://www.philosopiad.org/]site[/url]. Enjoy :)",
  "Solution_1": "I actually didn't know that such an olympiad exists until now !!  :( And anyway, it might be sometime before India begins to participate..."
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "If the word REPETITION is repeatedly written until there are 1000 letters, what percent of the letters are E's?",
  "Solution_1": "Well, there are exactly $ 10$ letters, so you can divide the 1000 letters by 10 and by 10 again.  Thus, there are only $ \\boxed{20}$ percent E's.",
  "Solution_2": "Note that $ \\dfrac{\\frac{1000}{10}}{10}\\equal{}10$ :wink:",
  "Solution_3": "No, ernie is correct. There are 2 e's. We write the word REPETITION 100 times. We have 20 e's. Thus the percent that we want is $ \\boxed{\\%20}$.\r\n\r\nEdit: Oh...",
  "Solution_4": "AIME15 wasn't stating an answer, just stating an observation.",
  "Solution_5": "Sorry if I'm being ignorant but I don't quite understand what AIME15's observation was about.  What does it have to do with the problem?",
  "Solution_6": "I just looked at the problem and saw it was 2 e's out of 10. no matter what multiple of 10, it will always be 20%."
}
{
  "Tag": [
    "inequalities",
    "inequalities open"
  ],
  "Problem": "hey friends!\r\n\r\ngoing thru many resources i hav come across the symmetric and cyclic summation ....symmetric mostly...\r\n\r\nthe expression of a symmetric summation is generally done in a single step as if it were on their tips...\r\n\r\ni find it tedious and irritating to find these summations by the beginners' method..\r\n\r\nso if any of u r aware of any ways tricks or shortcuts to find these expressions please post them here..\r\n\r\nsfsreporter",
  "Solution_1": "does no one want 2 help or is it that there not any known methods 2 do this?",
  "Solution_2": "http://www.mathlinks.ro/Forum/viewtopic.php?t=103773",
  "Solution_3": "i think u got me wrong i have problems expanding an expression in the symmetric\\cyclic forms\r\n\r\nfor example---\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=32470&start=20\r\n\r\npost #22\r\n\r\nhow did this man got the expression in a single step...... fine he must have done some computations on paper beforehand ...but this seems 2 be genius.\r\n\r\neven in assignments like Kiran Kedlaya and Mildrof they do this saying \"after some [b]simple computations[/b] we get.....\"\r\n\r\ni keep trying these \"simple computations\" and it seems that i am a fool!! who finds this difficult.\r\n\r\nso i wanted 2 know if there is any method to do all this by \"SIMPLE COMPUTATIONS\"???",
  "Solution_4": "By \"simple computations\" they mean something that, say, a computer could do for you. It isn't hard to keep applying the distributive law over and over and over and over and over (as long as you don't make a mistake in the middle).\r\n\r\nIt doesn't mean that they found a really fast way to do it by magic. It means that they actually sat down and expanded and expanded and expanded, and didn't feel like writing all of the expanding down so they skipped right to the end of all the expanding in their explanation for us.\r\n\r\nAn analogy: somebody asks you what 534/716 is to eight decimal places. You tell them, \"after some simple calculations, I got 0.74581005.\" They say, \"ohmigod how did you figure that out, I kept guessing 0.7457 or maybe 0.74583??? How did you do it in one step??\" The truth is you spent three hours with a pencil and paper doing long division...",
  "Solution_5": "thx for that kindda consolidation.\r\n\r\ni feel better now..\r\n\r\nbut then if i get Iran 1996... i mean a similar problem in an olympiad ...so do i need to expand it all sitting dere in da examination hall?? \r\n\r\nis there no other possible way?",
  "Solution_6": "sorry cant edit but i meant consolation...not consolidation",
  "Solution_7": "[quote=\"sfsreporter\"]thx for that kindda consolidation.\n\ni feel better now..\n\nbut then if i get Iran 1996... i mean a similar problem in an olympiad ...so do i need to expand it all sitting dere in da examination hall?? \n\nis there no other possible way?[/quote]\r\nFirst of all: Skip the abbreviations. It is very difficult to read what you have written, when you use such bad english.\r\nIn general there is a lot of other ways than expanding.\r\nLet's take an example:\r\n$ \\sum_{cyc} \\frac {x \\plus{} z }{2y} \\ge \\sum_{cyc} \\frac {2y}{x \\plus{} z}$ for positive $ x,y,z$.. Now you could multilply with $ xyz(x \\plus{} y)(y \\plus{} z)(z \\plus{} x)$ expand it and then use muirhead and schur or something like that.\r\nA easy way to solve this is to notice that $ x \\plus{} z \\ge 2\\sqrt {xy}$ and thus $ \\frac {y}{\\sqrt {xz}} \\ge \\frac {2y}{x \\plus{} z}$.\r\nBut the inequality $ \\sum_{cyc} \\frac {x \\plus{} z}{2y} \\ge \\sum_{cyc} \\frac {y}{\\sqrt {xz}}$ is very easy to prove. It comes from muirhead since $ (1,0, \\minus{} 1) \\succ (1, \\minus{} \\frac {1}{2}, \\minus{} \\frac {1}{2})$. Also $ \\frac {y}{x} \\plus{} \\frac {y}{z} \\ge \\frac {y}{\\sqrt {xz}}$, and so on with all three variables. Addition yields the desired.\r\nSometimes you can also make a minor change in the inequality like i did, and then expand. It comes from experience:)",
  "Solution_8": "[quote]Skip the abbreviations. It is very difficult to read what you have written, when you use such bad english. [/quote]\r\n\r\nsorry but i was in a hurry.. i think it was not that much of sms language..it was quite moderate but its fine...i'll prefer the other way now..\r\n\r\n\r\ni'm afraid but u hav made a typo in the problem ...cud you plz edit it so that i can have a better look at the thing.\r\n\r\nthanks!",
  "Solution_9": "[quote=\"sfsreporter\"][quote]Skip the abbreviations. It is very difficult to read what you have written, when you use such bad english. [/quote]\n\nsorry but i was in a hurry.. i think it was not that much of sms language..it was quite moderate but its fine...i'll prefer the other way now..\n\n\ni'm afraid but u hav made a typo in the problem ...cud you plz edit it so that i can have a better look at the thing.\n\nthanks![/quote]\r\nI sound more arrogant than I actually am, when I'm writing. I just get confused when I see \"R\" instead of \"are\" :)\r\n\r\nAnyway. The example is fixed."
}
{
  "Tag": [],
  "Problem": "I am sorry for asking so much but:\r\nHow many positive two-digit integers are there in which each of the two digits is prime?\r\nI got 4 but the answer says 16...\r\n\r\nakhil0422",
  "Solution_1": "There are $ 4$ prime digits: $ 2$, $ 3$, $ 5$, and $ 7$.  Thus, there are $ 4$ choices for the first prime digit, and $ 4$ choices for the second prime digit.  Multiplying yields $ 4 \\times 4 \\equal{} \\boxed{16}$."
}
{
  "Tag": [
    "MATHCOUNTS",
    "AMC",
    "AMC 8",
    "email",
    "articles",
    "search"
  ],
  "Problem": "Sigh...you people who left early to have time to trick-or-treat or whatever missed out on sooo much. Oh well. I hope you have the info necessary to do the homework. If not I can explain stuff here I guess. If you stayed, thanks for being patient. Either way, discuss the problem sets here. Last week's answers will be up eventually.\r\n\r\nOh yeah--AMC-8 NEXT TUESDAY! Sign up! Yay.",
  "Solution_1": "hey, I don't think you're doing a good job of making the mc kids come here.\r\nthis forum is like, pretty inactive. No one(from jls) posts here cept susan, linda and that other slt person that I dont know. \r\nand do everyone who sign up for AMC8 get to take it? that's nice....",
  "Solution_2": "susan, linda, slt don't post here anymore. this forum is dead. now i can't post since there aren't any new posts. also can i take the amc 8 for fun?",
  "Solution_3": "Huh, I think I'll start demanding that everyone post at least one solution to a problem beginning the week after next. I don't know if anyone will listen to me but whatever. David, we've only got 30 tests. I strongly suggest you not waste one that could have been used by a JLSer. Though I suppose you could go in case less than 30 people show up.",
  "Solution_4": "Wait a sec...the AMC-8 is NEXT week?! I thought it was today! Darn it. That means I shouldn't have gone last week, when I did, and should have gone today, when I didn't. I'm so easily confused. Darn it. I hope Ms. Beddoes didn't wait too long for me. Better email her about this.",
  "Solution_5": "Uh... As amusing as that was...no more spam.\r\n\r\nIt's a good thing that the AMC wasn't today: now I can recruit that former 6th grader that totally pwned math class to math club. :)",
  "Solution_6": "[quote=\"ccy\"]It's a good thing that the AMC wasn't today: now I can recruit that former 6th grader that totally pwned math class to math club. :)[/quote]\r\n\r\nWho's that? Does (s)he go to Terman?",
  "Solution_7": "according to ccy, its jeffrey wang's brother",
  "Solution_8": "[quote=\"moogra\"]according to ccy, its jeffrey wang's brother[/quote]\r\nand according to you, you're a retard",
  "Solution_9": "[quote=\"junggi\"][quote=\"moogra\"]according to ccy, its jeffrey wang's brother[/quote]\nand according to you, you're a retard[/quote]\r\n\r\nno thats you. ur the only one here trying to screw around. im answering questions.",
  "Solution_10": "[quote=\"moogra\"][quote=\"junggi\"][quote=\"moogra\"]according to ccy, its jeffrey wang's brother[/quote]\nand according to you, you're a retard[/quote]\n\nno thats you. ur the only one here trying to screw around. im answering questions.[/quote]\r\nno you weren't.\r\nyou were giving answers that made no sense.\r\naccording to you, ppl throw helmets at each other when theyre a tiny bit pissed. of course that's complete nonsense. even if that was true, you still haven't justified your action, because from what I saw from your face before you threw the fing shiit, you weren't just pissed. You were EXTREMELY angry. Just because i threw a piece of paper. wtf is that? (Remember that you tried to punch my head off before you threw that helmet.)\r\n\r\njust to remind you what happened, bc, given your intelligence level, I doubt that you remember.\r\n\r\nI tried to get you to come to mandelbrot because I thought you would have liked to(because you spend a lot of time on mathcounts and all that. and of course I was wrong to think that you would still have interest left in math. oh and btw, you coming to this forum is a complete waste of your time, because what do you do? YOU POST SOLUTIONS FOR MATHCOUNTS AND MOEMS PROBLEMS. I'd be fine with it if you were trying to teach them something, but no, you post either wrong solutions, or completely inelegant solutions. I wouldn't be saying this if posting solutions there actually helped you. But it doesn't. You only do the trivial problems. You'd be better off playing Maple story or some other games for retards.). Then you keep insulting me, and how I'm superlame because I like math and that bs.\r\nSo I get pissed, like everyone would when theyre insulted. So I throw a piece of paper. Then you suddenly turn into this beast thingy with an extremely angry face on, and you try to punch my head off. Then you stop for some reason, so I try to pick up the paper. Then when I had my head down, you, like an fing coward, pick up the helmet, and threw it at my head. WTF",
  "Solution_11": "You do have a point. and i never said math was lame, and i don't post wrong solutions. Also good job on the essay. It was interesting from your point of view. \r\n-edit-\r\n(even if I do post wrong solutions (might be 1 somewhere), i don't try to and that shows I actually tried to do the problems)",
  "Solution_12": "This is weird.",
  "Solution_13": "I agree.",
  "Solution_14": "Yes this is weird.  :D This scares everyone away",
  "Solution_15": "er...hi...am I interrupting anything??????\r\n\r\nOK. SO bookaholic. I'm gonna drop out of Analysis soon. Another thing. Tell me. Just HOW do you avoid idoticly retarded mistakes, like the one I did on the practice Chapter Round and the five retardedly idiotic mistakes I made on the Analysis test?\r\n\r\nBy the way, junggi, nice er...essay...you have there....\r\n\r\nSomething else I forgot........",
  "Solution_16": "[quote=\"moogra\"]Yes this is weird.  :D This scares everyone away[/quote]\nok f u, retard.\nand seriously, stop coming here, and play maple story or something\n\n[quote]By the way, junggi, nice er...essay...you have there.... [/quote]\r\nlol yea did you actually read that? of course not lol",
  "Solution_17": "[quote]Quote:\nBy the way, junggi, nice er...essay...you have there....\n\nlol yea did you actually read that? of course not lol\n[/quote]\r\n\r\nWRONG junggi, I actually read it. Shows how bored I got. 0.0\r\n\r\nBOOKAHOLIC WHENEVER YOU'RE ON I NEED SOME AMC 8, 10. and 12 PRACTICE PROBLEMS!!! (Er...I don't exactly know where to find them so.... :?:  :| )",
  "Solution_18": "[quote=\"Bluestalk\"][quote]Quote:\nBy the way, junggi, nice er...essay...you have there....\n\nlol yea did you actually read that? of course not lol\n[/quote]\n\nWRONG junggi, I actually read it. Shows how bored I got. 0.0\n\nBOOKAHOLIC WHENEVER YOU'RE ON I NEED SOME AMC 8, 10. and 12 PRACTICE PROBLEMS!!! (Er...I don't exactly know where to find them so.... :?:  :| )[/quote]\r\n................................................................................\r\nTRY THE AOPS FORUMS? :?:  :?:\r\n(THAT MEANS MATHCOUNTS,AND HSB FORUMS)",
  "Solution_19": "Drop out of Analysis?! Why?! You're an eighth grader, your grades this year won't be recorded. Even if you de-skip a grade, shouldn't you at least wait until it actually matters? Or is it taking too much time or something?\r\n\r\nAs for stupid mistakes...I'm not sure. I think there's an article about eliminating stupid mistakes in the Resources section. You should try the ideas in there. I didn't find it very helpful but that's probably just me. My carelessness is too ingrained in my nature. Um. If the problem is longish and complicated, I would suggest running through the solution in your head or even writing it out (very abbreviated of course) if you have time, and stopping at each step to make sure it's logical. Keep your scratch work very organized if possible. Make sure you're reading the problems right, I think misreading is the most preventable of stupid mistakes. What else...if you're making arithmetic mistakes or something, that should theoretically get better with practice. I don't know if it actually does.\r\n\r\nOh. Practice problems? They're all over AoPS. http://www.artofproblemsolving.com/Forum/resources.php?c=182 is the official page for all the former AMC series contest problems, but the unofficial problems in the forums are almost as useful. If you search in the AMC forum, people have been writing a lot of mock AMCs and AIMEs. I think there have been mock MathCounts and AMC-8 contests in the MathCounts forum. You can also buy some from http://www.unl.edu/amc . They're not really all that expensive.",
  "Solution_20": "[quote=\"Bluestalk\"]\nI'm gonna drop out of Analysis soon. \n[/quote]\n  :o whoa i didnt see that before...\nas for stupid mistakes, this might help\n[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]\n[quote=\"moogra\"]and i never said math was lame, and i don't post wrong solutions. Also good job on the essay. It was interesting from your point of view.\n[/quote]\ndude, stop lying. \nI clearly remember you saying \"YOURE TRYING TO MAKE TO COME TO A [i]MATH [/i]CONTEST! HOW LAME CAN YOU GET?\" and similar stuff\n[quote=\"moogra\"]\n(even if I do post wrong solutions (might be 1 somewhere), i don't try to and that shows I actually tried to do the problems)\n[/quote]\r\ndude, I saw you posting SUPER inelegant(means retarded soln in case you didnt know) solns to trivial problems. Also, PLEASE dont post solns after 1028084107 people already posted the exact same sonl you have",
  "Solution_21": "[quote=\"bookaholic\"]Drop out of Analysis?! Why?! You're an eighth grader, your grades this year won't be recorded. Even if you de-skip a grade, shouldn't you at least wait until it actually matters? Or is it taking too much time or something?\n\nAs for stupid mistakes...I'm not sure. I think there's an article about eliminating stupid mistakes in the Resources section. You should try the ideas in there. I didn't find it very helpful but that's probably just me. My carelessness is too ingrained in my nature. Um. If the problem is longish and complicated, I would suggest running through the solution in your head or even writing it out (very abbreviated of course) if you have time, and stopping at each step to make sure it's logical. Keep your scratch work very organized if possible. Make sure you're reading the problems right, I think misreading is the most preventable of stupid mistakes. What else...if you're making arithmetic mistakes or something, that should theoretically get better with practice. I don't know if it actually does.\n\nOh. Practice problems? They're all over AoPS. http://www.artofproblemsolving.com/Forum/resources.php?c=182 is the official page for all the former AMC series contest problems, but the unofficial problems in the forums are almost as useful. If you search in the AMC forum, people have been writing a lot of mock AMCs and AIMEs. I think there have been mock MathCounts and AMC-8 contests in the MathCounts forum. You can also buy some from http://www.unl.edu/amc . They're not really all that expensive.[/quote]\r\n\r\nUh bookaholic...  Next time you see bluestalk, ask her about the retarded (for the lack of a better word) way our teacher grades tests and quizzes.  I got points knocked off because I didn't round correctly :( and for some other trivial mistakes I can't remember.\r\n[hide=\"testimony by junggi\"][20:05:37] me: do you agree that beck grades retardedly?\n[20:05:43] junggi: of course\n[20:05:49] junggi: she takes off 3 pts for one mistake\n[20:06:00] me: thanks im quoting you\n[20:06:05] junggi: ok[/hide]\r\n\r\n@bluestalk: Here's a nice resource for AMC practice problems: http://www.unl.edu/amc/mathclub/04,1-problems.html (click the IAP links).  What will you do if you drop analysis?  School service?  :maybe:",
  "Solution_22": "Gee, people, FYI, I'm stupidder than you think. Stupidder. Stupider. Which one...?\r\n\r\nOK, thanks for all those links....I'm only allowed like 10 minutes on the computer at home, so that helps. =P",
  "Solution_23": "It's probably stupider. But at least you probably have never written \"2n^2-2n+1=2n^2-1\" on a contest problem sheet...\r\n\r\n(See, the thing is, I distributed 2n(n-1)+1 as 2n^2-2+1, so I wrote 2n(n-1)+1=2n^2-2+1=2n^2-1 on my paper. Later I realized I'd left out an n and scribbled it into the expression in the middle but FORGOT that that made the expression on the right invalid and just used it anyway. Argh.)",
  "Solution_24": "Er...OK.....I do that periodically...\r\n\r\nBookaholic, I pmed you about some Mock AMC questions I didn't get.....\r\n\r\nIf you have the time, could you post solutions? Thanks!!\r\n\r\nOH YEAH. One of the orchestra flute players is the president of PALY Math Club. Adrian Sanborn or somethinglikethat...anyway he says you're good at math. =P Don't deny it....\r\n\r\nDeny it in your head if you must then...."
}
{
  "Tag": [
    "combinatorics solved",
    "combinatorics"
  ],
  "Problem": "We have a regular polygon with 12 sides. How many edges can we draw between 2 of the vertex in such a way that there are no triangles???",
  "Solution_1": "Do you mean triangles whose vertices are vertices of the 12-gon? Or whose vertices are allowed to be intersection points of some diagonals?\r\n\r\nPierre.",
  "Solution_2": "I mean triangles whose vertices are vertices of the 12-gon.",
  "Solution_3": "So, the maximum is 12 <sup>2</sup> /4 = 36 as a direct consequence of Turan's theorem on graph without triangle.\r\n\r\nPierre."
}
{
  "Tag": [
    "modular arithmetic",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $ n$ be a natural number, $ n \\geq 2$.Prove that if $ \\frac{a^n\\minus{}1}{a\\minus{}1}$ is a prime power for some positive integer $ a$, then $ n$ is a prime number.",
  "Solution_1": "Assume that n is a composite number. We can write it as n=x.y x and y are positive integers greater than 1. (Not neccesarily distinct.) W.l.o.g. we may assume that x is greater than or equal to y. (a^(x.y) - 1)/(a-1) = p^k for some prime p and positive integer k. We can rewrite this equation as (a^x - 1)((a^x)^(y-1) + (a^x)^(y-2) + ... + 1)/(a-1). For the sake of the proof we will use the following notation. a^x = r. (a^(x-1) + a^(x-2) + ... + 1)(r^(y-1) + r^(y-2) + ... + 1) =p^k. r=1 (mod a^(x-1) + ... + 1). (Since r=(a-1).(a^(x-1)+ ... 1) +1.) So the first is smaller than the second. a^(x-1) + a^(x-2) + ... + 1 divides r^(y-1) + ... +1. r=1 (mod a^(x-1) + ... +1) So r^(y-1) + ... + 1= y (mod a^(x-1) + ... + 1) So a^(x-1) + ... + 1 should divide y. But at first, we assumed that x is greater than or equal to y. a^(x-1) + ... + 1 is greater than x. (It can be shown by using Bernoulli's Inequality.) So our assumption was wrong hence n must be a prime.  :P",
  "Solution_2": "$ n = kl$ $ \\implies$ $ (k,l > 1)$\r\n$ \\frac {a^{n} - 1}{a - 1} = p^{s}$  $ \\implies$\r\n$ (a^{k})^{l} - 1 = p^{s}.(a - 1)$ $ \\implies$ \r\n$ (a^{k} - 1)((a^{k})^{l - 1} + ... + a^{k} + 1) = p^{s}.(a - 1)$\r\n$ p|a^{k} - 1,p|(a^{k})^{l - 1} + ... + a^{k} + 1$  $ \\implies$  $ p|l$\r\nsimilarlay; $ (a^{l})^{k} - 1 = (a - 1).p^{s}$  $ \\implies$  $ a^{l}\\equiv 1\\pmod{p}$ \r\n$ \\implies$  $ p|k$  $ k = p.k_{1},l = p.l_{1}$  $ \\implies$\r\n$ k_{1}.l_{1} = q$  $ \\implies$  $ p|q$ $ \\implies$\r\n$ n = p^{\\alpha}.x$ $ (p\\nmid x)$ $ \\implies$\r\n$ (a^{p^{\\alpha})^{x} - 1 = (a - 1)p^{s}}$ $ \\implies$\r\n$ p|x$ contradiction! $ \\implies$ if $ n$ is composite number we get contradiction.",
  "Solution_3": "it can be easily proved by the Lemma $ v_p(a^n\\minus{}1)\\equal{}v_p(a\\minus{}1)\\plus{}v_p(n)$  :)  :D"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "A coin is flipped, a 6 sided die numbered 1-6 is rolled, and a 10 sided die numbered 0-9 is rolled.\r\nWhat is the probability that the coin is heads and the sum of the numbers on the faces is 8?\r\n[hide]\nprobability coin is heads = 1/2\nthere are 7 ways to have a sum of eight: (1,7) (2,6) (6,2) (3,5) (5,3) (4,4)\nthe chance of roling each of these is 1/6 x 1/10 = 1/60\nthe chance of rolling an eight is 6/60, multiply that by 1/2 to get 6/120\nso the answer is [b]1/20[/b][/hide]",
  "Solution_1": "Six possibilites on the first die, ten on the second. Sixty in all. Only six combinations sum to eight, so we have $ p_2\\equal{}\\frac{1}{10}$. Of course, $ p_1\\equal{}\\frac{1}{2}$. Looks like the probability is $ \\frac{1}{20}$.",
  "Solution_2": "Thanks for the problem.",
  "Solution_3": "I wonder why people post solutions on the first post...\r\n\r\nWouldn't that just allow people to cheat off of it? :maybe: \r\n\r\nHmmm... :|"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "Perhaps more realistic. Which individual do you think will win at nationals?\r\nArgh, spelled Neil Gurram wrong, can a mod correct that, please?\r\n\r\n[size=75][color=green][LadyKn1ght: Fixed the spelling][/color][/size]",
  "Solution_1": "Someone else from CA, TX, MI, or IN <- That's condescending considering Washington tied MI in cutoff and OK tied TX for cutoff.\r\n\r\nI voted Nathan btw but I think these people should also be included in your poll\r\n\r\nSachin Shinde (Indiana)    \r\nAnand Oza (Maryland)    \r\nMatthew Vengalil (Michigan) \r\nDaniel Li (Virginia)",
  "Solution_2": "BIff actually does have a good chance of making it. Have you seen some of his recent scores on nats rounds.",
  "Solution_3": "Obviously he can't include all prospective winners. However, I think the major favorites are Andrew Ardito, Neal Wu, and Nathan Benjamin, for their performance last year.",
  "Solution_4": "Argh, forgot Karan Batra, too. Hmm.",
  "Solution_5": "[quote=\"Ignite168\"]Someone else from CA, TX, MI, or IN <- That's condescending considering Washington tied MI in cutoff and OK tied TX for cutoff.\n\nI voted Nathan btw but I think these people should also be included in your poll\n\nSachin Shinde (Indiana)    \nAnand Oza (Maryland)    \nMatthew Vengalil (Michigan) \nDaniel Li (Virginia)[/quote]\r\n\r\nYou were just taking the first places from those states right?\r\n\r\nMatthew Vengalil wrote 7th in Michigan, and used countdown to gain first place, so there probably isn't a really high chance he'll make countdown, unless that was a choke.",
  "Solution_6": "[quote=\"kyyuanmathcount\"][quote=\"Ignite168\"]Someone else from CA, TX, MI, or IN <- That's condescending considering Washington tied MI in cutoff and OK tied TX for cutoff.\n\nI voted Nathan btw but I think these people should also be included in your poll\n\nSachin Shinde (Indiana)    \nAnand Oza (Maryland)    \nMatthew Vengalil (Michigan) \nDaniel Li (Virginia)[/quote]\n\nYou were just taking the first places from those states right?\n\nMatthew Vengalil wrote 7th in Michigan, and used countdown to gain first place, so there probably isn't a really high chance he'll make countdown, unless that was a choke.[/quote]\r\n\r\nNo, Sachin Shinde got 4th, Anand Oza got 3rd or 4th. Daniel Li keeps getting 30 on practice sprints. If I was just taking 1sts, I'd put first place in Texas up there or even 2nd as Kevin Chen was 3rd.",
  "Solution_7": "Wow, thanks for putting me on your poll.",
  "Solution_8": "Where is all your self-esteem, Klebian? I see that you did not put yourself up there. (this doesn't include someone else from IN, TX...)",
  "Solution_9": "[quote=\"Ignite168\"]  \nMatthew Vengalil (Michigan) \nDaniel Li (Virginia)[/quote]\r\n^^^ know them in real life.\r\n\r\nThey are both very good compeditors. If Matthew V. makes it to CD, he will surely go far. I don't know Daniel Li's ability in CD, but I am sure he is proficient at it.",
  "Solution_10": "Yeah, this is true.\r\nI am also pretty good at countdown, but I just choked at state countdown.  Usually, I do best for some reason when I am 3rd Place or 4th Place.\r\n\r\nIntrepid Math, did you qualify for nats?",
  "Solution_11": "I feel so happy to be included in the poll! haha....I only have 2 votes though  :( \r\n\r\nIt's going to be an interesting year...",
  "Solution_12": "[quote=\"samath\"]I feel so happy to be included in the poll! haha....I only have 2 votes though  :( \n\nIt's going to be an interesting year...[/quote]\r\n\r\nNow you have 3 :P I just voted for you.",
  "Solution_13": "Thanks.  Yay now I'm in 4th place...not counting other.",
  "Solution_14": "I'm both None of the Above and Person from MI, TX,indiana so I have 8 votes.",
  "Solution_15": "Actually, since you are from TX, IN, MI, etc, you can't be none of the above, since that category is \"above\".",
  "Solution_16": "SOMEONE ACTUALLY VOTED FOR ME!!!\r\n\r\nOh wait that was me.\r\n\r\nI think that we are overrating Neal and Nathan. I know Neal has godly speed, but nathan really needs to sharpen up if he wants to win. I figure i have a decent chance as long as i stop misreading. So, my money's on Neal. Sorry, gurram, never even heard of you. I'd like to meet you though.\r\n\r\n[size=75][color=orange][Around here we don't worship Aphrodite and Zeus and Hermes. And we consider references to their *cough* practices rather uncalled-for. Would you say OM** to a girl you liked in front of your pastor and principal? -Billy][/color][/size]",
  "Solution_17": "Actually as far as Daniel Li in VA, I've seen him the past 2 years at state.  I don't deny his written round abilities, but I personally beat him in the countdown round at state this year. (And I'm not half as good as some of you guys!)",
  "Solution_18": "[quote=\"aznness\"]SOMEONE ACTUALLY VOTED FOR ME!!!\n\nOh wait that was me.\n\nI think that we are overrating Neal and Nathan. I know Neal has godly speed, but nathan really needs to sharpen up if he wants to win. I figure i have a decent chance as long as i stop misreading. So, my money's on Neal. Sorry, gurram, never even heard of you. I'd like to meet you though.\n\n[size=75][color=orange][Around here we don't worship Aphrodite and Zeus and Hermes. And we consider references to their *cough* practices rather uncalled-for. Would you say OM** to a girl you liked in front of your pastor and principal? -Billy][/color][/size][/quote]\r\n\r\nOverrating? In 7th grade, one of them got #2 written and the other was National Champion. There is no question that they are the big favorites heading into the competition.",
  "Solution_19": "I voted Gurram because Michigan people are KO0o0OLer than everybody else.\r\n\r\nP.S. How many of those people up there are AoPSers?",
  "Solution_20": "Everyone but Neal Wu. :D\r\nNathan Benjamin (I_Would_Say)\r\nSam Keller (samath)\r\nAndrew Ardito (hexahedron)\r\nKevin Chen (bionuniquineist (sp?))\r\nNeil Gurram (mathgeniuse^ln(x))\r\nKevin Yang (aznness)",
  "Solution_21": "Go Nathan!! Woo!  :D",
  "Solution_22": "Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.\r\n\r\nThis could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)\r\n\r\nBut don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.",
  "Solution_23": "[quote=\"jrshoch\"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.\n\nThis could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)\n\nBut don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]\r\n\r\nAbsolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power. \r\n\r\nAll of these examples are from another competition - NOT MC! : From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.) Remember, this ISN'T from MC results.\r\n\r\nSo don't be overconfident. You are not the best in the world.",
  "Solution_24": "I would also point out this goes for state competition as well. Last year, a 7th grader placed 5th at our State Competition and was there as an individual. Top 4 were all 8th graders. (including jrshoch who won) This year, that 7th grader's team was there as well and he placed 6th while 7th graders who placed 8th and 11th went to 3rd and 1st respectively and the two other people who made the team were seventh graders.",
  "Solution_25": "[quote=\"solafidefarms\"][quote=\"jrshoch\"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.\n\nThis could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)\n\nBut don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]\n\nAbsolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power. More examples of falling: From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.)\n\nSo don't be overconfident. You are not the best in the world.[/quote]\r\nI agree, however you are ignoring people who do well after returning. Looking ONLY at 04 and 05 I see that both david and sergei were in lower parts of countdown. Pardha got 30th in 04. Mark Zhang got 40th in 04.\r\nAlmost half of last years countdown made the top 57 in 04. Half of the people making it into countdown are returning. You only named 4 drops in the past many years.",
  "Solution_26": "[quote=\"pgpatel\"][quote=\"solafidefarms\"][quote=\"jrshoch\"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.\n\nThis could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)\n\nBut don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]\n\nAbsolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power. More examples of falling: From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.)\n\nSo don't be overconfident. You are not the best in the world.[/quote]\nI agree, however you are ignoring people who do well after returning. Looking ONLY at 04 and 05 I see that both david and sergei were in lower parts of countdown. Pardha got 30th in 04. Mark Zhang got 40th in 04.\nAlmost half of last years countdown made the top 57 in 04. Half of the people making it into countdown are returning. You only named 4 drops in the past many years.[/quote]\r\n\r\nOoops, I didn't make it clear I wasn't talking about MC at all in those examples.",
  "Solution_27": "[quote=\"solafidefarms\"][/quote][quote=\"pgpatel\"][quote=\"solafidefarms\"][quote=\"jrshoch\"]Hey just wanted to let everyone know they shouldn't base 8th grade performance very much on 7th grade performance..... speaking from personal experience  :oops: I got 31st as a 7th grader (5th best for 7th grade), and then got 48th as an 8th grader.  It was a bad day... and harder problems than I was expecting.\n\nThis could also serve as a warning to those contending for countdown round (don't get less than 3 hours of sleep the night before the competition, even if it means using sleeping pills)\n\nBut don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]\n\nAbsolutely. Always sleep :blush: before the competition. And if you make Nats in 7th grade, you need to know that not many people improve in 8th, judging from Mathcounts and Word Power. More examples of falling: From the first year of going to nats to the second (in another competition), one person went from 4th to 7th, 12th to 13th, 7th to 13th, and 11th to not making nats at all. How many people improved enough to get in the top 10 the second year who had returned? Nobody. (Of course, 3rd year people, having gotten disappointed the second year, generally do much much better; in that competition, two different people have made it to nats one year, didn't make the finals or nats the second, and won 3rd the third year.)\n\nSo don't be overconfident. You are not the best in the world.[/quote]\nI agree, however you are ignoring people who do well after returning. Looking ONLY at 04 and 05 I see that both david and sergei were in lower parts of countdown. Pardha got 30th in 04. Mark Zhang got 40th in 04.\nAlmost half of last years countdown made the top 57 in 04. Half of the people making it into countdown are returning. You only named 4 drops in the past many years.[/quote][quote=\"solafidefarms\"]\n\nOoops, I didn't make it clear I wasn't talking about MC at all in those examples.[/quote] :idea:",
  "Solution_28": "[quote=\"jrshoch\"]\nBut don't drink coffee to wake yourself up either.  When I was in 7th grade my teammate did that and crashed mid-Target round and only managed to solve 1 Team round.[/quote]\r\n\r\nI like coffee.  :D  Yeah, my friend and I both went to the nationals last year as 7th graders, but this year, my friend got 7th place I think at states. And I dropped from 2nd to 3rd at states. :oops: Oh well.",
  "Solution_29": "Thanks for not including my name in your list of other potential winners :( Just because MathCounts was mean and didn't post my 44th placing last year on their returning competitors list doesn't mean i didn't get 44th :( :( :(. (Btw, Daniel Li got 47th. I beat him on tiebreaker :D).\r\n\r\nAnyway, both his written and CD skills are good, but I've beaten him at countdown more times than he's beaten me (if I remember correctly), so as long as my written round is good enough to get into CD i'll have a good shot, but i mainly wanna get top 4 for masters round and top 12 for being on TV ;)\r\n\r\nA few other points:\r\n1) Daniel's only getting 30s on practice sprints cuz he's done em all recently and remembers them. There's no way of knowing how he'll do on a new thing considering he's doing better at these nats competitions than the 2006 state one.\r\n2) Everyone from VA is getting 2x on all of their practice sprints with 14/16 on targets. \r\n3) Written isn't everything.\r\n\r\nI voted for someone else meaning someone from VA.",
  "Solution_30": "[quote=\"lotrgreengrapes7926\"]I like coffee.  :D  Yeah, my friend and I both went to the nationals last year as 7th graders, but this year, my friend got 7th place I think at states. And I dropped from 2nd to 3rd at states. :oops: Oh well.[/quote]\r\n\r\nSame in Washington.  I won again, but he dropped from 3rd to 7th and so he's not on my team.  It's because he's not an AOPSer, I bet  :D"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "trigonometry",
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $a,b,c$ be non-negative numbers such that \r\n                   $a^2 + b^2 + c^2 + abc = 4$\r\nFind the least and the greatest values of $a+b+c$",
  "Solution_1": "I think it is a USAMO question.\r\n\r\nSubsitute $a = 2\\cos{\\alpha}$, $b = 2\\cos{\\beta}$ and $c = 2\\cos{\\gamma}$ where $\\alpha, \\beta, \\gamma$ are the angles of an acute triangle. It is then very easy using the Hardy-Littlewood-Polya majorization inequality. I haven't checked it properly but it seems the values are 2 and 3, am I right?",
  "Solution_2": "I doubt if the min is 2.\r\nI don't think equality case can be attained.",
  "Solution_3": "Yes!!\r\n$2 \\le a+b+c \\le 3$\r\nsory, Arne! I'd like to know Hardy-Littlewood-Polya majorization inequality :)",
  "Solution_4": "[quote=\"minhkhoa\"]Yes!!\n$2 \\le a+b+c \\le 3$\nsory, Arne! I'd like to know Hardy-Littlewood-Polya majorization inequality :)[/quote]\r\n\r\nminhkhoa, here's majorization inequality, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975\r\nmay i know when a+b+c=2?",
  "Solution_5": "Dear Suihochung!\r\n(a,b,c)=(2,0,0) :)",
  "Solution_6": "[quote=\"minhkhoa\"]Dear Suihochung!\n(a,b,c)=(2,0,0) :)[/quote]\r\n\r\nah! stupid me...  :blush:",
  "Solution_7": "Sorry, this is not a USAMO question.\r\n\r\nThe USAMO question I had in mind is similar:\r\n\r\nIf $a, b, c$ are positive reals such that $a^2 + b^2 + c^2 + abc = 4$, prove that $0 \\leq ab + bc + ca - abc \\leq 2$.",
  "Solution_8": "[quote=\"Arne\"]\nSubsitute $a = 2\\cos{\\alpha}$, $b = 2\\cos{\\beta}$ and $c = 2\\cos{\\gamma}$ where $\\alpha, \\beta, \\gamma$ are the angles of an acute triangle.[/quote]\r\n\r\nI'm not exactly a stranger to trigonometric identities, but I didn't know this one. \r\n\r\nAre you saying that for a triangle,  $\\sum cos^2(\\alpha) + 2 \\Pi cos(\\alpha) = 1$?\r\n\r\n(so that triangles parametrize the $a,b,c$ satisfying the identity)",
  "Solution_9": "[quote=\"fleeting_guest\"]I'm not exactly a stranger to trigonometric identities, but I didn't know this one. \n\nAre you saying that for a triangle,  $\\sum cos^2(\\alpha) + 2 \\Pi cos(\\alpha) = 1$?[/quote]\n\nYes, exactly. We have:\n\n[b]Theorem 1.[/b] Given a triangle with angles A, B, C, then, denoting, x = cos A, y = cos B, z = cos C, we have $x^2+y^2+z^2+2xyz=1$.\n\nThere exists a kind of converse of this theorem for positive numbers and acute-angled triangles:\n\n[b]Theorem 2.[/b] If x, y, z are three positive reals such that $x^2+y^2+z^2+2xyz=1$, then there exists an acute-angled triangle whose angles A, B, C satisfy x = cos A, y = cos B, z = cos C.\n\nI have proven this theorem in http://www.mathlinks.ro/Forum/viewtopic.php?t=23888 post #8.\n\nNow, using Theorem 2, we can prove the original inequality:\n\n[quote=\"minhkhoa\"]Let $a,b,c$ be non-negative numbers such that \n                   $a^2 + b^2 + c^2 + abc = 4$\nFind the least and the greatest values of $a+b+c$[/quote]\r\n\r\nIn fact, we will show that $2\\leq a+b+c\\leq 3$. Equality in $2\\leq a+b+c$ is obtained for a = 2, b = 0, c = 0, and equality in $a+b+c\\leq 3$ is obtained for a = 1, b = 1, c = 1; thus, the problem of finding the least and the greatest values of a + b + c will be solved.\r\n\r\nIn order to prove $2\\leq a+b+c\\leq 3$, we consider two cases:\r\n\r\n[b]Case 1.[/b] All three numbers a, b, c are positive.\r\n[b]Case 2.[/b] At least one of the numbers a, b, c is 0.\r\n\r\n(In fact, we know that the three numbers a, b, c are nonnegative, so that each of them is either positive or 0.)\r\n\r\nIn Case 1, define the numbers $x=\\frac{a}{2}$, $y=\\frac{b}{2}$, $z=\\frac{c}{2}$. Then, a = 2x, b = 2y, c = 2z, and thus, the equation $a^2 + b^2 + c^2 + abc = 4$ rewrites as $\\left(2x\\right)^2+\\left(2y\\right)^2+\\left(2z\\right)^2+2x\\cdot 2y\\cdot 2z=4$, what simplifies to $x^2+y^2+z^2+2xyz=1$. Furthermore, since the three numbers a, b, c are positive, the numbers $x=\\frac{a}{2}$, $y=\\frac{b}{2}$, $z=\\frac{c}{2}$ are positive, too. Thus, after Theorem 2, there exists an acute-angled triangle whose angles A, B, C satisfy x = cos A, y = cos B, z = cos C. Now, a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21188]really well-known inequality[/url] yields that $1<\\cos A+\\cos B+\\cos C\\leq\\frac32$; since $\\cos A=x=\\frac{a}{2}$, $\\cos B=y=\\frac{b}{2}$ and $\\cos C=z=\\frac{c}{2}$, this rewrites as $1<\\frac{a}{2}+\\frac{b}{2}+\\frac{c}{2}\\leq\\frac32$. Multiplication with 2 yields $2<a+b+c\\leq 3$, and thus, our inequality $2\\leq a+b+c\\leq 3$ is proven.\r\n\r\nIn Case 2, we cannot apply Theorem 2, but this case is really easy. At least one of the numbers a, b, c is 0; we can WLOG assume that a = 0. Then, the condition $a^2 + b^2 + c^2 + abc = 4$ rewrites as $b^2+c^2=4$. Hence, $\\left(b+c\\right)^2=b^2+c^2+2bc\\geq b^2+c^2=4=2^2$, so that $b+c\\geq 2$; in other words, $2\\leq b+c$. On the other hand, the AM-QM inequality yields $\\frac{b+c}{2}\\leq\\sqrt{\\frac{b^2+c^2}{2}}=\\sqrt{\\frac{4}{2}}=\\sqrt2$, so that $b+c\\leq 2\\sqrt2$, and since $2\\sqrt2<3$, we thus get $b+c<3$. Combining this with $2\\leq b+c$, we get $2\\leq b+c<3$. Since a = 0, we can rewrite this as $2\\leq a+b+c<3$. Thus, again, the inequality $2\\leq a+b+c\\leq 3$ is proven.\r\n\r\nProblem solved.\r\n\r\n  darij",
  "Solution_10": "[quote=\"darij grinberg\"] [In $\\Delta ABC$], x = cos A, y = cos B, z = cos C, we have $x^2+y^2+z^2+2xyz=1$.  [/quote]\r\n\r\nThank you.  Any nice proof or interpretation?",
  "Solution_11": "I think there must be some elegant ways, but direct substitution works in most cases.\r\n\\[\r\n\\cos C =  - \\cos A\\cos B + \\sin A\\sin B\r\n\\]\r\nSubstitute this into x^2+y^2+z^2+2xyz\r\n\\[\r\n\\begin{array}{l}\r\n \\cos ^2 A + \\cos ^2 B + \\cos ^2 C + 2\\cos A\\cos B\\cos C \\\\ \r\n  = \\cos ^2 A + \\cos ^2 B + \\left( {\\cos A\\cos B - \\sin A\\sin B} \\right)^2  - 2\\cos A\\cos B\\left( {\\cos A\\cos B - \\sin A\\sin B} \\right) \\\\ \r\n  = \\cos ^2 A + \\cos ^2 B - \\cos ^2 A\\cos ^2 B + \\sin ^2 A\\sin ^2 B \\\\ \r\n  = \\cos ^2 A + \\cos ^2 B - \\cos ^2 A\\cos ^2 B + \\left( {1 - \\cos ^2 A} \\right)\\left( {1 - \\cos ^2 B} \\right) \\\\ \r\n  = 1 \\\\ \r\n \\end{array}\r\n\\]",
  "Solution_12": "Nice solution Darij Grinberg !!!\r\nHere is the main idea of my solution(i'll post it later)\r\n$\\sqrt{4-a^2} \\le b+c \\le 2\\sqrt{2-a}$",
  "Solution_13": "[b]for the least value[/b]\r\nFrom $a^2 + b^2 + c^2 + abc = 4$ we get $a,b,c \\le 2$\r\nHence,\r\n$4-a^2 = b^2 + c^2 + abc \\le b^2 + c^2 + 2bc = (b+c)^2 $\r\n\r\n$\\Rightarrow b+c \\ge \\sqrt{4-a^2}  $\r\n\r\n$\\Rightarrow a+b+c \\ge \\sqrt{4-a^2}+a = \\sqrt{(2-a)(2+a)}-(2-a)+2 $\r\n$= \\sqrt{2-a}(\\sqrt{2+a}-\\sqrt{2-a})+2 \\ge 2  $\r\n\r\n[b]for the greatest value[/b]\r\nIt is easy to prove that $b^2+c^2+abc \\ge \\frac{1}{2}(b+c)^2 + \\frac{a}{4}(b+c)^2$\r\nHence,\r\n$4=a^2 + b^2 + c^2 + abc \\ge a^2 + \\frac{1}{2}(b+c)^2 + \\frac{a}{4}(b+c)^2$\r\n\r\n$\\Rightarrow   (b+c)^2 \\le 4(2-a)$\r\n\r\n$\\Rightarrow   b+c \\le 2\\sqrt{2-a} \\le 1+2-a   (AM-GM)$\r\n\r\n$\\Rightarrow   a+b+c \\le 3$\r\nit's not so clear!!! ;)",
  "Solution_14": "[quote=\"Arne\"]Sorry, this is not a USAMO question.\n\nThe USAMO question I had in mind is similar:\n\nIf $a, b, c$ are positive reals such that $a^2 + b^2 + c^2 + abc = 4$, prove that $0 \\leq ab + bc + ca - abc \\leq 2$.[/quote]\r\nThe ineq. is nice!\r\nWhere can i find its solution?",
  "Solution_15": "http://www.kalva.demon.co.uk/usa/usoln/usol013.html .\r\n\r\n  darij",
  "Solution_16": "[quote=\"darij grinberg\"]http://www.kalva.demon.co.uk/usa/usoln/usol013.html .\n\n  darij[/quote]\r\nbut why i can not open it",
  "Solution_17": "[quote=\"i am a chinese\"][quote=\"darij grinberg\"]http://www.kalva.demon.co.uk/usa/usoln/usol013.html .\n\n  darij[/quote]\nbut why i can not open it[/quote]\n\nBecause you are a chinese  :(  and for some reason, the chinese \"firewall\" blocks Kalva's domain.\n\nI hope Kalva has nothing against me quoting his solution:\n\n[quote=\"Kalva\"]30th USAMO 2001\n------\n\n[b]Problem A3[/b]\n\nNon-negative reals x, y, z satisfy $x^{2}+y^{2}+z^{2}+xyz=4$. Show that $xyz\\leq xy+yz+zx\\leq xyz+2$.\n\n[b]Solution[/b]\n\nAssume $x \\geq y \\geq z$. If z > 1, then $x^{2}+y^{2}+z^{2}+xyz > 1+1+1+1 = 4$. Contradiction. So $z \\leq 1$. Hence $xy+yz+zx \\geq xy \\geq xyz$.\n\nPut x = u + v, y = u - v, so that $u, v \\geq 0$. Then the equation given becomes $u^{2}\\left(2+z\\right)+\\left(2-z\\right)v^{2}+z^{2}= 4$. So we we keep z fixed and reduce v to nil, then we must increase u. But $xy+yz+zx-xyz = \\left(u^{2}-v^{2}\\right)\\left(1-z\\right)+2zu$, so decreasing v and increasing u has the effect of increasing xy + yz + zx - xyz. Hence xy + yz + zx - xyz takes its maximum value when x = y. But if x = y, then the equation gives $x = y = \\sqrt{2-z}$. So to establish that $xy+yz+zx-xyz \\leq 2$ it is sufficient to show that $2-z+2z\\sqrt{2-z}\\leq 2+z\\left(2-z\\right)$. Evidently we have equality if z = 0. If z is non-zero, then the relation is equivalent to $2\\sqrt{2-z}\\leq 3-z$ or $\\left(z-1\\right)^{2}\\geq 0$. Hence the relation is true and we have equality only for z = 0 or 1.[/quote]\r\n\r\n  darij"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "find all values of $ x/x \\plus{} y \\plus{} y/y \\plus{} z \\plus{} z/x \\plus{} y$,x,y,z>0",
  "Solution_1": "Do you mean $ \\frac{x}{x\\plus{}y}\\plus{}\\frac{y}{y\\plus{}z}\\plus{}\\frac{z}{x\\plus{}y}$?",
  "Solution_2": "yes. exactly that",
  "Solution_3": "\\[ \\left(\\frac{x}{x\\plus{}y}\\plus{}\\frac{y}{y\\plus{}z}\\plus{}\\frac{z}{z\\plus{}x}\\right)\\in (1,2)\\]",
  "Solution_4": "please post your solution :!:",
  "Solution_5": "You should do the same in the Inequalities Marathon . \r\n\r\n\\[ \\sum_{cyc}\\frac{x}{x\\plus{}y}\\ge \\sum_{cyc}\\frac{x}{x\\plus{}y\\plus{}z}\\equal{}1\\]\r\n\r\nand \\[ \\sum_{cyc}\\frac{x}{x\\plus{}y}\\plus{}\\sum_{cyc}\\frac{y}{x\\plus{}y}\\equal{}3\\] and use the fact that \\[ \\sum_{cyc}\\frac{y}{x\\plus{}y}\\ge 1\\]"
}
{
  "Tag": [
    "calculus",
    "trigonometry",
    "parameterization",
    "quadratics",
    "algebra"
  ],
  "Problem": "A driver in a car is in the desert, 10 km from the nearest point, $ N$, on a long, straight road.  The driver can cover 30 km/h in the desert, and 50 km/h on the road.  What is the shortest time for the driver to reach a point on the road that is:\r\n\r\n(a) 20 km from $ N$?\r\n(b) 30 km from $ N$?",
  "Solution_1": "I would imagine the road is perpendicular to the path from the driver to $ N$?",
  "Solution_2": "Yes, since the $ N$ is the nearest point on the road, 10 km is the perpendicular distance from the driver to the road.",
  "Solution_3": "[hide=\"Ugly Solution\"]While on the desert, the path will contain only lines, because the shortest distance between any two points is a line, and once we hit the road, it is the most efficient to continue to the finish line by the Triangle Inequality. Now assume that the path while on the desert is not just one line, but a composition of them. Start from the last two lines; we can make this shorter with only one line by the Triangle Inequality. We can continue like this to show that the shortest desert path to some given point on the road is always a line.\n\nSo let the point that the desert path ends be $ x$ away from $ N$, and thus $ 20 \\minus{} x$ from the finish point. The time taken would then be $ \\frac {\\sqrt {x^2 \\plus{} 100}}{30} \\plus{} \\frac {20 \\minus{} x}{50} \\equal{} \\frac25 \\plus{} \\frac1{150}(5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x)$. Now we want to find the maximum $ k$ such that $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x\\ge k$ for all $ 0\\le x\\le20$. So moving $ 3x$ to the RHS and simplifying yields $ (16x \\minus{} 3k)^2 \\plus{} 25(1600 \\minus{} k^2)\\ge0$. Keeping in mind that $ k$ is positive, if $ k \\equal{} 40$, $ 0\\le x \\equal{} \\frac {120}6 \\equal{} \\frac {15}2\\le20$ is where the minimum occurs, if $ k < 40$, the minimum is not achievable for any $ x$ since $ 1600 \\minus{} k^2 > 0$ so certainly not $ 0\\le x\\le20$, and if $ k > 40$, then for $ 0\\le x \\equal{} \\frac {15}2\\le20$, $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x < k$, which is a contradiction. Thus the minimum time occurs when $ x \\equal{} \\frac {15}2$, which makes the time $ \\frac23$ of an hour or $ 40$ minutes.\n\nSimilarly, for the second part, we want to minimize $ \\frac35 \\plus{} \\frac1{150}(5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x)$. But this obviously also occurs when $ x \\equal{} \\frac {15}2$, giving a time of $ \\frac {13}{15}$ of an hour or $ 52$ minutes.[/hide]\r\n\r\nAhh there should be a nicer solution...",
  "Solution_4": "[quote=\"math154\"] and if $ k > 40$, then for $ 0\\le x \\equal{} \\frac {15}2\\le20$, $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x < k$, which is a contradiction. [/quote] How was this contradiction proved (without circular reasoning of $ 5\\sqrt{x^{2}\\plus{}100}\\minus{}3x < k$ for $ k>40$ since the maximum $ k$ is $ k\\equal{}40$)?",
  "Solution_5": "OK, so if $ k > 40$, then by definition, $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x\\ge k > 40$ for all $ 0\\le x\\le20$. But $ x \\equal{} \\frac {15}2$ (which is between $ 0$ and $ 20$) yields $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x \\equal{} 40$, which is less than $ k$, contradicting the role of $ k$.",
  "Solution_6": "OK I see... for some strange reasons I misunderstood the definition of $ k$... :huh:",
  "Solution_7": "Now we want to find the maximum $ k$ such that $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x\\ge k$ for all $ 0\\le x\\le20$.\r\n\r\nThis is the first sentence of the solution that I don't understand.  Could you please explain it differently?  Among other things, I'm not sure why the sentence calls for a maximum when we're trying to minimize the driving time.\r\n\r\nThanks.",
  "Solution_8": "[quote=\"math154\"]Now we want to find the maximum $ k$ such that $ 5\\sqrt {x^2 \\plus{} 100} \\minus{} 3x\\ge k$ for all $ 0\\le x\\le20$.[/quote]\r\n\r\nThis is the first sentence of the solution that I don't understand.  Could you please explain it differently?  Among other things, I'm not sure why the sentence calls for a maximum when we're trying to minimize the driving time.\r\n\r\nThanks.",
  "Solution_9": "Of course. What if we picked, say $ k\\equal{}\\minus{}100$? Clearly all $ 5\\sqrt{x^2\\plus{}100}\\minus{}3x\\ge k$, but it's not [i]achievable[/i], which is what we need for $ k$ to really be a minimum. So we need to find $ k$ such that $ 5\\sqrt{x^2\\plus{}100}\\minus{}3x\\ge k$ for all $ 0\\le x\\le20$, and so that for some $ x$ in the range $ 0\\le x\\le20$, $ 5\\sqrt{x^2\\plus{}100}\\minus{}3x\\equal{}k$. This corresponds to finding the maximum possible $ k$ with these conditions. For example, here $ k\\equal{}40$ is the maximum possible lower bound, achievable when $ 0\\le x\\equal{}7.5\\le20$, because if $ k<40$, then there wouldn't exist any $ x$ in the range achieving the minimum, and if $ k>40$, then when $ x\\equal{}7.5$, we would have $ 5\\sqrt{x^2\\plus{}100}\\minus{}3x\\equal{}40<k$, but that contradicts the definition that $ 5\\sqrt{x^2\\plus{}100}\\minus{}3x\\ge k$ for all $ 0\\le x\\le20$!",
  "Solution_10": "[b]math154[/b]'s argument clearly shows how the problem is really to find the most economical way of getting to the highway given which direction on the highway the driver ultimately wants to go. That is, we want to minimize the time required to get to the highway minus the highway time saved. It is also clear that we should take a straight path to the highway.\r\n\r\nI would instead suggest setting up the problem using angles/trigonometry, since the problem scales.\r\n\r\nSuppose the driver going directly to the highway means an angle of zero, and the driver traveling in the direction of the highway means an angle of 90, and generally, the angle the driver makes from the normal route to the highway an angle of theta.\r\n\r\nWe're a distance $ d$ away. Let the velocity on the road be $ 1$, and the relative velocity on the sand be $ v \\in [0,1]$. We're interested in the time required to get to the highway minus the highway time saved. This is given by:\r\n\r\n$ \\frac{d}{v \\cos \\theta} \\minus{} d \\tan \\theta \\equal{} \\frac{d}{v} \\left( \\frac{1 \\minus{} v \\sin \\theta}{\\cos \\theta} \\right)$\r\n\r\n[hide=\"Minimizing\"]The trick is now to minimize $ \\frac{1 \\minus{} v \\sin \\theta}{\\cos \\theta}$ without calculus. I would suggest using the parametrization $ \\sin \\theta \\equal{} \\frac{1\\minus{}t^2}{1\\plus{}t^2}$ and $ \\cos \\theta \\equal{} \\frac{2t}{1\\plus{}t^2}$, with the order of the two strategically chosen.\n\nWe thus get $ \\frac{(1\\minus{}v)\\plus{}(1\\plus{}v)t^2}{2t} \\equal{} \\frac{(1\\minus{}v) t^{\\minus{}1} \\plus{} (1\\plus{}v) t}{2}$\n\nBy AM-GM, we get this to be minimized when $ (1\\minus{}v) t^{\\minus{}1} \\equal{} (1\\plus{}v) t \\implies t^2 \\equal{} \\frac{1\\minus{}v}{1\\plus{}v}$\n\nSticking this value of $ t^2$ back into our expression for $ \\sin \\theta$, we find that:\n\n$ \\sin \\theta \\equal{} \\frac{1\\minus{}\\frac{1\\minus{}v}{1\\plus{}v}}{1\\plus{}\\frac{1\\minus{}v}{1\\plus{}v}} \\equal{} v$\n\nSo now, we know exactly what angle to approach the road. If the desired destination is at a larger angle than $ \\arcsin v$, then we should go in at an angle of $ \\arcsin v$ and take the highway. If the desired destination is at a smaller angle, we should just go directly to that point.[/hide]",
  "Solution_11": "Refer to the diagram. The point C is the present location of the car in the desert, D is the point to be reached with $ ND \\equal{} d$. Let the car first go to point P ($ NP \\equal{} x$) and then continue on the highway. Let $ u$ be the velocity of the car in the desert and $ v$ on the highway. Also, let $ CN \\equal{} a$. The time taken is \r\n$ t \\equal{} \\dfrac{CP}{u} \\plus{} \\dfrac{PD}{v} \\equal{} \\dfrac{\\sqrt {x^2 \\plus{} a^2}}{u} \\plus{} \\dfrac{d \\minus{} x}{v} \\equal{} \\dfrac{v\\sqrt {x^2 \\plus{} a^2} \\minus{} ux}{uv} \\plus{} \\dfrac{d}{v}$\r\nIn order to minimize $ t$, we need to minimize $ v\\sqrt {x^2 \\plus{} a^2} \\minus{} ux$. Let\r\n$ y \\equal{} v\\sqrt {x^2 \\plus{} a^2} \\minus{} ux$\r\n$ \\Rightarrow\\ y^2 \\plus{} u^2x^2 \\plus{} 2uxy \\equal{} v^2x^2 \\plus{} a^2v^2$\r\n$ \\Rightarrow\\ (v^2 \\minus{} u^2)x^2 \\minus{} 2uxy \\plus{} a^2v^2 \\minus{} y^2 \\equal{} 0\\qquad(1)$\r\nSince $ x$ must be real, the roots of the above quadratic equation in $ x$ must be real. That happens if the discriminant of (1) is non-negative; i.e.\r\n$ 4u^2y^2 \\minus{} 4(v^2 \\minus{} u^2)(a^2v^2 \\minus{} y^2)\\ge 0$\r\ni.e.\r\n$ y\\ge a\\sqrt {v^2 \\minus{} u^2} \\equal{} y_\\mathrm{min}$\r\nWhen the equality is attained, the discriminant of (1) is zero, and the double root becomes\r\n$ x_m \\equal{} \\dfrac{uy_\\mathrm{min}}{v^2 \\minus{} u^2} \\equal{} \\dfrac{au}{\\sqrt {v^2 \\minus{} u^2}}$\r\nAs such the minimum time is \r\n$ t_\\mathrm{min} \\equal{} \\dfrac{y_\\mathrm{min}}{uv} \\plus{} \\dfrac{d}{v} \\equal{} \\dfrac{a\\sqrt {v^2 \\minus{} u^2}}{uv} \\plus{} \\dfrac{d}{v} \\equal{} \\dfrac{d \\plus{} a\\sqrt {(v/u)^2 \\minus{} 1}}{v}$",
  "Solution_12": "Thanks to all of you.  These are all superb explanations, and I'm trying to work through them now.  I'm very grateful for your help."
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "f:R->R , $f(x)= \\frac{x^{2}+mx+n }{x^{2}+1}$ , with $m$, $n$ real. find m,n so as $f ( \\mathbb{R})=(1,3]$",
  "Solution_1": "You have $f(x)>1$ for all $x$. What condition does this impose on $m$?",
  "Solution_2": "0  :blush:"
}
{
  "Tag": [
    "vector",
    "function",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Could someone help me with this problem:\r\n\r\nf is a conformal mapping from the unit disk onto a convex set.\r\nShow that the image of any disk centered at the origin is convex.",
  "Solution_1": "[url=http://www.mathlinks.ro/viewtopic.php?t=280607]this[/url] is one of the most beautiful problems in complex analysis I have ever seen.",
  "Solution_2": "The image of the circle |z|=r is convex iff u(z)= Re(1+zf\"(z)/f'(z)) is non-negative on this circle.\r\n(This is an easy computation: parametrize the circle by r.exp(it) and then write the condition\r\nthat the argument of the tangent vector to the image of the circle is increasing as a function\r\nof t.)\r\nNow u is a harmonic function (because f'(z) is never 0), so if the image of some circle |z|=r\r\nis convex then the image of any smaller circle |z|=r'<r is also convex, by Maximum Principle.\r\n\r\nPym.\r\n\r\nP.S. This class of functions is called convex univalent; it was studied a lot, see, for\r\nexample Hayman's Multivalent Functions, chap. I"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "vector",
    "trigonometry",
    "rhombus",
    "angle bisector"
  ],
  "Problem": "oxy: angle\r\nod: bisector\r\nAB=A'B'\r\nAM=MA'\r\nBN=NB'\r\n\r\nPROVE THAT: MN//od\r\n\r\nTHERE IS AN IMAGE IN OREDER TH HELP:[url=http://img236.imageshack.us/my.php?image=graphij5.png][img]http://img236.imageshack.us/img236/5430/graphij5.th.png[/img][/url]",
  "Solution_1": "No hard.\r\nLet K on AB' so KA=KB' and .... can you do continue . :lol:",
  "Solution_2": "I DON;T GET IT.  :wink:  ANY CHANCE FOR FURTHERMORE INFORMATION? :)",
  "Solution_3": "Denote the points $ M',$ $ N',$ as the intersection points of $ AA',$ $ BB'$ respectively, from the angle bicector of $ \\angle BOB'$ and as $ C,$ $ C',$ the intersection points of $ BB',$ from the lines through $ A,$ $ A'$ resprctively and parallel to $ ON'.$\r\n\r\nBy [b][size=100]Thales theorem[/size][/b] and the [b][size=100]angle bisector theorem,[/size][/b] we can now to prove that $ BC = B'C'$ and then $ NC = NC'.$ \r\n\r\nHence, we conclude that $ MM'\\parallel AC\\parallel A'C'\\parallel ON'$ and the proof is completed.\r\n\r\nKostas Vittas.\r\n\r\nPS. I will post here later more details.",
  "Solution_4": "[quote=\"Virgil Nicula\"][color=darkred][b][u]Lemma[/u] (well-known).[/b] Let $ ABC$ be a triangle and let $ M\\in AB$, $ N\\in AC$ be two points so that $ BM=CN$.\nDenote the middlepoints $ X$, $ Y$ of the segments $ [BC]$, $ [MN]$ respectively. Prove that :\n[b]1.[/b] If the line $ BC$ doesn't separate the points $ M$, $ N$ then the line $ XY$ is parallel with the internal bisector of the angle $ \\widehat{BAC}$.\n[b]2.[/b] If the line $ BC$ separates the points $ M$, $ N$ then the line $ XY$ is parallel withe the external bisector of the angle $ \\widehat{BAC}$.[/color][/quote]\r\n[color=darkblue][b]Proof.[/b] Denote $ \\{D,D'\\}\\subset BC$ so that the rays $ [AD$, $ [AD'$ are respectively the internal and external bisectors of the angle $ \\widehat{B AC}$. Define\nthe reflection $ P$ of the point $ N$ w.r.t. the point $ X$. Observe that $ MP\\parallel XY$, $ PB\\parallel AC$, $ MB=NC=PB$, i.e. $ \\triangle BMP$ is $ B$- isosceles.  \n\n$ \\{\\begin{array}{cccccccc}1\\blacktriangleright & m(\\widehat{MBP})=B+C & \\implies & m(\\widehat{BMP})=\\frac{A}{2}& \\implies & MP\\parallel AD & \\implies & XY\\parallel AD\\ .\\\\\\\\ 2\\blacktriangleright & m(\\widehat{MBP})=A & \\implies & m(\\widehat{BMP})=\\frac{B+C}{2}& \\implies & MP\\parallel AD' & \\implies & XY\\parallel AD'\\ .\\end{array}$\n\n[b]Remark.[/b] Let $ M\\in AB$, $ N\\in AC$ be two mobile points so that $ BM=CN$.\nThen the geometrical locus of the midpoint of the segment $ [MN]$ is $ AD\\cup AD'$.[/color]",
  "Solution_5": "[quote=\"Virgil Nicula\"][color=darkred][b][u]Lemma[/u] (well-known).[/b] Let $ ABC$ be a triangle and let $ M\\in AB$, $ N\\in AC$ be two points so that $ BM=CN$.\nDenote the middlepoints $ X$, $ Y$ of the segments $ [BC]$, $ [MN]$. Prove that :\n[b]1.[/b] If the line $ BC$ doesn't separate the points $ M$, $ N$ then the line $ XY$ is parallel with the internal bisector of the angle $ \\widehat{BAC}$.\n[b]2.[/b] If the line $ BC$ separates the points $ M$, $ N$ then the line $ XY$ is parallel withe the external bisector of the angle $ \\widehat{BAC}$.[/color][/quote]\r\nNice one...I suppose the nicest way to do this is what cuopbientcd said?  Create the point $ K$, use the midsegment theorem, and then use vectors?",
  "Solution_6": "There is also an easy solution using complex numbers. Let $ A=e^{-i\\theta}$, and $ A'=a e^{i\\theta}$, and $ B=b e^{-i\\theta}$. A short calculatin gives $ N-M=2(b-1)\\cos\\theta$, which is real.",
  "Solution_7": "[quote=\"Virgil Nicula\"][color=darkred][b][u]Lemma[/u] (well-known).[/b] Let $ ABC$ be a triangle and let $ M\\in AB$, $ N\\in AC$ be two points so that $ BM=CN$.\nDenote the middlepoints $ X$, $ Y$ of the segments $ [BC]$, $ [MN]$ respectively. Prove that :\n[b]1.[/b] If the line $ BC$ doesn't separate the points $ M$, $ N$ then the line $ XY$ is parallel with the internal bisector of the angle $ \\widehat{BAC}$.\n[b]2.[/b] If the line $ BC$ separates the points $ M$, $ N$ then the line $ XY$ is parallel withe the external bisector of the angle $ \\widehat{BAC}$.[/color][/quote]\r\n[color=darkblue][b]Proof I.[/b] Denote $ \\{D,D'\\}\\subset BC$ so that the rays $ [AD$, $ [AD'$ are respectively the internal and external bisectors of the angle $ \\widehat{B AC}$. Define\nthe reflection $ P$ of the point $ N$ w.r.t. the point $ X$. Observe that $ MP\\parallel XY$, $ PB\\parallel AC$, $ MB=NC=PB$, i.e. $ \\triangle BMP$ is $ B$- isosceles.  \n\n$ \\{\\begin{array}{cccccccc}1\\blacktriangleright & m(\\widehat{MBP})=B+C & \\implies & m(\\widehat{BMP})=\\frac{A}{2}& \\implies & MP\\parallel AD & \\implies & XY\\parallel AD\\ .\\\\\\\\ 2\\blacktriangleright & m(\\widehat{MBP})=A & \\implies & m(\\widehat{BMP})=\\frac{B+C}{2}& \\implies & MP\\parallel AD' & \\implies & XY\\parallel AD'\\ .\\end{array}$\n\n[b]Remark.[/b] Let $ M\\in AB$, $ N\\in AC$ be two mobile points so that $ BM=CN$.\nThen the geometrical locus of the midpoint of the segment $ [MN]$ is $ AD\\cup AD'$.\n\n[b]Proof II.[/b] Let $ \\{\\begin{array}{c}\\{M,M'\\}\\subset AB\\\\\\ \\{N,N'\\}\\subset AC\\end{array}$ be four different points so that $ \\{\\begin{array}{c}M\\in (AB)\\ ,\\ N\\in (AC)\\\\\\ BM=BM'\\ ,\\ CN=CN'\\end{array}$. Denote the midpoints $ X$, $ Y$, $ U$, $ V$ $ Z$ of the segments $ [BC]$, $ [MN]$, $ [M'N]$, $ [MN']$, $ [M'N']$ respectively. Prove easily that the quadrilateral $ UYVZ$ is a rhombus ($ UY\\parallel VZ\\parallel AB$, $ YV\\parallel UZ\\parallel AC$) and $ X\\in UV\\cap YZ$. Thus, $ \\overline{UXV}\\perp \\overline{YXZ}$, i.e. $ \\overline{YXZ}\\parallel AD$ and $ \\overline{UXV}\\parallel AD'$.\n\n[b]Remark.[/b] Let $ ABCD$ be a convex quadrilateral so that $ E\\in AB\\cap CD$ and $ AB=CD$. Denote the midpoints $ X,Y,U,V$ of the segments $ [BC]$, $ [AD]$, $ [AC]$, $ [BD]$ respectively. Then the lines $ XY$, $ UV$ are parallel respectively to the internal and external bisectors of the angle $ \\widehat{AED}$.\n[/color]",
  "Solution_8": "I'M LOOKING FOR A SIMPLE SOLUTION (WITH DETAILS IF IT'S POSSIBLE) THANKS :)",
  "Solution_9": "I think that Virgil Nicula's proof is quite very simple and very clear. Try to read his proof then.",
  "Solution_10": "[b]HINT(nothing the new !!!)[/b]\r\n\r\nWell, \r\n\r\n take two segments $ MK,ML$ equal to $ AB$ and parallele to $ AB, A'B'$respectively. Then $ MK = ML$ and $ BKB'L$ is a parallelogram.But then $ KL$ passes through N and MN is the bisector of $ \\angle KML$ in the isosceles triangle $ MKL$. Obviously $ MN$ is parallele to the bisector of angle $ BOB'$  too ect.",
  "Solution_11": "[quote=\"vittasko\"]....\nBy [b][size=100]Thales theorem[/size][/b] and the [b][size=100]angle bisector theorem,[/size][/b] we can now to prove that $ BC = B'C'$ and then $ NC = NC'.$\n...... [/quote]\r\n\r\nFrom $ AC\\parallel ON',$ we have that $ \\frac{BA}{BO}= \\frac{BC}{BN'}$ $ ,(1)$\r\n\r\nFrom $ A'C'\\parallel ON',$ we have that $ \\frac{B'A'}{B'O}= \\frac{B'C'}{B'N'}$ $ ,(2)$\r\n\r\nFrom $ (1),$ $ (2),$ $ \\Longrightarrow$ $ \\frac{BA}{BO}\\cdot \\frac{B'O}{B'A'}= \\frac{BC}{BN'}\\cdot \\frac{B'N'}{B'C'}$ $ ,(3)$\r\n\r\nFrom $ (3)$ $ \\Longrightarrow$ $ \\frac{B'O}{BO}= \\frac{BC}{B'C'}\\cdot \\frac{B'N'}{BN'}$ $ \\Longrightarrow$ $ \\frac{BC}{B'C'}= 1$ $ ,(4)$ $ ($ because of $ \\frac{B'O}{BO}= \\frac{B'N'}{BN'}$ by [b][size=100]angle bisector theorem[/size][/b] $ ).$\r\n\r\nSo, from $ (4)$ $ \\Longrightarrow$ $ BC = B'C'$ $ \\Longrightarrow$ $ CN = C'N$ $ ,(5)$\r\n\r\nFrom $ (5)$ we conclude that $ MN\\parallel AC\\parallel A'C'\\parallel ON'$ and the proof is completed.\r\n\r\nKostas Vittas.",
  "Solution_12": "transfer $ AB$ with $ \\overrightarrow{AM}$ and $ A'B'$ with $ \\overrightarrow{A'M}$.let $ B\\rightarrow B_{1}$ and $ B'\\rightarrow B_{1}'$.also we know $ \\overrightarrow{AB}+\\overrightarrow{A'B'}=2\\overrightarrow{MN}$ and $ MB_{1}=MB_{1}'$ then $ MN\\parallel \\text{angle bisector of}\\ \\angle A$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "vector",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Given a polynomial $ P \\in \\mathbb{R}[X]$, find a norm $ \\|. \\|$ on $ \\mathbb{R}[X]$ such that $ \\|X^n \\minus{} P \\| \\rightarrow 0$ when $ n \\rightarrow \\infty$",
  "Solution_1": "Without extra assumptions, it is easy: given any algebraic basis $ e_1,e_2,\\dots$ in an infinite dimensional vector space, we can always define a norm such that $ \\|e_n\\|\\to 0$ (say, by $ \\Bigl\\|\\sum_j \\alpha_j e_j\\Bigr\\|\\equal{}\\sum_j\\frac{|\\alpha_j|}j$). It remains to note that $ 1,x,\\dots,x^m,x^{m\\plus{}1}\\minus{}P, x^{m\\plus{}2}\\minus{}P,\\dots$ is a basis in $ \\mathbb R[X]$ where $ m$ is the degree of $ P$. Here is a more challenging question: can we find such a norm with the extra property that $ \\|QR\\|\\le\\|Q\\|\\cdot\\|R\\|$ for any polynomials $ Q,R$? ;)",
  "Solution_2": "thanks fedja, I'll think about your question  :wink:"
}
{
  "Tag": [],
  "Problem": "Yeah... I dunno how this could be applied to real life, but maybe one day, someone will make use to it... i don't know how, but I hope someone considers my theory and give feedbacks...\r\n\r\nTo be honest, I thought mathematicians are always open to new ideas, and not become dependent on the facts that they already know.  I always thought that mathematicians are ready to analyze the unthinkable, and consider whether it's valid or not.\r\n\r\nMaybe how I worded my idea, may throw people off... I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value) (oddest oxymoron).  Every contradiction stated in this forum thus far is relied on what I believe to be \"lies.\"\r\n\r\nIf I offend you even more, I'll take it all back... Honestly, I should have not stated this, because apparently no one is considering my theory, which is backed up with true information as well...  I should have find a better place to post this -.-;",
  "Solution_1": "sorry Peter, may you delete this... I pushed the wrong button -.-;"
}
{
  "Tag": [
    "geometry",
    "trapezoid"
  ],
  "Problem": "\u5728\u68af\u884cABCD\u4e2d AB\u5e73\u884cCD \r\n\u904eA\u9ede\u5728BC\u908a\u4e0a\u53d6\u4e00\u9edeE\r\n\u9023\u63a5AE\u53caDE \u4f7f\u89d2AED\u70ba\u76f4\u89d2\r\n\u8acb\u554f\u5982\u4f55\u8b49\u660e\u5f97\u5230\u4e09\u89d2\u5f62ADE\u9762\u7a4d\u70ba\u68af\u5f62\u9762\u7a4d\u7684\u4e00\u534a\r\n\r\nTranslate:\r\n\r\nIn a trapezium $ ABCD , AB//CD ,$ take a point $ E$  on $ BC$ such that $ \\angle AED$ is a right angle. \r\n\r\nProve that the area of $ \\triangle ADE$ is as half as that of trapezium $ ABCD$ .",
  "Solution_1": "anybody has idea?",
  "Solution_2": "The problem is obviously wrong: basically there are 2 points $E$ on $BC$ with $AE\\bot DE$ and, as per your 'problem' we need each of them to have area half of that of the trapezoid, i.e. $AD\\parallel BC$, or $ABCD$ a parallelogram.\n\nBest regards,\nsunken rock",
  "Solution_3": "[color=darkred][b][u]PP[/u].[/b] Let $ ABCD$ be a trapezoid, where $\\left\\{\\begin{array}{c}\nAB\\parallel CD\\\\\\\nAB\\perp AD\\end{array}\\right|$ so that  exists unique $E\\in (BC)\\ ,\\ EA\\perp ED$ . Prove that $\\frac {[ADE]}{[ABCD]}\\le\\frac 12$ .[/color]"
}
{
  "Tag": [
    "induction",
    "algebra",
    "polynomial",
    "logarithms",
    "function",
    "calculus",
    "derivative"
  ],
  "Problem": "Given $x_1 + .. + x_n = a_1 + .. + a_n$ with $x, a > 0$\r\n\r\nProve that the product $P = {x_1}^{a_1}..{x_n}^{a_n}$ attain max value when $x_i = a_i$.\r\n\r\nNote: $a$ is fixed, $x$ can be changed.",
  "Solution_1": "just a thought, could you use strong induction on $n$? i am not sure that it would be rigirous through, assume for n and 1, multiply them for n+1",
  "Solution_2": "That's what I tried, but it looks ugly since I had to used the polynomial expansion.  I wonder if there's another way that look better.",
  "Solution_3": "I tried Jensen's, but it only allows you to conclude $x_n, a_n$ similarly sorted, but they aren't necessarily equal...",
  "Solution_4": "[quote=\"t0rajir0u\"]I tried Jensen's, but it only allows you to conclude $x_n, a_n$ similarly sorted, but they aren't necessarily equal...[/quote]\r\n\r\nHmm, can you show me a case when they are not equal?",
  "Solution_5": "Note that $\\ln{P} = \\ln{(x_1^{a_1} \\cdots x_n^{a_n})} = \\sum a_i\\ln{x_i}$.\r\n\r\nTake the concave function $f(x) = \\ln{x}$. Let $\\sum x_i = \\sum a_i = A$ for convenience.\r\n\r\nBy Jensen's with weights $\\frac{a_i}{A}$, we have\r\n\r\n$\\frac{\\ln{P}}{A} = \\sum \\frac{a_i}{A}\\ln{x_i} \\le \\ln{\\left(\\frac{1}{A}\\sum a_ix_i\\right)}$.\r\n\r\nSo it is only necessary to maximize $\\sum a_ix_i$. But by Cauchy,\r\n\r\n$\\sum a_ix_i \\le \\sqrt{(a_1^2+\\cdots+a_n^2)(x_1^2+\\cdots+x_n^2)}$\r\n\r\nwhich has equality at $a_i = x_i$ (because $a_i$'s are constant).\r\n\r\nEDIT: On second thought, since $\\sum x_i^2$ isn't constant, that Cauchy step might not work...\r\n\r\nEDIT2: Maybe this approach will work better... Assume it has a maximum at $x_i = a_i$. Then show that by changing any two terms, we have $(a_j-\\delta)^{a_j}(a_k+\\delta)^{a_k} \\le a_j^{a_j}a_k^{a_k}$. I think we can show this by analyzing the derivative of the function $f(\\delta) = (a_j-\\delta)^{a_j}(a_k+\\delta)^{a_k}$ and showing that it's always negative...\r\n\r\n$f'(\\delta) = -\\ln{a_j} \\cdot (a_j-\\delta)^{a_j} - \\ln{a_k} \\cdot (a_k+\\delta)^{a_k}$ if I did that correctly.",
  "Solution_6": "Thank you, the problem is finally solved.  Taking the derivative is a good approach :)",
  "Solution_7": "I already considered the 'changing two terms' possibility, but I don't think it works because the maximum might be attainable by changing more than two terms simultaneously... I'm also very suspicious of the Cauchy solution because I don't think you can show that the Jensen's and Cauchy parts have the same maximum case.  :(\r\n\r\nLaputa:  For example $x_1 = 1, x_2 = 3, x_3 = 6$ and $a_1 = 1, a_2 = 2, a_3 = 7$ have the same sum and are similarly sorted, but $x_i \\neq a_i$.\r\n\r\nEdit:  [b]WAIT.[/b]  The problem as stated is incomplete!  In the above example, $\\sum x_i = \\sum a_i = 10$ and we are supposed to show that the maximum comes for $x_i = a_i$, however, $1^1 \\times 3^2 \\times 6^7 > 1^1 \\times 3^3 \\times 6^6$ !  There has to be some extra condition.  (In fact, the maximum comes when $x_1 = x_2 = 1, x_3 = 8$ and $a_1 = a_2 = 1, a_3 = 8$, I think.  But even so, now the wording of the problem confuses me.)",
  "Solution_8": "I don't think you're understanding the problem quite correctly... The given $a_i$ are FIXED. As in numbers. Then we want to show to maximize $P$ we have to set all the $x_i = a_i$.\r\n\r\nFor instance, if $a_1 = 1, a_2 = 2, a_3 = 7$ as in your example, then we want to show that $x_1 = 1, x_2 = 2, x_3 = 7$. I think it's true.\r\n\r\nI'm pretty sure the Cauchy/Jensen solution doesn't work; I'm still working out the specifics of the derivative solution...\r\n\r\nMaybe if we take $f(\\delta) = \\prod (a_i-m_i\\delta)^{a_i}$ for constants $m_i$ such that $\\sum m_i = 0$. That should cover all possibilities. The derivative of that is a beast, but I think it's always negative (of course given the condition that $a_i > m_i\\delta$).",
  "Solution_9": "Well, if you're going to use derivatives, it's easiest to use partial derivatives having in mind the given conditions. Then you directly get $x_i=a_i.$",
  "Solution_10": "Couldn't we use the rearrangemente inequality between base and exponent, saying that P is maximum when $x_i>x_j\\rightarrow a_i>a_j$?",
  "Solution_11": "There's actually an easy proof of this.  You're given a sum, and you're trying to maximize a product.  Does anything come to mind?",
  "Solution_12": "Apply the weighted AM-GM inequality to this?",
  "Solution_13": "I have a solution like this, waiting for any comments:\r\n\r\n[b]1. [/b] Case ${a^a}{b^b}$\r\n\r\nConsider the the function $f(x) = {a*ln(a-x) + b*ln(b+x)}$ with $0 \\leq x \\leq a$\r\n\r\nwe want to maximize the product so $a*ln(a-x) + b*ln(b+x)$ should be maximized as well.\r\n\r\n$f'(x) = \\frac{b}{b+x} - \\frac{a}{a-x}$\r\n\r\nSince $\\frac{b}{b+x} \\leq 1$ & $\\frac{a}{a-x}\\geq 1, f'(x) \\leq 0$, so $x = 0$ to attain the maximum value.\r\n\r\n[b]2.[/b] Consider the case of of ${a^a}{b^b}{c^c}$:\r\n\r\nlet $f(x_1, x_2) = {a^a}{b^b}{c^c} - {(a-x_1-x_2)^a}{(b+x_1)^b}{(c+x_2)^c}$\r\n\r\nSo $f(x_1, x_2) = [{a^a}{b^b}{c^c} - {(a-x_1)^a}{(b+x_1)^b}{c^c}] + [{(a-x_1)^a}{(b+x_1)^b}{c^c} - {(a-x_1-x_2)^a}{(b+x_1)^b}{(c+x_2)^c}]$\r\n\r\nFrom 1. above, we know that the derivative of the first square bracket is $\\leq 0$, so consider the second bracket:\r\n\r\nLet $a\\\" = a-x_1$ then\r\n\r\n$(2nd Bracket) = {(b+x_1)^b}[{(a\\\")^a}{c^c} - {(a\\\"-x_2)^a}{(c+x_2)^c}]$\r\n\r\nAgain, consider the function $g(x_2) = a*ln(a\\\"-x_2) + c*ln(c+x_2)$\r\n\r\n$g'(x) = \\frac{c}{c+x} - \\frac{a}{a\\\"-x}$ with $0 \\leq x \\leq a\\\"$\r\n\r\nSince $a = a\\\" + x \\geq a\\\"-x$, $\\frac{a}{a\\\"-x} \\geq 1$ while $\\frac{c}{c+x} \\leq 1$ so $g'(x) \\leq 0$.\r\n\r\n--> The derivative of $f(x_1, x_2) \\leq 0$\r\n\r\nSo we can conclude that if we decrease from one variable and increase in other variables then the product will be less than before.  So this problem is proven.",
  "Solution_14": "Note: Please put any comment on my proff on this page:\r\n\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=419269#p419269[/url]",
  "Solution_15": "[quote=\"zeus_three\"]Apply the weighted AM-GM inequality to this?[/quote]\r\n\r\nYes."
}
{
  "Tag": [
    "algebra",
    "difference of squares",
    "special factorizations",
    "AMC"
  ],
  "Problem": "How many right triangles have integer leg lengths $ a$ and $ b$ and a hypotenuse of length $ b\\plus{}1$, where $ b<100$?\r\n\r\n$ \\textbf{(A)}\\ 6 \\qquad\r\n\\textbf{(B)}\\ 7 \\qquad\r\n\\textbf{(C)}\\ 8 \\qquad\r\n\\textbf{(D)}\\ 9 \\qquad\r\n\\textbf{(E)}\\ 10$",
  "Solution_1": "[hide]the difference of squares:\nbetween b^2 and (b+1)^2\n(a+b)(a-b)\na-b is always 1\na+b=2b+1\n\na^2+b^2=(b+1)^2\n\nso basically... a^2=2b+1\nyou just have to find all the b's possible where 2b+1=a square number... 2b+1 = odd number so all the odd squares from 3-199 work.\n\n3^2,5^2,7^2,9^2,11^2,13^2\n\nthat's 6\nAns:A[/hide]",
  "Solution_2": "[hide=\"Solution\"]\n(Inspired by one of the Proofs Without Words things on the AoPS sidebar)\n\nSay $ b$ is the side length of a square, and the hypotenuse, $ b \\plus{} 1$, is the side length of another square.\n\nSince $ (b \\plus{} 1)^2 \\minus{} b^2 \\equal{} a^2$,\n\n$ \\frac{a^2 \\minus{} 1}{2} \\equal{} b$.\n\nNow all we have to do is determine for which positive values of $ a$ that $ 0 < b < 100$\n\nNote that $ a$ needs to be an odd number for $ b$ to be an integer\n\n$ 3 \\leq a \\leq 13$\n\n$ \\frac{13 \\minus{} 3}{2} \\plus{} 1 \\equal{} \\boxed{6}$, so $ A$.\n[/hide]",
  "Solution_3": "[hide=\"Lucky Way\"]\nIf you know some of your pythagorean triplets, then you know:\n3, 4, 5\n5, 12, 13\n7, 24, 25\nif you watch, the numbers seem increasing by a significant amount so I assumed that there would be no more than 6, which is the smallest answer choice.\nthe next ones would somewhere around 50-55 for b and then 70-75, and then 90-95\n(as a guess)\ntoo bad I didn't realize this simple and lucky way DURING the test  :wallbash: \n[/hide]",
  "Solution_4": "[quote=\"madshock\"][hide=\"Lucky Way\"]\nIf you know some of your pythagorean triplets, then you know:\n3, 4, 5\n5, 12, 13\n7, 24, 25\nif you watch, the numbers seem increasing by a significant amount so I assumed that there would be no more than 6, which is the smallest answer choice.\nthe next ones would somewhere around 50-55 for b and then 70-75, and then 90-95\n(as a guess)\ntoo bad I didn't realize this simple and lucky way DURING the test  :wallbash: \n[/hide][/quote]\r\n\r\nWow, great solution!",
  "Solution_5": "[hide]$ a^2\\plus{}b^2 \\equal{} b^2\\plus{}2b\\plus{}1 \\implies a^2 \\equal{} 1\\plus{}2b$, where $ 0<b<100$.\n\nThus, $ a^2$ can be any odd square where b is in that range. $ 1<a^2<199$, so squares $ 3^2,...,13^2$ work, or 6 squares, A.[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "hi everyone\r\n\r\nI have this question :\r\n\r\nCan anyone explain me why when we have for example the function $ f(x)\\equal{}\\int^{x\\plus{}1}_{x}e^{t^2}dt$ then it holds \r\n\r\n$ x\\leq t \\leq x\\plus{}1$ ?",
  "Solution_1": "Because we are taking the integral on the interval $ [x,x\\plus{}1]$  :lol:"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "let I,J be 2 ideals of ring R.show residue of any element of IJ in R/I inter. J\r\n is nilpotent.",
  "Solution_1": "In fact every such element is 0 since IJ is a subset of $I\\cap J$"
}
{
  "Tag": [
    "inequalities",
    "search",
    "inequalities unsolved"
  ],
  "Problem": "1.$ (x,y,z > 0)$            prove that     $ x^2 \\plus{} y^2 \\plus{} z^2\\ge\\sqrt {2}(xy \\plus{} yz)$\r\n2.$ (c\\ge b\\ge a\\ge 0)$  prove that     $ (a \\plus{} 3b)(b \\plus{} 4c)(c \\plus{} 2a)\\ge 60abc$",
  "Solution_1": "to 1. It's easy!   \r\n$ L\\equal{}(x^2\\plus{}\\frac{y^2}{2})\\plus{}(\\frac{y^2}{2}\\plus{}z^2)\\ge \\sqrt{2}(xy\\plus{}yz)\\equal{}R$ (by AM-GM)\r\nIt's good!\r\n\r\n\r\nbut 2.?",
  "Solution_2": "[quote=\"sxgfly\"]to 1. It's easy!   \n$ L \\equal{} (x^2 \\plus{} \\frac {y^2}{2}) \\plus{} (\\frac {y^2}{2} \\plus{} z^2)\\ge \\sqrt {2}(xy \\plus{} yz) \\equal{} R$ (by AM-GM)\nIt's good!\n\n\nbut 2.?[/quote]\r\n\r\n\r\n\r\nit is done by AM-GM and the fact that $ c\\ge b\\ge a\\ge 0$, $ LHS \\ge 60 a^{\\frac{11}{12}}b^{\\frac{19}{20}}c^{\\frac{17}{15}} \\ge 60 abc$",
  "Solution_3": "http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1289497002&t=104592\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?search_id=1289497002&t=208321"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "tetrahedron",
    "circumcircle",
    "geometry unsolved"
  ],
  "Problem": "Let $ ABCD$ is a tetrahedron and let $ AB\\equal{}a, CD\\equal{}b, AC\\equal{}AD\\equal{}BC\\equal{}BD\\equal{}1$. Find radius of the circumsphere of $ ABCD$.",
  "Solution_1": "There are \"ready made\" expressions for that question but, once the tetrahedron is not \"very irregular' since all the faces are isosceles, we'll try to make it from basics. \r\n\r\nThe center of the circumsphere, $ O$ , will be the intersection of the perpendiculars $ JO$ and $ KO$ to the faces $ BCD$ and $ ACD$ , with $ J$ and $ K$ as their circumcenters, in the medians $ BM$ and $ AM$ of length $ h \\equal{} \\sqrt {1 \\minus{} c^2}$ where $ c \\equal{} {b\\over2}$.\r\nThe distance from $ J$ and $ K$ to $ M$ , is $ y \\equal{} {{h^2 \\minus{} c^2}\\over{2h}}$.\r\nProlonging $ MO$ it'l meet $ BA$ , perpendicularly , at it's center $ N$ , so that the rectangular triangles, $ JMO$ and $ MNB$ are similar, and the dinstance between $ M$ and $ O$ is $ x \\equal{} {{hy}\\over{\\sqrt {h^2 \\minus{} d^2}}} \\equal{} {{h^2 \\minus{} c^2}\\over{2\\sqrt {h^2 \\minus{} d^2}}}$, here $ d \\equal{} {a\\over2}$.\r\n\r\nNow the distance from $ O$ to any vertex will be the circumradius; for instance $ R \\equal{} \\sqrt {c^2 \\plus{} x^2} \\equal{} {{4(h^2 \\minus{} d^2)c^2 \\plus{} {(h^2 \\minus{} c^2)}^2}\\over{4(h^2 \\minus{} d^2)}}$."
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "Below are a number of statements: \r\n1.Precisely one of these statements is untrue. \r\n2. Precisely two of these statements are untrue. \r\n3. Precisely three of these statements are untrue. \r\n4. Precisely four of these statements are untrue. \r\n5 .Precisely five of these statements are untrue. \r\n6. Precisely six of these statements are untrue. \r\n7. Precisely seven of these statements are untrue. \r\n8. Precisely eight of these statements are untrue. \r\n9. Precisely nine of these statements are untrue. \r\n10. Precisely ten of these statements are untrue. \r\n \r\n\r\n\r\n Which of these statements is true?",
  "Solution_1": "since untrue=false,[hide]only the ninth one is true[/hide]",
  "Solution_2": "[hide=\"Explanation\"]\n\n  Firstly, its impossible for there to be 0 untrue statements, it would mean everyone is telling the truth, a contradiction.  Now, we know that the number of untrue statements varies from:\n\n1 - 10\n\n  We also know one and only one number of untrue statements is correct.  Thus, only one statement is true, and the other nine are false.  Which means the 9th statement is true. [/hide]",
  "Solution_3": "But [u][i][b]noooooooooooooooooooooooo[/b][/i][/u]. The statement that \"below are a number of statements\" is true. Therefore there are eight untrue statements.",
  "Solution_4": "[quote=\"mathnerd314\"]But [u][i][b]noooooooooooooooooooooooo[/b][/i][/u]. The statement that \"below are a number of statements\" is true. Therefore there are eight untrue statements.[/quote]\r\n\r\nNice try, but there's still 9 untrue statements ;)",
  "Solution_5": "Oh. So... uh... what's the correct answer?!",
  "Solution_6": "[quote=\"mathnerd314\"]So... uh... what's the correct answer?![/quote]\r\n\r\n#9."
}
{
  "Tag": [
    "arithmetic sequence"
  ],
  "Problem": "What is the value of $ 100\\minus{}81\\plus{}64\\minus{}49\\plus{}36\\minus{}25\\plus{}16\\minus{}9\\plus{}4\\minus{}1$?",
  "Solution_1": "hello, we have $ \\sum_{i\\equal{}1}^{10}(\\minus{}1)^i\\cdot i^2\\equal{}55$.\r\nSonnhard.",
  "Solution_2": "Which doesn't really explain the solution.\r\n\r\nConsider the terms in pairs.\r\n\r\n19+15+11+7+3, this is an arithmetic sequence with 5 terms and a middle term of 11. Therefore the sum is 5*11=55.",
  "Solution_3": "hello, this will explain my solution it comes from the sum $ \\sum_{i\\equal{}1}^n(\\minus{}1)^i\\cdot i^2\\equal{}\\frac{1}{2}(\\minus{}1)^nn(n\\plus{}1)$.Sonnhard."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "This is from ASHME 1954 (also problem #80 chapt. 4 in aops book 1) \r\n\r\nIf x varies as the cube of y, and y varies as the fifth root of z, then x varies as the nth power of z. what is n?\r\n\r\ndoes varies mean that x is inversly proportional to y?  what does varies mean.  \r\n\r\nThanks",
  "Solution_1": "hmmm... I think it means x is proportional to y^3, etc.\r\n[hide=\"so here is my solution\"]$x=ky^3; y=nz^{1/5} \\implies x=k\\cdot n^3 \\cdot z^{3/5} = mz^{3/5}$ So x varies as the 3/5 power of z, I belive[/hide]",
  "Solution_2": "o i get it!  thanks"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Prove that for any group $ G, \\, \\, |G/Z(G)| \\neq 91.$ \r\n\r\n[hide=\"The hint in the book says ...\"] that assuming $ |G/Z(G)| = 91$, prove that $ G/Z(G)$ is cyclic.[/hide]",
  "Solution_1": "if $ H$ is a group of order $ 7*13$, sylow theorems yield normal subgroups of order $ 7$ and $ 13$. they have to intersect trivially, thus their product is $ H$. then $ H \\cong \\mathbb{Z}/7 \\times \\mathbb{Z}/13 \\cong \\mathbb{Z}/7*13$, i.e. $ H$ is cyclic.\r\n\r\nif $ G/Z(G)$ is cyclic, then $ G$ is abelian (easy exercise), i.e. $ G/Z(G)$ is trivial.",
  "Solution_2": "How does one get  $ H \\cong \\mathbb{Z}/7 \\times \\mathbb{Z}/13 \\cong \\mathbb{Z}/91$?\r\nThanks.",
  "Solution_3": "well think over it, then the only non-obviousl fact is that the elements of the two normal subgroups commute with each other. this comes from a general fact:\r\n\r\nlet $ N,M$ be normal subgroups of a group $ G$ with trivial intersection. then $ mn=nm$ for all $ m \\in M, n \\in N$ (and thus $ NM \\cong N \\times M$).\r\n\r\nthis follows from $ mnm^{-1}n^{-1}\\in M \\cap N$."
}
{
  "Tag": [
    "linear algebra"
  ],
  "Problem": "Given $ \\pi_{v} \\in L(V \\oplus W, V)$, $ S \\in L(U,V)$, and $ T \\in L(U,W)$, show that there is a natural linear transformation $ R \\equal{} S \\times T \\in L(U, V \\oplus W)$ such that $ \\pi_{v} \\circ R \\equal{} S$. Also, $ \\pi_{v}(x,y) \\equal{} x$, where $ x$ and $ y$ are vectors.\r\n\r\nI proceeded with this problem by  toying around with $ S$ and $ T$. Since we know that $ S: U \\longrightarrow V$ and $ T: U\\longrightarrow W$ then i just wrote \r\n\r\n$ S(u) \\equal{} ( v_{1},v_{2})$ and $ T(u) \\equal{} (w_{1}, w_{2}).$ where $ u \\in U$ and $ (v_{1}, v_{2}) \\equal{} v \\in V$ and similarly with $ w\\in W$\r\n\r\nSince i m trying to construct $ R: U \\longrightarrow V \\oplus W$ I added $ S(u)$ and $ T(u)$ to get \r\n\r\n$ R \\equal{} S(u) \\plus{} T(u) \\equal{} (v_{1} \\plus{} w_{1}, v_{2} \\plus{} w_{2})$.\r\n\r\nthen to show $ \\pi_{v} \\circ R \\equal{} S$, I simply applied $ \\pi_{v}$  on $ R$.Thus,\r\n\r\n$ \\pi_{v}(R) \\equal{} \\pi_{v}(v_{1} \\plus{} w_{1}, v_{2} \\plus{} w_{2}) \\equal{} v_{1} \\plus{} w_{1} \\in S(u)$???\r\n\r\nI m new to linear transformations, so can anyone tell me how to actually do this correctly? better yet, if it's all wrong, please tell me the best way to start. Thanks.",
  "Solution_1": "it's all wrong (remember the definitions), you have to define R(u)=(S(u),T(u))."
}
{
  "Tag": [
    "Asymptote",
    "algebra",
    "function",
    "domain",
    "vector",
    "LaTeX"
  ],
  "Problem": "(Maybe this belongs in a lower level of mathematics but still found it in a high school textbook)\r\n\r\nWhen we are required to draw graphs (e.g. using excel), do we have to show all the\r\n-asymptotes\r\n-breaking points in the graph if it is not continuous at at least one point?\r\n\r\nIf so how much should we show of each?  If the graph is starting to curve down at one side and up the other side at a certain value, can we just show the start and say that it is an asymptote?\r\n\r\nAlso if the graph stops at a point, (i.e. Domain of the function being drawn is x<10)  Do we have to draw the graph upto x=10 or upto x=10.1 to show that the graph in discontinuous at x=10?",
  "Solution_1": "Usually it depends on the requirement. What are the graphs for? A class assignment? Personal clarification? \r\n\r\nIf you wish to leave out the asymptote, you may, but you'll have to note somewhere that there is an asymptote and whatever value of $ x$ or $ y$. Generally, you should always display holes as a simple open circle in your graph.",
  "Solution_2": "but can we actually show those using excel and nothing else?",
  "Solution_3": "I'm not sure; I'm terrible at Excel other than graphing simple data graphs for science labs. \r\n\r\nYou can however, learn [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]Asymptote[/url], which is a programming language similar to LaTeX, in which you can easily express asymptotes and such on graphs."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $ a,b,c$ be positive real numbers. Prove that:\r\n$ \\sum_{cyclic}\\frac{a^3}{a^3\\plus{}b^3\\plus{}abc} \\ge 1$\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n-Nguyen Van Thach",
  "Solution_1": "I can't find any better way than expanding  :( \r\nExpanding, we get\r\n$ a^6b^3 \\plus{} b^6c^3 \\plus{} c^6a^3 \\ge a^5b^2c^2 \\plus{} a^2b^5c^2 \\plus{} a^2b^2c^5$\r\nLet $ a^2b \\equal{} x$, $ c^2a \\equal{} y$, $ b^2c \\equal{} z$, then this is equivalent to\r\n$ x^3 \\plus{} y^3 \\plus{} z^3 \\ge x^2y \\plus{} y^2z \\plus{} z^2x$\r\nBut $ x^3 \\plus{} x^3 \\plus{} y^3 \\ge 3x^2y$, so adding this cyclically gives the desired result.",
  "Solution_2": "Normalize to $ abc\\equal{}1,a^3\\equal{}x,b^3\\equal{}y,c^3\\equal{}z,xyz\\equal{}1$\r\n\r\n$ \\sum \\frac{x}{x\\plus{}y\\plus{}1}\\ge 1$\r\n\r\nThis is better for expanding, get to similar as dgreen, and finish it off.",
  "Solution_3": "[quote=\"dgreenb801\"]I can't find any better way than expanding ....[/quote]\r\nthis what i did substitute $ x\\equal{}\\frac{a}{b}$ and similarly define $ x,y,z$ the inequality becomes direct cauchy also i think it can be solved by writing the one in the RHS as $ \\frac{a^3\\plus{}b^3\\plus{}c^3}{a^3\\plus{}b^3\\plus{}c^3}$ and taking it to the LHS and using sos.",
  "Solution_4": "[hide=\" solution\"]\n\nthe $ LHS$ is homogeneous so let $ a^3b^3c^3 \\equal{} 1$\n\nlet $ a^3 \\equal{} \\frac {x}{y}, b^3 \\equal{} \\frac {y}{z}, c^3 \\equal{} \\frac {z}{x}$\n\ntherefore\n$ \\sum_{cyclic}\\frac {a^{3}}{a^{3} \\plus{} b^{3} \\plus{} abc} \\equal{} \\sum_{cyclic}\\frac {xz}{xz \\plus{} y^2 \\plus{} yz} \\equal{} \\sum_{cyclic}\\frac {(xz)^2}{(xz)^2 \\plus{} xy^2z \\plus{} xyz^2}$\n\nNo by cauchy\n\n$ (xy \\plus{} yz \\plus{} zx)^2 \\left(\\sum_{cyclic}\\frac {(xz)^2}{(xz)^2 \\plus{} xy^2z \\plus{} xyz^2} \\right) \\ge (xy \\plus{} yz \\plus{} zx)^2$\n\nand we're done\n\n[/hide]"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $H$ be a subgroup of $G$ with $h$ elements. Suppose $\\exists a\\in G,\\ \\forall x\\in H\\ (xa)^{3}=1$. Let $P=\\{x_{1}ax_{2}a\\dots x_{n}a|x_{i}\\in H, n\\in \\mathbb N\\}$. Find an upper bound for $|P|$.\r\nPS This is a simple problem. It might be helpful if you consider the case $G$ being abelian first.",
  "Solution_1": "Ok, here is the final [hide=\"answer\"]$3h^{2}$[/hide].Now, any idea?"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "parallelogram"
  ],
  "Problem": "[color=red]Three equal circles with a comon point H, are intersecting two by two in the points A,B,C.Prove that the circumcircle of the triangle ABC is equal with the three circles mentioned.(This problem is known as the Titeica's theorem[/color])\r\n[i]Proof 1[/i] $ BH$ is the segment of intersection $ C(O_3,R)$ and $ C(O_1,R)$. The triangle $ O_3BH$ is isoscel, and alsow  $ O_1BH$ is isoscel,and results that $ O_3BO_1H$ romb$ \\Longrightarrow$ and therefore $ O_3B_1\\equal{}O_1B_1$ where $ B_1$ is the midle of the segment $ BH$.In the same way $ O_2HCO_1$ romb $ \\Longrightarrow$ $ O_2C_1\\equal{}O_1C_1$ where $ C_1$ is the midle of the segment $ CH$,and $ O_2A_1\\equal{}O_3A_1$.  $ C(O_1)\\equal{}C(O_2)$ and $ C(O_1) \\cap C(O_2)\\equal{}CH$ $ \\Longrightarrow$ $ AB$ is the comon tangent to this circles $ \\Longrightarrow$ $ O_1B \\bot AB$ , $ O_2A \\bot AB$ \\Longrightarrow AO_2O_1B$ is a paralelogram with$<B=90$ .In the same way$O_3O_2CB is a paralelogram with <C=90, the same as $ AO_3O_1C$ As a consequence H becomes the ortocenter of ABC, and it was the center of $ C(O_1O_2O_3)$.But $ \\Delta ABC \\equiv \\Delta O_1O_2O_3 \\Longrightarrow C(ABC)\\equal{}C(O_1O_2O_3)$. But $ C(O_1O_2O_3)\\equal{}C(O_1)\\equal{}C(O_2)\\equal{}C(O_3)$ and results that $ C(ABC)\\equal{}C(O_1)\\equal{}C(O_2)\\equal{}C(O_3)$\r\n.This was a clasic demonstration.\r\n[i]Proof 2[/i] Here's an elegant one:D. $ A_1 , B_1 , C_1$ are on the Euler's circle in $ \\Delta O_1O_2O_3$. We had proved that  H is the ortocenter of the triangle $ ABC \\Longrightarrow A_1, B_1, C_1$ are on the Euler's circle in $ \\Delta ABC$.But $ A_1,B_1, C_1$ are on the Euler's circle of $ \\Delta O_1O_2O_3$ and in the same time on the Euler's circle of $ \\Delta ABC$ $ \\Longrightarrow C(ABC)\\equal{}C(O_1O_2O_3)\\equal{}C(O_1)\\equal{}C(O_2)\\equal{}C(O_3)$\r\n[i]Proof 3[/i] \r\nHere's a complex solution:D. Consider a cartezian system, and $ O_1(z_1), O_2(z_2), O_3(z_3)$",
  "Solution_1": "Continuing: results that $ A,B,C$ will have the afixez $ z_2\\plus{}z_3,z_1\\plus{}z_3,z_1\\plus{}z_2$. \\[ AB\\equal{} | (z_1\\plus{}z_3)\\minus{}(z_2\\plus{}z_3)|\\equal{}|z_1\\minus{}z_2|\\equal{}O_1O_2\\] In the same way we obtain $ BC\\equal{}O_2O_3$, $ AC\\equal{}O_1O_3$ .Therefore $ \\Delta ABC \\equiv \\Delta O_1O_2O_3$ so their circumcircles will be equal.\r\nIf we note $ \\omega\\equal{}z_1\\plus{}z_2\\plus{}z_3$ and let Q be a point with the afix $ \\omega$.We have $ |z_1|\\equal{}|z_2|\\equal{}|z_3|\\equal{}r$ we can prove that Q is the center of the circumscribe circle of the triangle ABC.\r\n\r\n\r\nI'm expecting more solutions to this beautifull problem!",
  "Solution_2": "The second problem is a problem of Mihail Popescu.\r\n[color=darkblue]If three equal equilateral triangles $ OA_1B_1, OA_2B_2,OA_3B_3$ have point O in comon then:\na). $ C(OA_1B_1),C(OA_2B_2),C(OA_3B_3)$ are intersectating two by two in three points who form a triangle and the circumcircle of that triangle is equal to the circles mentioned.\nb). $ C(OA_1B_2), C(OA_2B_3), C(OA_3B_1)$ are intersectating two by two In three points who form a equilateral triangle equal whith those that are mentioned.The same property  takes place  for the circles $ (OA_1B_3),(OA_2B_1),(OA_3B_2)$.[/color]\r\n\r\nI will post the solution this night.I'm interested in a complex solution wich I haven't discovered yet:D",
  "Solution_3": "I'm back with the solution.For the point a, from the equality of the triangles results the equality of the circumcircles and the problem reduces to Titeica's theorem.\r\nFor point b,let $ C(O,R)$ the circle who passes the rest of the top points of the equilateral triangles mentioned.Let A\",B\",C\" the centers of $ C(OA_3B_2)$ , $ C(OA_1B_3)$,$ C(OA_2B_1)$ and $ A$,$ B$,$ C$ the points of intersection of these circles. $ D$,$ E$,$ F$ the midle of the segments $ B_2A_3$, $ B_3A_1$, $ B_1A_2$ with $ M$,$ N$,$ P$,$ Q$ the midles of segments $ OC,OB,OA_3,OB_3$ and $ <A_3OB_2\\equal{}\\alpha$,$ <A_1OB_3\\equal{}\\beta$,$ <A_2OB_1\\equal{}\\gamma$, $ \\alpha\\plus{}\\beta\\plus{}\\gamma\\equal{}\\pi/2$.\r\n$ <B'FD\\equal{}<B'DF\\equal{}<B_3B_1A_3\\equal{}\\pi/6$ and results that B' is the center of the circumscribe circle of the equilateral triangle build on DF as a side.Same A' and C'. We all know that the centers of the circumscribe circles of the equilateral triangles build in the exterior by the sides of a triangle,are forming a equilateral triangle, such as A'B'C'.\r\nThe two quadrilaterals $ A'A_3PA''$ and $ B'B_3QB''$ are inscriptibles, therefore $ OA' \\cdot OA''\\equal{}OB' \\cdot OB''\\equal{}R^2/2$ from where A'B' and A\"B\" are antiparalels in the $ \\Delta OA'B'$, from where $ <OB''A''\\equal{}<OA'B'$\r\n.In the same way $ <OC''A''\\equal{}<OA'C'$ and $ <OB''A''\\plus{}<OC''A''\\equal{}\\pi/3$.In the triangle A\"B\"C\" , $ <A''\\equal{}\\beta\\plus{}\\gamma$.Consider the circle of diameter OA\" and M,N,P points on this circle.$ <MA''N\\equal{}<PA''O$ therefore $ MN\\equal{}OP\\equal{}R/2$ and $ BC\\equal{}2MN\\equal{}R$.In the same way $ AB\\equal{}AC\\equal{}R$ and it results that the triangle ABC is equilateral, equal with the three triangles given."
}
{
  "Tag": [
    "conics",
    "parabola",
    "geometry",
    "geometric transformation",
    "reflection",
    "analytic geometry",
    "graphing lines"
  ],
  "Problem": "how do you find the focus of a parabola?  I know what it means, but how do you find it in a parabola?  it would really help if you told me how to do this.  Thanks!!!",
  "Solution_1": "It depends on how your parabola is constructed.\r\n\r\n$ \\blacktriangleright$ If your parabola is in the Cartesian plane and has an central axis parallel to either the x or y axis of the plane, then you can write the equation in [url=http://en.wikipedia.org/wiki/Parabola#Analytic_geometry_equations]Focus-Directrix[/url] form. This is likely what you are talking about.\r\n\r\n$ \\blacktriangleright$ In general, the parabola has an interesting property that any ray parallel to the axis of the parabola that is coming from the inside will reflect off the parabola and travel in a new ray that contains the focus. Thus, if two different rays parallel to the parabola reflect off the inside, then the intersection of the two reflected rays will be the focus.",
  "Solution_2": "[hide=\"Answer\"]If a parabola is in the form: \n\n$ y\\equal{}a(x\\minus{}h)^2\\plus{}k$\n\nThen the focus is the point:\n\n$ (h,k\\plus{}\\frac{1}{4a})$\n\nIf a parabola is in the form:\n\n$ x\\equal{}a(y\\minus{}k)^2\\plus{}h$\n\nThen the focus is the point:\n\n$ (h\\plus{}\\frac{1}{4a},k)$[/hide]\n[hide=\"Note\"] Also note that a parabola will not have a focus if it is degenerate (it will either be a line or two parallel lines). An example of a degenerate parabola would be $ x^2\\minus{}x\\equal{}0$, it is degenerate because there isn't a $ y$ value therefore the graph would be the line $ x\\equal{}0$.\n\n\n[/hide]",
  "Solution_3": "In conics, the equation of a parabola could also be $ (y\\minus{}k)^2\\equal{}4a(x\\minus{}h)$ with the vertex at $ (h\\plus{}a, k)$. (Interchange x and h with y and k for the 2nd equation, with vertex at $ (h, k\\plus{}a)$.)\r\n\r\nIn the form $ f(x)\\equal{}ax^2\\plus{}bx\\plus{}c$, the vertex is at $ (\\minus{}\\frac{b}{2a},f(\\minus{}\\frac{b}{2a}))$.",
  "Solution_4": "I'm going to throw out three claims without proof -- you're invited to explore/text/prove them yourself.\r\n\r\n[b]Claim 1:[/b] It is possible to arrange the Cartesian equation of any parabola into the form $ (ax \\plus{} by)^2 \\equal{} cx \\plus{} dy \\plus{} e$\r\n[i][u]Hint:[/u] Use the [url=http://en.wikipedia.org/wiki/Conic#Cartesian_coordinates]discriminant[/url] of a conic section[/i]\r\n\r\n$ \\blacktriangleright$ I define the [i]family[/i] of a line to be the set of all lines on the Cartesian plane that have the same slope.\r\n\r\n[b]Claim 2:[/b] The family of lines that contains the axis of a particular parabola is the only family in which every member line intersects that parabola only once.\r\n\r\n[b]Claim 3:[/b] If a parabola's equation is written in the form $ (ax \\plus{} by)^2 \\equal{} cx \\plus{} dy \\plus{} e$, then the line $ ax \\plus{} by \\equal{} 0$ is in the same family as the axis.\r\n\r\n\r\n\r\n\r\nFrom here, we know the slope of the axis, which means that we can emulate the light-ray approach suggested in the second triangular bullet of my first post.\r\n\r\n[b]Example:[/b] The equation of a parabola is $ (x \\minus{} y)^2 \\equal{} x \\plus{} y \\plus{} 2$. Find the focus.\r\n\r\nThe family that contains the axis is $ x \\minus{} y \\equal{} C_1$, where $ C_1$ is an arbitrary constant. Thus, the family that contains the tangent to the vertex is $ x \\plus{} y \\equal{} C_2$. Substituting this form into our original equation and looking for a value of $ C_2$ that gives only one point of contact, we find that the vertex is $ ( \\minus{} 1, \\minus{} 1)$. Since the axis contains the vertex, the axis must be $ x \\minus{} y \\equal{} 0$, or $ x \\equal{} y$. We know the focus is on this line.\r\n\r\nNow, consider the line $ x \\minus{} y \\equal{} \\minus{} 1$. This intersects the parabola at the point $ ( \\minus{} 1,0)$. We can use calculus to find that the slope of the normal line at this point is $ \\minus{} \\frac {1}{3}$, so the normal line at this point is $ y \\equal{} \\minus{} \\frac {1}{3}(x \\plus{} 1)$. The reflected ray is a portion of the line formed by reflecting $ x \\minus{} y \\equal{} \\minus{} 1 \\iff y \\equal{} (x \\plus{} 1)$ across the line $ y \\equal{} \\minus{} \\frac {1}{3}(x \\plus{} 1)$, which results in $ y \\equal{} \\minus{} 7 (x \\plus{} 1)$.\r\n\r\nFinally, we seek the intersection between $ y \\equal{} \\minus{} 7(x \\plus{} 1)$ (reflected ray) and $ x \\equal{} y$ (axis) to find that the focus is $ \\left( \\minus{} \\frac {7}{8} , \\minus{} \\frac {7}{8} \\right)$\r\n\r\n[b]Note:[/b] In this case, it would have also been straightforward to rotate the parabola 45\u00ba, since the rotation of the parabola is clear from the equation.",
  "Solution_5": "Here's how I would \"prove\" claim 2. Because the slope of any parabola (when rotated so the axis of symmetry is parallel with the y-axis) is infinitely growing, any line other than a vertical one would eventually intersect the parabola.\r\n\r\nI'm sure that there are \"better\", more algebraic ways to prove it, but this is a simple intuitive proof."
}
{
  "Tag": [
    "logarithms",
    "trigonometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Answer the following Question.\r\n\r\n[1] $\\frac{d}{dx} \\ln \\sqrt{1+\\cos ^ 2 \\frac{x}{2}}$\r\n\r\n[2] ${\\frac{d}{dx} (\\sqrt{x})^x}$\r\n\r\n[3] $\\frac{d}{dx} \\ln _x (\\ln x)$\r\n\r\n[4] $e^x\\frac{d}{dx}\\left(x^2\\frac{d}{dx}(x^2e^{-x})\\right)$\r\n\r\n[5] Find the values of $a,b$ such that $\\frac{d}{dx}\\{x\\sqrt{x^2+3}+a\\ln (x+\\sqrt{x^2+3})\\}=b\\sqrt{x^2+3}$.",
  "Solution_1": "1\r\n$\\frac{-\\sin{x}}{4(1+\\cos^2{(\\frac{x}{2})}}$\r\n2\r\n$(\\sqrt{x})^x(\\ln{\\sqrt{x}}+\\frac{1}{2})$\r\n3\r\ni don't get what that x is behind the second ln, is just x times the ln?\r\n4\r\n$6x^2+2x^3-x^4$",
  "Solution_2": "[quote=\"amirhtlusa\"]\ni don't get what that x is behind the second ln, is just x times the ln?\n[/quote]\r\n\r\nHe means $\\frac{d}{dx} \\left(\\frac{\\ln(\\ln x)}{\\ln x}\\right)$\r\n\r\n$ = \\frac{1-\\ln(\\ln x)}{x\\ln^2 x}$",
  "Solution_3": "[quote=\"amirhtlusa\"]1\n$\\frac{-\\sin{x}}{4(1+\\cos^2{(\\frac{x}{2})}}$ :)$ =-\\frac{\\sin x}{2(\\cos x +3)}$\n2\n$(\\sqrt{x})^x(\\ln{\\sqrt{x}}+\\frac{1}{2})$ :)$=\\frac{1}{2}(\\sqrt{x})^x (\\ln x+1)$\n3\ni don't get what that x is behind the second ln, is just x times the ln? x [color=blue]expresses the base of logarithm.\n[/color]\n4\n$6x^2+2x^3-x^4$ :( [/quote]\r\n\r\nHow about for 5?",
  "Solution_4": "[quote=\"Jimmy\"][quote=\"amirhtlusa\"]\ni don't get what that x is behind the second ln, is just x times the ln?\n[/quote]\n\nHe means $\\frac{d}{dx} \\left(\\frac{\\ln(\\ln x)}{\\ln x}\\right)$\n\n$ = \\frac{1-\\ln(\\ln x)}{x\\ln^2 x}$[/quote]\r\n\r\nYou are right.",
  "Solution_5": "5)\r\n\r\nAfter differentiating, we have \r\n\r\n$\\sqrt{x^2+3} + \\frac{a+x^2}{\\sqrt{x^2+3}} = b\\sqrt{x^2+3}$\r\n\r\n$\\frac{2x^2+a+3}{\\sqrt{x^2+3}} = b\\sqrt{x^2+3} \\Rightarrow 2x^2+a+3=bx^2+3b$\r\n\r\nSo, now we have $b=2$, and $a+3=3b\\Rightarrow a = 3$.",
  "Solution_6": "That's right, Jimmy.",
  "Solution_7": "4)\r\n\r\n$x^4 - 6x^3 + 6x^2$",
  "Solution_8": "[quote=\"Jimmy\"]4)\n\n$x^4 - 6x^3 + 6x^2$ :) [/quote]"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Let $ A\\in \\mathcal{M}_n(\\mathbb{R})$ prove that : \r\n\r\n1) if $ A\\equal{}A^T$ then $ \\lambda_k \\in \\mathbb{R}$, for $ k\\equal{}\\overline{1,n}$\r\n2)if $ A\\equal{}\\minus{}A^T$ then $ \\lambda_k \\in \\mathbb{C\\minus{}R}$ ,for $ k\\equal{}\\overline{1,n}$\r\n\r\nNote : $ A^T$ is the transpose of the matrix $ A$ ,and $ \\lambda_k ,k\\equal{}\\overline{1,n}$ are its eigenvalues.",
  "Solution_1": "You can use $ <AX,Y>$=$ <X,A^*Y>$ where $ X,Y\\in{M_{n,1}(C)}$ , $ A^*$ is the adjunct matrix(the transpose of the conjugate), and$ <X,Y>$ is the scalar product of $ 2$ column vectors.\r\nIn the both case for the demonstration $ Y=X$",
  "Solution_2": "2) should say that if $ A\\equal{}\\minus{}A^T,$ then $ \\lambda\\equal{}i\\mathbb{R}$ - in other words, all of the eigenvalues are pure imaginary.\r\n\r\nSaying $ \\mathbb{C}\\setminus\\mathbb{R}$ is simultaneously too weak - \"pure imaginary\" is (for the most part) a much stronger condition than \"not real\" - while at the same time being incorrect, since an antisymmetric matrix could have $ 0$ as an eigenvalue, and $ 0$ is real.",
  "Solution_3": "Another approach would be the following : \r\n\r\n[hide=\"Notations\"]\nLet $ z \\equal{} a \\plus{} ib \\in \\mathbb{C}$ , $ a,b \\in \\mathbb{R}$ .I denote $ \\overline{z} \\equal{} a \\minus{} ib$.\nIf $ X \\in \\mathcal{M}_n(\\mathbb{C})$ , $ X \\equal{} (x_{i,j})_{1 \\leq i,j \\leq n }$ .By $ \\overline{X}$, I understand $ \\overline{X} \\equal{} ( \\overline{x}_{i,j})_{1 \\leq i,j \\leq n}$\n[/hide]\r\n\r\n1.\r\n\r\nLet $ \\lambda$ be an eigenvalue of $ A$ and $ X \\in \\mathcal{M}_{n,1}(\\mathbb{C})$ be the corresponding eigenvector. \r\nFrom $ AX \\equal{} \\lambda X$ we get $ \\overline{A}\\cdot \\overline{X} \\equal{} \\overline{\\lambda}\\cdot \\overline{X}$ $ \\Rightarrow$ \r\n$ \\overline{X}^T \\cdot \\overline{A}^T \\equal{} \\overline{\\lambda}\\cdot \\overline{X}^T$ \r\nBut because $ A\\in \\mathcal{M}_n(\\mathbb{R})$ we have that $ \\overline{A} \\equal{} A$ ,therefore $ \\overline{A}^T \\equal{} A^T \\equal{} A$ .(the last equality is from our hypothesis i.e. $ A^T \\equal{} A$)\r\nSo $ \\overline{X}^T \\cdot A \\equal{} \\overline{\\lambda}\\cdot \\overline{X}^T$ $ \\Rightarrow$ \r\n$ \\overline{X}^T \\cdot AX \\equal{} \\overline{\\lambda}\\cdot \\overline{X}^TX$ $ \\Leftrightarrow$ \r\n$ \\overline{X}^T \\cdot \\lambda X \\equal{} \\overline{\\lambda}\\cdot \\overline{X}^TX$ and from here we conclude that  $ \\lambda \\equal{} \\overline{\\lambda}$ $ \\Rightarrow$ $ \\lambda \\in \\mathbb{R}$\r\n\r\nWe can prove the second part analogously."
}
{
  "Tag": [
    "trigonometry",
    "complex numbers"
  ],
  "Problem": "Prove that cos 36 - cos 72 = 0.5, and find the numerical values of cos 36 and cos 72. I couldn't solve this problem without looking at the answers.  :wink:",
  "Solution_1": "[quote=vishalarul]Prove that cos 36 - cos 72 = 0.5[/quote]\r\n\r\n\r\nMaybe use complex numbers.\r\n :)",
  "Solution_2": "[hide]\nConsider the 5th roots of unity. They are of the form $\\cos{\\frac{2k\\pi}{5}}+i\\sin{\\frac{2k\\pi}{5}}$, and we're looking for $\\cos{\\frac{2\\pi}{5}}$.\n\nWe find them by solving $x^{5}-1=0$. Find, divide by $x-1$ to get $x^{4}+x^{3}+x^{2}+x+1=0$. Then, divide by $x^{2}$ to get $x^{2}+x+1+\\frac{1}{x}+\\frac{1}{x^{2}}=0$. Let $w: =x+\\frac{1}{x}$ This becomes $w^{2}+w-1=0$, so $w=\\frac{-1\\pm\\sqrt{5}}{2}=-\\phi,\\frac{1}{\\phi}$.\n\nIf we want x, we know that $x^{2}-wx+1=0$ or $x=\\frac{w \\pm \\sqrt{w^{2}-4}}{2}$. The real part, which is $\\cos{\\frac{2k\\pi}{5}}$ (depending on the root), is $\\frac{w}{2}$. If we want $\\cos{72}$, we use the positive value for w, so $\\cos{72}=\\frac{\\sqrt{5}-1}{4}=\\frac{1}{2\\phi}$.\n\nWe know that $\\cos{36}$ is a positive number such that $\\cos^{2}{36}-1=\\cos{72}=\\frac{1}{2\\phi}$.\n\nWe see that $\\cos{72}+\\frac{1}{2}= \\frac{1}{2\\phi}+\\frac{1}{2}=\\frac{1}{2}(\\frac{1}{\\phi}+1)=\\frac{\\phi}{2}$. \nThen, $2(\\cos{72}+\\frac{1}{2})^{2}-1=2\\frac{\\phi}{2}^{2}-1=\\frac{\\phi^{2}}{2}-1=\\frac{\\phi+1}{2}-1=\\frac{\\phi-1}{2}=\\cos{72}$.\nSo $\\cos{36}-\\cos{72}=\\frac{1}{2}$.\n[/hide]",
  "Solution_3": "[hide=\"Nice solution for cos36-cos72:\"]\n$\\cos(36)=2\\cos^{2}(18)-1=1-2\\sin^{2}(18)=1-2\\cos^{2}(72)$\n$\\cos(72)=2\\cos^{2}(36)-1$\n$\\cos(36)+\\cos(72)=2(\\cos^{2}(36)-\\cos^{2}(72))$\n$\\frac12=\\cos(36)-\\cos(72)$[/hide]",
  "Solution_4": "Nice solutions. The first solution was an ingenious example of using the roots of unity, while the other was similar to the solution in the answer key. Thanks.  :lol:"
}
{
  "Tag": [
    "search",
    "number theory theorems",
    "number theory"
  ],
  "Problem": "Could you tell me if $ \\left(a,b,c\\right)=\\left(m^2 + n^2 - p^2 - q^2,2\\left(mp + nq\\right),2\\left(mp - nq\\right)\\right)$ and any cyclyc permutation and $ d=m^2 + n^2 + p^2 + q^2$ for integer $ m,n,p,q$ are all integer solutions of $ a^2 + b^2+ c^2 = d^2$?",
  "Solution_1": "Yes.  See for example [url=http://en.wikipedia.org/wiki/Pythagorean_quadruple]the Wikipedia article[/url] (interestingly, the Mathworld article does not contain the complete parameterization).  I gave a sketch of one proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=729475814&t=176802]here[/url] and scorpius119 gave another which is quite nice."
}
{
  "Tag": [
    "percent",
    "geometry",
    "3D geometry"
  ],
  "Problem": "What percent of the volume of a $ 10 \\times 10 \\times 10$ box can be filled with $ 4 \\times 4 \\times 4$ wooden cubes? Express your answer as a decimal to the nearest tenth.",
  "Solution_1": "Does this have to be as many as possible?  Or is this just $ \\frac{4^3}{10^3}$?  If the latter is correct, then the answer is $ \\boxed{25.6}$.",
  "Solution_2": "We can fill it up to 8x8x8=512/1000=51.2%",
  "Solution_3": "Why 8x8x8? I think ernie's correct.",
  "Solution_4": "Maybe because it says cube[b]s[/b].",
  "Solution_5": "Think about it. With 4x4x4, you can fill it to axbxc , but a b and c must be multiples of 4.",
  "Solution_6": "Mew's right.  You can definitely fit 8x8x8 into a 10x10x10x :P",
  "Solution_7": "[quote=\"AIME15\"][size=200][b]Mew's right.[/size][/b]  You can definitely fit 8x8x8 into a 10x10x10x :P[/quote]\r\n\r\nNow that's not something you see every day....",
  "Solution_8": ":spam: mew....does post count really matter that much?  And yes, you don't see that every day, especially from MOI\r\n\r\n\r\nErnie, the reason why your wrong is it says What percent of the volume of a 10x10x10 box CAN BE FILLED, so it implies it's looking for the MAXXXXX (too much caffeine!)",
  "Solution_9": "I don't suppose you guys read the stickies too often, do you? I suggest you do so.\r\n\r\nLocked."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "If $a,b,c$ are nonnegative numbers such that $a\\le b\\le c$ and $ab+bc+ca=3$, then\r\n\r\n$\\frac 1{2a+b}+\\frac 1{2b+c}+\\frac 1{2c+a}\\ge 1$.",
  "Solution_1": "Thank you, VASC,\r\n\r\nBy my result post in your topic\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=144140\r\n\r\nI refer that it suffices to consider the inequality in case $\\min(a,b,c)=0$. Moreover, I already proved that for all $k\\ge 1$ and $a\\ge b\\ge c$ then\r\n\r\n\\[\\frac{1}{ka+b}+\\frac{1}{kb+c}+\\frac{1}{kc+a}\\ge \\frac{3}{k+1}.\\]\r\n\r\nHowever, I will try to solve your problem without my result. Thank you again,",
  "Solution_2": "[quote=\"hungkhtn\"] I already proved that for all $k\\ge 1$ and $a\\ge b\\ge c$ then\n\\[\\frac{1}{ka+b}+\\frac{1}{kb+c}+\\frac{1}{kc+a}\\ge \\frac{3}{k+1}. \\]\nHowever, I will try to solve your problem without my result. Thank you again,[/quote]\r\n\r\nThe inequality is not true for $k=1$. See Iran inequality.",
  "Solution_3": "I am sorry. I misunderstand at the first time. I did prove that for all $k\\ge 1$ and $a\\le b\\le c$ then\r\n\\[\\frac{1}{ka+b}+\\frac{1}{kb+c}+\\frac{1}{kc+a}\\ge \\min\\left(\\frac{3}{k+1},\\frac{1}{k\\sqrt{3}}+\\frac{1}{\\sqrt{3}}+\\frac{1}{(k+1)\\sqrt{3}}\\right). \\]\r\nThe equality must holds for $a=b=c=1$ or $a=0,b=c=\\sqrt{3}$. It is my implication.\r\n\r\nNotice that for $k=2$, we have\r\n\r\n\\[\\frac{1}{\\sqrt{3}}+\\frac{1}{2\\sqrt{3}}+\\frac{1}{3\\sqrt{3}}=1.05>1.\\]\r\n\r\nSo your problem is really sharp, VASC.",
  "Solution_4": "There is a nice solution to the original inequality. Who find it?",
  "Solution_5": "[quote=\"Vasc\"]If $a,b,c$ are nonnegative numbers such that $a\\le b\\le c$ and $ab+bc+ca=3$, then\n\n$\\frac 1{2a+b}+\\frac 1{2b+c}+\\frac 1{2c+a}\\ge 1$.[/quote]\r\nUsing the known $(x+y+z)^{2}\\ge 3(xy+yz+zx)$, it suffices to prove that\r\n\\[\\sum\\frac{1}{(2a+b)(2b+c)}\\ge \\frac{1}{ab+bc+ca}\\]\r\nOr\r\n\\[\\frac{3(a+b+c)}{(2a+b)(2b+c)(2c+a)}\\ge \\frac{1}{ab+bc+ca}\\]\r\nOr\r\n\\[ab^{2}+bc^{2}+ca^{2}\\ge a^{2}b+b^{2}c+c^{2}a \\]\r\nIt is trivial since $a \\le b \\le c$.\r\nWe are done. :)",
  "Solution_6": "Nice (only nice and not very nice, because this is also my solution  :lol: ).",
  "Solution_7": "Much simpler than I though! Thanks for VASC and Can,  :lol:",
  "Solution_8": "Toanhocmuonmau, you are really very clever.  :)"
}
{
  "Tag": [],
  "Problem": "1) Outline a synthesis of $ C_6H_5\\minus{}C \\equiv C\\minus{}CH_2OH$.\r\n\r\n2) What product will be obtained when the alcohol of 1) is treated first with $ Br_2/CH_2Cl_2$, and then with $ Ph_3P, Br_2$?",
  "Solution_1": "Probably start with C6H6:\r\n[hide=\"click to reveal the hidden content\"]1.Gattermann Hydroformylation --> Benzaldehyde\n2.Treat with ylide PPh3=CH2 --> Styrene\n3.Treat with Br2/CCl4-->dibromo product\n4.Treat with alc.NaNH2 to get phenyl ethyne\n5.Treat it again with NaNH2 to get the carbanion\n6.Treat it with formaldehyde(Nef synthesis)-->Required product[/hide]",
  "Solution_2": "My method:\r\n1. Get benzaldehyde from benzene\r\n2. Perkins rn to get cinnamic acid\r\n3. $ \\alpha$ halogenation (?doubtful)\r\n4. Alc. $ \\ce{KNH2}$\r\n5. Use LAH to reduce the COOH to $ \\ce{CH2OH}$\r\n :)",
  "Solution_3": "U can't alpha chlorinate cinnamic acid :)  :)",
  "Solution_4": "Chemrock's synthesis is correct, but:\r\n\r\n[quote=\"Chemrock\"]alc.NaNH2[/quote]\n\n?\n\n[quote=\"Chemrock\"]Treat it with formaldehyde(Nef synthesis)[/quote]\r\n\r\n?",
  "Solution_5": "strong base to do 2 dehalogenation to get alkyne.\r\n\r\nNef synthesis is treating metal alknyl with carbonyl compound. It is a nucleophilic addition reaction.",
  "Solution_6": "I know that. But what does it mean \"alc.NaNH2\"?\r\n\r\nAnd how about a shorter synthesis.",
  "Solution_7": "What is alc.KOH then?\r\nIt is just that in alcoholic medium it acts as a good base and hence eliminate them.\r\nProbably from benzaldehyde:\r\n[hide=\"click to reveal the hidden content\"]\n1.Do claisen smidt reaction with CH3-CHO-->cinnamaldehyde\n2.Then brominate and then dehalogenate\n3.reduce it with LAH[/hide]",
  "Solution_8": "[quote=\"Chemrock\"]What is alc.KOH then?[/quote]\r\n\r\nNo. Now you are further away from the correct reasoning.",
  "Solution_9": "[quote=\"Carcul\"]I know that. But what does it mean \"alc.NaNH2\"?\n\nAnd how about a shorter synthesis.[/quote]\r\n\r\nDo you want him to say that its a strong base  :maybe:",
  "Solution_10": "[quote=\"chemrock\"]U can't alpha chlorinate cinnamic acid [/quote]\r\n\r\nhey y cant u alpha chlorinate it??\r\nu can do HVZ rite??",
  "Solution_11": "no because it will prefer the benzylic carbon rather than the carbon alpha to teh acid.",
  "Solution_12": "Excuse me the mechanism of HVZ doesn't permit that. :maybe: \r\n[quote=\"valeriummaximum\"]no because it will prefer the benzylic carbon rather than the carbon alpha to teh acid.[/quote]\r\nthat also is a sp2-s bond and is stable mr.valerium :huh:",
  "Solution_13": "then can u giove the reason why u cant a-chlorinate cinnamic acid...",
  "Solution_14": "I have given it just above urs.\r\nHere's a better explanation with the mechanism below:\r\nThe enol content of cinnamyl chloride is very low so that's why the reaction gives very poor yields",
  "Solution_15": "where have u given the reason?>??",
  "Solution_16": "I also do not see the reason...\r\n\r\nWhat you have there is a mechanism for the HVZ reaction.",
  "Solution_17": "So does that mean you can alpha chlorinate?  :maybe:   :maybe:",
  "Solution_18": "[quote=\"chemrock\"]\nThe enol content of cinnamyl chloride is very low so that's why the reaction gives very poor yields[/quote]\r\nHere!! :|  :huh:",
  "Solution_19": "[quote=\"Valeriummaximum\"]then can u giove the reason why u cant a-chlorinate cinnamic acid...[/quote]\r\n\r\nThe reason is simple: the alpha carbon of cynnamic acid is a sp2 carbon, not a sp3.",
  "Solution_20": "I also gave that before here see:\r\n[quote=\"chemrock\"]It is a sp2-s bond and is stable mr.valerium[/quote]"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let O be the center of a circle on the plane.\r\nWe define a fix point M on a circle. We define a fix chord AB.\r\nLet CD be a variable chord (not AB) such that <CMD = < AMB.\r\nProve that point P = {AB \u2229 CD} is a fix point.",
  "Solution_1": "Unless I'm misunderstanding something, that can't be true.. What did you really mean?",
  "Solution_2": "I also have some troubles taking it in....",
  "Solution_3": "I am very sorry Grobber, Sailor and others.\r\nActually I was thinking about another probelm, which I interpret wrong, because I can't understand it.\r\nThis is Fregier theorem :\r\nIf through the point M of two fascicols of lines that are in involution we draw a circle, such that lines connect points which are on the circle goes through the fix point. (please explain solution clearly)\r\n\r\nGrobber maybe you will explain to everybody involution and omografy.\r\n\r\nP.S I hope you will for give me.",
  "Solution_4": "I think the best solution is to google \"involution\", \"homography\" etc. I'm definitely not an expert in the field, and all I did was apply methods such as the ones you have mentioned to a couple of problems. Besides, I'm really bad at explaining stuff I don't master myself :D.\r\n\r\n[url=http://www.euclidraw.com/Eng_fls/EucliDraw_htmls/Vocabulary.htm]Here[/url]'s a very good place to start.\r\n\r\nI'd also like to add that I didn't quite understand your formulation of the Fregier theorem.",
  "Solution_5": "Many thanks, I think I mustn't to go in the depth of geometry.\r\nBy the way just for you again Grobber condition in Romanian:\r\n\r\nDaca prin virful M a doua fascicole de drepte in involutie se duce un cerc, dreptele ce unesc punctele situate pe cerc a doua raze omoloage trec printr-un fix punct."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "This is a problem taken from kalva, but it 's not easy:\r\nThe chord $ CD$ is perpendicular to the diameter $ AB$ and meets it at $ H$.Let $ O$ be the center of the circle. The distances $ AB$ and $ CD$ are integral. The distance $ AB$ has 2 digits and the distance $ CD$ is obtained by reversing the digits of $ AB$. The distance $ OH$ is a non-zero rational. Find $ AB$.",
  "Solution_1": "Sorry, i've found the solution in kalva, thanks for paying attention.",
  "Solution_2": "just because i want to practice LaTexing, Iwould put the solution.\r\n\r\nif $AB = 10a + b$, then $CD = 10b + a$\r\n$(OH)^2 = (CO)^2 - (CH)^2$\r\n$= \\left(\\frac{1}{2}(AB) \\right)^2 - \\left(\\frac{1}{2}(CD) \\right)^2$\r\n$= (5a + \\frac{1}{2}b)^2 - (5b + \\frac{1}{2}a)^2$ \r\n$= \\frac{99}{4} \\left(a^2 - b^2 \\right)$\r\nsince OH is a rational, then $a^2 - b^2$ has to be 11\r\nthen $a+b = 11, a-b = 1$ so $a = 6, b = 5,$ yielding $AB = 65.$\r\n\r\nI ommitted the possibility of $a+b = 1, a-b = 1$, becuase it gives negative values.\r\n\r\n\r\nTHAT WAS A NICE PRACTICE OF LATEXING btw."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "perimeter",
    "incenter"
  ],
  "Problem": "If there are n lines that can divide triangle ABC into two parts, so the ratio of the areas of two parts equals to the ratio of the divided perimeter of triangle ABC. (the perimeter doesn't include the line segment that divides the triangle.) prove these n lines pass through the same point.( hint: this same point is G of triangle ABC.)",
  "Solution_1": "Consider a arbitrary line cut $ AB,AC$ at $ D,E$. We have:\r\n$ \\frac{S_{ADE}}{S_{BDEC}}\\equal{}\\frac{AD\\plus{}AE}{BD\\plus{}EC\\plus{}CB}\\Leftrightarrow \\frac{S_{ADE}}{S_{ABC}}\\equal{}\\frac{AD\\plus{}AE}{AB\\plus{}BC\\plus{}CA}$\r\n$ \\Leftrightarrow \\frac{AD.AE}{AB.AC}\\equal{}\\frac{AD\\plus{}AE}{AB\\plus{}BC\\plus{}CA}\\Leftrightarrow \\frac{1}{AD}\\plus{}\\frac{1}{AE}\\equal{}const$\r\nLet $ DE$ cut bisector of angle $ A$ at $ I$ and construct $ D'E'$ pass through $ I$ ang perpendicular with $ AI$. ($ D',E'$ on $ AB,AC$)\r\nApply Menelaus theorem for triangle $ AD'E'$ we have:\r\n$ \\frac{DA}{DD'}.\\frac{ID'}{IE'}.\\frac{EE'}{EA}\\equal{}1\\Leftrightarrow \\frac{AE\\minus{}AE'}{EA}\\equal{}\\frac{AD'\\minus{}AD}{DA}$\r\n$ \\Leftrightarrow \\frac{1}{AD}\\plus{}\\frac{1}{AE}\\equal{}\\frac{2}{AD'}$ (Because $ ID'\\equal{}IE',AD'\\equal{}AE'$)\r\nBecause $ \\frac{1}{AD}\\plus{}\\frac{1}{AE}\\equal{}const$ we have $ AD'\\equal{}const$, it mean $ I$ is fix point!\r\nSimilarly then all lines pass through incentre $ I$ of triangle $ ABC$!\r\nRecheck we have solution!",
  "Solution_2": "good solution! thx. \r\nSry I gave the wrong hint. It should be the incenter of ABC."
}
{
  "Tag": [
    "search",
    "national olympiad"
  ],
  "Problem": "Will there be any results?\r\nWhich countries participated?\r\nCan you tell me if possible?",
  "Solution_1": "[quote=\"Loser\"]Will there be any results?\nWhich countries participated?\nCan you tell me if possible?[/quote]\r\n\r\nThe contest is currently running. Problems will be posted in a few days. Russians usually participate in ARO though there might some guests invited, e.g. Chinese contenders.  :)",
  "Solution_2": "As I know, Blgarians, Chinese and 1 Ukrainian participate.",
  "Solution_3": "Do you know complete results?",
  "Solution_4": "No, I do not.",
  "Solution_5": "Now I know results of SPb guys. They may be found at http://spbolymp.hut.ru forum.",
  "Solution_6": "It is very useful info for me! Thank you! You are so kind.",
  "Solution_7": "Sorry, but I really do not know any other results. This results are gotten by mobile phone. You may call to your children and ask them aswell, yes?\r\n\r\nFinally, you are not the onliest, who is interested in results.",
  "Solution_8": "You are not correct here :(\r\nObviously, the only man here (except you) who partially knows children on ARO and really interested in results is me. For all others, this info is just curious.",
  "Solution_9": "I definitely know some people from SPb, who are interested too. Fedor Bakharev, for example. They may be rarely online, but can search some information about ARO here.",
  "Solution_10": "It is a strange place for that... Especially for SPb guys.",
  "Solution_11": "Why? It seems to be the only mathematical olympiads forum, known in Russia. We do not have local forum in SPb, but if we would have, it will be a subforum of Russian subforum of Mathlinks.",
  "Solution_12": "[quote=\"Fedor Petrov\"]Why? It seems to be the only mathematical olympiads forum, known in Russia. We do not have local forum in SPb, but if we would have, it will be a subforum of Russian subforum of Mathlinks.[/quote]By the way, if you want that I can arrange it for you ;)",
  "Solution_13": "Thank you. But now even general Russian forum is not active. Thank you in any case, may be somewhen in future, ok?",
  "Solution_14": "[quote=\"Fedor Petrov\"]Thank you. But now even general Russian forum is not active. Thank you in any case, may be somewhen in future, ok?[/quote]\r\n\r\nI guess you should tell the SPb math talents about the Russian subforum on ML.  ;)",
  "Solution_15": "Well it's never going to be active unless people start using it :D",
  "Solution_16": "Finally, I have a file with results. Ask me by PM or e-mail and I may send it.",
  "Solution_17": "Fedor, why don't you attach the file with the results here? :? :)",
  "Solution_18": "[quote=\"Valentin Vornicu\"]Fedor, why don't you attach the file with the results here? :? :)[/quote]\r\n\r\nYEs,I wanna know what our team did there.I've heard they have participated",
  "Solution_19": "OK. It is.",
  "Solution_20": "[quote=\"Fedor Petrov\"]OK. It is.[/quote]\r\n10x again."
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "ratio",
    "angle bisector",
    "exterior angle",
    "geometry proposed"
  ],
  "Problem": "Let $ ABC$ be a triangle.\r\nThe $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$.\r\nThe $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$.\r\nThe $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$.\r\nThe lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$.\r\nProve that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram.\r\n\r\n[i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$.\r\nSimilarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined.\r\n\r\n[hide=\"Source of the problem\"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]",
  "Solution_1": "$ O_aB \\parallel A_bC_b,$ both perpendicular to the bisector $ BO_b$ of the $ \\angle B,$ and $ O_aC \\parallel A_cB_c,$ both perpendicular to the bisector $ CO_c$ of the $ \\angle C.$ The $ \\triangle O_aBC \\sim \\triangle A_bA_cX$ are centrally similar, having parallel sides. (Since $ BA_c \\equal{} s \\minus{} a \\equal{} CA_b$ and $ A_bA_c \\equal{} b \\plus{} c,$ the similarity center is the common midpoint $ A'$ of their corresponding sides $ BC \\sim A_bA_c$ and the similarity coefficient $ \\frac {a}{b \\plus{} c}.$) Let $ D \\in BC$ be foot of the A-altitude and let $ K \\in BC$ be foot of the $ \\angle A$ bisector $ AO_a$ of the $ \\triangle ABC.$ A is the internal similarity center of the excircles $ (O_b),\\ (O_c),$ $ \\frac {AO_b}{AO_c} \\equal{} \\frac {r_b}{r_c} \\equal{} \\frac {DA_b}{DA_c}.$ The points $ A_a,\\ K$ divide the side $ BC$ of the $ \\triangle O_aBC$ in the ratios\r\n\r\n$ \\frac {A_aB}{A_aC} \\equal{} \\frac {s \\minus{} c}{s \\minus{} b},\\ \\ \\frac {KB}{KC} \\equal{} \\frac {c}{b},$\r\n\r\nwhile the points $ D,\\ A_a$ divide the corresponding side $ A_bA_c$ of the $ \\triangle XA_bA_c$ in the ratios\r\n\r\n$ \\frac {DA_b}{DA_c} \\equal{} \\frac {r_b}{r_c} \\equal{} \\frac {s \\minus{} c}{s \\minus{} b},\\ \\ \\frac {A_aA_b}{A_aA_c} \\equal{} \\frac {s \\minus{} a \\plus{} s \\minus{} b}{s \\minus{} a \\plus{} s \\minus{} c} \\equal{} \\frac {c}{b}.$\r\n\r\nThus $ A_a \\sim D,\\ K \\sim A_a$ are corresponding points of the centrally similar $ \\triangle O_aBC \\sim \\triangle A_bA_cX,$ therefore $ O_aA_a \\parallel XD,\\ O_aK \\parallel XA_a.$ $ A \\in O_aK$ by definition, $ O_aA \\equiv O_aK.$ $ A \\in XD,$ $ XA \\equiv XD,$ because $ AD \\perp BC$ by definition and $ O_aA_a \\perp BC,$ hence $ XD \\perp A_bA_c \\equiv BC.$",
  "Solution_2": "[quote=\"yetti\"]$ O_aB \\parallel A_bC_b,$ both perpendicular to the bisector $ BO_b$ of the $ \\angle B,$ and $ O_aC \\parallel A_cB_c,$ both perpendicular to the bisector $ CO_c$ of the $ \\angle C.$ [/quote]\r\nAnother solution of using this.\r\nDenot $ P,Q\\equal{}l\\cap XA_{b},XA_{c}$ where $ l$ is line that pass $ A$ and parallel to $ BC$. Then triangle $ O_{a}BC$ and $ XPQ$ are homothety. $ BA_{b}\\equal{}BC_{b}$ imply $ AP\\equal{}AC_{b}\\equal{}A_{a}B$, and similarly $ AQ\\equal{}A_{a}C$. Hence $ PQ\\equal{}BC$. Therefore, triangle $ O_{a}BC$ and $ XPQ$ are congruent and homothety.Since $ AP\\equal{}A_{a}B$ and $ AQ\\equal{}A_{a}C, A_{a},A$ are corresponding points.So $ XA\\equal{}O_{a}A_{a}$ and $ XA\\parallel O_{a}A_{a}$.Hence $ AO_{a}A_{a}X$ is parallelogram."
}
{
  "Tag": [
    "inequalities",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "Let $X, Y\\in M_{2}(\\mathbb{C})$ such that \\[ det(X^{2}+Y^{2})+det(XY+YX)=\\frac{1}{2}. \\]\r\n  Prove that \r\n                                            \\[ det(X+Y)+det(X-Y)\\leq\\sqrt{2}. \\]",
  "Solution_1": "i'll make it short because i'm in a rush. i have a biology test paper tomorrow. i will explain tomorrow more, if necesarry. here we go . let's calculate\r\n\r\n$det{(X+Y)^2}=det(X^2+Y^2+XY+YX)=det(X^2+Y^2)+det(XY+YX)+\\alpha=\\frac 12 + \\alpha$ , and also\r\n\r\n$det{(X-Y)^2}=det(X^2+Y^2-(XY+YX))=det(X^2+Y^2)+det(XY+YX)-\\alpha =\\frac 12 - \\alpha$\r\n\r\nadd them , and u get $det{(X+Y)^2}+det{(X-Y)^2}=1$ , so ${[det(X-Y)]}^2+{[det(X+Y)]}^2=1$\r\n\r\nnow we'll apply that inequality $\\sqrt{\\frac{a^2+b^2}{2}} \\geq \\frac{a+b}{2}$ for $det(X+Y)$ and $det(X-Y)$\r\n\r\nso u get that $\\sqrt{\\frac 12} \\geq \\frac{det(X+Y)+det(X-Y)}{2}$ from which u get the conclusion\r\n\r\n\r\ni'd like to thank my fans, appologyse for the awful writing, and admire Cezar Lupu cause he's cool and does Math in the same time"
}
{
  "Tag": [],
  "Problem": "problem 1\r\n\r\n\r\n$ \\sqrt {x^{\\rm{2}}  \\minus{} 16}  \\minus{} \\left( {x \\minus{} 4} \\right) \\equal{} \\sqrt {x^{\\rm{2}}  \\minus{} 5x \\plus{} 4}$\r\n\r\nproblem2\r\n\r\n\r\n\\[ 2\\left( {x^2  \\plus{} \\frac{1}{{x^2 }}} \\right) \\minus{} 9\\left( {x \\plus{} \\frac{1}{x}} \\right) \\plus{} 14 \\equal{} 0\r\n\\]\r\n\r\n\r\nmembers please submit more sums of same type with solutions\r\nSTUDENTS FROM INDIA\r\nONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE\r\nFOR DETAILS MAIL sanbittu@hotmail.com",
  "Solution_1": "[hide=\"#1\"]\n$ \\sqrt{(x\\minus{}4)(x\\plus{}4)}\\minus{}\\sqrt{(x\\minus{}4)(x\\minus{}1)}\\equal{}x\\minus{}4$\n$ x^2\\minus{}16$ is concave upwards with solutions of $ x\\equal{}\\pm4$, thus $ x\\le\\minus{}4$ & $ x\\ge4$\nClearly, $ \\sqrt{0}\\minus{}\\sqrt{0}\\equal{}0\\Longrightarrow\\boxed{x\\equal{}4}$[/hide]",
  "Solution_2": "[hide=\"Another Solution for 1\"]$ y \\equal{} x \\minus{} 4$\n[list]\n$ \\sqrt {y(x \\plus{} 4)} \\minus{} y \\equal{} \\sqrt {y(x \\minus{} 1)} \\implies \\sqrt {x \\plus{} 4} \\minus{} \\sqrt {x \\minus{} 1} \\equal{} \\sqrt {y}$\n\n$ y \\equal{} (x \\plus{} 4) \\minus{} 2\\sqrt {(x \\plus{} 4)(x \\minus{} 1)} \\plus{} (x \\minus{} 1) \\implies$\n\n$ (x \\minus{} 4) \\equal{} (2x \\plus{} 3) \\minus{} 2\\sqrt {(x \\plus{} 4)(x \\minus{} 1)}$\n\n$ 2\\sqrt {(x \\plus{} 4)(x \\minus{} 1)} \\equal{} x \\plus{} 7 \\implies 4(x \\plus{} 4)(x \\minus{} 1) \\equal{} x^2 \\plus{} 14x \\plus{} 49$\n\n$ 3x^2 \\minus{} 2x \\minus{} 65 \\equal{} 0 \\implies x \\equal{} \\frac {1\\pm14}{3} \\implies x \\equal{} \\boxed{5}, x\\not\\equal{}\\minus{}\\frac{13}{3}$\n\n[/list] \n[/hide]\n\n[hide=\"2\"]$ 2\\left(x^2 \\plus{} \\frac {1}{x^2}\\right) \\minus{} 9\\left(x \\plus{} \\frac {1}{x}\\right) \\plus{} 14 \\equal{} 0 \\implies 2x^4 \\minus{} 9x^3 \\plus{} 14x^2 \\minus{} 9x \\plus{} 2 \\equal{} 0$\n\n$ \\equal{} (x \\minus{} 1)^2(2x \\minus{} 1)(x \\minus{} 2)$\n\nSo $ x \\equal{} \\boxed{1, 1, \\frac {1}{2}, 2}$[/hide]"
}
{
  "Tag": [],
  "Problem": "Find mn if 3 times the sum of m and n is 24 and the difference of m and n is 12.",
  "Solution_1": "First equation: $3(m+n)=24$\r\nSecond equation: $m-n=12$\r\n\r\nNow:\r\n\r\n$m-n = 12 \\implies m+n = 12+2n \\implies 12+2n = 8 \\implies n =-2 \\implies m = 10 \\implies m \\cdot n =-20$\r\n\r\n :mad:",
  "Solution_2": "m+n=8\r\nm-n=12\r\n2m=20\r\nm=10\r\nn=-2\r\nmn=-20",
  "Solution_3": "Here is a worse solution:\r\n$3(m+n)=24$, so $m+n=8$ and we are given that $m-n=12$.  Therefore, $(m+n)^{2}-(m-n)^{2}=4mn=8^{2}-12^{2}=64-144=-80$.  Therefore, $4mn=-80\\Rightarrow mn=-20$",
  "Solution_4": "[quote=\"xxxxx\"]Find mn if 3 times the sum of m and n is 24 and the difference of m and n is 12.[/quote]\r\n[hide]$3(m+n)=24;\\,m-n=12$\n\n$m+n=8;\\,m-n=12$\n\n$2m=20\\Rightarrow m=10$\n\n$10+n=8\\Rightarrow n=-2$\n\n$mn=\\boxed{-20}$[/hide]"
}
{
  "Tag": [],
  "Problem": "Prove that given a circle and a tangent to that circle, the tangent line and the radius that meets the tangent line form a right angle.",
  "Solution_1": "can we use the fact that angles in a semicircle are 90 degress?",
  "Solution_2": "I believe that comes from this theorem.",
  "Solution_3": "[quote=\"arsemesomething\"]can we use the fact that angles in a semicircle are 90 degress?[/quote]\r\n\r\nI thought there were $180^\\circ$ in a semicircle, which isn't that just gotten from the fact that there are $360^\\circ$ in a circle?",
  "Solution_4": "i meant that if you have a semicircle, and a point on its arc, the angle at that point subtended by the diameter is 90 degrees.",
  "Solution_5": "can the tangen theorem be considered a special case of the chord theorem, where the two ends of the chord is the same point?"
}
{
  "Tag": [
    "modular arithmetic",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find alls integers n and k : \r\n$k^n\\equiv 0\\pmod{n^k +1}$",
  "Solution_1": "$(2,3)$ is the only positive solution. This is because for any $n>2, n^k > k^n$ if and only if $k > n$. So $n^k - k^n$ would be smallest if $k = n + 1$, and $n^{(n+1)} - (n+1)^n$ increases with $n$. \r\n\r\n$(0,k)$ and $(n,0)$ work for all positive $n$ and $k$. $(0,0)$ doesn't work. These are trivial.\r\n\r\nIf $n<0$, the LHS is a fraction, and a fraction mod an integer is a fraction (if you even consider it to exist), so that doesn't work.\r\n\r\nIf $k<0$, you've got fractions again, so if you want to worry about fractional mods, it seems like there ought to be a solution. Namely, $(2,-3)$. For positive $n$, that should be the only solution, by the same argument as above. \r\n\r\nFor both $n$ and $k$ less than 0, we've got $(-2,-1)$. I believe that's the only one, but I'm tired and I don't want to think about how to prove that."
}
{
  "Tag": [],
  "Problem": "If we count by 3's starting with 1, the following sequence is obtained: 1, 4, 7, 10,...\r\nWhat is the 100th number in the sequence?",
  "Solution_1": "[hide=\"Answer\"]$300$[/hide]",
  "Solution_2": "i think that's wrong because the sequence starts with 1.",
  "Solution_3": "Forgot to subtract 2  :( \r\n[hide=\"Corrected answer\"]$298$[/hide]",
  "Solution_4": "[hide]1 + 3(100 - 1) = 1 + 297 or 298.[/hide]",
  "Solution_5": "99*3+1=297+1=298",
  "Solution_6": "298. We see that the 100th number is obviously 3x100 which is 300. Subtracting 2 by the first and last digits according to a formula which I forget what it is.",
  "Solution_7": "pplz u just revived so many old topic.\r\nplz stop as nick42 has already mentioned"
}
{
  "Tag": [
    "inequalities",
    "analytic geometry",
    "trigonometry",
    "Pythagorean Theorem",
    "geometry",
    "triangle inequality"
  ],
  "Problem": "Given:  isosceles triangle $ PAB$, with $ |AP|\\equal{}|PB|$ and $ C$ on side $ PB$, strictly between $ P$ and $ B$.\r\n\r\nProve:  $ |CB|<|AC|.$",
  "Solution_1": "On the coordinate plane, let $ A \\equal{} ( \\minus{} m,0), B \\equal{} (m,0), P \\equal{} (0,n), C \\equal{} (x,y)$. Drop a perpendicular from $ C$ onto $ AB$ and call the intersection point at $ (x,0)$ $ D$. Since $ C$ is on side $ PB$, strictly between $ P$ and $ B, 0 < x < m$. Therefore, $ AD > DB$ and from the Pythagorean theorem, it follows that $ AC > BC$.",
  "Solution_2": "[quote=\"gooeymeister\"]On the coordinate plane, let $ A \\equal{} ( \\minus{} m,0), B \\equal{} (m,0), P \\equal{} (0,n), C \\equal{} (x,y)$. Drop a perpendicular from $ P$ onto $ AB$ and call the intersection point $ D$. Since $ C$ is on side $ PB$, strictly between $ P$ and $ B, 0 < x < m$. Therefore, $ AD > DB$ and from the Pythagorean theorem, it follows that $ AC > BC$.[/quote]\r\n\r\nI'm not following this -- could you clarify?  Specifically:\r\n\r\nFirst, since the triangle is isosceles with apex $ P$, $ D$ is the midpoint of $ A$ and $ B$, and is thus the origin.\r\n\r\nSecond, for the same reason, $ AD \\equal{} DB$, not $ AD > DB$.\r\n\r\nThird, in your last sentence, if the proof is still viable despite the first two comments, to what triangle are you applying Pythagoras's theorem?\r\n\r\n(Note that my intuition is that there must be a way to solve this through repeated application of the triangle inequality, but I'm not seeing it.)",
  "Solution_3": "[quote=\"Peterhi\"]Given:  isosceles triangle $ PAB$, with $ |AP| \\equal{} |PB|$ and $ C$ on side $ PB$, strictly between $ P$ and $ B$.\n\nProve:  $ |CB| < |AC|.$[/quote]\r\n\r\nIt is clear that $ \\angle CAB < \\angle PAB$ , because $ AC$ is an internal ray to the angle $ \\angle PAB$ but $ \\angle PAB \\equal{} \\angle PBA \\longrightarrow \\angle PBA > \\angle CAB \\longrightarrow CA > CB$ as desired  :)",
  "Solution_4": "[quote]First, since the triangle is isosceles with apex P, D is the midpoint of A and B, and is thus the origin. [/quote]\r\n\r\nSorry, I miswrote $ C$ as $ P$ and have now corrected it.",
  "Solution_5": "[quote=\"gooeymeister\"][quote]First, since the triangle is isosceles with apex P, D is the midpoint of A and B, and is thus the origin. [/quote]\n\nSorry, I miswrote $ C$ as $ P$ and have now corrected it.[/quote]\r\n\r\nGooey, could you explain your use of the Pythagoras's theorem in the last point of your original post?\r\n\r\nThanks.",
  "Solution_6": "Since $ AD \\bot CD$, $ AD^2\\plus{}CD^2\\equal{}AC^2$. Similarly, $ BD^2\\plus{}CD^2\\equal{}BC^2$. Therefore, $ AD>BD \\Rightarrow AC^2>BC^2 \\Leftrightarrow AC>BC$.",
  "Solution_7": "[quote=\"gooeymeister\"]Since $ AD \\bot CD$, $ AD^2 \\plus{} CD^2 \\equal{} AC^2$. Similarly, $ BD^2 \\plus{} CD^2 \\equal{} BC^2$. Therefore, $ AD > BD \\Rightarrow AC^2 > BC^2 \\Leftrightarrow AC > BC$.[/quote]\r\n\r\nVery nice, thanks.",
  "Solution_8": "Trigonometric Solution:\r\n\r\n [geogebra]dc33a2b03ce2ab5b74b9c06b2776cc4dc4452079[/geogebra]\r\n\r\nBy the Sine Law, it follows that $ \\frac{CB}{AC}\\equal{}\\frac{\\sin{\\angle{CAB}}}{\\sin{\\angle{CBA}}}$. Since $ \\angle{CAB} < \\angle{CBA} < 90$, and because $ \\angle{CAB}$ increases so does $ \\sin{\\angle{CAB}}$ (by the graph of sine) it also follows that $ \\sin{\\angle{CAB}} < \\sin{\\angle{CBA}}$ therefore $ \\frac{\\sin{\\angle{CAB}}}{\\sin{\\angle{CBA}}}\\equal{}\\frac{CB}{AC}<1$ and $ CB<AC$."
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "modular arithmetic",
    "factorial",
    "induction",
    "relatively prime",
    "number theory unsolved"
  ],
  "Problem": "Let $ n$ and $ q$ be positive integers. Prove that the sum\r\n\r\n$ \\sum_{k \\equal{} 0}^{n \\minus{} 1} \\binom{\\left(n,k\\right)q}{\\left(n,k\\right)}$ \r\n\r\nis divisible by $ nq$.\r\n\r\nHere, for any two integers $ x$ and $ y$, the term $ \\left(x, y\\right)$ denotes the greatest common divisor of $ x$ and $ y$, and the term $ \\binom{x}{y}$ denotes the binomial coefficient $ \\frac {x\\left(x \\minus{} 1\\right)...\\left(x \\minus{} y \\plus{} 1\\right)}{y!}$.",
  "Solution_1": "[quote=\"snake_pissken\"]Let $ n$ and $ q$ be positive integers. Prove that\n\nthe sum of binomial coefficients\n\n$ \\sum_{k \\equal{} 0}^{n \\minus{} 1} \\binom{(n,k)q}{(n,k)}$ \n\nis divisible by $ nq$, where $ (x, y)$ denotes the greatest common divisor of $ x$ and $ y$.[/quote]\r\n\r\nThe sum can be easily rewritten as $ \\sum_{d|n}\\phi(\\frac nd)\\binom{dq}d \\equal{} q\\sum_{d|n}\\phi(\\frac nd)\\binom{dq \\minus{} 1}{d \\minus{} 1}$, so all we have to do is to prove that $ n|\\sum_{d|n}\\phi(\\frac nd)\\binom{dq \\minus{} 1}{d \\minus{} 1}$. Now, using the well known identity $ \\phi(m) \\equal{} \\sum_{d|m}\\mu(d)\\frac md$ we can rewrite again the sum as $ \\sum_{m|n}\\frac nm\\sum_{d|m}\\mu(d)\\binom{\\frac mdq \\minus{} 1}{\\frac md \\minus{} 1}$, so that to prove that it is congruent to $ 0$ modulo $ n$ it is enough to prove that\r\n\\[ m|\\sum_{d|m}\\mu(d)\\binom{\\frac mdq \\minus{} 1}{\\frac md \\minus{} 1}\\]\r\nbut this last is a trivial task, because if $ p^e$ is a prime power factor of $ m$, you can group the summands $ \\binom{\\frac mdq \\minus{} 1}{\\frac md \\minus{} 1}$ and $ \\binom{\\frac m{d'}q \\minus{} 1}{\\frac m{d'} \\minus{} 1}$ in couples such that $ \\frac md$ and $ \\frac m{d'}$ differ in a power of $ p$, and they are congruent modulo $ p^e$.\r\n\r\nNOTE: I will clarify the last assertion with an example. Suppose we want to prove that $ 300$ divides $ \\sum_{d|300}\\mu(d)\\binom{\\frac {300}dq \\minus{} 1}{\\frac {300}d \\minus{} 1} \\equal{} \\binom{300q \\minus{} 1}{300 \\minus{} 1} \\minus{} \\binom{150q \\minus{} 1}{150 \\minus{} 1} \\minus{} \\binom{100q \\minus{} 1}{100 \\minus{} 1} \\minus{} \\binom{60q \\minus{} 1}{60 \\minus{} 1} \\plus{} \\binom{50q \\minus{} 1}{50 \\minus{} 1} \\plus{} \\binom{30q \\minus{} 1}{30 \\minus{} 1} \\plus{} \\binom{20q \\minus{} 1}{20 \\minus{} 1} \\minus{} \\binom{10q \\minus{} 1}{10 \\minus{} 1}$, then it suffices to prove that it is a multiple of $ 4$, of $ 3$ and of $ 25$.\r\nTo prove that $ 25$ divides the above sum it suffices to prove that $ 25$ divides $ \\binom{300q \\minus{} 1}{300 \\minus{} 1} \\minus{} \\binom{60q \\minus{} 1}{60 \\minus{} 1}$, $ \\binom{150q \\minus{} 1}{150 \\minus{} 1} \\minus{} \\binom{30q \\minus{} 1}{30 \\minus{} 1}$, and so on, grouping together the summands who have coefficients as $ 300, 60$ and $ 150, 30$ differing in a $ 5$ factor.\r\nNow, as an example, we prove that $ 25$ divides $ \\binom{300q \\minus{} 1}{300 \\minus{} 1} \\minus{} \\binom{60q \\minus{} 1}{60 \\minus{} 1}$, what is inmediate because\r\n\\[ \\binom{300q \\minus{} 1}{300 \\minus{} 1}\\equiv\\frac {300q \\minus{} 1}{300 \\minus{} 1}\\frac {300q \\minus{} 2}{300 \\minus{} 2}\\frac {300q \\minus{} 3}{300 \\minus{} 3} \\frac {300q \\minus{} 4}{300 \\minus{} 4}\\frac {300q \\minus{} 5}{300 \\minus{} 5}\\cdots\\equiv\\frac {300q \\minus{} 5}{300 \\minus{} 5} \\frac {300q \\minus{} 10}{300 \\minus{} 10}\\frac {300q \\minus{} 15}{300 \\minus{} 15}\\cdots\\equiv\\frac {60q \\minus{} 1}{60 \\minus{} 1} \\frac {60q \\minus{} 2}{60 \\minus{} 2}\\frac {60q \\minus{} 3}{60 \\minus{} 3}\\cdots\\equiv\\binom{60q \\minus{} 1}{60 \\minus{} 1}\\pmod{25}\\]",
  "Solution_2": "Does $(n,k) = gcd(n,k)$?",
  "Solution_3": "Yes, $(a,b)=\\gcd (a,b)$",
  "Solution_4": "Maybe you should rename the subject \"quite fun\".  I liked this problem.\r\n\r\nI want to give a little motivation to the solution I am posting.  There seemed to be two parts of the problem to me, one of them was how to control the gcd thingy and the other was how to finish it after you can phrase that question in the right way.  To provide some insight into how the second part would work out, I tried the simpler cases of $n = p^{k}$, where it is clear exactly how the gcds will work out.  The trick that fell out from there ended up being the final trick in the full solution.  And it was because that trick existed, then I decided to consider one prime power at a time and to provide all the conditions necessary for that one final trick.  \r\n\r\nHere is the actual solution after that introductory junk:\r\n\r\nWe first clean up a little clutter and restate the problem.  To get rid of $q$ we remember that $\\binom{aq}{a}= q \\binom{aq-1}{a-1}$.  Secondly, in order to more clearly understand what is happening, we sum in a different way.  We consider all of the possible $k$ for a given $(n,k)$ and realize that we are going to have $\\phi(n/(n,k))$ duplicates.  Thus we can rewrite the entire problem as: \\[n|\\sum_{d|n}\\phi(\\frac{n}{d})\\binom{dq-1}{d-1}\\] Now, we consider one prime power at a time.  That is to say we let $n = p^{k}t$ where $(p,t) =1$ \\[p^{k}|\\sum_{d|p^{k}t}\\phi(\\frac{p^{k}t}{d})\\binom{dq-1}{d-1}\\] Now is the big trick.  We hope that this problem is not that bad and hope that this big sum actually decomposes into many smaller reasonable sums that are also all zero in mod $p^{k}$. In fact, this doesn't require that much faith after having checked the cases when $n= p^{k}$.  We now consider divisors $d'$ of $t$   and for each divisor we notice that we have the divisors $d', d'p, \\cdots, d'p^{k}$ of $n$.  So we only consider one of these sums at a time: \\[\\sum_{a=0}^{k}\\phi(\\frac{p^{k-a}t}{d'})\\binom{p^{a}d'q-1}{p^{a}d'-1}\\] We now remember that $\\phi(x)$ is multiplicative so we factor out the useless (it turns out) $\\phi(\\frac{t}{d'})$  So we have: \\[(1)\\binom{p^{k}d'q-1}{p^{k}d'-1}+(p-1)\\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}+\\cdots+(p^{k}-p^{k-1})\\binom{d'q-1}{d'-1}\\] Now remember that we need $p^{k}$ to divide this junk.  Our first step is to prove that: \\[\\binom{p^{k}d'q-1}{p^{k}d'-1}\\equiv \\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}\\pmod{p^{k}}\\] But if you just write out the terms in the factorials for the LHS, you notice that all the terms on the top and on the bottom are the same in mod $p^{k}$.  Thus we expect it to be 1. However, we must be careful that some of the terms are not relatively prime to $p$, so division is a little tricky.  We save all of those leftover terms and cancel a factor of $p$ from each of them.  What we have left over is exactly equal to the RHS.  \r\nSo now all we need is that: \\[p^{k}|(p)\\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}+(p^{2}-p)\\binom{p^{k-2}d'q-1}{p^{k-2}d'-1}+\\cdots+(p^{k}-p^{k-1})\\binom{d'q-1}{d'-1}\\] But we can know divide out one factor of $p$ and continue the process through induction or whatever. \\[p^{k-1}|(1)\\binom{p^{k-1}d'q-1}{p^{k-1}d'-1}+(p-1)\\binom{p^{k-2}d'q-1}{p^{k-2}d'-1}+\\cdots+(p^{k-1}-p^{k-2})\\binom{d'q-1}{d'-1}\\] At the very end we will just have that we need: \\[1|\\binom{d'q-1}{d'-1}\\] which is obvious.  This works for all the prime power divisors of $n$, so we are done by CRT or whatever.  WOOHOO!",
  "Solution_5": "I think you are expert at Number Theory. Thank you for your solution.",
  "Solution_6": "Wow, I'm flattered.  You give me too much credit, sometimes I solve the problems, other times I don't.  All I can say is that I really like number theory.  That's why it was no problem at all for me to post this solution, it's fun for me."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "algebra",
    "polynomial",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Prove that: $\\exists A \\in M_n (R) $ invertible, such that \\[ :A + A^{ - 1}  = I_n  \\Leftrightarrow n = 2k,k \\in N. \\]",
  "Solution_1": "We need an invertible matrix $A$ with minimal polynomial $x^2-x+1$. The characteristic polynomial has the roots $\\omega,\\bar{\\omega}$, and they must have the same multiplicity, because the polynomial is real. This means that the degree is even.\r\n\r\nIf $n$ is even, we must find an example. I bet that's not hard to do :).",
  "Solution_2": ":) indead\r\n\r\nby using companion matrics :\r\nconsider A=  [0   -1\r\n                     1    1]\r\n\r\nits inverse is [1    1\r\n                     -1   0]\r\n\r\nnow just put these matrices as blocks on the diagonal of the 2n dimensional matrix and u have a good A",
  "Solution_3": "fredbel6's $2\\times2$ blocks:\r\n\r\n$A=\\begin{bmatrix} 0 & -1 \\\\ 1 & 1 \\end{bmatrix}$\r\n\r\n$A^{-1}=\\begin{bmatrix} 1 & 1 \\\\ -1 & 0 \\end{bmatrix}$\r\n\r\nOf course, any matrix that satisfies the condition of this problem also satisfies $A^3=-I$ and thus also $A^6=I.$"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "AoPS Books"
  ],
  "Problem": "Hey!\r\n\r\nI know the AOPS books are great for AMC and AIME, but what are some other good books\r\n\r\nand is Book 2 AOPS good for the AIME\r\n\r\nWhat are some good books for AMC AND AIME??",
  "Solution_1": "AoPS books seem to be the \"standard\" books for preparing for the AMC and AIME.  If AoPS Volume 1 and 2 aren't helping you for one reason or another, you could check out the AoPS Intro and Intermediate books that are geared toward specific subjects.  Most other books that are geared toward contest math are closer to USAMO/IMO level.  One book that many people recommend, although I don't like it that much, is Art and Craft of Problem Solving, which is somewhere between AIME and USAMO level.\r\n\r\nI suggest you stick with the AoPS Volume 1 and 2 books for learning new concepts, and focus on solving a lot of problems.",
  "Solution_2": "You should master the 4 introductory books and have a good understanding of the new intermediate books coming out this year.",
  "Solution_3": "[quote=\"mathswimmer\"]You should master the 4 introductory books and have a good understanding of the new intermediate books coming out this year.[/quote]This is not good advice. Besides being prohibitively expensive, buying *all* the AoPS books is unnecessary. I would be very surprised if *anyone* had the time to really master all the books. I think they're best if you are weak in a particular subject and/or actually do need an introduction to it.",
  "Solution_4": "So what did people use to prepare for the amc's (or ahsme) before aops came out?",
  "Solution_5": "I think that's why AoPS has been so successful as *the* introductory math problem solving books, there really wasn't anything comparable before them. Does it matter what people used before? All I could think of is past tests and math circles (if you're lucky in that regard).",
  "Solution_6": "[quote]What are some good books for AMC AND AIME??[/quote]\n\n[quote]what did people use to prepare for the amc's (or ahsme) before aops came out[/quote]\r\n\r\nConsider the Contest Problem Books I-IX and all the other books in the Problem Books category available from the \r\nMAA on-line bookstore and major on-line book retailers.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions"
}
{
  "Tag": [
    "USAMTS"
  ],
  "Problem": "Is it sufficient to find integers (c1, c2, c3, c4, c5, c6) that satisfy the conditions and prove it? Do we have to prove that it is the only such representation, or anything like that?",
  "Solution_1": "[quote=\"mathclass\"]Is it sufficient to find integers (c1, c2, c3, c4, c5, c6) that satisfy the conditions and prove it? Do we have to prove that it is the only such representation, or anything like that?[/quote]\r\nDoes the problem ask for that? If you are uncertain, reread the problem. Solutions are also subject to Rule 1 on the first page (\"1. You must show your work and prove your answers on all problems.\")\r\n\r\nErin Schram\r\nUSAMTS grader"
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be a group such that $ G'$ is abelian and each normal and abelian subgroup of $ G$ is finite. Prove that $ G$ is finite.",
  "Solution_1": "Sorry what is $ G'$?",
  "Solution_2": "[quote=\"Ahwingsecretagent\"]Sorry what is $ G'$?[/quote]\r\n\r\n[url=http://en.wikipedia.org/wiki/Commutator_subgroup]here[/url], note that this is fairly difficult to deal with.",
  "Solution_3": "Isn't $ \\mathbb{Z}/p^{\\infty}$ a counterexample? Is the problem supposed to be worded differently?",
  "Solution_4": "If $ G$ is itself abelian, it's one of those subgroups that has to be finite.",
  "Solution_5": "Suppose G is a metabelian group in which every abelian normal subgroup is finite.\r\n\r\nSince G'=[G,G] is abelian it is finite.  By Zorn's lemma, choose some maximal abelian normal subgroup N containing [G,G].  Consider z in the centralizer of N. Since G/N is abelian, the subgroup <z,N> is normal, and since z centralizes itself and N, <z,N> is abelian normal.  Since N is maximal, <z,N>=N, and N is its own centralizer. Hence the map from G into Aut(N) has kernel N, and G is an extension of the finite abelian group N by a finite abelian subgroup of the finite group Aut(N)."
}
{
  "Tag": [],
  "Problem": "Fie $f: R\\to R$, $f\\neq 0$ o functie continua, periodica, F o primitiva a sa si g un polinom de grad cel putin 2. Sa se arate ca $Fog$ nu este periodica.\r\n-----\r\nAm reusit sa rezum problema la cazul F periodica. Altfel zis, cum pot demonstra ca daca compunem o periodica cu o polinomiala de grad cel putin 2, nu mai obtinem o periodica?",
  "Solution_1": "La asta am dat o demonstratie pe forumu de calculus. Era un post a lu perfect-radio. Vezi si tu daca e buna solutia.  Poate gasesti una mai simpla  :)"
}
{
  "Tag": [],
  "Problem": "The product of the base seven numbers $ 24_7$ and $ 30_7$ is expressed in base seven. What is the base seven sum of the digits of this product?",
  "Solution_1": "Converting to base 10, $ 18\\cdot{21}\\equal{}378$ which is $ 1050$ in base 7, so $ 6$",
  "Solution_2": "Or, you can say\r\n[code]\n 1\n  24\n x30\n-----\n1050\n[/code]"
}
{
  "Tag": [
    "probability",
    "quadratics",
    "MATHCOUNTS",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "algebra"
  ],
  "Problem": "This round isnt going to be hard, for this is the first of the series of competition and I wanted to open it up to even the least skilled people (no offence :))\r\n\r\nOK, Question 1:\r\nCorrect Answer: 1 points\r\nEach different ways of arriving at the answer: 1/4 points\r\n\r\nWhat is the probability of rolling 2 dices and getting (2^(1/3)+1)(4^(1/3)-2^(1/3)+1) as the sum of the two numbers rolled?\r\n\r\nQuestion 2:\r\nCorrect Answer: 2 points\r\nEach different ways of arriving at the answer: 1/4 points\r\n\r\nFind the equation of a line going through the origin and (2, (5^(1/3)+25^(1/3)-(2*40^(1/3))/(5^(1/3)-1)) in point-intercept form\r\n\r\nQuestion 3:\r\nCorrect Answer: 3 points\r\nEach different ways of arriving at the answer: 1/2 points\r\n\r\nFind all quadratic equation which goes through (a,b), (b,c), and (c,b) where as a=(5^(1/3)+25^(1/3)-(2*40^(1/3))/(5^(1/3)-1), b=8^(1/3)+64^(1/3), and c=((((8/27)^(-1/3))^(3/2))/(3/2)^(-1/2))^1/2\r\n\r\nEnjoy! and remember, use spoilers!",
  "Solution_1": "Finally I get an answer!  \n\nAs for the question, [hide]your correct, but can you explain the (2^1/3+1)(2^2/3-2^1/3+1)=3 part a bit more?[/hide]\n\n\n\nYou know, I never realized how hard it is to select a phrase or an equation from a Spoiler text",
  "Solution_2": "I should give you partial credit because you double-posted, but oh well. Oh, and by the way, [hide]you're correct![/hide]",
  "Solution_3": "i feel confused and betrayed...",
  "Solution_4": "Yup, that's definitely one of the key points to this problem.",
  "Solution_5": "why does he only get partial credit? he was busy",
  "Solution_6": "note that I was not competing in the first place...",
  "Solution_7": "[quote=\"Syntax Error\"]why does he only get partial credit? he was busy[/quote]\r\n :?: Why does busy people gets to double-post? I mean, those two things aren't connected...(or so I think? :) )\r\n\r\nand MysticTerminator, sorry.",
  "Solution_8": "im not participating either, just postin mass amounts of problems",
  "Solution_9": "Okey dokey. I'd still love it if you refrain from posting anymore questions until I have sufficient participants for the competition. I mean, your problems can be solved anytime but the competition has to end by September 30th, you know. :) \r\n\r\nAm I being too self-centered? :P",
  "Solution_10": "Waiit..then who IS going to participate?",
  "Solution_11": "i dunno, but ill keep on posting problems cause it draws MORE attention.  :twisted:",
  "Solution_12": "Personally, I think there should be different divisions for member of the month...\r\n\r\nOne for the MathCounts participants\r\n\r\nOne for the AMC people\r\n\r\nOne for the USAMO making people\r\n\r\nand One for Simon and MysticTerminator  :wink:",
  "Solution_13": "um...you realize simon would easily destroy me in such a competition...but yeah, besides that i think that's a good idea",
  "Solution_14": "I should learn how to write problems, so maybe I should write such contests for others rather than solving other problems.",
  "Solution_15": "[quote]I should learn how to write problems, so maybe I should write such contests for others rather than solving other problems.[/quote]\r\n\r\nSimon, are you majoring in math in college? Anyways, you still have the Putnam.",
  "Solution_16": "I wouldn't consider not majoring in math for a second.",
  "Solution_17": "Oh, come on, Syntax!\r\nWe're just dividing those precious attention amongst ourselves.\r\n\r\nCan you do this, as a favor?",
  "Solution_18": "[quote=\"TheSouthpaw\"]Personally, I think there should be different divisions for member of the month...\n\nOne for the MathCounts participants\n\nOne for the AMC people\n\nOne for the USAMO making people\n\nand One for Simon and MysticTerminator  :wink:[/quote]\r\nThe last one's no good *booing and geering from the audience* but as for the rest, I'll post my \"plans\" in October. Till then, you'll just have to speculate :)",
  "Solution_19": "[quote=\"MysticTerminator\"]i feel confused and betrayed...[/quote]\r\n\r\nAgain, I publically apologize.\r\nI mean, what else can I do? harakiri? Chinese water torture? The ripply thing!? :)",
  "Solution_20": "[hide]1.  (2^(1/3)+1)(4^(1/3)-2^(1/3)+1)=8^(1/3)-4^(1/3)+2^(1/3)+4^(1/3)-2^(1/3)+1=3\n\nTotal possiblities of 2 dice = 6*6 = 36\n\nCombinations that yield 3 = 1,2 and 2,1 = 2 combinations\n\n2/36 = 1/18, the answer\n\n\n\n2.  The equation of the line will be y = m*x + b.\n\nThe line passes through 0,0 so that when x = 0, and y = 0, b must equal 0.  So, 2*m = (5^(1/3) +25^(1/3) - (2*(40^(1/3))))/(5^(1/3)-1)\n\n\n\nSimplify\n\n2*m = (25^(1/3)-(3*5^(1/3)))/(5^(1/3)-1)\n\n\n\nso, y = 2*(25^(1/3)-(3*5^(1/3)))/(5^(1/3)-1)*x, or\n\ny = 2*(5^(1/3)*(5^(1/3)-3))/(5^(1/3)-1)\n\n\n\n\n\n3.  You can get three equations, with a, b, and c currently not simplified.\n\n\n\nd*a^2+e*a+f = b\n\n\n\nd*b^2+e*b+f = c\n\n\n\nd*c^2+e*c+f = b\n\n\n\nSolving for f in the first equation, pluging it into the second and solving for e, pluging both of them into the third and solving for d gives that:\n\nd = 1/(a-b)\n\ne = -(a+c)/(a-b)\n\nf = (a*(b+c)-b^2)/(a-b)\n\n\n\nWhen simplified, b comes out as 6, and c as 3/2.\n\na simplifies to (5^(1/3)*(5^(1/3)-3))/(5^(1/3)-1)\n\n\n\nLeaving a as a (otherwise the final equation is very long), I get:\n\n\n\ny = (2*x^2-(2*a+3)*x+3*(5*a-24))/(2*(a-6))[/hide]",
  "Solution_21": "OK,. be patient with me here bookworm, because I'm making some tough decisions to make here...I'll be done in a minute. \r\n\r\nValo, (OK in Spanish, fsi (for somebody's info  :)) here is the official score: 4 1/2 points \r\nExplanation: The first one is correct. 1 and a quarter points. \r\nThe second one is correct, but it could be improved upon, so half credit for the answer. However, you're on the right track, so I'll give you the quarter points. Total of 1 and a quarter points. \r\nThe third one is the same as the second. Good answer, not the best. 2 points. \r\nMaking up the total of 4 and a half points.\r\n\r\n\r\n'Tis truly sad that there are less people actually participating than the people who said \"Count me in!\" :("
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope",
    "AMC",
    "AIME",
    "quadratics",
    "calculus"
  ],
  "Problem": "=) Hello. This is a question from the 2005 Fermat contest, which is written in Ontario, Canada. It was the last, most difficult question, and nobody in the school solved it (although some guessed). Apparently it took the entire math department and a few grade 12's about half an hour working together to solve it (the contest itself was only an hour long). A full solution written for someone at a Grade 11 level (so that I can understand this interesting problem :)) would be greatly, greatly appreciated! An interesting problem for all, I hope ^_^\r\n\r\nA 'triline' is a line with the property that three times its slope is equal to the sum of its $x$-intercept and its $y$-intercept. For how many integers $q$ with 1<= q <= 10,000 is there at least one integer $p$ so that there is exactly one triline through $(p,q)$?\r\n\r\nA) 60\r\nB) 57\r\nC) 58\r\nD) 61\r\nE) 59",
  "Solution_1": "btw, Fermat is written in more than just Ontario. Students from all over Canada write it.",
  "Solution_2": "I'll sketch a solution, the problem isn't that hard (if I'm reading it correctly).\r\n\r\nConsider the line through $p,q$ with slope $m$.  By the triline equation, we get:\r\n\r\n$3m=q-pm+p-q/m\\Rightarrow$\r\n$(p+3)m^2-(p+q)m+q=0\\Rightarrow$ two solutions unless $\\triangle=0$, therefore\r\n$(p+q)^2-4pq-12q=0\\Rightarrow$\r\n$(p-q)^2=12q$\r\n\r\nThis is equivalent to the number of integers between $1,346$ divisible by $6$; the answer is thus $57$.",
  "Solution_3": "Hey... My Solution was actually correct.. That's surprising..\r\n\r\nNow I should just hope that I didn't do anything silly..  :D",
  "Solution_4": "By the way, I am in grade 11, and blahblahblah's solution is exactly the same as mine..",
  "Solution_5": "[quote=\"blahblahblah\"]I'll sketch a solution, the problem isn't that hard (if I'm reading it correctly).\n\nConsider the line through $p,q$ with slope $m$.  By the triline equation, we get:\n\n$3m=q-pm+p-q/m\\Rightarrow$\n$(p+3)m^2-(p+q)m+q=0\\Rightarrow$ two solutions unless $\\triangle=0$, therefore\n$(p+q)^2-4pq-12q=0\\Rightarrow$\n$(p-q)^2=12q$\n\nThis is equivalent to the number of integers between $1,346$ divisible by $6$; the answer is thus $57$.[/quote]\r\n\r\nI'm not sure I understand what you did here -- could you expand on your solution a bit more for a relative beginner?  :lol: \r\n\r\nI especially don't understand your last part... \"the number of integers between $1,346$ divisible by $6$\"... did you mean the number of integers between $1,346$ *and* $1,000$? If not, then there are more than 57 integers divisible by six between 1-1,346. If so, where'd that come from? \r\n\r\nSorry for all these questions  :blush: I'd just like to understand what you did here....",
  "Solution_6": "In general, I never use commas when writing out numbers, so that should be read as the number of integers between $1$ AND $346$.\r\n\r\nYou should be able to follow the first part pretty easily, for the last part, notice that $12\\vert p^2$ if and only if $6\\vert p$.",
  "Solution_7": "I'd say this is a pretty difficult problem for a high school competition.  I wouldn't expect to see it before the end of the AIME, and maybe not even there.",
  "Solution_8": "good AIME problems are much harder. This is a very straightforward problem. I'll expand slightly on blah^3's soln (which is in fact the way I worked it, and really seems to be the most logical way to work it)\r\n\r\nSo we are given this fairly random condition for trilines. Let's try to quantify this. First, we know any line can be written in the form $y = mx + b$, where m is the slope and b is the y-intercept.  It's also pretty easy to find the x-intercept in this general case - simply plug 0 in for y and solve for x to get $-\\frac{b}{m}$. So, the condition for \"trilinity\" is that thrice the slope is the sum of the intercepts, i.e. $3m = b + -\\frac{b}{m}$. Cleaning this up a little, we have $b = \\frac{3m^2}{m-1}$, so the general form of a triline is $y = mx + \\frac{3m^2}{m-1}$. Now, we are interested in the trilines that pass through some point $(p,q)$. In specific, we want there to be a single solution for $m$ (the slope). So, plugging $(p,q)$ in for $(x,y)$, we have $q = mp + \\frac{3m^2}{m-1}$. We want to solve for $m$ (as we have a condition on $m$), so doing so, we find $(p+3)m^2 - (p+q)m + q = 0$. So, as this is a quadratic in $m$, if we want this to have a single solution, the discriminant must be $0$. So, we have $(p+q)^2 - 4(p+3)q = 0$. We want to solve this for $p$ (as we have a condition for $p$ in terms of a specific $q$). Doing so, we find $(p-q)^2 = 12q$. As $q$ is an integer, there will be an integral solution for $p$ if and only if $12q$ is a perfect square. We can make this a bit easier on ourselves by noting $12q$ will be a perfect square if and only if $q$ is $3$ times a perfect square. From here, all we have to note is that $3 = 3 \\cdot 1^2$ is the smallest integer of this form in the given range, and $9747 = 3 \\cdot 57^2$ is the largest integer of this form in the range, for a total of $57$ eligible values of $q$.",
  "Solution_9": "MysticTerminator, you have made my day. I fully understand that solution, and now I see where blahblahblah is coming from :P Thank you so much for your patience with that solution!\r\nI love this place  :lol:",
  "Solution_10": "How do you know that this:\r\n\r\n[quote=\"MysticTerminator\"]From here, all we have to note is that $3 = 3 \\cdot 1^2$ is the smallest integer of this form in the given range, and $9747 = 3 \\cdot 57^2$ is the largest integer of this form in the range, for a total of $57$ eligible values of $q$.[/quote]\n\nis the same as this:\n\n[quote=\"blahblahblah\"]This is equivalent to the number of integers between $1,346$ divisible by $6$; the answer is thus $57$.[/quote]",
  "Solution_11": "Okay, so $346$ is the largest integer whose square is less than $120000$\r\n\r\nWe want integers whose square is divisible by $12$.  $12=2^3\\cdot 3$, and the extra factor of $2$ doesn't help.  So we're just looking for integers divisible by $6$.",
  "Solution_12": "By the way, dr. rice mentioned that there's something funny about this problem.\r\n(not to worry, the answer is 57) anybody sees anything interesting or extremely funny?"
}
{
  "Tag": [
    "vector",
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "Let $V$ be a 10-dimensional real vector space and $U_1,U_2$ two linear subspaces such that $U_1 \\subseteq U_2, \\dim U_1 =3, \\dim U_2=6$. Let $\\varepsilon$ be the set of all linear maps $T: V\\rightarrow V$ which have $T(U_1)\\subseteq U_1, T(U_2)\\subseteq U_2$. Calculate the dimension of $\\varepsilon$. (again, all as real vector spaces)",
  "Solution_1": "Trivial number 1 as always: pick a basis of $U_1$, extend it to a basis of $U_2$, extend that to a basis of $V$. Then we have fixed $3(10-3)+(6-3)(10-6)=33$ elements, and the rest can be chosen completely free. So $\\dim\\varepsilon = 100-33=67$."
}
{
  "Tag": [
    "email",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "LaTeX",
    "geometry"
  ],
  "Problem": "I emailed a school in Kobe, Japan (I'm moving there by early February) asking if someone could proctor the AMCs for me to take. Here is the reply:\r\n\r\n\r\nDear Soo-Hyun Yoo,\r\n\r\nI am the high school counselor at Canadian Academy on Rokko Island in\r\nKobe, Japan.  Our headmaster forwarded your request for a test proctor\r\nto me.   Before I can look for a proctor I will need more information.\r\n How much does the test cost to take and how will we get it?  Are\r\nstudents such as yourself allowed to take the test independent of\r\nattending our school?  I haven't had time yet to look at the websites.\r\n Please make sure that you have reviewed the conditions necessary for\r\nyou to take the test as an individual and then let me know exactly\r\nwhat you will need from us.\r\n\r\n[hide=\"Original email\"]Dear Mr. Wesson,\n \nI am a high school freshman (9th grade in high school) currently living in the United States of America. I will be moving to Kobe, Japan by early February.\n \nI have regularly taken the American Mathematics Competitions (AMC) since 6th grade. This is a huge contest in which over 250,000 students participate annually. I am seeking a place to take this year's AMCs and subsequent tests in Kobe, and a proctor to administer the test.\n \n \nTest format:\nAMC 10/12 - taken by over 250,000 students - high scorers qualify to take the AIME\nAIME - score calculated with AMC score - qualification step for USAMO\nUSAMO - taken by only about 500 U.S. residents, qualification step for International Math Olympiad team\n \n \nCould you please see if one of the teachers at your school can offer his/her time to proctor the AMC? A quiet classroom after school is sufficient. Even a corner in the school office where I can be monitored (to ensure the authenticity of my answers) is a fine place to take the test. I would greatly appreciate it.\n \nMore information on the AMC, offered by the MAA, can be found on their website: http://www.unl.edu/amc/\n \nInformation for the AMC 10/12, as well as the order forms, can be found in their information packet: http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf\n \nThank you so much for taking the time to read this. Please reply as soon as possible.\n \nSincerely,\nSoo-Hyun Yoo\n[/hide]\r\n\r\nI'm aware that there is a \\$40 registration/shipping fee per school with each bundle of 10 exams sold at \\$14 or \\$16 for the AMC 10 and 12, respectively.\r\n\r\nThe overnight shipping cost is \\$60.\r\n\r\nThe completed order form needs to be faxed to the AMC office by the deadline.\r\n\r\n[b]However,[/b] I am not sure if I am allowed to take the test officially, registered as a student at a school that I don't attend.\r\n\r\nAlso, how should I pay the fees? Is mailing cash okay?\r\n\r\nCould someone confirm the conditions necessary and any additional information I should provide for the school?",
  "Solution_1": "If you aren't sure, just take the AMC B, since by that time you'll be in Japan.  I think you can register online, but I'm not sure if you can pay by credit card or not.  If you want to ask the MAA if they will allow your condition, you should try to contact them directly.\r\n\r\nBy the way, to use Latex, surround your code with dollar symbols.  For example, instead of \"x^2+z \\$10\" you should type\r\n[code]$x^2+z$ $\\$10$[/code]\r\nwhich will give you $ x^2\\plus{}z$ $ \\$10$.",
  "Solution_2": "It wasn't if if or not I could take the A test that I wasn't sure about - it's if or not I can take the tests officially registered as a school I don't attend. I'll be homeschooled - but registered as a student at a public school.\r\n\r\nI've already contacted the MAA weeks ago, but I haven't gotten a reply yet. Should I email them again?\r\n\r\nThanks though.\r\n\r\nEDIT: I've just emailed to amcinfo@maa.org my original post above.",
  "Solution_3": "The AMC staff member who formerly answered email at amcinfo@maa.org retired effective December 22, then the AMC office was closed for the holidays.  Then the new person who answers amcinfo@maa.org was at the Joint Math Meetings in San Diego almost immediately after returning.  That probably accounts for the delay in answering.  As of late last week, information requests were caught up again. \r\n\r\nPlease email amcinfo@maa.org and state your question simply and clearly.  From the posts above I do not understand your request.  \r\n\r\nFurther pursuit of the question here on the Forum will probably not help, your request appears to be too specfiic to your individual situation to make general advice possible.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions",
  "Solution_4": "My school only offers the AMC A tests. Would it be possible to take the B test at another school in the area (a school that I am not enrolled in)?",
  "Solution_5": "You can take the B tests at a local university or higher education site:\r\n\r\n[url]http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-CU-list.shtml[/url]\r\n\r\nprovided that you are lucky enough to be near one."
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "Galois Theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "I thought this was a nice problem. \r\n\r\n[b]Problem. [/b] A [i]minimal non-abelian group[/i] $G$ is a group such that $G$ is not abelian, but every proper subgroup of $G$ is abelian. Likewise, a minimal non-nilpotent group is a group that is not nilpotent, but every proper subgroup is nilpotent.\r\n\r\n(a) Prove that a minimal non-abelian group cannot be simple;\r\n(b) Prove that a minimal non-nilpotent group cannot be simple;\r\n(c)* Can a minimal non-solvable group be simple?\r\n\r\n[I have solved (a) and (b), but not (c). ]\r\n\r\n--Vesselin",
  "Solution_1": ":D  Just in case ... (c) is a joke  ;) .",
  "Solution_2": "First (c):\r\n\r\n$A_{5}$ is minimal non-solvable and it is simple. I believe this is what you meant when you said that it was a joke :).\r\n\r\n\r\n(a)\r\n\r\nIn [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=abelian&t=54910&sid=5bc9072c1c1cc41498c0e44e1a303d6d]this[/url] other thread there is a proof (which can be simplified) of the stronger fact that if a group has a maximal subgroup which is abelian then it is solvable. Here's a proof of the statement (stronger than the problem (a) but weaker than the problem in that other topic):\r\n\r\nIf a finite group (non-abelian) $G$ has a maximal subgroup $A$ which is abelian, then $G$ cannot be simple.\r\n\r\nAssume $G$ is simple. Let $g$ be an element outside $A$. If we were to have an element $1\\ne x\\in A\\cap gAg^{-1}$, then the centralizer of $x$ in $G$ would be a subgroup of $G$ containing the maximal subgroup $A$ strictly, so $x$ would be central in $G$, which contradicts the simplicity of $G$. This means that $gAg^{-1}\\cap A=\\{1\\}$ whenever $g\\in G\\setminus A$. I claim that this implies $(|A|,[G: A])=1\\ (*)$.  \r\n\r\nMake $A$ act by multiplication to the left on the set of $[G: A]$ left cosets of $A$. Let $a\\in A$ be an element of prime order $p\\ |\\ |A|$. $a$ fixes $A$ (regarded as the coset of $1$), and if $g\\not\\in A$, then, because $g^{-1}ag\\not\\in A$, we have $agA\\ne gA$ (i.e. $a$ does not fix $gA$). This means that the cycle structure of $a$ as a permutation of the left cosets of $A$ consists of one cycle of length $1$, and several cycles of length $p$. This means that $p|[G: A]-1$, which proves $(*)$. \r\n\r\nNow let $P$ be a Sylow subgroup of $A$. By $(*),\\ P$ is  Sylow subgroup of $G$. It\u2019s not normal in $G$ because we assumed $G$ to be simple, so the normalizer of $P$ is $A$. Clearly, it\u2019s also its centralizer, so $P$ is a Sylow subgroup of $G$ contained in the center of its normalizer. By Burnside\u2019s Normal Complement theorem (discussed on the forum several times), $P$ has a normal complement in $G$. In particular, $G$ is not simple, and we have our contradiction.\r\n\r\n\r\n(b)\r\n\r\nI don\u2019t really have a proof to this, except for one (which sort of feels like cheating :)) using the following theorem of Hall (which I\u2019ve also cited in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47434&sid=807169e25f4b3b935f52c90249cadc15]this[/url] other thread started by vess):\r\n\r\nLet $G$ be a finite group, and let $P$ be a Sylow $p$-subgroup of $G$ with the property that for every subgroup $Q$ of $P$, every element of $G$ of order prime to $p$ which normalizes $Q$ centralizes $Q$. Then $G$ has a normal complement of $P$. \r\n\r\nUsing this, the problem becomes easy. Let $P$ be a Sylow $p$-subgroup of our minimal non-nilpotent group $G$, assumed simple, and let $Q\\le P$. $Q$ is not normal, so its normalizer is contained in some nilpotent subgroup $N$ of $G$. $N$ is the direct product of its Sylow subgroups, so every element in $G$ with order prime to $p$ which normalizes $Q$ certainly centralizes $Q$. The hypotheses of the theorem above are satisfied, so $G$ has a normal subgroup, contradicting the assumption that it\u2019s simple."
}
{
  "Tag": [
    "trigonometry",
    "algebra",
    "polynomial",
    "complex numbers",
    "algebra unsolved"
  ],
  "Problem": ":arrow: [color=red]Let k be a positive integer. Prove that sqrt(k+1)-sqrt(k) is not the real part of the complex number z with z^n=1 for some positive integer n.[/color]",
  "Solution_1": "If $\\cos{\\alpha}=\\sqrt{k+1}-\\sqrt{k}$, here $z=\\cos{\\alpha}+i\\sin{\\alpha}$ and $z^{n}=1$. Then $\\cos{n\\alpha}=1$ therefore $\\sqrt{k+1}-\\sqrt{k}$ is a root of $g(x)-1$, here $g(\\cos{\\alpha})=\\cos{n\\alpha}$. But minimal polynomial of $\\sqrt{k+1}-\\sqrt{k}$ is $f(x)=x^{4}-(4k+2)x^{2}+1$, so $f(x)|[g(x)-1]$ contradiction because product of moduls of all roots of $f(x)$ is equal to $1$.",
  "Solution_2": "And now try this problem : \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=113831"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14613[/img]\r\n\r\n :(",
  "Solution_1": "You posted this problem at [url]http://www.mathlinks.ro/viewtopic.php?t=205642[/url]",
  "Solution_2": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14623[/img]\r\n\r\n :thumbup:"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "[b](1) Solve for x,y If \n\nx^2+xy = a^2\nx^2+xy = b^2\nx^2+y^2 = c^2[/b]",
  "Solution_1": "are you sure about the second equation :?:",
  "Solution_2": "i think the second equation must be y^2 + xy = b^2 \r\nthen it reduces to the most trivial thing ..\r\njust add (i) and (ii) .. get 2xy first add and then subtract and get x+y , x-y ... solve them :)[/img]"
}
{
  "Tag": [
    "search",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Show that for any $ a > 0$ there exist infinitely many $ n\\in N$ such that $ n^2 \\plus{} 1$ divides $ [an]!$. \r\n(where $ [x]$ is integer part of x.)",
  "Solution_1": "Posted before. \r\nI will solve it again :\r\nConsider the Pell's equation :\r\n$ x^2\\plus{}1\\equal{}(d^2\\plus{}1)y^2$ where $ d>0$\r\nIt has smallest  positive solution : \r\n$ (x,y)\\equal{}(d,1)$\r\nSo it has infinite solution : \r\n$ y_1\\equal{}$\r\n$ y_{n\\plus{}2}\\equal{}2(d^2\\plus{}1)y_{n\\plus{}1}\\minus{}y_n$\r\nThis is a sequence of root of this equation. \r\nSo $ \\gcd(y_{2n\\plus{}1},d^2\\plus{}1)\\equal{}1$\r\nConsider the sequence $ \\{y_{2n\\plus{}1}\\}$\r\nChose d,n large enough we have $ ax_{2n\\plus{}1}>2y_{2n\\plus{}1},\\forall n\\geq n_0$\r\nSo $ [ax_{2n\\plus{}1}]\\geq y_{2n\\plus{}1}$ and $ d^2\\plus{}1<[ax_n\\plus{}1]$ for all $ n\\geq n_0$\r\nSo $ y_{2n\\plus{}1}^2|[ax_{2n\\plus{}1}]!$ and $ d^2\\plus{}1|[ax_n]!$\r\nBut from $ \\gcd(y_{2n\\plus{}1},d^2\\plus{}1)\\equal{}1$ so \r\n$ x_{2n\\plus{}1}^2\\plus{}1|[ax_{2n\\plus{}1}]!$\r\nProof is complete . \r\nNote : please use search to find the general problem I have post . \r\nExist infinite $ x$ such that \r\n$ ax^2\\plus{}b|[cx]!$ :wink:"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all functions $ f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that \r\n$ f(x\\plus{}y)\\plus{}f(x)f(y)\\equal{}f(xy)\\plus{}f(x)\\plus{}f(y)$ for all $ x,y \\in \\mathbb{R}.$",
  "Solution_1": "$ P(x,0): (f(x) \\minus{} 2)f(0) \\equal{} 0$\r\nIf $ f(0)\\ne0$, we get the first solution is $ f(x) \\equal{} 2$.\r\nNow $ f(0) \\equal{} 0$.\r\n$ P(2,2): f(2)^2 \\equal{} 2f(2)$. So $ f(2) \\equal{} 0$ or $ f(2) \\equal{} 2$.\r\n$ P(1,1): f(1)^2 \\minus{} 3f(1) \\plus{} f(2) \\equal{} 0$. We can solve $ f(1)$ for all possible $ f(2)$ and there are four cases.\r\n\r\nCase 1) $ f(2) \\equal{} 2,f(1) \\equal{} 2$\r\n$ P(x,1): f(x \\plus{} 1) \\equal{} 2$, No solution.\r\n\r\nCase 2) $ f(2) \\equal{} 0,f(1) \\equal{} 3$\r\n$ P(x,1): f(x \\plus{} 1) \\equal{} 3 \\minus{} f(x)\\Rightarrow f(x \\plus{} 2) \\equal{} f(x)$\r\n$ P(x,2): f(2x) \\equal{} f(x \\plus{} 2) \\minus{} f(x) \\equal{} 0$. No solution\r\n\r\nCase 3) $ f(2) \\equal{} 0,f(1) \\equal{} 0$\r\n$ P(x,1): f(x \\plus{} 1) \\equal{} 2f(x)\\Rightarrow f(x \\plus{} n) \\equal{} 2^nf(x)$, where $ n$ can be any integer.\r\nWe also see $ f(n) \\equal{} 0$\r\n$ P(x,n): f(nx) \\equal{} (2^n \\minus{} 1)f(x)$\r\n$ P(nx,2x)$ implies $ f(x^2) \\equal{} \\frac3{2^n \\minus{} 1}(f(x)^2 \\plus{} f(x))$ for any integer $ n$.\r\nWe can let $ n$ be arbitrarily large and we get the second solution $ f(x) \\equal{} 0$.\r\n\r\nCase 4) $ f(2) \\equal{} 2,f(1) \\equal{} 1$\r\n$ P(x,1): f(x \\plus{} 1) \\equal{} f(x) \\plus{} 1\\Rightarrow f(x \\plus{} n) \\equal{} f(x) \\plus{} n$ and $ f(n) \\equal{} n$.\r\n$ P(x,n): f(nx) \\equal{} nf(x)\\Rightarrow f(rx) \\equal{} rf(x),\\forall r\\in \\mathbb Q$ and $ f(r) \\equal{} r$ \r\n$ P(x,r): f(x \\plus{} r) \\equal{} f(x) \\plus{} r$\r\n$ P(x,x): f(x^2) \\equal{} f(x)^2\\ge0$\r\nFor any $ x$, we find rational sequences $ r_n\\uparrow x,R_n\\downarrow x$.\r\n$ f(x) \\equal{} f(x \\minus{} r_n) \\plus{} r_n\\ge r_n$ and $ f(x) \\equal{} R_n \\minus{} f(R_n \\minus{} x)\\le R_n$.\r\nSo $ f(x) \\equal{} x$.\r\n\r\nConclusion: $ f(x) \\equal{} x$, $ f(x) \\equal{} 0$ and $ f(x) \\equal{} 2$."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Suppose that $ f$  $ : (0,\\infty) \\to (0,\\infty)  $ is continuous differentiable and   strict monotonic  on  $ (0,\\infty) ,$ such that $ \\displaystyle f^{-1}(x)=\\displaystyle f^{\\prime}(x) \\; ,\\; \\forall x\\in (0,\\infty) .$\r\n[b]  If $ \\varphi:=\\displaystyle \\frac{1+\\sqrt{5}}{2} , $  find $ \\displaystyle  f(\\varphi)\\; .$ [/b]\r\n[b]Remark : [/b] My solution need some inequalities.",
  "Solution_1": "I just love this problem. I saw it for the first time at oral ENS and after more than one month I finally posted a solution on the analysis forum. Fedja gave a more elegant solution. It is a very very difficult problem, I think."
}
{
  "Tag": [
    "vector",
    "analytic geometry",
    "Functional Analysis",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $ a,b$ be unit vectors in real Banach space such that $ \\parallel{}2a \\plus{} b\\parallel{} \\equal{} \\parallel{}a \\minus{} 2b\\parallel{} \\equal{} 3$. prove that there exists continuous linear functionals $ f,g$ with the norm $ 1$ such that $ f(a) \\equal{} f(b) \\equal{} 1$, $ g(a) \\equal{} \\minus{} g(b) \\equal{} 1$.",
  "Solution_1": "Are the functionals required to have unit norm too? Otherwise the problem is easy: just check that $ a$ and $ b$ cannot be collinear.",
  "Solution_2": "[quote=\"mlok\"]Are the functionals required to have unit norm too? Otherwise the problem is easy: just check that $ a$ and $ b$ cannot be collinear.[/quote]\r\n\r\nYes,yes, sorry. they are both required to have norm $ 1$.",
  "Solution_3": "Actually, such a and b cannot exist. Mark the points (1,0), (0, 1), (2,1), (1,-2) on a coordinate plane, then mark the points symmetric to them about (0,0). There is no convex region that contains all these points on the boundary.",
  "Solution_4": "Once again, sorry, i've edited the problem conditions: there must be $ \\parallel{}2a \\plus{} b\\parallel{} \\equal{} \\parallel{}a \\minus{} 2b\\parallel{} \\equal{} 3$.",
  "Solution_5": "Mark the points (1,0), (0, 1), (2/3,1/3), (1/3,-2/3) on a coordinate plane, then mark the points symmetric to them about (0,0). The only convex region that contains all these points on the boundary is the square with vertices $ (\\pm 1,0)$, $ (0,\\pm 1)$. So the norm on the span of $ a,b$ is the $ \\ell_1$ norm, and the required functionals can be given explicitly (and extended to the entire space by Hahn-Banach)."
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "function",
    "inequalities unsolved"
  ],
  "Problem": "Find the largest value for real $k$ such that $a^2+b^2+c^2+3k\\geq (k+1)(a+b+c)$ for all $a,b,c\\in \\mathbb{R^{+}}$ satisfy $abc=1$.\r\n(Clearly $k\\geq 1$)",
  "Solution_1": "And clearly $k<3$, because if we set $c$ close to $1$, and $a=b$, we get the inequality become false for $k=3$.\r\nCan anyone help me?",
  "Solution_2": "very nice problem.\r\nI guess the largest value of $k$ is $\\frac 8 3$ ;) \r\nbut have no proof  :(",
  "Solution_3": "Nice problem !  :)  \r\n We use mixing variables and calculus to solved this problem : \r\n The ineq is equivalent to : $\\frac{a^2+b^2+c^2-a-b-c}{a+b+c-3} \\geq k$ \r\n Let $f(a,b,c)= \\frac{a^2+b^2+c^2-a-b-c}{a+b+c-3}$ \r\n [b]Step 1:[/b] prove  $f(a,b,c) \\geq f(a,\\sqrt{bc},\\sqrt{bc})$ , where $a= min (a,b,c)$ \r\n  Indeed ,  \r\n$f(a,b,c) - f(a,\\sqrt{bc},\\sqrt{bc}) =(\\sqrt{b}-\\sqrt{c})^2[3-3(\\sqrt{b}+\\sqrt{c})^2+a(\\sqrt{b}+\\sqrt{c})^2-a^2+2\\sqrt{bc}(b+c+\\sqrt{bc})] \\geq (a+2\\sqrt{bc})(\\sqrt{b}+\\sqrt{c})^2-3(\\sqrt{b}+\\sqrt{c})^2+3-a-2\\sqrt{bc} =(a+2\\sqrt{bc}-3)((\\sqrt{b}+\\sqrt{c})^2-1) \\geq 0$ \r\n \r\n [b]Step 2:[/b] we find the minimum value of $f(a,t,t)$ , where $at^2=1$ and $t\\geq 1$ \r\n We have : \r\n $f(a,t,t)=g(t)=\\frac{2t^6-2t^5-t^2+1}{2t^5-3t^4+t^2}$\r\n $g'(t)=\\frac{2t(t+1)(t-1)^4(t^2+t+1)(2t^2-2t-1)}{(2t^5-3t^4+t^2)^2}$ \r\n $g'(t)=0 \\iff t=\\frac{1+\\sqrt{3}}{2}$\r\n So $g(t) \\geq g(\\frac{1+\\sqrt{3}}{2}) = 2,803847578$ \r\n So the maximum value of $k$ is $g(\\frac{1+\\sqrt{3}}{2}) = 2,803847578$",
  "Solution_4": "Great solution, thanks so much!",
  "Solution_5": "Hmm but the solution using calculus bothers me. Is there a solution without using calculus?\r\nBecause the function simplifies to $g(t) = (t+\\frac{1}{2}) + \\frac{\\frac{3}{4}}{t+\\frac{1}{2}} + \\frac{1}{t^2}$\r\nAnd also, if $x = \\frac{\\sqrt{3}+1}{2}$ then $x^2 = x+\\frac{1}{2}$\r\nSo maybe there is a no-calculus way to find the minimum of $g(t)$?\r\n\r\nBtw, I calculated that $g(\\frac{\\sqrt{3}+1}{2}) = 8 - 3\\sqrt{3}$"
}
{
  "Tag": [
    "logarithms",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Compute  $ \\sum_{n\\equal{}1}^{\\infty}\\frac{(\\minus{}1)^{n\\plus{}1}n^2}{n^3\\plus{}1}$",
  "Solution_1": "Let $ \\xi = e^{i\\pi/3}$. Then\r\n\r\n\\begin{eqnarray*}\r\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}n^2}{n^3+1}\r\n& = & \\frac{1}{3} \\sum_{n=1}^{\\infty} (-1)^{n-1} \\left[ \\frac{1}{n+1} + \\frac{1}{n-\\xi} + \\frac{1}{n+\\xi^2} \\right] \\\\\r\n& = & \\frac{1}{3} (1 - \\log 2) + \\frac{1}{3} \\sum_{n=-\\infty}^{\\infty} \\frac{(-1)^n}{\\xi - n}\\\\\r\n& = & \\frac{1}{3} [ 1 - \\log 2 + \\pi \\, \\text{sech} ( \\sqrt{3} \\pi / 2 ) ]\r\n\\end{eqnarray*}"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Find all real solutions (x,y,z) to the following system of equations.\r\n\r\nx + y =  :sqrt: (4z - 1)\r\ny + z = :sqrt:(4x - 1)\r\nx + z = :sqrt:(4y - 1)\r\n\r\nI know it's symmetric but I haven't gotten anywhere past squaring everything and adding to get 2(x :^2:  + y:^2: + z:^2:) + 2(xy + yz + xz) = 4(x + y + z) - 3.  From there though I'm stuck.  Any help'd be appreciated!  :-)",
  "Solution_1": "First thing I thought of was subtract the first two equations to get rid of y. However that didn't quite work so instead I [hide] let x+y = a and y+z = b. Then to get rid of square roots I took a^2 and b^2. So b^2 - a^2 = 4(x-z). But what does b-a equal?\n\n[/hide]",
  "Solution_2": "WarpedKlown1335 wrote:Find all real solutions (x,y,z) to the following system of equations.\n\nx + y = sqrt(4z - 1)\ny + z = sqrt(4x - 1)\nx + z = sqrt(4y - 1)\n\nI know it's symmetric but I haven't gotten anywhere past squaring everything and adding to get 2(x^2 + y^2 + z^2) + 2(xy + yz + xz) = 4(x + y + z) - 3.  From there though I'm stuck.  Any help'd be appreciated!  \n\n[hide] square the first two => x:^2:+y:^2:+2xy=4z-1, y:^2:+z:^2:+2yx=4x-1, subtract => x:^2:-z:^2:=4(z-x)=> either x=z or x+z=4. We can do a similar analysis on other pairs of equations, so now we have either x=y=z OR x+y=4, x+z=4, y+z=4 => x=y=z=2 OR x=y, x+z=y+z=4 (w/ permutations among the variables in this case). The last case (the interesting one) gives us 16=4x-1, and 4x:^2:=4z-1 => x=y=17/4, z=293/16 (and permutations). The second case doesn't work. Period. The first case gives us 4x:^2:=4x-1=> x=y=z=1/2. So, I get all solutions as (17/4, 17/4, 293/16), (17/4, 293/16, 17/4), (293/16, 17/4, 17/4), (1/2, 1/2, 1/2). Correct?[/hide]",
  "Solution_3": "Ummm...(y+z) :^2:  isn't y:^2: + 2xy + z:^2:...sorry :( .  And the only solution I got when I solved it on my own time (didn't use your method) was (.5,.5,.5)  I didn't try out your answers yet though.",
  "Solution_4": "What about this ?\r\n\r\nIf x < z then x + y < y + z so :sqrt:(4z - 1) < :sqrt:(4x - 1) so z < x.\r\nContradiction. So x, y, z must be equal.\r\nNow we have 2x = :sqrt:(4x - 1) iff (2x - 1) = 0.\r\nSo the only solution is (x, y, z) = (1/2, 1/2, 1/2).",
  "Solution_5": "[color=cyan]Mystic, you should have checked the \"interesting\" solutions.[/color]",
  "Solution_6": "[quote=\"Arne\"]What about this ?\n\nIf x < z then x + y < y + z so :sqrt:(4z - 1) < :sqrt:(4x - 1) so z < x.\nContradiction. So x, y, z must be equal.\nNow we have 2x = :sqrt:(4x - 1) iff (2x - 1) = 0.\nSo the only solution is (x, y, z) = (1/2, 1/2, 1/2).[/quote]\r\nThat was ridiculously elegant. Wow.",
  "Solution_7": "[quote=\"Arne\"]What about this ?\n\nIf x < z then x + y < y + z so :sqrt:(4z - 1) < :sqrt:(4x - 1) so z < x.\nContradiction. So x, y, z must be equal.\nNow we have 2x = :sqrt:(4x - 1) iff (2x - 1) = 0.\nSo the only solution is (x, y, z) = (1/2, 1/2, 1/2).[/quote]\r\n\r\nHmm, is this a trick that is used a lot? That is, when given a bunch of equations that, when added, are symmetrical, trying to let one variable be < than another and then finding out that all of the variables are equal? Because I seem to recall a few other solutions that were previously posted around with forum using that strategy...I think...",
  "Solution_8": "Try USAMO #2 2002.",
  "Solution_9": "Hey Arne, that is incredibly beautiful.  I got (1/2, 1/2, 1/2) also methodically, but it was a bit longer than what you did.  Eventually I figured out that x=y=z so then I solved the system by substitution to get (1/2, 1/2, 1/2).   :)",
  "Solution_10": "Well this trick can be very helpful when given a symmetric system.\r\n\r\nDo you remember the system \r\nx = 3y - 2\r\ny = 3z - 2\r\nz = 3x - 2 ?\r\n\r\nThis (selfmade) problem was solved using the same method.\r\n\r\nAnother idea when solving symmetric systems is using inequalities.\r\nAs an example, British Olympiad 1998 :\r\n\r\nDetermine all positive real solutions to\r\nxy + yz + zx = 2\r\nxyz = x + y + z + 2."
}
{
  "Tag": [],
  "Problem": "How many triangles are in the figure to the right?\n\n[asy]draw((-10,0)--(0,13)--(13,0)--cycle);\ndraw((0,0)--(0,13));\ndraw((7,0)--(0,13));\ndraw((7,0)--(11,2));[/asy]",
  "Solution_1": "Here is the diagram again:\r\n\r\n\r\n[asy]draw((-10,0)--(0,13)--(13,0)--cycle);\ndraw((0,0)--(0,13));\ndraw((7,0)--(0,13));\ndraw((7,0)--(11,2));[/asy]\r\n\r\n\r\nIt is easier to at first disregard the smallest line of the figure (appears to be cutting the rightmost triangle into two parts).  Once we can count the number of triangles without that little line, we can add back in the number of extra triangles resulting from the line.\r\n\r\nThe number of triangles in the diagram w/o the little line is actually C(4,2).  You may notice that the number of triangles will be the number of ways we can randomly choose 2 vertices of the total 4 on the bottom of the diagram since 1 vertex of the triangle is at the top.  (Triangles have 3 vertices).  Since the order in which we choose the vertices does not matter, we use Combination, not Permutations (P). \r\n\r\nC(4,2) = [u]4![/u]\r\n               2!2!\r\n\r\nAfter calculating that out, you will eventually arrive at the answer 6.\r\n\r\nBut wait, 6 is not the answer!  Now we must count the number of extra triangles that little line creates!\r\n\r\nThe little line creates 2 more triangles by cutting the rightmost big triangle into 2 parts: a small part and a bigger part.\r\n\r\nThus, the answer is 6+2= [b][i][u]8[/u][/i][/b][/u][/i]",
  "Solution_2": "Oops.  C(4,2)=4!/(2!)(2!), just in case anyone gets confused.  I did it wrong.\r\n\r\n4!=24, 2!=2, 24/(2)(2)=6.  There we go.\r\n\r\nAnswer is still 8."
}
{
  "Tag": [
    "videos"
  ],
  "Problem": "[youtube]giAy8wEvoVsv[/youtube]\r\nHAHA\r\nThis video is awesome",
  "Solution_1": "i saw this video so long ago\r\nbut is still funny\r\ni love the asian mother part and the rapping part\r\n*yo yo yo\r\nmy name's macbeth\r\nmy wife wants me to kill the king (etc.)*\r\n :rotfl:  :rotfl:  :rotfl: \r\nis hilarious..."
}
{
  "Tag": [
    "inequalities",
    "function",
    "inequalities proposed"
  ],
  "Problem": "Imagine a function f on the naturals. \r\nSuppose n = a<sub>k</sub>a<sub>k-1</sub>...a<sub>2</sub>a<sub>1</sub> in base 2.\r\nLet f(n) be the sum of all indices i for which a<sub>i</sub> = 1. For example, \r\n21 = (10101)<sub>2</sub> => f(21) = 5 + 3 + 1 = 9.\r\n29 = (11101)<sub>2</sub> => f(29) = 5 + 4 + 3 + 1 = 13.  \r\n13 = (1101)<sub>2</sub> => f(13) = 4 + 3 = 1 = 8.\r\nNow prove that for any positive integer n:\r\nf(3n) \\leq 2f(n) + \\sqrt(f(n))\r\nand determine when equality occurs.",
  "Solution_1": "Hey, no solution? :?"
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "derivative",
    "function",
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Let $n$ be a positive integer, and $a_1,...,a_n, b_1,..., b_n$ be $2n$ positive real numbers such that \r\n$a_1 + ... + a_n = b_1 + ... + b_n = 1$.\r\n\r\nFind the minimal value of \r\n$ \\frac {a_1^2} {a_1 + b_1} + \\frac {a_2^2} {a_2 + b_2} + ...+ \\frac {a_n^2} {a_n + b_n}$.",
  "Solution_1": "this has surely been posted on the forum before, because I have seen it :P it's trivial indeed using cauchy.",
  "Solution_2": "Lets denote with E that expression!\r\n\r\nIt's easy to see that we have: $ E \\geq \\frac{ (\\sum a_{1})^{2}}{\\sum a_{1}+ \\sum b_{1}} =\\frac{1}{2}$\r\nSo the minimum is $\\frac{1}{2}$.\r\n\r\nright?",
  "Solution_3": "We can easily generalize this:\r\n\r\nLet n be a positive integer and $a_{1},a_{2}, \\ldots,a_{n},b_{1},b_{2},\\ldots, \\b_{n}$ be 2n positive real numbers so that $ \\sum a_{1}= \\sum b_{1}=k$ with k>0 and q $\\in N$ ($ q\\geq2$).\r\n\r\nFind the minimal value of:\r\n\r\n$ \\sum \\frac{a_{1}^{q}}{(a_{1}+b_{1})^{q-1}}$.\r\n\r\nIn the same way we get that: $ \\sum \\frac{a_{1}^{q}}{(a_{1}+b_{1})^{q-1}} \\geq \\frac{( \\sum a_{1})^{q}}{( \\sum a_{1}+ \\sum b_{1})^{q-1}}= \\frac{k^{q}}{(2k)^{q-1}}$. for k=1 and q=2 we have this problem.\r\n\r\ncheers! :D :D",
  "Solution_4": "This was problem 4...But a TST problem with an one-line solution, as noticed by Valentin, it is maybe really too easy... :( \r\n\r\nPierre.",
  "Solution_5": "a1               a1b1                 a1+b1 \r\n______=a1- ______>=a1- ________ \r\na1+b1            a1+b1                 4 \r\n\r\n\r\n\r\nwhen u sum u obtain 1/2 \r\nso easy man\r\nsorry guys",
  "Solution_6": "[quote=\"pbornsztein\"]Let $n$ be a positive integer, and $a_1,...,a_n, b_1,..., b_n$ be $2n$ positive real numbers such that \n$a_1 + ... + a_n = b_1 + ... + b_n = 1$.\n\nFind the minimal value of \n$ \\frac {a_1^2} {a_1 + b_1} + \\frac {a_2^2} {a_2 + b_2} + ...+ \\frac {a_n^2} {a_n + b_n}$.[/quote]\r\n\r\nIt's really easy. But let's try another solution.\r\n\r\nLet $ \\displaystyle X=\\frac {a_1^2} {a_1 + b_1} + \\frac {a_2^2} {a_2 + b_2} + \\cdots + \\frac {a_n^2} {a_n + b_n}$ and define it's dual $ \\displaystyle Y=\\frac {b_1^2} {a_1 + b_1} + \\frac {b_2^2} {a_2 + b_2} +  \\cdots + \\frac {b_n^2} {a_n + b_n}$. Then\r\n\r\n$\\displaystyle X-Y = \\sum_{i=1}^n \\frac {a_i^2-b_i^2} {a_i + b_i} = \\sum_{i=1}^n (a_i-b_i) = 0$. On the other hand\r\n\r\n$\\displaystyle X+Y = \\sum_{i=1}^n \\frac {a_i^2+b_i^2} {a_i + b_i} \\geq \\sum_{i=1}^n\\frac{1}{2}(a_i+b_i) = 1$.\r\n\r\nTherefore $X=Y \\geq \\frac{1}{2}$.",
  "Solution_7": "Yes, it has been solved like that by some of the contestants. :) \r\nMost of the solvers did with Cauchy.\r\nOne uses partial derivatives for a 3 pages solution :D \r\nOther uses convexity for a function with more than one variable to get a wrong solution. :maybe: \r\nAnd many didn't solve it at all... :( \r\n\r\nPierre.",
  "Solution_8": "[quote=\"pbornsztein\"]Yes, it has been solved like that by some of the contestants. :) \nMost of the solvers did with Cauchy.\nOne uses partial derivatives for a 3 pages solution :D \nOther uses convexity for a function with more than one variable to get a wrong solution. :maybe: \nAnd many didn't solve it at all... :( \n\nPierre.[/quote]\r\n\r\nLOL  :D  :D \r\n\r\nBy the way, how many are allowed to take TST? In US, only USAMO winners (max. 12) can take TST.\r\n\r\nLook at problem 11 of http://www.bath.ac.uk/~masgcs/ukimo2004/team_sheet2.pdf\r\n\r\nand at the top of the solutions  http://www.bath.ac.uk/~masgcs/ukimo2004/ts2.pdf\r\n\r\nThe same problem was being used for preparation last year by the top nation at IMO 2003\r\n\r\nAnd it was being reused by UK and France for IMO 2004 preparation and TST!",
  "Solution_9": "Another solution:\r\n\r\nIt's easy to verify that $\\displaystyle \\frac{x^2}{x+y} \\geq \\frac{3}{4}x - \\frac{1}{4}y$ for $x,y>0$.\r\n\r\nSo $\\displaystyle \\sum_{i=1}^n \\frac {a_i^2} {a_i + b_i} \\geq \\frac{3}{4} \\sum_{i=1}^n a_i - \\frac{1}{4} \\sum_{i=1}^n b_i  = \\frac{1}{2}$.\r\n\r\nPierre, did anybody use this method? :-)",
  "Solution_10": "There were around 25 contestants (I do not remember exactly) :\r\nThe main part comes from our correspondance program. We add some from the 'Concours gnral' (which is now a big problem with a lot of questions for a lot of pages, and absolutely not an olympiad type competition), and some students from the lyce Louis-le-Grand, which is one of the best lycees in France (and because the test was organised there).\r\n\r\nFrom what I've said, there is no surprise that each of the 6 members of the french Imo-team were coming from our correspondance program (the others had never seen an olympiad problem).\r\n\r\nPierre.",
  "Solution_11": "[quote=\"math_sipo\"]a1               a1b1                 a1+b1 \n______=a1- ______>=a1- ________ \na1+b1            a1+b1                 4 \n\n\n\nwhen u sum u obtain 1/2 \nso easy man\nsorry guys[/quote] you can use latex to write up your stuff. Check out the latex tutorial in the latex help. :D",
  "Solution_12": "1/2 .easy :lol:",
  "Solution_13": "By Titu's lemma we have that:\n$$\\sum_{k=1}^{n} \\dfrac{a_k^2}{a_k+b_k} \\ge \\dfrac{\\left(\\sum_{k=1}^{n} a_k \\right)^2}{\\sum_{k=1}^{n} a_k+\\sum_{k=1}^{n} b_k}=\\dfrac{1}{2}$$",
  "Solution_14": "This problem must be a joke. Using Cauchy-Schwarz we are done!\nI wonder why this problem was problem 4....."
}
{
  "Tag": [
    "inequalities",
    "trigonometry"
  ],
  "Problem": "The last one for now  :D :\r\n\r\nSuppose a, b, c > 0 such that ab + bc + ca  = 1. Prove that\r\n\r\n(1 - a)/(1 + a) + (1 - b)/(1 + b) + (1 - c)/(1 + c) <= 3/2.\r\n\r\nHave fun ! :D",
  "Solution_1": "A quick try, before grobber, Maverick or an other guy from the belgian-romanian very high-skilled gang comes back  :mrgreen:  :mrgreen:  :wink: \r\n\r\n\r\nAs said in a previous message, in a triangle\r\ntgA+tgb+tgC = tgA tgb tgC. \r\nif x = 1/tgA, y = 1/tgB, z=1/tgC,\r\nxy + yz + zx = 1\r\n\r\nAfter a little trigonometry what we have to prove is\r\n\r\n3 - 2(cosA+cosB+cosC) <= 3/2\r\n\r\nor (cosA+cosB+cosC) >= 3/4\r\nor (sinA+sinB+sinC) <= 9/4\r\n\r\n\r\nAlas, sin is [b]not[/b] convex   :evil: \r\n\r\nC =  :pi: - (A+B)\r\nsinA+sinB+sinC = sinA+sinB+sin(A+B)\r\n\r\nNow I used sin(a+b), sin=1-cos, cos(a+b) to get\r\n\r\nsinA+sinB+sinC = 2 + 2cosAcosBcosC\r\n\r\nAs shown before, cosA+cosB+cosC<=3/2\r\nSo by AM-GM, cosAcosBcosC <= (cosA+cosB+cosC)/3) :^3: <= 1/8   \r\n\r\nPhew !",
  "Solution_2": "A somewhat shorter way to prove that cos A + cos B + cos C >= 3/4 :\r\n\r\nIn every triangle we have\r\ncos A + cos B + cos C\r\n= 1 - 2 cos A cos B cos C\r\n>= 1 - 2((cos A + cos B + cos C)/3)\r\n>= 1 - 2((3/2)/3) \r\n= 3/4\r\n\r\nwhere cos A + cos B + cos C <= 3/2 follows from Jensen's inequality. \r\n(The fact that\r\ncos A + cos B + cos C + 2 cos A cos B cos C = 1\r\nin every triangle is quite well - known)\r\n\r\nNice one huh ?  :D \r\n\r\nTrigonometric substitution can be VERY useful !",
  "Solution_3": "[quote=\"Arne\"]cos A + cos B + cos C + 2 cos A cos B cos C = 1 in every triangle is quite well - known)\n[/quote]\r\n\r\nQuite well known ?\r\nI didn't know that just 4 hours ago  :!:",
  "Solution_4": "OK then ... You know, there are lots of useful identities like this.\r\nIt's no bad idea to study them for a while ...",
  "Solution_5": ":Sigma: (1+a^2)/(1+a^2)- :Sigma: 2a^2/(1+a^2)<= 3/2\r\n3-3/2<= :Sigma: 2a^2/(1+a^2)\r\n3/2<=2 :Sigma:a^2/(a^2+1)\r\n3/4<=.... but\r\n :Sigma: a^2/(1+a^2)>=(a+b+c)^2/(3+a^2+b^2+c^2)>=3/4\r\n4(a^2+b^2+c^2)+8(ab+bc+ca)>=9+3(a^2+b^2+c^2)\r\na^2+b^2+c^2>=ab+bc+ca=1\r\n\r\n\r\nHope my soln is  good! :lol:",
  "Solution_6": "That's also correct ... a non - trig solution ! Nice work !"
}
{
  "Tag": [
    "number theory",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $x,y,z,w >0$ positive numbers ;)\r\n\r\nSuch that: \r\n$x^2 + y^2 + z^2 + w^2= 3(x + y + z + w).$\r\n\r\n Fin all solution:$(x,y,z,w)$.\r\n\r\n-----------------\r\nFabiola Bravo    :rotfl:    \r\nLima -PERU",
  "Solution_1": "Ani body ??\r\n\r\nAny idea to solve this question??\r\n\r\n\r\n\r\nFabiola Bravo  :pilot:\r\nLIma- PERU",
  "Solution_2": "Looks like number theory but indeed it's more algebra.\r\n\r\n\r\nWlog $x \\leq y \\leq z \\leq w$\r\n$3*x - x^2 \\leq 2$ $\\forall x \\in 1,2,....$\r\nSo $w^2 \\leq 3w+6$ or $w \\leq 4$\r\n\r\nI find 4 solutions with $x \\leq y \\leq z \\leq w$"
}
{
  "Tag": [
    "Diophantine equation",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "$ n$ is a positive integer.Prove that  $ (2^n \\minus{} 1)(3^n \\minus{} 1)$ isn't a  square.",
  "Solution_1": "Hey,no one is interested in my questions?It's a nice problem! :blush:",
  "Solution_2": "lemma 1: If $ v_2(3^n\\minus{}1)$ is even then $ n\\equal{}2^k$\r\nLemma 2: $ 3$ divides $ 2^{2^n}\\minus{}1$ but $ 9$ does not.\r\nTwo lemmas above solve your problem!",
  "Solution_3": "[quote=\"Allnames\"]lemma 1: If $ v_2(3^n \\minus{} 1)$ is even then $ n \\equal{} 2^k$\nLemma 2: $ 3$ divides $ 2^{2^n} \\minus{} 1$ but $ 9$ does not.\nTwo lemmas above solve your problem![/quote]\r\nNo ,the first lemma is wrong . I have a solution for this problem but it 's quite  complex . It is very clear that $ n$ must be even ,if not then $ 2\\parallel{}3^n \\minus{} 1$ , so we can write in the form $ n \\equal{} 2k$ . Suppose the contradiction this product is a perfect square ,then exist d,x,y such that :\r\n\\[ (3^k)^2 \\minus{} 1 \\equal{} dx^2 ; (2^k)^2 \\minus{} 1 \\equal{} dy^2\\]\r\nIf d is a perfect square , it is not hard to show that $ k \\equal{} 0$ ,which gives contradiction to the fact that $ n > 0$ . So d is not a perfect square and two equations show that $ 3^k$ and $ 2^k$ are solutions of Pell's equation  :\r\n\\[ X^2 \\minus{} dY^2 \\equal{} 1\\]\r\nSuppose the smallest solution of this Pell equation is $ (a,b)$ then all roots of this equation are :\r\n\\[ X_{0} \\equal{} 1,X_1 \\equal{} a ; X_{n \\plus{} 2} \\equal{} 2aX_{n \\plus{} 1} \\minus{} X_n (*)\\]\r\nOur purpose is showing that this sequence can't both contain $ 2^k$ and $ 3^k$ . Suppose exist $ i,j$ such that $ X_i \\equal{} 2^k ,X_j \\equal{} 3^k$ . \r\n[u]Claim 1 [/u]: a must be an even number . \r\nProof : If not then by the relation (*) , all terms of this sequence are odd , which means that it doesn't contain a power of 2 . \r\n[u]Claim 2 : [/u] $ X_{2k \\plus{} 1}\\equiv 1 (\\mod 2 )$and $ X_{2k}\\equiv 0 (\\mod 2)$ for all positive integer $ k$ . \r\nThis fact is so obvious . Notice that we have $ X_i \\equal{} 2^k,X_j \\equal{} 3^k$ , by claim 2 we have $ i$ is even and $ j$ is odd . \r\n[u]Claim 3 [/u]: $ X_{2n \\plus{} 1}\\equiv 0 (\\mod a ),\\forall n\\geq 0$ and $ x_{2k}\\equiv 1, \\minus{} 1 (\\mod a )$ .\r\nProof :We only need to notice that $ X_{n \\plus{} 2}\\equiv \\minus{} X_n (\\mod a)$ and $ X_1\\equiv 0 (\\mod a)$ . \r\nBy claim 2 and claim 3 we have $ x_i\\equiv 0 (\\mod a )$ or $ a|2^k$ ,which means that exist $ t\\leq k$ such that $ a \\equal{} 2^t$ . \r\n[u]Claim 4 :[/u] 3 is a divisor of $ a$ . If not we consider two cases : \r\ni) $ a\\equiv 1 (\\mod 2 )$ then the sequence $ x_n(\\mod 3)$ is {1,1,1,..} ,and it doesn't contain any power of 3 . \r\nii) $ a\\equiv 2 (\\mod 3)$ then the sequence $ x_n(\\mod 3)$ is {1,2,1,2...} , and it also doesn't contain any power of 3 . \r\nSo we  must have $ 3|a$ ,which gives contradiction to the fact that $ a \\equal{} 2^k$ .",
  "Solution_4": "TTsphn has the right idea(very nice one, I must say) and almost the right proof (I think. Of course, the mistake is probably mine).\r\n\r\nBy claim 2 and claim 3 we have that $ a \\equal{} 3^{t}$ and not $ a \\equal{} 2^{t}$  (as $ a | X_{2n\\plus{}1}$) which shortens the proof a little, as that immediately contradicts claim 1.",
  "Solution_5": "[hide=\"Edit, this doesnt work\"] Nvm, I realized that my solution doesnt work.  :oops: [/hide]",
  "Solution_6": "[quote=\"shenyixin\"]$ n$ is a positive integer.Prove that  $ (2^n \\minus{} 1)(3^n \\minus{} 1)$ isn't a  square.[/quote]\r\nthis problem is China TST 2002.\r\nYou can read offical solution(I have  :D ).",
  "Solution_7": "To math 10\r\nCan you post your official soluton of it?I think it will be nice material.\r\nThanks.",
  "Solution_8": "[quote=\"gb2124\"]To math 10\nCan you post your official soluton of it?I think it will be nice material.\nThanks.[/quote]\r\nYes,I will posst them in tomorrow.Becausse Now I verry busy :) \r\nPlease wait.\r\n\r\nPs.Offical solution be a nice proof :D",
  "Solution_9": "[quote=\"math10\"]this problem is China TST 2002.\nYou can read offical solution(I have  :D ).[/quote]\r\nI knew this problem from my teacher,with a solution different from TTsphn's.(TTsphn's is also a nice one :) )\r\nMaybe it's just the official solution.",
  "Solution_10": "[quote=\"Allnames\"]lemma 1: If $ v_2(3^n \\minus{} 1)$ is even then $ n \\equal{} 2^k$[/quote]\r\nActually\uff0clemma 1 should be:$ v_2(3^n \\minus{} 1)\\equal{}v_2(n)\\plus{}2$",
  "Solution_11": "While working on this problem I made a MAPLE-assisted observation: if p is a 'large' prime divisor of m^n - 1, it is VERY often the case that p - 1 is divisible by n.\r\n\r\nIs this a well known theorem (with the addition of extra conditions -- like the obvious p > n -- as appropriate)?",
  "Solution_12": "[quote=\"shenyixin\"][quote=\"Allnames\"]lemma 1: If $ v_2(3^n \\minus{} 1)$ is even then $ n \\equal{} 2^k$[/quote]\nActually\uff0clemma 1 should be:$ v_2(3^n \\minus{} 1) \\equal{} v_2(n) \\plus{} 2$[/quote]\r\n\r\nHello shenyixin can you prove this lemma  :) \r\n\r\nthank you !! :lol:",
  "Solution_13": "This is a direct consequence of the [url=http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf]lifting the exponent lemma[/url]. We have $ v_2(\\frac{3^2 \\minus{} 1}{2}) \\equal{} 2$, so by the lemma $ v_2(3^n \\minus{} 1) \\equal{} 2 \\plus{} v_2(n)$.",
  "Solution_14": "[quote=\"Zhero\"]This is a direct consequence of the [url=http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf]lifting the exponent lemma[/url]. We have $ v_2(\\frac {3^2 \\minus{} 1}{2}) \\equal{} 2$, so by the lemma $ v_2(3^n \\minus{} 1) \\equal{} 2 \\plus{} v_2(n)$.[/quote]\r\n\r\nThank you \"Zhero\"  :lol:",
  "Solution_15": "See also http://www.mathlinks.ro/viewtopic.php?p=1389586#1389586",
  "Solution_16": "Regarding #12 (not central to this discussion anyway) I got some [url=http://www.groupsrv.com/science/about443292.html]help[/url] at sci.math that might be of some interest.",
  "Solution_17": "EDIT: This doesn't work",
  "Solution_18": "[quote]The number $ (a^n \\minus{} 1)(3^n \\minus{} 1)$ is not a perfect square $ \\forall a \\geq 2,n\\geq1$. [/quote]\r\n\r\nWell, you also need $ a \\neq 3$, for sure! In fact for your use of the Pythagorean triplets to work it is good to have $ a$ even ... but for sure it's a short and sweet proof for $ a \\equal{} 2$ (the original problem). However, at second thought, I think that the co-primality of the elements of that Pythagorean triple also needs to be argued."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "quadratics",
    "number theory",
    "relatively prime"
  ],
  "Problem": "There is a polynomial $f(x)$ of degree $n$. The remainder is $1$ when $(x-1)$ divides it. When $(x-3)$ divides $f(x)$, the remainder is $3$. When $(x-5)$ divides $f(x)$, the remainder is $21$. \r\n\r\nWhat would be the remainder when $f(x)$ is divided by $(x-1) \\cdot (x-3) \\cdot (x-5)$ ?",
  "Solution_1": "[hide=\"Method 1\"] The remainder is some quadratic polynomial, and you can plug in $x = 1, 3, 5$ to solve for its coefficients. [/hide] [hide=\"Method 2\"] The Chinese Remainder Theorem applies to polynomials. ;) [/hide]",
  "Solution_2": "[quote=\"t0rajir0u\"][hide=\"Method 1\"] The remainder is some quadratic polynomial, and you can plug in $x = 1, 3, 5$ to solve for its coefficients. [/hide] [hide=\"Method 2\"] The Chinese Remainder Theorem applies to polynomials. ;) [/hide][/quote]\n\n[hide]Do you mean to say that the problem cannot be solved definitively? I tried with CRM, but could not complete it since $n$ is variable and unknown.  :( [/hide]",
  "Solution_3": "well...when you divide it by a first degree polynomial, the remainder is constant.  what does that say about division by a third degree polynomial?\r\n\r\n\r\nedit:  tor already mentoined that its second degree",
  "Solution_4": "[quote=\"xxxyyyy\"]I tried with CRM, but could not complete it since $n$ is variable and unknown.  :([/quote]\r\n\r\n$n$ is irrelevant.  (In fact, you can assume WLOG that $n = 2$, which makes your life [b]very[/b] easy ;) )\r\n\r\nThe Chinese Remainder Theorem is a theorem about two things:\r\n\r\n- A modulus (rather, a product of relatively prime moduli)\r\n- A remainder (rather, a set of remainders)\r\n\r\nBoth are definite in this problem.",
  "Solution_5": "[hide]\nUm, if I remember and if I have time, I'll type my solution in latex. Anyway, I just did what tor hinted at. I got $2x^{2}-7x+4$.[/hide]"
}
{
  "Tag": [
    "LaTeX",
    "calculus",
    "calculus computations"
  ],
  "Problem": "If f(x)= m (subscript 1)x+b(subscript 1) and g(x)= m(subscript 2)x+b(subscript 2), prove (f(g(x)))'= f '(x) times g '(x).\r\n\r\n\r\n\r\nI found out that (fog)'= f ' g(x) times g ' (x) and that f '(x)= m (subscript 1) and g '(x)= m (subscript 2) but not everything works out right.\r\n\r\nPlease help me. Thanks!",
  "Solution_1": "[quote]If $ f(x) \\equal{} m_1 x \\plus{} b_1$ and $ g(x) \\equal{} m_2 x \\plus{} b_2$, then prove that $ \\left(f(g(x))\\right)' \\equal{} f'(x) \\cdot g'(x)$.[/quote]\r\n$ f'(x) \\equal{} m_1$, $ g'(x) \\equal{} m_2$, thus $ f'(x) \\cdot g'(x) \\equal{} m_1m_2$.\r\n\r\n$ \\frac{d}{dx}\\left[f(g(x))\\right] \\equal{} f'(g(x))g'(x) \\equal{} m_1 m_2$.\r\n\r\nNote that $ f'(g(x)) \\equal{} m_1$ as $ f'(x) \\equal{} m_1$.\r\n\r\nWhat's not working out?",
  "Solution_2": "wouldn't f'(g(x))= m(subscript 2)?\r\n\r\nbecause it would be f'(m (subscrpt 2)x+b(subscript 2)) which then equals m (subscript 2)",
  "Solution_3": "If you don't want to use $ \\text{\\LaTeX}$, commonly accepted shorthand for \"a subscript b\" is a_b. Wrapping this in dollar signs gives you $ a_b$. Just for future reference.\r\n\r\nAnyway, $ f(x)=m_1x+b_1$, so $ f'(x)=m_1$. This is a constant function. It is a horizontal line. Regardless of what value of $ x$ you put in there, you're going to get $ m_1$ as the output. Therefore, $ f'(g(x))$ is simply $ m_1$.",
  "Solution_4": "$ \\frac{d}{dx}\\left[f(g(x))\\right] \\neq f'(g(x))$"
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "least common multiple",
    "prime factorization"
  ],
  "Problem": "it can be proven that where (a,b) means the gcd of a and b, and [a,b] means the lcm of a and b\r\n\r\n$(a,b)[a,b]=ab$\r\n\r\n$(a,b,c)[ab,bc,ca]=abc$\r\n\r\n$(ab,bc,ca)[a,b,c]=abc$\r\n\r\ncan you find more complex identites like these, or ones which generalize to arbitratily many variables?",
  "Solution_1": "[hide=\"Solution\"] Let $e_{p}(n)$ denote the greatest $k$ such that $p^{k}| n$.  \n\n[b]Lemma:[/b]  $n = m \\Leftrightarrow e_{p}(n) = e_{p}(m) \\forall p \\in \\mathbb{P}$\n\n[b]Lemma:[/b]  $e_{p}(nm) = e_{p}(n)+e_{p}(m)$\n\n[b]Lemma:[/b]  $gcd(a, b, c, ...) = \\prod_{p \\in \\mathbb{P}}p^{min(e_{p}(a), e_{p}(b), e_{p}(c), ...)}$\n\n[b]Lemma:[/b]  $lcm(a, b, c, ...) = \\prod_{p \\in \\mathbb{P}}p^{max(e_{p}(a), e_{p}(b), e_{p}(c), ...)}$\n\nHence the following gcd identities reduce to the simple identities\n\n$min(x, y)+max(x, y) = x+y$\n$min(x, y, z)+max(x+y, y+z, z+x) = x+y+z$\n$max(x, y, z)+min(x+y, y+z, z+x) = x+y+z$\n\nWhere $x = e_{p}(a), y = e_{p}(b), z = e_{p}(c)$ for some unspecified prime $p$.  It is easy to see where these identities come from; simply assume WLOG $x \\ge y \\ge z$.\n\nSimilarly, arbitrary generalizations can be had in the following way:\n\nGiven $x_{1}\\ge x_{2}\\ge ... \\ge x_{n}$ with sum $S$ we have the identities\n\n$min(x_{i})+max(S-x_{i}) = S$\n$min(x_{i}+x_{j})+max(S-x_{i}-x_{j}) = S$\n...\n\nThis chain of identities is obvious because by our WLOG assumption we have\n\n$min(x_{i}) = x_{1}, max(S-x_{i}) = S-x_{1}$\n$min(x_{i}+x_{j}) = x_{1}+x_{2}, max(S-x_{i}-x_{j}) = S-x_{1}-x_{2}$\n...\n\nAnd so forth. [/hide]",
  "Solution_2": "i don't understand all that fancy notation  :blush: \r\n\r\ncan you write out the identities with the (a,b) and [a,b] notation i used?",
  "Solution_3": "They come out to what you expect.  For example, the four-variable identities are given by\r\n\r\n$(a, b, c, d)[abc, bcd, cda, dab] = abcd$\r\n$(ab, ac, ad, bc, bd, cd)[ab, ac, ad, bc, bd, cd] = abcd$\r\n$(abc, bcd, cda, dab)[a, b, c, d] = abcd$\r\n\r\nEdit:  Let me explain myself a little more.\r\n\r\n[hide=\"Slower this time\"] Let me explain the two-variable case using $e_{p}(n)$.\n\n$e_{p}(n)$ is just a convenient notation for the exponents in the prime factorization of $n$.  That is, we have\n\n$n = 2^{e_{2}(n)}3^{e_{3}(n)}5^{e_{5}(n)}7^{e_{7}(n)}...$\n\nGiven two numbers $a, b$ with some values of $e_{p}(a), e_{p}(b)$, $gcd(a, b)$ is the unique positive integer that satisfies\n\n$e_{p}( gcd(a, b) ) = min(e_{p}(a), e_{p}(b))$\n\nFor each prime $p$ (because $x | y$ if and only if $e_{p}(x) \\le e_{p}(y)$ for each prime; hence we require that $e_{p}( gcd(a, b)) \\le e_{p}(a), e_{p}(b)$ for each prime).  \n\nOn the other hand, and for the same reason, $lcm(a, b)$ is the unique positive integer that satisfies\n\n$e_{p}( lcm(a, b) ) = max(e_{p}(a), e_{p}(b))$\n\nWhen multiplying two numbers we simply add the exponents in their prime factorizations; in other words,\n\n$e_{p}(xy) = e_{p}(x)+e_{p}(y)$\n\nHence, taking the identity\n\n$(a, b) [a, b] = ab$\n\nAnd comparing the exponents of the prime factorizations of both sides (which must be equal) we get the identity\n\n$min(e_{p}(a), e_{p}(b))+max(e_{p}(a), e_{p}(b)) = e_{p}(a)+e_{p}(b)$\n\nWhich is easy to show for the reasons I gave in my first post, and easy to generalize. \n\nThe sequence of min-max identities corresponds to a sequence of $gcd, lcm$ identities. [/hide]"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "AMC 10",
    "AMC 12 A",
    "AMC 10 A"
  ],
  "Problem": "At least 5% of students participating in the 2007 AMC 12 A scored 97.5 or more, so the AIME qualifying score on the 2007 AMC 12 A is 97.5.\r\n\r\nAt least 1% of students participating in the 2007 AMC 10 A scored 117.0 or more, so the AIME qualifying score on the 2007 AMC 10 A is 117.0.\r\n\r\nWe still have much to do in the AMC office with the scoring and tabulating.  AMC 10 A and AMC 12 A school reports will go out to contest managers approximately Friday, Feb 23, 2007 assuming all continues to go well with no problems.\r\n\r\nPlease note:   This is an extremely busy time at the AMC office.  As a result, I will not answer any inquiries, specific or general at any time.  Any inquiry will be immediately deleted.  Your contest manager will receive your official score report when we  are able to send them.\r\n\r\nSteve Dunbar\r\nAMC Director",
  "Solution_1": "A couple more things regarding this:\r\n\r\n- a thread for discussion of this announcement (though I'm not sure what there is to \"discuss\") is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133662]here[/url].  \r\n\r\n- When Mr. Dunbar says \"Any inquiry will be immediately deleted\", he means any inquiry that you send to the AMC offices.  You can still post questions here, and though I doubt Mr. Dunbar will have time to respond to them, an AoPS user or admin might know the answer.  But [b]please[/b], do at least a minimal amount of research before asking a question; for most questions that I see here, the answer is readily obtainable by visiting the [url=http://www.unl.edu/amc]AMC website[/url].    \r\n\r\n- If you doubt how busy the AMC is, check out [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-1-1012-dir,pic.shtml]these photos[/url].",
  "Solution_2": "By the way, thanks very much to the AMC Director for posting this information!"
}
{
  "Tag": [
    "vector",
    "function"
  ],
  "Problem": "Find all $f(x)$ that map the reals to the reals such that $f(2x)=2f(x)-2x^{2}$.",
  "Solution_1": "Substitute $f(x)=g(x)-x^{2}$ to get $g(2x)=2g(x)$. Hence $g(x)$ is linear, giving $f(x)=ax-x^{2}, a\\in\\mathbb{R}$",
  "Solution_2": "[quote=\"Farenhajt\"]Substitute $f(x)=g(x)-x^{2}$ to get $g(2x)=2g(x)$. Hence $g(x)$ is linear, giving $f(x)=ax-x^{2}, a\\in\\mathbb{R}$[/quote]\r\n\r\nThat is not necessarily true. Only if $g$ is continous at 0. Because the solution to the Cauchy's equation $f(x+y) = f(x)+f(y)$ has non-trivial (non-linear solutoions). If one accepts the axiom of choice. He can contruct something called the \"Hamel Basis\". It is the basis for the vector space of the reals over the rationals. So it does not need to be linear.",
  "Solution_3": "Let $h(x)$ and $j(x)$ be two functions that map $[0,1)$ to the reals.\r\n\r\nThen, a generic solution is given by $g(x)=\\left \\{ \\begin{array}{ll}0&{ \\text{if}\\ \\ x = 0}\\\\ h( \\{ log_{2}x \\})x&{ \\text{if}\\ \\ x>0}\\\\ j( \\{ log_{2}(-x) \\})x&{ \\text{if}\\ \\ x < 0}\\end{array}\\right.$\r\n\r\n$g(x)$ is equivalent to a linear function iff. $j(x)=h(x)=k$ for some real number $k$. However, the linear case is the only case in which $g$ (and therefore $f$) is continuous at $0$."
}
{
  "Tag": [
    "geometry",
    "geometry unsolved"
  ],
  "Problem": "The incircle of an isosceles triangle $ABC$ with $AB = AC$ is centered at $O$ and touches $BC, CA, AB$ at $K, L, M$ , respectively. Lines $KM$ and $OL$ meet at $N$ , and line $BN$ meets $CA$ at $Q$. Let $P$ be the projection of $A$ onto $BQ$. Suppose that $BP = AP+2PQ$. Find all possible values of $\\frac{AB}{BC}.$",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=5830"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "We have recently made some changes on our servers, and while we tried to make sure we haven't missed anything, given the size of our site, we might have. So if you see any missing images (that were here before) or files, please let us know.\r\n\r\nAlso, the wiki is back online.",
  "Solution_1": "It looks like transcripts of recent Math Jams are missing.\r\n\r\n(Separate point: On the transcripts page, it would be nice if the Jam Date listed the year in addition to the month and day.)"
}
{
  "Tag": [
    "number theory",
    "relatively prime",
    "Diophantine equation",
    "number theory proposed"
  ],
  "Problem": "For every positive integer $ n$, $ b(n)$ denote the number of positive divisors of $ n$ and $ p(n)$ denote the sum of all positive divisors of $ n$. For example, $ b(14)\\equal{}4$ and $ p(14)\\equal{}24$. Let $ k$ be a positive integer greater than $ 1$.\r\n\r\n(a) Prove that there are infinitely many positive integers $ n$ which satisfy $ b(n)\\equal{}k^2\\minus{}k\\plus{}1$.\r\n\r\n(b) Prove that there are finitely many positive integers $ n$ which satisfy $ p(n)\\equal{}k^2\\minus{}k\\plus{}1$.",
  "Solution_1": "(a) it is enough to take $ n\\equal{}p^{k^2\\minus{}k}$,where $ p$ is a prime number.\r\n(b) Obviously if $ n>k^2\\minus{}k\\plus{}1$,then $ p(n)>n>k^2\\minus{}k\\plus{}1$,am i miss something?",
  "Solution_2": "[quote=\"Erken\"](b) Obviously if $ n > k^2 \\minus{} k \\plus{} 1$,then $ p(n) > n > k^2 \\minus{} k \\plus{} 1$,am i miss something?[/quote]Why is $ n>k^2\\minus{}k\\plus{}1$? Did you notice that I wrote [b]finitely many[/b], not infinitely many? :wink:",
  "Solution_3": "[quote=\"Johan Gunardi\"][quote=\"Erken\"](b) Obviously if $ n > k^2 \\minus{} k \\plus{} 1$,then $ p(n) > n > k^2 \\minus{} k \\plus{} 1$,am i miss something?[/quote]Why is $ n > k^2 \\minus{} k \\plus{} 1$? Did you notice that I wrote [b]finitely many[/b], not infinitely many? :wink:[/quote]\r\nI take enough large $ n$,in our case it is $ n>k^2\\minus{}k\\plus{}1$,so that our equality can not be true,so there is no $ n>k^2\\minus{}k\\plus{}1$,such that $ p(n)\\equal{}k^2\\minus{}k\\plus{}1$,that's why there are finitely many $ n$.",
  "Solution_4": "[quote=\"Erken\"][quote=\"Johan Gunardi\"][quote=\"Erken\"](b) Obviously if $ n > k^2 \\minus{} k \\plus{} 1$,then $ p(n) > n > k^2 \\minus{} k \\plus{} 1$,am i miss something?[/quote]Why is $ n > k^2 \\minus{} k \\plus{} 1$? Did you notice that I wrote [b]finitely many[/b], not infinitely many? :wink:[/quote]\nI take enough large $ n$,in our case it is $ n > k^2 \\minus{} k \\plus{} 1$,so that our equality can not be true,so there is no $ n > k^2 \\minus{} k \\plus{} 1$,such that $ p(n) \\equal{} k^2 \\minus{} k \\plus{} 1$,that's why there are finitely many $ n$.[/quote]Okay. Then why did you write 'am i miss something?'?",
  "Solution_5": "dont we have to show the existance of atleast one such n ?",
  "Solution_6": "[quote=\"gagan_goku\"]dont we have to show the existance of atleast one such n ?[/quote]Actually I think so. But the official solution does not mention it. Erken's solution is essentially the same as the official solution.",
  "Solution_7": "but if there are no such n , shouldnt we say there is no such n\r\ndont know , you can also count 0 in finite numbers. anyways , someone please tell me how to increase my brain activity.\r\ni feel really good when im able to solve some nice problem. but im too dumb  :(",
  "Solution_8": "$ 0$ is a finite number.  \r\n\r\nThere are solutions for some $ k$:  $ p(p^2) \\equal{} p^2 \\plus{} p \\plus{} 1 \\equal{} (p \\plus{} 1)^2 \\minus{} (p \\plus{} 1) \\plus{} 1$ for $ p$ a prime.  There aren't solutions for every $ k$; I believe there are no solutions for $ k \\equal{} 5$, for example.  (Additionally, $ p(k^2 \\minus{} k \\plus{} 1) \\equal{} k^2 \\minus{} k \\plus{} 1$ whenever $ k^2 \\minus{} k \\plus{} 1$ is prime, but it's an open question whether infinitely many such primes exist.)\r\n\r\nEdit:  Two other sporadic solutions.  $ p(36) \\equal{} 91 \\equal{} 10^2 \\minus{} 10 \\plus{} 1$ and $ p(16) \\equal{} 31 \\equal{} 6^2 \\minus{} 6 \\plus{} 1$.  If two consecutive values $ k, k \\plus{} 1$ have (relatively prime) solutions, then so does $ k^2 \\plus{} 1$.\r\n\r\nEdit 2:  When $ n$ is a power of two, we are trying to solve the Diophantine equation $ 2^a \\minus{} 1 \\equal{} k^2 \\minus{} k \\plus{} 1 \\implies 2^a \\equal{} k^2 \\minus{} k \\plus{} 2 \\implies 2^{a \\plus{} 2} \\equal{} 4k^2 \\minus{} 4k \\plus{} 8 \\equal{} (2k \\minus{} 1)^2 \\plus{} 7$.  The solutions here are given by the solutions to [url=http://mathworld.wolfram.com/RamanujansSquareEquation.html]Ramanujan's Square Equation[/url], which here gives the rather large solution $ k \\equal{} 91, n \\equal{} 2^{a \\minus{} 1} \\equal{} 2^{13}$ (in addition to what we found above).  This is a difficult equation to solve, so it seems that even the case that $ n$ is a power of a prime is not easy."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $ x,y,z>0$, show\r\n$ (\\frac{x}{2x\\plus{}y})^3\\plus{}(\\frac{y}{2y\\plus{}z})^3\\plus{}(\\frac{z}{2z\\plus{}x})^3\\geq \\frac{1}{9}$.",
  "Solution_1": "Bump...\r\n\r\nAlthough I'm not the original poster, I would very much like to know the solution. :)",
  "Solution_2": "The problem is equivalent to show $ (\\frac {1}{2 \\plus{} a})^3 \\plus{} (\\frac {1}{2 \\plus{} b})^3 \\plus{} (\\frac {1}{2 \\plus{} b})^3\\ge 1/9$, where $ a,b,c\\in R_ \\plus{}$ and $ abc \\equal{} 1$. \r\nIt remains to show $ 2(\\frac {1}{2 \\plus{} a})^3 \\plus{} (\\frac {1}{2 \\plus{}1/a^2} )^3 \\ge 1/9$  for $ a > 0$.",
  "Solution_3": "[quote=\"miwalin\"]The problem is equivalent to show $ (\\frac {1}{2 \\plus{} a})^3 \\plus{} (\\frac {1}{2 \\plus{} b})^3 \\plus{} (\\frac {1}{2 \\plus{} b})^3\\ge 1/9$, where $ a,b,c\\in R_ \\plus{}$ and $ abc \\equal{} 1$. \nIt remains to show $ 2(\\frac {1}{2 \\plus{} a})^3 \\plus{} (\\frac {1}{2 \\plus{} 1/a^2} )^3 \\ge 1/9$  for $ a > 0$.[/quote]\r\nIt should be \r\n$ (\\frac {1}{2 \\plus{} a})^3 \\plus{} (\\frac {1}{2 \\plus{} b})^3 \\plus{} (\\frac {1}{2 \\plus{} c})^3\\ge 1/9$, where $ a,b,c\\in R_ \\plus{}$ and $ abc \\equal{} 1$. \r\nIt remains to show $ 2(\\frac {1}{2 \\plus{} a})^3 \\plus{} (\\frac {1}{2 \\plus{} 1/a^2} )^3 \\ge 1/9$  for $ a > 0$. \r\nSorry for carelessness.",
  "Solution_4": "The inequality can be proved in the following way:\r\n$ \\large\\begin{array}{rcl}\r\n\\frac{2}{(2\\plus{}a)^3}\\plus{}\\frac{1}{\\left (2\\plus{}\\frac{1}{a^2}\\right )^3}&\\ge &\\frac{1}{9}\\qquad\\Leftrightarrow\\\\\r\n& & \\\\\r\n\\frac{a^9\\plus{}6a^8\\plus{}12a^7\\plus{}24a^6\\plus{}24a^4\\plus{}12a^2\\plus{}2}{(2\\plus{}a)^3(2a^2\\plus{}1)^3}&\\ge &\\frac{1}{9}\\qquad\\Leftrightarrow\\\\\r\n& & \\\\\r\n9\\left (a^9\\plus{}6a^8\\plus{}12a^7\\plus{}24a^6\\plus{}24a^4\\plus{}12a^2\\plus{}2\\right )\\minus{}(2\\plus{}a)^3(2a^2\\plus{}1)^3&\\ge &0\\qquad\\Leftrightarrow\\\\\r\n& & \\\\\r\na^9\\plus{}6a^8\\plus{}80a^6\\minus{}150a^5\\plus{}84a^4\\minus{}73a^3\\plus{}54a^2\\minus{}12a\\plus{}10&\\ge &0\\qquad\\Leftrightarrow\\\\\r\n& & \\\\\r\n(a\\minus{}1)^2\\cdot (a^7\\plus{}8a^6\\plus{}15a^5\\plus{}102a^4\\plus{}39a^3\\plus{}60a^2\\plus{}8a\\plus{}10)&\\ge &0.\r\n\\end{array}$\r\nBut this is true for all $ a>0$."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Find all positive real solutions to x^3 + y^3 + z^3 = x + y + z, x^2 + y^2 + z^2 = xyz.",
  "Solution_1": "Such systems of equations having more variables than equations are usually solved using inequalities, especially when you are given the information that x, y, z must be positive.\r\n\r\nIn fact, from your two equations\r\n\r\n$x^{3}+y^{3}+z^{3}=x+y+z$\r\n\r\nand\r\n\r\n$x^{2}+y^{2}+z^{2}=xyz$\r\n\r\nit follows that\r\n\r\n$\\left( x^{3}+y^{3}+z^{3}\\right) \\left( x^{2}+y^{2}+z^{2}\\right) ^{2}=\\left( x+y+z\\right) \\left( xyz\\right) ^{2}$.\r\n\r\nBut for any three positive numbers x, y, z we have\r\n\r\n$\\left( x^{3}+y^{3}+z^{3}\\right) \\left( x^{2}+y^{2}+z^{2}\\right) ^{2}>\\left( x+y+z\\right) \\left( xyz\\right) ^{2}$,\r\n\r\nsince\r\n\r\n$\\left( x^{3}+y^{3}+z^{3}\\right) \\left( x^{2}+y^{2}+z^{2}\\right) ^{2}$\r\n$=\\left( x^{3}+y^{3}+z^{3}\\right) \\left( x^{4}+y^{4}+z^{4}+2y^{2}z^{2}+2z^{2}x^{2}+2x^{2}y^{2}\\right) $\r\n$>\\left( x^{3}+y^{3}+z^{3}\\right) \\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\\right) $\r\n$=x^{3}\\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\\right) +y^{3}\\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\\right) +z^{3}\\left( y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}\\right) $\r\n$>x^{3}y^{2}z^{2}+y^{3}z^{2}x^{2}+z^{3}x^{2}y^{2}=\\left( x+y+z\\right) \\left( xyz\\right) ^{2}$.\r\n\r\nTherefore, we get contradiction, and your system does not have any positive solutions.\r\n\r\n  Darij"
}
{
  "Tag": [
    "Stanford",
    "college",
    "function",
    "geometry",
    "algebra",
    "floor function"
  ],
  "Problem": "Did anyone attend this contest today?",
  "Solution_1": "yea i did, took the general, didn't stick around for the awards and solutions though, so i don't know what i got",
  "Solution_2": "yeah, i put n^2 instead of m^2 too...grrrr... and on the floor function one... i did the exact opposite function and got x to equal 3 instead of 2. how uncool. i was surprised i actually got 10th though....\r\n\r\ni bombed geometry too... haha.. but we won't go into that..."
}
{
  "Tag": [
    "probability",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Q1 Find  probability of getting 3 consecutive sixes in 10 throws of a die.\r\n\r\nQ2 A die is thrown again and again until we obtain 2 consecutive sixes are  obtained .find the probabiltiy of getting 2 consecutive sixes",
  "Solution_1": "Neither of these are calculus problems, and the second is nonsensical.  For the first, a standard attack with recursions or equivalently state diagrams works nicely to compute the probability of the complement. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190619]here[/url] for one example of the method; searching for the words \"state diagram*\" (e.g., in the High School Forums) should find more."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "superior algebra",
    "superior algebra open"
  ],
  "Problem": "This is a shameless reposting from MathOverflow but the question is extremely beautiful and I really want to know what mathlinkers can make of it. I have already spent 3 days thinking of it myself but with very little progress :wallbash:.\r\n\r\nDo there exist an infinite field $ k$ and a polynomial $ p\\in k[x]$ such that the image of $ k$ under the mapping $ x\\mapsto p(x)$ is not all $ k$ but its complement in $ k$ is finite?\r\n\r\nI can only show that if $ \\text{char\\,} k\\ne 2,3$, then we must have $ \\text{deg\\,}p\\ge 4$ and handle a few very special polynomials of degree $ 4$ and $ 5$. :(",
  "Solution_1": "I agree it's a great question!  Just for the record the MO question is [url=http://mathoverflow.net/questions/6820/can-a-non-surjective-polynomial-map-from-an-infinite-field-to-itself-miss-only-fi]here[/url] and the comments there are certainly helpful, even if they don't give a complete answer."
}
{
  "Tag": [
    "counting",
    "derangement",
    "complementary counting"
  ],
  "Problem": "hi again,\r\n\r\nLast time i posted a qstn on a derangement problem. BUt how about a variation like this?. I have two vatying answers from two different methods. Can anyone confirm either of them? \r\n\r\nLetters A,B, C D E F are arranged in order such taht A is not in the 1st positon, B is not in the 2nd position, and F is not in the 6th position. How many possible ways are there of arranging the letters?",
  "Solution_1": "[hide=\"Method of attack\"]For each of C, D and E, consider whether it is in its \"natural place.\"  This gives you eight (or four with an obvious simplification) cases (representative example: C and E are in their natural places, but D is not), and each is an instance of the derangement problem for some number of letters.[/hide]\nA more general version of this problem is the following:\n\nSuppose we have $ n$ letters with $ k$ special letters.  How many permutations are there of these $ n$ letters such that none of the $ k$ special letters occurs in its natural place?\n[hide=\"Answer to the general version of this problem\"]$ \\sum_{j \\equal{} 0}^{n \\minus{} k} \\binom{n \\minus{} k}{j} D_{k \\plus{} j}$.  Here $ j$ counts the number of non-special letters that don't go to their natural place.  $ \\binom{n \\minus{} k}{j}$ chooses every possible subset of them, while $ D_{k \\plus{} j}$ arranges them and the $ k$ special letters so that none goes to the natural place.  (The other $ n \\minus{} k \\minus{} j$ letters have only one choice each, their natural places.)[/hide]\r\nDoes this answer simplify?",
  "Solution_2": "The way I'm reading the problem (which is most likely wrong, considering this is in the pre-olympiad forum  :P) is that we could fix A,B,F and check the number of permutations that way ($ 3!$) and subtract that from the total number of permutations. So our answer would be $ 6!\\minus{}3!$. What is it that I'm missing that leads JBL to his solution?",
  "Solution_3": "[quote=\"max_tm\"]we could fix A,B,F[/quote]\r\n\r\nNope.  The complement of the case that none of $ A, B, F$ are in their natural position is that [b]at least one of[/b] $ A, B, F$ is in its natural position, not all of them.",
  "Solution_4": "Sorry torajirou but could you elaborate on that a bit? It's not coming as immediately intuitive to me.",
  "Solution_5": "With no conditions, there are $ 6!$ permutations.  We are considering the subset of these permutations that satisfies [b]all[/b] of the following three conditions:\r\n\r\n- $ A$ is not in the first position.\r\n- $ B$ is not in the second position.\r\n- $ F$ is not in the sixth position.\r\n\r\nThe number of permutations fixing $ A, B, F$ in the first, second, and sixth positions is the number of permutations that satisfies [b]none[/b] of the above three conditions.  However, your logic completely ignores the permutations that satisfy [b]some, but not all[/b] of the above three conditions.",
  "Solution_6": "Ah, I see. Could we patch up my logic? As in counting:\r\n\r\n- When A is fixed\r\n- When B is fixed\r\n- When C is fixed\r\n- When A,B is fixed\r\n- When B,C is fixed\r\n- When A,C is fixed\r\n- When all three of A,B,C are fixed\r\n\r\n.... Although I would never be proud of a solution like that  :P.",
  "Solution_7": "A slight modification of that idea (to avoid overcounting) works, but each of those individual cases is just as hard as the original problem (I think, except for the last).  Look at JBL's idea.",
  "Solution_8": "max_tm's idea should work for a perfectly respectable solution by inclusion-exclusion.  If we interpret \"A is fixed\" as \"A is fixed and we permute the others however we like,\" then each individual case is simple and inclusion-exclusion takes care of the overcounting.  It might even result in a simpler expression for the final sum -- give it a try!",
  "Solution_9": "yehh guys. i think i have it. I seem to be getting 426 with 3 different approaches. \r\n\r\nAs you guys mentioned, one method is complementary counting. It is not as complicated as it seems. Working through each of the cases, we want to subtract all subsets with at least one of A B or F in their correct position (ie, only A, B, F is in their correct position, only, AB, AF, and BF are in their correct positions, and when all A E F are in their correct positions). \r\n\r\n1.  So now, for the case where all ABF are in correct position is simple; if we treat positions 1,2 and 6 as one group and then the rest as another, we get that there is only one way to permute  ABF in positions 1,2 ,6 such that they are all in the correct position, and then there are 3! ways of permuting the others in the other 3 positions).  \r\n\r\n2.   Now, when only one of A E or F are in their correct positions, there are 3 ways of choosing which to be in their correct positions only. Now for the other two who are not to be in their fixed position (ie say i chose A to be in its fixed position, so B and F  can either be in the group positon 1,2 6 and not in their position in various permutations)\r\ni) when all 2 others that are not to be in their fixerd positions are not to be in the 1,2 6  position group, the number of ways is 3*2*3!. (since two of the other 3 letters C D or E must be placed in the remaining 2 unoccupied positions in 1,2 6)\r\nii) when only 1 other that are not to be in their fixed positions are not to be in 1,2 6 position group, the  number of ways is 2*3*3! (this time, the 2 meaning there are two choices of which of the 2 unfixed numbers are placed in the 126 positonal group)\r\nii) when all the other 2 unfixed letters are to be placed in the 1,2,6 group --> 3! (since there is only one way of derranging the 2 unfixed letters such that they dont appear in their forbidden positions)\r\n\r\nNow for this case, (from i), ii), iii)), you get 3(2*3*3! + 2*3*3! +3!)\r\n\r\n3.  Now, when all 2 other unfixed letters are to be in their correct positions, then there would 3(3*3!) ways\r\n\r\nHence the total number of arrangements are  6! - 3! - 3(3*3!) - 3(2*3*3! + 2*3*3! +3!) = 426. \r\n\r\nI understand this is a rather longwinded and elaborate method, which is not very desirable. But a better method is applying the Principle of Inclusion and Exclusion. and u simply get:  6! - (3C1)(5!) + (3C2)(4!) - (3C3) (3!) = 426.\r\n\r\nA THIRD METHOD, can follow along the lines of the first method, but not do complementary counting, but rather look at the 4 cases, again treating positions 1,2 and 6 as one group and the other positions as another (ie. when all A B F are in the 126 group, when only 2 of them are in the 126 group, when only 1 of them are in the 126 group, and when all of them are out of the 126 group). THis method givse:\r\n\r\n2*3! + (3!)(3!) + (3C1)(2)(3*2)(3!) + (3C2)(3)(3)(3!) = 426\r\n\r\nI would assume from 3 different methods, the answer should be 426. Funnily enough, from my first attempt, i achieved answers of 372 from my third method, and 480 for my 1st method. And then i did my second, and got 426, and realised how my previous 2 methods, the first was down by 54 and the other was up by 54. I thought this could not be a coincidence, and so i thought since one method (the first) was subtracting cases, and the last was adding cases (as methods 1 and 3 are mirrors of each other), I realised i must have made the same mistake for both! ....and eventually i found it LOL!).",
  "Solution_10": "Just for your satisfaction (if you need it), your answer is definitely right :)\r\n\r\n[hide][code]p = Permutations[{1, 2, 3, 4, 5, 6}];\nTotal[Table[Which[(p[[n]][[1]] - 1)*(p[[n]][[2]] - 2)*(p[[n]][[6]] - 6) == 0, 0,True, 1], {n, 1, Length[p]}]][/code][/hide]"
}
{
  "Tag": [
    "function",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "[color=darkred]$ 1\\blacktriangleright a\\in\\mathcal R$ and a continue monoton function $ f: [0,1]\\rightarrow\\mathcal R$ $ \\implies$ $ \\int_{0}^{1}\\left|f(x)\\minus{}a\\right|\\ dx\\ge\\int_{0}^{1}\\left|f(x)\\minus{}f\\left(\\frac{1}{2}\\right)\\right|\\ dx$ ([b]level ****, i.e. difficult ![/b]).\n$ 2\\blacktriangleright$ A continue function $ f: [a,b]\\rightarrow\\mathcal R$ $ \\implies$ exists $ c\\in (a,b)$ so that $ \\int_{a}^{c}f(x)\\ dx\\plus{}\\int_{b}^{c}f(x)\\ dx\\equal{}(a\\plus{}b\\minus{}2c)\\cdot f(c)$ ([b]level **, i.e. easy ![/b]).[/color]",
  "Solution_1": "$ 1.$ First we assume $ f$ is an increasing function (if $ f$ satisfies $ \\rightarrow\\minus{}f$ satisfies the conditions with $ \\minus{}a$).  Then it's easy to see that it's enough to prove the statement for $ a\\in[f(0),f(1)]$. Then let $ f(b) \\equal{} a$, and suppose $ b\\ge\\frac{1}{2}$ (for $ b <\\frac{1}{2}$ it's the same). Then we have:\r\n$ \\int_{0}^{\\frac{1}{2}}f(b)\\minus{}f(x)dx\\plus{}\\int_{\\frac{1}{2}}^{b}f(b)\\minus{}f(x)dx\\plus{}\\int_{b}^{1}f(x)\\minus{}f(b)dx\\ge$\r\n$ \\int_{0}^{\\frac{1}{2}}f(\\frac{1}{2})\\minus{}f(x)dx\\plus{}\\int_{\\frac{1}{2}}^{b}f(x)\\minus{}f(\\frac{1}{2})dx\\plus{}\\int_{b}^{1}f(x)\\minus{}f(\\frac{1}{2})dx$\r\nThis is equivalent with $ (b\\minus{}\\frac{1}{2})f(b)\\ge\\int_{\\frac{1}{2}}^{b}f(x)dx$, but $ \\int_{\\frac{1}{2}}^{b}f(x)dx \\equal{} (b\\minus{}\\frac{1}{2})f(c), c\\in(\\frac{1}{2},b)$, so we get\r\n$ f(b)\\ge f(c)$ which is obviously true($ b\\ge c$).",
  "Solution_2": "$ 2.$ We apply Rolle to the following function:\r\n$ G(t) \\equal{} (a\\plus{}b\\minus{}2t)(\\int_{a}^{t}f(x)dx\\plus{}\\int_{b}^{t}f(x)dx)$\r\n\r\nThe first problem can easily be generalized:\r\n$ f: [a,b]\\rightarrow\\mathbb{R}$,$ a < b$ continuous and monotonic function,$ c\\in\\mathbb{R}$ then:\r\n$ \\int_{a}^{b}\\left|f(x)\\minus{}c\\right|\\ dx\\ge\\int_{a}^{b}\\left|f(x)\\minus{}f\\left(\\frac{a\\plus{}b}{2}\\right)\\right|\\ dx$\r\n\r\nThis can easily be reduced to the original problem with the substitution $ t \\equal{}\\frac{x\\minus{}a}{b\\minus{}a}$. \r\nIt's a well-known problem, but thank you for reminding me of it. Va multumesc! :)"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra proposed"
  ],
  "Problem": "Show in the neatest possible way that \\[\\frac{c(a - b)^3 + a(b - c)^3 + b(c - a)^3}{c^2(b - a) + a^2(c - b) + b^2(a - c)} = a + b + c.\\]",
  "Solution_1": "Let $f(a,b,c)=c(a-b)^3+a(b-c)^3+b(c-a)^3$, $g(a,b,c)=c^2(b-a)+a^2(c-b)+b^2(a-c)$,we have $f(b,b,c)=0,f(c,b,c)=0,f(-(b+c),b,c)=0$, so we obtain $f(a,b,c)=(a-b)(b-c)(c-a)(a+b+c)$ and then \r\n\r\n$g(b,b,c)=0,g(a,b,a)=0$, so we obtain $g(a,b,c)=(a-b)(b-c)(c-a)$,yielding $\\frac{f(a,b,c)}{g(a,b,c)}=a+b+c$.Q.E.D.",
  "Solution_2": "Another way to factor the numerator: consider the polynomial $P$ whose roots are $a-b,b-c,c-a$. Since they add up to zero, $P$ has the form\r\n\\[P(x)=x^3+mx+n,\\]\r\nwhere $n=-(a-b)(b-a)(c-a)$.\r\nBecause $cP(a-b)+aP(b-c)+bP(c-a)=0$, we obtain\r\n\\[\\sum {c(a-b)^3}+m\\sum {c(a-b)}-(a+b+c)(a-b)(b-c)(c-a)=0.\\]\r\nObvously, $\\sum {c(a-b)}=0$, hence the result."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "homothety",
    "search"
  ],
  "Problem": "Prove that $ E$ is midpoint of major arc $ \\stackrel{\\frown}{AB}$.",
  "Solution_1": "What is $ E$?\r\n\r\nAh, that $ E$ looked like an $ F$ to me",
  "Solution_2": "Click on figure above :idea:",
  "Solution_3": "[hide=\"Hint\"]\nDraw perpendicular bisector of AB, let it hit the circle at F, and prove that E and F coincide.[/hide]",
  "Solution_4": "[quote=\"Ubemaya\"][hide=\"Hint\"]\nDraw perpendicular bisector of AB, let it hit the circle at F, and prove that E and F coincide.[/hide][/quote]\r\n\r\nSomebody can show how to do that? No other approach?\r\n\r\nMaybe the figure helps.",
  "Solution_5": "If $ l$ is the line that is tangent to the circumcircle of triangle $ ABE$ at $ E$, then $ l//AC$ since a homothety center $ D$ maps one circle (and tangent line) to another. Then let $ A'$, $ B'$ be on $ l$ so $ A'XB'$ is a line, and $ AB$, $ A'B'$ are similarly oriented.\r\n\r\nThen $ \\text{arc}\\, AE\\equal{}\\angle ABE\\equal{}\\angle BEB'\\equal{}\\text{arc}\\, BE$.",
  "Solution_6": "Thank you for your answer. \r\nI would like to know if there are another solution. The book where i got this problem don't say anything about homothetic center.",
  "Solution_7": "See  http://www.mathlinks.ro/viewtopic.php?search_id=1711576097&t=155800",
  "Solution_8": "[hide=\"Alternate Solution\"]\nCall the center of the circle containing chord $ AB$ to be $ O$ and that containing point $ C$ to be $ Q$.  So, we call $ \\angle AOB \\minus{} 2\\theta$.  Thus, we have that $ \\angle OAB \\equal{} \\angle OBA \\equal{} 90 \\minus{} \\theta\\rightarrow \\angle OBC \\equal{} 90 \\plus{} \\theta$.  Now, we call $ \\angle QCD \\equal{} \\phi \\equal{} \\angle QDC$, so $ \\angle BCD \\equal{} 90 \\minus{} \\phi$ and $ \\angle CDO \\equal{} 180 \\minus{} \\phi$.  Thus, $ \\angle BOD \\equal{} 360 \\minus{} (\\angle OBC \\plus{} \\angle BCD \\plus{} \\angle CDO) \\equal{} 2\\phi \\minus{} \\theta$.  Also, $ \\angle CDQ \\equal{} \\angle ODE \\equal{} \\phi \\equal{} \\angle OED$, so $ \\angle DOE \\equal{} 180 \\minus{} 2\\phi$.  Thus, $ \\angle AOE \\equal{} 360 \\minus{} (\\angle AOB \\plus{} \\angle BOD \\plus{} \\angle OED) \\equal{} 180 \\minus{} \\theta$.  Also, $ \\angle BOE \\equal{} \\angle BOD \\plus{} \\angle DOE \\equal{} 180 \\minus{} \\theta \\equal{} \\angle AOE$, so arcs $ AE$ and $ BE$ have the same angle, and thus $ E$ is the midpoint of major arc $ AB$.   [/hide]\r\nEDIT: Oh, sorry, that should be $ \\angle QCD \\equal{} \\phi \\equal{} \\angle QDC$, edited.",
  "Solution_9": "[quote=\"The QuattoMaster 6000\"]Now, we call $ \\angle QCD \\equal{} \\phi \\equal{} \\angle ODC$  [/quote]\r\n\r\nMaybe $ \\angle QCD \\equal{} \\phi \\Longrightarrow \\angle ODC\\equal{}90^{\\circ}\\plus{}\\phi$?"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "are the phase curves of the following differential equation closed?\r\n\r\n$ x''\\plus{}100x\\equal{}0$",
  "Solution_1": "Yes. The general solution is $ x(t) \\equal{} \\dots$, and the parametric equation $ t\\mapsto (x(t),x'(t))$ describes an ellipse.",
  "Solution_2": "There are more challenging questions of whether phase curves are closed. For instance, this well-known classic:\r\n\r\n$ x'\\equal{}ax\\minus{}bxy$\r\n$ y'\\equal{}\\minus{}cy\\plus{}dxy$\r\n\r\nwith $ a,b,c,d>0.$ And the only phase curves we care about are in the first quadrant, $ x>0,y>0.$",
  "Solution_3": "Topic-changing replies to this have been moved into the Round Table forum."
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "1)Let T:V-->V be any linear map and V* be its dual.Let W be a k-dimensional subspace of V.(V is finite dimensional).Then prove that there exits a Linearly independent subset {f1,f2,...f(n-k)} subset of V* such that W=inetsection of Kernel(fi), where i=1,2,...,n-k.\r\n\r\n\r\n2)Let T* denote the adjoint of T. If T is surjective then show that T* is injective",
  "Solution_1": "1) $ dim(V)\\equal{}n$. Let $ \\{w_1,\\cdots,w_k,v_1,\\cdots,v_{n\\minus{}k}\\}$ be a basis of $ V$ s.t. $ \\{w_1,\\cdots,w_k\\}$ is a basis of $ W$. We define $ f_i$ as follows: if $ x\\equal{}\\sum\\lambda_rw_r\\plus{}\\sum\\mu_jv_j$ then $ f_i(x)\\equal{}\\mu_iv_i$.\r\n2) \"Let $ V^*$ be its dual\" is a nonsense. You must define a basis of $ V$: $ \\{e_1,\\cdots,e_n\\}$.\r\nIf $ T$ is onto then $ T$ is one to one. Then the correct question is: if $ T$ is invertible then $ T^*$ is invertible.\r\nAssume $ T^*(f)\\equal{}0$. Then for all $ i$, $ T^*(f)e_i\\equal{}0\\equal{}f(T(e_i))$. Thus $ f(Im(T))\\equal{}0$ or $ Im(T)\\subset{k}er(f)$. Conclusion: $ f\\equal{}0$ and $ T^*$ is invertible.",
  "Solution_2": "I wrote to quickly. There are 2 corrections:\r\n1) Replace \"let $ V^*$ be its dual \" is a nonsense with \"Let T* denote the adjoint of T\" is a nonsense. Indeed there do not exist any canonical isomorphisms between $ V$  and $ V^*$.\r\n2) Replace $ f_i(x)\\equal{}\\mu_iv_i$ with $ f_i(x)\\equal{}\\mu_i$.",
  "Solution_3": "Third correction !\r\nThe adjoint, or the transpose, of $ T$ is well-defined. -I thought about the transpose of a matrix (Too much champagne during the festivals !)- \r\nIf $ T: V\\rightarrow{W}$ then $ T^*: W^*\\rightarrow{V^*}$ is defined by $ T^*(f)\\equal{}foT$.\r\nAssume that $ T$ is onto and $ T^*(f)\\equal{}0$ that is $ foT\\equal{}0$. \r\nThen $ f(W)\\equal{}0$ and $ f\\equal{}0$ and $ T^*$ is one to one."
}
{
  "Tag": [
    "logarithms",
    "pigeonhole principle",
    "continued fraction",
    "absolute value"
  ],
  "Problem": "Prove or disprove that any positive real number $ n$ may be represented to an arbitrary degree of precision (so, any degree of precision as high as you desire - pi could be represented to 10 digits, 100, 1000...) as $ 2^j*3^k$, where j and k are integers.",
  "Solution_1": "Maybe I'm understanding this wrong, but $ 5\\neq2^k3^j$ because the only prime factor is 5.",
  "Solution_2": "$ k,j$ are not necessarily positive integers, and by \"degree\" sponge means \"degree of precision\", i.e. that for any $ \\epsilon>0$, you can choose $ j,k$ so that $ |2^j3^k\\minus{}n|<\\epsilon$.",
  "Solution_3": "Thanks for the clarification, Temperal; I seem to be remarkably bad at writing down mathematical thoughts translated to English. (fixed the first post)",
  "Solution_4": "You can rewrite the relation as \r\n\r\n$ a\\log_2 2\\plus{}b\\log_2 3\\equal{}\\log\\plus{}2 n$ where $ a,b\\in\\mathbb{Z}$.\r\n\r\nConsider all $ m^2\\minus{}1$ pairs $ (a,b)$ with $ a,b\\in \\{1,2,...,m\\}$\r\n\r\nNow $ 1<a\\plus{}b\\log_2 3\\le m\\log_2 6$. \r\n\r\nNow break the interval $ (1,m\\log 6]$ into $ m^2\\minus{}2$ intervals. By the pidgeonhole principle, there must be two pairs $ (a_1,b_1)$ and $ (a_2,b_2)$ that lie in the same interval of width $ \\frac{m\\log_2 6}{m^2\\minus{}2}<\\frac{3m}{m(m\\minus{}1)}\\equal{}\\frac{3}{m\\minus{}1}$ for sufficiently large $ m$. \r\n\r\nThen it follows that $ (A_m,B_m)\\equal{}(a_1\\minus{}a_2,b_1\\minus{}b_2)$ is in the interval $ (0,\\frac{3}{m\\minus{}1}]$. (The difference cannot be zero $ a\\plus{}b\\log_2 3\\ne 0$ for any $ a,b$ as $ \\log_2 3$ is irrational. ).\r\n\r\nNow if we consider $ (l A_m,l B_m)$ it is easy to show that we get to $ n$ with an arbitrary degree of accuracy where $ l$ is an integer (I'm being lazy...)",
  "Solution_5": "[hide]Lemma: $ \\log 2$ and $ \\log 3$ are linearly independent over the rationals.\nProof: If this were not true, then $ a\\log 2 \\plus{} b\\log 3 \\equal{} 0$ or, by the properties of logs, $ \\log 2^a3^b \\equal{} 0$.  We take the antilog of both sides to get $ 2^a3^b \\equal{} 1$ which is absurd (unless $ a \\equal{} b \\equal{} 0$ which is the trivial solution).  So the lemma is proven.\n\nWe take the log of $ n\\approx 2^k3^j$ and get $ log n\\approx k\\log2 \\plus{} j\\log3$ where $ \\approx$ means to whatever degree is needed.  Now we need to express $ \\log n$ as a linear combination of $ \\log 2$ and $ \\log 3$ to arbitrary degree.  We can do this by taking the continued fraction expansion of $ \\frac {\\log 3}{\\log 2}$ to arbitrary degree: let $ \\frac {\\log 3}{\\log 2}\\approx \\frac {p}{q}$.  Then $ |q\\log 3 \\minus{} p\\log 2| \\equal{} \\epsilon$ where $ \\epsilon$ is very small.  Now let $ \\frac {\\log n}{\\epsilon} \\equal{} r$.  ($ \\epsilon$ is never 0, because the definition of $ \\epsilon$ and the fact that $ \\log 2$ and $ \\log 3$ are linearly independent over the rationals combine to show this.)  Then $ \\log n \\equal{} r\\epsilon \\equal{} r|q\\log 3 \\minus{} p\\log 2|$.  WLOG we can drop the absolute value signs: if $ p\\log 2\\geq q\\log 3$, then switch the two in the following.\n\nWe have $ \\log n \\equal{} (rq)\\log 3 \\plus{} ( \\minus{} rp)\\log 2$.  We round $ rq$ and $ \\minus{} rp$ to the nearest integers $ s$ and $ t$.  Take the antilog of both sides and get $ n\\approx 3^s2^t$.  Note that the higher the degree of $ p$ and $ q$, the higher the degree of accuracy of $ s$ and $ t$.  Thus $ s, t$ are the integers we are looking for.[/hide]\r\nEdit: Ninja'd again."
}
{
  "Tag": [
    "quadratics",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $p = a^2 + 4b^2$ be a prime. Show that $(\\frac ap) = +1$",
  "Solution_1": "Look at the equation $\\mod a$:\r\nWe get that $\\left( \\frac{p}{a} \\right) = 1$, and since $p \\equiv 1 \\mod 4$ it follows by quadratic reciprocity (for Jacoby symbols here) that $\\left( \\frac{a}{p} \\right) = 1$."
}
{
  "Tag": [
    "algorithm",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Show that there exists infinite triples $(x,y,z) \\in N^3$ such that $x^2+y^2+z^2=3xyz$.",
  "Solution_1": "First we see that $(x,y,z)=(1,1,1)$ is a solution.\r\n\r\nNow let $(a,b,c)$ is a solution of the equation with $1 \\leq a\\leq b \\leq c$.\r\n\r\nThen $a$ is a root of the equation $x^2-3bcx+b^2+c^2=0$ so \r\n$x^2-3bcx+b^2+c^2=(x-a)(x-(3bc-a))=0 \\Longrightarrow (3bc-a,b,c)$\r\n is also a solution.\r\n\r\nSince every permutation of a solution is also a solution we take that\r\npermutation of $(3bc-a,b,c), \\, (a_{1},b_{1},c_{1})$ with $a_{1} \\leq b_{1} \\leq c_{1}$ \r\nand we do the same with this solution.\r\n\r\nIf we begin with the solution $(1,1,1)$ and apply the above algorithm\r\nwe take infinite distinct solutions .",
  "Solution_2": "Well, I think it is an algorithm which generates ALL solutions. In fact it resembles the more general problem to find all numbers such that $\\frac{x^2+y^2+z^2}{xyz}=S$ is integer. And $S=3$!",
  "Solution_3": "[quote=\"Ramanujan\"]First we see that $(x,y,z)=(1,1,1)$ is a solution.\n\nNow let $(a,b,c)$ is a solution of the equation with $1 \\leq a\\leq b \\leq c$.\n\nThen $a$ is a root of the equation $x^2-3bcx+b^2+c^2=0$ so \n$x^2-3bcx+b^2+c^2=(x-a)(x-(3bc-a))=0 \\Longrightarrow (3bc-a,b,c)$\n is also a solution.\n\nSince every permutation of a solution is also a solution we take that\npermutation of $(3bc-a,b,c), \\, (a_{1},b_{1},c_{1})$ with $a_{1} \\leq b_{1} \\leq c_{1}$ \nand we do the same with this solution.\n\nIf we begin with the solution $(1,1,1)$ and apply the above algorithm\nwe take infinite distinct solutions .[/quote]\nDid you apply the algorithm?\nFirst we have $(1,1,1)$.Then we get $(2,1,1)$ and after that we'll have $(1,1,1)$ again!",
  "Solution_4": "No, [color=#0000FF]Atlas[/color]. After you get $(2,1,1)$, you permute to $(1,2,1)$ before getting the next, which will be $(5,2,1)$, then permute again, etc. Funny (and pity) the famous [b]Markoff equation [/b]found its way in such a recent Brazilian test.",
  "Solution_5": "I forgot the permutation!\nThanks",
  "Solution_6": "using quadratic equation formula in solving x this is solved when you will find (x,y,z) = (1,1,1) is a solution"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ f: [a,b] \\to \\mathbb{R}$ a function continuous on $ [a,b]$ and diferentiable on $ (a,b)$ with $ f(a)\\equal{}f(b)$.Prove that $ \\exists c \\in (a,b)$ such that \\[ f'(c)\\cdot (c\\minus{}a)\\equal{}f(a)\\minus{}f(c)\\]",
  "Solution_1": "Let $ g: [a,b] \\to \\mathbb{R}: \\forall x\\in[a,b]: \\ h(x)\\equal{}(f(x)\\minus{}f(a))(x\\minus{}a)$\r\n$ g$ is a  continuous function on $ [a,b]$ and diferentiable on $ (a,b)$ with $ h(a) \\equal{} h(b)\\equal{}0$\r\nthen $ \\exists c\\in(a,b): \\ h'(c)\\equal{}0$ and $ h'(c)\\equal{}f'(c)(c\\minus{}a)\\plus{}f(c)\\minus{}f(a)$  :wink:"
}
{
  "Tag": [
    "inequalities",
    "vector",
    "analytic geometry",
    "induction",
    "geometry",
    "Putnam",
    "algebra"
  ],
  "Problem": "Hi!  I recently derived a relatively simple proof of the Cauchy Inequality, but it relies on the fact that the triangle inequality is true for any pair of vectors in n-space.  I know that it works in the coordinate plane, and it seems logical that it works in all dimensions.  Is this true?",
  "Solution_1": "So you want\r\n\r\n$\\|v\\|+\\|w\\| \\geq \\|v+w\\|$?\r\n\r\nIf you saw my previous two attempts at a post, they were wrong lol.",
  "Solution_2": "isn't that obvious from vector addition (i think)?  im used to calling that the triangle inequality for complex numbers.  i used that to get the form $(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+...+y_{n}^{2})\\ge(x_{1}y_{1}+x_{2}y_{2}+...+x_{n}y_{n})^{2}$\r\nisn't that what people are referring to when they say \"cauchy?\"",
  "Solution_3": "Well, the way I've seen it written is $(u \\cdot v)^{2}\\leq{\\|u\\|}^{2}{\\|v\\|}^{2}$.\r\n\r\nThe statement \"the product of the sum of the squares is greater than or equal to the square of the sum of the products\" follows directly from this, by looking at this inequality in n-space and working out the dot product/length.\r\n\r\nThe triangle inequality does hold in n-space, and this can easily be shown with Cauchy-Schwarz as expressed in vector form (although this proof is useless to you, as you can't use Cauchy-Schwarz to prove the Triangle Inequality after using the Triangle Inequality to prove Cauchy-Schwarz!).",
  "Solution_4": "Look here:\r\n[url]http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality[/url] :roll:",
  "Solution_5": "oh, so basically what I didn't know was that the Cauchy inequality is essentially the same as the triangle inequality.",
  "Solution_6": "wait a minute... isn't the fact $|v|+|w|\\ge|v+w|$ obvious from vector addition and the triangle inequality?\r\nso then my proof [i]would[/i] work  :maybe: \r\nam i missing something?",
  "Solution_7": "[quote=\"daermon\"]wait a minute... isn't the fact $|v|+|w|\\ge|v+w|$ obvious from vector addition and the triangle inequality?\nso then my proof [i]would[/i] work  :maybe: \nam i missing something?[/quote]\r\n\r\nThat is the Triangle Inequality for vectors.",
  "Solution_8": "[quote=\"rjt\"]you can't use Cauchy-Schwarz to prove the Triangle Inequality after using the Triangle Inequality to prove Cauchy-Schwarz!.[/quote]\r\n\r\nA subtle point that is not always easy to grasp.  daermon:  this implies that we need to show Cauchy-Schwarz by other means to even define the concept of angles (through the dot product) in n-space.\r\n\r\nEdit:  The algebraic expression of the Triangle Inequality in n-space is known as Minkowski's Inequality, and alternately you can prove that inequality algebraically (through induction?).",
  "Solution_9": "There are several things to realize:\r\n\r\nmost (meaning all I can think of, ie no exception in mind) inequalities used in algebra (not number theoretic, etc.) are a consequence of sums of squares. That is the essence of Cauchy; it is merely a heuristic tool which tells us when a nice sum of squares exists without having to do all that work.\r\n\r\nAlso, any proof of Cauchy utilizes this in some way (except possibly the interesting geometric area proof I once saw). An induction proof (used to prove many famous inequalities) are another way to hide messiness in sum of squares but that is all that is being done...\r\n\r\nsorry if this doesn't make sense or I've said something blatantly incorrect (btw a putnam problem once asked to prove that for any polynomial always greater than zero, a sum of squares exists to prove this... basically, it is well known that all classical inequalities have a sum of squares solution)"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Place 2n points arbitrarily in the plane such that no three are collinear. Color each point with one of k colors. For each color, there must be at least one point which is painted in that color. A \"good line\" is a line that goes through two points of the same color such that there are equal number of points in either one of the half-planes, created by the line. Show that for every positive integer 1<=k<=2n, there exists a coloring such that the number of good lines is even.  :)",
  "Solution_1": "no one? guess everyone is caught up with the IMO...understandable. :)",
  "Solution_2": "bump bump... :?",
  "Solution_3": "I think I have an idea though it might be a little hard to explain. \r\n\r\nFirstly take some random good line (it is obvious that some must exist) - we will construct our set inductively. Call the endpoints of the chosen line $a_1$, $b_1$ - we have two half-planes. Take one point from the one half-plane and one from the other half-plane such that they have the property that line through them creates the smallest angle between $a_1b_1$. As no 3 points are collinear it is clearly possible - call this two ponts $a_2$ and $b_2$. Analogically name all the points as to $a_n$, $b_n$ so that there are two halfplanes - $a_i$'s belong to one while $b_i$'s to other. Easy observation leads to the fact that only $a_ib_i$ can be a good line (if colored properly) - lines $a_pb_q$ where $p \\not = q$ divide set into two unequal subsets. So we have (almost  :P ) full control of how many good lines we want to have  - anyway obtaining even number of good lines is clearly possible.",
  "Solution_4": "[quote=\"Megus\"]I think I have an idea though it might be a little hard to explain. \n\nFirstly take some random good line (it is obvious that some must exist) - we will construct our set inductively. [/quote]\r\n\r\nhmm, my confidence about your proof already dropped to 50% at the first sentence, when I made up the problem, I purposely considered a lot of cases where nothing else is possible EXCEPT to have 0 good lines, which is even. :D For a trivial example, take k = 2n, then clearly you have 0 good lines. For sufficiently large k, it is only possible to prove that we can have 0 good lines. So it's not at all obvious to me why good lines must exist.",
  "Solution_5": "Think about construction above in other way - you pick up not exactly good lines but \"could-be\" good lines - at first you don't color them. After you have all such \"could be\" good lines you begin coloring of endpoints. If $k$ is in range $n \\leq k \\leq 2n$ we can easily create no good lines - just color (using different color for each endpoint) randomily all $a_i$'s and few (if $n < k$) $b_i$ and then color remaining (if there are any) $b_i$'s such that each of $b_i$ has different color than corresponding $a_i$. For $k<n$, create $k$ good lines, if $k$ is odd, add one more (it is possible since $2k \\leq 2n-2$) Remaining endpoints color in such way that corresponding $a_i$ and $b_i$ have different colors.  \r\n\r\nDoes it make you more confident about my proof?   ;)"
}
{
  "Tag": [],
  "Problem": "what is cyclic notation ?\r\ni read about it in Count Down  when they were explaining Ian Le's solution when he used Jensen's  Inequality.",
  "Solution_1": "Cyclic notation is a shorthand for writing down the sum of a bunch of terms that \"cycle\" through the variables. For example, $ x_1 \\plus{} x_2 \\plus{} ... \\plus{} x_n$ can be written as $ \\sum_{cyc} x_1$.\r\n\r\nA more rigorous explanation is at [[Cyclic sum]]."
}
{
  "Tag": [
    "function",
    "probability",
    "geometry",
    "geometric transformation",
    "calculus",
    "derivative",
    "algebra"
  ],
  "Problem": "I would like to check that I am understanding this correctly.\r\n\r\nThe Laplace Transform for the function $ g(t) \\equal{} H(t\\minus{}\\frac{\\pi}{6}) \\ \\sin(t\\minus{}\\frac{\\pi}{6})$ is simply $ \\frac{e^{\\frac{\\pi}{6}s}}{s^2\\plus{}1}$, where $ H(t)$ is the Heaviside function.\r\n\r\nIs that the correct Laplace Transform?\r\n\r\nThanks.",
  "Solution_1": "That should be $ \\frac{e^{\\minus{}\\frac{\\pi}6}s}{s^2\\plus{}1}.$\r\n\r\nOther than that, you're on the right track.",
  "Solution_2": "Thanks for catching that mistake!  I misread my notes.\r\n\r\nBy the way, are Laplace Transforms used for methods other than solving differential equations?  That's the only use I've found so far.\r\n\r\nThanks again!",
  "Solution_3": "There are a whole family of transforms that are closely related: Fourier transform (including the mapping from periodic functions to Fourier series coefficients and the discrete Fourier transform), the Laplace transform, the Z-transform, the moment generating function and characteristic function of probability theory, and many of the ideas surrounding generating functions.\r\n\r\nThe Laplace transform is one-sided (that is, you usually consider the interval $ [0,\\infty)$) and it has provisions for incorporating initial conditions. That makes it the member of this family directed most towards solving initial value problems for ODE's.\r\n\r\nThere are a number of identities or principles that hold broadly across this family, wherever they make sense:\r\n\r\n* Translation on one side gives multiplication by an exponential (a \"character\") on the other side. This is the principle you're using in this particular problem.\r\n\r\n* Differentiation on one side gives multiplication by a variable on the other; more complicated constant-coefficient differential operators on one side give multiplication by polynomials on the other side. (In the usual way Laplace transforms are used, there's an initial condition adjustment to this.)\r\n\r\n* Pointwise multiplication on one side gives convolution on the other side.\r\n\r\nThe whole family of transforms has a vast array of uses. Laplace transforms are the member of the family optimized for ODE uses.\r\n\r\nMy favorite Laplace transform identity: Suppose $ A$ is a square matrix. Then the Laplace transform of $ e^{tA}$ is $ (sI\\minus{}A)^{\\minus{}1}.$"
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "$ x^3\\plus{}px^\\plus{}qx\\plus{}72 polynom divided by x^2\\plus{}ax\\plus{}b and x^2\\plus{}bx\\plus{}a so find x^3\\plus{}px^\\plus{}qx\\plus{}72 polynom root$",
  "Solution_1": "Restated\r\n$ x^{3} \\plus{} px^{2}\\plus{}qx \\plus{} 72$ polynomial is divided by $ x^{2} \\plus{} ax \\plus{} b$ and $ x^{2} \\plus{} bx \\plus{} a$\r\n so find$ x^{3} \\plus{} px^{2} \\plus{} qx \\plus{} 72$ polynomial roots\r\nFrankly, I still don't understand.\r\nIs it \r\n$ x^{3} \\plus{} px^{2}\\plus{}qx \\plus{} 72$ polynomial [b]could be[/b] divided by $ x^{2} \\plus{} ax \\plus{} b$ and $ x^{2} \\plus{} bx \\plus{} a$?"
}
{
  "Tag": [],
  "Problem": "has no friends",
  "Solution_1": "huh? what's this about???",
  "Solution_2": "Absolutely no flaming."
}
{
  "Tag": [
    "function",
    "set theory",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Does there exist the sequence of continious real functions $f_{n}(x)$ s.t. $f_{n}(x)\\to+\\infty$ ($n\\to\\infty$) at rational points $x$ only.",
  "Solution_1": "Since nobody has answered for a month and since it is a good problem, I'll post a \r\n[hide=\"hint\"]\nThe answer is \"Yes\".\n[/hide]",
  "Solution_2": "It seems to me that I found a solution, if I understand the problem correctly.\r\n\r\nFirst of all, denote by $ \\Gamma $ the set of dyadic rationals, and by $ \\mathbb{Q} $\r\nthe set of all rationals. Then from a theorem in set theory, there exists a bijection\r\n$ \\mathbb{Q} \\rightarrow \\Gamma $, which is monotonic increasing. Then, it is easy to\r\nshow, that this function extends to a continuous function $ \\mathbb{R} \\rightarrow\r\n\\mathbb{R} $. Therefore it is enough to solve the problem, taking instead set\r\nof rationals $ \\mathbb{Q} $, the set of dyadic rationals $ \\Gamma $ ( just after finding\r\nthe desired sequence of continuous functions for $ \\Gamma $, compose them with the function\r\nwe have found).\r\n\r\nNow, denote by $ <x> := dist(x, \\mathbb{Z}) $ for $ x \\in \\mathbb{R} $. Then define\r\n$ f(x):= max(1 - 4<x>, 0) $. Then $ f $ is of course a continuous function\r\n$ \\mathbb{R} \\rightarrow \\mathbb{R} $, $ f(x) = 1 $ for $ x \\in \\mathbb{Z} $, and\r\n$ f(x) > 0 \\Leftrightarrow dist(x, \\mathbb{Z}) < \\frac{1}{4} $. Now define\r\n$ f_{n}(x) := nf(2^{n}x) $ for $ n \\in \\mathbb{N} $.\r\nThen of course for every $ x \\in \\Gamma $, we have for $ n \\in \\mathbb{N} $, that\r\n$ 2^{n}x \\in \\mathbb{Z} $, so for $ n $ large enough, we have $ f_{n}(x) = n $, so\r\n$ lim_{n\\rightarrow \\infty} f_{n}(x) = \\infty $. Now suppose that we have a number\r\n$ x \\in \\mathbb{R} $, and $ lim_{n\\rightarrow \\infty} f_{n}(x) = \\infty $.\r\nThen in particular there exists $ n_{0} $, such that for every $ n \\geqslant n_{0} $,\r\nwe have $ f_{n}(x) > 0 $, so $ f(2^{n}x) > 0 $, so for $ n \\geqslant n_{0} $ we have that\r\n$ dist(2^{n}x, \\mathbb{Z}) < \\frac{1}{4} $. Now it is easy to see that if we have\r\na real number $ y \\in \\mathbb{R} $, such that $ dist(y, \\mathbb{Z}) < \\frac{1}{4} $,\r\nthen $ dist(2y, \\mathbb{Z}) = 2dist(y, \\mathbb{Z}) $. Therefore for $ n \\geqslant n_{0} $\r\nwe have that $ dist(2^{n+1}x, \\mathbb{Z}) = 2 dist(2^{n}x, \\mathbb{Z}) $, but on the\r\nother hand, all $ dist(2^{n}x, \\mathbb{Z}) $ are bounded by $ \\frac{1}{2} $, so the only\r\nconclusion is that $ dist(2^{n_{0}}x, \\mathbb{Z})  = 0 $, so $ 2^{n_{0}}x \\in \\mathbb{Z} $,\r\nso $ x \\in \\Gamma $. Therefore we see that $ lim_{n\\rightarrow \\infty} f_{n}(x) = \\infty $\r\nif and only if $ x \\in \\Gamma $.",
  "Solution_3": ":(  :(  :(  :( \r\nAll my work is vain.\r\n\r\nThe complete answer for this question is in http://www.mathlinks.ro/Forum/viewtopic.php?t=22177",
  "Solution_4": "Ops... I solved [i]slighly[/i] another problem in that topic.",
  "Solution_5": "I feel Leva's proof is very illuminating to say the least....",
  "Solution_6": "What do you mean by that?  :?",
  "Solution_7": "I meant the Myth's reply ...",
  "Solution_8": "Myth, what another problem did you solved?",
  "Solution_9": "I have erased everything I had written, because it was stupid :(.",
  "Solution_10": "[quote=\"eugene\"]Does there exist the sequence of continious real functions $f_{n}(x)$ s.t. $f_{n}(x)\\to+\\infty$ ($n\\to\\infty$) at rational points $x$ only.[/quote]\r\n\r\nFor an enumeration of rationals $q_i$, let $f_n(x)$ be the n'th partial sum of $a_1|x - q_1| + a_2|(x-q_1)(x-q_2)| + a_3|(x-q_1)(x-q_2)(x-q_3)| + ...$. \r\nIt is possible to choose $a_i$ and the enumeration to guarantee that the sum is infinite at irrational points.",
  "Solution_11": "[quote=\"Leva1980\"]Myth, what another problem did you solved?[/quote]\r\nFollow the link I wrote above.\r\nI proved that if $f_n(x)$ is unbounded for all rationals, then there is irrational $s$ s.t. $f_n(s)$ is unbounded too. :?",
  "Solution_12": "[quote=\"fleeting_guest\"][quote=\"eugene\"]Does there exist the sequence of continious real functions $f_{n}(x)$ s.t. $f_{n}(x)\\to+\\infty$ ($n\\to\\infty$) at rational points $x$ only.[/quote]\n\nFor an enumeration of rationals $q_i$, let $f_n(x)$ be the n'th partial sum of $a_1|x - q_1| + a_2|(x-q_1)(x-q_2)| + a_3|(x-q_1)(x-q_2)(x-q_3)| + ...$. \nIt is possible to choose $a_i$ and the enumeration to guarantee that the sum is infinite at irrational points.[/quote]\r\nWhy did you said that?",
  "Solution_13": "[quote=\"eugene\"]Does there exist the sequence of continious real functions $f_{n}(x)$ s.t. $f_{n}(x)\\to+\\infty$ ($n\\to\\infty$) at rational points $x$ only.[/quote]\r\n\r\nI think the following works as well: \r\n\r\nIf $(q_n)_{n\\ge 1}$ is an enumeration of the rationals, we take $f_n$ to be $n$ in $q_1,q_2,\\ldots,q_n$, and $0$ outside $A_n=\\bigcup_{k=1}^n(q_k-\\varepsilon_n,q_k+\\varepsilon_n)\\ (*)$. If $\\varepsilon_n>0$ is small enough s.t. the intervals in $(*)$ are disjoint for every $n$, and $(\\varepsilon_n)_n$ decreases (to $0$) fast enough s.t. $A_n\\cap A_{n+1}=A_{n+1}\\setminus(q_{n+1}-\\varepsilon_{n+1},q_{n+1}+\\varepsilon_{n+1}),\\ \\forall n$, then every irrational will lie outside infinitely many $A_n$'s: \r\n\r\nLet $t\\in\\mathbb R$ and $n_0$ is large enough s.t. $t\\in A_n,\\ \\forall n\\ge n_0$. Since $t\\in A_n$, it lies in some $(q_k-\\varepsilon_n,q_k+\\varepsilon_n)$. $t\\in A_{n+1}$ as well, so, from the conditions imposed on the $A_n$'s, we must have $t\\in(q_k-\\varepsilon_{n+1},q_k+\\varepsilon_{n+1})$ (the same $k$ as before). We repeat this for $n\\to\\infty$, and we get $t=q_k$ because $\\varepsilon_n\\to 0$, so $t$ is rational.\r\nIs there anything wrong with this? :?"
}
{
  "Tag": [
    "logarithms",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Which number is bigger: $3{}^{3{}^{3^{...}^3}}$ (2005 3's) or $2^{2^{{{{...}{}^2}{}^6}}}$ (2004 2's and one 6).\r\n\r\nThanks for help",
  "Solution_1": "Nice problem!\r\nSuppose $3^m<2^{\\frac{5}{6}n}$, and $n\\geq 6$. Then $3^{3^m}<2^{\\frac{5}{6}2^n}$ :!:.\r\nIndeed, \\[3^{3^m}<2^{\\frac{5}{6}2^n}\\iff 3^m<\\frac{5}{6}\\cdot {2^n}\\log_32,\\] but $2^{\\frac{5}{6}n}<\\frac{5}{6}\\cdot{2^n}\\log_32$ for $n\\geq 6$ !!!\r\nWe initially have $3^3=27<32=2^{\\frac{5}{6}\\cdot 6}$. Using :!: we obtain that $3^{3^{...^3}}<2^{\\frac{5}{6}2^{...2^6}}$. Hence the result."
}
{
  "Tag": [],
  "Problem": "$\\frac{x+x^{2}+x^{3}....+x^{50}}{x^{-1}+x^{-2}+...+x^{-50}}$",
  "Solution_1": "Mods, can this be put to Intermediate or High School Basics?",
  "Solution_2": "[quote=\"ashrafmod\"]$\\frac{x+x^{2}+x^{3}....+x^{50}}{x^{-1}+x^{-2}+...+x^{-50}}$[/quote]\r\n$x^{51}$ :P  :P"
}
{
  "Tag": [
    "greatest common divisor"
  ],
  "Problem": "For those of you who don't know, Bezout's Lemma states that for any two integers $ a$ and $ b$, there exist integers $ m$ and $ n$ such that $ am \\plus{} bn \\equal{} (a, b)$, where $ (a, b)$ denotes the greatest common factor of $ a$ and $ b$. \r\n\r\nHow can we show that there exist infinitely many ordered pairs $ (m, n)$ satisfying $ am \\plus{} bn \\equal{} (a, b)$?",
  "Solution_1": "$ am \\plus{} bn \\equal{} a(m \\plus{} bk) \\plus{} b(n \\minus{} ak)$.",
  "Solution_2": "Thanks. Part two: \r\n\r\nLet's say $ (m, n)$ is one solution to the equation. Then do all other solutions $ m'$ and $ n'$ necessarily take the form $ m \\plus{} bk$ and $ n \\minus{} ak$ for some integers $ k$? If so, prove it. If not, give a counterxample.",
  "Solution_3": "Thats incorrect, but it is true if you write $ m\\plus{}\\frac{b}{(a,b)}\\cdot k$, etc\r\n\r\nWe prove this by saying that $ (m,n)$ is a solution so $ am\\plus{}bn\\equal{}(a,b)$.\r\n\r\nThen suppose that $ (x\\plus{}m,y\\plus{}n)$ is another solution...\r\n\r\n$ a(x\\plus{}m)\\plus{}b(y\\plus{}n)\\equal{}(a,b)$\r\n$ ax\\plus{}by\\plus{}(am\\plus{}bn\\minus{}(a,b))\\equal{}0$\r\n$ ax\\equal{}\\minus{}by$. \r\n\r\nThen it is easy to see that $ \\frac{b}{(a,b)}$ must divide $ x$, and such..."
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14780[/img]\r\n\r\n :(",
  "Solution_1": "Let's solved with your figure.\r\nLet M,N be on AC such that AM=MN=NC.\r\nS(AHDB)=S(ABM)=S(ABC)/3, so S(BDH)=S(BMH), thus BH is parallel to DM. H have just been constructed.\r\nThe line through N and parallel to BH meets BC at I. S(IHN)=S(IBN), so S(IHC)=S(BNC)=S(ABC)/3=S(CEKH), which follows that S(IHE)=S(KHE), hence IK is parallel to EH.",
  "Solution_2": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14791[/img]\r\n\r\n\r\n[b]\"The line through N and parallel to BH meets BC at I\" not correct\n\n\"The line through N and parallel to AB meets BC at I\"   correct[/b]\r\n\r\n\r\n :thumbup:"
}
{
  "Tag": [],
  "Problem": "I have trouble understanding this... (excerpted from The Art and Craft of Problem Solving by Paul Zeitz, pg. 224 - 225)\r\n\r\n[quote](d)If $ g|a$ and $ g|b$, then $ g|ax \\plus{} by$, where x and y can be any integers. WE call a quantity like ax + by a linear combination of a and b. [...] 7.1.5 An important consequence of the division algorithm, plus 7.1.3 (d), is that [i]The greatest common divisor of a and b is the smallest positive linear combination of a and b.[/i] [...] Let's prove 7.1.5; it is rather dry but a great showcase for the use of the extreme principle pls arguement by contradiction. Define u to be the smallest positive linear combination of a and b and let g := (a,b). We know from 7.1.3 (d) that g divides any linear combination of a and b, and so certainly g divides u. This means that $ g \\le u$. We would like to show that in fact, $ g\\equal{}u$. We will do this by showing that /u is a common divisor of a and b. Since u is greater than or equal to g, and g is the greatest common divisor of a and b, this would force g and u to be equal. \nSuppose, on the contrary, that u is not a common divisor of a and b. Without loss of generality, suppose that u does not divide a. Then yb the division algorithm, there exists a quotient $ k /ge 1$ and the positive remainder $ r < u$ such that \\[ a \\equal{} ku \\plus{} r\\]\nBut then $ r \\equal{} a \\minus{} ku$ is positive and less than u. [b]This is a contradiction, because r is also a linear combination of a and b, yet u was defined to be the smallest positive linear combination![/b] Consequently, u divides a, and likewise u divides b. So u is a common divisor; thus $ u \\equal{} g$. [/quote]\r\n\r\nCan anyone please explain the bolded part - namely, \"This is a contradiction, because r is also a linear combination of a and b, yet u was defined to be the smallest positive linear combination!\" \r\n\r\nI mean, can anyone explain in detail how r is a linear combination of a and b???????\r\n\r\nI thank you for your time and patience.",
  "Solution_1": "just for purposes of clarification: I understand everything up to the bolded point. Thanks.",
  "Solution_2": "$ u$ is a linear combination of $ a,b$; hence so is $ a \\minus{} ku$.",
  "Solution_3": "sorry, but apparently i can't think striaght today for some reason  :( \r\n\r\nCan you explain in detail? (some algebra would be nice)",
  "Solution_4": "$ u$ is a linear combination of $ a,b$; hence we may write $ u \\equal{} ax \\plus{} by$, where $ x,y$ are integers. Then $ r \\equal{} a \\minus{} ku \\equal{} a \\minus{} k(ax \\plus{} by) \\equal{} (1 \\minus{} kx)a \\plus{} (\\minus{}ky)b$, where $ 1\\minus{}kx, \\minus{}ky$ are obviously integers."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "LaTeX",
    "Olimpiada de matematicas"
  ],
  "Problem": "[color=green]disculpen la molestia, si alguien me pudiera proporcionar los enunciados de la olimpiada de mayo 2005 y 2006 se los agradeceria mucho, he pasado miles de horas buscando en internet y he encontrado hasta la 2004, y los estoy necesitando ahorita. gracias de antemano x su ayuda![/color] :)",
  "Solution_1": "[color=blue]Ya me pidieron tambi\u00e9n algunos chicos, y en ning\u00fan lado se consiguen, as\u00ed como los enunciados de la XVI Olimpiada Matem\u00e1tica del Cono Sur.\n\u00bfQui\u00e9n podr\u00eda explicar el motivo?[/color]",
  "Solution_2": "aqui tem tanto da cone sul, como da olimpiada de maio, mas est\u00e1 em portugu\u00eas!  :blush: \r\n\r\nhttp://www.obm.org.br/frameset-provas.htm",
  "Solution_3": "Aqui tengo el examen del 2006",
  "Solution_4": "muchas gracias a todos x su ayuda :lol: se los agradezco mucho, la verdad q estos examenes no son muy faciles de encontrar :)",
  "Solution_5": "Buenas tardes,\r\n\r\naki tengo las fuentes de latex q generaron este archivo ..\r\n\r\nclaro si alguien las desea para editar otros exams de este formato ..\r\n\r\nsaludos,",
  "Solution_6": "[color=blue]Aviso que mis \"disc\u00edpulos\" realizar\u00e1n el examen de la mencionada olimpiada este s\u00e1bado 11 de mayo.\nEs bien sabido que est\u00e1 prohibido divulgar hasta el mes siguiente.[/color]",
  "Solution_7": "Buenas noches,\r\n\r\neso es cierto!!\r\n\r\nNo publicar los exams de Mayo no al menos dentro de 1 mes\r\n\r\nGracias!!",
  "Solution_8": "Para ser m\u00e1s exactos no se puede divulgar las pruebas hasta el 25 de Mayo.\r\n\r\n$Tipe$",
  "Solution_9": "[quote=\"tipe\"]Para ser m\u00e1s exactos no se puede divulgar las pruebas hasta el 25 de Mayo.\n\n$Tipe$[/quote]\r\n\r\nPerfecto Jorge,\r\n\r\nGracias por la aclaracion...\r\n\r\n\r\nSaludos,",
  "Solution_10": "[quote=\"tipe\"]Para ser m\u00e1s exactos no se puede divulgar las pruebas hasta el 25 de Mayo.\n\n$Tipe$[/quote]\r\n\r\nYa es 26 de mayo :P , \u00bfalguien tiene los enunciados? Gracias y saludos!",
  "Solution_11": "yo los tengo  :)  solo q para tener un poco mas de orden mejor los voy a poner en topicos separados! en un momento los subo"
}
{
  "Tag": [
    "Pythagorean Theorem",
    "geometry"
  ],
  "Problem": "I was doing team round problems, I could do all but 7 and 8, could some 1 help?\r\n\r\nhttp://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Team.pdf\r\n\r\n\r\nEdit: I found out 7",
  "Solution_1": "Next time, please post the problems\r\n\r\n[hide=\"hint for number 8\"]\nUse Pythagorean Theorem (or memorize some triples) to find AG, and then AG=AC because they are both radii and then figure it out from there\n[/hide]",
  "Solution_2": "Ok so um heres some hints.\r\nFor number 7, use the fact that the radius of the circle is the same as a line from the center of the hexagon to one of it's vertexes (Plural form.... :| ), which means that it is also the length of one of the sides of the hexagon.\r\nFor number eight, just use the Pythagorean Theorem to find out the radius of the circle, and subtract 7 from it."
}
{
  "Tag": [
    "college",
    "geometry",
    "Stanford",
    "Berkeley Math Circle"
  ],
  "Problem": "Hi, \r\nI'm at UC Berkeley right now for the summer (until august 14) for the summer college program.\r\nI'm a high school student, and I'm hoping that I can get involved in some math clubs or activities nearby.\r\nIf there's any program or meeting held during the summer near the UC berkeley campus, and if it's not too late, can anyone provide me with information regarding nearby math circles etc.?\r\n(My goal for this is to prepare for upcoming olympiads next year.)\r\n\r\nThanks\r\n\r\nDavid Lee",
  "Solution_1": "I'm also going to be around UC Berkeley for the summer. There's Berkeley Math Circle on campus, but that won't start until September. \r\n\r\nThe only math circle in the area that has summer sessions is the Stanford Math Circle http://www.stanfordmathcircle.org/Main_Schedule.html It's from 7-9 on Tuesday nights - It's fairly far from Berkeley - if you want to get there by public transportation you'll have to take BART then transfer to Caltrain. Then its a walk to campus.",
  "Solution_2": "Sounds fun, Looks fun, Seems fun!\r\nBut, yea.. I don't think I'll be able to make it.. \r\nStill, I'll try my best to get there if I can, if I have time and no writing homework :D"
}
{
  "Tag": [
    "geometry solved",
    "geometry"
  ],
  "Problem": "I want some easy inversion problem,if anybody have plz send. :D",
  "Solution_1": "Have you proved that The inverse of a line that doesn't pass by the inversion center is a circunferencia thought the center?",
  "Solution_2": "[quote=\"hucht\"]Have you proved that The inverse of a line that doesn't pass by the inversion center is a circunferencia thought the center?[/quote]\r\nof course. :D",
  "Solution_3": "A problem of mine is a example of using inversion to solve problems easily.\r\n\r\nSee\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=55206",
  "Solution_4": "http://www.mathlinks.ro/Forum/viewtopic.php?t=14958\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=56658\r\n\r\n  Darij"
}
{
  "Tag": [
    "vector",
    "function",
    "geometry",
    "parallelogram",
    "search",
    "linear algebra"
  ],
  "Problem": "A norm on a vector space U is a function from U to nonnegative real such that ||u||>=0, ||au|| = |a|||u|| for all a in F and all u in U, and ||u+v|| is less than or equals to ||u||+||v|| for all u,v in U. Prove that a norm satisfying the parallelogram equality comes from an inner product(that there exists an inner product < , > that ||u||=$ < u,u > ^\\frac {1}{2}$ for all u in U)",
  "Solution_1": "See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1961927883&t=224293]here[/url].",
  "Solution_2": "[quote=\"t0rajir0u\"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1961927883&t=224293]here[/url].[/quote]\r\n\r\nI got confused, the problem want to show that the function is the norm of an inner product space, but the first step of the proof just say $ {\\parallel{}x\\plus{}y\\parallel{}}^2\\equal{}<x\\plus{}y,x\\plus{}y>$, which I think only holds if the function is  the norm of an inner product space. Is this a logical error?",
  "Solution_3": "You're reading the .pdf file downloaded from that, right? What you're quoting is from a \"discovery\" phase of the project: suppose there were a valid inner product. How would it be related to the norms?\r\n\r\nThe proof then starts about halfway down the page, with the statement (at least for the real case) that \"Our goal is to show that $ \\langle x,y\\rangle\\equal{}\\frac{\\|x\\plus{}y\\|^2\\minus{}\\|x\\minus{}y\\|^2}{4}$ does indeed define an inner product.\"",
  "Solution_4": "Oh,by bad\r\nI'm so careless!",
  "Solution_5": "I'm sorry, but I'm not able to prove the continuity of <ax,z> for a.(last step for proving real case) Can anyone help?",
  "Solution_6": "At this point we've already proved the linearity for all real and homogeneity for all rational. So we can use these results. \r\n\r\nThe last step is Use Parallelogram identity to prove that , for any x,z, the function a - <ax,z> is a continuous function of a. (a is real)\r\n\r\nHelp me please :maybe:"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let n be a natural number grater or equal to 2. Determine the numbers a_k naturals and different from 0, k={1,2,...,n}, distinct two by two, knowing that\r\n$ a_1\\plus{}a_2\\plus{}...\\plus{}a_n\\equal{}\\dfrac{n^2\\plus{}n\\plus{}2}{2}$",
  "Solution_1": "[quote=\"St. Elena\"]Let n be a natural number grater or equal to 2. Determine the numbers a_k naturals and different from 0, k={1,2,...,n}, distinct two by two, knowing that\n$ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n \\equal{} \\dfrac{n^2 \\plus{} n \\plus{} 2}{2}$[/quote]\r\n\r\n$ a_i\\equal{}i$ $ \\forall i\\in[1,n\\minus{}1]$\r\n$ a_n\\equal{}n\\plus{}1$"
}
{
  "Tag": [
    "Princeton",
    "college"
  ],
  "Problem": "I bought Barron's SAT II Math IIC book and like people said, it was much harder than some of the practice tests I took. But how much harder is it? Will questions like: what is the 3rd term of (8x-y)^(1/3) be in the test?\r\n\r\nThanks for any answer.",
  "Solution_1": "Ok let's put it this way. Under practice-test (meaning extremely distracted) conditions, not bothering to double-check, etc, I got a best score of 45 out of 50 on the Barron's practice tests (I took all 6). \r\n\r\nI went into the test thinking I would get a 750 at best.\r\n\r\nI got an 800 and thought it was extremely easy. \r\n\r\nBarron's is notorious for making their prep books harder than the real test to \"over-prepare\" you because they assume that there's no way that you can know all the material in a review book anyways.",
  "Solution_2": "I find the practice tests in the SparkNotes Math 1C and 2C books to be pretty close in level to the real thing.",
  "Solution_3": "Out of curiosity, how do you have a third term for $\\sqrt[3]{8x-y}$?",
  "Solution_4": "haha, i think he probably meant to the third power, not third root.",
  "Solution_5": "Or it could have been referring to the infinite [url=http://mathworld.wolfram.com/BinomialSeries.html]binomial series[/url], which is indeed too advanced for SAT II math.",
  "Solution_6": "It has some errors, but they aren't too much of a burden. However, it doesn't parallel the actual test at all - nowhere on the test you'll have to provide multiple answers for a trigonometric equation, for example. It's actually nice for improving one's knowledge of Pre-Calc concepts.",
  "Solution_7": "Hmmm I always thought those books were routine, but I guess I did learn Algebra from Barron's \"Algebra the Easy Way.\"  I have to say from all that I've seen/heard, Barron has some good books.",
  "Solution_8": "[quote=\"Ravi B\"]Or it could have been referring to the infinite [url=http://mathworld.wolfram.com/BinomialSeries.html]binomial series[/url], which is indeed too advanced for SAT II math.[/quote]Wow I never knew that.",
  "Solution_9": "[quote=\"eryaman\"][quote=\"Ravi B\"]Or it could have been referring to the infinite [url=http://mathworld.wolfram.com/BinomialSeries.html]binomial series[/url], which is indeed too advanced for SAT II math.[/quote]Wow I never knew that.[/quote]\r\n\r\nDiscovered by Newton. :lol:",
  "Solution_10": "Don't stress too much over the IIC.  It's not that hard.",
  "Solution_11": "im taking it this saturday, june 4th.  i've been using the princeton review though.",
  "Solution_12": "best of luck to everybody including my brother (he's taking bio) :)",
  "Solution_13": "[quote=\"bubala\"]If you don't make careless errors--read the question--and go at the right pace, then you will get an 800 easily.[/quote]\r\n\r\nI'm sure that there're a lot of people who do those things, but don't get an 800. What's easy to you may not be easy to everyone else. You have to respect people's individuality.",
  "Solution_14": "well, I think that most, if not all people, on THIS FORUM in particular, are capable of getting an 800 or close to that.  \r\n\r\nI wouldn't say something like that to all my friends in school, for example.\r\n\r\nIt depends on the environment.\r\n\r\nHowever, I apologize for such comments because the above statement may not indeed be true.",
  "Solution_15": "I took the IIC on May 7th and scored an 800....it was extremely easy, and I'm not surprised that 800 is only 90th percentile.\r\n\r\nIf you are in even the second highest level of 11th grade math offered at your school, this test should be a breeze, no book required",
  "Solution_16": "What is the lowest raw score that guarantees a 800?",
  "Solution_17": "[quote=\"ffdbzathf\"]im taking it this saturday, june 4th.  i've been using the princeton review though.[/quote]\r\n\r\ni used the princeton review book too.  i just toke it 1hr and 15mins ago :D",
  "Solution_18": "[quote=\"anhnguyen\"]What is the lowest raw score that guarantees a 800?[/quote]\r\n\r\nyou can miss like 5 or 6 and still get a 800 on IIC",
  "Solution_19": "Each administration of the test is scaled differently, so the number of questions that equals an 800 will vary from administration to administration.",
  "Solution_20": "So it's been concluded that a question like the oneproposed in the topic will not be present in the SAT II Math IIC?\r\n\r\nIf not, how would you work it out? I saw the link, but something doesn't sink in unless I see it being done.",
  "Solution_21": "No it will not be in the SAT II. I asked the question and I just took the test this morning. It was much easier than Barron's. Don't worry about those types of questions. They are too advance to be in the test.",
  "Solution_22": "Awesome"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "AMC"
  ],
  "Problem": "AMC 1976 #30.\r\n\r\nHow many distinct ordered triples (x,y,z) satisfy the equations\r\nx+2y+4z=12\r\nxy + 4yz + 2xz = 22\r\nxyz = 6\r\n\r\nShow your work.",
  "Solution_1": "[hide] x+2y+4z=12, 2xy+8yz+4xz=44, and x(2y)(4z)=48, so polynomial with roots x, 2y, and 4z is x^3 -12x^2 +44x +48= 0[/hide]\n\n\n\nmore later gotta run.",
  "Solution_2": "Since x+2y+4z=12, xy+4yz=12y-2y\u22272\r\nSince xy+4yz+2xz=22, xy+4yz=22-2xz\r\nSo 12y-2y\u22272=22-2xz\r\n\t6y-y\u22272=11-xz\r\n\ty\u22272-6y=9=xz-2\r\n\t(y-3) \u22272=xz-2\tsince xyz=6\r\nso\t(y-3) \u22272=6/y-2\r\n\ty\u22273-6y\u22272+9y=6-2y\r\n\ty\u22273-6y\u22272+11y-6=0\r\n\ty\u22273-y-(6y\u22272-12y+6)=0\r\n\t(y-1)(y-2)(y-3)=0\r\nso y can be 1, 2, or 3\r\nsubstitute the values of y into the original equation and get the answer"
}
{
  "Tag": [],
  "Problem": "[url=http://www.imo-official.org/year_reg_team.aspx?code=ROU]Participantii[/url]",
  "Solution_1": "Subiectul de anul trecut:",
  "Solution_2": "Rezultatele finale de la baraje:",
  "Solution_3": "[url=http://www.imo-official.org/year_info.aspx?year=2009]Rezultatele oficiale[/url]"
}
{
  "Tag": [
    "inequalities",
    "symmetry",
    "geometric series"
  ],
  "Problem": "[color=blue][b] Problem.[/b]\n\n  Let $a, b, c\\in (0,1)$. Prove that the following inequality holds true:\n\n                        \\[ \\frac{a}{1-a}+\\frac{b}{1-b}+\\frac{c}{1-c}\\geq\\frac{3\\sqrt [3] {abc}}{1-\\sqrt [3] {abc}}. \\]\n\n  [/color]",
  "Solution_1": "$\\\\\\frac{a}{1-a}+\\frac{b}{1-b}+\\frac{c}{1-c}\\\\ =\\sum_{k=1}^{\\infty}(a^k+b^k+c^k)\\geq 3\\sum_{k=1}^{\\infty}\\sqrt[3]{abc}^k$\r\nIt is suffiient enough, due to symmetry, to prove that $a^k+b^k+c^k\\geq 3\\sqrt[3]{abc}^k$, which is true by AM-GM.\r\nQ.E.D.\r\n\r\nMasoud Zargar",
  "Solution_2": "[quote]$\\frac{a}{1-a}+\\frac{b}{1-b}+\\frac{c}{1-c} =\\sum_{k=1}^{\\infty}(a^k+b^k+c^k)$[/quote]\r\nCan you please tell me why this is true?\r\n\r\nI think I also have a proof for the inequality:\r\n\r\nBy Chebyshev inequality, we obtain: \\[ \\displaystyle\\frac{a}{1-a}+\\frac{b}{1-b}+\\frac{c}{1-c} \\geq \\frac{a+b+c}{3}.\\left(\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}\\right) \\]\r\n\r\nBy AM-HM and AM-GM, we know that \\[ \\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c} \\ge \\frac{9}{3-(a+b+c)} \\ge \\frac{3}{1-\\sqrt[3]{abc}} \\]\r\nUsing this, we get \\[ \\displaystyle\\frac{a+b+c}{3}.\\left(\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}\\right) \\ge \\frac{a+b+c}{1-\\sqrt[3]{abc}} \\ge \\frac{3\\sqrt[3]{abc}}{1-\\sqrt[3]{abc}} \\]\r\n\r\n$\\mathbb{QED}$",
  "Solution_3": "[quote=\"Jan\"][quote]$\\frac{a}{1-a}+\\frac{b}{1-b}+\\frac{c}{1-c} =\\sum_{k=1}^{\\infty}(a^k+b^k+c^k)$[/quote]\nCan you please tell me why this is true?\n[/quote]\r\nHe used $\\frac{1}{1-x}=1+x+x^2+x^3+\\cdots =\\sum_{k=0}^\\infty x^k$ for any real number $\\in (0,1)$. Hmm... strange solution. :)",
  "Solution_4": "Well, since $a,b,c\\in (0,1)$ and we have $\\frac{a}{1-a}$ and such terms, I thought of using geometric series; $\\sum_{k=1}^{\\infty} x^k=\\frac{x}{1-x}$ :)",
  "Solution_5": "[hide]Divide out the numerators: $\\frac{a}{1-a}=\\frac{1}{1-a}-1$ (similar for the others) and the right side is $\\frac{3}{1-\\sqrt[3]{abc}}-3$.\n\nSo this inequality is equivalent to $\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}\\geq \\frac{3}{1-\\sqrt[3]{abc}}$.\n\nBy AM-HM, $\\frac{\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}}{3}\\geq \\frac{3}{(1-a)+(1-b)+(1-c)}$.\n\nThis becomes $\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}\\geq \\frac{9}{3-(a+b+c)}=\\frac{3}{1-\\frac{a+b+c}{3}}$.\n\nNow, from AM-GM, $\\frac{a+b+c}{3}\\geq \\sqrt[3]{abc}\\Rightarrow 1-\\frac{a+b+c}{3}\\leq 1-\\sqrt[3]{abc}\\Rightarrow \\frac{3}{1-\\frac{a+b+c}{3}}\\geq \\frac{3}{1-\\sqrt[3]{abc}}$\n\n(where the last inequality is true because $a,\\ b,\\ c\\in (0,1)\\Rightarrow \\frac{a+b+c}{3},\\ \\sqrt[3]{abc}<1$.\n\nSince $\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}\\geq \\frac{3}{1-\\frac{a+b+c}{3}}\\geq \\frac{3}{1-\\sqrt[3]{abc}}$, the inequality is true.[/hide]"
}
{
  "Tag": [
    "Gauss"
  ],
  "Problem": "Esmeralda has a bottle with 9 liters of a mixture that has 50% alcohol and 50% water. She wants to put water in the bottle so that only 30% of the mixture is alcohol. How many gallons of water will put it?\r\n\r\nPortuguese question (if necessary):\r\nEsmeralda tem uma garrafa com 9 litros de uma mistura que tem 50% de \u00e1lcool e 50% de \u00e1gua. Ela quer colocar \u00e1gua na garrafa de tal forma que apenas 30% da mistura seja de \u00e1lcool. Quantos litros de \u00e1gua ela ir\u00e1 colocar?\r\n\r\n\r\n\r\nHow to do ?  :|",
  "Solution_1": "We see there are 9 liters of the mixture and half of it is water, so there's $ 4.5$ liters of water.\r\n\r\nLet $ x$ be the amount of water in liters to produce a 30% alcohol (70% water) solution.\r\n\r\n$ \\frac{4.5\\plus{}x}{9\\plus{}x}\\equal{}\\frac{7}{10}$\r\n\r\nCross multiplying givexs\r\n\r\n$ 45\\plus{}10x\\equal{}63\\plus{}7x$\r\n$ 3x\\equal{}18$\r\n$ x\\equal{}6$\r\n\r\nSo $ 6$ liters, or $ \\frac{6}{3.78541178} \\approx \\boxed{1.585}$ gallons need to be added.",
  "Solution_2": "Bom dia luiseduardo.\r\n\r\nSuppose she puts in $ x$ litres of water. Then she has $ 4.5$ litres of alcohol and $ 4.5 \\plus{} x$ litres of water. \r\n\r\n$ \\frac{4.5}{9\\plus{}x} \\equal{} \\frac{3}{10}$\r\n\r\n$ 27 \\plus{} 3x \\equal{} 45$\r\n\r\n$ x \\equal{} 6$",
  "Solution_3": "Hello Andrew and Gauss, Thank you very much ;)"
}
{
  "Tag": [
    "inequalities",
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "this is similar to an USAMO problem i did\r\n\r\nProve that:\r\n\r\n$a^ab^bc^c \\ge a^bb^cc^a$\r\n\r\nfor $a,b,c >0$",
  "Solution_1": "it seems kinda like rearrangement. \r\nif u know the prove of rearrangment, here is the idea:\r\n\r\n$a\\ge b$\r\n$a^{a-b}\\ge b^{a-b}$\r\n$a^a\\cdot b^b\\ge a^b\\cdot b^a$\r\n\r\n;) im 2 lazy to write up the whole prove but u only have to look at 2 cases because WLOG $a\\ge b\\ge c$ or $a\\le b\\le c$ so u can do the thing above first with $a$ and $b$ and then with $b$ and $c$",
  "Solution_2": "heh nice job\r\n\r\nthis is pretty similar to the USAMO problem:\r\n\r\n$a^ab^bc^c \\ge (abc)^{(a+b+c)/3}$\r\n\r\nfor $a,b,c>0$\r\n\r\ni think thats the problem at least",
  "Solution_3": "If you take the log of both sides, it is direct rearrangement.  Actually if you apply your inequality cyclically you get the USAMO inequality.",
  "Solution_4": "Here is a nice proof of the USAMO inequality\r\nIt comes from John Scholes -his solution to Question 2 Canadian MO 1995.\r\n\r\nAssume  a>= b >= c.\r\n\r\nThen a/c, a/b  and b/c  are all at least 1 and \r\n\r\n(a-c)/3 , (a-b)/3  and (b-c)/3  are all positive. \r\n\r\nTherefore  (a/c) ^([a-c]/3) ,  (a/b)^([a-b]/3)  and ( b/c)^([b-c]/3) are all at least 1.\r\nHence their product is at least 1, which leads to the required inequality.",
  "Solution_5": "[quote=\"gotztahbeazn\"]heh nice job\n\nthis is pretty similar to the USAMO problem:\n\n$a^ab^bc^c \\ge (abc)^{(a+b+c)/3}$\n\nfor $a,b,c>0$\n\ni think thats the problem at least[/quote]\r\nI got the generialization from this USAMO inequality.  :P \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=46247"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Prove that a left annihilitor of a right ideal of a ring $ R$ with unity is generated by an idempotent",
  "Solution_1": "I have no idea what you could mean,\r\nbut take $ R\\equal{}\\mathbb{Z}/4\\mathbb{Z}$ and ideal $ I\\equal{}2\\mathbb{Z}/4\\mathbb{Z}$."
}
{
  "Tag": [
    "puzzles",
    "WindClan for the win"
  ],
  "Problem": "1. A man had a book that was worth $\\$40,000$. He threw it in the furnace, reducing it to a pile of soot. Why did he do this?\r\n[hide=\"hints\"] \nA) He is not mentally disordered or retarded. He did this for a very good reason.\nB) He does care about money.\n[/hide]\r\n2) The police arive at a house. There is a fresh corpse on the floor. Other than that, there is only a tape recorder in the room. When the police played it, they heard this: \"I can't go on. I have nothing to live for\" After this, they hear a gunshot. How did they know the person had been [i]murdered?[/i]",
  "Solution_1": "My educated guess for the 2nd situation:\r\n[hide]Those quoted words are the words of the killer.  He recorded them (his own words) and the gunshot's sound on purpose.  He turned off the tape recorder right after that so that sounds following that would not be heard on the playing of the tape.[/hide]",
  "Solution_2": "not quite. Youre on the right track though.",
  "Solution_3": "[hide=\"My Thoughts on the situation\"][hide=\"1)\"]He did not care about money, so he just used the book to burn and keep himself warm.[/hide][hide=\"2)\"]The question reads [quote=\"SoxFan34\"]How did they know the person had been killed?[/quote]Uhhh...maybe because he's lying there dead? :rotfl: [/hide][/hide]",
  "Solution_4": "[quote=\"kstan013\"][hide=\"My Thoughts on the situation\"][hide=\"2)\"]The question reads [quote=\"SoxFan34\"]How did they know the person had been killed?[/quote]Uhhh...maybe because he's lying there dead? :rotfl: [/hide][/hide][/quote]\r\n\r\nnice thought on #2 :rotfl: ...but the person can be dead even if it was suicide. :ninja:",
  "Solution_5": "[hide=\"2\"]\nIf it had been a suicide, the tape would never have stopped. The fact that it was not only stopped but also [b]rewound[/b] indicates that someone else was involved.\n[/hide]",
  "Solution_6": "Contributing to the first one:\r\n\r\n[hide] Perhaps there were only two copies of the book in the entire world, and he had both...so by destroying one, he made the other significantly more valuable?[/hide]\nOr\n[hide]In the form of soot, it turns out to be more valuable than in its original form[/hide]",
  "Solution_7": "[hide=\"1\"]the book was worth and insured for 40,000 dollars but he payed a lot less for it[/hide]",
  "Solution_8": "I made a topic exactly like number 2 with the tape recorder[/hide]",
  "Solution_9": "[hide] \n\nK81o7's first solution is right; the man destroys it because he has two, and by only having one, the value goes way up.\nKstan, mathmeister22, youre both wrong, but that's partly my fault... :rotfl:   Mathnerd is right: If he'd commited suicide, then the tape recorder wouldve kep going. It was Rewound. [/hide]",
  "Solution_10": "16 year bump"
}
{
  "Tag": [
    "AMC",
    "AMC 8"
  ],
  "Problem": "I took the AMC 8 on tues. and I was wondering when we get our \"Offical\" scores. Does anyone know?",
  "Solution_1": "My teacher says that our official scores don't come out until like Late January/Early February.",
  "Solution_2": "Oh they come out normally around Dec. 10th.",
  "Solution_3": "The AMC office will begin mailing official scores and reports in early to mid-December, precise start of mailing depending on the scoring and processing.  Most schools in the US should have results before the end-of-year holidays.\r\n\r\nPlease also read Section VI, Report of Results, page 5 in the 2009 AMC 8 Teachers' Manual,\r\nhttp://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC8/2009-amc8TM.pdf\r\n\r\nSteve Dunbar\r\nDirector, MAA American Mathematics Competitions",
  "Solution_4": "Since the teachers are allowed to open the solutions on November 25, and they can keep the actual test booklet, aren't we able to know unofficial results as early as next week? I remember circling every answer on the test booklet, so that should be enough, right?",
  "Solution_5": "yes\r\nwhen I took the AMC8, my math coordinator checked the answers on our test booklets with the official answers, so we had the unofficial results within a week",
  "Solution_6": "is that even allowed?",
  "Solution_7": "[quote=\"AlphaBetaTheta\"]is that even allowed?[/quote]\r\n\r\nYes, if you only discuss within the school. My teacher did that too.",
  "Solution_8": "[quote]is that even allowed?[/quote]\r\n\r\nTeachers' Manual, Section VI, Item 4. page 5.  Pay attention to items 4.a. and 4.b. too.\r\n\r\nSteve Dunbar\r\nDirector, MAA American Mathematics Competitions",
  "Solution_9": "ya my school also disussed",
  "Solution_10": "aren't the teachers unable to discuss the answers until next week.\r\ni mean she said that she couldn't tell us the answers yet because other people take it at different times.",
  "Solution_11": "i'm not certain, but don't all competitors take the test at the same time?\r\nthat's how we did it at my school, anyway\r\n\r\nthere is an AMC8B that some people may take, but it should be different enough from AMC8A that discussion of the answers should not affect the score of those taking AMC8B that much",
  "Solution_12": "I'm pretty sure they are the same test, but there is an alternate date for taking it or something like that, in case of conflicts.",
  "Solution_13": "really?\r\nwell, idk\r\nmy coordinator never ordered the second one =)\r\n\r\ni would think there would be some difference tho\r\notherwise, couldn't they just give one test with a range of dates to administer them?",
  "Solution_14": "There is only one AMC 8.",
  "Solution_15": "Go to the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC8/2009-10-AMC8-bro.pdf]AMC8 brochure[/url], the second page, at the top it says \"Administration\". That explains it a little.\r\n\r\nHow it works is there is a single test that is mailed to all the proctors. They are supposed to open them for the first time and give the tests to the students on Tuesday, November 17. If a school has a conflict or problem with that date, they can request special permission to administer the test on another date. Solutions (which are mailed to proctors with the tests) should not be opened until November 25, when everybody should have taken the test. Also, students may not discuss the test with anyone but the people who took the test at the same place and time (you signed something at the bottom of the answer sheet pledging that). On November 24th, the proctors mail the answer sheets back to the AMC headquarters, where the official grading is done. Since the actual question pamphlets don't have to be mailed, if the person who took the test circled all the answers he/she had in the test pamphlet, the proctor can compare the solutions with the circled answers and give an unofficial grade. The only time when this is different than the official grade is if you bubbled incorrectly or your proctor graded incorrectly."
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "quadratic formula",
    "special factorizations"
  ],
  "Problem": "I need help with the steps turning the standard form into the quadratic formula can someone please help me with the proof and the steps involved. Thanks",
  "Solution_1": "One way to prove the quadratic formula by completing the square - since the general composition of a quadratic equation is of the form $ ax^2 + bx + c$, we can divide both steps of the equation by $ a.$\r\n\r\nSo far we have:\r\n\r\n$ ax^2 + bx + c = 0$\r\n(Note that $ a\\neq0.$)\r\n$ x^2 + \\dfrac{bx}{a} + \\dfrac{c}{a} = 0$\r\n$ x^2 + \\dfrac{bx}{a} = - \\dfrac{c}{a}$\r\nCompleting the square, we add $ \\dfrac{b^2}{4a^2}$ to both sides of the equation.\r\nWe now have the following:\r\n$ x^2 + \\dfrac{bx}{a} + \\dfrac{b^2}{4a^2} = - \\dfrac{c}{a} + \\dfrac{b^2}{4a^2}$.\r\n$ (x + \\dfrac{b}{2a})^2 = \\dfrac{ - 4ac}{4a^2} + \\dfrac{b^2}{4a^2}$.\r\n$ (x + \\dfrac{b}{2a})^2 = \\dfrac{b^2 - 4ac}{4a^2}$\r\nTherefore, \r\n$ x + \\dfrac{b}{2a} = \\pm \\sqrt {\\dfrac{b^2 - 4ac}{4a^2}}$\r\n$ x + \\dfrac{b}{2a} = \\pm \\dfrac{\\sqrt {b^2 - 4ac}}{2a}$\r\n$ x = - \\dfrac{b}{2a}\\pm \\dfrac{\\sqrt {b^2 - 4ac}}{2a}$\r\nAdding the two like terms we get $ \\dfrac{ - b\\pm \\sqrt {b^2 - 4ac}}{2a}$.\r\n\r\nDid you understand my steps?  If you need any help, just ask.",
  "Solution_2": "Of course though, $ a\\neq0.$ Otherwise the expression wouldn't be a quadratic and you'd be dividing by 0. The rest is okay.",
  "Solution_3": "Oh right, forgot about that.  :)",
  "Solution_4": "You should have also explained why the step $ \\sqrt {4a^2} \\equal{} 4a$ is valid. Remember that in general $ \\sqrt {x^2} \\equal{} |x|$. Using a bit of casework ($ a < 0$ and $ a > 0$),you'll notice that regardless of the sign of $ a$, the final expression will ultimately remain the same.",
  "Solution_5": "Which step are you talking about?  $ \\sqrt{4a^2} \\neq 4a$.",
  "Solution_6": "He meant $ \\sqrt{4a^2}\\equal{}2a$. Also, remember that since were working with variables, $ a$ could be negative."
}
{
  "Tag": [
    "inequalities",
    "algebra",
    "polynomial",
    "calculus",
    "2-variable inequality"
  ],
  "Problem": "$a,b$ are non negative reals not both equal to $0$. Prove that\n\\[\\frac{a^4+b^4}{(a+b)^4}  + \\frac{\\sqrt{ab}}{a+b}   \\geq \\frac58\\]",
  "Solution_1": "Without loss of generality we may assume that $a+b=1$. We must prove that $a^4+b^4+\\sqrt{ab} \\geq \\frac58$.\nOn the boundary $a=0$ the inequality is trivial. Than by the Lagrange principle there exists $\\lambda$ such that the following equalities hold:\n\\[4a^3+\\frac12\\sqrt{\\frac{b}a}=\\lambda, \\  \\ \\ \\ 4b^3+\\frac12\\sqrt{\\frac{a}b}=\\lambda\\]\n\nthen $8(a^3-b^3)=\\frac{a-b}{\\sqrt{ab}}$. So $a-b=0$ (then $a=b=\\frac12$ and inequality become to equality), or $8(1-ab)\\sqrt{ab}=1$. It is easy to see that \n$\\sqrt{ab} \\in \\left(0;\\frac12\\right)$, but there is no root of the polynomial $t^3-t+\\frac18$ in that interval.",
  "Solution_2": "Your proof is good, Fedor.\r\nI am thinking about a proof without calculus but I don't know if it exists.",
  "Solution_3": "We can prove a stronger result: \n\\[\\frac{a^4 + b^4}{(a+b)^4} +  \\frac{3ab}{(a+b)^2}  \\geq \\frac78 \\ \\ (1)\\]\n(by an easy computation it is equivalent to $(a-b)^4  \\geq 0$.\n\nOur desired inequality now follows from $(1)$ because  \n\\[ \\frac12 \\geq \\frac{\\sqrt{ab}}{a+b}  \\geq \\frac{2ab}{(a+b)^2}\\]",
  "Solution_4": "Congratulation Namdung, your proof is very nice and very neat.\r\nA beautiful reasoning!!"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "MATHCOUNTS"
  ],
  "Problem": "Figure A is a single wooden cube that has edge length 3 meters.  Square holes with side length 1 meter centered in each face are cut through to the opposite face.  The edges of the holes are paralell to the edges of the cube.  Figure B is formed by connecting 27 of these cubes.  What is the number of square meters in the total surface area of Figure B?\r\n\r\nI don't have the original diagram, but if anyone would like to post it, feel free to do so. :lol: \r\nI've never understood this question for some reason, though the rest of the questions from that year are EASY.",
  "Solution_1": "this problem isn't particularly difficult...im pretty sure aops vol. 1 has a diagram. was this a target?\r\nanyways, basically just imagine a cube with a hole in each center square on each face..the way i do it is i find the surface area of the outside faces first (which in this case is $ (9\\minus{}1)\\cdot 6$ then imagine what the inside of the cube looks like...that should help.  :)",
  "Solution_2": "But wouldn't that only apply for Figure A?  Figure B's 27 smaller squares would have some of the faces covered by other squares, right?",
  "Solution_3": "Im assuming that fig B is a $ 3 \\times 3 \\times 3$ of Fig A.\r\nThus you can see the interior surface from the holes is always counted.\r\nThere are 54 faces of Fig A on the cube.\r\n\r\ninterior surface area $ 8*27\\equal{}216$\r\nExterior $ 54*8\\equal{}432$ \r\nSurface Area$ \\equal{}216\\plus{}432\\equal{}648$",
  "Solution_4": "firgure B has 9*6=54 faces of figure A on it. each face of figure A has a surface area of 9+4=13. then do 54*13=702. \r\n\r\n[b]frankel-[/b]where did you get you interior surface area?",
  "Solution_5": "[quote=\"abacadaea\"]firgure B has 9*6=54 faces of figure A on it. each face of figure A has a surface area of 9+4=13. then do 54*13=702. \n\n[b]frankel-[/b]where did you get you interior surface area?[/quote]\r\n\r\nFigure A has holes in it.",
  "Solution_6": "abacadaea, that's what I got, but the answer's [hide]1080[/hide] :huh:",
  "Solution_7": "Ahh I miscounted.\r\nThe exterior surface is the same.\r\nThe interior surface of Fig A is $ 27*24\\equal{}648$\r\nTotal Surface Area$ \\equal{}648 \\plus{}54*8\\equal{}1080$",
  "Solution_8": "[quote=\"Fraenkel\"][quote=\"abacadaea\"]firgure B has 9*6=54 faces of figure A on it. each face of figure A has a surface area of 9+4=13. then do 54*13=702. \n\n[b]frankel-[/b]where did you get you interior surface area?[/quote]\n\nFigure A has holes in it.[/quote]\r\nyes i realize that. the surface area of each hole is 4 more thean the side would be without a square. then there would be 5*54+8*54\r\n\r\nEDIT: you beat me to the post i probably misinterpreted the diagram",
  "Solution_9": "I think this was #6 on a handout (I vaguely remember looking over some kids on my team's paper...and they were a bit confused since they got it wrong)\r\n\r\nso just do\r\n\r\n[hide]\n6(72) for outer part + 27(24) for inner part\n\n432+648=1080\n\nso the 72 is just each side with 9 'blocks' of 8 squares in each so 72 (and 6 of them so 1080)\n\nand the 24 is just the surface area for each 'small' cube (then you just multiply by 27 because all the holes are connected, so it doesn't decrease the area amount)\n\nthere's probably an easier way  :oops: [hide]\n\nhope that helps![/hide][/hide]",
  "Solution_10": "I don't understand.  How did you get 24?  I keep getting 30 for the surface area of the small cube.    I think that there is a 1x1x1 cube in the center of figure A, so it's faces should be included in the surface area, right? :blush:",
  "Solution_11": "[quote=\"Math Geek\"]I don't understand.  How did you get 24?  I keep getting 30 for the surface area of the small cube.    I think that there is a 1x1x1 cube in the center of figure A, so it's faces should be included in the surface area, right? :blush:[/quote]\r\n\r\nok so the surface area for the inner part of a 3x3x3 cube is 24 for the following reason (imagine this...). alright, so you have 6 'tunnels' each from a different side of the cube, and in these tunnels you have open space surrounded by 4 1x1 'walls', thus the surface area of each of these is 4...multiply that by 6 and you have 24. the reason why you don't have any surface area in the middle is because it's empty space surrounded by those 'tunnels' (aka no 'walls' -> no surface area)...so 24 is it...hope that helps \r\n(and you multiply by 27 because none of the inner surface area is blocked off by other 3x3x3 cubes)\r\n\r\ni hope my visual (without being visual) explanation was ok\r\n\r\nmaybe an experienced mathcounts person can help on the image if you still don't understand",
  "Solution_12": "im lost could someone post a diagram?",
  "Solution_13": "It's a bit hard to post any more than this...\r\n\r\n(i'm attempting to attach it)",
  "Solution_14": "oh whoops. i thought there was a little indent on each side of a figure A",
  "Solution_15": "so, umm, the answer is 1080?  that's what i got.....i get dizzy when i see this problem.   :D",
  "Solution_16": "If you read some of the other entries, you would know that the answer is 1080. Please don't revive  :wink:"
}
{
  "Tag": [],
  "Problem": "how do we factor this? \r\n\r\nx^2 - 7x + 12\r\n\r\nand\r\n\r\nx^2 - 7x - 60\r\n\r\n\r\nThank you.",
  "Solution_1": "[quote] Knowledge is power. [/quote]\r\n\r\nYou are absolutely right, thus you'd better do your homework by yourself.  :P \r\n\r\nPierre.",
  "Solution_2": "Factors of 12 that add up to -7, (-3 and -4) (x-3)(x-4)\r\nFactors of -60 that add up to -7, (-12 and 5) (x-12)(x+5)",
  "Solution_3": "i would do this by: \r\n\r\n1.  What multiplie to 12\r\n     and adds to -7\r\n well -4 and -3 do.\r\nso it would be (x-4)(x-3)\r\n\r\n2.what multiplies to 60\r\n   and adds to -7\r\nwell, -12 and 5 do.\r\nso it would be (x-12)(x+7)",
  "Solution_4": "Just like me  :lol:",
  "Solution_5": "[quote=\"nat mc\"]Just like me  :lol:[/quote]\r\nyea, i'm just trying to get better at typing.",
  "Solution_6": "hey good job on trying!  :lol:"
}
{
  "Tag": [
    "function",
    "induction"
  ],
  "Problem": "Let the function $f: \\mathbb{R}\\to\\mathbb{R}$ satisfy:\r\n$f(1)=2007$ and $f(1)+f(2)+\\cdots+f(n)=n^2f(n), \\forall n \\in \\mathbb{N}$.\r\nFind $f(2006)$.",
  "Solution_1": "It is similar to British Mathematical Olympiad, 1996, question 2.\r\n[hide]\n$f(1)+f(2)=2^2f(1)$ implies $f(2)=\\frac{1}{(2^2-1)}\\cdot f(1)$.\nNext,\n$f(3)=\\frac{1}{3^2-1}\\cdot (f(1)+f(2))=\\frac{1}{3^2-1}\\cdot [ f(1) + \\frac{1}{2^2-1}] \\cdot f(1) \\\\ = \\frac{1}{2^2-1}\\cdot \\frac{2^2}{3^2-1} \\cdot f(1)$\n$f(4) = ........$\nGood luck!\n\n[/hide]",
  "Solution_2": "$f(1)+f(2)+\\cdots+f(n)=n^2f(n)$\r\n$f(1)+f(2)+\\cdots+f(n)+f(n+1)=(n+1)^2f(n+1)$\r\nBy subtracting first equation from the second we obtain:\r\n$f(n+1)=(n+1)^2f(n+1)-n^2f(n)$\r\n$f(n+1)=\\frac{n}{n+2}f(n)$\r\nBy induction we have:\r\n$\\boxed{f(n)=\\frac{4014}{n(n+1)}}$",
  "Solution_3": "omg...its the OMK2006 question!",
  "Solution_4": "whats OMK?",
  "Solution_5": "Olympiad Matematik Kebangsaan. Means Malaysia National Mathematic Olympiad"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "LaTeX"
  ],
  "Problem": "we drop small balls (can considered to be points) on a concave half sphere (that is the half sphere is open upwards) of radius $ R$. We drop the balls from height $ h$ with respect to the point it hits in the half sphere. find $ h$ so that all the balls will bounce after one single fully elastic collision to the very bottom point of the sphere.",
  "Solution_1": "Probably i misunderstood the problem, anyway, let's try..\r\nThe balls hit the half sphere in the point P and rebounce elastically in the direction of the radius (OP). The velocity is [b]v[/b] with an angle of [b]90 - t[/b] relative to the horizontal axis (look at the picture!). Now i consider the parabolic motion of a single ball and i use the parametric equation.\r\n\r\n[i]y = y(0) + tg(90 - t) * x - (g*x^2)/(2*(v*cos(90-t)^2)[/i]\r\n\r\nThe ball has a horizontal distance from the bottom point H equal to [b]R*sin(t)[/b], while it stands at the height [b]R*(1 - cos(t))[/b]. Setting y=0 and x=[b]R*sin(t)[/b], we calculate the velocity v\r\n\r\n0 = R(1-cos(t)) + R sin(t)*cos(t)/sin(t) - (g*(Rsin(t))^2)/(2*(v*sin(t))^2)\r\n\r\n1 - cos (t) + cos(t) - g*R/(2*(v^2)) = 0\r\n\r\nv^2 = g*R/2         (1)\r\n\r\nThe ball is dropped from the height h with respect to point P, with an inizial velocity=0, so\r\nv^2 = 2*g*h\r\n\r\nFrom the (1) we have:  h= R/4\r\n\r\nSorry, but i haven't latex yet...let me know if it's right, even if i think as i understood the problem is quite easy..\r\nenjoy :D",
  "Solution_2": "That's not quite right.\r\n\r\nAfter elastic collision the angle with the normal (t) is the same. So if that is correct what you say, then t=90-t which is clearly wrong...  :wink:",
  "Solution_3": "so the balls don't fall in the normal direction...well, i considered the balls' direction after they hit the sphere as the direction of reflected rays...\r\ni'm not sure of what u say, the direction should be normal to the surface, so radial!\r\nCan you post a picture? If the direction is the same before and after they hit the sphere, we only need to substitue t to 90-t...\r\nsee u",
  "Solution_4": "ok here is a rough diagram  :) \r\n\r\nso after the collision the angle of the velocity with the radius is $ \\phi$ hence the angle of the velocity with the horizontal is $ 90\\minus{}2\\phi$.",
  "Solution_5": "ok!! Now it's clear..the ball rebounces with the same angle relative to the radial direction, because the collision is elastically (i didn't understood how toe express this fact)\r\nSubstituing 90 - 2t in my wrong solution, we have:\r\n\r\nh= R/(8*cos(t)*(2cos(t)-1))\r\nor something like that...\r\nlet's see if u can post harder problems :lol:  :P",
  "Solution_6": "yes thats correct :) \r\n\r\nWell I can post harder problems, check my previous ones. some of them are really tough. Anyway I will post another hard one if you wish :)"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "geometry",
    "rhombus",
    "national olympiad"
  ],
  "Problem": "\\[\\text{Day 1}\\]\r\n1. Polynomial $P(x)$ has integer coefficients. Prove, that if polynomials $P(x)$ and $P(P(P(x)))$ have common real root, they also have a common integer root.\r\n\r\n2. $ABCDE$ is a convex pentagon and:\r\n\\[BC=CD, \\;\\;\\; DE=EA, \\;\\;\\; \\angle BCD=\\angle DEA=90^{\\circ}\\]\r\nProve, that it is possible to build a triangle from segments $AC$, $CE$, $EB$. Find the value of its angles if $\\angle ACE=\\alpha$ and $\\angle BEC=\\beta$.\r\n\r\n3. We are given equilateral triangle with side $n$ which is divided into $n^{2}$ congruent equilateral triangles with side $1$. Let's call these small triangles plates. Each plate is either black or white. We perform the following operation:\r\nwe choose a plate which has common sides with at least two plates of diiferent colour from chosen plate and then we change the colour of our plate. For every $n\\geq 2$ decide whether exists a configuration of colours that enables to perform infinite sequence of operations.\r\n\\[\\text{Day 2}\\]\r\n4. $a$, $b$, $c$, $d$ are positive integers and\r\n\\[ad=b^{2}+bc+c^{2}\\]\r\nProve that\r\n\\[a^{2}+b^{2}+c^{2}+d^{2}\\]\r\nis a composed number.\r\n\r\n5. A convex quadrilateral $ABCD$ with $AB\\neq CD$ is cyclic. Quadrilaterals $AKDL$ and $CMBN$ are rhombs with side $a$. Prove that quadrilateral $KLMN$ is cyclic.\r\n\r\n6. $a$, $b$, $c$, $d$ are positive real numbers satisfying the following condition:\r\n\\[\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}=4 \\]\r\nProve that:\r\n\\[\\sqrt[3]{\\frac{a^{3}+b^{3}}{2}}+\\sqrt[3]{\\frac{b^{3}+c^{3}}{2}}+\\sqrt[3]{\\frac{c^{3}+d^{3}}{2}}+\\sqrt[3]{\\frac{d^{3}+a^{3}}{2}}\\leq 2(a+b+c+d)-4. \\]\r\nThanks robpal and barasawala  :wink:",
  "Solution_1": "1. http://www.mathlinks.ro/Forum/viewtopic.php?t=136487\r\n2. http://www.mathlinks.ro/Forum/viewtopic.php?t=136488\r\n3. http://www.mathlinks.ro/Forum/viewtopic.php?t=136490\r\n4. http://www.mathlinks.ro/Forum/viewtopic.php?t=136491\r\n5. http://www.mathlinks.ro/Forum/viewtopic.php?t=136489\r\n6. http://www.mathlinks.ro/Forum/viewtopic.php?t=135374",
  "Solution_2": "This is version pdf"
}
{
  "Tag": [],
  "Problem": "What's your favorite fruit?\r\n\r\nSee Fav. fruit 1 and Fav. fruit 3 if your favorite fruit is not here. The \"other\" option is in Fav. fruit 3. Please only vote in one of the polls.\r\n\r\nFav. fruit 1 is the main thread for discussion.\r\n\r\nSorry if your favorite fruit is bunched up in a category and you don't like the other fruits in that category.",
  "Solution_1": "*bump* for those who might not know about this and number 3..."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "function",
    "calculus computations"
  ],
  "Problem": "Hi, i have two problem that i cant resolve \"please help me\"  :P\r\n\r\n1.-  Change in temperature along a circunference\r\n\r\nA particle is moving in the direction of a clockwise around the circle of 1m radius centered at the origin. Rate of 2 m/s.\r\n\r\nHow fast the temperature experimented by the particle change in degrees Celsius per meter, in the point\r\n\r\n[url=http://imageshack.us][img]http://img17.imageshack.us/img17/7596/pointuo6.jpg[/img][/url]\r\n\r\n2.- I have to prove the wave equation:\r\n\r\n [url=http://img145.imageshack.us/my.php?image=waveecuationqm0.jpg][img]http://img145.imageshack.us/img145/2910/waveecuationqm0.jpg[/img][/url]\r\n\r\n\r\nThanks !",
  "Solution_1": "The first problem doesn't make any sense as written. I suspect that you just forgot half of the conditions.\r\n\r\nAs to the second, you cannot \"prove the wave equation\". I suspect that you were asked to prove that the given function is a solution to it. Then all you need to do is to differentiate a few times using the chain rule.",
  "Solution_2": "Thanks Fedja for the reply.\r\n\r\nThe first problem was taken from a book, it must be rigth :S \r\n\r\nThe only thing i could do is find the ecuation of the particle movement but i don know how to find a conection with the temperature ecuation.\r\n\r\nThe particle ecuation -> r(t) = sen (2t) i + cos (2t) j   \r\n\r\nThis ecuation move the particle in the direction of a clockwise around a circle of 1m radius centered at the origin with rate of 2 m/s.\r\n\r\n\r\nI have to find: \r\n\r\ndt = (Directional Derivative in the point) . df\r\n\r\nI think the direction of the derivative is the direction of the particle velocity in the point, but i dont know how to find 'df'\r\n\r\n\r\nAnd in the second problem i have to 'prove that the given function is a solution',  :P but when i tried to prove that i find some complications, if someone can teach me the steps to resolve that  :oops:"
}
{
  "Tag": [],
  "Problem": "What about if the question is like this:\r\n\r\n$ 1. 3x_{1}+5x_{2}+x_{3}+x_{4}= 10$\r\n$ 2. 3x_{1}+5x_{2}+x_{3}+x_{4}= 100$",
  "Solution_1": "What is your question? Would you like to solve this system in integers?\r\nIf I was not failed to understand the question it is extremely easy.\r\n\r\nSubtraction gives $ x_{1}=90$ and the system becames $ 5x_{2}+x_{3}+x_{4}=-107$. \r\nTake any $ x_{2}$ and $ x_{3}$ and calculate $ x_{4}$ from this.\r\nAnswer:\r\n$ x_{1}=90$\r\n$ x_{2}=u$\r\n$ x_{3}=v$\r\n$ x_{4}=-107-5u-v$\r\nfor arbitrary integers $ u$ and $ v$.",
  "Solution_2": "[quote=\"digger\"]What is your question? Would you like to solve this system in integers?\nIf I was not failed to understand the question it is extremely easy.\n\nSubtraction gives $ x_{1}=90$ and the system becames $ 5x_{2}+x_{3}+x_{4}=-107$. \nTake any $ x_{2}$ and $ x_{3}$ and calculate $ x_{4}$ from this.\nAnswer:\n$ x_{1}=90$\n$ x_{2}=u$\n$ x_{3}=v$\n$ x_{4}=-107-5u-v$\nfor arbitrary integers $ u$ and $ v$.[/quote]\r\n\r\nIts finding the number of non-negative solutions for each equation. the two are independent from each other. I just tried number 2 if its easy to solve given a higher sum. Can't seem to find an easy way to do it."
}
{
  "Tag": [
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "How can we compute $\\underset{k=0}{\\overset{+\\infty }{\\sum }}\\frac{(-1)^{k}x^{2k+1}}{(2k+1)k!}$ for a certain real $x$ using fourier series ? (if possible of course  :lol: )",
  "Solution_1": "This is $\\int_0^xe^{-t^2}\\,dt.$\r\n\r\nWe're not going to find a closed form formula; we know that $e^{-t^2}$ has no elementary antiderivative.\r\n\r\nThe limit as $x\\to\\infty$ is well known. If we want numerical approximations, for $x$ not too large, the power series itself converges satisfactorily rapidly. There's a useful asymptotic expression we could use for large $x.$\r\n\r\nI don't see a natural or straigtforward link to Fourier series.",
  "Solution_2": "Ok, I just wanted to know if there is a way to link it with fourier series, because my objectif was to compute that well known integral through it. But as it seems, it's not possible  :D \r\nThanks"
}
{
  "Tag": [],
  "Problem": "CuSO4(aq) + Fe(s) --> FeSO4(aq) + Cu(s)\r\nFeSO4(aq) + Cu(s) --> CuSO4(aq) + Fe(s)\r\n\r\n[img]http://img106.imageshack.us/img106/6140/chem3fp.png[/img]\r\n\r\nWhy does the iron plate becomes brown?\r\nI have no precise idea, but I think it might be related to the loss of electrons of the iron plate. Could someone help, please?",
  "Solution_1": "It is because the iron is going into solution and the copper is coming out of solution forming the brown you see on the iron plate.",
  "Solution_2": "CuSO4(aq) + Fe(s) --> FeSO4(aq) +  [b]Cu(s)[/b]\r\n\r\nthats why..."
}
{
  "Tag": [
    "algebra",
    "function",
    "domain"
  ],
  "Problem": "For what values of $x$ is this defined?\r\n\r\n$\\frac{23}{(x-1)(x+2)(x-3)(x+4)}$\r\n\r\nWrite in interval  notation if you can.",
  "Solution_1": "[quote=\"Drunken_Math\"]For what values of $x$ is this defined?\n\n$\\frac{23}{(x-1)(x+2)(x-3)(x+4)}$\n\nWrite in interval  notation if you can.[/quote]\r\n\r\n[hide]The only values not in the domain are -4,-2,1,3\n\nSo the values in this domain are $(-\\infty,-4)U(-4,-2)U(-2,1)U(1,3)U(3, \\infty)$.[/hide]",
  "Solution_2": "[quote=\"mathgeniuse^ln(x)\"][quote=\"Drunken_Math\"]For what values of $x$ is this defined?\n\n$\\frac{23}{(x-1)(x+2)(x-3)(x+4)}$\n\nWrite in interval  notation if you can.[/quote]\n\n[hide]The only values not in the domain are -4,-2,1,3\n\nSo the values in this domain are $(-\\infty,-4)U(-4,-2)U(-2,1)U(1,3)U(3, \\infty)$.[/hide][/quote]\r\n\r\nGood job!  :D"
}
{
  "Tag": [
    "trigonometry",
    "inequalities",
    "function",
    "calculus",
    "derivative",
    "geometry",
    "inequalities unsolved"
  ],
  "Problem": "prove that for x in (0,pi/2) we have sin(2x)<2/(3x-x^3)",
  "Solution_1": "can anybody tell me the proof of this inequality? and similar ones, that's sinx > a function of x. I have no idea how to prove this kind of inequality. Or are there some inequalities that are well-known to prove this kind of inequalities? Thx",
  "Solution_2": "would differentation help?  or one could try a geometrical approach to compare areas or side lengths.\r\n\r\nlike the proof of tan x > x > sin x and the proof of (tan x + sin x)/2 > x (i think the second's correct its an old hungarian problem)",
  "Solution_3": "x^3+2>=3x   and   x<pi/2<sqrt(3)\r\nwhich means that  2/(3x-x^3)>=1>=sin(2x)\r\nCases of equality are different so equality never holds."
}
{
  "Tag": [],
  "Problem": "Suppose you could uniformly heat a liquid in a container under the normal gravitational pull of the Earth. Assume that liquid exerts a pressure due to its own weight . Will the liquid start to boil on the bottom, at the top, or uniformly throughout? \r\n\r\nMy theory is that the liquid at the bottom must bear the extra pressure of not only atmospheric but also the water above it, so it boils at a higher temperature (b/c boiling point happens when vapor presure equals atmospheric pressure. So is my logic right? Or can someone explain this in a more clear way?",
  "Solution_1": "[quote=\"Bythecliff\"]Assume that liquid exerts a pressure due to its own weight[/quote]\n\nA liquid always exerts a pressure on its container. Are you taking into account in your argument the currents of convection?\n\n[quote=\"Bythecliff\"]boiling point happens when vapor presure equals atmospheric pressure[/quote]\r\n\r\nIt would be more correct to say \"... when vapour pressure equals the external pressure\", since, for example, when cooking in a \"pressure apparatus\" (I apologize, I don't know how to say this in English) the boiling doesn't happen when the vapour pressure equals atmospheric pressure since the external pressure is quite greater than atmospheric pressure - because the container is closed.",
  "Solution_2": "The English for that is \"pressure cooker.\"\r\n\r\nBythecliff's thought experiment might be difficult to do in practice, but if you could, the fluid [i]wouldn't[/i] convect. The ambient pressure would be lowest at the top, and boiling would happen first at the top. Furthermore, it wouldn't bubble. It wouldn't even look like it was boiling - but vapor would rise from it and the fluid level would drop.\r\n\r\nOur experience of boiling, with pots of water on the stove, involves situations in which all, or almost all, of the heat is added to the lower surface of the container. So the water boils from the bottom, bubbling and convecting furiously. Even a coffee cup stuck in a microwave oven isn't that close to the heat being added uniformly - the rate of heat transfer should be greater towards the outside of the cup and lower in the interior.",
  "Solution_3": "[quote=\"Kent Merryfield\"]The English for that is \"pressure cooker.\"[/quote]\n\nOk, thanks.\n\n[quote=\"Kent Merryfield\"]the fluid wouldn't convect. The ambient pressure would be lowest at the top[/quote]\r\n\r\nYes, I agree with that. With uniform heating, convenction don't make sense. The external pressure is lowest at the top, that's true, since in the interior of the liquid we have to take into account the hydrostatic pressure. However, note that the hydrostatic pressure contibution is neglegible: for example, for a column of water with 50 cm tall the pressure at the bottom would be about 1.05 atm, just 0.05 atm higher than atmospheric pressure. So, I think it would be also reasonable to consider that the liquid would start to boil is a very close uniform manner throughout the liquid."
}
{
  "Tag": [
    "irrational number"
  ],
  "Problem": "take two irrationals q1 and q2, with q1 does not equal q2. is it true that there is always a rational r1 such that q1<r1<q2? prove this if true.\r\n\r\nnew problem -\r\ntake two rationals x1 and x2, with x1 doesnt equal x2. is it true that there is always an irrational number i1 such that x1<i1<x2? prove this if true.",
  "Solution_1": "a) In the decimal representation of q_1 and q_2, go from right to left, and find the first time that a digit differs (this must happen, otherwise $ q_1\\equal{}q_2$). Take all of the digits to the right of q_2 and change them all to 0.\r\n\r\ne.g. 314.141285689... and 314.141285694... the first time they differ is 314.14128568 and 314.14128569. 314.14128569 is you number.\r\n\r\nb)$ i_1\\equal{}x_1\\plus{}\\frac{x_2\\minus{}x_1}{\\pi}$ it's clearly larger than x_1 and $ x_1\\plus{}\\frac{x_2\\minus{}x_1}{\\pi}<x_2$\r\n\r\n<-->$ \\frac{x_2\\minus{}x_1}{\\pi}<x_2\\minus{}x_1$ <--> $ 0<(x_2\\minus{}x_1)(\\pi\\minus{}1)$ which is true.",
  "Solution_2": "Both of these statements are true because,\r\n\r\n[i]The set of rational numbers is dense in the reals.[/i]\r\n\r\nHere's a proof for the first statement,\r\n\r\n[hide=\"Proof\"]\n(i) Let $ q_1,q_2\\in\\mathbb{R}: q_1<q_2$ and $ r\\in\\mathbb{Q}$.\n\n(ii) Therefore $ q_2\\minus{}q_1>0$ and $ \\frac{1}{q_2\\minus{}q_1}>0$.\n\nLemma: Let $ x\\in\\mathbb{R}$. Then there exists $ n\\in\\mathbb{N}$ such that $ n>x$.\n\n(iii) By the lemma, there exists $ n\\in\\mathbb{N}$ such that $ n>\\frac{1}{q_2\\minus{}q_1}>0$.\n\n(iv) Therefore, $ 0<\\frac{1}{n}<q_2\\minus{}q_1$\n\n(v) By the lemma, there exists $ m\\in\\mathbb{N}$ such that $ \\frac{m_1}{n}>q_1$.\n\n(Note: This is true because, by the lemma, there is an $ m_1\\in\\mathbb{N}: m_1>nq_1$. Dividing through by $ n$ yields $ \\frac{m_1}{n}>q_1$.)\n\n(vi) With the same logic as in (v), we know there exists $ m_2\\in\\mathbb{N}$ such that $ \\minus{}\\frac{m_2}{n}<q_1<\\frac{m_1}{n}$.\n\n(vii) Therefore, the set $ m_k\\in\\mathbb{Z}$ for which $ \\frac{m_k}{n}>q_1$ is non-empty, and bounded below. \n\n(viii) Thus is has a smallest element $ m$. Therefore, $ \\frac{m\\minus{}1}{n}<q_1$.\n\n(ix) $ \\frac{m}{n}\\equal{}\\frac{m\\minus{}1}{n}\\plus{}\\frac{1}{n}<q_1\\plus{}\\frac{1}{n}<q_1\\plus{}(q_2\\minus{}q_1)\\equal{}q_2$\n\n(x) We have found a value of $ m\\in\\mathbb{Z}$ for which $ \\frac{m}{n}>q_1$ and $ \\frac{m}{n}<q_2$.\n\nQ.E.D.\n[/hide]\r\n\r\nActually, what we have shown is more general. We have shown that there exists a rational number between any two distinct real numbers, which statement number one is a special case of. \r\n\r\nIf you want to see the proof of the lemma used in the proof, I will be glad to post it.",
  "Solution_3": "What I would suggest is switch $ q$ and $ r$, so $ r_1, r_2 \\in \\mathbb R$ and $ q \\in \\mathbb Q$. It just seems more natural that way. :)\r\n\r\nBecause when I go to the middle of the proof, when I see $ q_1$ and $ q_2$, I get confused and think \"Are they the rational numbers or the real numbers?\"\r\n\r\nAlso:\r\n[quote=\"JZambrano201\"]\n(v) By the lemma, there exists $ \\color{red} m_1\\in\\mathbb{N}$ such that $ \\frac {m_1}{n} > q_1$.\n[/quote]\r\n\r\nJust very minor tecnhicalities, the proof overall is great :D",
  "Solution_4": "Thanks for the insight. I kind of confused myself too :) . Trying to make a proof readable is almost as hard as the proof itself sometimes."
}
{
  "Tag": [
    "algebra",
    "binomial theorem",
    "complex numbers",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Show that if n is a positive multiple of 6,\r\n\r\n1) nCr(n, 1) - 3 nCr(n,3) + 3^2 nCr(n,5) - .... = 0\r\n\r\n2) nCr(n, 1) - 1/3 nCr(n,3) + 1/3^2 nCr(n,5) - .... = 0\r\n\r\nCombintorial argument or whatever cleverness is welcome.",
  "Solution_1": "Such statements can be proved using complex numbers. By binomial theorem it's quite obvious, taht the first one is equivalent to showing that the imaginary part of $(1+i\\sqrt 3)^n$ is equal to 0, which is true, because $(1+i\\sqrt 3)^6=2^6$, so if 6 divides n, the imginary part is 0. The second one is equivalent of showing that imaginary part of $\\left(1+i\\frac{\\sqrt 3} 3\\right)^n$ is equal to 0, \r\n$\\left(1+i\\frac{\\sqrt 3} 3\\right)^6=\\frac{2^6} {3^3}$, so the imaginary part is  0 again."
}
{
  "Tag": [
    "Harvard",
    "college",
    "calculus",
    "Princeton",
    "percent",
    "Stanford",
    "MIT"
  ],
  "Problem": "I am interested in majoring in computer science/mathematics.\r\n\r\nI have reasearching the top schools and I know that Harvard has a good math department. However, I was wondering if anyone could give me any information regarding the quality of Harvard's CS department.\r\n\r\nAlso, I am currently enrolled in a Community college and will earn an AS degree and high school diploma concurrently. Does anyone know weather or not I can transfer my credits? I have read the site, but the information seems a little vague.\r\n\r\nAny information would be greatly appreciated.",
  "Solution_1": "You cannot transfer community college credits to most of the highly selective elite colleges, such as Harvard. Are you planning to enter Harvard as a first-year undergraduate, or as a transfer student?",
  "Solution_2": "Tokenadult is 100% correct. I've talked to two Harvard representatives and they both told me the same thing. (:()\r\n\r\nA good idea might be to take the corresponding AP test to get the credit. That's what I did for Calculus BC and I had to do about 2 hrs of self-study and got a 5, and I plan to take a few sciences to get some credit also. You might need to pay some money out of your pocket depending on your school but I'm sure that it'll cost you less than a semester or two of college.\r\n\r\nGood luck!\r\n\r\nQuestion:\r\n\r\nIt would probably be at my disadvantage to apply to Harvard as a transfer from a community college rather than a highschool applicant, correct?",
  "Solution_3": "Odds of admission to most of the elite colleges are much lower for transfer applicants than for freshman applicants. Princeton claims to not accept transfer applications at all. So the smart way to take college-level courses before applying to an elite college is to take them as part of your \"high school\" program, which is something done by about 5 percent of the high school students in the United States. \r\n\r\n[url=http://www.ed.gov/news/pressreleases/2005/04/04062005a.html]High School Students Using Dual-Enrollment Programs[/url]",
  "Solution_4": "[quote=\"tokenadult\"]Odds of admission to most of the elite colleges are much lower for transfer applicants than for freshman applicants. [/quote]\r\n\r\nThat statistic is not a reliable indicator.   Transfer students are subject to different application procedures and the standards may be higher or (sometimes dramatically) lower than for the regular population.",
  "Solution_5": "What are successful examples of transfer applications known to readers of this thread that fit the pattern described by the original poster?",
  "Solution_6": "[quote=\"tokenadult\"]What are successful examples of transfer applications known to readers of this thread that fit the pattern described by the original poster?[/quote]\r\n\r\nThe original poster was talking about Harvard CS.  I know several successful transfer applicants to Harvard from community college (or state institutions comparable to community college),  including at least one in CS.    Whether or not they received one-to-one transfer credits, the transfer generally did not increase the number of semesters it took them to graduate.   Transfer of credits is judged on arrival by professors from the relevant departments.",
  "Solution_7": "Thank you for your information regarding transferability of credits. Now how about some information regarding the quality of Harvard's computer science department",
  "Solution_8": "I should say, having reviewed your first message, that the issue really is that if you do your college credits as part of a high school program, that will help you be more likely to be admitted, but you won't get any transfer credits at all. For sure. As long as you are getting a high school diploma as part of the program you are doing, you are a high school student in Harvard's eyes, irrespective of the associates degree you simultaneously earn. Advanced standing at Harvard is based on test scores, not on previous courses taken. \r\n\r\n[url=http://www.fas.harvard.edu/~fdo/publications/advancedstanding0607/]Advanced Standing at Harvard College[/url]  \r\n\r\nBut you will always be placed in appropriate courses, based on your background, and Harvard takes care to make sure that it's courses are challenging even for very well prepared students. \r\n\r\nAs for whether you should apply to Harvard if you are very interested in computer science, sure, why not? The traditional list of \"top five\" computer science undergraduate departments in the United States includes Stanford, MIT, Carnegie Mellon, and one or two other non-Harvard colleges that I'm sure someone will fill in here, but Harvard's department is not bad.",
  "Solution_9": "[quote=\"tokenadult\"] you will always be placed in appropriate courses, based on your background [/quote]\r\n\r\nThe top US universities, including Harvard,  tend to have plain vanilla placement exams to calibrate matriculants' level within the basic calculus, French, etc course sequences.   Apart from that and some perfunctory supervision by academic advisors, students can enroll in a nearly unrestricted set of classes.   Attempts to take a third-year PhD class by somebody just out of high school might require some permission, but the general philosophy, particularly at Harvard, was to \"give them enough rope to hang themselves\".   There has been a trend toward increased supervision in the past decade, but also increased awareness that some students enter with a specialized background in science and should be given relatively free rein. \r\n\r\nHarvard CS in the past had some notable faculty such as Rabin and Valiant, but otherwise was not a strong department.   More recently they invested a lot of money to upgrade the department, and the quality is closer to top-level now.",
  "Solution_10": "Harvard faculty is strong across the board, including CS, however, the CS dept is clearly not as strong as the biggies like MIT.  Perhaps the biggest difference at Harvard is that the CS dept is much much smaller than at the elite CS schools.  \r\n\r\nFrom an undergrad point of view, the biggest differences would be: (a) far fewer courses to choose from (although cross-enrollement at MIT is possible, of course), and (b) smaller, more intimate atmosphere.  You'll probably see the same people in your classes and know everyone who is concentrating in CS.  The students in the Harvard CS dept are probably as good as those anywhere.  (Again, because of numbers, you probably don't hear much about them.)\r\n\r\nAlso important is how sure you are that you want to do CS.",
  "Solution_11": "[quote=\"yenlee\"]Also important is how sure you are that you want to do CS.[/quote] \r\n\r\nYes, that is very important. Harvard offers many other majors (\"concentrations,\" in Harvard terminology) and most of those are academically strong and well regarded. Some other schools with notable CS departments have few other strong departments, so Harvard has value as a place where you can change your mind and still end up with a good major program."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Find all positive integers $ n$ such that the following inequality holds for positive reals $ a_{1},a_{2},\\cdots,a_{n}$:\r\n\r\n$ \\left(\\sum^{n}_{i=1}a^{2}_{i}\\right)\\left(\\sum^{n}_{i=1}a_{i}\\right)-\\left(\\sum^{n}_{i=1}a^{3}_{i}\\right)\\geq6\\;a_{1}a_{2}\\cdots a_{n}$",
  "Solution_1": "This problem purposed in Polish Mathematical Contests And: n=3.",
  "Solution_2": "[quote=\"navid\"]This problem purposed in Polish Mathematical Contests And: n=3.[/quote]\r\n\r\nwell nice Navid, would you please send you solution too! :wink:",
  "Solution_3": "If $ n>3$, then since the degree of right side is larger than 3 and both right side and left side is positive, for big positive real, the inequality flips. In the similar way, if $ n<3$, contradiction. So, the remain case is $ n=3$, and the inequality changes to $ \\sum_{sym}a_{1}^{2}a_{2}\\geq 6a_{1}a_{2}a_{3}$, which is true by muirhead."
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find a closed-form formula for the following sequence: $ a_n\\equal{}\\frac{2}{n}(a_1\\plus{}\\cdots\\plus{}a_{n\\minus{}1})\\plus{}kn$ (in terms of $ k$ and $ a_1$).",
  "Solution_1": "Obviosly $ na_n \\minus{}(n\\plus{}1)a_{n\\minus{}1}\\equal{}k(2n\\minus{}1),n>1$.\r\nLet $ f(x)\\equal{}\\sum_n a_nx^n$. Then we get $ f'(x)(1\\minus{}x)\\minus{}2f(x)\\equal{}a_1\\plus{}\\frac{1\\plus{}x}{(1\\minus{}x)^2}.$\r\nIt give $ f(x)\\equal{}\\frac{a_1(x\\minus{}x^2/2)\\minus{}kx\\minus{}2ln|1\\minus{}x|}{(1\\minus{}x)^2}$. It give closed form for $ a_n\\equal{}\\frac{f^{(n)}(0)}{n!}$.",
  "Solution_2": "That's not really a closed-form answer; otherwise any generating function would work as a closed-form answer\r\nfor a sequence. (Although the g.f. is often the best you can do...)",
  "Solution_3": "I think Rust's point is that this particular generating function has a Taylor series which is relatively straightforward to compute.",
  "Solution_4": "[quote=\"bambaman\"]Find a closed-form formula for the following sequence: $ a_n \\equal{} \\frac {2}{n}(a_1 \\plus{} \\cdots \\plus{} a_{n \\minus{} 1}) \\plus{} kn$ (in terms of $ k$ and $ a_1$).[/quote]\r\n\r\nI dont think there is a closed form for this sequence :\r\n\r\n$ \\forall n > 1$ : $ na_n \\equal{} 2\\sum_{i \\equal{} 1}^{n \\minus{} 1}a_i \\plus{} kn^2$\r\n$ \\forall n > 2$ : $ (n \\minus{} 1)a_{n \\minus{} 1} \\equal{} 2\\sum_{i \\equal{} 1}^{n \\minus{} 2}a_i \\plus{} k(n \\minus{} 1)$ and so $ (n \\plus{} 1)a_{n \\minus{} 1} \\equal{} 2\\sum_{i \\equal{} 1}^{n \\minus{} 1}a_i \\plus{} k(n \\minus{} 1)^2$\r\n\r\nSo, $ \\forall n > 2$ : $ na_n \\minus{} (n \\plus{} 1)a_{n \\minus{} 1} \\equal{} k(2n \\minus{} 1)$\r\n\r\nSo, $ \\forall n > 2$ : $ \\frac {a_n}{n \\plus{} 1} \\minus{} \\frac {a_{n \\minus{} 1}}{n} \\equal{} k\\frac {2n \\minus{} 1}{n(n \\plus{} 1)}$ $ \\equal{} k(\\frac {3}{n \\plus{} 1} \\minus{} \\frac 1n)$\r\n\r\nAdding all these equalities (for index greater than 2), we get :\r\n\r\n$ \\forall n\\geq 3$ : $ \\frac {a_n}{n \\plus{} 1} \\minus{} \\frac {a_{2}}{3}$ $ \\equal{} k(\\sum_{i \\equal{} 3}^n\\frac {3}{i \\plus{} 1} \\minus{} \\sum_{i \\equal{} 3}^n\\frac 1i)$\r\n\r\n$ \\forall n\\geq 3$ : $ \\frac {a_n}{n \\plus{} 1} \\equal{} \\frac {a_{2}}{3} \\plus{} k(\\sum_{i \\equal{} 3}^n\\frac {3}{i \\plus{} 1} \\minus{} \\sum_{i \\equal{} 3}^n\\frac 1i)$\r\n\r\nAnd so, since $ a_2 \\equal{} a_1 \\plus{} 2k$ and simplifying the sums :\r\n\r\n$ \\forall n\\geq 3$ : $ \\boxed{a_n \\equal{} (a_1 \\minus{} 10k)\\frac {n \\plus{} 1}{3} \\plus{} 3k \\plus{} 2k(n \\plus{} 1)\\sum_{i \\equal{} 1}^n\\frac 1i}$\r\n\r\nAnd finding a closed form for this expression would mean finding a closed form for $ \\sum_{i \\equal{} 1}^n\\frac 1i$ ...",
  "Solution_5": "Hello Bambaman !\r\n\r\nSince you posted in the \"proposed and own problems\", you have the solution to this problem.\r\n\r\nCould you kindly give us this solution, or the hint to find the closed form since it seems most of us think there is no such closed form.\r\nOr at least tell us where we are wrong.\r\n\r\nThanks a lot.",
  "Solution_6": "Actually, I considered $ H_n$ as a legitimate element in a closed-form formula. My solution was along the lines of Rust's solution (he has some calculation mistakes, though). I was delighted to see your elegant solution!",
  "Solution_7": "OKi :)\r\nThanks for your answer"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "A jet fighter takes off at an angle of 27.0 degrees w/ the horizontal, accelerating at 2.62 m/s*s. The plane weighs 79,300 N. I need to find the thrust T of the engine on the plane and the lift force L exerted by the air perpendicular to the wings. \r\n\r\nShould i switch the axis around, and then the Force of the thrust = (2.62cos27-gsin27)m and L  equal to (2.62sin27+gcos27)m?",
  "Solution_1": "I'm not sure how you came to those numbers, could you explain? According to my calculations, it should be\r\n\r\n[hide]\n$T=ma+mg\\sin\\phi$, $L=mg\\cos\\phi$[/hide]"
}
{
  "Tag": [
    "modular arithmetic",
    "quadratics",
    "Vieta",
    "number theory",
    "relatively prime",
    "algebra"
  ],
  "Problem": "Find all pairs of positive integers $ a,b$ such that \\begin{align*} b^2 + b+ 1 & \\equiv 0 \\pmod a \\\\ a^2+a+1 &\\equiv  0 \\pmod b . \\end{align*}",
  "Solution_1": "First I will show that $ a\\ne b\\plus{}1$. If instead $ a\\equal{}b\\plus{}1$, then we would have\r\n\r\n$ b\\plus{}1\\equal{}a|b^2\\plus{}b\\plus{}1$\r\n$ b\\plus{}1|b(b\\plus{}1)\\equal{}b^2\\plus{}b$\r\n$ b\\plus{}1|(b^2\\plus{}b\\plus{}1)\\minus{}(b^2\\plus{}b)\\equal{}1$\r\n\r\nwhich is impossible since $ b\\ge 1$ and hence $ b\\plus{}1\\ge 2$.\r\n\r\nNow, I will show the problem is equivalent to $ a^2 \\plus{} a \\plus{} b^2 \\plus{} b \\plus{} 1\\equiv 0\\mod ab$.\r\n\r\nWe know that $ a$ and $ b$ must be relatively prime. Otherwise, $ \\gcd(a,b)|a^2 \\plus{} a$ so $ \\gcd(a,b)\\not| a^2 \\plus{} a \\plus{} 1$ if $ \\gcd(a,b)>1$. Now $ a^2 \\plus{} a \\plus{} b^2 \\plus{} b \\plus{} 1\\equiv 0\\mod a$ and $ a^2 \\plus{} a \\plus{} b^2 \\plus{} b \\plus{} 1\\equiv 0\\mod b$, and combining these, since $ a$ and $ b$ are relatively prime, gives $ a^2 \\plus{} b^2 \\plus{} a \\plus{} b \\plus{} 1\\equiv 0\\mod ab$.\r\n\r\nThis goes the other way to. If $ a^2 \\plus{} b^2 \\plus{} a \\plus{} b \\plus{} 1\\equiv 0\\mod ab$ then $ a^2\\plus{}a\\plus{}b^2\\plus{}b\\plus{}1\\equiv 0\\mod a\\implies b^2\\plus{}b\\plus{}1\\equiv 0\\mod a$ and $ a^2\\plus{}a\\plus{}b^2\\plus{}b\\plus{}1\\equiv 0\\mod b\\implies a^2 \\plus{} a \\plus{} 1\\equiv 0\\mod a$.\r\n\r\nSo now we just need to solve $ a^2\\plus{}a\\plus{}b^2\\plus{}b\\plus{}1\\equiv0\\mod ab$. Let\r\n\r\n$ a^2 \\plus{} a \\plus{} b^2 \\plus{} b \\plus{} 1 \\equal{} kab$\r\n\r\nfor some integer $ k$. Also, assume that $ a>b$ and that $ a,b>1$.\r\n\r\nThen\r\n\r\n$ a^2 \\minus{} (kb\\minus{}1)a \\plus{} b^2 \\plus{} b \\plus{} 1 \\equal{} 0$\r\n\r\nThis is a quadratic equation, so if $ a$ is a solution, then by Vieta's, $ a'\\equal{}kb\\minus{}1\\minus{}a\\equal{}\\frac{b^2\\plus{}b\\plus{}1}{a}$ is also a solution. Since $ b^2\\plus{}b\\plus{}1$ and $ a$ are postive, $ a'$ must be positive. Now (by assumption that $ a>b$),\r\n\r\n$ a'\\equal{}\\frac{b^2\\plus{}b\\plus{}1}{a}<\\frac{b^2\\plus{}b\\plus{}1}{b}\\equal{}b\\plus{}1\\plus{}\\frac{1}{b}$\r\n$ a'\\le b\\plus{}1$.\r\n\r\n(because $ b>1$ by assumption, we know that $ 0<\\frac{1}{b}<1$, but $ a'$ is an integer). Also, $ a'\\ne b\\plus{}1$ by above, so $ a'\\le b$. So $ (a',b)$ is another solution if we already had $ (a,b)$. Since $ a>b$ and $ a'\\le b$, our $ (a',b)$ solution is smaller. So, we know that we can always produce a smaller solution as long as $ a,b>1$ and $ a\\ne b$. Now let's look at the cases where either $ a\\equal{}b$ or one of $ a,b$ is $ 1$.\r\n\r\n$ a\\equal{}b$. We know that $ a$ and $ b$ are relatively prime so for $ a$ to equal $ b$ we must have $ a\\equal{}b\\equal{}1$.\r\n\r\n$ a\\equal{}1$ or $ b\\equal{}1$. If $ a\\equal{}1$ then $ b$ divides $ 1^2\\plus{}1\\plus{}1\\equal{}3$, so the only pairs with $ 1$ in them are $ (1,3)$ and $ (1,1)$.\r\n\r\nSo we have $ (1,1)$ and $ (1,3)$. Note than $ (1,3)$ can be reduced, as $ 5ab\\equal{}1^2\\plus{}1\\plus{}3^2\\plus{}3\\plus{}1$, so $ k\\equal{}5$ and $ a'\\equal{}kb\\minus{}1\\minus{}a\\equal{}5(1)\\minus{}1\\minus{}3\\equal{}1$, so $ (1,3)$ is reduced to $ (1,1)$. Thus, all pairs besides $ (1,1)$ can be reduced in this manner.\r\n\r\nSince this reduction process does not change $ k$, and $ k\\equal{}5$ for $ (1,1)$, we have $ k\\equal{}5$ for all solutions. So since all solutions can be reduced to $ (1,1)$ by $ (a,b)\\to (5b\\minus{}a\\minus{}1,b), (a>b)$ any solution can be obtained by the reverse procedure $ (a,b)\\to (5b\\minus{}a\\minus{}1,b), (a\\le b)$. Therefore if we define the sequence $ a_0\\equal{}1,a_1\\equal{}1$, and $ a_{n\\plus{}2}\\equal{}5a_{n\\plus{}1}\\minus{}a_{n}\\minus{}1$, all solutions $ (a,b)$ are consecutive terms of this sequence."
}
{
  "Tag": [
    "ratio",
    "puzzles"
  ],
  "Problem": "Can you generate Golden Ratio $\\phi$ using $4$ number of $4's$ .\r\n\r\nUse any mathematical operator you can, but use exactly $4$ number of $4's$.\r\n\r\ncan you?",
  "Solution_1": "[hide]\\[ 4^{-1/2} + \\sqrt{(4/4 + 4^-1)} \\][/hide]",
  "Solution_2": "[quote=\"agolsme\"][hide]\\[ 4^{-1/2} + \\sqrt{(4/4 + 4^-1)}  \\][/hide][/quote]\n\n[hide]your solution uses $1$ and $2$. Remember only $4$ number of $4$'s but any number of operators are allowed.\n\n$\\sqrt{4}$ can also be written as $4^{1/2}$, so this is OK.... but others are not valid.\n\n?[/hide]",
  "Solution_3": "[hide]$\\frac{\\sqrt{4}+\\sqrt{4!-4}}{4}$[/hide]"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hello,\r\nI want this problem again suggest.Maybe we found a good solution. The problem is:\r\n\r\nFind all $ f: \\mathbb{R} \\to \\mathbb{R}$ continuous functin with $ F(f(x))\\plus{}f(F(x))\\equal{}x^2 \\forall x \\in \\mathbb{R}$, where $ F(x)\\equal{}f'(x), F(0)\\equal{}0$",
  "Solution_1": "Perhaps you meant $ f(x) \\equal{} F'(x)$, $ F(0) \\equal{} 0$?",
  "Solution_2": "Yes! Sorry :blush:",
  "Solution_3": "So the problem is\r\nFind all $ f: \\mathbb{R}\\to \\mathbb{R}$ continuous finction such that $ F(f(x))\\plus{}f(F(x))\\equal{}x^2$ $ (\\forall) x\\in \\mathbb{R}$, where $ F'(x)\\equal{}f(x)$, and $ F(0)\\equal{}0$"
}
{
  "Tag": [],
  "Problem": "$ \\binom{3n}{0}\\plus{}\\binom{3n}{3}\\plus{}\\binom{3n}{6}\\plus{}...\\plus{}\\binom{3n}{3n}\\equal{}?$",
  "Solution_1": "It's been posted many times.  See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url] and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url]."
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "Show that every power of (\u221a2 - 1) can be written in the form \u221a(k+1) - \u221ak.",
  "Solution_1": "Induction? \r\n\r\n  Let $\\large{(\\sqrt{2}-1)^n\\,=\\,\\sqrt{k+1}-\\sqrt{k}\\,\\Longrightarrow{\\,(\\sqrt{2}-1)^{n+1}\\,=\\,(\\sqrt{2}-1)(%Error. \"sqty\" is a bad command.\n{k+1}-\\sqrt{k})\\,=\\,\\sqrt{2(k+1)}+\\sqrt{2k}-\\sqrt{(k+1)}-\\sqrt{k}\\,=\\,X}}$\r\n\r\n   See that $\\large{\\sqrt{a}+-\\sqrt{b}\\,=\\,\\sqrt{a+b+-2\\sqrt{ab}}}$ , then $\\large{X\\,=\\,\\sqrt{2(k+1)+k-2\\sqrt{2k(k+1)}}\\,-\\,\\sqrt{2k+(k+1)-2\\sqrt{2k(k+1)}\\,}}$\r\n\r\n        $=\\,\\sqrt{[3k-2\\sqrt{2k(k+1)}]+1}\\,-\\,\\sqrt{(3k-2\\sqrt{2k(k+1)}}\\,=\\,\\sqrt{k_1+1}-\\sqrt{k_1}$\r\n\r\n   Proved",
  "Solution_2": "[hide]\nLet $s_n$ be the $n$th power of $\\sqrt{2}-1$. We will proceed by induction. For the base case, $\\sqrt{2}-1 = \\sqrt{2}-\\sqrt{1}$ so that works. Suppose\n\n$s_n = a+b\\sqrt{2} = \\sqrt{k+1}-\\sqrt{k}$.\n\nThen $|a^2-2b^2| = 1$. So\n\n$s_{n+1} = (\\sqrt{2}-1)(a+b\\sqrt{2}) = (2b-a)+\\sqrt{2}(a-b)$.\n\nBut since $|(2b-a)^2-2(a-b)^2| = |2b^2-a^2| = |a^2-2b^2| = 1$, we know\n\n$s_{n+1} = (2b-a)+\\sqrt{2}(a-b)$ can be written as $\\sqrt{k+1}-\\sqrt{k}$ for some $k$ as well, completing the induction.[/hide]\r\n\r\nEDIT: Boo, someone beat me to it again."
}
{
  "Tag": [
    "function",
    "calculus",
    "integration",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Suppose that a function $ f$ is analytic throughout the finite plane except for a finite number of singular points $ z_1, z_2, \\ldots, z_n$. Show that\r\n\r\n$ \\text{Res}_{z\\equal{}z_1}f(z)\\plus{}\\text{Res}_{z\\equal{}z_2}f(z)\\plus{} \\cdots \\plus{} \\text{Res}_{z\\equal{}z_n}f(z)\\plus{}\\text{Res}_{z\\equal{}\\infty}f(z)\\equal{}0$.\r\n\r\nI don't see how to show this right now.  Any hints on where to go would be nice.  Thank you.",
  "Solution_1": "It's practically the definition of the residue at $ \\infty$. Consider the integral of $ f$ around a very large circle, which encloses all singularities.",
  "Solution_2": "This is not true. A counterexample is 1/z.\r\n\r\nPym.",
  "Solution_3": "Sorry, I did not notice that the residue at infinity is also included.\r\nThe reply of jmerry is of course correct.\r\n\r\nPym."
}
{
  "Tag": [
    "rate problems"
  ],
  "Problem": "let v(t) be the amount of water in a tank at time t. If the average rate of change of V between times t = 3 and t = 6 is 3, and the average rate between t = 6 and t = 9 is 4, what is the average rate of change of V between t =3 and t = 9?",
  "Solution_1": "Average rate = :Sigma: change /  :Sigma: time",
  "Solution_2": "[quote=\"blahblahblah\"]Average rate = :Sigma: change /  :Sigma: time[/quote]\r\n\r\nAre you allowed to do that, considering we are working with averages?",
  "Solution_3": "[quote=\"mikewd\"]Are you allowed to do that, considering we are working with averages?[/quote]\r\n\r\nYou know the average, though, and how many numbers you averaged, so you can multiply to get the sums and add the sums together, then divide by the time."
}
{
  "Tag": [
    "vector",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "The vertex $P$ of a rectangular block with edges $a,b,c$ is selected. Consider the plane passing through the vertices adjacent to $P$. Denote the distance of the plane from $P$ is $m$. Prove that \\[\\frac{1}{m^{2}}=\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\]",
  "Solution_1": "Put origin at P and x, y, z-axes along the edges a = PA, b = PB, c = PC. Equation of the plane (ABC) is $\\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}=1.$ Vector product of any 2 vectors parallel to (ABC) is perpendicular to (ABC).\r\n\r\n$\\vec{AB}\\times \\vec{AC}= (-a, b, 0) \\times (-a, 0, c) = (bc, ca, ab) = abc \\left(\\frac{1}{a},\\frac{1}{b},\\frac{1}{c}\\right)$\r\n\r\n$\\vec{PN}= k \\left(\\frac{1}{a},\\frac{1}{b},\\frac{1}{c}\\right) \\perp (ABC)$ for any $k \\neq 0$. If $N \\in (ABC),$ then $k \\left(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\right) =1,$ yielding k, and $m^{2}= \\vec{PN}\\cdot \\vec{PN}= k^{2}\\left(\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}\\right) = k,$ $\\frac{1}{m^{2}}= \\frac{1}{k}= \\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}.$",
  "Solution_2": "Thank for your help."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "$n$ distinct points are on a plane.\r\n\r\nProve that the number of pairs of points \r\n\r\nwhich have a unit distant is less than $2n\\sqrt{n}$.",
  "Solution_1": "Using cauchy will help you.",
  "Solution_2": "can the number of pairs of points which have a unit distance can be more than $2n-3$ (for $n\\geq 3$)?"
}
{
  "Tag": [],
  "Problem": "Question attached.\r\n\r\nI know that it involves simple stoichiometry, but this is the first time I am applying my knowledge of stoichiometry on redox reactions (and the sort), and it is overly complicated for me to comprehend.\r\n\r\nPlease provide an outline of what to do, and why each step is done, so I may solve similar problems in the future.\r\n\r\nThanks",
  "Solution_1": "The problem gives you an equation as balanced as you need it, and asks you to do the stochiometry calculations. The fact that there is oxidation-reduction does not matter once you have a balanced equation. They are not asking for grams and making you use molecular weights. There is no change of units, just moles, liters and mililiters. It is an easy problem if you can interpret the meaning.\r\n\r\nThe problem is in the wording. I'll try to translate.\r\nIt seems like they used the same 25.00 mL sampling through the whole procedure.\r\nThe first titration was to determine the moles of $ Fe^{+2}$ ion.\r\nThe second titration was to determine the total moles of $ Fe$ (found all as $ Fe^{+2}$ after the zinc treatment).\r\nYou are only concerned with the titrations. How al the iron was reduced to $ Fe^{+2}$ does not matter. What matters is that it was all reduced to $ Fe^{+2}$ before the second titration.\r\nIn the titration each mole of permanganate oxidizes 5 moles of $ Fe^{+2}$ ion. (Each milimole of permanganate oxidizes 5 milimoles of $ Fe^{+2}$ ion).\r\nYour permanganate titrant contains 0.0200 moles/L (0.0200 mmoles/mL).\r\nHow many milimoles of permanganate titrant were used in the first titration?\r\n(0.0200 mmoles/mL) X (23.0 mL) = 0.46 mmoles\r\nHow many milimoles of $ Fe^{+2}$ ion were oxidized in the first titration?\r\n(5 mmoles $ Fe^{+2}$/1 mmol permanganate) X (0.46 mmoles permanganate) = 2.3 mmoles $ Fe^{+2}$\r\nWhat was the $ Fe^{+2}$ ion concentration if those milimoles were in a 25.0 mL volume?\r\n(2.3 mmoles ) / (25.0 mL) = 0.092 mmol/mL = 0.092 M = 92 mM\r\nThe second titration would be calculated the same way to give total $ Fe$.You can find the ${ Fe^[+3}$  concentration by difference.\r\n\r\nThere's no magic and no need to memorize formulas or multistep procedures. You only need to understand the problem. It's just a \"word problem.\" It helped to have been an analytical chemist back in the early seventies when all we could do was titrations, though."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c>0$ and $ a^2\\plus{}b^2\\plus{}c^2\\equal{}1$ prove that\r\n\\[ a\\plus{}b\\plus{}c\\plus{}\\frac{1}{abc}\\ge \\frac{4\\sqrt{3}}{9}(a\\plus{}b\\plus{}c)(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c})\\]",
  "Solution_1": "[quote=\"M.A\"]Let $ a,b,c > 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$ prove that\n\\[ a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc}\\ge \\frac {4\\sqrt {3}}{9}(a \\plus{} b \\plus{} c)(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})\n\\]\n[/quote]\r\nLet $ a\\plus{}b\\plus{}c\\equal{}3u,$ $ ab\\plus{}ac\\plus{}bc\\equal{}3v^2$ and $ abc\\equal{}w^3.$ Then your inequality is equivalent to\r\n$ uw^3\\plus{}3(3u^2\\minus{}2v^2)^2\\geq4uv^2\\sqrt{3u^2\\minus{}2v^2}.$\r\nBut by Schur we obtain: $ 2uw^3\\geq15u^2v^2\\minus{}9u^4\\minus{}4v^4.$\r\nThus, it remains to prove that $ \\frac{1}{2}(15u^2v^2\\minus{}9u^4\\minus{}4v^4)\\plus{}3(3u^2\\minus{}2v^2)^2\\geq4uv^2\\sqrt{3u^2\\minus{}2v^2}.$\r\nLet $ u^2\\equal{}tv^2.$ Then $ t\\geq1.$\r\nHence, $ \\frac{1}{2}(15u^2v^2\\minus{}9u^4\\minus{}4v^4)\\plus{}3(3u^2\\minus{}2v^2)^2\\geq4uv^2\\sqrt{3u^2\\minus{}2v^2}\\Leftrightarrow$\r\n$ \\Leftrightarrow45t^2\\minus{}57t\\plus{}20\\geq8\\sqrt{t(3t\\minus{}2)}\\Leftrightarrow(t\\minus{}1)^2(2025t^2\\minus{}1080t\\plus{}672)\\plus{}272(t\\minus{}1)\\geq0.$\r\nDone!",
  "Solution_2": "[quote=\"M.A\"]Let $ a,b,c > 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1$ prove that\n\\[ a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc}\\ge \\frac {4\\sqrt {3}}{9}(a \\plus{} b \\plus{} c)(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})\n\\]\n[/quote]\r\n\r\nI kill problem by mixing variable \r\n$ a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc}\\ge \\frac {4\\sqrt {3}}{9}(a \\plus{} b \\plus{} c)(\\frac {1}{a} \\plus{} \\frac {1}{b} \\plus{} \\frac {1}{c})$\r\n\r\n$ \\leftrightarrow a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc} \\minus{} \\frac {4\\sqrt {3}}{9}(\\frac {b \\plus{} c}{a} \\plus{} \\frac {a \\plus{} c}{b} \\plus{} \\frac {a \\plus{} b}{c})\\ge \\frac {4\\sqrt {3}}{3}$\r\n\r\nsetting $ f(a,b,c) \\equal{} a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc} \\minus{} \\frac {4\\sqrt {3}}{9}(\\frac {b \\plus{} c}{a} \\plus{} \\frac {a \\plus{} c}{b} \\plus{} \\frac {a \\plus{} b}{c})$\r\n\r\n$ \\rightarrow f(a,b,c) \\minus{} f(a,\\sqrt {\\frac {b^2 \\plus{} c^2}{2}},\\sqrt {\\frac {b^2 \\plus{} c^2}{2}})\\ge (b \\minus{} c)^2[\\frac {1}{abc(b^2 \\plus{} c^2)} \\plus{} \\frac {4\\sqrt {3}}{9a(b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)})} \\minus{} \\frac {1}{b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)}} \\minus{} \\frac {4\\sqrt {6(b^2 \\plus{} c^2)} \\plus{} 4\\sqrt {3}a}{9bc\\sqrt {2(b^2 \\plus{} c^2)}}]$\r\nwe needs prove that:\r\n\r\n$ \\frac {1}{abc(b^2 \\plus{} c^2)} \\plus{} \\frac {4\\sqrt {3}}{9a(b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)})} \\minus{} \\frac {1}{b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)}} \\minus{} \\frac {4\\sqrt {6(b^2 \\plus{} c^2)} \\plus{} 4\\sqrt {3}a}{9bc\\sqrt {2(b^2 \\plus{} c^2)}}]\\ge 0$\r\n\r\n$ A \\equal{} \\frac {4\\sqrt {3}}{9a(b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)})} \\minus{} \\frac {1}{b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)}} \\equal{} \\frac {4\\sqrt {3} \\minus{} 9a}{9a(b \\plus{} c \\plus{} \\sqrt {2(b^2 \\plus{} c^2)})}$\r\nassume that $ a \\equal{} min{a,b,c}$ $ \\rightarrow a\\le \\frac {1}{\\sqrt {3}}$  $ \\rightarrow 9a\\le 3\\sqrt {3} < 4\\sqrt {3}$ $ \\rightarrow A\\ge 0$\r\nthen, we needs proof\r\n\r\n$ B \\equal{} \\frac {1}{abc(b^2 \\plus{} c^2)} \\minus{} \\frac {4\\sqrt {6(b^2 \\plus{} c^2)} \\plus{} 4\\sqrt {3}a}{9bc\\sqrt {2(b^2 \\plus{} c^2)}}\\ge 0$\r\n\r\n$ \\leftrightarrow 4\\sqrt {6}a(b^2 \\plus{} c^2) \\plus{} 4\\sqrt {3}a^2\\sqrt {b^2 \\plus{} c^2} \\le 9\\sqrt {2}$\r\n\r\n$ 4\\sqrt {6}a(b^2 \\plus{} c^2) \\equal{} 4\\sqrt {6}a(1 \\minus{} a^2)\\ge 4\\sqrt {2}$ and \r\n\r\n$ 4\\sqrt {3}a^2\\sqrt {b^2 \\plus{} c^2} \\equal{} 4\\sqrt {3}a^2\\sqrt {1 \\minus{} a^2}\\le 4\\sqrt {3}a\\frac {1}{2}\\le 2$ (am-gm)\r\n\r\n$ \\rightarrow 4\\sqrt {6}a(b^2 \\plus{} c^2) \\plus{} 4\\sqrt {3}a^2\\sqrt {b^2 \\plus{} c^2} \\le 2 \\plus{} 4\\sqrt {2} < 9\\sqrt {2}$\r\n$ \\rightarrow B\\ge 0$\r\n\r\n$ \\rightarrow f(a,b,c)\\ge f(a,\\sqrt {\\frac {b^2 \\plus{} c^2}{2}},\\sqrt {\\frac {b^2 \\plus{} c^2}{2}})$\r\n\r\nwe needs prove that $ f(a,b,c)\\ge \\frac {4\\sqrt {3}}{3}$ with $ a \\equal{} \\sqrt {1 \\minus{} 2t^2}$ and $ b \\equal{} c \\equal{} t (t \\equal{} \\sqrt {\\frac {b^2 \\plus{} c^2}{2}})$\r\nDone!",
  "Solution_3": "there is a really simple solution to this one.\r\nhint:\r\nit also holds $ a\\plus{}b\\plus{}c\\plus{}\\frac{1}{abc}\\geq \\frac43 (\\frac1a\\plus{}\\frac1b\\plus{}\\frac1c)$.",
  "Solution_4": "[quote=\"Albanian Eagle\"]there is a really simple solution to this one.\nhint:\nit also holds $ a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc}\\geq \\frac43 (\\frac1a \\plus{} \\frac1b \\plus{} \\frac1c)$.[/quote]\r\nYou are right Albanian Eagle this is an easy problem :lol: \r\n\\[ a \\plus{} b \\plus{} c \\plus{} \\frac {1}{abc}\\geq \\frac43 (\\frac1a \\plus{} \\frac1b \\plus{} \\frac1c)\\Leftrightarrow\\]\r\n\\[ (a\\minus{}b)^4\\plus{}(b\\minus{}c)^4\\plus{}(c\\minus{}a)^4\\plus{}(a^4\\plus{}b^4\\plus{}c^4\\minus{}a^2bc\\plus{}b^2ca\\plus{}c^2ab)\\ge 0\\]\r\nso clear."
}
{
  "Tag": [
    "probability",
    "expected value",
    "probability and stats"
  ],
  "Problem": "Hi all again, here with another problem that I get:\r\nbuscar\r\n\r\nA medical office patients arrive at a constant rate of $ 1/10$ \t\r\npatients per minute, following a Poisson process. If the doctor does not begin to patients unless they meet in the waiting room has three such. What is the expected time for the doctor see the first patient? What is the probability that the doctor did not see \t\r\nany patient during the first hour?\r\n\r\n\t\r\nI've done this, point b): $ Pr[X<\\equal{}2] \\equal{} Pr(X\\equal{}0) \\plus{} Pr(X\\equal{} 1) \\plus{} Pr(X\\equal{}2)$ calculated with the Poisson distribution.\r\n\r\nThe a) does not come out yet, I appreciate any help.\r\n\r\n    \t\r\nGreetings and many thanks  :)",
  "Solution_1": "Just answer a simple question: \r\n\r\nFor Poisson process with intensity $ \\lambda$, what is the distribution of the times between arrivals?",
  "Solution_2": "hello, I think I did well like this:\r\n\r\nThe question what is the expected time for the doctor to see the first patient?\r\nMay arise as follows: what is the expected time until they reach three patients?,\r\n\r\n\t\r\nSince the arrival of patients is a Poisson process of rate \u03bb = 1 / 10 patients per minute the expected time till the third patient $ E(S_3) \\equal{} n/\\lambda \\equal{} 3/1/10 \\equal{} 30min$. $ S_n$ follows a Gamma distribution.",
  "Solution_3": "You're right, but you put too much mathematics into this problem. \r\n\r\nYes, $ S_3$ is Gamma distributed. \r\nBut to solve this problem, one needs not to know what is Gamma distribution. Times between arrivals are exponentially distributed, expected value of sum = sum of expected values = 3*1/$ \\lambda$ =30.",
  "Solution_4": "Ok. Thank you for your help.\r\n\r\nYou know anything about this?: \r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=298329\r\n\r\n\t\r\nthank you very much again and greetings",
  "Solution_5": "[quote=\"extranger\"]\nYou know anything about this?: \nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=298329\n[/quote]\r\nI'm not sure I know. I will be able to read your text only if you make the formulae readable.",
  "Solution_6": "Ok. I appreciate if suddenly I can help, especially with code R."
}
{
  "Tag": [],
  "Problem": "Anyone here ever play the game Rummy-Q, I think there is a lot of strategy involved there.",
  "Solution_1": "Is Rummy-Q the card game where you have like 13 cards and you have to try get them in order or something?",
  "Solution_2": "Something like that, but they are not cards, yeah u need like a combination of at least 3 things depending on color or order.",
  "Solution_3": "[quote=\"hello\"]Anyone here ever play the game Rummy-Q, I think there is a lot of strategy involved there.[/quote]\r\nAre you talking about Rummikub or is it something else?\r\n\r\nEDIT: I guess we're talking about the same thing :)",
  "Solution_4": "ive played Rummy-Q, or something very like it.  Your description sounds a lot like what i played."
}
{
  "Tag": [
    "geometry",
    "geometry proposed"
  ],
  "Problem": "Prove that in any convex polygon with $4n+2$ sides ($n\\geq 1$) there exist two consecutive sides which form a triangle of area at most $\\frac 1{6n}$ of the area of the polygon.",
  "Solution_1": "Hmm... I can't prove it for $n=1$  :blush: (Kuba said me that it is the hardest thing in this problem :P )\r\nAll over cases can be derived from $n=1$.\r\nLet's show it.\r\nSuppose the contrary, i.e. there are no triangle of area $\\leq 1/6n$.\r\nConsider $4n+2$-gon with $n\\geq 2$ and cut a 6-gon from it by mean of a diagonal ($AB$ on figure). Let's prove that the triangle of area $<1/6n$ can't adjoin the diagonal $AB$. Otherwise if $ABC$ is a such triangle then draw a line throw $A$ parallel to $BC$. We will necessary have $D$ and $E$ lies in the other halfplane than $BC$. So $4n+2$ is not convex.Therefore we can divide $4n+2$-gon into $n$ 6-gons by mean of diagonals and no triangle which adjoins these diagonals will have area $<1/6n$.",
  "Solution_2": "[quote=\"Myth\"]  Hmm... I can't prove it for n=1 (Kuba said me that it is the hardest thing in this problem  :P )  [/quote] In my humble opinion, it is not harder than the part done by Myth and it uses pretty much the same idea. Here is a little [hide=\"hint\"] Assume for a moment that the diagonals AD, BE and CF of a hexagon ABCDEF intersect at one point O. Then one of the six triangles into which they divide the hexagon has the area at most $1/6$ of the area of the hexagon. If it is, say, the triangle AOB, what can you say about the areas of ACB and AFB?  [/hide] Nice problem, by the way! :)",
  "Solution_3": "Nice idea, Myth! I can prove the n=1 case:\r\n\r\nIn hexagon ABCDEF, $P=AD\\cap CF$, $Q=CF\\cap EB$, $R=EB\\cap AD$. The six triangles APB, BPC, CQD, DQE, ERF, FRA cover less than the whole hexagon (if the three diagonals are not concurrent) or the whole triangle (if they are). In either case, one of those six triangles will have area $\\le \\frac{1}{6}(\\text{area})$, so say it is APB. Then one of AFB and ACB has area at most the area of APB, so we're done.",
  "Solution_4": "Sorry, I don't see it. \r\n\r\nIndeed, the case described by Fedor is clear. But common case... I am missing something.",
  "Solution_5": "I really don't understand what Myth did, especially the part:\r\n[quote=\"Myth\"]We will necessary have $D$ and $E$ lies in the other halfplane than $BC$.[/quote]\r\nAnd also, how exactly is the case $n>1$ derived from $n=1$?",
  "Solution_6": "If $E$ lies in the same halfplane, then $S_{EBC}\\leq S_{ABC}\\leq \\frac{1}{6n}$, so we find a small requared triangle $EBC$. Contradiction.\r\n\r\nThe general cases can be done in the following way: cut the polygon into $n$ hexagons by mean of diagonals. One of this hexagon have area $\\leq 1/n$ and we can cut a tringle of area $\\leq 1/6$ from it, therefore this triangle is at most $\\frac{1}{6n}$ of the area of the initial polygon. And we prooved that this triangle can't be \"inner\" triangle, so it is a required triangle.",
  "Solution_7": "okay. now i understand the general case.\r\n\r\n[quote=\"Myth\"]If $E$ lies in the same halfplane, then $S_{EBC}\\leq S_{ABC}\\leq \\frac{1}{6n}$, so we find a small requared triangle $EBC$. Contradiction.\n[/quote]\r\n\r\ni still don't understand the first part.\r\nWhy $S_{EBC}\\leq S_{ABC}\\leq \\frac{1}{6n}$ yields a contradiction?",
  "Solution_8": "Because we supposed the contrary, i.e. there is no such triangle.",
  "Solution_9": "Case n=1 is also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27331]here[/url]"
}
{
  "Tag": [
    "pigeonhole principle",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "A positive integer is [i]dapper[/i] if at least one of its multiples begins with $ 2008$. For example, $ 7$ is dapper because $ 200858$ is a multiple of $ 7$ and begins with $ 2008$. Observe that $ 200858 \\equal{} 28694\\times 7$.\r\n\r\nProve that every positive integer is dapper.",
  "Solution_1": "Let m be a positive integer. Let $ n \\equal{} [log_{10}{m}] \\plus{} 1$. Consider $ 2008 \\cdot 10^n \\plus{} a, 0 \\le a \\le m \\minus{} 1 ( < 10^n)$. These are m numbers that begin with 2008 and have a different residue when divided by m, thus one of them is a multiple of m. We conclude that all positive integers are dappers.",
  "Solution_2": "Consider the m following numbers:\r\n\r\n2008, 20082008, 200820082008, ..., 20082008...20082008\r\n\r\nThe i-th term have i 2008's.\r\nIf one of them is multiple of m, ok.\r\nSuppose that there aren't multiples of m.\r\nBy the Pigeonhole Principle, there are 2 numbers with the same remainder at the divison by m.So, take the difference beetween this 2 numbers.It's obviously that this new number is multiple of m and begins with 2008. (q.e.d.)\r\n\r\n\r\nCya people,\r\nSecco",
  "Solution_3": "let n < $ 10^{m}$\r\n\r\nlet us consider 2008 X $ 10^{m \\plus{} 1}$. let it give r as remainder when divided by n.\r\nthen 2008 X $ 10^{m \\plus{} 1}$ +n-r is certainly a multiple of n and hence proved"
}
{
  "Tag": [
    "trigonometry",
    "limit",
    "geometric series",
    "complex numbers",
    "geometric sequence"
  ],
  "Problem": "$\\displaystyle\\sum_{n=1}^{\\infty} 3^{n-1}\\cdot\\sin^3 \\left(\\frac{\\pi}{3^{n+1}}\\right)$\r\n\r\nI tried doing it by changing the sin expression into the Im part of $e^{\\frac{\\pi}{3^{n+1}}}$, then putting it into the inf geometric series formula and then taking the imaginary part of that at the end.\r\n\r\nHowever, I got $\\frac{\\sqrt{3}}{14}$ which is approximately .12, while Mathematica approximates this sum to 0.045293.  Anybody care to look at it?",
  "Solution_1": "Use $\\sin 3\\theta=3\\sin \\theta-4\\sin ^ 3 \\theta\\Longleftrightarrow \\sin ^ 3 \\theta=\\frac{1}{4}(3\\sin \\theta-\\sin 3\\theta),$ then telescoping method and $\\lim_{\\theta \\longrightarrow 0} \\frac{\\sin \\theta}{\\theta}=1.$",
  "Solution_2": "i am pretty sure that this is not a geometric sequence\r\n\r\n$\\Im \\left(\\sum_{n=1}^{\\infty} 3^{n-1}*e^{\\frac{\\pi i}{3^n}}\\right)$\r\n\r\nnotice that it would if it were $e^{n}$, instead of $e^{3^{n}}$, it would be a geometric sequence",
  "Solution_3": "The answer is $\\frac{1}{4}\\left(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{2}\\right).$",
  "Solution_4": "can you show exactly what you did. i am not able to completely follow.",
  "Solution_5": "I'm able to get everything except that $\\frac{\\pi}{3}$ term to come up.",
  "Solution_6": "$\\sum_{k=1}^n 3^{k-1}\\sin ^ 3 \\frac{\\pi}{3^{k+1}}$\r\n\r\n${=\\frac{1}{4}\\sum_{k=1}^n 3^{k-1} (3\\sin \\frac{\\pi}{3^{k+1}}-\\sin \\frac{\\pi}{3^k}})$\r\n\r\n${=\\frac{1}{4}\\sum_{k=1}^n (3^k\\sin \\frac{\\pi}{3^{k+1}}-3^{k-1}\\sin \\frac{\\pi}{3^k}})$\r\n\r\n$=\\frac{1}{4}(3^n\\sin \\frac{\\pi}{3^{n+1}}-\\sin \\frac{\\pi}{3})$\r\n\r\n\r\n$=\\frac{1}{4}(\\frac{\\pi}{3} \\frac{\\sin \\frac{\\pi}{3^{n+1}}}{\\frac{\\pi}{3^{n+1}}}-\\frac{\\sqrt{3}}{2})$\r\n\r\n$\\longrightarrow \\frac{1}{4}(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{2})\\ (n\\rightarrow \\infty)$",
  "Solution_7": "Ah, I see that now. Very inginuitive.",
  "Solution_8": "very nice!  :D"
}
{
  "Tag": [],
  "Problem": "This is an easy, in fact, very easy problem...\r\n\r\nI spent 3 hours this morning to find a better formula....\r\n\r\n1 coconut... 1st column\r\n\r\n3 coconuts   2nd column\r\n\r\n6 coconuts   3rd colum\r\n\r\n10 coconuts  4th colum\r\n\r\n\r\n\r\nnow find the number of coconuts in 100th colum....\r\n\r\nI just posted it because there weren't enough posts in this forum...",
  "Solution_1": "[hide=\"solution\"]\nformula is:\nn(n+1)/2\n\n100(100+1)/2\n100(101)/2\n5050\n(triangle numbers formula)\n[/hide]",
  "Solution_2": "[hide=\"to Treething\"]yup, you got it right, but I hid it so people could still try to solve the problem...[/hide]\r\n\r\n\r\n :lol:  ;)  :blush:  :P",
  "Solution_3": "[quote=\"shinwoo\"]This is an easy, in fact, very easy problem...\n\nI spent 3 hours this morning to find a better formula....\n\n1 coconut... 1st column\n\n3 coconuts   2nd column\n\n6 coconuts   3rd colum\n\n10 coconuts  4th colum\n\n\n\nnow find the number of coconuts in 100th colum....\n\nI just posted it because there weren't enough posts in this forum...[/quote]\r\n\r\n[hide]x(x+1)/2\n100*101/2=5050[/hide]",
  "Solution_4": "[hide]\nn(n+1)/2\n=100*101/2\n=5050[/hide]",
  "Solution_5": "[hide]5050[/hide]\r\n\r\n[color=darkred]<_<\n>_>\n\nNext time don't check the \"Disable BBCode in this post\" box.[/color]",
  "Solution_6": "yuppy do guys!!!\r\n\r\ngood work!! :D",
  "Solution_7": "[hide] 5050[/hide]"
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "trigonometry",
    "calculus computations"
  ],
  "Problem": "I'm getting a wildly different answer from the back of the book on this one, and looking at the solutions manual, it (the book) seems to be wrong. Can someone verify? I hope this is the appropriate forum.\r\n\r\nThe function $T(x)=\\frac{200}{1+x^2}$ represents the temperature in degrees Celsius percieved by a person standing $x$ meters from the fire. If the person moves away from the fire at 2 m/s, how fast is the temperature changing when the person is 5 meters away?\r\n\r\nThanks in advance :)",
  "Solution_1": "So we're talking temperature in relation to distance from fire. Then we're looking at a \"change in temperature\" when there's a \"change in distance\" at a certain distance...what does this shout to you? \"DERIVATIVE!!!!! DO ME!!!!!\" So we do it:\r\n\r\n$dT=\\frac{-400x}{(1+x^{2})^{2}}dx$\r\n\r\nAt $5$m away and moving at $2$m/s, we substitute $x=5$ amd $dx=2$ and we get $dT=-2.96$\r\n\r\n\r\nMy calculation might be wrong but I suck at arithmetic.",
  "Solution_2": "$\\frac{dT}{dx} = \\frac{-400x}{(1+x^2)^2} \\rightarrow \\frac{dT}{dt} = \\frac{dT}{dx}\\frac{dx}{dt} = \\frac{-400x}{(1+x^2)^2} \\frac{dx}{dt}$\r\n\r\nFrom the problem statement, $\\frac{dx}{dt} = 2 m/s$ and $x=5m$.  This should yield the same answer.  Note also that this formula is dimensionally incorrect.",
  "Solution_3": "[quote=\"FMako\"]Note also that this formula is dimensionally incorrect.[/quote]\r\n\r\nThen ... doesn't this mean the formula is incorrect...? Or am I completely misunderstanding what you meant there?  :blush:",
  "Solution_4": "FMako is talking about units of measurement whether stated or implied.\r\n\r\nThe biggest problem is that denominator, $1+x^2.$ We have that $x^2$ is the square of a distance; we measure it in $\\text{m}^2.$ What does that make $1?$ We can only add or subtract two quantities if they have the same units.\r\n\r\nIf we assume that the given $1$ is actually meant to be $1\\text{m}^2$ and the $200$ in the numerator is actually $200\\text{ deg}\\cdot\\text{m}^2$, then we can make the whole thing make sense.\r\n\r\nThat's not very satisfying. I like to use implied units of measurement - \"dimensional analysis\" - as a continuing check for the accuracy of computations, perhaps especially the intermediate steps. It's hard to get that kind of help from a formula like this.\r\n\r\nI'd be happier with $T=\\frac{k}{a^2+x^2}.$\r\n\r\nThen I read it as $a$ being a distance and $k$ being a constant of proportionality - and constants of proportionality carry whatever units they need to carry.\r\n\r\nOne little principle from this dimensional analysis outlook: suppose you see a formula like $e^{-kt}.$ Then if $t$ is a time and is measureed in units of measurement appropriate for a time, then $k$ must be the reciprocal of a time. Why? Because the argument of a \"transcendental function\" like the sine* or the exponential must be a pure unitless number.\r\n\r\n--\r\n\r\n* Footnote: an angle given in radians [i]is[/i] a \"pure unitless number.\""
}
{
  "Tag": [
    "trigonometry",
    "calculus",
    "ceiling function",
    "algebra",
    "polynomial",
    "induction",
    "complex analysis"
  ],
  "Problem": "Let $\\theta=\\frac{\\pi}{2n+1}\\/$ where $n$ is a positive integer.\r\nProve that \r\n1. $\\prod_{r=1}^{n} \\sin r\\theta =  2^{-n}\\sqrt{2n+1}\\/$\r\n2. $\\prod_{r=1}^{n} \\cos r\\theta = 2^{-n}\\/$\r\n\r\n(I don't know if it is suitable to post this question here :?  :?  :? )",
  "Solution_1": "i am sure that here is a elementary way and a complex analysis way to solve it. \r\nbut for calculus , \r\ni tried to break sin into  a series and add them up  , but got screwed up , \r\n\r\ncan anyone help ?",
  "Solution_2": "[quote=\"lisai\"]i am sure that here is a elementary way and a complex analysis way to solve it. \nbut for calculus \n[/quote]\r\nAny way is good as long as it yields a solution. So feel free to post the proof you know regardless of whether you consider it \"calculus\" or not. :)",
  "Solution_3": "[quote=\"pipi\"]Let $\\theta=\\frac{\\pi}{2n+1}\\/$ where $n$ is a positive integer. Prove that \n1. $\\prod_{r=1}^{n} \\sin r\\theta = 2^{-n}\\sqrt{2n+1}\\/$\n2. $\\prod_{r=1}^{n} \\cos r\\theta = 2^{-n}\\/$ [/quote]\r\n Denote $\\Theta= \\frac{\\pi}{2n+1}$, $P_1=\\prod\\limits_{k=1}^{n}\\sin{\\frac{k\\pi}{2n+1}},\\; \\; Q_1=\\prod\\limits_{k=n+1}^{2n}\\sin{\\frac{k\\pi}{2n+1}}$ and observe that $P_1>0 .$ Because $(*)\\hspace*{0.5cm} \\ds \\prod\\limits_{k=n+1}^{2n}T_k=\\ds \\prod\\limits_{k=1}^{n}T_{2n+1-k}\\; ,$ we have\r\n$Q_1=\\prod\\limits_{k=1}^{n}\\sin(2n+1-k)\\Theta= \\prod\\limits_{k=1}^{n}\\sin(\\pi-k\\Theta)=P_1\\; .$ Further we try to find $P_1^2=P_1Q_1=\\prod\\limits_{k=1}^{2n}\\sin{k\\theta}.$\r\n Let $z_k=\\cos{(2k\\theta)}+ i\\cdot\\sin{(2k\\theta)}.$ If we select $x=1$ in the  equalities $\\ds \\frac{x^{2n+1}-1}{x-1}=\\sum\\limits_{k=0}^{2n}x^k=\\ds \\prod\\limits_{k=1}^{2n}\\left(x-z_k\\right) ,$ we find\r\n\\[ (**)\\; \\; \\; \\; \\begin{array}{c}\\ds \\ds 2n+1=\\ds \\prod\\limits_{k=1}^{2n}2\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)\\sin{k\\Theta}=\\\\ =\\ds 2^{2n}P_1^2 \\prod\\limits_{k=1}^{2n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)=2^{2n}P_1^2\\; . \\end{array}  \\]\r\nThe last equality may be justified in the following way : \r\n\\[ \\begin{array}{c} \\prod\\limits_{k=1}^{2n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)=\\\\ =\\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)\\right)\\cdot \\left( \\prod\\limits_{k=n+1}^{2n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)\\right)=\\\\ =\\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)\\right)\\cdot \\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{(2n+1-k)\\Theta}-i\\cdot\\cos{(2n+1-k)\\Theta}\\right)\\right)=\\\\ =\\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)\\right)\\cdot \\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{(\\pi-k\\Theta)}-i\\cdot\\cos{(\\pi-k\\Theta)}\\right)\\right)=\\\\ =\\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{k\\Theta}-i\\cdot\\cos{k\\Theta}\\right)\\right)\\cdot \\left( \\prod\\limits_{k=1}^{n}\\left(\\sin{k\\Theta}+i\\cdot\\cos{k\\Theta}\\right)\\right)=1 \\; . \\end{array}  \\]\r\nNow (**) enables us to write $\\ds P_1=\\ds \\frac{\\sqrt{2n+1}}{2^n}\\; .$ \r\nFurther try to find $P_2: =\\prod\\limits_{k=1}^{n}\\cos{k\\Theta}$ in  following manner: observe that $P_2>0$ and  $P_1P_2=\\ds \\frac{1}{2^n}\\prod\\limits_{k=1}^{n}\\sin{2k\\Theta}$ . Consider further the product $Q_2: =\\ds\\prod\\limits_{k=n+1}^{2n}\\sin{2k\\Theta}$ ... \r\n[b]Remark:[/b] In above hint/solution, equality (*) has an important  role .",
  "Solution_4": "I have another idea.\r\nfor 2,I think you can use the conclusion posted  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40503]here[/url]\r\nfor1,you can find a solution similar way(but a little different)\r\nsorry for my short post,if who want to know more,i will try to explain again :)",
  "Solution_5": "Thanks for your solution, flip2004! \r\nI get the idea but I think there is something not allowed if we write:\r\n[quote=\"flip2004\"]\n Let $ z_k=\\cos{(2k\\theta)}+ i\\cdot\\sin{(2k\\theta)}. $ If we select $x=1$ in the  equalities $\\ds \\frac{x^{2n+1}-1}{x-1}=\\sum\\limits_{k=0}^{2n}x^k=\\ds \\prod\\limits_{k=1}^{2n}\\left(x-z_k\\right)$,...[/quote]\r\nthanks again.",
  "Solution_6": "[quote=\"zhaobin\"]I have another idea.\nfor 2,I think you can use the conclusion posted  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40503]here[/url]\nfor1,you can find a solution similar way(but a little different)\nsorry for my short post,if who want to know more,i will try to explain again :)[/quote]\r\nHi zhaobin, it is possible to share your solution (for  1)?",
  "Solution_7": "Here are some questions I wish to post:\r\nLet $\\theta=\\frac{\\pi}{2n+1}\\/$ where $n$ is a positive integer. Prove that:\r\n(i)  \\[\\sum_{r=1}^{n} cos \\  r \\theta = - \\frac{1}{2} \\] \r\n(ii)  \\[\\sum_{r=1}^{n} cos^{2} \\ r \\theta =  \\frac{2n-1}{4} \\] \r\n(iii) $r_1,r_2 \\in \\{1,2, \\ldots, n\\}\\/$ \\[\\sum_{r_1 >r_2} cos \\ r_1 \\theta cos \\ r_2 \\theta=  \\frac{1-n}{4} \\] \r\n(iv)  $\\forall n \\geq k \\geq 1\\/$ (more gerenal case for (i)),  \\[\\sum_{r=1}^{n} (cos \\ r \\theta)^{2k-1} = - \\frac{1}{2} \\] \r\n\r\n\r\nI manage to get the solution for the above questions, and I make a conjecture (base on (i) and (ii) above), as follow:\r\nLet $\\theta = \\frac{2 \\pi}{2n+1}, n \\in N\\/$ and $r_1,r_2,\\ldots ,r_k \\in \\{1,2, \\ldots, n\\}\\/$.\r\n\\[\\sum_{r_1 >r_2> \\ldots >r_k} cos \\ r_1 \\theta cos \\ r_2 \\theta \\ldots  cos \\ r_k \\theta = (-1)^l \\frac {{n- \\left\\lceil \\frac{k}{2} \\right\\rceil \\choose \\left\\lceil \\frac{k-1}{2} \\right\\rceil }}{2^k} \\]\r\n\r\nwhere $l= \\left\\lceil \\frac{k}{2} \\right\\rceil\\/$\r\n\r\nCan somebody help me on this  :?  :?  :?  \r\nThanks a lot!!",
  "Solution_8": "[quote=\"pipi\"]Thanks for your solution, flip2004! \nI get the idea but I think there is something not allowed if we write:\n[quote=\"flip2004\"]\n Let $ z_k=\\cos{(2k\\theta)}+ i\\cdot\\sin{(2k\\theta)}. $ If we select $x=1$ in the  equalities $\\ds \\frac{x^{2n+1}-1}{x-1}=\\sum\\limits_{k=0}^{2n}x^k=\\ds \\prod\\limits_{k=1}^{2n}\\left(x-z_k\\right)$,...[/quote]\nthanks again.[/quote]\r\nsure :) .\r\nfirst,we can now $\\frac{\\sin{(2n+1)x}}{\\sin x}$ is a polynomial of $\\sin x$\r\nlet $P_n(2\\sin x)=\\frac{\\sin{(2n+1)x}}{\\sin x}$\r\nwe can know $P_n(x)$ have 2n roots which are $2\\sin{\\frac{k}{2n+1}}$ or $-2\\sin{\\frac{k}{2n+1}}$,$(k=1,2,\\cdots,n)$\r\nthen find the relation of $P_n(x),P_{n-1}(x),P_{n-2}(x)$\r\nwith induction you will find the answer :)"
}
{
  "Tag": [
    "function",
    "trigonometry",
    "symmetry",
    "complex numbers",
    "complex analysis"
  ],
  "Problem": "I have been taught that the roots of complex numbers are multivalued, for example $ \\sqrt{3\\plus{}4i}\\equal{}\\pm (2\\plus{}i)$. What to do I have to add two complex roots, for example to compute $ \\sqrt{3\\plus{}4i}\\plus{}\\sqrt{3\\plus{}4i}$? Is the correct way to do it by take any branch of the first square root and add it to any branch of the second square root, so $ \\sqrt{3\\plus{}4i}\\plus{}\\sqrt{3\\plus{}4i}\\equal{}4\\plus{}2i, 0, \\text{ or }\\minus{}4\\minus{}2i$? Or does such situation arise anywhere in mathematics?",
  "Solution_1": "$ \\sqrt {3 \\plus{} 4i} \\plus{} \\sqrt {3 \\plus{} 4i} \\equal{} 2\\sqrt {3 \\plus{} 4i} \\equal{} \\pm 2(2 \\plus{} i)$, never $ 0$, otherwise you'd be saying $ \\sqrt {3 \\plus{} 4i} \\neq \\sqrt {3 \\plus{} 4i}$. I'd recommend using the principal square root for cases like this one, though (then again, what do I know? XD).",
  "Solution_2": "$ \\sqrt{3 \\plus{} 4i}$ is not a well-defined expression.  You should never have to evaluate it unless the branch is explicitly given.",
  "Solution_3": "I wish to disagree:  Can it not be all of them depending on what branches are chosen?  Consider the funtion $ \\displaystyle f(z)\\equal{} \\sqrt{z}\\plus{}\\sqrt{z}$.  Cannot $ f(1)\\equal{}0$ if the branches are chosen appropriately?  How about $ \\displaystyle g(z)\\equal{}z^{1/3}\\plus{}z^{1/4}$?  If $ \\displaystyle g(1)\\equal{}0$ then what is $ \\displaystyle g(\\minus{}1)$?  I think $ \\displaystyle \\sqrt{3\\plus{}4i}\\plus{}\\sqrt{3\\plus{}4i}\\equal{}2\\sqrt{3\\plus{}4i}$  when the underlying square root multifunctions are being added symmetrically:  first branch of the first expression added to the first branch of second one and second branch of the first with the second of the second expression.  That's confusing I know.  Never use the principal branch unless you need to use it or the context implies or warrants it: approach roots, logs, inverse trigs, other multifunctions as multivalued quantities until specific values are needed, specific branches are specified,  or if the symmetry above is implied, that is if $ \\sqrt{z}\\plus{}\\sqrt{z}\\equal{}2\\sqrt{z}$ is assumed. I think $ \\sqrt{3\\plus{}4i}$ is perfectly well defined.  It's $ \\displaystyle \\sqrt{3\\plus{}4i}\\equal{}r^{1/2}e^{i/2(\\Theta\\plus{}2k\\pi)}\\equal{}\\pm r^{1/2}e^{i/2\\Theta}$.   If this was a test question, I'd stick to my guns and hope mine is on top the next day . . ."
}
{
  "Tag": [
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be an abelian finite group such that $ n$ divides $ |G|$.\r\nLet $ H\\equal{}\\{ x^n, x\\in G\\}$. There are many ways to prove that $ |H|\\equal{}|G|/n$, but I am trying to find an elementary proof that would not use the notion of quotient or factorization of homomorphism (i.e. $ G/ker (f) \\simeq f(G)$), let alone Cauchy or Sylow.\r\nAny idea?",
  "Solution_1": "[quote=\"Fleury\"]Let $ G$ be an abelian finite group such that $ n$ divides $ |G|$.\nLet $ H \\equal{} \\{ x^n, x\\in G\\}$. There are many ways to prove that $ |H| \\equal{} |G|/n$.[/quote]\r\nAre you sure of that? Because, if $ G \\equal{} \\{\\minus{}1,1\\} \\times \\{\\minus{}1,1\\}$ and $ n \\equal{} 2$, it appears that $ H \\equal{} \\{(1,1)\\}$ and that $ |H| \\equal{} 1 \\neq \\frac{|G|}{2} \\equal{} 2$.\r\nOr I may have misundertood what you had written...",
  "Solution_2": "[quote=\"LaChouetteRieuse\"]Are you sure of that? [/quote]\r\nNot anymore :blush:",
  "Solution_3": "There are some values of $ n$ when your statement is true (although I don;t have an \"elementary solution\"), namely if $ |G|\\equal{}\\prod_{i\\equal{}1}^{r}p_i^{a_i}$ ($ p_i$ primes and $ a_i>0$) and $ n\\equal{}\\prod_{i\\equal{}1}^{r}p_i^{b_i}$ with $ b_i\\equal{}a_i$  or $ b_i\\equal{}0$."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "If $ a,b,c,d \\ge 0$  then prove\r\n$ \\frac {a}{b\\plus{}c}\\plus{}\\frac {b}{c\\plus{}d}\\plus{}\\frac {c}{d\\plus{}a}\\plus{}\\frac {c}{d\\plus{}a} \\ge \\frac {5}{2}$.",
  "Solution_1": "[quote=\"soruz\"]If $ a,b,c,d \\ge 0$  then prove\n$ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} d} \\plus{} \\frac {c}{d \\plus{} a} \\plus{} \\frac {c}{d \\plus{} a} \\ge \\frac {5}{2}$.[/quote]\r\n\r\nI love inequalities that fail for $ a\\equal{}b\\equal{}c\\equal{}d\\equal{}1$...\r\n\r\nSee http://www.mathlinks.ro/viewtopic.php?t=140344 for the right one (I hope).\r\n\r\n  darij",
  "Solution_2": "If $ a_1,a_2,...,a_n\\ge 0$  then prove\r\n\r\n$ \\frac {a_1}{a_2\\plus{}a_3}\\plus{}\\frac {a_2}{a_3\\plus{}a_4}\\plus{}...\\frac {a_n}{a_n\\plus{}a_n} \\ge \\frac{n}{2}$\r\n\r\nfor $ 0 \\ge a_1 \\ge a_2 ...\\ge a_n$.",
  "Solution_3": "Actually, the following nice statement holds:\r\nIf $ 1 \\le k \\le {n\\minus{}1}$ is an integer number and $ a_1,a_2,...,a_n\\ge 0$  \r\nsuch that  $ a_1 \\le a_2\\le...\\le a_n$   then\r\n\r\n$ \\frac {a_1}{ka{_1}\\plus{}a_{k\\plus{}1}}\\plus{}\\frac {a_2}{ka{_3}\\plus{}a_{k\\plus{}2}}\\plus{}...\\frac {a_n}{ka_n\\plus{}a_k} \\ge \\frac{n}{k\\plus{}1}$",
  "Solution_4": "If $ a,b,c,d,e,f \\ge 0$    then prove\r\n \r\n$ \\frac {a}{b\\plus{}c}\\plus{}\\frac {b}{c\\plus{}d}\\plus{}\\frac {c}{d\\plus{}e} \\plus{}\\frac{d}{e\\plus{}a}\\plus{}\\frac{f}{a\\plus{}b} \\ge \\frac{5}{2}$.",
  "Solution_5": "[quote=\"soruz\"]If $ a,b,c,d,e,f \\ge 0$    then prove\n \n$ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} d} \\plus{} \\frac {c}{d \\plus{} e} \\plus{} \\frac {d}{e \\plus{} a} \\plus{} \\frac {f}{a \\plus{} b} \\ge \\frac {5}{2}$.[/quote]\r\nTry Cauchy - Schwarz. :wink:",
  "Solution_6": "[quote=\"soruz\"]If $ a,b,c,d,e,f \\ge 0$    then prove\n \n$ \\frac {a}{b \\plus{} c} \\plus{} \\frac {b}{c \\plus{} d} \\plus{} \\frac {c}{d \\plus{} e} \\plus{} \\frac {d}{e \\plus{} a} \\plus{} \\frac {f}{a \\plus{} b} \\ge \\frac {5}{2}$.[/quote]\r\n\r\nThis statement is quite as correct as the spelling of the word \"correct\" in your post...\r\n\r\n  dg",
  "Solution_7": "I think the \"correct\" statement should be\r\n\r\n$ \\frac{a}{b\\plus{}c} \\plus{} \\frac{b}{c\\plus{}d} \\plus{} \\frac{c}{d\\plus{}e} \\plus{} \\frac{d}{e\\plus{}a} \\plus{} \\frac{e}{a\\plus{}b} \\geq \\frac{5}{2}$\r\n\r\nIs this right?\r\n\r\n[hide=\"Solution\"]\n\nAssume WLOG $ a\\plus{}b\\plus{}c\\plus{}d\\plus{}e \\equal{} 1$ and let $ f(x)\\equal{}\\frac{1}{x}$ which concaves up for $ x \\in \\mathbb{R}^{\\plus{}}$. By Jensen's Inequality:\n\n$ af(b\\plus{}c) \\plus{} bf(c\\plus{}d) \\plus{} cf(d\\plus{}e) \\plus{}df(e\\plus{}a) \\plus{} ef(a\\plus{}b) \\geq f(ab\\plus{}ac\\plus{}ad\\plus{}ae\\plus{}bc\\plus{}bd\\plus{}be\\plus{}cd\\plus{}ce\\plus{}de) \\equal{} \\frac{(a\\plus{}b\\plus{}c\\plus{}d\\plus{}e)^{2}}{ab\\plus{}ac\\plus{}ad\\plus{}ae\\plus{}bc\\plus{}bd\\plus{}be\\plus{}cd\\plus{}ce\\plus{}de}$\n\nSo we need to prove that: \n\n$ \\frac{(a\\plus{}b\\plus{}c\\plus{}d\\plus{}e)^{2}}{ab\\plus{}ac\\plus{}ad\\plus{}ae\\plus{}bc\\plus{}bd\\plus{}be\\plus{}cd\\plus{}ce\\plus{}de} \\geq \\frac{5}{2} \\iff (a\\minus{}b)^{2} \\plus{} (a\\minus{}c)^{2} \\plus{} (a\\minus{}d)^{2} \\plus{} (a\\minus{}e)^{2} \\plus{} (b\\minus{}c)^{2} \\plus{} (b\\minus{}d)^{2} \\plus{} (b\\minus{}e)^{2} \\plus{}(c\\minus{}d)^{2} \\plus{}(c\\minus{}e)^{2} \\plus{}(d\\minus{}e)^{2} \\geq 0$.[/hide]",
  "Solution_8": "[quote=\"BanishedTraitor\"]\n\n\n\nSo we need to prove that: \n\n$ \\frac {(a \\plus{} b \\plus{} c \\plus{} d \\plus{} e)^{2}}{ab \\plus{} ac \\plus{} ad \\plus{} ae \\plus{} bc \\plus{} bd \\plus{} be \\plus{} cd \\plus{} ce \\plus{} de} \\geq \\frac {5}{2}$[/quote]\r\nThis is just Cauchy. :wink:"
}
{
  "Tag": [
    "Gauss",
    "algebra",
    "polynomial",
    "function",
    "integration",
    "calculus",
    "derivative"
  ],
  "Problem": "Let A the Gauss ring of integers and $z_1,...,z_n$ in A such that the distance from $z_i$ to $z_1$ is greater than 2 for all $i>1$. Prove that the polynomial $ 1+(x-z_1)...(x-z_n)$ is irreducible in $A[X]$.",
  "Solution_1": "What is Gauss ring of integers? Is it $a+bi$, $a,b\\in\\mathbb{Z}$?",
  "Solution_2": "Yes, it is",
  "Solution_3": "Then it is very standart.\r\nConsider polynomials $f(z)=(z-z_1)(z-z_2)...(z-z_n)$ and $g(z)=(z-z_1)(z-z_2)...(z-z_n)$. We see that $|f(z)-g(z)|=1<|f(z)|+|g(z)|$ for all $z$ lying on the circle $|z-z_1|=1$. So $f(z)$ has the same number of roots inside of $|z-z_1|=1$ as $g(z)$, but $g(z)$ has the only such root $z_1$. \r\n\r\nSuppose the contrary, i.e. $f(z)=a(x)b(x)$. We may WLOG assume $z_1=0$. Then we have $a(0)b(0)=1$, but both $|a(0)|,|b(0)|\\geq 1$ and at least one of these numbers is greater than 1, since one of the polynomials $a$ or $b$ has all roots are greater than 1 at absolute value. Contradiction.",
  "Solution_4": "Yes, Myth, this was the idea. Probably for you it is standard, but I do not think that proving Rouche's theorem in an oral examination is so easy.",
  "Solution_5": "And who ask to prove it in an oral examanation? Just use it!",
  "Solution_6": "This is also my opinion, Myth, but the problem is that sometimes the examinator does not accept high theorems during the examination if you do not know a proof. This is the problem with oral examinations! But anyway, in this case I know a nice proof of this particular case of Rouche's theorem, thus I do not have problems.",
  "Solution_7": "is there any elementary proof (no complex analysis) of Rouche's theorem ?\r\nHarazi, what is the proof for this particular case you are talking about? I'm very interested. Thanx  :)",
  "Solution_8": "We made in a \"travaux dirig\u00e9\" something about \"indice d'un lacet\" and the teacher has shown to us that if we consider the function $ f(t)=P(re^{it})$ then the \"indice\" of this function, that is $\\int_{0}^{2\\pi}{\\frac{f'(t)}{2\\pi\\cdot (t)}}$ is equal to the number of roots of the polynomial P that lie inside the disk $|z|<r$, assuming that P does not have roots of modulus r. This is simple to show. The hard part is to show that if a \"lacet\" f verifies $|f(t)-1|<1$ for all t then it has \"indice\" 0. It follows that if $|P(z)-Q(z)|<|Q(z)|$ for all z of modulus r, then P and Q have the same number of roots in the disk of radius r (just use the lacet $\\frac{P(re^{it})}{Q(re^{it})}$). Alekk, if you come to my room (I forgot the number of your room), I can give you the TD where he proved the hard part.",
  "Solution_9": "Nice private conversation...",
  "Solution_10": "Sorry, Myth!  :blush:  :P  :P  :P  :P",
  "Solution_11": "Harazi, do you, by any chance, mean to say something like what I wrote in post number $16$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150&postorder=asc&highlight=curve&start=15]this topic[/url]? I think your \"indice\" is the degree of that function wrt the unit circle (of course, we can use any other circle instead of the unit circle, with the same results).",
  "Solution_12": "Yes, grobber, it is the same concept. Yet, I do not find a proof of the fact that the index of such a \"lacet\" is 0 of the trajectory is in a circle does does not contain the origin. With the interpretation of the \"lacet\" it is clear, but is it that easy to prove it rigorously?",
  "Solution_13": "I don't really know what \"lacet\" means :). I understood \"indice\" :).",
  "Solution_14": "Ok, sorry, I do not know how we call this in english:\r\n   A \"lacet\" is a derivable function f from $[0,2\\pi]$ into the complex numbers, such that the derivative is continuous and $f(0)=f(2\\pi)$. The indice of the lacet is the number (integer ;) ) $\\frac{1}{2\\cdot i\\pi}{\\int_{0}^{2\\cdot\\pi}{\\frac{f'(t)}{f(t)}dt}}$. The idea is to prove that if the lacet has no zeroes and $ |f(t)-1|<1$ for all t then its indice is 0. Do you have a simple solution for this?",
  "Solution_15": "No, not really :).\r\n\r\nHere's how I would translate it, in the language used in that post I mentioned: given a polynomial $P$ which has no roots on or inside the unit circle, the order of $\\frac{P(z)}{|P(z)|}$ is $0$. If we use the fact that the fundamental group of the circle is $\\mathbb Z$, the order can be defined without resorting to integrals, and then it would become a pretty standard exercise in algebraic topology (of which I know next to nothing ;)). \r\n\r\nFirst we would show that the order is invariant to homotopy, and then we would use the homotopy $H(z,t)=\\frac{P(tz)}{|P(tz)|}$. This takes our path $\\frac{P(z)}{|P(z)|}$ to the constant path $\\frac{P(0)}{|P(0)|}$, which has order $0$ (as every constant path does).",
  "Solution_16": "Yes, my teacher used the same kind of made, only he translated them in the language of mere mortals that we are.  ;)  :(",
  "Solution_17": "ok, thank you very much harazi. I remember we did the same last year. I think \"indice\" is called winding number in english ."
}
{
  "Tag": [
    "HMMT",
    "Harvard",
    "college",
    "MIT",
    "AMC",
    "AIME",
    "geometry"
  ],
  "Problem": "Hi, I was wondering how useful the Harvard-MIT math tourney problems were and if their level of difficulty corresponded to that of the AIME tests. Are there particular sections (algebra/ combo/ geo) that you found to be more useful/realistic of the AMC/AIME tests? Any years specifically that helped you or any helpful info is appreciated.",
  "Solution_1": "I think the HMMT problems are good practice problems for the AIME -- even though they have some differences with AIME problems. The HMMT problems are (ideally) meant to be completed in an average of 5 minutes per problem, whereas AIME problems are (ideally) meant to be completed in an average of 12 minutes per problem. The HMMT problems are more geared towards one subject at a time, whereas AIME problems are notorious for crossing subjects or topics. So a problem that starts out as a geometry problem on the AIME may require some skills from algebra or complex numbers or something.\r\n\r\nBut the HMMT problems correspond in difficulty to maybe #1-10 on the AIME, and they make good practice in solving problems that require skills or knowledge beyond your regular AMC-12 or local math contest questions. They're also good because they have solutions available to check your work.\r\n\r\nAlong with ARML problems, HMMT problems are probably the best practice problems for AIME (aside from AIME problems themselves - or problems found in the Intermediate Forum here).",
  "Solution_2": "What book/contest has problems for #11-15 on AIME?",
  "Solution_3": "The AIME would be your best source. You could also try some of the early problems on some of the easiest problems like the CMO (canada) and the JBMO."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "vector"
  ],
  "Problem": "Is it true that every singular matrix $ A \\in \\mathcal M_n ( \\mathbb C )$ can be written as $ TN$, where $ T$ is invertible and $ N$ is nilpotent?",
  "Solution_1": "For a moment, I had myself convinced that $ \\begin{bmatrix}1&1\\\\0&0\\end{bmatrix}$ was trouble. But\r\n\r\n$ \\begin{bmatrix}1&1\\\\0&0\\end{bmatrix}\\equal{}\\begin{bmatrix}1&0\\\\1&1\\end{bmatrix}\\begin{bmatrix}1&1\\\\\\minus{}1&\\minus{}1\\end{bmatrix},$ so we're OK there.",
  "Solution_2": "Yes, this is true. In the first place, there are invertible matrices $P, R$ such that $P^{-1}AR^{-1} = \\begin{bmatrix} I_r & O \\\\ O & O \\end{bmatrix}$, with $r = rank(A) \\leq n-1$. Now define the invertible matrix $X = [e_{r+1} \\,\\, ... \\,\\, e_n \\,\\, e_1 \\,\\, ... \\,\\, e_r]$, where the $e_i$ are the column vectors of the canonical basis. Then,\n\n$P^{-1}AR^{-1}X = \\begin{bmatrix} I_r & O \\\\ O & O \\end{bmatrix} X = \\begin{bmatrix} O & I_r \\\\ O & O \\end{bmatrix} = H$,\n\nwhere $H$ is a nilpotent matrix (since it is strictly upper triangular). Making $R^{-1}X = S$, we have \n\n$A = PHS^{-1} = PS^{-1} \\cdot SHS^{-1}$.\n\nNow let's make $PS^{-1} = T$ and $SHS^{-1} = N$: then $T$ is invertible, $N$ is nilpotent, and $A = TN$. $\\diamondsuit$"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "search",
    "linear algebra unsolved"
  ],
  "Problem": "Give A is a matrix satisfy $ a_{ij} \\plus{} a_{ji} \\equal{} 0, 1 \\leq i,j \\leq n$\r\nProve that  $ det(A) \\geq 0$",
  "Solution_1": "Try to guess: how many different topics do we have with titles \"det(A)\", \"det A\", etc. ;)\r\n\r\nThis one was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=143797 , though I wouldn't mind someone posting a proof for the Pfaffian formula. ;)\r\n\r\n  darij",
  "Solution_2": "Some may want to check [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1077620069&t=25572]this[/url] out."
}
{
  "Tag": [
    "floor function"
  ],
  "Problem": "Prove that these two sequences (called \"Beatty sequences\") of integers:\r\n\\[a_{n}= \\left\\lfloor n \\alpha \\right\\rfloor \\] and\r\n\\[b_{n}= \\left\\lfloor n \\beta \\right\\rfloor \\] \r\nfor $n \\in \\mathbb{Z^{+}}$, contain together all the positive integers, without repetition, if $\\alpha+\\beta = \\alpha \\cdot \\beta$ and both $\\alpha$ and $\\beta$ are positive, irrational numbers.\r\n\r\n([url=http://mathworld.wolfram.com/BeattySequence.html]See this page[/url] on Wolfram. I have basically no idea how to go about proving this. Anyone have ideas?)",
  "Solution_1": "I think the following argument shows that they contain all integers:\r\n\r\nOne of $\\alpha, \\beta$ must be in the interval $(1, 2)$.  Let's say that $\\beta$ is.  Then the sequence $\\{\\lfloor n\\beta \\rfloor\\}_{n \\in{\\bf Z}_{>0}}$ has gaps of size 1 or less.  So, suppose we have a gap somewhere, $b_{n}= m$ and $b_{n+1}= m+2$.  Thus $m < n\\beta < m+1$ and $m+2 < (n+1)\\beta < m+3$ and $n \\leq m$.\r\n\r\nThis means that $\\frac{m+2}{n+1}< \\beta < \\frac{m+1}{n}$, so if I write $n = m-k+1$ we have $k > 0$ and\r\n\\[\\frac{m+2}{m-k+2}< \\beta < \\frac{m+1}{m-k+1}\\]\r\n\\[\\frac{k}{m-k+2}< \\beta-1 < \\frac{k}{m-k+1}\\]\r\n\\[m-k+1 < \\frac{k}{\\beta-1}< m-k+2 \\]\r\n\\[m+1 < k\\cdot \\frac{\\beta}{\\beta-1}< m+2 \\]\r\nBut $\\frac{\\beta}{\\beta-1}= \\alpha$, so $\\lfloor k \\cdot \\alpha \\rfloor = m+1$ and the two sequences together cover all the integers."
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "determine k so that f(x) = 2x^2 +(k-1)x - k(k+1) is divisible by x+2 and the value of k is an exact divisor of \r\n\r\n9\r\n15\r\n27\r\n49\r\nnone of these",
  "Solution_1": "[hide=\"HINT\"]\nVieta's.\n[/hide]",
  "Solution_2": "[hide=\"Hint\"]\nFactor theorem.\n\nIf $ f(x)$ is a polynomial then $ (x\\minus{}a)$ is a factor iff $ f(a)\\equal{}0$\n[/hide]",
  "Solution_3": "could be Taylor's????"
}
{
  "Tag": [
    "function",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Give two fintes set $X$ and $Y$ such that $|X|=n,|Y=m|$ where $n \\geq m >0$.\r\nFind the number of onto function mapping form $X$ to $Y$  :)",
  "Solution_1": "By 'onto' you mean surjective?\r\nIn that case, just look to another thread in this section.\r\n\r\nPierre.",
  "Solution_2": "Anyway ;\r\nlet\r\n$f: X \\rightarrow Y,|X|=x,|Y|=y$ define $A_{i}: =\\{f: X \\rightarrow Y;j\\notin ImY\\},j=1,2,\\dots ,y$\r\nThen $|A_{1}^{c}\\cap A_{2}^{c}\\cap \\dots \\cap A_{y}^{c}|=y^{x}-{y \\choose 1}(y-1)^{x}+\\dots+(-1)^{y}{y \\choose y}(y-y)^{x}=\\sum_{i=0}^{y}(-1)^{i}{y \\choose i}(y-i)^{x}$"
}
{
  "Tag": [
    "inequalities",
    "inequalities open"
  ],
  "Problem": "let a,b,c  and k are possitive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 3$  . Prove the inequality:\r\n $ \\sqrt {ka^2 \\plus{} b} \\plus{} \\sqrt {kb^2 \\plus{} c} \\plus{} \\sqrt {kc^2 \\plus{} a} \\geq 3\\sqrt {k \\plus{} 1}$",
  "Solution_1": "It isn't usually trues with all $ k \\ge 0$, so you should write \"Find all the calue of $ k$ for the following ineq still trues\"  :maybe:\r\nMy YM: langtukhongnha1012",
  "Solution_2": "[quote=\"AngelHolder\"]let a,b,c  and k are possitive real numbers such that $ a \\plus{} b \\plus{} c \\equal{} 3$  . Prove the inequality:\n $ \\sqrt {ka^2 \\plus{} b} \\plus{} \\sqrt {kb^2 \\plus{} c} \\plus{} \\sqrt {kc^2 \\plus{} a} \\geq 3\\sqrt {k \\plus{} 1}$[/quote]\r\n\r\n\r\n\r\n$ k\\geq \\frac{1}{2}$\r\n\r\nThe inequality  holds."
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "1. A car with tires having radius 33cm was driven on a trip and the odometer indicated that the distance travelled was 640km. Two weeks later, with snow tires installed, the odometer indicated that the distance for the return trip over the same route was 625 km. Find the radius of the snow tires.\r\n\r\nI keep thinking this is $\\frac{640}{625}=\\frac{33}{x}$ But it does not agree with the answer given:\r\n[hide]33.8[/hide]\n2. Bob and Jim, next-door neighbours, use hoses from both houses to fill Bob's swimming pool. They know it takes eighteen hours using both hoses. They also know that Bob's hose, used alone, can fill the pool in six hours less than Jim's hose. How much time is required by each hose alone?\n\nThis one is just plain confusing, I've declared about 5 variables in the first 15 seconds of trying to solve it, the answer is:\n[hide]33.25h with Bob's hose and 39.25h with Jim's hose[/hide]\r\n\r\n3. Prove that $\\frac{\\sqrt{2}+\\sqrt{6}}{\\sqrt{2}+\\sqrt{3}}=2$\r\nI'm really confused on this one, as my calculator tells me its not 2 lol.",
  "Solution_1": "[b]1[/b]. $t_{1}$ - time of travel of the car with tires having radius $R_{1}= 33$cm. $S_{1}= 640$km - distance of this car .\r\n$t_{2}$ - time of travel of the car with tires having radius $R_{2}= x$cm. $S_{2}= 625$km - distance of this car .\r\n\r\nThen: $t_{1}\\approx \\frac{1}{R_{1}}$, or $t_{1}=\\alpha \\frac{1}{R_{1}}$ and $t_{2}\\approx \\frac{1}{R_{2}}$, or $t_{2}=\\alpha \\frac{1}{R_{2}}$, $\\alpha$ - aspect ratio.\r\n\r\nAnd $S_{1}\\approx t_{1}$, $S_{2}\\approx t_{2}$.\r\n\r\nThen $\\frac{S_{1}}{S_{2}}=\\frac{\\alpha \\frac{1}{R_{1}}}{\\alpha \\frac{1}{R_{2}}}$.\r\n$\\Rightarrow \\frac{S_{1}}{S_{2}}=\\frac{R_{2}}{R_{1}}$\r\n$\\Rightarrow \\frac{640}{625}=\\frac{x}{33}$\r\n$\\Rightarrow x = 33,792$ or $x \\approx 33,8$\r\n\r\n[b]2[/b]$x$h - time, use Bob's hose to fill one swimming pool.\r\n$(x+6)$h - time, use Jim's hose to fill one swimming pool.\r\nFor $x$h Jim's hose to fill $\\frac{x}{x+6}$ part swimming pool.\r\nUse hoses from both houses, then for $x$h  fill $(1+\\frac{x}{x+6})$ pools. \"We know it takes eighteen hours using both hoses\".\r\nProportion:\r\n$\\frac{1}{1+\\frac{x}{x+6}}=\\frac{18}{x}$\r\n$18(1+\\frac{x}{x+6})=x$\r\n$18\\frac{x}{x+6}=x-18$\r\n$18x=(x+6)(x-18)$\r\n$x^{2}-30x-108=0$\r\n$x \\approx 33,25$h - Bob's hose, $(x+6) \\approx 39,25$;// Excuse me for my English",
  "Solution_2": "3)\r\n\r\nMultiply top and bottom by $\\sqrt(3)-\\sqrt(2)$ to get:\r\n$\\frac{(\\sqrt(3)-\\sqrt(2))*(\\sqrt(2)+\\sqrt(6))}{3-2}$\r\n$= (\\sqrt(3)-\\sqrt(2))*(\\sqrt(2)+\\sqrt(6))$\r\n$= \\sqrt(6)+\\sqrt(18)-2-\\sqrt(12)$\r\n$= \\sqrt(6)+3\\sqrt(2)-2-2\\sqrt(3)$.\r\n\r\n\r\nNow, if this is $2$, then $\\sqrt(6)+3\\sqrt(2)-2\\sqrt(3) = 4$,\r\nwhich is not true."
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "An arbitrary point $B$ is chosen on the line $AC$ and on the segments $AB,BC$ and $CA$ as diameters, circles $k_1,k_2$ and $k$ are constructed. An arbitrary line through the point $B$ intersects $k$ in the points $P$ and $Q$ , and circles $k_1$ and $k_2$ (besides the point $B$ )in the points $R$ and $S$ , respectively. Prove that $PR=QS$.",
  "Solution_1": "Let AR and CQ meet the circle k at M and N. \r\nSince <ARB=<CSB=$90^{o}$. So AR || CQ. \r\nHence arc AN = arc CM.  So arc MAN = arc AMC. Which means AC=MN. So MN is the diameter of the circle k. \r\nRest is easy congruents.",
  "Solution_2": "$RS$ is the projection of $AC$ on $PQ$, hence $R$ and $S$ are equally apart of $O$, midpoint of the diameter $AC$, but so are $P$ and $Q$, done.\n\nBest regards,\nsunken rock"
}
{
  "Tag": [
    "search",
    "calculus",
    "algebra",
    "polynomial",
    "probability",
    "geometry",
    "ARML"
  ],
  "Problem": "how much of pi do you know? put all posts in spoilers please.  this is pi to the something digit. (look if you really want to know it) 900th digit (don't use the internet! i wanna see what other people know. [hide]3.1415926535897932384626433832795028841971693993751058209749  \n\n445923078164062862089986280348253421170679821480865132823066 470938446095505822317253594081284811174502841027019385211055 596446229489549303819644288109756659334461284756482337867831 652712019091456485669234603486104543266482133936072602491412 737245870066063155881748815209209628292540917153643678925903 600113305305488204665213841469519415116094330572703657595919 530921861173819326117931051185480744623799627495673518857527 248912279381830119491298336733624406566430860213949463952247 371907021798609437027705392171762931767523846748184676694051 320005681271452635608277857713427577896091736371787214684409 012249534301465495853710507922796892589235420199561121290219 608640344181598136297747713099605187072113499999983729780499 510597317328160963185950244594553469083026425223082533446850 352619311881710100031378387528865875332083814206171776691473 ...[/hide]",
  "Solution_1": "3\r\n\r\nThat's it. \r\nNice job stretching the whole screen all the way so the text won't warp anymore.\r\nAlso, this topic has been posted before a lot.",
  "Solution_2": "it has? i didn't know that...\r\n\r\nu only know 3 digits? oh...\r\n\r\nand also i didn't know how to \"unstretch\" my post.  i don't know how to fix it.",
  "Solution_3": "You didn't have to post all those digits. \r\nNext time use the search option.\r\nI do not know 3 digits of pi, but only know it rounded to the nearest integer. I believe that's 3.",
  "Solution_4": "what's the search option?  and how do you become a mod?",
  "Solution_5": "the question is now unstretched. happy?",
  "Solution_6": "3.14159265358979323846 (im not sure about the last 2 digits)\r\n\r\ni dont kno anymore... and i bet u used the internet (unless ur lyk pifreak)",
  "Solution_7": "no i'm not like pifreak (who was that person) and yes i did use the internet otherwise i would've only known 3.141592653589793238462643383279 (i think) so yes.",
  "Solution_8": "I know not much...\n\n\n\n\n\n\n\n[hide]pi  :downto: 3.141592653589793238462643383279502884197169399...[/hide]",
  "Solution_9": "hmm...pifreak knew way way way way more than that cardwacko.",
  "Solution_10": "Personally, if I were to memorize anything for a hobby, it'd probably be the Periodic Table.  It's more useful, and more practical.  Plus it's more multifaceted, since there's atomic masses, charges, oxidation numbers, and properties to remember in addition to names and symbols.",
  "Solution_11": "the periodic table is too long...i memorized up to i think only 40 elements.  i need to get a new hobby...",
  "Solution_12": "I only know [hide]3.1415926...[/hide].",
  "Solution_13": "an u are A+MATH? ... i judge people by their usernames. sorry.",
  "Solution_14": "dont judge people by how many digits of pi they know",
  "Solution_15": "why $pi$ why not $pi*e$? apparantly it is not shown that it is transcendental, so you might be able to find a polynomial which ends (= and memorise that!",
  "Solution_16": "[quote=\"jli\"]here are some pi songs to sing on March 14:\n\nHappy Pi Day \n\nHappy Pi day to you,\nHappy Pi day to you,\nHappy Pi day everybody,\nHappy Pi day to you. \n\n(to the tune of \"Happy Birthday\") \n\n\nOh Number PI \n\nOh, number Pi\nOh, number Pi\nYour digits are unending,\nOh, number Pi\nOh, number Pi\nNo pattern are you sending.\nYou're three point one four one five nine,\nAnd even more if we had time,\nOh, number Pi\nOh, number Pi\nFor circle lengths unbending. \n\nOh, number Pi\nOh, number Pi\nYou are a number very sweet,\nOh, number Pi\nOh, number Pi\nYour uses are so very neat.\nThere's 2 Pi r and Pi r squared,\nA half a circle and you're there,\nOh, number Pi\nOh, number Pi\nWe know that Pi's a tasty treat. \n\n(to the tune of \"Oh Christmas Tree\") \n\n\nPi Day Song \n\nRefrain:\n\nPi day songs\nAll day long.\nOh, what fun it is,\nTo sing a jolly pi day song\nin a fun math class\nlike this. (Repeat )\n\nVerse:\nCircles in the snow,\nAround and round we go.\nHow far did we have to run?\nDiameter times pi! (Refrain )\n\n(to the tune of \"Jingle Bells\")[/quote]\r\n\r\njust wondering... did you make that up?\r\n(I haven't been on for a while!... but now I'M BACK - why don't I hear any cheering? :D )",
  "Solution_17": "[quote=\"sunnyray\"][quote=\"jli\"]here are some pi songs to sing on March 14:\n\nHappy Pi Day \n\nHappy Pi day to you,\nHappy Pi day to you,\nHappy Pi day everybody,\nHappy Pi day to you. \n\n(to the tune of \"Happy Birthday\") \n\n\nOh Number PI \n\nOh, number Pi\nOh, number Pi\nYour digits are unending,\nOh, number Pi\nOh, number Pi\nNo pattern are you sending.\nYou're three point one four one five nine,\nAnd even more if we had time,\nOh, number Pi\nOh, number Pi\nFor circle lengths unbending. \n\nOh, number Pi\nOh, number Pi\nYou are a number very sweet,\nOh, number Pi\nOh, number Pi\nYour uses are so very neat.\nThere's 2 Pi r and Pi r squared,\nA half a circle and you're there,\nOh, number Pi\nOh, number Pi\nWe know that Pi's a tasty treat. \n\n(to the tune of \"Oh Christmas Tree\") \n\n\nPi Day Song \n\nRefrain:\n\nPi day songs\nAll day long.\nOh, what fun it is,\nTo sing a jolly pi day song\nin a fun math class\nlike this. (Repeat )\n\nVerse:\nCircles in the snow,\nAround and round we go.\nHow far did we have to run?\nDiameter times pi! (Refrain )\n\n(to the tune of \"Jingle Bells\")[/quote]\n\njust wondering... did you make that up?\n(I haven't been on for a while!... but now I'M BACK - why don't I hear any cheering? :D )[/quote]\r\n\r\nWooohoooo!!!!! ;)",
  "Solution_18": "You can lord all you want over pi-memorizers but memorizing pi is like exercising for some mathematicians, it keeps their brains in shape so they can think qucker.\r\n\r\nThis is not personal experience, but someone told it to me.\r\n\r\nWhy dont other numbers get memorized, like e?",
  "Solution_19": "Also, I heard somewhere that the probability of finding a seven digit number in pi (phone number) is 1",
  "Solution_20": "[quote=\"mcalderbank\"]Also, I heard somewhere that the probability of finding a seven digit number in pi (phone number) is 1[/quote]\r\n\r\n\r\nhaha wow... so i'm gonna randomly call 314-1592... and we'll see who picks up! :-D hahah\r\n\r\nin geometry, on pi day, our teacher gave us one bonus point on a test for each 10 digits of pi you knew (up to 5 points)... hence i memorized the first 50 after the decimal... [hide]3.141592653589793238462643383279502884197169399375105[/hide]",
  "Solution_21": "oh yeah, i found this too.\r\n\r\n[u][b]Pi vs e[/b][/u]\r\nPi goes on and on and on ...\r\nAnd e is just as cursed.\r\nI wonder: Which is larger\r\nWhen their digits are reversed?",
  "Solution_22": "[quote=\"sunnyray\"]just wondering... did you make that up?\n(I haven't been on for a while!... but now I'M BACK - why don't I hear any cheering? :D )[/quote]Hmmmm....if I remember right weren't you one of those people from Miller Middle School that spammed all over the place?",
  "Solution_23": "[quote=\"gameworld7\"]the periodic table is too long...i memorized up to i think only 40 elements.  i need to get a new hobby...[/quote]\r\n\r\nI've memorized Tom Lehrer's \"The Elements.\" Does that count?  :lol:",
  "Solution_24": "I memorized Pi up to 3.141592653589793238462643383279502884 and then i think 197128944 or something back in like 3rd grade and it's stuck with me since :-)",
  "Solution_25": "Sorry for topping old threads but can't resist. :P I'm looking at frankthetank's previous posts.\r\n\r\nFor the 8th grade talent show, I memorized the entire periodic table. All the way to meitnerium. I still remember the tune of it. I also memorized the list of US Presidents from 1st to last and may consider memorizing all Nobel Prize Physics winners from 1900 to present.",
  "Solution_26": "I know:\r\n3.1415926535897932384626433832795......",
  "Solution_27": "120 digits...i got hooked when someone did a pi song at san jose arml 2004  :D",
  "Solution_28": "3.14159265??",
  "Solution_29": "I remember pi upto 3.14159 and never get into trouble. Memorizing these things is the job of machines. My calculator has it built in."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "invariant",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all sequences of positive integers $a_n$ such that \r\n$a_na_{n+3} = a_{n+1}a_{n+2} + 1 $ for all integers n",
  "Solution_1": "All integers n? I suppose it doesn't really matter, but is that meant to be whole numbers? Or natural numbers?",
  "Solution_2": "It doesn't really matter, but I meant all integers. The equation is symmetric so we can solve for either the next term or the previous term in the recursion, unless a particular term is 0. The restriction of values to the positive integers is much more important- it gets very complicated when 0 is a term.",
  "Solution_3": "After seeing the new site organization, I think this would fit best in number theory/proposed and own problems. It's not a perfect match, but at least the problem is based in integer arithmetic. This is also clearly an olympiad-level problem.\r\n\r\nI am the source here- I've never seen any version of this problem elsewhere, and I first posed it to others in 1999 (at MOSP).",
  "Solution_4": "I'm reviving this problem, since no solutions were posted the first time.\r\n\r\nUp to translation (which I don't care about), there are finitely many solutions. If the sequence took integer values, there would be uncountably many solutions- but almost all would take 0 as a value infinitely often.\r\n\r\nIt isn't that hard to find the solutions just by playing with the recursion; that's how I came up with this problem in the first place.",
  "Solution_5": "I think a crucial fact is writing $\\frac{a_{n+3}}{a_{n+1}}=\\frac{a_{n+2}+a_{n+4}}{a_n+a_{n+2}}$, and then multiplying these. We get $a_{2k+2}=-a_{2k}+\\frac{a_0+a_2}{a_1}\\cdot a_{2k+1}$ (I think; something like that, anyway). I'll complete it after I get some sleep :).",
  "Solution_6": "The sequence is uniquely determined by a triplet of consecutive terms, $a_0,a_1,a_2$ s.t. $a_0|a_1a_2+1,\\ a_1|a_0+a_2,\\ a_2|a_1+a_0$. I mean that these conditions are necessary and sufficient (all triplets of consecutive terms satisfy these). \r\n\r\nNow take the triplet with the minimal sum. We have $(a_0,a_1)=(a_1,a_2)=(a_2,a_0)=1$, and $a_0\\le \\frac{a_1a_2+1}{a_0},\\ a_2\\le\\frac{a_0a_1+1}{a_2}$. These relations show that $\\max(a_0,a_1,a_2)=a_1\\Rightarrow a_0+a_2\\in\\{a_1,2a_1\\}$. If it\u2019s $2a_1$, we get $a_0=a_1=a_2=1$. \r\n\r\nLet\u2019s look at the case $a_1=a_0+a_2$ now. We apply two transformations, and go to the triplet $(a_2,a_3,a_4)$. It\u2019s easy to show that $a_3=a_2+a_4$, so the minimaltiy of the sum of the first triplet is equivalent to $a_1\\le a_3\\iff a_1\\le \\frac{a_1a_2+1}{a_0}$. Assume WLOG that $a_2<a_0$ (otherwise we use the triplet $(a_{-2},a_{-1},a_0)$ instead of $(a_2,a_3,a_4)$). We thus get $a_1\\le a_1-\\frac{a_1-1}{a_0}\\Rightarrow a_1=1$, which is absurd. We must thus have $a_0=a_2$, and since $(a_0,a_2)=1$, we get $(a_0,a_1,a_2)=(1,2,1)$.\r\n\r\nThese are the two solutions: $(a_0,a_1,a_2)\\in\\{(1,1,1),(1,2,1)\\}$. \r\n\r\nSince $\\frac{a_{2k}+a_{2k+2}}{a_{2k+1}}$ and $\\frac{a_{2k-1}+a_{2k+1}}{a_2k}$ are invariants of the sequence and the pairs of such invariants of the sequences generated by the two triplets are different, the two sequences are, indeed, different.\r\n\r\nI know there\u2019s some stuff I used without proof, but I can make it clear in case someone wants me to. \r\n\r\nIs it Ok?",
  "Solution_7": "It's related to: http://www.mathlinks.ro/Forum/viewtopic.php?p=15723&highlight=&sid=cb803f607f5f513e4b0956fe9f83075c#15723\r\n\r\nMore general , let $a_n$ be the numbers produced by a machine such that  :\r\n\r\n$a_{n+3} = \\frac {a_{n+1}a_{n+2} + T}{a_n}$ , where $T$ is an integer ,  $a_0,a_1,a_2$ are non zero and coprime integers with $T$ ( the common divisors with $T$ are only 1 and -1)\r\n\r\n If  $a_3,a_4,a_5$ are integers numbers then  the machine produces only coprime integers with $T$,  or it stopes (because division by 0 :D ) .\r\n\r\n :cool: Very interesting, it's a reverse machine !:cool:",
  "Solution_8": "Your proof looks good, grobber; maybe the algebra you started with should be expanded, including the proof that (1,2,1) and (1,1,1) actually work.\r\n\r\nMy own proof used a minimal pair (either in sum or product); it is easy to show that one of these must be a 1, giving the sequence a (1,x,1) block. If $a_{-1}=1, a_0=x,a_1=1$, then $a_3=\\frac{x+2}{x}$, so either $x=1$ or $x=2$.\r\nFor the rest, I used a lemma that if six consecutive terms are nonzero integers, the next term on each side is an integer. In the positive case, this forces all terms to be integers (this lemma would also work for the problem Diogene posted, with no modifications to its proof).\r\n\r\nA few terms of these sequences:\r\n$\\dots, 1,1,1,2,3,7,11,26,41,97,153, \\dots$\r\n$\\dots, 1,2,1,3,2,7,5,18,13,47,34, \\dots$\r\n\r\nA sequence with a zero in it, just as a taste of what that allows:\r\n$\\dots, -7,3,-5,2,-3,1,-1,0,1,-1,-2,3,5,-8,-13, \\dots$",
  "Solution_9": "In order to explain why those solutions work, it's enough to observe that if $a_n,a_{n+1},a_{n+2}$ satisfy those three relations, then so do $a_{n-1},a_n,a_{n+1}$ and $a_{n+1},a_{n+2},a_{n+3}$ (the three relations I'm talking about are: $a_n|a_{n+1}a_{n+2}+1,\\ a_{n+1}|a_n+a_{n+2},\\ a_{n+2}|a_na_{n+1}+1$)."
}
{
  "Tag": [
    "modular arithmetic",
    "quadratics",
    "number theory",
    "relatively prime",
    "Divisibility Theory",
    "pen"
  ],
  "Problem": "Let $m$ and $n$ be natural numbers such that \\[A=\\frac{(m+3)^{n}+1}{3m}\\] is an integer. Prove that $A$ is odd.",
  "Solution_1": "[quote=\"PDatK40SP\"]We can also prove that $m \\equiv 2 \\ (mod \\ 12)$ ( and replace $m$ by $12t+2$ in the formula of $A$, we will have $A$ is odd immediately ). Use this lemma: ( I think it's well-known ) Let $n$ is an odd integer, and $k$ is an odd divisor of $3^{n}+1$. Then, $k \\equiv 1 \\ (mod \\ 6)$\n$m|(m+3)^{n}+1 \\Rightarrow m|3^{n}+1$\nOf course $3 \\not | m$, so if $n$ is even, then $(m+3)^{n}+1 \\equiv 2 \\ (mod \\ 3)$, which is impossible because $3m|(m+3)^{n}+1$. \nSo $n$ is even, then, from the lemma, we have $m = 2^{k}(6t+1) \\ \\ (k,t \\in \\mathbb{N})$\nIf $k$ is even, we have $(m+3)^{n}+1 \\equiv 2 \\ (mod \\ 3)$, which is impossible. Moreovers, easy to see that $8 \\not | 3^{n}+1 \\ \\forall n \\in \\mathbb{N}$ so $k \\geq 3$ is impossible.\nSo, $k=1 \\Rightarrow m=12t+2$[/quote]",
  "Solution_2": "[quote=\"Peter\"][quote=\"PDatK40SP\"] Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n}\\plus{}1$. Then, $ k\\equiv 1\\ (mod\\ 6)$[/quote][/quote]\r\nWhy?",
  "Solution_3": "[quote=\"\u0421\u0442\u0443\u0434\u0435\u043d\u0442\"][quote=\"Peter\"][quote=\"PDatK40SP\"] Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n}\\plus{}1$. Then, $ k\\equiv 1\\ (mod\\ 6)$[/quote][/quote]\nWhy?[/quote]\r\n\r\nIt's equivalent to showing that none of its prime divisors can be 5 modulo 6.\r\n\r\n\r\nLet $ a \\equal{} 3^{(n\\plus{}1)/2}$ and suppose $ 3^{n}\\plus{}1$ is divisible by a prime $ p\\equiv 5\\pmod{6}$. Then $ a^{2}\\equiv\\minus{}3\\pmod{p}$. At this point, Quadratic Reciprocity would work. You could also show that\r\n\\[ x\\equal{}2^{\\minus{}1}(a\\minus{}1)\\Rightarrow x^{2}\\plus{}x\\plus{}1\\equiv 0\\pmod{p}\\]\r\nSo $ x^{3}\\equiv 1\\pmod{p}$ and $ x^{p\\minus{}1}\\equiv 1\\pmod{p}$. We have $ p\\minus{}1$ and 3 relatively prime, so $ x\\equiv 1\\pmod{p}$. But as $ x^{2}\\plus{}x\\plus{}1\\equiv 0\\not\\equiv 3\\pmod{p}$, this is a contradiction.",
  "Solution_4": "More :\r\n1) There exist infinite $ (m,n)\\in N$ satisfy the condition.\r\n2)$ m\\equiv 2(\\mod 12)$ (take from my teacher).",
  "Solution_5": "[quote=\"Peter\"][quote=\"PDatK40SP\"] Use this lemma: ( I think it's well-known ) Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n} \\plus{} 1$. Then, $ k \\equiv 1 \\ (mod \\ 6)$[/quote][/quote]\r\n\r\nwhat's well known lemma?",
  "Solution_6": "i will fakesolve ur problem",
  "Solution_7": "[quote=TTsphn]More :\n1) There exist infinite $ (m,n)\\in N$ satisfy the condition.\n2)$ m\\equiv 2(\\mod 12)$ (take from my teacher).[/quote]\n\nI agree with your Teacher. Except of some ifs and buts.  According to the mysterious Lemma the solution of the problem A17 is:\n\nm = 2 + 12t  for ANY natural t, including t = 0. Those t's may also be perfect squares like t = 4, 9, 16, 25,... so that m = 50, 110, 194, 302,.. \n\nAll those m's make A an odd number, at least for certain n's.  Right?   If you agree, let me offer you my CONJECTURE A17. :\n\n                                        If  t > 1  is a perfect square, then A  is not  an integer, for ANY n.\n\nUnable to prove or disprove my conjecture.  If I'm right, then maybe Peter's Lemma has to be modified.(?)  Please help. Thanks.\n\n\n\n\n",
  "Solution_8": "Peter,\nThe first sentence of your post.  You \u201c\u2026will have A odd immediately\u201d(?)\nBut what if t = 4, 9, 16, 25, 36 , 49\u2026or  12, 5, 19 and n is any number. Any!  \nThen A will be proper or improper fraction - not an odd as you claim.   \n    If  t = 0  AND  n = 1, 3, 5, 7,...    or   t = 7 AND  n = 21, 63, 105, 147, \u2026. or   t = 22 AND  n = 9, 27, 45, 63, 81\u2026.\nthen in each case we get infinite number of odd A\u2019s  \u201cimmediately\u201d and  \"m = 2 + 12t\" works.  Nothing is always\n   You ignore(?) that A is not A(t), but  A(t, n).   t is parameter, n is a variable (see below)\nTrivial: A can be either  a fraction (for any n) or an odd  (for certain t\u2019s and certain unlimited  n\u2019s). \n     \nLet\u2019s subdivide the set of 26 consecutive t\u2019s (from 0 to 25)  by two subsets:  \u2018bad\u2019  t\u2019s  and  \u2018good\u2019  t\u2019s.\n\nProposition 1.         \u2018Bad\u2019 t\u2019s do not make A an odd. For any n. \n\nHere they are:   t =    4,  9,  12,  14,  15,  16,  18,  19,  20,  21,  23,  24,  25.\n\nProposition 2.       \u2018Good\u2019 t\u2019s do make A an odd.  But not for any n.\n\nGood  t\u2019s:            t =     0,  1,  2,  3,  5,  6,  7,  8,  10,  11,  13,  17,  22 \n\nBad t\u2019s do not solve the problem A17.  We\u2019ll ignore them from now on.\n\nNow we\u2019ll find all unlimited odd A\u2019s (all of them) for first 13 good t\u2019s and and predefined n\u2019s\nWe \u2018ll show that A can not be just any odd  but unlimited strictly defined ones.\nIn the table below k runs through all odds:   k = 1, 3, 5, 7, \u2026.\n\nIn A(t, n):  t is parameter,  n is variable (a linear function of odd k\u2019s for each \u2018good\u2019 t). Some examples:\nt = 0     A(0, k) is  an  odd  for any n  =   k  =  1, 3, 5, 7, \u2026\u2026.             then A(0,  k) = 1,  21,  521,  13 021,   325 521, \u2026\nt = 1     A(1, 3k) is an odd   for any n = 3k   =   3, 9, 15, 21\u2026..           then A(1, 3k) = 117,   2 823 520 869,  \u2026\u2026\u2026           \nt = 7     A(7, 21k) is an odd  for any  n = 21k = 21,  63,  105,  147,\u2026 then A(7,21k) =\u2026..infinite number of huge odds        \nt = 13   A(13, 39k) is an odd  for any n = 39k =  39,  117,  195,\u2026      then A(13, 39k) = \u2026..          \u2018\u2019\n\n\u2018good\u2019 t___m=2+12t_____n_______A(t,n)  =  ((5+12t)^n + 1)/3(2+12t)_____A(t, n),  all odds.\n============ ==================================================================\n                                                                                                                                                                                                         \n0_________2__________k________A (0, k) = (5^k + 1)/6_____________A(0, k)  = 1,  21,  521,  13021,  325521,.                                                                                                                                                                                                                                   \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t   \n1________14_________ 3k_______A (1, 3k) = (17^3k + 1)/42_________ A(1, 3k) =  117,  2823520869, \u2026..          \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\n2________26__________3k_______A(2, 3k) = (29^3k + 1)/78_________ A(2, 3k) = Infinite number of  huge odds \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t \t\n3________38__________9k_______A(3, 9k) = (41^9k + 1)/114________ A(3, 9k) = _______ same\n\n5________62_________15k_______A(5,15k) = (65^15k + 1)/186_______A(5, 15k)________ same                                                                                   \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\n6________74__________9k_______A (6, 9k) =  (77^9k + 1)/222_______ A(6, 9k)_________ same\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t     \n7________86_________ 21k______ A(7, 21k)  =  (89^21k + 1)/258_____ A(7, 21k)________ same\n\n8________98__________21k______A(8, 21k)  = (101^21k + 1)/294_____A(8, 21k)________ same\n\n10______122___________5k______A(10, 5k) = (125^5k + 1)/366______ A(10, 5k)__________\"\n\n11______134__________11k______A(11, 11k) = (137^11k + 1)/402_____A(11, 11k)__________\" \n\n13______158__________39k______A(13, 39k) = (161^39k + 1)/474_____A(13, 39k)_________\u2018\u2019                                                                                                                                                                                          \n \n17______206__________17k______A(17, 17k) = (209^17k + 1)/618_____A(17, 17k)_________\u2018\u2019      \n\n22______266___________9k_____ A(22, 9k ) = (269^9k  + 1)/798______A(22, 9k)___________\u2018\u2019\n\nThere are 13 good and 13 bad t's. Just coincidence.\nThat\u2019s all I can say about that. As an exercise: for  25 < t < 51 find all \u2018good\u2019 t\u2019s and corresponding n\u2019s.",
  "Solution_9": "all odd prime factors of m are $\\equiv 1 \\pmod{3}$\nwrite $m=2^k \\cdot j$, $k \\geq 0$ and $j\\equiv 1 \\pmod{3}$\ntaking modulo 3, $m\\equiv 2\\pmod{3}$ and n is odd\n$2^k \\cdot j \\equiv 2 (mod 3) \\implies$ k is odd\nbut $k\\geq 3 \\iff 8|m \\implies 8|3^n+1$ creates contradiction\nso $k=1 \\iff 2||m$\nso $m\\equiv 2\\pmod {12}$\n\n$(m+3)^n+1\\equiv 2\\pmod4$ , so A is odd\none example is $(m,n,A)=(2,1,1)$",
  "Solution_10": "[quote=Peter]Let $m$ and $n$ be natural numbers such that \\[A=\\frac{(m+3)^{n}+1}{3m}\\] is an integer. Prove that $A$ is odd.[/quote]\n\nWhen $m$ is odd ,$ A$ is always odd because both numerator and denominator are odd.\n\n [b]Claim:[/b]  When $  m,n$ are both even ,or  when $ m$ is divisible by $3$ then $ A$ is never an integer.\n[b]Proof[/b] If $m \\equiv 0 \\mod {3}$ then $$(m+3)^n+1 \\equiv m^n+1 \\equiv 1 \\mod 3 $$ wich contradicts the fact that $ A$ is integer .\nWhen $m  \\not\\equiv 0 \\mod 3$ and $n $ even we have that$$(m+3)^{2k}+1 \\equiv m^{2k}+1 \\equiv 2 \\mod 3 $$wich again contradicts that $A$ is integer \n\n[b]Claim:[/b] When $m$ is even , then we need $m=2k $ or $m= 4k$ where $k$ is odd integer for $A $ to be integer. \n[b]Proof:[/b] We must have $$ (m+3)^n+1 \\equiv 3^n+1\\equiv 0 \\mod m$$ or $$ m| 3^n+1$$ since $n$ is odd\n\nBy LTE we can say that $$v_2(m) \\leq v_2 (3^n+1)=v_2(4)+v_2(n)=2 $$\n\n[b]Claim: [/b] When $m=2k$ or $m=4k$ then $v_2(A)=0$\n[b]Proof[/b] $$v_2(A)=v_2 ((m+3)^n+1)-v_2(3m)=v_2(2k+4)-v_2(6k)=0$$ for $ m=2k$ and \n$$v_2(A)=v_2(4k+4)-v_2(12k)=0$$ for $m=4k$\n\nAfter checking all cases we can say that $A$ is always odd. $\\blacksquare$",
  "Solution_11": "[quote=XbenX]\n[b]Claim: [/b] When $m=2k$ or $m=4k$ then $v_2(A)=0$\n[b]Proof[/b] $$v_2(A)=v_2 ((m+3)^n+1)-v_2(3m)=v_2(2k+4)-v_2(6k)=0$$ for $ m=2k$ and \n$$v_2(A)=v_2(4k+4)-v_2(12k)=0$$ for $m=4k$\n[/quote]\n\nThis part is wrong. Suppose we have $m \\equiv 4 \\text{ (mod }8 \\text{)}$ and $n$ an odd integer, then $(m+3)^n$ is congruent to $7$ (mod $8$). then $v_2(A)$ might be positive, and in that case we need to prove the divisibility relation can't hold, instead of computing $v_2(A)$ more. ",
  "Solution_12": "Legandre symbol"
}
{
  "Tag": [
    "permutations",
    "combinatorics",
    "IMO",
    "IMO 1963"
  ],
  "Problem": "Five students $ A, B, C, D, E$ took part in a contest. One prediction was that the contestants would finish in the order $ ABCDE$. This prediction was very poor. In fact, no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order $ DAECB$. This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.",
  "Solution_1": "[hide]\nWe use casework on the first place student.\n\nIt is impossible for $ A$ to be first, as this would agree with Prediction 1.\n\nIf $ B$ is first, then $ (CB)$ could not have finished consecutively.  So the possible pairs from Prediction 2 are $ (DA)$, $ (AE)$, and $ (EC)$.  Only $ (DA)$ and $ (EC)$ are disjoint.  We cannot start with $ BEC$, as this would place $ C$ as predicted in Prediction 1.  So our order is $ BDAEC$, which has no agreement with Prediction 2.  So $ B$ cannot be first.\n\nIf $ C$ is first, then the possible pairs from Prediction 2 are $ (DA)$, $ (AE)$, and $ (CB)$.  We cannot use $ (CB)$ because it would place $ B$ as predicted by Prediction 1.  This leaves only $ (DA)$ and $ (AE)$, which are not disjoint, so it is impossible for $ C$ to be first.\n\nIf $ D$ is first, we can either use $ (DA)$ or not.  If we do, then we must choose either $ (EC)$ or $ (CB)$ as our second pair.  If we use $ (EC)$, we cannot start $ DAEC$, as this agrees in four places with Prediction 2.  So we have $ DABEC$, which shares $ (AB)$ with Prediction 1, so it is impossible.  If we use $ (CB)$, we cannot start $ DACB$, as $ C$ agrees with Prediction 1.  So we are left with $ DAECB$, which agrees in five places with Prediction 2.\n\nSo we cannot use $ (DA)$.  The possible pairs from Prediction 2 are $ (AE)$, $ (EC)$, and $ (CB)$.  Only $ (AE)$ and $ (CB)$ are disjoint.  We cannot place $ (AE)$ right after $ D$, as $ (DA)$ is forbidden.  So we must have $ DCBAE$, in which $ E$ agrees with Prediction 1.  So $ D$ cannot be first.\n\nThus, $ E$ is first.  The possible pairs from Prediction 2 are $ (DA)$, $ (EC)$, and $ (CB)$. If we use $ (EC)$, then we must also use $ (DA)$.  But $ ECBDA$ and $ ECDAB$ have only zero and one correspondences with Prediction 2, respectively.  So we cannot use $ (EC)$   \n\nIf we don't use $ (EC)$, then we must have $ EDACB$ (preventing $ (EC)$ by not placing $ (CB)$ right after $ E$).  This clearly satisfies all requirements, so it is the desired order.\n[/hide]",
  "Solution_2": "Wouldn't that order be wrong, though? B and C finish consecutively, as do D and E.\n\nOr is it that they must finish consecutively in the order that was predicted? So A can't be right before B, but B can be right before A?",
  "Solution_3": "Why there are no pure combinatorics questions in the early IMOs?",
  "Solution_4": "[quote=sunjae62700]Why there are no pure combinatorics questions in the early IMOs?[/quote]\n\nDo you want to mean counting??"
}
{
  "Tag": [
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "1,Suppose $ A,B\\in F^{n\\times n},AB\\equal{}BA\\equal{}\\mathbf{0},rank(A^2)\\equal{}rank(A)$,show that $ rank(A\\plus{}B)\\equal{}rank(A)\\plus{}rank(B)$.\r\n2,Suppose $ A,B\\in F^{n\\times n},AB\\equal{}BA\\equal{}\\mathbf{0}$.Prove that,there exists a positive integer $ k$ such that $ rank(A^k\\plus{}B^k)\\equal{}rank(A^k)\\plus{}rank(B^k)$.\r\n3,Suppose $ A\\in F^{m\\times n},B\\in F^{n\\times m}$.Prove: $ rank(AB)\\equal{}rank(A)$ if and only if there exists $ C\\in F^{m\\times n}$ such that $ A\\equal{}ABC$. Furthermore ,Prove that if $ (AB)^2\\equal{}AB$ then $ (BA)^2\\equal{}BA$.\r\n\r\nI found these problems difficult for me. Any help will be appreciated.",
  "Solution_1": "I found on CMJ,  David Calan proved the following theorem:\r\n\r\nIf $ A$ and $ B$ are two matrices. Then\r\n\\[ \\text{Rank}(A) \\plus{} \\text{Rank} (B) \\equal{} \\text{Rank}(A \\plus{} B)\r\n\\]\r\nif and only if $ C(A)\\cap C(B) \\equal{} \\{0\\}$ and $ R(A)\\cap R(B) \\equal{} \\{0\\}$, where $ C(X)$ and $ R(X)$ mean the column and row space of $ X$ respectively.\r\n\r\nThe proof is quite elementary (I'll post the proof if anybody need it. :) ).\r\n\r\nThis theorem solves the first problem since form $ AB \\equal{} 0$ implies $ C(B) \\subset \\text{Ker}(A)$ and $ \\text{Rank}(A^2) \\equal{} \\text{Rank}(A)$ implies $ C(A) \\cap \\text{Ker}(A) \\equal{} \\{0\\}$ thus  $ C(A)\\cap C(B) \\equal{} \\{0\\}$. Similarly $ R(A)\\cap R(B) \\equal{} \\{0\\}$.",
  "Solution_2": "Wich issue of  CMJ did you find the result of David Calan ? thank",
  "Solution_3": "For those who have access to the jstor: [url=http://www.jstor.org/pss/2687710]When is rank additive?[/url]",
  "Solution_4": "For 2/\r\n\r\nThere exist $ q$ positive integer such that \r\n\r\n$ Im(A^k)\\equal{}Im(A^q)$ for any $ k\\geq q$\r\n\r\nWe have also $ Im(A^q) \\cap ker(A^q) \\equal{} 0$ \r\n\r\nSince A,B commute we have $ A^qB^q\\equal{}0$ \r\n\r\n$ Im(B^q)$ is included in $ ker(A^q)$ \r\n\r\nWe obtain \r\n\r\n$ Im(A^q) \\cap Im(B^q)$ is included in $ Im(A^q) \\cap ker(A^q) \\equal{} 0$ \r\n\r\nSo $ C(A^q) \\cap C(B^q) \\equal{} 0$  \r\n\r\nto finish apply the result of beginner \r\n\r\n$ rank(A^q\\plus{}B^q)\\equal{}rank(A^q)  \\plus{} rank(B^q)$",
  "Solution_5": "can some one post a proof of beginner theorem"
}
{
  "Tag": [
    "IMO"
  ],
  "Problem": "Greetings! I was the fourth deadly SIN at the Glasgow IMO.\r\nI am now a conscript. *sob*\r\nSadly, this happens to all male singaporean citizens who aren't exempted on medical grounds. It is a sad fate. So while my foreign peers are out there in university, I'm languishing in an undisclosed location, somewhere within 20km of the coast. If you look at the map of Singapore, you will realise that this is not saying much.\r\n\r\nThe reason why I am online the whole of this sunday is really because I'm on manning duty, which means that I'm sitting at my office computer waiting for a phone call that I hope will never come. (This is ******. ****** is activated. ****** will be at ******)\r\n\r\nMy path to the IMO:\r\n2001: Lose tie-breaker, get sent to SEAMEO-MO instead, first (but vietnam wasn't there)\r\n\r\n2002: Win tiebreaker (notice the pattern?). Get silver. Become extremely depressed a few months later upon getting conscripted. (and upon the realisation of the fact that I'll probably never have so much fun)",
  "Solution_1": "IMO2002\r\n\r\nZachary Leung (no longer a citizen - who'd want to be at our age? Ans. Females)\r\n21 bronze\r\nCalvin Lin (now a conscript)\r\n10\r\nMeng Dazhe (not a citizen)\r\n20 bronze\r\nKiat Chuan (conscript)\r\n24 silver\r\nJulius Poh (conscript)\r\n25 silver\r\nColin Tan (medically exempted.)\r\n12 HM\r\n\r\nNote that 3/6 are exempt from conscription. Clearly I have been less than cunning.\r\n\r\nWe were 30th. I would like to point out that there was only 1 team <20 pts below us, whereas there were 6 teams within 10 pts (more) of us, and 12 within 20.\r\n\r\nThe only thing I can say to this is - darn!\r\n\r\nBut this year our team's 18th! Makes an old fogey like me feel dull.\r\n*wave flag*"
}
{
  "Tag": [],
  "Problem": "From:   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  "Solution_1": "\u5f13\u867d\uff01"
}
{
  "Tag": [
    "topology",
    "function",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "1. Prove that in $2^{nd}$ countable space a non countable set has an accumulation point.\r\n2. Prove that a space with a closed basis is completely regular.\r\n\r\nThanks!\r\n\r\n[b]edit[/b]: Corrected the statement of the 1st problem.",
  "Solution_1": "What's the first question really like? :) As it is, it's clearly false: take $\\mathbb R$ with the usual topology. It's second countable, and there are lots of uncountable closed sets.",
  "Solution_2": "I'm really sorry grobber.  :blush: \r\n\r\nAnd now I see it's pretty easy..  :oops: \r\n[hide]\nLet [tex]<X,\\tau>[/tex] be [tex]2^{nd}[/tex] countable topological space. Let [tex]A \\subseteq X[/tex] s.t. [tex]|A| > \\aleph_{0}[/tex]. Let's assume there is no accumulation point of [tex]A[/tex]. Then for any [tex]p \\in A[/tex] we have an open set [tex]G_{p}[/tex] about [tex]p[/tex] s.t. it does not contain any other points from [tex]A[/tex]. Let's define [tex]f: P(A) \\rightarrow \\tau[/tex] like this:\n\n$f(I)= \\bigcup_{\\mu \\in I} G_{\\mu}$ \n\nLet $I_1 \\neq I_2$ be subsets of $A$. Then WLOG, there is $\\mu \\in I_1$ s.t. $\\mu \\notin I_2$. This clearly means that $\\mu \\in f(I_1)$ but $\\mu \\notin f(I_2)$. Therefore, $f(I_1) \\neq f(I_2)$. So $f$ is an injective map, therefore $|\\tau| \\geq 2^{|A|} > 2^{\\aleph_0}$. But $2^{nd}$ countable space can't have topology of cardinality above $2^{\\aleph_0}$! Therefore we got our contradiction.\n$\\blacksquare$\n[/hide]",
  "Solution_3": "You can prove a lot more: every isolated point of the set is the only point of the set in some properly chosen basis element. This means that we can establish an injection from the set of isolated points of our set to the basis. If the basis is countable, this implies that every set has countably many isolated points, and the first question is weaker than this.\r\n\r\nAs for the second one, what do you mean by closed basis? :? Usually, bases of open sets are considered. Do you mean here that you have some open sets which form a basis for the topology and are also closed?",
  "Solution_4": "I mean that all the members of the basis are closed and open at once.",
  "Solution_5": "grobber, I don't understand how are you building your injection from the set of isolated points to the members of a countable basis.. Can you explain please?",
  "Solution_6": "For every isolated point you consider a basis element which contains only that point (there must be such a basis element, by the definition of an isolated point). \r\n\r\nAs for the other question, if $x$ is a point and $F\\supset x$ is a closed set not containing $x$, then take any basis element $B$ around $x$ and disjoint from $F$, and define the function you're looking for to be $0$ in $B$ and $1$ in the complementary of $B$. It's easy to check that's it's continuous.",
  "Solution_7": "Thanks grobber!  :)"
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "trigonometry",
    "quadratics",
    "trapezoid"
  ],
  "Problem": "Let ABCD is a quarilateral , $\\ S_{ABCD}=8$, $\\ AB+AC+CD=8$.Prove that ABCD is a parallelogram and find its diagonal :lol:",
  "Solution_1": "[quote=\"TRAN THAI HUNG\"]Let ABCD is a quarilateral , $\\ S_{ABCD}=8$, $\\ AB+AC+CD=8$.Prove that ABCD is a parallelogram and find its diagonal :lol:[/quote]\r\nIn the solution below, we use the idea that $ax^{2}+bx+c \\geq 0$ for all $x \\in \\mathbb{R}$ iff $a \\geq 0$ and $b^{2}-4ac \\leq 0 .$\r\n\r\nLet $\\angle BAC = \\alpha$ and $\\angle ACD = \\beta.$ Then we have the relation $\\frac12 (AB)(AC) (\\sin \\alpha)+\\frac12 (CD)(AC)(\\sin \\beta) = S_{ABCD}= 8$\r\n\r\n$\\Rightarrow (\\sin \\beta ){AC}^{2}+[\\sin \\beta (AB-8)-AB \\sin \\alpha] AC+16 =0$ ... (I) (quadratic in $AC$ with real roots.)\r\n\r\n$\\Rightarrow{[(\\sin \\beta (AB-8)-AB \\sin \\alpha)]}^{2}-4 \\cdot 16 \\cdot \\sin \\beta \\geq 0$\r\n\r\n$\\Rightarrow{(\\sin \\beta-\\sin \\alpha)}^{2}{AB}^{2}-16 \\sin \\beta (\\sin \\beta-\\sin \\alpha) AB+64({\\sin}^{2}\\beta-\\sin \\beta ) \\geq 0$\r\n\r\nNow using the result noted in the beginning, we obtain\r\n\r\n${16}^{2}{\\sin}^{2}\\beta{(\\sin \\beta-\\sin \\alpha)}^{2}-4{(\\sin \\beta-\\sin \\alpha)}^{2}\\cdot 64 ({\\sin}^{2}\\beta-\\sin \\beta) \\leq 0$\r\n\r\n$\\Rightarrow{(\\sin \\beta-\\sin \\alpha)}^{2}\\sin \\beta \\leq 0$\r\n\r\n$\\because \\sin \\beta \\neq 0,$ we must have $\\sin \\beta = \\sin \\alpha$ ... (II)\r\n\r\nPlugging this back into equation (I), we get $AC(8-AC) \\sin \\beta =16.$ But note that $AC$ and $AC-8$ are both positive, and so applying $AM-GM,$ we get $AC(8-AC) \\leq 16 \\Rightarrow \\sin \\beta \\geq 1 \\Rightarrow \\beta = 90^{\\circ}= \\alpha.$\r\n\r\nThus, we have $AC(8-AC) = 16 \\Rightarrow{(AC-4)}^{2}= 0 \\Rightarrow AC=4$ (the length of the diagonal.)\r\n\r\nCan someone finish this off by proving that $ABCD$ is really a parallelogram? :)",
  "Solution_2": "[hide][quote=\"FieryHydra\"][quote=\"TRAN THAI HUNG\"]Let ABCD is a quarilateral , $\\ S_{ABCD}=8$, $\\ AB+AC+CD=8$.Prove that ABCD is a parallelogram and find its diagonal :lol:[/quote]\nIn the solution below, we use the idea that $ax^{2}+bx+c \\geq 0$ for all $x \\in \\mathbb{R}$ iff $a \\geq 0$ and $b^{2}-4ac \\leq 0 .$\n\nLet $\\angle BAC = \\alpha$ and $\\angle ACD = \\beta.$ Then we have the relation $\\frac12 (AB)(AC) (\\sin \\alpha)+\\frac12 (CD)(AC)(\\sin \\beta) = S_{ABCD}= 8$\n\n$\\Rightarrow (\\sin \\beta ){AC}^{2}+[\\sin \\beta (AB-8)-AB \\sin \\alpha] AC+16 =0$ ... (I) (quadratic in $AC$ with real roots.)\n\n$\\Rightarrow{[(\\sin \\beta (AB-8)-AB \\sin \\alpha)]}^{2}-4 \\cdot 16 \\cdot \\sin \\beta \\geq 0$\n\n$\\Rightarrow{(\\sin \\beta-\\sin \\alpha)}^{2}{AB}^{2}-16 \\sin \\beta (\\sin \\beta-\\sin \\alpha) AB+64({\\sin}^{2}\\beta-\\sin \\beta ) \\geq 0$\n\nNow using the result noted in the beginning, we obtain\n\n${16}^{2}{\\sin}^{2}\\beta{(\\sin \\beta-\\sin \\alpha)}^{2}-4{(\\sin \\beta-\\sin \\alpha)}^{2}\\cdot 64 ({\\sin}^{2}\\beta-\\sin \\beta) \\leq 0$\n\n$\\Rightarrow{(\\sin \\beta-\\sin \\alpha)}^{2}\\sin \\beta \\leq 0$\n\n$\\because \\sin \\beta \\neq 0,$ we must have $\\sin \\beta = \\sin \\alpha$ ... (II)\n\nPlugging this back into equation (I), we get $AC(8-AC) \\sin \\beta =16.$ But note that $AC$ and $AC-8$ are both positive, and so applying $AM-GM,$ we get $AC(8-AC) \\leq 16 \\Rightarrow \\sin \\beta \\geq 1 \\Rightarrow \\beta = 90^{\\circ}= \\alpha.$\n\nThus, we have $AC(8-AC) = 16 \\Rightarrow{(AC-4)}^{2}= 0 \\Rightarrow AC=4$ (the length of the diagonal.)\n\nCan someone finish this off by proving that $ABCD$ is really a parallelogram? :)[/quote][/hide]\r\nYou can use the another way for the easier.I will post the solution later :wink:",
  "Solution_3": "[quote=\"FieryHydra\"]\nLet $\\angle BAC = \\alpha$ and $\\angle ACD = \\beta.$ Then we have the relation $\\frac12 (AB)(AC) (\\sin \\alpha)+\\frac12 (CD)(AC)(\\sin \\beta) = S_{ABCD}= 8$\n[/quote]\r\nFrom this $8=\\frac{AB \\sin \\alpha+CD \\sin \\beta}{2}AC \\le \\frac{AB+CD}{2}AC = \\frac{(8-AC)AC}{2}\\le 8$ and the equality holds only when $AC=4$ and $\\alpha = \\beta =90^\\circ$\r\nso [b]ABCD is trapezoid[/b]. Consider the case $AB=3$, $CD=1$. Hope I'm not wrong but it satisfies all conditions, so its impossible to prove that ABCD is paralelogram.  :maybe:",
  "Solution_4": "[quote=\"delta\"][quote=\"FieryHydra\"]\nLet $\\angle BAC = \\alpha$ and $\\angle ACD = \\beta.$ Then we have the relation $\\frac12 (AB)(AC) (\\sin \\alpha)+\\frac12 (CD)(AC)(\\sin \\beta) = S_{ABCD}= 8$\n[/quote]\nFrom this $8=\\frac{AB \\sin \\alpha+CD \\sin \\beta}{2}AC \\le \\frac{AB+CD}{2}AC = \\frac{(8-AC)AC}{2}\\le 8$ and the equality holds only when $AC=4$ and $\\alpha = \\beta =90^\\circ$\nso [b]ABCD is trapezoid[/b]. Consider the case $AB=3$, $CD=1$. Hope I'm not wrong but it satisfies all conditions, so its impossible to prove that ABCD is paralelogram.  :maybe:[/quote]\r\nAhh! I should have seen that! :) Thanks! \r\n\r\nYeah, it seems, $ABCD$ need not be a parallelogram after all.",
  "Solution_5": "Sorry,that my mistake it's a trapezoid :blush:  :wallbash_red:"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $n$ be a positive integer which has more than $k$ different prime factors. Prove that there exists a convex $n$-gon in the plane such that\r\n\r\n[b]1.[/b] all angles of the $n$-gon are equal;\r\n[b]2.[/b] the lengths of the sides of the $n$-gon are the numbers $1^{k}$, $2^{k}$, ..., $n^{k}$ in some order.\r\n\r\n  Darij",
  "Solution_1": "As I remember there is a generalisation with $P(x)=\\sum_{i=1}^{k}a_{i}x^{i}$\r\nand the polygon with sides $P(1),P(2),...,P(n)$\r\nbut the proof is almost the same :wink:"
}
{
  "Tag": [],
  "Problem": "For a positive integer $n$, let $r(n)$ denote the sum of the remainders when $n$ is divided by $1, 2, \\ldots , n$ respectively. Prove that $r(k) = r(k-1)$ for\r\ninfinitely many positive integers $k$.\r\n\r\nI was just kind of screwing around with this problem and I guessed \r\n\r\n[hide]$r(2^{k})=r(2^{k}-1)$[/hide]\r\n\r\nwhich I don't know is right or not, and if it is I don't know how to prove it... :maybe:",
  "Solution_1": "[hide] If you let $s_{m}(n)$ be the remainder when $n$ is divided by $m$, we have $r(n) = \\sum_{m=1}^{n}s_{m}(n)$.\n\nNote that for $m > 1$\n\n$s_{m}(n) = s_{m}(n-1)+1$ if $m \\not| n$ and $s_{m}(n) = 0$ if $m | n$.\n\nSo\n\n$r(2^{k}) = \\sum_{m=2}^{2^{k}}s_{m}(2^{k}) = \\sum_{m=1}^{2^{k}-1}[s_{m}(2^{k}-1)+1]-\\sum_{i=0}^{k-1}2^{i}$.\n\nWe add $1$ to every $s_{m}$ term and subtract away the ones for which $s_{m}(2^{k}) = 0$, i.e. all powers of $2$ which were counted in the $2^{k}-1$ summation. This simplifies to\n\n$\\sum_{m=1}^{2^{k}-1}s_{m}(2^{k}-1)+\\sum_{m=1}^{2^{k}-1}1-\\sum_{i=0}^{k-1}2^{i}= r(2^{k}-1)+(2^{k}-1)-(2^{k}-1) = r(2^{k}-1)$.\n\nSo your answer is right :). [/hide]",
  "Solution_2": "This problem has been posted at least five times on this forum:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=125625\r\nhttp://www.mathlinks.ro/viewtopic.php?t=108343\r\nhttp://www.mathlinks.ro/viewtopic.php?t=37133\r\nhttp://www.mathlinks.ro/viewtopic.php?t=5866\r\n\r\nA follow up question: Prove that $ \\frac{n^2}{10}<r(n)<\\frac{n^2}{4}$ for all integers $ n\\ge7$."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ a,$ $ b,$ $ c$ be nonnegative real numbers such that $ a\\plus{}b\\plus{}c\\equal{}ab\\plus{}bc\\plus{}ca>0.$ If $ k \\ge \\frac{64}{105},$ then\r\n$ a\\plus{}b\\plus{}c \\ge \\sqrt{\\frac{a^2\\plus{}k}{1\\plus{}k}}\\plus{}\\sqrt{\\frac{b^2\\plus{}k}{1\\plus{}k}}\\plus{}\\sqrt{\\frac{c^2\\plus{}k}{1\\plus{}k}}.$",
  "Solution_1": "[quote=\"can_hang2007\"]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers such that $ a \\plus{} b \\plus{} c \\equal{} ab \\plus{} bc \\plus{} ca > 0.$ If $ k \\ge \\frac {64}{105},$ then\n$ a \\plus{} b \\plus{} c \\ge \\sqrt {\\frac {a^2 \\plus{} k}{1 \\plus{} k}} \\plus{} \\sqrt {\\frac {b^2 \\plus{} k}{1 \\plus{} k}} \\plus{} \\sqrt {\\frac {c^2 \\plus{} k}{1 \\plus{} k}}.$[/quote]\r\n\r\n\r\n\r\n$ k\\geq 2$\r\n\r\n$ (a \\plus{} b \\plus{} c)^2 \\ge 3\\sum {\\frac {a^2 \\plus{} k\\frac {(ab \\plus{} bc \\plus{} ca)^2}{(a \\plus{} b \\plus{} c)^2}}{1 \\plus{} k}}$."
}
{
  "Tag": [
    "FTW",
    "AMC",
    "AIME",
    "analytic geometry"
  ],
  "Problem": "after my failed last tourny... :(  i've decided to make another one\r\nhopefully one that'll do better\r\nanyway, here are the rules\r\n\r\nthis will be a 16-player tourny, open to all players\r\nwhen you sign up, plz give me your highest rating, as seeding will be done by this\r\nIt will be a normal game, 10 questions, 12 seconds, time based\r\nwhether it's rated or non-rated, will be up to the players\r\n\r\nplz sign up here and i wish all participants good luck!!!",
  "Solution_1": "pinkpiglet signs up!!!! highest: 1323 :blush:",
  "Solution_2": "WHEN WILL THE MADNESS END???? JUST SO YOU KNOW I, MEWTO55555, WILL BOYCOTT [b][u]ALL FUTURE TOURNAMENTS[/b][/u] UNTIL THE EXCESS AMOUNT OF TOURNAMENTS ENDS.",
  "Solution_3": "1367 highest\r\nWIILL GET HIGHER SOON!",
  "Solution_4": "[quote=\"mewto55555\"]WHEN WILL THE MADNESS END???? JUST SO YOU KNOW I, MEWTO55555, WILL BOYCOTT [b][u]ALL FUTURE TOURNAMENTS[/b][/u] UNTIL THE EXCESS AMOUNT OF TOURNAMENTS ENDS.[/quote]\r\n\r\n....didnt u start the first tourney...",
  "Solution_5": "Please ignore this post.",
  "Solution_6": "[quote=\"ernie\"]Please ignore this post.[/quote]\r\n\r\nu just [i]had[/i] to spam in my topic didn't u?\r\nwell, i'll do that 2 urs to :D \r\nMHAHAHAHAHA (i can't find an evil scientist's face...)",
  "Solution_7": "OK\r\nthis is the bracket\r\n\r\nROUND 1 \r\ngame 1: person 1 vs. person 16 \r\ngame 2: person 2 vs. person 15 \r\ngame 3: person 3 vs. person 14 \r\ngame 4: person 4 vs. person 13 \r\ngame 5: person 5 vs. person 12 \r\ngame 6: person 6 vs. person 11 \r\ngame 7: person 7 vs. person 10 \r\ngame 8: person 8 vs. person 9 \r\n\r\n\r\nWINNER'S BRACKET \r\nROUND 2 \r\ngame 9: winner of game 1 vs. winner of game 8 \r\ngame 10: winner of game 2 vs. winner of game 7 \r\ngame 11: winner of game 3 vs. winner of game 6 \r\ngame 12: winner of game 4 vs. winner of game 5 \r\n\r\n\r\nROUND 3 \r\ngame 13: winner of game 9 vs. winner of game 12 \r\ngame 14: winner of game 10 vs. winner of game 11 \r\n\r\n\r\nROUND 4 \r\ngame 15: winner of game 13 vs. winner of game 14 \r\n\r\nLOSER'S BRACKET \r\nROUND 2 \r\ngame 16: loser of game 1 vs. loser of game 8 \r\ngame 17: loser of game 2 vs. loser of game 7 \r\ngame 18: loser of game 3 vs. loser of game 6 \r\ngame 19: loser of game 4 vs. loser of game 5 \r\n\r\n\r\nROUND 2b \r\ngame 20: winner of game 16 vs. loser of game 9 \r\ngame 21: winner of game 17 vs. loser of game 10 \r\ngame 22: winner of game 18 vs. loser of game 11 \r\ngame 23: winner of game 19 vs. loser of game 12 \r\n\r\nROUND 3 \r\ngame 24: winner of game 20 vs. winner of game 23 \r\ngame 25: winner of game 21 vs. winner of game 22 \r\n\r\nROUND 3b \r\ngame 26: winner of game 24 vs. loser of game 13 \r\ngame 27: winner of game 25 vs. loser of game 14 \r\n\r\nROUND 4 \r\ngame 28: winner of game 26 vs. winner of game 27 \r\n\r\nROUND 4b \r\ngame 29: winner of game 28 vs. loser of game 15 \r\n\r\n\r\n\r\nCHAMPIONSHIP ROUND \r\ngame 30: winner of game 15 vs. winner of game 29",
  "Solution_8": "whoo hoo WOOT i join! 1357",
  "Solution_9": "ooh i wanna join\r\nTOURNEY FEVER :O\r\n\r\nhighest 1404",
  "Solution_10": "i'm joining 1369",
  "Solution_11": "ok, \r\nsoooo, pplz participating so far\r\n\r\n#1. Xoangieexo\u2026\u2026.1404\r\n#2. Johnnybeard\u2026\u20261367\r\n#3. Mathblitz\u2026\u2026\u2026.1357\r\n#4. Pinkpiglet\u2026\u2026\u2026.1323",
  "Solution_12": "you forgot sicarius",
  "Solution_13": "oops, right\r\nsrry sicarius\r\n\r\n#1. Xoangieexo\u2026\u2026.1404\r\n#2. Sicarius1029\u2026\u20261369\r\n#3. Johnnybeard\u2026\u20261367\r\n#4. Mathblitz\u2026\u2026\u2026.1357\r\n#5. Pinkpiglet\u2026\u2026\u2026.1323\r\n\r\nwe just need 11 more pplz to get started!!\r\ntell all ur friends!!!!",
  "Solution_14": "um...13someting...",
  "Solution_15": "changes in bracket\r\nwinners will be decided by the coin toss\r\n\r\nwinner's bracket round 3 and loser's bracket round 2 must be completed july 30th\r\nnext deadline will be after the weekend, during that time, i will not be here because i'm going to acadia(not sure if that's how u spell it)\r\ni will give out instructions and pms on July 31st, before i leave  \r\n\r\nROUND 1. \r\n#5. AIME_is_hard\u2026\u2026...1375 vs #9. Mathblitz(winner due to coin toss)\u2026\u2026\u2026..\u2026.1357 (game 1) \r\n#6. AIME15 (winner)\u2026\u2026....\u2026\u2026\u20261370 vs#10. Pinkpiglet\u2026\u2026\u2026\u20261323 (game 2) \r\n#7. Sicarius1029(winner due to ducky's FTW failing)\u2026\u2026..\u20261369 vs#11. Myyellowducky82...1200 (game 3) \r\n#8. Johnnybeard(winner due to coin toss)\u2026\u2026..\u20261367 vs#12. Pacman2812\u2026\u2026\u2026.1200 (game 4) \r\n\r\nWINNER'S BRACKET \r\nROUND 2. \r\nKl2836 vs. Johnnybeard(winner due to kl2836 not responding) (game 5) \r\nBudi713 vs. Sicarius1029(winner due to coin toss) (game 6) \r\nXoangieexo vs. AIME15(winner) (game 7) \r\nBisbmard9(winner due to coin toss) vs. Mathblitz (game 8) \r\n\r\nROUND 3 \r\n[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b](game 9) \r\n[b]Sicarius1029 [/b]vs. [b]AIME15[/b](game 10) \r\n\r\nROUND 4 \r\nwinner of game 9 vs. winner of game 10 (game 11) \r\n\r\nLOSER'S BRACKET \r\nROUND 2 \r\n[b]AIME_is_hard [/b]vs. [b]Mathblitz [/b](game 12) \r\nPinkpiglet(winner) vs. xoangieexo (game 13) \r\n[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b](game 14) \r\n[b]pacman2812 [/b]vs. [b]Kl2836 [/b](game 15) \r\n\r\nROUND 2a \r\nwinner of game 15 vs. winner of game 14 (game 16) \r\nwinner of game 12 vs. [b]Pinkpiglet [/b](game 17) \r\n\r\nROUND 3. \r\nwinner of game 16 vs. loser of game 9 (game 18) \r\nwinner of game 17 vs. loser of game 10 (game 19) \r\n\r\nROUND 3a \r\nwinner of game 18 vs. winner of game 19 (game 20) \r\n\r\nROUND 4. \r\nwinner of game 20 vs. loser of game 11 (game 21) \r\n\r\nCHAMPIONSHIP ROUND \r\n\r\nWINNER OF GAME 21 vs. WINNER OF GAME 11 \r\n\r\nIF THE WINNER OF THE LOSER'S BRACKET WINS THE CHAMPIONSHIP ROUND, THEN THE ROUND MUST BE PLAYED AGAIN \r\n\r\n[b][i][u]games that can be played right now[/u][/i][/b]\r\n[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b] (bisbmard9 has responded, if the game does not take place for any reason, and johnnybeard does not respond, then bisbmard wins)\r\n[b]Sicarius1029 [/b]vs. [b]AIME15[/b] (sicarius1029 has responded, if the game does not take place for any reason, and AIME15 does not respond, then sicarius1029 wins)\r\n[b]AIME_is_hard [/b]vs. [b]Mathblitz [/b] (BTW, i haven't seen AIME_is_hard on in a long time, so if this game doesn't get completed by the deadline, mathblitz wins)\r\n[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b]\r\n[b]pacman2812 [/b]vs. [b]Kl2836 [/b]\r\n\r\ni will send out reminders tomorrow as well, when i get on AoPS on july 30, any games not played will be decided by the above criteria and the coin toss",
  "Solution_16": "Vallon- if you do a coin toss, wouldn't that just defeat the whole purpose of an FTW tourney?  Unless you're doing a coin-tossing tourney...  :rotfl:",
  "Solution_17": "ya, to a certain extent, but i'm just trying to get the tourney to keep moving, that way, the people actually participating can play new pplz\r\nanyway, i have told pplz that if they respond, but their opponent doesn't, they get the win, so....\r\nit's not completely coin toss\r\nthe people that play move on, the people that don't play lose",
  "Solution_18": "[quote=\"vallon22\"]ya, to a certain extent, but i'm just trying to get the tourney to keep moving, that way, the people actually participating can play new pplz\nanyway, i have told pplz that if they respond, but their opponent doesn't, they get the win, so....\nit's not completely coin toss\nthe people that play move on, the people that don't play lose[/quote]\r\n\r\nBut that [b]completely[/b] would defeat the purpose of any math tourney.  This is (no offense) like a probablity/social reply tourney.",
  "Solution_19": "well, i just wanna get this tourney moving\r\nhopefullly, some pplz will actually play\r\n\r\non that note, i have postponed the deadline for the games in winner's bracket round 3 and loser's bracket round 2 until August 1\r\n\r\ngood luck with your games",
  "Solution_20": "Bah, I need to prepare to a selection test and make homeworks. No time to FTW\r\n\r\nkl(something) wins.",
  "Solution_21": "ok\r\npacman forfeits so....\r\n\r\nROUND 1. \r\n#5. AIME_is_hard\u2026\u2026...1375 vs #9. Mathblitz(winner due to coin toss)\u2026\u2026\u2026..\u2026.1357 (game 1) \r\n#6. AIME15 (winner)\u2026\u2026....\u2026\u2026\u20261370 vs#10. Pinkpiglet\u2026\u2026\u2026\u20261323 (game 2) \r\n#7. Sicarius1029(winner due to ducky's FTW failing)\u2026\u2026..\u20261369 vs#11. Myyellowducky82...1200 (game 3) \r\n#8. Johnnybeard(winner due to coin toss)\u2026\u2026..\u20261367 vs#12. Pacman2812\u2026\u2026\u2026.1200 (game 4) \r\n\r\nWINNER'S BRACKET \r\nROUND 2. \r\nKl2836 vs. Johnnybeard(winner due to kl2836 not responding) (game 5) \r\nBudi713 vs. Sicarius1029(winner due to coin toss) (game 6) \r\nXoangieexo vs. AIME15(winner) (game 7) \r\nBisbmard9(winner due to coin toss) vs. Mathblitz (game 8) \r\n\r\nROUND 3 \r\n[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b](game 9) \r\n[b]Sicarius1029 [/b]vs. [b]AIME15[/b](game 10) \r\n\r\nROUND 4 \r\nwinner of game 9 vs. winner of game 10 (game 11) \r\n\r\nLOSER'S BRACKET \r\nROUND 2 \r\n[b]AIME_is_hard [/b]vs. [b]Mathblitz [/b](game 12) \r\nPinkpiglet(winner) vs. xoangieexo (game 13) \r\n[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b](game 14) \r\npacman2812 vs. Kl2836(winner) (game 15) \r\n\r\nROUND 2a \r\n[b]Kl2836 [/b]vs. winner of game 14 (game 16) \r\nwinner of game 12 vs. [b]Pinkpiglet [/b](game 17) \r\n\r\nROUND 3. \r\nwinner of game 16 vs. loser of game 9 (game 18) \r\nwinner of game 17 vs. loser of game 10 (game 19) \r\n\r\nROUND 3a \r\nwinner of game 18 vs. winner of game 19 (game 20) \r\n\r\nROUND 4. \r\nwinner of game 20 vs. loser of game 11 (game 21) \r\n\r\nCHAMPIONSHIP ROUND \r\n\r\nWINNER OF GAME 21 vs. WINNER OF GAME 11 \r\n\r\nIF THE WINNER OF THE LOSER'S BRACKET WINS THE CHAMPIONSHIP ROUND, THEN THE ROUND MUST BE PLAYED AGAIN \r\n\r\n[u][i][b]games that can be played right now [/b][/i][/u]\r\n[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b](bisbmard9 has responded, if the game does not take place for any reason, and johnnybeard does not respond, then bisbmard wins) \r\n[b]Sicarius1029 [/b]vs. [b]AIME15 [/b](sicarius1029 has responded, if the game does not take place for any reason, and AIME15 does not respond, then sicarius1029 wins) \r\n[b]AIME_is_hard[/b] vs. [b]Mathblitz [/b](BTW, i haven't seen AIME_is_hard on in a long time, so if this game doesn't get completed by the deadline, mathblitz wins) \r\n[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b]",
  "Solution_22": "I forfeit. (spelling?)",
  "Solution_23": "[quote=\"myyellowducky82\"]I forfeit. (spelling?)[/quote]\r\n\r\nSpelling is correct. :wink:\r\n\r\nYAY!  My 800th post!\r\n\r\nAnd vallon, no offense, but this tourney is in serious need of cancelling.",
  "Solution_24": "ya, ur prob right\r\nbut right now, hardly any tournies are up and going\r\nhmm...\r\ni'll think of something l8r\r\n\r\nTHIS TOURNEY IS OFFICIALLY CANCELLED  :(  :(  :(",
  "Solution_25": "since its cancelled, admins (no mods for this forum) please lock this thread?",
  "Solution_26": "admins hardly come in here\r\nthe easiest way is for pplz to just stop posting\r\ndang, [i]i'm[/i] posting....",
  "Solution_27": "I'LL JOIN.THOUGH I'LL LOSE IN THE FIRST ROUND. HIGHEST RATING 1321",
  "Solution_28": "It's canceled, Andrew. At least read the post right before you.",
  "Solution_29": "This tourney is canceled.\r\n11111111111111111111111\r\n11111111111111111111111\r\n11111111111111111111111\r\n11111111111111111111111\r\n11111111111111111111111\r\n11111111111111111111111\r\n :D  :D  :D"
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "trig identities",
    "Law of Sines",
    "geometry unsolved"
  ],
  "Problem": "I need help, this is the only topic I am really struggling at.\r\nThis may seem rude as its my first post, but if anyone is kind enough! :) \r\n\r\n--------\r\n\r\n\r\n[img]http://i14.tinypic.com/2heatyd.jpg[/img]\r\n\r\nShow that, for triangle RST, t = r sin (alpha + beta) [b]/[/b] sin alpha\r\n\r\n\r\n\r\niv tried using the sine rule,\r\nt [b]/[/b] sin(180 - alpha - beta)   [b]=[/b]   r [b]/[/b] sin alpha\r\nt = r / sin alpha [b]x[/b] sin(180 - alpha - beta)\r\n\r\nAfter that, i get confused because usually i would expand the sin bracket. ie:\r\n[b]sin(45+45) = sin 45 cos 45 + cos 45 sin 45[/b]\r\n\r\nbut in this case, i have 180, alpha, and beta in one bracket.......\r\n\r\nAm i even taking the right steps?\r\nPlease help.",
  "Solution_1": "The angle measure of $T=180-(a+b)$\r\n\r\nBy law of sines,:\r\n$\\frac{r}{\\sin a}=\\frac{t}{\\sin (180-(a+b))}$\r\n\r\nFor any number $x$, $\\sin x = \\sin (180-x)$\r\n\r\nSo you get:\r\n$\\frac{r}{\\sin a}=\\frac{t}{\\sin (a+b)}$\r\n$\\implies \\boxed{t=\\frac{r\\sin (a+b)}{\\sin a}}$",
  "Solution_2": "thanks  :)"
}
{
  "Tag": [
    "function",
    "special factorizations"
  ],
  "Problem": "Please help me. I'm struggling over math.\r\n\r\n\r\n\r\nKindly show the complete solutions.. THANKS ;]",
  "Solution_1": "ok we can rewrite n(n) as $ n^2$. it is a well-known theorem that the minimum of a parabolic function (or maximum of a negative one) comes when $ x\\equal{}\\minus{}\\frac{b}{2a}$ which in this case is $ n\\equal{}\\boxed{600}$ plugging that in, we get $ c\\equal{}\\boxed{600}$ as well. coooooollll....",
  "Solution_2": "We see this by completing the square (never liked formulas anyway :) ):\r\n\\begin{align*}\r\nc &= n^2 - 120n + 4200 \\\\\r\n&= n^2 - 120n + 3600 + 600 \\\\\r\n&= (n - 60)^2 + 600.\r\n\\end{align*}\r\nSince something squared is always nonnegative, we easily see that the minimum value of c is 600, since the smallest $ (n-60)^2$ can be is 0."
}
{
  "Tag": [
    "search",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "A palindrome is a number or word that is the same when read forward and backward, for example, \"176671\" and \"civic.\" Can the number obtained by writing the number from 1 to $ n$ in order ($ n>1$) be a palindrome?",
  "Solution_1": "This is Russia 1996.  It has appeared many times on the forum, and it's easy enough to find with a search for the word \"palindrome.\"\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=128508\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=19872"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "induction",
    "extremal principle",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Let $ A_1A_2...A_n$ be a convex polygon. Show that there exists an index $ j$ such that the circum-circle of the triangle $ A_j A_{j \\plus{} 1} A_{j \\plus{} 2}$ covers the polygon (here indices are read modulo n).",
  "Solution_1": "[hide=\"Hint\"]Take a circle that covers the polygon and try to make it get smaller until it becomes a circumcircle of some triangle.[/hide]",
  "Solution_2": "I don't think that hint will work for me. I mean.... how will you show that that circle will always become some triangle's circumcircle?",
  "Solution_3": "i would say try to find a triangle (general) whose circumcircle covers the n gon and then try to create more triangles",
  "Solution_4": "[quote=\"aakansh92\"]Let $ A_1A_2...A_n$ be a convex polygon. Show that there exists an index $ j$ such that the circum-circle of the triangle $ A_j A_{j \\plus{} 1} A_{j \\plus{} 2}$ covers the polygon (here indices are read modulo n).[/quote]\r\n\r\n[hide=\"Hint to the solution\"]\nProve that the biggest circle possible through any 3 vertices is\ni) :arrow:  covering the polygon\nii) :arrow: passing through three consequtive points.\n\nboth can be proved by contradiction method. :) [/hide]",
  "Solution_5": "Please be more clear and give full solution.\r\n\r\nGaurav, could you do that problem during the postals? If yes, I am eager to look at the solution.",
  "Solution_6": "Aaknash, were u at the IMOTC last year?",
  "Solution_7": "Here's my (not well-written  :oops: ) solution. Hope you understand the idea.\r\n\r\n[hide]I don't know if the following requires proof. :| \n[b]Lemma 1:[/b] \nSuppose $ S$ is a circle which contains $ Y,Z$ but doesn't contain $ X$.\nThen the portion of the circumference of $ S$ that lies opposite $ X$ wrt $ YZ$ lies outside the circumcircle of $ XYZ$\n [geogebra]8311777ddf6748db7b929f40c8136d68ab2b9775[/geogebra]\n\nSuppose the problem statement is true for all convex $ n$-gon for some $ n\\ge 4$\n(Base case $ 4$ is easy to prove)\nAssume there exists a convex $ n \\plus{} 1$-gon $ A_1,A_2,\\cdots A_{n \\plus{} 1}$ such that the statement isn't true.\n\nBy induction hypothesis on the polygon $ A_1A_2\\cdots A_n$, there exists $ A_j,A_{j \\plus{} 1},A_{j \\plus{} 2}$ such that exactly one vertex of the polygon, $ A_{n \\plus{} 1}$ lies outside their circumcircle.\n\nBy induction hypothesis on the polygon $ A_1A_2\\cdots A_jA_{j \\plus{} 2}\\cdots A_{n \\plus{} 1}$, exactly one vertex of the polygon, $ A_{j \\plus{} 1}$ lies outside a circle $ S$ that has on it's circumference 3 consecutive points from the polygon.\n [geogebra]a1b4053b21d62e5195188b71d634b48bdaf91f91[/geogebra] \nBut none of these three points lie on the same side as  $ A_{j \\plus{} 1}$ wrt $ A_jA_{j \\plus{} 2}$. But since $ A_{j \\plus{} 1}$ lies outside $ S$, by lemma 1 such three points can't lie inside the circumcircle of $ A_jA_{j \\plus{} 1}A_{j \\plus{} 2}$. So these 3 points must be $ A_j,A_{j \\plus{} 2},A_{n \\plus{} 1}$. But these are consecutive only if $ n \\equal{} 3$. Which isn't true because $ n\\ge 4$[/hide]\r\nDo reply if you understand or not. :oops:",
  "Solution_8": "Ah... Sorry... When I replied, I was thinking in another problem.\r\n\r\nAkashnil, I think you should replace three of your $ A_{j \\plus{} 1}$'s by $ A_{j \\plus{} 2}$'s. Anyway, what is wrt?",
  "Solution_9": "wrt = with relation to, or something similar.",
  "Solution_10": "wrt=with respect to.\r\n$ A$ is opposite $ B$ wrt $ CD$ means the points $ A$ and $ B$ lie on oposite sides of the line $ CD$.\r\nYes I messed up some of $ A_{j\\plus{}1}$ and $ A_{j\\plus{}2}$s . Sorry. I've edited now. Thanks.\r\n\r\nAnd once again do reply if you understand it now or not.",
  "Solution_11": "[quote=\"aakansh92\"]Please be more clear and give full solution.\n\n[/quote]\r\nSorry for incomplete post, actually i dont like Latexing much. anyway [hide=\"hare goes the proof\"]\nAmong the finitely many circles through any 3 vertices, there is[b] [u]a maximal one, denote it by $ \\tau$[/u][/b].\n\nas said earlier we prove:\n1.  :arrow: $ \\tau$ covers n-gon.\n2.  :arrow: $ \\tau$ passes through 3 consequtive vertices.\n\n$ 1.$ Suppose point A' lies outside the maximal circle about $ \\triangle ABC$ where $ A,B,C,A'$\nare vertices of polygon so that they form a convex quadrilateral. Then circumcircle of $ \\triangle A'BC$ has larger radius than $ \\triangle ABC$. :!: CONTRADICTION. :) \n$ 2.$ Let $ A_i,A_j,A_k$ be vertices on the MAXIMAL circle. Let $ A_l$ lie between $ A_j \\& A_k$, not on $ \\tau$. Because of (1), it lies inside $ \\tau$, but then circle about $ \\triangle A_jA_lA_k$ is larger than maximal circumcircle. :!:  CONTRADICTION. :)  [/hide].\r\nplease verify the solution.",
  "Solution_12": "[quote=\"Akashnil\"]I don't know if the following requires proof. :| \n[b]Lemma 1:[/b] \nSuppose $ S$ is a circle which contains $ Y,Z$ but doesn't contain $ X$.\nThen the portion of the circumference of $ S$ that lies opposite $ X$ wrt $ YZ$ lies outside the circumcircle of $ XYZ$\n [/quote]\r\nP.S.- \r\nabove lemma can be easily proved since two circles(not identical) can intersect at only two points.\r\nP.P.S.- AKashnil, your proof is really very good. :coolspeak:",
  "Solution_13": "Yours too :coolspeak:",
  "Solution_14": "those two solutions are god damn BEAUTIFUL\r\nhats off to you both ith_power and Akashnil\r\nmy solution amounted to atleast 5 whole written pages in Ideas but yours(both) is just beauty",
  "Solution_15": "That's really wonderfull.Thanks!",
  "Solution_16": "It is very unfortunate that this problem was taken from the very well known Arthur Engel book's Extremal principle chapter. Infact it is an example problem. Too bad of the problem setters in India.\r\nAnyway, akashnil's solution is superb as it is a different one .\r\nith_power ur solution is superb as u have used the extremal principle well. I also solved it by that way and even in the book they solved it in the same way.\r\n\r\n\r\n\r\n\r\nModerators if this is an unnecessary post sorry but just wanted to say that it was copied thats all.",
  "Solution_17": "engel does not state consecative",
  "Solution_18": "Enegel does state consecutive.  :D"
}
{
  "Tag": [
    "calculus",
    "integration",
    "floor function",
    "calculus computations"
  ],
  "Problem": "Here's another one that was interesting:\r\n\r\nDetermine whether the improper integral\r\n\r\n\\[ \\int_0^\\infty (\\minus{}1)^{\\lfloor x^2 \\rfloor}dx\r\n\\]\r\n\r\nconverges or diverges.",
  "Solution_1": "[hide]The question is equivalent to the convergence of the series $ \\sum_{k \\equal{} 0}^{\\infty} ( \\minus{} 1)^k(\\sqrt {k \\plus{} 1} \\minus{} \\sqrt {k})$ (here $ k$ replaces $ \\lfloor x^2\\rfloor$. The series converges by the alternating series test.[/hide]"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "I want to make my reference goes like this:\r\n\r\nAmacc, D and JXIK, L. 2009, paper compilation, journal of xx, 202-241\r\n\r\nBxxx, L and  wxx, D. 2008, xxxx, .\r\n\r\nCxxx, ..\r\n\r\nBasically, I want the references ordered by the alphabetical order of authors' last name.(.Amacc/0 is last name..and then first name.  \r\n\r\nwhich bibliography style should I use with \\usepackage{natbib} ?\r\n\r\nI download the APA from CTAN..But when I tried to install it to my TexnicCenter, it fails..No ideas what I should do now..pls help me with the procedure of downloading new bibliography style.",
  "Solution_1": "[quote=\"weng\"]I download the APA from CTAN..But when I tried to install it to my TexnicCenter, it fails..No ideas what I should do now..pls help me with the procedure of downloading new bibliography style.[/quote]This makes no sense. TexnicCenter is merely an editor so you can't install anything to it. What is \"the APA\"? What does \"it fails\" mean? MiKTeX already comes with APA like styles in it such as apacite and apalike so there's nothing to install. See MiKTeX, Browse Packages for what is installed.\r\n\r\nAlso look at the sort-by-letters styles already included in MiKTeX. These are described [url=http://tug.ctan.org/tex-archive/biblio/bibtex/contrib/sort-by-letters/README]here[/url].",
  "Solution_2": "when i used \\bibliographystyle{apa}\r\n\r\nthe page of Bibliography just disappear..\r\n\r\nI tried to install apa into MikTek....",
  "Solution_3": "[quote=\"weng\"]when i used \\bibliographystyle{apa}\n\nthe page of Bibliography just disappear..[/quote]You should know by now that you must give a minimal example to have any chance of analyzing the problem.\n[quote=\"weng\"]I tried to install apa into MikTek....[/quote]Why? It's already installed assuming you have the full version. Again you haven't given any information to work with. What did you do to install it? Didy you use the MiKTeX install procedure in Browse Packages? Did you install it manually? If so, did you refresh the database?",
  "Solution_4": "Thank you for clarifying those to me!!!!! \r\nI get the problem fixed.."
}
{
  "Tag": [
    "ARML",
    "email"
  ],
  "Problem": "I pm-ed Paul about ARML but he hasn't replied. Does anyone know who I should contact and what the requirements are?",
  "Solution_1": "For Iowa ARML, you're looking for James Kirpes. Email is: kirpes.james@iccsd.k12.ia.us.  Hm...I'm not entirely sure of requirements.  Usually you have to have scored well (placed) in GPML meets.  AMC scores are also factored in.  I've never had too much trouble getting on a team.  If you're a regular on this site, I doubt you will either.",
  "Solution_2": "Oh, looks like Darren took care of it. I probably should have just given you the email from the beginning, Penguin. <_<;;\r\n\r\nOh well.",
  "Solution_3": "Sorry to call up an old post, but who in here is going to the Iowa City ARML this June?\r\n\r\nFor those of you that have done it before, how are the three teams divided up?",
  "Solution_4": "Well, I'm going.  :lol: \r\n\r\nThere's a Central Academy team, I'm pretty sure. And then the other two teams are divided according to how Kirpes views their mathematical ability. (based on experience with the team and AMC scores and such)\r\n\r\nI think. Darren or Nick, do you know anything?",
  "Solution_5": "woohoo! im going. hope to c u guys there   :)",
  "Solution_6": "Yeah, you're pretty much right Paul.  CA's got their team.  And then Kirpes divides people based on who's good and who's not.  More or less.",
  "Solution_7": "Generally speaking, what is the cutoff for the two teams (based on AMC)? This could be skewed, as many people (like me) who take a lot of time suck at speed-type contests (over 1 hour).",
  "Solution_8": "I know from central 3 8th graders are going (me being one of them :P ), 1 or 2 freshmen, and 6ish seniors and juniors.",
  "Solution_9": "I don't think there is a specific cutoff. It's mostly qualitative, rather than quantitative. Don't worry! If you guys aren't on the first team this year, then chances are that a decent performance at ARML will put you on the top team. At least, that's what I think is true.\r\n\r\nDarren, Allendorf, some help here?"
}
{
  "Tag": [],
  "Problem": "Each page number of a 599-page book is printed one time in the book. The first page is page 1 and the last page is page 599. When printing all of the page numbers, how many more 5's are printed than 8's?\r\n\r\nHide your solutions\r\n\r\nBilly",
  "Solution_1": "[color=indigo]\n[hide]There are 100 more 5's than 8's. First i counted the number of 5's: Between 1-99 there are 20. Between 100-199 there are 20. Same for the 200's, the 300's and the 400's. Then in the 500's there are 120 5's. THat mkes a total of 220 fives. Then the 8's are pretty much the same except they don't get the extra 100 in the 500's so there are 120 8's. :D [/color][/hide]",
  "Solution_2": "[hide]If you realize, the 5's have 500-599. 8 dont.\nOtherwise, they're the same\nThere are 100 pages in 500s\n>>100<<[/hide]",
  "Solution_3": "[hide]From 1-599, 8 shows up 9 * 6 times in the ones spot, 9 * 6 times in the 10s spot and 0 times in the hundreds spot. 5 shows up 9 * 6 times in the ones spot, 9 * 6 times in the 10s spot and 100 times in the hundreds spot, so its 100 more. [/hide]",
  "Solution_4": "so why don't you edit your post, or reenter another to fix it?"
}
{
  "Tag": [
    "function",
    "search",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Suppose that $f,g$ are  an analytic functions on $\\mathbb C$. Prove that if $f,g$ have no common roots, then there exist analytic functions $A,B$ whcih $Af+Bg=1$.",
  "Solution_1": "Leva proved it (on the forum) a while back. I'm sure it's easy to locate using the search function.",
  "Solution_2": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46388]The topic grobber was referring to.[/url]"
}
{
  "Tag": [
    "trigonometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "I need to solve a problem from physics, but first i need to solve these differential equations to find the frequence of two moving rods.\r\n$\\begin{array}{l}\\left\\{ \\begin{array}{l}\\varphi_{1}= \\frac{{\\varphi_{0}}}{2}[\\cos \\omega_{1}t+\\cos \\omega_{2}t] \\\\ \\varphi_{2}= \\frac{{\\varphi_{0}}}{2}[\\cos \\omega_{1}t-\\cos \\omega_{2}t] \\\\ \\end{array}\\right.\\\\ \\end{array}$\r\nwhere\r\n$\\omega_{1}^{2}= \\omega_{0}^{2}+\\omega^{'2}\\\\ \\omega_{2}^{2}= \\omega_{0}^{2}-\\omega^{'2}$",
  "Solution_1": "So, any idea?",
  "Solution_2": "Shouldn't this be done using the addition of fazors, in a fazorial graph?",
  "Solution_3": "What's fazorial graph? I need to solve the differential equations.",
  "Solution_4": "[quote=\"Nea\"]What's fazorial graph? I need to solve the differential equations.[/quote]\r\nYour original post does not contain any differential equations. It is unclear what you are asking.",
  "Solution_5": ":blush: \r\nI've erroneously written the previous equations. You arrive to them after some substitutions to these:\r\n$\\begin{array}{l}ml^{2}\\varphi_{1}^{''}+mgl\\varphi_{1}+ca^{2}(\\varphi_{1}+\\varphi_{2}) = 0 \\\\ ml^{2}\\varphi_{2}^{''}+mgl\\varphi_{2}+ca^{2}(\\varphi_{1}+\\varphi_{2}) = 0 \\\\ \\end{array}$\r\nwhere:\r\n$\\begin{array}{l}\\varphi_{1}+\\varphi_{2}= \\varphi \\\\ \\varphi_{1}-\\varphi_{2}= \\varphi \\\\ \\end{array}$\r\nAnyway only these equations, that I wrote now,  count. I've arrived at the previous equations because of the initial conditions I had in the problem.",
  "Solution_6": "I was wondering if you needed $mgl\\varphi_{1}'$ as opposed to $mgl\\varphi_{1}$.\r\nYou wanted to let people know what $m,l,c,a$ are (constants I guess). \r\nI was wondering if something is wrong with the following array because after solving it for $\\varphi$ something is not true about $\\varphi_{2}$\r\n$\\begin{array}{l}\\varphi_{1}+\\varphi_{2}= \\varphi \\\\ \\varphi_{1}-\\varphi_{2}= \\varphi \\\\ \\end{array}$",
  "Solution_7": "[quote=\"akech\"]I was wondering if you needed $mgl\\varphi_{1}'$ as opposed to $mgl\\varphi_{1}$.\nYou wanted to let people know what $m,l,c,a$ are (constants I guess). \nI was wondering if something is wrong with the following array because after solving it for $\\varphi$ something is not true about $\\varphi_{2}$\n$\\begin{array}{l}\\varphi_{1}+\\varphi_{2}= \\varphi \\\\ \\varphi_{1}-\\varphi_{2}= \\varphi \\\\ \\end{array}$[/quote]\r\n\r\n1. $mgl\\varphi_{1}$ is correct not $msgl\\varphi_{1}'$\r\n2. Yes $m,l,c,a$ are constants\r\n3. About the \"array problem\" you mentioned, the two equations are true after approximations."
}
{
  "Tag": [
    "Support"
  ],
  "Problem": "It seems police practices have gotten out of control in the US with  constant racial profiling always looking to pulling over black people to try and catch them ridin' dirty as well as tasering college students for not producing an ID or getting up on time when they want to use the library:\r\n\r\nhttp://helpychalk.blogspot.com/2006/11/ucla-student-repeatedly-tasered-for-not.html\r\n\r\n\r\nMore taser fun here\r\n\r\nhttp://portland.indymedia.org/en/2006/01/332829.shtml\r\n\r\nTasers have been common now for police to use probably because they are claimed not to be dangerous at all.\r\n\r\n\r\nHow brutal are the police in other countries? Would actions such as these be frowned down upon or are they common?",
  "Solution_1": "Our police are known for their alternating acts of heroism and callousness. Oh, and they have never heard of tasers.",
  "Solution_2": "Tasers are small weapons that disperse electricity to stun somebody, they are considered a \"safe\" alternative to standard practices and a good way to disarm people.\r\n\r\nIn Florida, it's even legal to taser minors.",
  "Solution_3": "They're either illegal in some 40 states (USA) or legal in some 40, can't remember which. I know that they are used commonly in prisons to break up fights and stuff. I think as long as they don't get so advanced that they start actually doing damage to people, they are fine, but should by no means be available to the public, even with a liscense (unlike handguns)",
  "Solution_4": "The media is out to sell stories, not report the truth.\r\n\r\nThe ones that have gotten confused about this are no longer in business, as the public won't support a press or something that isn't what is demanded.\r\n\r\nMost of the police in the USA, as elsewhere, are people just trying to do the best job they can.  Some are drawn to the job for poor reasons, such as: they like to have power over others, they are part of a gang that use the police power to further gang goals, etc.\r\n\r\nRacial profiling is used for one reason; it is effective.  For most purposes it isn't allowed to be used to directly or indirectly lead to a conviction of a criminal.\r\n\r\nBut let's think about this, what if your loved one was kidnapped by someone who was observed.  Wouldn't you want the police to stop those of the same race/height/weight/gender as the witness identified?  I would, but this is profiling.\r\n\r\nIf the police are stopping all black men drivers just because they are black, they are not being very productive.  This will also clearly show up in their arrest records, so I think the actual occurrances of this, while real, are perhaps overstated.  \r\n\r\nLet's be rational, how easy would it be for a defense counsel, looking for any way out for his client, to find out that a traffic officer issued 2 or 3 times as many tickets to purple people, or any other class than exist in the general population or than other officers did.  The feds and ACLU would both love to have such a case.  I don't believe that such things are still occurring.  I will believe that those with the same race as the officer might be treated a bit different, as would an attractive young driver, by an officer who happened to be of the other gender.  Also, what paper wouldn't want to run such a story?\r\n\r\nI won't argue that it doesn't occur, but I think the prevelence is far overstated by the press.\r\n\r\nA Taser is a less-lethal method of stopping someone.  Wouldn't you want of officer to Tase a kid before he shoots one?  I would.  It should be legal, but just like shooting -- only in appropriate circumstances.\r\n\r\n(I understand that a Taser fires two darts that penerate the skin and a battery that powers an electric shock that is very painful, ususally causes the equivalent of a short seizure in a person hit by one and is usually not lethal.)\r\n\r\nOur local police force is obtaining them.  Several officers asked to be shot themselves so they would know what they were subjecting people to if they decided to use it on a member of the public.  Admirable.  One was quoting saying that they wouldn't want to do anything they didn't experience themselves.  (I don't think he extended this to shooting.)\r\n\r\nI hope this helps.",
  "Solution_5": "[quote=\"wizard2378\"]\n Several officers asked to be shot themselves so they would know what they were subjecting people to if they decided to use it on a member of the public.  Admirable.  One was quoting saying that they wouldn't want to do anything they didn't experience themselves.  (I don't think he extended this to shooting.)[/quote]\r\n\r\nIn my state, i'm pretty sure it's required for police officers to go through the effects of thier various weapons, so that in court they can justify any action they may have taken with the weapon because they've felt it too. This includes tasers and pepper-spray, but of course not guns.",
  "Solution_6": "a) What you have provided is very small evidence for making a general statement about police practices in the U.S.\r\n\r\nb) Did you even read this part? [quote]Predictably, the coroner says the taser had nothing to do with his death.[/quote]\r\n\r\nc) Edit: after looking at it again, I change my mind - I think they did step over reach their boundaries and shouldnt' have gone that far."
}
{
  "Tag": [],
  "Problem": "\u039a\u03b1\u03bb\u03bf \u03a6\u03b8\u03b9\u03bd\u03cc\u03c0\u03c9\u03c1\u03bf \u03c3\u0384 \u03cc\u03bb\u03bf\u03c5\u03c2\r\n\r\n\u03a3\u03ba\u03b1\u03c1\u03c9\u03c3\u03b1 \u03bc\u03b9\u03b1 (\u0393\u03b9\u03b1 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b7 \u03c4\u03b7\u03bd \u03c0\u03b7\u03b3\u03b1)\r\n\r\n\u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\r\n \r\n$ (\\frac {\\sqrt {a^4 \\plus{} b^4} \\minus{} \\sqrt {b^4 \\plus{} c^4} \\minus{} \\sqrt {c^4 \\plus{} a^4}}{2})^2 \\plus{} \\frac {(|ab| \\minus{} |bc| \\minus{} |ca|)^2}{2}\\geq{c^4}$  \r\n\u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 $ a,b,c$ \u03c3\u03c4o $ R^{*}$",
  "Solution_1": "[quote=\"xp\"]\u039a\u03b1\u03bb\u03bf \u03a6\u03b8\u03b9\u03bd\u03cc\u03c0\u03c9\u03c1\u03bf \u03c3\u0384 \u03cc\u03bb\u03bf\u03c5\u03c2\n\n\u03a3\u03ba\u03b1\u03c1\u03c9\u03c3\u03b1 \u03bc\u03b9\u03b1 (\u0393\u03b9\u03b1 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b7 \u03c4\u03b7\u03bd \u03c0\u03b7\u03b3\u03b1)\n\n\u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\n \n$ (\\frac {\\sqrt {a^4 \\plus{} b^4} \\minus{} \\sqrt {b^4 \\plus{} c^4} \\minus{} \\sqrt {c^4 \\plus{} a^4}}{2})^2 \\plus{} \\frac {(|ab| \\minus{} |bc| \\minus{} |ca|)^2}{2}\\geq{c^4}$  \n\u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 $ a,b,c$ \u03c3\u03c4o $ R^{*}$[/quote]\r\n\r\nCounterexample: $ (a,b,c)\\equal{}(1,0.5,0.45)$",
  "Solution_2": "\u03c0\u03b1\u03c1\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c5\u03c0\u03cc\u03b8\u03b5\u03c3\u03b7 \u03c4\u03b1 \u03b1,b,c \u03c4\u03b5\u03c4\u03bf\u03b9\u03b1 \u03c9\u03c3\u03c4\u03b5 \u03bf\u03b9 \u03b2\u03b1\u03c3\u03b5\u03b9\u03c2 \u03c4\u03c9\u03bd \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03c9\u03bd \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03b5\u03c2.",
  "Solution_3": "[quote=\"xp\"]\u039a\u03b1\u03bb\u03bf \u03a6\u03b8\u03b9\u03bd\u03cc\u03c0\u03c9\u03c1\u03bf \u03c3\u0384 \u03cc\u03bb\u03bf\u03c5\u03c2\n\n\u03a3\u03ba\u03b1\u03c1\u03c9\u03c3\u03b1 \u03bc\u03b9\u03b1 (\u0393\u03b9\u03b1 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b7 \u03c4\u03b7\u03bd \u03c0\u03b7\u03b3\u03b1)\n\n\u039d\u03b1 \u03b1\u03c0\u03bf\u03b4\u03b5\u03b9\u03c7\u03b8\u03b5\u03af \u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1\n \n$ \\left[\\frac {\\sqrt {a^4 \\plus{} b^4} \\minus{} \\sqrt {b^4 \\plus{} c^4} \\minus{} \\sqrt {c^4 \\plus{} a^4}}{2}\\right]^2 \\plus{} \\frac {(|ab| \\minus{} |bc| \\minus{} |ca|)^2}{2}\\geq{c^4}$  \n\u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 $ a,b,c$ \u03c3\u03c4o $ \\mathbb{R^{*}}$[/quote]\r\n\r\n\u0398\u03b1 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03b9\u03ae\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03b1\u03ba\u03cc\u03bb\u03bf\u03c5\u03b8\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1:\r\n\r\n$ \\left(a^{p}_{1} \\minus{} a^{p}_{2} \\minus{} a^{p}_{3}\\right)^{\\frac {1}{p}} \\plus{} \\left(b^{p}_{1} \\minus{} b^{p}_{2} \\minus{} b^{p}_{3}\\right)^{\\frac {1}{p}}\\geq \\left[\\left(a_1 \\plus{} b_1\\right)^{p} \\minus{} \\left(a_2 \\plus{} b_2\\right)^{p} \\minus{} \\left(a_3 \\plus{} b_3\\right)^{p}\\right]^{\\frac {1}{p}}$,\r\n\u03bc\u03b5\r\n$ 0 < p < 1$\r\n\r\n\u039f\u03c0\u03cc\u03c4\u03b5 \u03b3\u03b9\u03b1:\r\n\r\n$ a_1 \\equal{} a^4 \\plus{} b^4,a_2 \\equal{} b^4 \\plus{} c^4,a_3 \\equal{} c^4 \\plus{} a^4\\wedge b_1 \\equal{} 2a^2b^2,b_2 \\equal{} 2b^2c^2,b_3 \\equal{} 2c^2a^2\\wedge p \\equal{} \\frac {1}{2}$ \r\n\r\n\u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03cc\u03c4\u03b9 \r\n\r\n$ \\left(\\sqrt {a^4 \\plus{} b^4} \\minus{} \\sqrt {b^4 \\plus{} c^4} \\minus{} \\sqrt {c^4 \\plus{} a^4}\\right)^{2} \\plus{} \\left[\\sqrt {2}\\left(\\left|ab\\right| \\minus{} \\left|bc\\right| \\minus{} \\left|ca\\right|\\right)\\right]^{2}\\geq \\left(\\sqrt {\\left(a^2 \\plus{} b^2\\right)^{2}} \\minus{} \\sqrt {\\left(b^2 \\plus{} c^2\\right)^{2}} \\minus{} \\sqrt {\\left(c^2 \\plus{} a^2\\right)^{2}}\\right)^{2}$$ (1)$\r\n\r\n\u0394\u03b9\u03b1\u03b9\u03c1\u03ce\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd $ (1)$ \u03bc\u03b5 \u03c4\u03bf $ 4$ \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ \\left[\\frac {\\sqrt {a^4 \\plus{} b^4} \\minus{} \\sqrt {b^4 \\plus{} c^4} \\minus{} \\sqrt {c^4 \\plus{} a^4}}{2}\\right]^{2} \\plus{} \\frac {\\left(\\left|ab\\right| \\minus{} \\left|bc\\right| \\minus{} \\left|ca\\right|\\right)^{2}}{2}\\geq \\frac {\\left(a^2 \\plus{} b^2 \\minus{} b^2 \\minus{} c^2 \\minus{} c^2 \\minus{} a^2\\right)^{2}}{4} \\equal{} c^{4}$",
  "Solution_4": "\u03a9\u03c1\u03b1\u03af\u03bf\u03c2 \r\n\u0397 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 \u03bb\u03b5\u03b3\u03b5\u03c4\u03b1\u03b9 \u0392elman \u03ba\u03b1\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03b8\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2 \u03b2\u03b1\u03c3\u03b5\u03b9\u03c2 .\u0392\u03bb\u03b5\u03c0\u03b5\u03b9\u03c2 \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03c0\u03bf\u03c5 \u03b5\u03af\u03c7\u03b5 \u03c4\u03b7\u03bd \u0392elman \u03c4\u03bf \u03b5\u03af\u03c7\u03b5 \u03be\u03b5\u03c7\u03ac\u03c3\u03b5\u03b9 \u03b1\u03c5\u03c4\u03bf, \u03b3\u03b9\u03b1 \u03b1\u03c5\u03c4\u03cc \u03ba\u03b1\u03b9 \u03b5\u03bc\u03c6\u03b1\u03bd\u03af\u03c3\u03c4\u03b7\u03ba\u03b5 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03b1\u03ba\u03b9......",
  "Solution_5": "\u0391\u03c2 \u03b4\u03ce\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bb\u03bf\u03b9\u03c0\u03cc\u03bd \u03ba\u03b1\u03b9 \u03c4\u03b7 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03bc\u03ad\u03bd\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1 $ Bellman$.\r\n\r\n\u0388\u03c3\u03c4\u03c9 $ 0<p<1$ \u03ba\u03b1\u03b9 $ a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}\\in \\mathbb{R}$ \u03c4\u03ad\u03c4\u03bf\u03b9\u03b1 \u03ce\u03c3\u03c4\u03b5 $ a^{p}_{1}\\geq a^{p}_{2}\\plus{}a^{p}_{3}\\plus{}...\\plus{}a^{p}_{n}$ \u03ba\u03b1\u03b9 $ b^{p}_{1}\\geq b^{p}_{2}\\plus{}b^{p}_{3}\\plus{}...\\plus{}b^{p}_{n}$. \r\n\r\n\u03a4\u03cc\u03c4\u03b5 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b7 \u03b1\u03ba\u03cc\u03bb\u03bf\u03c5\u03b8\u03b7 \u03b1\u03bd\u03b9\u03c3\u03cc\u03c4\u03b7\u03c4\u03b1:\r\n\r\n$ \\left(a^{p}_{1}\\minus{}a^{p}_{2}\\minus{}...\\minus{}a^{p}_{n}\\right)^{\\frac{1}{p}}\\plus{}\\left(b^{p}_{1}\\minus{}b^{p}_{2}\\minus{}...\\minus{}b^{p}_{n}\\right)^{\\frac{1}{p}}\\geq \\left[\\left(a_1\\plus{}b_1\\right)^{p}\\minus{}\\left(a_2\\plus{}b_2\\right)^{p}\\minus{}...\\minus{}\\left(a_n\\plus{}b_n\\right)^{p}\\right]^{\\frac{1}{p}}$"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "trigonometry",
    "floor function",
    "probability",
    "geometry",
    "trig identities"
  ],
  "Problem": "In triangle ABC, it is given that $\\angle B = \\angle C$. Points P and Q lie on AC and AB, respectively, so that $AP = PQ = QB = BC$. Angle ACB is r times as large as angle APQ, where r is a positive real number. Find the greatest integer that doesn't exceed 1000r.",
  "Solution_1": "Huh? This doesn't seem right... I obviously made a mistake somewhere, but where?\r\n\r\n$\\angle APQ$ is between two same-length segments. $\\angle QBC$ is also between two same-length segments, and these two segments are also the same length as the other two segments, so $\\angle QBC=\\angle APQ$. We already know that $\\angle QBC=\\angle ACB$, so $\\angle APQ=\\angle ACB$. Therefore, $r=1$, and the greatest integer that does not exceed that is $1000$...\r\n\r\nThe other problem with this is that, if $\\angle APQ=\\angle ACB$, then $PQ\\parallel BC$, but then they cannot be equal, or we wouldn't have a triangle.",
  "Solution_2": "\" \\angle APQ is between two same-length segments. \\angle QBC is also between two same-length segments, and these two segments are also the same length as the other two segments \"\r\n\r\n\r\nSo let's take some two arbitrary isosceles triangle ABC and XYZ.  You are only given the fact that AB=AC=XY=XZ.\r\n\r\nthis doesn't imply any relationship between angle BAC and angle YXZ. \r\n\r\na nice diagram will show this.\r\n\r\n\r\n\r\n\r\nnote: check out my exceptional paint and grammar skills",
  "Solution_3": "Angle chase.  Let $\\angle PAQ = x$.  Then $\\angle PQA = x \\implies \\angle QPC = 2x \\implies \\angle QCP = 2x \\implies \\angle CQB = 3x \\implies \\angle CBQ = 3x \\implies \\angle ACB = 3x \\implies \\angle QCB = x$.  Also note that from triangle QCB, $7x = 180$.  Now we have $\\frac{\\angle BCA}{\\angle APQ} = \\frac{3x}{180-2x} = \\frac{3x}{5x} = \\frac{3}{5}$ so the answer is 600.",
  "Solution_4": "I believe the answer is actually 571.\r\n\r\nlet $\\angle QPB=k$ \r\nlet $AP=l$\r\n$\\angle QPB=k \\Rightarrow \\angle PQA=2k$\r\n$AQ=2l\\cos2k \\Rightarrow PC=2l\\cos2k$ due to ABC being isosceles.\r\nnote that $\\angle PCB=90-k$ , $\\angle PBC=90-2k$, and $\\angle BPC=3k$\r\n\r\n$BP=2l\\cos k$\r\n$BC=l$ is given\r\n$BC=BP\\cos(90-2k)+PC\\cos(90-k)=2l\\cos k\\sin(2k)+2l\\cos(2k)\\sin k$\r\nso\r\n$l=2l\\cos k\\sin(2k)+2l\\cos(2k)\\sin k$\r\n$\\frac{1}{2}=\\cos k\\sin(2k)+\\cos(2k)\\sin k$\r\n$\\frac{1}{2}=\\sin(k+2k)=\\sin(3k)$\r\nso $3k=30$, $k=10$\r\n\r\nby angle chasing, we arrive at $\\angle APQ=180-4k$, and $\\angle PCB=90-k$\r\nsince k=10, we have $\\angle APQ=140$, and $\\angle PCB=80$\r\n\r\n$\\lfloor 1000\\times \\frac{80}{140} \\rfloor=571$",
  "Solution_5": "I don't understand those relationships. probability, how did you get any of these steps: $\\angle QPC=2x \\Rightarrow \\angle QCP=2x \\Rightarrow \\angle CQB=3x \\Rightarrow \\angle ACB=3x \\Rightarrow \\angle QCB=x$\r\n\r\nNosepuppy, how did you get: $\\angle QPB=k \\Rightarrow \\angle PQA=2k$",
  "Solution_6": "[quote=\"Nosepuppy\"]I believe the answer is actually 571.\n\nlet $\\angle QPB=k$ \nlet $AP=l$\n$\\angle QPB=k \\Rightarrow \\angle PQA=2k$\n$AQ=2l\\cos2k \\Rightarrow PC=2l\\cos2k$ due to ABC being isosceles.\nnote that $\\angle PCB=90-k$ , $\\angle PBC=90-2k$, and $\\angle BPC=3k$\n\n$BP=2l\\cos k$\n$BC=l$ is given\n$BC=BP\\cos(90-2k)+PC\\cos(90-k)=2l\\cos k\\sin(2k)+2l\\cos(2x)\\sin x$\nso\n$l=2l\\cos k\\sin(2k)+2l\\cos(2k)\\sin k$\n$\\frac{1}{2}=\\cos k\\sin(2k)+\\cos(2k)\\sin k$\n$\\frac{1}{2}=\\sin(k+2k)=sin(3k)$\nso $3k=30$, $k=10$\n\nby angle chasing, we arrive at $\\angle APQ=180-4k$, and $\\angle PCB=90-k$\nsince k=10, we have $\\angle APQ=140$, and $\\angle PCB=80$\n\n$\\lfloor 1000\\times \\frac{80}{140} \\rfloor=571$[/quote]\r\n\r\nIn my opinion, it is correct, although a mysterious $x$ slipped in there for $k$ at one point. Alternatively: Take $AP = PQ = QB = BC = 1$ and $\\angle{A} = x$ so that $\\angle{C} = 90-\\frac{x}{2}$. Now, by the law of sines, $\\frac{1}{\\sin(x)} = \\frac{1+2\\cos(x)}{\\sin\\left(90-\\frac{x}{2}\\right)}$. Use that $\\sin(x) = 2\\sin\\left(\\frac{x}{2}\\right)\\cos\\left(\\frac{x}{2}\\right)$, $\\cos(x) = 1 - 2\\sin^2\\left(\\frac{x}{2}\\right)$ and $\\sin\\left(\\frac{3x}{2}\\right) = 3\\sin\\left(\\frac{x}{2}\\right) - 4\\sin^3\\left(\\frac{x}{2}\\right)$ to obtain $\\frac{1}{2} = \\sin\\left(\\frac{3x}{2}\\right)$; whence, $x= 20$, $\\angle{C} = 80$, and $\\angle{APQ} = 140$, for an answer of $\\left\\lfloor \\frac{4000}{7}\\right\\rfloor = 571$.\r\n\r\nTo answer your question Jesusfreak, he used the external angle theorem on the isosceles $\\triangle{PQB}$.",
  "Solution_7": "What's the external angle theorem? Checked Mathworld and can't find it.",
  "Solution_8": "Perhaps that isn't its primary name. Anyway, it simply says that in a triangle $ABC$, the external angle at $A$ is equal in measure to $\\angle{B} + \\angle{C}$.",
  "Solution_9": "[quote=\"Mildorf\"]\n\nIn my opinion, it is correct, although a mysterious $x$ slipped in there for $k$ at one point. \n\n[/quote]\r\n\r\n:O\r\n\r\nman, i need to start proofreading my solutions. fixed",
  "Solution_10": "Oh, okay, that makes sense. Thanks."
}
{
  "Tag": [
    "FTW",
    "\\/closed"
  ],
  "Problem": "is it possible to send a pm to more than one person?\r\nmaybe a team for a FTW tourny or reminding a group of pplz to do something?\r\n\r\nedit: actually, now that i look at the site and gerneral information tab closer...\r\ni think this should go under questions, suggestions, and annoucements place\r\ncould someone move this?",
  "Solution_1": "No, there is no way for you to do that.",
  "Solution_2": "you could create a macro that logs in as you and then loops through a number of people...but that'd take like 50 years to make",
  "Solution_3": "[quote=\"vallon22\"]is it possible to send a pm to more than one person?\nmaybe a team for a FTW tourny or reminding a group of pplz to do something?\n\nedit: actually, now that i look at the site and gerneral information tab closer...\ni think this should go under questions, suggestions, and annoucements place\ncould someone move this?[/quote]\r\nI think there should be a feature that allows this.",
  "Solution_4": "No. It's a tool for spamming.",
  "Solution_5": "you're right , whoops im wrong."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation",
    "trigonometry",
    "AMC"
  ],
  "Problem": "I recently took a very hard math tournament and I encountered the following problem:\r\n\r\n In an equilateral triangle ABC, there exists a point P inside the triangle so that AP=3, BP=4, and CP=5. Find the side length of the triangle.\r\n\r\nIf anyone could please help me solve this I would be very grateful. Thank you very much",
  "Solution_1": "Ahhh this is a \"classic.\"\r\n\r\n[hide=\"Hint 1: bananas\"]$3^2 + 4^2 = 5^2$[/hide]\n[hide=\"Hint 2: plantains\"]Use a rotation[/hide]",
  "Solution_2": "I've worked over and over on this problem and still can't get it.\r\n\r\nCan someone please post a solution?\r\n\r\nThanks a lot.",
  "Solution_3": "this problem and its variations have been posted millions of times...search",
  "Solution_4": "Consider a rotation about B taking C to A.  It takes A to A' and P to P'.  Then PP' = 4, because PBP' = 60 deg by our rotation, and BP = BP' = 4.\r\n\r\nSo PAP' is a 3-4-5 triangle, and thus P'PA is 90 degrees.  So in APB, we have side lengths 3 and 4 contained by a 150 deg angle.  We can find answer w cosine law."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "In any convex $ 2n$-gon, there is a diagonal not parallel to any side.\r\n\r\n[hide=\"A bit confused\"] I am a bit confused, why should each side can have at most $ n\\minus{}2$ diagonals parallel to it. :maybe: [/hide]",
  "Solution_1": "Moonmathpi496 wrote\r\n[quote]I am a bit confused, why should each side can have at most n-2 diagonals parallel to it.[/quote]\r\n\r\nDraw a picture may help you understand it .\r\nLet the polygon be $ CDA_1 A_2 \\dots A_{2n\\minus{}2}$, where $ CD$ is the side, then for any vertice $ A_x$you can find at most 1 vertice in $ \\{A_n\\}$ say $ A_k$ such that $ A_x A_k // CD$ so there are at most $ n\\minus{}1$ diagonals $ // CD$ when all the vertices are utilized.Now let $ A_p A_q$ be the parallel diagonals having the longest distances from $ CD$ , if it's one of the sides of the polygon, then done.If it isn't , there must be a point beyond the parallel line , not all of the vertices are utilized, done also."
}
{
  "Tag": [
    "Mafia",
    "pre100"
  ],
  "Problem": "Okay guys, finally it's here. To mark the quarter century I am bringing in a brand new set of roles. Hope it turns out great, eh? :) \n_____________________________________________________\n\n\n[u][b]Story:[/b][/u]\n\nSo there is this settlement with all its cheerful people, and they go about doing things that cheerful people in settlements usually do. Let's not bother about the details - as they are irrelevant, and, more importantly, thoroughly uninteresting.\n\nHowever, in the province in which our small settlement is situated, there are two rival gangs of buddies ... er, i mean, baddies (sorry, I have just watched the awesome movie Air Buddies, it ended around ten minutes ago). These gangs go by the fanciful names \"The Cobras\" and \"The Scorpions\" (cliched, no doubt, but very strong nevertheless, and anyway you can't reasonably expect more innovation from these baddies can you?). As always happens in mafia game stories, they decide to send their representatives to our settlement, three each.\n\nNow what do these newcomers do? Elementary, my dear Watson. They look at the list of inhabitants and each independently chooses to bump one off and then impersonate him. Now, fortunately for them, (and to keep the game straight,) no two choose the same person. However none of the six have any idea at all as to which of the other people are baddies in disguise.\n\nIn the days to come, as the two gangs try to take control of our settlement, it is going to turn into one huge bloody mess ... ooh, I can't even imagine it. Thankfully I am not in there, so for all those of you who are, best of luck!\n_____________________________________________________\n\n\n[b][u]Roles:[/u] [/b]\n\n3 cobras\n3 scorpions\n2 inspectors\n2 doctors \n2 vigilante (should I put an 's' here to indicate the plural? :maybe: )\n10 civilians\n_____________________________________________________\n\n\n[b][u]Player list:[/u] [/b]\n\nb-flat\nPeterVDD\nFanatic\nmustafa\nLena\nprobability1.01\nJashper\njackdillon\nArs\njunggi\npianoforte\negghead91\nperfect628\nVihang\nlife=tennis+math\nkheg_k\nSly Si\nbaseballfreak2\nCaptain Sugar\nweiquan\nlaughinghead505\nMelissa\n\n_____________________________________________________\n\n[u][b]Rules:[/b][/u]\n\n1) Each night, each of the special roles sends in an independent choice.\n2) If two people of the same group visit the same person on a particular night, they will be connected to each other.\n3) Two rival baddies visiting each other on the same night will both fail, but will get to know each other.\n4) Each day will be exactly 96 hours long, irrespective of what goes on during the day. Each night will be 48 hours long. (You can think of time having been slowed down by a factor of six.) The ONLY exception to this rule will be if the site's server is down or the forum is locked, in either of which case I'll make appropriate adjustments.\n5) Anybody who has been accused and seconded TWICE (all by different persons) is declared to be on the block.\n6) Any number of people can be put on the block on a single day.\n7) Once you are on the block, you don't get off it till the day is over.\n8) At the end of the day, a guy on the block is executed if and only if more than half the current living population have voted on him, and there are more ayes than nays.\n9) The accusers and the seconders are automatically assumed to vote unretractable ayes, irrespective of their statements later on. (Note that very late in the game this may imply that if you get on the blocked, you are automatically lynched. This is a bug that has made me unhappy, so I may change some of the rules later.)\n10) No one is allowed to accuse oneself, secodn oneself's accusation, or vote on oneself.\n11) Everyone else may change their votes or even retract them any number of times before the days is over.\n12) Since time is limited, you need not wait for someone's defense to start voting on him.\n13) Any number of people can be executed at the end of the day.\n14) Since there is no way I can stop you from interacting through PMs during the day, I make it legal.\n_____________________________________________________\n\n\nI am sure I have not covered everything, and you will have a lot of questions. So go ahead, the game doesn't start immediately (I have not even decided upon the final list of players), have your appetite full of questioning. However, I am not changing the above rules, with the possible exception of 4, where I am ready to make each day to 144 hours, if may of you wish to have it so.\n\nLooking forward to a grand game! :)",
  "Solution_1": "w00t first post!\r\n\r\nI'm kidna confused about the powers of the baddies? help?\r\n\r\n[quote=\"bubka\"]\n14) Since there is no way I can stop you from interacting through PMs during the day, I make it legal.[/quote]\r\ner what do you mean by this? we can discuss the game through PMs? :huh:",
  "Solution_2": "If two baddies are conencted, they can discuss the game.\r\nEach baddy gets one kill per night, he'll pray it's not one of his own gang.",
  "Solution_3": "so each night there's going to be 6 kills, one from each baddie?\r\nerm... :maybe: and do they know who the fellow baddies are?\r\nand what's that part about impersonating?",
  "Solution_4": "Vihang: what's that part about impersonating?\r\nso each night there's going to be 6 kills, one from each baddie?\r\n\r\nRitam: plus two from the vigilante\r\n\r\nVihang: eeeks\r\n8 !\r\n\r\nRitam: add to it around four lynches a day\r\n\r\nVihang: is the game 2 -3 days long ?\r\n\r\nRitam: do you want it to be longer?\r\nand die eventually?\r\n\r\nVihang: nope\r\nthis is good fast and furious\r\n\r\nRitam: no mafia game lasts beyond three days and remains full of activity\r\n_____________________________________\r\n\r\n\r\nThat answers a lot. As for knowing each other, read through the story carefully again.\r\n_____________________________________\r\n\r\nRitam: as for impersonating, the baddies start the game impersonating the people who play them\r\n\r\nVihang: ok\r\n\r\nRitam: the idea is that these people were inhabitants of the settlement\r\nso now nobody notices the difference\r\n\r\nVihang: yeah got it\r\n_____________________________________\r\n\r\n\r\nThat answers the rest.",
  "Solution_5": "So when do we get our role PM's ?",
  "Solution_6": "Role PMs?? \r\nDream on boy!\r\nI don't even have the final list of players yet!\r\n\r\nWhy, they maybe even a week off!",
  "Solution_7": "erm...\r\nif they haven't confirmed by now, might as well kick them out...it just means that they're going to be inactive in the game",
  "Solution_8": "Firstly, this game is a bit robust to inactives.\r\nSecondly, the situation's a bit different.\r\nI can't explain it.\r\nDon't worry, I'll try and hurry!",
  "Solution_9": "I would request that days be made at least 144 hours, if not longer, given that there will be very few of them.",
  "Solution_10": "96 hours? What the hell? How about one week?\r\n\r\nAlso, when do we find out the roles of people?\r\n\r\nThis game seems way too swingy.",
  "Solution_11": "Why don't we just start now? The people who haven't confirmed yet can be moved up to 26.",
  "Solution_12": "[quote=\"bubka\"]Role PMs?? \nDream on boy!\nI don't even have the final list of players yet!\n\nWhy, they maybe even a week off![/quote]I really suggest making the current list final... this looks so good that we want to start right away! :)\r\n\r\nAnd I agree 144 hours is much better.",
  "Solution_13": "I know I haven't signed up, but THESE NEW ROLES SOUND GREAT!!! :) \r\n\r\nP.S. This doesn't mean I'm signing up... :|",
  "Solution_14": "Okay 144 hours it be.\r\nAnd I'll try and hurry up the role PMs.\r\nBut lazy as I am, they are a torture. :( \r\n\r\nBy the way life=tennis+math is away on a ten-day easter break and he has asked me not to replace him. What do I do with him?",
  "Solution_15": "Okay now for the night events:\r\n\r\nThe purple cobra made short work of perfect628, a mere civilian.\r\nThe red scorpion got the eluding green cobra Sly Si finaly in the small of the back, and squared things up.\r\nThe second vigilante went back seething after finding perfect628 dead already.\r\nAs a parting shot, Sly Si disposed of the second fairy laughinghead505 who has long since bid this forum goodbye.\r\nFinally, the impatient purple cobra killed me.  :( \r\n\r\nIt shows the lengths to which I go for you people, that even after being ungratefully killed by one of you, I am continuing the game for you.\r\n\r\n'It's a new day' - [i]Bryan Adams, Here I am, Spirit: Stallion of the Cimarron soundtrack[/i]",
  "Solution_16": "Next night you should agree on who to kill, this was a waste of 2 kills. =)",
  "Solution_17": "Omg Peter VDD has dropped VDD and become simply Peter.",
  "Solution_18": "oh i was still in this...",
  "Solution_19": "hmm the purple cobra....\r\nsounds familiar",
  "Solution_20": "I'm closing this game too.",
  "Solution_21": "Sorry but it is my game. It's still open.",
  "Solution_22": "who are the people who are still playing?",
  "Solution_23": "i am the purple cobra\r\n\r\nhissssssssssssssss",
  "Solution_24": "everyone alive is still playing",
  "Solution_25": "but it's hardly playing...",
  "Solution_26": "this is really confusing\r\n\r\ngame title said game abandoned",
  "Solution_27": "So when does 26 start?",
  "Solution_28": "[color=red]I accuse lena[/color]",
  "Solution_29": "last post"
}
{
  "Tag": [],
  "Problem": "I had \"Contests Around the World \" volumes: 1998-1999; 1999-2000; and 2000-2001. I searched (by google ; and some sites in Vietnam ) but I couldn't found another volumes ( since 2001-2002 till now). Who have one of them (e-book versions  :) ) so please help me!\r\nThank you!",
  "Solution_1": "[quote=\"Allnames\"]I had \"Contests Around the World \" volumes: 1998-1999; 1999-2000; and 2000-2001. I searched (by google ; and some sites in Vietnam ) but I couldn't found another volumes ( since 2001-2002 till now). Who have one of them (e-book versions  :) ) so please help me!\nThank you![/quote]\r\n\r\nI want to purchase the original copy of :\r\n\r\nMathematical Olympiads: Problems and Solutions from around the World, 1995-1996\r\n\r\nCan any one tell me where i can find. I already tried Amazon and all other online book stores no one has this book in their database.",
  "Solution_2": "I don't claim to know whether these books are still copyrighted or not (however I believe they are), but do heed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126338]this[/url]."
}
{
  "Tag": [
    "ratio",
    "vector",
    "algebra",
    "polynomial",
    "limit",
    "induction",
    "absolute value"
  ],
  "Problem": "In this infinite tree, degree of each vertex is equal to 3. A real number $ \\lambda$ is given. We want to assign a real number to each node in such a way that for each node sum of numbers assigned to its neighbors is equal to $ \\lambda$ times of the number assigned to this node.\r\nFind all $ \\lambda$ for which this is possible.",
  "Solution_1": "What have I misunderstood? $ \\lambda/3$ to every node means any $ \\lambda$ seems to work. EDIT: Oh, ratios of sums.",
  "Solution_2": "That gives a ratio of $ 3$, not $ \\lambda$.  In fact, this is an eigenvalue problem:  if $ \\mathbf{A}$ denotes the adjacency operator of the graph (acting on the vector space consisting of linear combinations of its vertices), we want to find the set of real eigenvalues corresponding to real eigenvectors.  \r\n\r\nNow, I'm also getting that every real $ \\lambda$ works:  assign every vertex a weight $ w_n$ depending only on its depth $ n$ (where the root has depth $ 0$, each neighbor has depth $ 1$, and so forth) such that\r\n\r\n$ w_0 \\equal{} 1$\r\n$ w_1 \\equal{} \\frac {\\lambda}{3}$\r\n$ w_n \\plus{} 2w_{n \\plus{} 2} \\equal{} \\lambda w_{n \\plus{} 1} \\Leftrightarrow$\r\n$ w_{n \\plus{} 2} \\equal{} \\frac {\\lambda w_{n \\plus{} 1} \\minus{} w_n}{2}.$\r\n\r\n[b]Edit:[/b]  Did you mean to add the condition that the weights must be bounded?  The growth rate of $ w_n$ is controlled by the roots of the characteristic polynomial $ 2x^2 \\minus{} \\lambda x \\plus{} 1$, which are\r\n\r\n$ x \\equal{} \\frac {\\lambda \\pm \\sqrt {\\lambda^2 \\minus{} 8}}{4}.$\r\n\r\nWe want both roots to have absolute value less than or equal to $ 1$.  The larger root is less than or equal to $ 1$ if and only if $ \\lambda \\le 3$, and the smaller root is greater than $ \\minus{} 1$ if and only if $ \\lambda \\ge \\minus{} 3$, so the spectrum by this construction is $ [ \\minus{} 3, 3]$.  (Note that $ x$ doesn't have to be real.)  We claim that this is best possible.  \r\n\r\nBy the spectral radius formula, the spectrum of $ \\mathbf{A}$ is contained in the ball of radius $ \\lim_{n \\to \\infty} \\parallel{} \\mathbf{A}^n \\parallel{}^{\\frac {1}{n} }$, where $ \\parallel{} \\mathbf{A} \\parallel{}$ is the operator norm on $ \\ell^{\\infty}(V)$.  But since the degree of each vertex is $ 3$, by analogy with the finite-dimensional case it follows that $ \\parallel{} \\mathbf{A}^n \\parallel{} \\equal{} 3^n$, so the conclusion follows.  (I can give a more fleshed-out argument, but I hope it's not really necessary.)\r\n\r\n[b]Edit #2:[/b]  Okay, so a more elementary argument is possible.  Namely, if the weights are distributed such that they do not depend only on depth, show by induction that for every $ n$ there exists a vertex of depth $ n$ whose weight has absolute value greater than or equal to the absolute value of $ w_n$ from the above construction (essentially, from each vertex pick the child with the larger weight in absolute value).  If $ \\lambda \\not \\in [\\minus{}3, 3], w_n$ is unbounded and the conclusion follows."
}
{
  "Tag": [
    "function",
    "limit",
    "calculus",
    "derivative",
    "integration",
    "trigonometry",
    "algebra"
  ],
  "Problem": "What is l'hospital? I tried googling it, but I still don't understand what it is. Thanks in advance! \r\n\r\n(Also, could someone please provide a few examples? Thank you.)",
  "Solution_1": "St. Andrews history site link for the [url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/De_L%27Hopital.html]Marquis de L'H\u00f4pital.[/url]\r\n\r\nVariations in both the spelling of French over the centuries and for fashion in transliteration into English mean you may see it as L'Hospital, L'H\u00f4pital, L'Hopital, and possibly some other variants.\r\n\r\nHere's the [url=http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule]Wikipedia link.[/url]\r\n\r\nOK, neither link is posting correctly. I'll put them up as is while I explore what's wrong.",
  "Solution_2": "I think it's a matter of underscores not interacting well with the url tags.  i recall that darij knows a way around it but i don't remember what it is offhand, so you may want to pm him.",
  "Solution_3": "[quote=\"Kent Merryfield\"]St. Andrews history site link for the [url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/De%5FL%27Hopital.html]Marquis de L'H\u00f4pital.[/url]\n\nVariations in both the spelling of French over the centuries and for fashion in transliteration into English mean you may see it as L'Hospital, L'H\u00f4pital, L'Hopital, and possibly some other variants.\n\nHere's the [url=http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s%5Frule]Wikipedia link.[/url]\n\nOK, neither link is posting correctly. I'll put them up as is while I explore what's wrong.[/quote]\r\n\r\nOK, hex ASCII rules. Just substitute apostrophe with %27 and underscore with %5F.",
  "Solution_4": "Any quote ' should be escaped as %27. You cannot have quotes in an URL (legally.)",
  "Solution_5": "I'm going to assume you want to know what the rule is?  Suppose you have two functions, $ f(x)$ and $ g(x)$, and\r\n$ \\lim_{x\\rightarrow a}f(x)\\equal{}0$ and $ \\lim_{x\\rightarrow a}g(x)\\equal{}0$.  Then what is\r\n$ \\lim_{x\\rightarrow a}\\frac{f(x)}{g(x)}$?\r\nIf you just stick in yourt two limits, you're going to get $ 0/0$, which is called an [i]indeterminant[/i].  So to evaluate this limit, simply take the derivative of $ f$ and $ g$, and then take the limit.  That is,\r\n$ \\lim_{x\\rightarrow a}\\frac{f(x)}{g(x)}\\equal{}\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$\r\nNote, however, that you may still have an indeterminant integral so you might have to do it again.  This technique also holds for when the quotient is $ \\infty/\\infty$.  But if you don't have such an indeterminant expression, the rule will not work.\r\n\r\nHere's an example:\r\nFind $ \\lim_{x\\rightarrow 1}\\frac{x^{3}\\plus{}x\\minus{}2}{x^{2}\\minus{}3x\\plus{}2}$.\r\n\r\n[hide=\"Solution\"]\n$ \\lim_{x\\rightarrow 1}\\frac{x^{3}\\plus{}x\\minus{}2}{x^{2}\\minus{}3x\\plus{}2}\\equal{}\\lim_{x\\rightarrow 1}\\frac{3x^{2}\\plus{}1}{2x\\minus{}3}\\equal{}\\frac{4}{\\minus{}1}\\equal{}\\minus{}4$\nIf you try to use it again, you will get the wrong answer since what we have now is not an indeterminant.[/hide]",
  "Solution_6": "I generally argue for the minimization of the use of L'H\u00f4pital's rule. There are a few places in which it is the most convenient thing to use, and I'll use it there, but in many other cases, I will use other arguments. These arguments include:\r\n\r\n[b]1. Knowing familiar limits related to derivatives:[/b]\r\n\r\nExamples here would include $ \\lim_{x\\to 0}\\frac {\\sin x}{x},$ which you would need to know before you compute the derivative of the sine in the first place. It's circular reasoning to use anything involving the derivative to obtain it.\r\n\r\n[b]2. Algebra:[/b]\r\n\r\nI would not have used L'H\u00f4pital for JRav's example. Once I have verified that this polynomial divided by a polynomial is a $ \\frac00$ case as $ x\\to 1,$ then I know that $ (x \\minus{} 1)$ is a factor of both the numerator and denominator. Since I know that is a factor, I should be able to factor both quite easily (by synthetic division if I have to) and eliminate the common factor.\r\n\r\n[b]3. The heirarchy of size:[/b]\r\n\r\nI do not want to refer to L'H\u00f4pital every time I face $ \\lim_{x\\to\\infty}x^3e^{ \\minus{} x/2}$ or $ \\lim_{x\\to0^ \\plus{} } \\sqrt {x}\\,\\ln x.$ I know that both are zero because I know that exponentials grow faster than powers of $ x$ and logarithms grow slower than powers of $ x.$\r\n\r\nSimilarly, we can decide that $ \\lim_{n\\to\\infty}\\frac {3n^4 \\plus{} 3n \\minus{} 5}{n^4 \\plus{} n^2 \\plus{} 1} \\equal{} 3$ and $ \\lim_{n\\to\\infty}\\frac {n^4 \\minus{} 3^n}{n^4 \\plus{} 3^n} \\equal{} \\minus{} 1$ by simply paying attention to what parts of each numerator and denominator are important.\r\n\r\n[b]4. Taylor polynomials and power series:[/b]\r\n\r\nA famous geometry limit problem leads to this: $ \\lim_{\\theta\\to0}\\frac {\\theta \\minus{} \\sin\\theta}{(1 \\minus{} \\cos\\theta)\\sin\\theta}.$ Is this a good place to use L'H\u00f4pital? Not really: L'H\u00f4pital will be quite messy, and will need to be done more than once. Given my familiarity with the power series for sine and cosine, I can do this [i]in my head[/i]. Try it! See if you can do it in your head. [hide=\"Informal solution.\"]$ \\theta \\minus{} \\sin\\theta$ looks like $ \\frac {\\theta^3}6$ and $ 1 \\minus{} \\cos\\theta$ looks like $ \\frac {\\theta^2}2.$ Throw in $ \\sin\\theta\\approx\\theta$ and we have $ \\frac {\\theta^3}{\\theta^3}$ - the exponents match. (And I know why L'H\u00f4pital would have to be done several times.) Paying attention to the coefficients I see $ \\frac16$ divided by $ \\frac12.$ Do that, and the limit is $ \\frac13.$[/hide]\r\n\r\nSome posters around here are fond of posting things like \"Find $ \\lim_{x\\to 0}\\frac {e^x \\minus{} 1 \\minus{} x}{x^2}$ and don't use Taylor series or L'H\u00f4pital.\" And I immediately go away - why tell me not to use Taylor series when I can see [i]at a glance[/i] that that limit is $ \\frac12?$\r\n\r\nHere's another one: $ y \\equal{} \\lim_{m\\to\\infty}\\left(1 \\plus{} \\frac rm\\right)^{mt},$ from a compound interest problem. Since this is a $ 1^{\\infty}$ indeterminate case, we take the logarithm:\r\n\\[ \\ln y \\equal{} \\lim_{m\\to\\infty}mt\\ln\\left(1 \\plus{} \\frac rm\\right).\r\n\\]\r\nThis is now an $ \\infty\\cdot 0$ indeterminate case. Do we find a way to turn it into a quotient so we can use L'H\u00f4pital? I'd rather not - it's fairly easy to make mistakes that way in handling the extraneous variables $ r$ and $ t.$ Instead, let's use the Taylor series for $ \\ln(1 \\plus{} x) \\equal{} x \\minus{} \\frac {x^2}2 \\plus{} \\cdots.$\r\n\r\n$ \\ln y \\equal{} \\lim_{m\\to\\infty}mt\\left(\\frac rm \\minus{} \\frac {r^2}{2m^2} \\plus{} \\cdots\\right)$\r\n\r\n$ \\equal{} \\lim_{m\\to\\infty}\\left(rt \\minus{} \\frac {r^2t}{2m} \\plus{} \\cdots\\right) \\equal{} rt,$ from which $ y \\equal{} e^{rt}.$\r\n\r\n===================\r\n\r\nThere are cases in which L'H\u00f4pital is the most efficient method. Here's one, which makes use of the fact that L'Hopital works for complex analytic functions:\r\n\r\nIf $ f(z) \\equal{} \\frac {e^{iz}}{z^4 \\plus{} 4},$ find the residue of $ f$ at $ z \\equal{} 1 \\plus{} i.$ This amounts to finding $ \\lim_{z\\to 1 \\plus{} i}\\frac {e^{iz}(z \\minus{} 1 \\minus{} i)}{z^4 \\plus{} 4} \\equal{} e^{i(1 \\plus{} i)}\\lim_{z\\to 1 \\plus{} i}\\frac {z \\minus{} 1 \\minus{} i}{z^4 \\plus{} 4}.$\r\n\r\nNow, I could say about this what I said in #2: it's a rational function, I know the factors, and I can do this by algebra. But there are several factors with some complex mulitplications. Let's just use L'H\u00f4pital:\r\n\r\n$ \\equal{} e^{ \\minus{} 1 \\plus{} i}\\lim_{x\\to1 \\plus{} i}\\frac {1}{4z^3} \\equal{} \\frac {e^{ \\minus{} 1 \\plus{} i}}{4(1 \\plus{} i)^3} \\equal{} \\frac {e^{ \\minus{} 1 \\plus{} i}}{8 \\minus{} 8i},$ which can then be simplified further.\r\n\r\n(Further simplification will be dictated by the larger problem this is part of. That larger problem might be the computation of $ \\int_{\\minus{}\\infty}^{\\infty}\\frac{\\cos x}{x^4\\plus{}4}\\,dx.$ )",
  "Solution_7": "[quote=\"Kent Merryfield\"]\nI would not have used L'H\u00f4pital for JRav's example.\n[/quote]\r\nI probably would not use it either but that is a problem I took from the Spivak book, that emphasizes the fact the the rule doesn't work when we don't have an indeterminant."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let a,b,c are positive real numbers such that abc=1\r\nprove that $a^{2}+b^{2}+10(ab+bc+ca) \\geq 11(a+b+c)$",
  "Solution_1": "[quote=\"nkht-tk14\"]Let a,b,c are positive real numbers such that abc=1\nprove that $a^{2}+b^{2}+10(ab+bc+ca) \\geq 11(a+b+c)$[/quote]\r\nIt's wrong.\r\nMaybe $a^{2}+b^{2}+c^{2}+10(ab+ac+bc)\\geq11(a+b+c)$ $?$\r\nIf so, see here\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197"
}
{
  "Tag": [
    "geometry",
    "trigonometry",
    "circumcircle",
    "AoPS Books"
  ],
  "Problem": "I am currently in a math class and I get all of the  material, but the tests are SOOOO HARD!They are almost always curved and I got a 74, 68, and 88 for the quiz grades. I got 79 and 70 for tests grades so my grade is like a C+. I just took another insane quiz today and I will tell you about it Monday.\r\n\r\nSo if the tests are really hard, and the homework and classwork are [i]easy[/i] in comparison to the test problems, what should I do? I bought a AOPS textbook and it seems to help but I think I should do more\r\n\r\nAnd is it possible to still get an A-?The semester ends January 21st.",
  "Solution_1": "What math class are you in?  Does your school average the quarter grades or is it cumulative?",
  "Solution_2": "And to prepare for your tests, think about this:\r\n\r\nIt's highly unlikely that your teacher would give you something on the test that you had not even learned. (Right???  :huh: )\r\nSo the most likely reason that your tests are harder than the homework is that they apply the same things, but in different ways. So make sure you understand how and when to apply the concepts, in [i]all[/i] situations.  :wink:",
  "Solution_3": "In most cases, solving the \"C\" level exercises (the last 3-4 problems in each exercise test) helps you prepare for the harder problems on most tests.",
  "Solution_4": "I had one of these insanely hard teachers before for geometry. She would give out the most insanely hard tests for us. What i did to get prepared was that i did all of the C level problems and i also got my mom to help me on some of the proofs i didnt get  :P",
  "Solution_5": "Which AoPS book did you buy?  Geometry?",
  "Solution_6": "Volume 1 would really help, because it has a lot of information on Trig, and it has lots of problems that they don't teach you the solutions to. You have to kind of think about what you know and think out of the box. The solutions to the problems would probably help you a lot with some of the problems on your tests.",
  "Solution_7": "...solutions manual?",
  "Solution_8": "I suggest he get Volume 1, and the solution manual, if that's what you were wondering AIME.",
  "Solution_9": "No, I was saying that all the problems have solutions in the solutions manual...\r\n\r\nSome stuff you learn in AoPS  books won't help you much in school, but will hlep you in contests.",
  "Solution_10": "for example, some circumcircle properties and basic cyclic quadrilateral's are almost never covered in some geometry courses.",
  "Solution_11": "[quote=\"AIME15\"]Which AoPS book did you buy?  Geometry?[/quote]\r\nme? i just used the one our teacher gave us"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "What is the formula for gcd(ab,cd)?",
  "Solution_1": "In terms of what?\r\n\r\nPierre.",
  "Solution_2": "Look:  http://mathworld.wolfram.com/GreatestCommonDivisor.html\r\nHere is a formula for $gcd(ab,cd)$, but I found counterexample. Could you explain how should I read what is written there.",
  "Solution_3": "Talking about the 14th formula?\r\n\r\nThey might have forgotten a \"GCD\"..",
  "Solution_4": "yes, I guessed, but try according to this formula calculate\r\n$gcd(2\\cdot5, 4\\cdot6)$"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "AMC",
    "AMC 12",
    "MATHCOUNTS"
  ],
  "Problem": "Please do not discuss difficulty, your scores, problems, or anything else of the sort [EXCEPT ONLY your and your team's  placement - NOT the placements of, say, Jonathan Horton's or Wenyu Cao's unless you are one of those people] until after Feb. 25. Remember that some schools do determine their MC chapter participants by the School Round, and therefore if you inform a contestant who has not taken the test that the problems were easy/medium/hard and that you got a xx they have an unfair advantage. \r\n\r\n(Feb. 25 is the date of the last chapter competition; to be safe we may assume that schools may choose their competitors by the school round administered anytime up to the day before their chapter competition. It's only 20 days from now!)\r\n\r\nThanks for cooperating and making AoPS a great place to be!\r\n\r\nBilly\r\n\r\n[size=75][Edited to update the date to the present year.][Edited to update the title to reflect chapter too.][Edited to clarify that your rank is all right][/size]",
  "Solution_1": "[quote=\"solafidefarms\"]if you inform a contestant who has not taken the test that the problems were easy/medium/hard and that you got a xx they have an unfair advantage. \n\nBilly[/quote]\r\n\r\n[size=150]I do not think so. [/size]If Joe Jia or Alberta Ni informed you that AMC 12 problems are easy and they all got perfect scores two or three times, OR  if  Patricia Li  or Gregory Gauthier informed you that the school rounds are very easy and they all got 46, do you really have an unfair advantage when you take AMC12 or Mathcounts school rounds?",
  "Solution_2": "If they took those tests before you did and told you those things about that test, you would know that the tests were not super-super-impossibly hard. And if someone else said so, and they were a somewhat active poster, one could judge how good they were and thus how hard the test is.",
  "Solution_3": "If I told you that the school round was super-impossibly-hard, would it really give them an unfair advantage to know that information?",
  "Solution_4": "They would know to study really hard and to work really hard, whereas if you told them it was really easy they'd know that they didn't need to study. Let's say that your friend takes it and tells you that you don't need to study, it's really easy, and you know that he is truthful and also pretty stupid. You know that it's a really easy test. On the other hand, if your really super smart friend, let's say Andrew Ardito, tells you that the school round is really hard and you really ought to study for it. You know that you really have to study hard for it and you need to make sure the ones you do are right, not answer all of them. But in both cases you have an unfair advantage.",
  "Solution_5": "Whether you agree with solafidefarms is irrelevant.  The rule of this forum is that there shall be NO discussion of any MATHCOUNTS Competition until the window for taking the competition is closed.  End of discussion.\r\n\r\nDo not say anything about the 2006 School Round until after February 26.",
  "Solution_6": "I also agree with rcv and solafidefarms.\r\nIf you see a scores discussion, PM the moderators quick, as it gets along very fast.\r\nAlso, aren't we allowed to discuss our place in the school and whether we are a team member or individual?",
  "Solution_7": "[quote=\"BFG\"][quote=\"solafidefarms\"]if you inform a contestant who has not taken the test that the problems were easy/medium/hard and that you got a xx they have an unfair advantage. \n\nBilly[/quote]\n\n[size=150]I do not think so. [/size]If Joe Jia or Alberta Ni informed you that AMC 12 problems are easy and they all got perfect scores two or three times, OR  if  Patricia Li  or Gregory Gauthier informed you that the school rounds are very easy and they all got 46, do you really have an unfair advantage when you take AMC12 or Mathcounts school rounds?[/quote]\r\n\r\nThere's no point in making an argument.  The discussion to prevent discussion is one made by AoPS and requested by the MATHCOUNTS organization on a yearly basis.\r\n\r\nThere are plenty of past problems that can be discussed for math's sake.",
  "Solution_8": "Resurrected for this year.",
  "Solution_9": "Whoa, feb 26? That's weeks after my Chapters.",
  "Solution_10": "Some schools give the school test immediately before the chapter test. The last chapter test is Feb. 25. Ergo some schools may give the school test Feb. 25. Thus you should not discuss the school round until Feb. 26.",
  "Solution_11": "We should have a chapters topic up as well, I know our region had their chapter today.",
  "Solution_12": "Can we discuss our ranking?\r\nAlso, is it normal for someone who made nationals to not make states the year after :oops:",
  "Solution_13": "Can we say whether we're making the next level, even though our regionals are like 2 weeks from now?",
  "Solution_14": "I pm'ed nebula and he said that we can say our place, just not our score",
  "Solution_15": "is it after the 25th in case there is \"inclement\" weather and it will have to be held the 25th, not the 24th in some areas?\r\n\r\n-jorian",
  "Solution_16": "[quote=\"bpms\"]Can we discuss our ranking?\nAlso, is it normal for someone who made nationals to not make states the year after :oops:[/quote]\r\n\r\nI guess, as long as you do not discuss scores, it is perfectly fine to discuss the ranking.",
  "Solution_17": "[quote=\"jhredsox\"]is it after the 25th in case there is \"inclement\" weather and it will have to be held the 25th, not the 24th in some areas?\n\n-jorian[/quote]\r\nNo, it is after mathcounts posts county on their site.",
  "Solution_18": "This thread has been superseded by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132111]my official announcement[/url] of the discussion rules."
}
{
  "Tag": [
    "real analysis",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let A and B be two disjoint sets in the interval (0, 1) . Denoting by \u03bc the Lebesgue measure on the real line, let \u03bc(A) > 0 and \u03bc(B) > 0 . Let further n be a positive integer and \u03bb=$ \\frac{1}{n}$. Show that there exists a subinterval (c, d) of (0, 1) for which \u03bc(A$ \\cap$ (c, d)) = \u03bb\u03bc(A) and\r\n\u03bc(B $ \\cap$(c, d)) = \u03bb\u03bc(B) . Show further that this is not true if \u03bb is not of the form $ \\frac{1}{n}$.",
  "Solution_1": "As a general preliminary remark we should additionally assume $ A$ and $ B$ be measurable sets. [quote=\"Mimmzy\"]Show further that this is not true if \u03bb is not of the form $ \\frac {1}{n}$.[/quote]I don't understand, since there is an easy counterexample for every real $ \\lambda\\in(0,1)$:\r\n\r\nThen let $ A\\equal{}(a_1,a_2)$ and $ B\\equal{}(b_1,b_2)$ be two intervals with $ 0<a_1<a_2<b_1<b_2<1$. For every real $ \\lambda\\in(0,1)$ the claim can be fulfilled with  $ c_\\lambda\\equal{}\\lambda a_1\\plus{}(1\\minus{}\\lambda)a_2\\in A$ and $ d_\\lambda\\equal{}(1\\minus{}\\lambda)b_1\\plus{}\\lambda b_2\\in B$.\r\n\r\nOr am I missing something here?",
  "Solution_2": "Where is the counterexample? :maybe:\r\nThe statement claims that there $ exists$ an interval but not for $ all$ intervals or I'm missing something.",
  "Solution_3": "I interpret the last sentence as that if there are $ c$ and $ d$ with the property given above given $ A$, $ B$ and $ \\lambda$  then necessarily $ \\lambda\\in\\frac {1}{\\mathbb N}$, in particular $ \\lambda\\in\\mathbb Q$. For this I provided a counterexample that $ \\lambda$ need not necessarily be rational. \r\n... or is the last statement to be read in another way?  :(",
  "Solution_4": "I guess it should be readed that if for every pair $ A,B$ the statement is true then $ \\lambda$ is of the form $ \\frac {1}{n}$ (I did not think about the problem so maybe I'm wrong).",
  "Solution_5": "I had the same puzzle as Third Edition, that's why I post it here.",
  "Solution_6": "OK, if I understand it correctly then it is to be shown that, given a real number $ \\lambda\\in(0,1)$, the following assertions are equivalent (the case $ \\lambda\\equal{}1$ is trivial, hence excluded):\r\n\r\n1 - $ \\lambda\\in\\frac{1}{\\mathbb N}$;\r\n\r\n2 - For every pair $ A,B$ of disjoint measurable subsets of $ (0,1)$ with positive Lebesgue measure, there is an interval $ C\\equal{}(c,d)\\subset(0,1)$ such that $ \\mu(A\\cap C)\\equal{}\\lambda\\mu(A)$ and $ \\mu(B\\cap C)\\equal{}\\lambda\\mu(B)$."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "absolute value",
    "calculus computations"
  ],
  "Problem": "Integrate the following \\[\\int{\\sqrt{1-\\sin x}}\\,dx\\]",
  "Solution_1": "$\\sqrt{1-\\sin x}=\\frac{|\\cos x|}{\\sqrt{1+\\sin x}}.$",
  "Solution_2": "Try making the substitution $u=\\frac{\\pi}2-x,$ which should turn the problem into\r\n\r\n$\\int\\sqrt{1-\\cos u}\\,du$\r\n\r\nThen use a half-angle identity.",
  "Solution_3": "Also note $1-\\sin x = 1-2\\sin \\frac{x}{2}\\cos \\frac{x}{2}= (\\sin \\frac{x}{2}-\\cos \\frac{x}{2})^{2}.$",
  "Solution_4": "All of these things work. Also note something that is explicit in kunny's post (with the absolute value) and hidden but still implicitly there in the other two posts: you will have to be careful about signs in different quadrants. In a definite integral, you will probably need to break the integral into separate pieces.",
  "Solution_5": "many thanks for all ,\r\n\r\n[quote=\"Kent Merryfield\"]will probably need to break the integral into separate pieces.[/quote]  \r\n\r\n[b]How?[/b]",
  "Solution_6": "Make de substitution\r\n\r\n$\\sin x = u$,\r\n\r\nthus\r\n\r\n$\\int \\sqrt{1 \\pm \\sin x}\\, dx \\, = \\, \\int \\sqrt{1 \\pm u}\\, \\frac{du}{\\sqrt{1-u^{2}}}\\, = \\, \\int \\frac{du}{\\sqrt{1 \\mp u}}\\, =$\r\n\r\n$= \\, \\mp 2 \\sqrt{1 \\mp u}+C \\, = \\, \\mp 2 \\sqrt{1 \\mp \\sin x}+C$"
}
{
  "Tag": [
    "geometry solved",
    "geometry"
  ],
  "Problem": "hello to everybody\r\n\r\nthere is a circle S with radius r and centre O.\r\nit has a chord(not diameter) with points A,B on S.\r\nThe altitude from O to AB cuts AB and S respectively \r\nat the points P and Q.\r\nTake an another chord(not diameter and AB) which\r\ngoes through P and cuts S at M and N.( PM<PN )\r\nif PM=a and PQ=b \r\nprove that a=b.cos(ANB)\r\n\r\nthanks...",
  "Solution_1": "Are you sure? I'm pretty sure PM>PQ, so PM can't be PQ*cos(something)... Maybe I'm wrong.",
  "Solution_2": "i am sure you are wrong \"maniaq\"",
  "Solution_3": "What this wrong problem is still doing here???? :?",
  "Solution_4": "So you agree with me, Fedor? Did ne1 else check this out?",
  "Solution_5": "Fatih....I guess Grobber is correct\r\n\r\nIf we start-off with the assumption that a < b......\r\n\r\nWe have a.PN=b.(2r-b) as a < b\r\n\r\nWe have PN> 2r-b...results in stating that diameter is not the longest chord in a circle.\r\n\r\nhence....a=b*cos(ANB) is not possible.....\r\n\r\nPlease check the problem.",
  "Solution_6": "It's easy to see that if A and B are 2 pts on a circle and M is their midpt, then max{MX|X on the circle}=MS and min{MX|X on the circle}=MT, where S and T are the mdidpts of the larger/smaller arc AB, respectively."
}
{
  "Tag": [
    "analytic geometry",
    "Euler",
    "topology",
    "geometry",
    "3D geometry",
    "sphere",
    "invariant"
  ],
  "Problem": "1.  Start with the unit disc $ \\mathbb{D}$ and identify points on the boundary if their angles, thought of in polar coordinates, differ by a multiple of $ \\frac {\\pi}{2}$. Let $ X$ be the resulting space, use the Van Kampen's theorem to compute $ \\pi_{1}(X,*)$\r\n\r\n2. Let $ X \\equal{} S^{2}/\\{p_{1} \\equal{} \\ldots \\equal{} p_{k}\\}$  be the topological space obtained by identifying $ k$ distinct points $ p_{k}$  on it ( $ k\\geq 2$) :\r\na) find the fundamental group of $ X$\r\nb) Find the Euler characteristics of $ X$\r\nc) find the homology groups of $ X$\r\n\r\n\r\n\r\nI am still a beginner  of learning AT, I am learning it by myself so  it would be great for me if you could give me as details as possible.  Thanks  so much . \r\n\r\nBy the way, I have a question : is there any problem - solution book in Algebraic topology ? if there is, would you please give me some names?",
  "Solution_1": "1. Cover $ X$ with open sets $ A$ and $ B$, where $ A$ is the disc centered at the origin of radius $ 3/4$ and $ B$ is the complement of the closed disc centered at the origin with radius $ 1/2$. Then $ A$ is contractible, so by the van kampen theorem $ \\pi_1(X) \\equal{} \\pi_1(B) / \\pi_1(A \\cap B)$, where the quotient is induced by the inclusion $ \\pi_1(A \\cap B) \\to \\pi_1(B)$. \r\n\r\n$ B$ deformation retracts by a map $ r$ onto the boundary of the disc, consisting of four copies of a loop $ \\alpha$ which traverses precisely $ 1/4$ of the boundary of the disc. $ A \\cap B$ is a thin ribbon, which itself deformation retracts to a circle, which is carried by $ r$ to the concatenation $ \\alpha^4$. Therefore $ \\pi_1(X) \\equal{} \\langle \\alpha \\rangle / \\langle \\alpha^4 \\rangle \\equal{} \\mathbb{Z}/4$. (This is to be expected, since $ \\pi_1(\\mathbb{R}P^2) \\equal{} \\mathbb{Z}/2$ by a similar argument.)",
  "Solution_2": "thanks  for your solution for #1 Nukular , it is great",
  "Solution_3": "let's try and break down number 2 also...\r\n\r\nso..\r\nfor point a) we can notice that our space deformation retracts on the wedge of a 2-sphere and a bouquet $ k\\minus{}1$ circles, so fundemental group, by van kampen, is free over $ k\\minus{}1$ generators.\r\nfor point b) we can triangulate the sphere in such a way that the points we identify are vertices, so euler characteristic just decreases by $ k\\minus{}1$, and then it's $ 3\\minus{}k$.\r\nfor point c), we have three ways: either we use point a) and b) and notice that the space is connected, so we can deduce rational homologies (and they have rank $ 1,k\\minus{}1,1$ respectively over $ \\mathbb{Q}$), or we use mayer-vietoris sequence with the decomposition we already used in point a), or we use another one, which is seeing identification as \"summing\" with a segment having their endpoints on distinct points on the sphere and then using mayer-vietoris again.\r\n\r\nsince you're a beginner, i'll leave you the details :)",
  "Solution_4": "b) and c) should not require triangulation; that's too much work;\r\n\r\nthat deformation retract is all you need. It tells you directly that $ H_1(X) = H_1(S^2) \\times \\prod_{i = 1}^{k - 1} H_1(S^1) = \\mathbb{Z}^{k - 1}$. Furthermore $ H_2(X) = \\mathbb{Z}$, and $ H_0(X) = \\mathbb{Z}$. Therefore the Euler characteristic is ${ \\mathrm{Rank}(H_2(X)) - \\mathrm{Rank}(H_1(X)} + \\mathrm{Rank}(H_0(X)) = 3 - k$.\r\n\r\nma_go, is there an easy way to see that deformation retract? Seems plausible, but it's a bit weird lookin.",
  "Solution_5": "thanks Nukular and ma_go :)",
  "Solution_6": "Hi all. One way to see the homotopy equivalence to $ {\\mathbb S}^2\\vee({\\mathbb S}^1)^{\\vee (k\\minus{}1)}$ formally is as follows: for $ 1\\leq i<k$, attach a line segment $ s_i$ connecting $ p_i$ and $ p_{i\\minus{}1}$ to $ {\\mathbb S}^2$. Call the resulting space $ X$. The union $ S$ of the $ s_i$ is contractible, and since we are working with \"nice\" spaces, the map $ X\\to X/S$ contracting this segment to a point is a homotopy equivalence. Thus we can concentrate on the computation of the invariants of $ X$: Next (for $ 1\\leq i<k$) consider arcs $ c_i$ on $ {\\mathbb S}^2$ connecting $ p_i$ and $ p_{i\\plus{}1}$. Again, the union $ C$ of these arcs is contractible, and thus $ X\\to X/C$ is a homotopy equivalence. But $ X/C\\cong{\\mathbb S}^2\\vee({\\mathbb S}^1)^{\\vee (k\\minus{}1)}$.",
  "Solution_7": "[quote=\"Nukular\"]b) and c) should not require triangulation; that's too much work;[/quote]\nright, but for b) it has the advantage of not requiring to compute any homology group, so i thought it was nice :)\n[quote=\"Nukular\"]ma_go, is there an easy way to see that deformation retract? Seems plausible, but it's a bit weird lookin.[/quote]\r\nquite pointless after hanno's post, but you can also picture it in another way. join the \"center\" of the sphere with the $ k$ points on the surface with $ k$ segments and \"pull the strings\" towards the center: this tells you that this sort of spider web is homotopy equivalent to our quotient space. then you can just move your $ k$ points to a single point without changing homotopy type, and you see your wedge there.\r\n(the center was actually useless, but it was nice seeing it that way)."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "How many positive integers x satisfy: \r\n\r\n$x+6\\ge2x$",
  "Solution_1": "[hide]6 must be greater than or equal to x, so, the only non-negative integers which work are 1,2,3,4,5, and 6.[/hide]",
  "Solution_2": "is 0 positive?  :roll:",
  "Solution_3": "i edited the last post. you are very selective in your answers.(just kidding) :D",
  "Solution_4": "[quote=\"236factorial\"]How many positive integers x satisfy: \n\n$x+6\\ge2x$[/quote]\r\n[hide]\nI did this problem by chosing a large number and narrowing the numbers down. 8 is too big, so is 7.\n6 works, so there are 6 possible integers, since 0 isn't positive.[/hide]",
  "Solution_5": "[quote=\"236factorial\"]How many positive integers x satisfy: \n\n$x+6\\ge2x$[/quote]\r\n[hide]\n6+6=2x6\nso 6 is the highest\n\nso 6 numbers satisfy this condition\n\n[/hide]"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Hi, I have a problem with the bibliography of my document. I'm writing the document in a language different  from english and I'm using BibText to create de bibliography. My problem is that when I have a reference with many authors, in the pdf it joins the  last 2 authors with \"and\" instead of translating \"and\" to the language of the document. Is this normal???  :huh:  or do I have to do something to specify the language of the bibliography??\r\nthanks in advance!",
  "Solution_1": "Are you using the Babel package? Try putting \\usepackage[nameoflanguage]{babel} at the top of your document and see if that works.",
  "Solution_2": "yes, I'm using the Babel package and the \\usepackage[nameoflanguage]{babel}, but it's not working....\r\n\r\nthanks!!"
}
{
  "Tag": [],
  "Problem": "Bacteria have the following biological rhythm for spreading: each one lives an hour and in each half of hour of its life produces a new one. Starting from one, how many bacteria are after 6 hours?\r\n\r\n[hide]I know the answer:$ 377$ but I can't get it :( [/hide]",
  "Solution_1": "Look at it this way.\r\nIn 1 hour, 1 bacteria will make 1 new bacteria, but won't die.",
  "Solution_2": "[quote=\"sunnykim\"]Bacteria have the following biological rhythm for spreading: each one lives an hour and in each half of hour of its life produces a new one. Starting from one, how many bacteria are after 6 hours?\n\n[hide]I know the answer:$ 377$ but I can't get it :( [/hide][/quote]\r\n\r\nlet's do every 30 minutes\r\n\r\n0 -> 1\r\n30 -> 1\r\n60 -> 2\r\n90 -> 3\r\n120 -> 5\r\n150 -> 8\r\n\r\nNotice that if we let $ B_{n}$ be the number of bacteria after $ n$ minutes, then $ B_{30k}$ seems to be equal to $ F_{k\\plus{}1}$.  In fact, there is a reason.",
  "Solution_3": "At 30 minutes, there are 2 bacteria.",
  "Solution_4": "then the answer should be $ 610$, unless I can't count to 14 correctly",
  "Solution_5": "Hint: Try thinking about the bacteria in having two phases, phase 1 is the first 30 min, then phase 2 is the second 30 min, \r\nIn phase 1, they make a new bacteria\r\nIn phase 2, nothing happens.\r\nRepeat.\r\nEach phase takes 30min\r\nAnd then you have two types of bacteria, type 1 has phase 1 on the hour, and phase 2 on the half hours.\r\nType 2 does the opposite.",
  "Solution_6": "um, did you read at all or did you copy?",
  "Solution_7": "Read what?",
  "Solution_8": "[quote=\"alanchou\"]then the answer should be $ 610$, unless I can't count to 14 correctly[/quote]\r\n\r\nthe 14th fibonacci number is 377.\r\nYour reasoning is correct.\r\n\r\nIn fact this problem is the same problem that led fibonacci to the discovery of the pattern.\r\n\r\nThe only difference in his problem was he had bunnies.",
  "Solution_9": "Yeah the question was like the bunnies reproduce every month, and die in two months."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "$ 2{}^1{}^3{}^1{}^0 \\equiv n \\mod {999}$ and $ 0\\leq n < 999$\r\nFind $ n$",
  "Solution_1": "[quote=\"NapoleonXIV\"]$ 2{}^1{}^3{}^1{}^0 \\equiv n \\mod {999}$ and $ 0\\leq n < 999$\nFind $ n$[/quote]\r\n\r\n[hide]\n$ \\phi (999) = 999*2/3*36/37=648.$\nSo, $ 2^{648}\\equiv 1 \\mod {999}$\n$ 2^{1310} \\equiv 2^{14} = 16384 \\equiv 400 \\mod {999}$\n[/hide]"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Do there exist a convex polygon which can be partitioned into non-convex quadrilaterals?",
  "Solution_1": "I would say no, if I did not miss something in the following :\r\n\r\nAssume that $P$ is a convex $n \\geq 3$-gon for which such a dissection does exists. Let $m$ be the number of non-convex quadrilaterals used. The vertices of these quadrilaterals are :\r\n- vertices of $P$\r\n- or other points on the boundary of $P$, say that there are $r$ such vertices,\r\n- or interior points in $P$ which are between two vertices of some of the non-convex quadrilaterals (they are on the boundary of a quadrilateral but not a vertex of it), say that there are $q$ such points,\r\n- or interior points in $P$ which are not between two vertices of some of the non-convex quadrilaterals, say that there are $p$ such points.\r\n\r\nWe will compute the sum $S$ of the interior angles of all the quadrilaterals in two ways.\r\n\r\nBecause of the $r$points, $P$ has to be considered as a $(n+r)$ convex polygon, so that the sume of its interior angles is $(n+r-2) \\pi.$ Moreover, each of the $p+q$ interior points contributes for $2 \\pi$. Thus, $S = (n+r-2) \\pi + 2(p+q) \\pi.$ (1)\r\n\r\nIn another hand, each of the $q$ points defined above contributes for $\\pi$ in the sum of the interior angles in exactly one of the quadrilaterals (which in that case has to be considered as a degenerated $t$-gon, where $ t >4$). It means that if $Q$ is one of the quadrilaterals with $\\alpha$ of the $q$ points on its boundary (but not as one of its verteices), the total sum of the interior angles of $Q$ is $(4+ \\alpha - 2) \\pi$. Thus, summing over $Q$, it leads to $S = q \\pi + 2m \\pi.$ (2)\r\n\r\nFrom (1) and (2), we deduce that :\r\n$p = \\frac {2m-q+2-n-r} 2 \\leq \\frac {2m- n +2} 2 < m.$ (3)\r\n\r\nBut, since each of the quadrilaterals is non-convex it has exactly one interior angle greater than $\\pi$. This angle is necessarly at some interior point of $P$ (since $P$ is convex) and more precisely at some of the $p$ points defined above (since each of the $q$ others used an angle of $\\pi$). Obviously, any of the $p$ points can be used in that way by only one of the quadrilaterals. Therefore, we must have $m \\leq p$, which contradicts (3), and we are done.\r\n\r\nPierre."
}
{
  "Tag": [
    "Gauss",
    "Columbia",
    "MATHCOUNTS",
    "percent"
  ],
  "Problem": "Greetings,\r\n\r\nWe have been gradually building up our [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]contest directory[/url] and would like more information about Canadian Math Competitions.  I know there are tests called Pascal, Fermat, etc., but know little about them (grade level, what they are used for, are they national?).  I would appreciate any information about those and any other Canadian competitions we could add to our directory.\r\n\r\nThank you in advance!",
  "Solution_1": "Mathew,\r\n\r\nHave you seen this?\r\n\r\nhttp://cemc.uwaterloo.ca/english/contests/index.shtml\r\n\r\nThey are all national.  The PCF contests (pascal cayley fermat) are all pretty basic.  Good students don't have any problem scoring 140+ on then (e.g, students who can consistently score over 120 on the AMC).  The Euclid and COMC are both written contests.  They are the most important ones for better students (and obviously the CMO).  The COMC is a feeder contest for the CMO (top 50 or so make it)  It is mostly easy, although the last few problems are hard.",
  "Solution_2": "I have seen that site in passing.  Thanks for the nice summary.",
  "Solution_3": "Hello, info about Contests in Canada:\r\n\r\nGrades 4-6\r\nhttp://www.mathematica.ca/eng/home.htm\r\n\r\nGrades 7-8\r\nGauss Contest - http://www.cemc.uwaterloo.ca/english/contests/gauss.shtml\r\n\r\nGrades 9\r\nPascal Contest - http://www.cemc.uwaterloo.ca/english/contests/pascal.shtml\r\nFryer Contest - http://www.cemc.uwaterloo.ca/english/contests/fryer.shtml\r\n\r\nGrades 10\r\nCayley Contest - http://www.cemc.uwaterloo.ca/english/contests/cayley.shtml\r\nGalois Contest - http://www.cemc.uwaterloo.ca/english/contests/galois.shtml\r\n\r\nGrades 11\r\nFermat Contest - http://www.cemc.uwaterloo.ca/english/contests/fermat.shtml\r\nHypatia Contest - http://www.cemc.uwaterloo.ca/english/contests/hypatia.shtml \r\n\r\nGrades 12\r\nEuclid Contest - http://www.cemc.uwaterloo.ca/english/contests/euclid.shtml\r\n\r\nAdditional Contests \u2013\r\nLakehead University: http://math.lakeheadu.ca/math_competition.htm \r\nNova Scotia: http://www.mathstat.dal.ca/~mathleague/problems.html\r\nBC Math Contest- http://www.sci.ouc.bc.ca/instructors/clee/bcchsmc/BCCHSMC.htm\r\nAB Math Contest- http://www.math.ualberta.ca/~ahsmc/\r\nCalgary Math Contest- http://www.math.ucalgary.ca/education/JrMath/\r\nMaritime Math Contest  - http://ace.acadiau.ca/math/MMC/mmc.htm\r\nProblem of the Month (WLU) - http://www.wlu.ca/page.php?grp_id=43&s_id=71\r\nNewfoundland Math Contest - http://www.math.mun.ca/index.php3?content=includes/outreach/blundon.html\r\nMathleague- For grades 4 \u2013 12 http://www.mathleague.com/\r\n\r\nOlympiads-\r\nCOMC/CMO/APMO/MOCP - http://camel.math.ca/CMS/Olympiads/",
  "Solution_4": "Hmm, if you're including all the little local contests, then there might be more.\r\n\r\nLike the Edmonton Junior High Math Competition\r\n\r\n[Edit]  Heya lightrhee, do you have the site for the edmonton contest?  I sure don't",
  "Solution_5": "Thanks for all of the great information!  We have added them to [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]our list[/url].\r\n\r\nIf you have more to suggest (with weblinks), please post them here.",
  "Solution_6": "[quote=\"Sunny\"]\nHeya lightrhee, do you have the site for the edmonton contest?  I sure don't[/quote]\r\nNo idea. I'm not even sure whether there is one.",
  "Solution_7": "In British Columbia, about two years ago, the MATHCOUNTs group terminated their association with MATHCOUNTS and renamed themselves MATHChallengers. I believe the licencing and fees charged by MATHCOUNTS became too great, so the organizers decided to start their own competition pretty much from scratch. There was also talk about the Saskatchewan group joining MATHChallengers, but I don't know if that plan ever went anywhere.\r\nFYI: Here is the link to the MATHChallengers site\r\n\r\nhttp://www.apeg.bc.ca/mathchallengers/index.htm\r\n\r\nRegards,",
  "Solution_8": "Hey, is there anyone here in British Columbia? I heard that if you have high mark for Euclid, you will be accepted to the math honour class of UBC right? Do you know exactly how many percent do they need?"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "If $a,b,c$ are the sidelengths of a triangle, then \\[ \\frac{b}{c+1} + \\frac{c}{b+1} > \\frac{2a}{a+2} .  \\]\r\n\r\n[i]Victor Tudor[/i]",
  "Solution_1": "Hello,\r\n[hide=\"Here is a proof\"]$\\frac{b}{c+1} + \\frac{c}{b+1} > \\frac{2a}{a+2}$ \n$ab^2+ab+2b^2+2b+ac^2+ac+2c^2+2c>2abc+2ab+2ac+2a$\n$2(b+c-a)+a(b-c)^2+2(b^2+c^2)-a(b+c)>0$\n$2(b+c-a)+a(b-c)^2+2(b+c)^2-4bc-a(b+c)>0$\n$2(b+c-a)+a(b-c)^2+(b+c)(b+c-a)+(b-c)^2>0$\nWhich is true.[/hide]"
}
{
  "Tag": [
    "analytic geometry"
  ],
  "Problem": "[b]A circle is drawn with center at the origin and radius $ 3$. Find the coordinates of all intersections of the circle with an origin-centered square of side length 4 whose sides are parallel to the coordinate axes[/b]",
  "Solution_1": "This sounds pretty simple. We have circle\r\n\r\n$ x^{2}+y^{2}= 3^{2}$ and square\r\n\r\n$ y=-2$ $ y = 2$ $ x=-2$ $ x=2$ (the intersection of those 4 lines forms a square). Now just plug in those values into the circle equation to find our vertices.\r\n\r\nFor instance,\r\n\r\n$ (-2)^{2}+y^{2}= 9$\r\n$ y^{2}= 5$\r\n$ y = \\pm \\sqrt{5}$\r\n\r\nUpon similar calculations, our intersection points are:\r\n\r\n$ (-2, \\;-\\sqrt{5})$\r\n$ (-2, \\; \\sqrt{5})$\r\n$ (2, \\;-\\sqrt{5})$\r\n$ (2, \\; \\sqrt{5})$\r\n\r\nOr, simply written, $ (\\pm 2, \\pm \\sqrt{5})$ and $ (\\mp 2, \\pm \\sqrt{5})$\r\n\r\nHopefully I didn't overlook anything  :blush:",
  "Solution_2": "[quote=\"Prime factorization\"]This sounds pretty simple. We have circle\n\n$ x^{2}+y^{2}= 3^{2}$ and square\n\n$ y=-2$ $ y = 2$ $ x=-2$ $ x=2$ (the intersection of those 4 lines forms a square). Now just plug in those values into the circle equation to find our vertices.\n\nFor instance,\n\n$ (-2)^{2}+y^{2}= 9$\n$ y^{2}= 5$\n$ y = \\pm \\sqrt{5}$\n\nUpon similar calculations, our intersection points are:\n\n$ (-2, \\;-\\sqrt{5})$\n$ (-2, \\; \\sqrt{5})$\n$ (2, \\;-\\sqrt{5})$\n$ (2, \\; \\sqrt{5})$\n\nOr, simply written, $ (\\pm 2, \\pm \\sqrt{5})$ and $ (\\mp 2, \\pm \\sqrt{5})$\n\nHopefully I didn't overlook anything  :blush:[/quote]\r\n\r\nthat's a very reasonable answer and i believe its correct..its just i have some doubts about something...since the square is centered at the origin and has side length 4, does that mean that the coordinates of its vertices are (2,2), (2,-2), (-2,-2), (-2,2)..wouldn't that mean that the square intersects the circle at its vertices..im probably wrong.",
  "Solution_3": "[quote=\"mihail911\"\r\n\r\nthat's a very reasonable answer and i believe its correct..its just i have some doubts about something...since the square is centered at the origin and has side length 4, does that mean that the coordinates of its vertices are (2,2), (2,-2), (-2,-2), (-2,2)..wouldn't that mean that the square intersects the circle at its vertices..im probably wrong.[/quote]\r\n\r\nAre any of those points on the circle?  :wink:",
  "Solution_4": "yeah, I don't think they intersect, since the length of the diagonals of the square= $ 4\\sqrt{2}$, which is $ <$ the diameter of the circle = $ 9$",
  "Solution_5": "Yes, but they intersect in 8 places, not 4 as mihail thought.\r\n\r\nAlso, the diameter of the circle is not 9. :P",
  "Solution_6": "oops, I somehow got confused with the numbers somewhere in there.  :blush: \r\nOk, I agree with Prime Factorization now."
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Prove: If $ f(x)$ is differentiable at $ x\\equal{}a$ then $ f(x)$ is continuous at $ x\\equal{}a$.",
  "Solution_1": "See [url=http://www-math.mit.edu/~djk/18_01/chapter02/proof04.html]here[/url].  Note that the converse is false."
}
{
  "Tag": [
    "symmetry",
    "linear algebra",
    "matrix"
  ],
  "Problem": "Let $ A$ be a $ n\\times n$ matrix with $ a_{ij}\\equal{}i\\in\\mathbb{N}$  $ \\forall i\\equal{}j$ and $ a_{ij}\\equal{}k\\in\\mathbb{R}$  $ \\forall i\\neq j$.Find $ \\det A$.",
  "Solution_1": "Let $ X$ be the $ n\\times n$ matrix with every entry equal to $ k$.  Let $ Y$ be the diagonal matrix with diagonal entries $ 1\\minus{}k$, $ 2\\minus{}k$, ..., $ n\\minus{}k$.  Then the given matrix $ A$ is equal to $ X\\plus{}Y$.  To compute the determinant of a sum, we will use the formula developed in the post\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=591700#591700[/url].\r\n\r\nWe need to handle only the cases when 0 or 1 rows are taken from $ X$.  As long as $ k$ is not equal to 1, 2, ..., and $ n$, we get the answer\r\n\\[ \\prod_{i \\equal{} 1}^{n}(i\\minus{}k)\\left[ 1\\plus{}k\\sum_{i \\equal{} 1}^{n}\\frac{1}{i\\minus{}k}\\right ].\\]\r\nIf $ k$ happens to equal 1, 2, ..., or $ n$, we get the answer $ (\\minus{}1)^{k\\minus{}1}k! (n\\minus{}k)!$."
}
{
  "Tag": [
    "abstract algebra",
    "function",
    "geometry",
    "geometric transformation",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "we say that R is a euclidean ring,if R be abelian and unity.and exist a function f: R_{0}   $ %Error. \"display\" is a bad command.\n\\to$(positive integer numbers) so that:1)if a and b be nonzero in the R and a$ %Error. \"display\" is a bad command.\n\\mid$    b then f(a)   $ %Error. \"display\" is a bad command.\n\\le$f(b)   2)for every a and b   $ %Error. \"display\" is a bad command.\n\\neq$    0,exist q and r so that a=bq+r that r=0 or f(r)<f(b).now suppose R={a+b$ %Error. \"display\" is a bad command.\n\\sqrt {3} i$ $ %Error. \"display\" is a bad command.\n\\mid$  a&b$ %Error. \"display\" is a bad command.\n\\in$ Z or 2a&2b be odd integer numbers},show it is a euclidean ring.",
  "Solution_1": "[quote=\"m_jackson\"]we say that R is a oghlidosy ring,if R be abelian and unity.and exist a function f: R_{0}   $ %Error. \"display\" is a bad command.\n\\to$(positive integer numbers) so that:1)if a and b be nonzero in the R and a$ %Error. \"display\" is a bad command.\n\\mid$    b then f(a)   $ %Error. \"display\" is a bad command.\n\\le$f(b)   2)for every a and b   $ %Error. \"display\" is a bad command.\n\\neq$    0,exist q and r so that a=bq+r that r=0 or f(r)<f(b).now suppose R={a+b$ %Error. \"display\" is a bad command.\n\\sqrt {3} i$ $ %Error. \"display\" is a bad command.\n\\mid$  a&b$ %Error. \"display\" is a bad command.\n\\in$ Z or 2a&2b be odd integer numbers},show it is a oghlidosy ring.[/quote]\r\n\r\nHello.\r\nYou can take the function $ f: s + ti \\mapsto N(s + ti) = s^{2} + 3 t^{2}$.\r\nIf a is in $ R$ and $ b$ in $ R_{0}$, put $ a/b = x + y \\sqrt {3} i$, with $ x$ and $ y$ rational numbers. Let $ \\beta$ denote the closest integer to $ 2y$ and $ \\alpha$ the closest integer to $ 2x$ which has the same parity as $ \\beta$. Put $ q = (\\alpha + \\beta \\sqrt {3} i)/2$ and $ r = a - bq$.\r\nWell, I must confess that I copy this in the French book \"Th\u00e9orie des Nombres\" of Daniel Duverney, 2007, p. 55, and that I didn't check it carefully.\r\nIf there remain problems, we can have a closer look.\r\n\r\nFriendly,\r\nP.",
  "Solution_2": "\"oghlidosy ring\" sounds like a failed translation of \"euclidean ring\" ;)."
}
{
  "Tag": [
    "ARML"
  ],
  "Problem": "Could someone please explain angle chasing to me and give me a basic example?  Thanks.",
  "Solution_1": "angle chasing is basically when you have a diagram and you're given certain angles and you use angle properties in order to find another angle.",
  "Solution_2": "There is a chapter in AoPS volume 1 called angle chasing, which lists all of the basic angle chasing techniques and gives lots of nice problems :wink:",
  "Solution_3": "angle chasing is basically finding the measures of angles with various \"tools\" that nclude:\r\n\r\nsum of angles in a triangle\r\nsum of angles in a quadrilateral\r\nangles intercepting arcs\r\nangles with together form a line\r\nangles in a right traingle\r\ninscribed angles subtending the same arc angle bisectors\r\nthe various kinds of lines, angles, and triangles\r\n\r\n\r\nThisis what I understand about angle chasing.",
  "Solution_4": "As I understand it, angle chasing can be used in any problem when you want to find the measure of an angle.  Many times you need to find the measures of other angles first before you find the measure of the angle you are looking for.\r\n\r\n2006 ARML Individual #3 is a good example of angle chasing.  You can find more ARML problems and solutions at [url=http://www.arml.com]the ARML website[/url].",
  "Solution_5": "Thanks, guys.  BTW, I went to ARML this past June."
}
{
  "Tag": [
    "function",
    "algebra",
    "polynomial",
    "ratio"
  ],
  "Problem": "1- a,b,c ar positive real numbers and a+b+c=1 prove:\r\n5(a^2+b^2+c^2)<=6(a^3+b^3+c^3)+1\r\n\r\n2- for a real x we say <x> i the distance of x from the closest number to it\r\nprove for any irrational a and e>0 prove there exists natural n that\r\n<n^2a> is smaller than e.\r\n\r\n3-find all f from natural numbers to natural numbers that\r\n2n+2001<=f(f(n))+f(n)<=2\r\n\r\n4-does exist  a {bi} of positive numbers that for any positive m\r\nb(m)+b(2m)+b(3m)+...+1/m\r\n\r\n5-find all f from Q to Q that \r\n2f(f(x))=f(2x)           f(x)+f(1/x)=1\r\n\r\n6-find all functions on R-{0} that\r\nxf(x+1/y)+yf(y)+y/x=yf(y+1/x0+xf(x)+x/y\r\n\r\n7- We call a polynomial good if and only if p(x)^2+c=p(x^2+c) for a constant complex number c\r\na)if c is ratio and not integer find all such polynomials\r\nb)if c= a+bi (i^2=-1) and a^2+B^2>4  :)",
  "Solution_1": "hi! \r\n\r\nnice of you to supply these questions, but please put each problem in a separate topic into its proper subfolder :) also specify the source (country, olympiad type, year -> there is a topic that explains how to do that :) )."
}
{
  "Tag": [
    "search",
    "function",
    "modular arithmetic",
    "quadratics"
  ],
  "Problem": "Source: Polish MO 2006\r\n1. Determine all nonnegative integers $n$ for which $2^n+105$ is a perfect square.\r\n\r\n[/hide]",
  "Solution_1": "It's been posted before.  If the search function were working properly...  :|",
  "Solution_2": "Anyway, it's a simple problem, cause we see that $n$ must be even (looking $\\mod 3$), so\r\n\r\n$2^{2k}+105=m^2 \\rightarrow (m-2^k)(m+2^k)=105$",
  "Solution_3": "but u haven't fiinshed yet...does for every even n exist integer m?",
  "Solution_4": "We can solve it considering that $105=3 \\cdot 5 \\cdot 7$. Since $2^k-m$ and $2^k+m$ are integers and factors of 105, the easiest way is to try all possible \"combinations\" of factors, for example\r\n\r\n$2^k-m=35$\r\n$2^k+m=3$\r\n\r\nor\r\n\r\n$2^k-m=105$\r\n$2^k+m=1$\r\n\r\nand check which numbers $n$ work for the problem.\r\nI didn't write it because it's a \"standard way\" to solve this kind of problems, and it's not really interesting",
  "Solution_5": "Assume there is an integer m such that \r\n\r\n$(1) \\; 2^n + 105 = m^2.$\r\n\r\nIf $n$ is an odd integer, then \r\n\r\n$m^2 = 2^n + 105 \\equiv 2 \\pmod{3}$\r\n\r\nwhich is impossible because 2 is quadratic nonresidue of 3. Hence $n$ is even, i.e. $n = 2p$ where $p$ is a non-negative integer. So \r\n\r\n$(m - 2^p)(m + 2^p) = 105$\r\n\r\nby (1), thus \r\n\r\n$(2) \\; m - 2^p = d$ \r\n\r\n$(3) \\; m + 2^p = 105/d,$ \r\n\r\nwhere $d$ is an positive divisor of 105 and $d < 105/d$. In other words, $d \\leq 10$ and $d\\,|\\,105 = 3 \\cdot 5 \\cdot 7$. The only possibilities are $d = 1, \\, 3, \\, 5, \\, 7$. Subtracting (2) from (3) we get \r\n\r\n$(4) \\; 2^{p+1} = 105/d - d$.\r\n\r\nInserting $d = 1, \\, 3, \\, 5, \\, 7$ in (4), the results are $2^{p+1} = 104, \\, 32 = 2^5, \\, 16 = 2^4, \\, 8 = 2^3$. Consequently $p = 2, \\, 3, \\, 4$, which implies that (1) has 3 solutions, namely\r\n\r\n(m,n) = (11,4), (13,6), (19,8)."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "analytic geometry",
    "modular arithmetic"
  ],
  "Problem": "I don't where to post this problem , so I am posting it right here , so please help me in solving this problem...\r\n\r\nThere is a tenstory building having 10\u00d710 number of rooms on each floor (consider the building\r\nlike a 10\u00d710\u00d710 lattice with each 1\u00d71\u00d71 cube as a room).The electrician has made such an\r\narrangement for lighting that if you switch on alight in a room all the rooms above and below,all\r\nthe rooms to its right and left,and all the rooms in its front and back are also lighted. So as to say\r\nif there is a building with 3 \u00d73\u00d7 3 number of rooms and if welight the center room all but the\r\ndiagonally neighboring rooms are lighted.To optimize on the cost of electricity bill find out the\r\nleast number of rooms required to light the whole building.",
  "Solution_1": "[hide=\"Hint\"] Note that the number of other rooms that a single light can light is independent of the location of the light.  That gets you a lower bound. [/hide]",
  "Solution_2": "I do not have time, but the answer is\r\n[hide=\"Answer\"] 50 [/hide]",
  "Solution_3": "[b]Edit:[/b] Thanks diophantient, yeah, I don't know what I was on",
  "Solution_4": "[quote=\"The Zuton Force\"]Mmm?\n\nYou have 10 in the column, plus 8 others around it on the same floor, so that makes 18 rooms lit per room powered, and 50 = 1000/20 < 1000/18...\n\nAlso, in response to the hint by t0rajir0u, the hint you gave isn't completely true:\n[hide=\"Clarifier\"] It's true as long as the room isn't on a side-face of the building, or an edge (vertical); in these cases it gets cut down a bit[/hide][/quote]\r\nDoesn't each room light up 27 (9 in the row, 9 in the column, 9 in the pillar)?",
  "Solution_5": "[hide]50 lights is possible.  Consider splitting up the building into eight $ 5\\times 5\\times 5$ cubes.  Then we choose two small cubes opposite each other and light up 25 rooms in each cube such that no two of the 25 rooms lie on a line parallel to one of the edges of the cube.  If we can do this, then each of the 25 vertical lines throught the small cube we chose contain exactly one light, so all the rooms in the vertically adjacent small cube are lit.  Similarly all the rooms in the rest of the eight small cubes will be lit since each small cube is adjacent to one of the small cubes we chose initially (or is one of the ones we chose initially).\n\nTo prove this, give the individual cubes inside a $ 5\\times 5\\times 5$ cube coordinates from $ (0,0,0)$ to $ (4,4,4)$.  We want to pick 25 points such that no two points share more than one coordinate.  Consider the set of points $ \\{(x,y,z)|x\\plus{}y\\plus{}z\\equiv 0\\pmod{5}\\}$.  This set contains one point for each choice of $ x$ and $ y$, so it contains 25 points.  Since we can't change one variable without changing another, it also satisfies the condition that no two points share more than one coordinate.  Therefore we can light the whole building with 50 lights.\n\nI think you can prove that 49 lights is impossible by considering a floor with only 4 lights, but it's a little messy.[/hide]"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "So apparently Mathcounts is coming up (along with the somewhat less important to me this year AMCs).\r\n\r\nAre there still any middle schoolers at [i]all[/i] who come on this forum? Bleh I apparently have chapter on Febuary 7th, three days before the AMC As...meh.\r\n\r\nI was also going to create an AMC topic, but there's still a bit of time.",
  "Solution_1": "lol, I have chapter Mathcounts on feb. 7 as well. :)",
  "Solution_2": "Argh dude\r\nthis means that the AMC is Feb 10?\r\n\r\nOkay I'm going to an Academic challenge tourney this Saturday dang\r\nI'll take next Monday and prep for those dudes..\r\n\r\n\r\nhmm\r\nI should not make 3-2=2 errors",
  "Solution_3": "MY school does the 10B I think, so it's the 25th for me. According to my teacher...",
  "Solution_4": "Hm my school apparently signed up for the 10A on the deadline, so I'm taking it there. Yay.\r\n\r\nPossibly someone should create an AMC thread.",
  "Solution_5": "hm..Lol.\r\nAMC thread\r\nEDIT: sorry. revived a dead thread..."
}
{
  "Tag": [],
  "Problem": "In $\\triangle ABC$, an angle bisector $AD$ is drawn. Write $BD$ in term of side lengths $a,b,c$.",
  "Solution_1": "We have :\r\n$\\frac{c}{b}=\\frac{x}{y}$\r\n$\\frac{c}{b}=\\frac{x}{a-x}$\r\n$(a-x)c=bx$\r\n$x(b+c)=ac$\r\n$x=\\frac{ac}{b+c}$\r\n\r\nClick on the picture, then it'll be more clear."
}
{
  "Tag": [
    "special factorizations"
  ],
  "Problem": "I forgot how to do completing the square.  Can someone show me how to do it?",
  "Solution_1": "Completing the square is when you have an equation like $x^2+bx$ and you add $(\\frac{1}{2}b)^2$ to the equation.\r\n\r\nFor example, in the equation $x^2+2x$, you add $(\\frac{1}{2}(2))^2=1$ to get $x^2+2x+1$. Completing the square makes it a perfect square. MAke sure the coefficient of $x^2$ is 1!",
  "Solution_2": "thanks, i got my math packet yesterday.  It has 72 problems and its due next wednesday.\r\nit has about 10 completing the square problems.  Thanks 236!"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "A sequence $ \\{a_n\\}$ is said to be [i]of bounded variation[/i] if the series\r\n\\[ \\sum |a_n\\minus{}a_{n\\plus{}1}|\\]\r\nis convergent.\r\nProve that if $ \\{ b_n\\}$ is of bounded variation, and the partial sums of the series $ \\sum a_n$ are bounded, then the series $ \\sum a_n\\cdot b_n$ is convergent.",
  "Solution_1": "Use summation by parts (also often called Abel's formula). Difference the $ b_n$ and sum the $ a_n$ in your summation by parts; the hypotheses look custom-fitted to make that work.\r\n\r\nExcept there's a potential issue involving the boundary terms. Note that the hypothesis that $ b_n$ is of bounded variation does not necessarily imply that $ b_n\\to 0.$ It does imply that $ b_n\\to b$ for some fixed $ b.$ The statement is actually false unless we supply an additional hypothesis.\r\n\r\nExample: Let $ a_n\\equal{}(\\minus{}1)^n,$ and $ b_n\\equal{}1$ for all $ n.$ Then $ \\{b_n\\}n$ is of bounded variation, and $ \\sum_{n\\equal{}1}^Na_n$ is bounded but $ \\sum (\\minus{}1)^n\\cdot 1$ is divergent.\r\n\r\nIt looks like we must add the hypothesis that $ b_n\\to 0$ in order to make this work."
}
{
  "Tag": [],
  "Problem": "How many ways are there to label 999 points arranged around in a circle, using labels A,B, and C, in such a way that between any two occurrences (clockwise and anti-clockwise) of one label the number of labels of the other two types is even (so between any two A's the sum of B's and C's is even).",
  "Solution_1": "[hide=\"Uh\"]$ 0$, right? Consider all points marked $ A$ or $ B$. Between each consecutive pair is an even number of $ C$s. Then the total number of $ C$s must be even. Similarly for $ A,B$. But there are an odd number of points total. Contradiction.[/hide]",
  "Solution_2": "The wording is a little tricky and you've misread: between each consecutive pair of A, the [i]total number[/i] of B and C must be even.  You also missed three trivial colorings (all A, all B and all C).",
  "Solution_3": "Welll when I first approached this problem I thought there was only three possible solutions like JBL said. However suppose we had something like\r\n AAAAAAAAA.........AAAAABCAAAAAA....AAAAA So there would be 997 A's which satifies the property. So I ask some teachers and they recommended me to try smaller example... however I still have not made much progress."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\\equal{}2.$ Prove that:\r\n\\[ a^2b\\plus{}b^2c\\plus{}c^2a\\geq a\\sqrt{b\\plus{}c}\\plus{}b\\sqrt{c\\plus{}a}\\plus{}c\\sqrt{a\\plus{}b}\\]",
  "Solution_1": "[quote=\"arqady\"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \\equal{} 2.$ Prove that:\n\\[ a^2b \\plus{} b^2c \\plus{} c^2a\\geq a\\sqrt {b \\plus{} c} \\plus{} b\\sqrt {c \\plus{} a} \\plus{} c\\sqrt {a \\plus{} b}\n\\]\n[/quote]\r\n\r\nCauchy Schwarz works\r\n\\[ a\\sqrt {b \\plus{} c} \\plus{} b\\sqrt {c \\plus{} a} \\plus{} c\\sqrt {a \\plus{} b} \\le \\sqrt {\\left( {a^2 b \\plus{} b^2 c \\plus{} c^2 a} \\right)\\left( {\\frac {{b \\plus{} c}}{b} \\plus{} \\frac {{c \\plus{} a}}{c} \\plus{} \\frac {{a \\plus{} b}}{a}} \\right)}\r\n\\]\r\n\r\n\\[ \\left( {a^2 b \\plus{} b^2 c \\plus{} c^2 a} \\right)\\left( {\\frac {{b \\plus{} c}}{b} \\plus{} \\frac {{c \\plus{} a}}{c} \\plus{} \\frac {{a \\plus{} b}}{a}} \\right) \\le \\left( {a^2 b \\plus{} b^2 c \\plus{} c^2 a} \\right)^2\r\n\\]\r\n\r\n\\[ \\frac {{b \\plus{} c}}{b} \\plus{} \\frac {{c \\plus{} a}}{c} \\plus{} \\frac {{a \\plus{} b}}{a} \\le a^2 b \\plus{} b^2 c \\plus{} c^2 a\r\n\\]\r\n\r\n\\[ 6 \\le a^2 b \\plus{} b^2 c \\plus{} c^2 a\r\n\\]\r\nNice inequality.\r\n\r\nProve http://www.mathlinks.ro/viewtopic.php?t=248093 with the same.",
  "Solution_2": "I think your solution is not correct... You haven't proved $ a^2b\\plus{}b^2c\\plus{}c^2a \\ge 6 \\ge \\frac{a\\plus{}b}{c}\\plus{}\\frac{b\\plus{}c}{a}\\plus{}\\frac{c\\plus{}a}{b}$ correctly, while this inequality is obviously wrong since $ RHS \\ge 6$.",
  "Solution_3": "[quote=\"Honey_S\"]\n\\[ \\frac {{b \\plus{} c}}{b} \\plus{} \\frac {{c \\plus{} a}}{c} \\plus{} \\frac {{a \\plus{} b}}{a} \\le a^2 b \\plus{} b^2 c \\plus{} c^2 a\n\\]\n[/quote]\r\nI think it is true because it was equivalent to \\[ 3 \\le \\frac{a}{c} \\plus{} \\frac{b}{a} \\plus{} \\frac{c}{b}\r\n\\]",
  "Solution_4": "I thinks that he just haven't proof that : sigma(b+c)/b<=sigma(a\u00b2b) \r\nwich is equivalent to : 3abc+sigma(a\u00b2b)<=2sigma(a\u00b2b) (multiplying by : abc)\r\n<=> sigma(a\u00b2b)>=3abc wich is true by AM-GM"
}
{
  "Tag": [
    "analytic geometry",
    "induction",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $n$ be a positive integer and $\\mathcal S$ be the set of points $(x, y)$ with $x, y \\in \\{1, 2, \\ldots , n\\}$. Let $\\mathcal T$ be the set of all squares with vertices in the set $\\mathcal S$. We denote by $a_k$ ($k \\geq 0$) the number of (unordered) pairs of points for which there are exactly $k$ squares in $\\mathcal T$ having these two points as vertices. Prove that $a_0 = a_2 + 2a_3$.\n\n[i]Yugoslavia[/i]",
  "Solution_1": "We shall first prove that $ |\\mathcal{T}| \\equal{} \\frac {n^2(n^2 \\minus{} 1)}{12}$\r\n\r\nLet's first count the number of squares that have their sides parallel to coordinate axis. It is not to difficult to see that if a square has a side of length $ i$ there are exactly $ (n \\minus{} i)^2$ sqaures of such size. So the number of such squares is equal to $ \\sum_{i \\equal{} 1}^{n \\minus{} 1}i^2$.\r\nFor any other square draw parallels to coordinate axis through the vertecies of the square-they intersect at latice points- therefore any other square that doesn't have sides parallel to the axis is inscriebed in some(exactly one) square that has sides parallel to coordinate axis. We can inscribe $ i \\minus{} 1$ squres into a sqaure with side length of $ i$. Combining that with previous result we get that the total number of squares is $ \\sum_{i \\equal{} 1}^{n \\minus{} 1}i^2 \\plus{} \\sum_{i \\equal{} 1}^{n \\minus{} 1}i^2(n \\minus{} i \\minus{} 1) \\equal{} \\sum_{i \\equal{} 1}^{n \\minus{} 1}i^2(n \\minus{} i)$. It can be proven by the means of induction that this equals $ \\frac {n^2(n^2 \\minus{} 1)}{12}$.\r\n\r\nWe continue our proof with observation that $ a_k \\equal{} 0$ for every $ k \\ge 4$. So $ a_0 \\plus{} a_1 \\plus{} a_2 \\plus{} a_3 \\equal{} \\frac {n^2(n^2 \\minus{} 1)}{2}$, since every pair of points gets counted in one of the first four terms.\r\n\r\nWe can also note that $ |\\mathcal{T}| \\equal{} \\frac {a_1 \\plus{} 2a_2 \\plus{} 3a_3}{6}$.(Each square correspondes to six pairs of points each gets counted the appropriate number of times in the right term).\r\n\r\nWe see that $ 6|\\mathcal{T}| \\equal{} a_0 \\plus{} a_1 \\plus{} a_2 \\plus{} a_3$ (combinig first two results). By putting in the new value for $ |\\mathcal{T}|$ we get. $ a_1 \\plus{} 2a_2 \\plus{} 3a_3 \\equal{} a_0 \\plus{} a_1 \\plus{} a_2 \\plus{} a_3$ form where the desiered result $ a_2 \\plus{} 2a_3 \\equal{} a_0$ follows.",
  "Solution_2": "In a \"20 years after\" fashion, check Problem 4 at this Balkan MO 2014 (see Contests section).",
  "Solution_3": "[hide]$a_{0}+a_{1}+a_{2}+a_{3}=a_{1}+2a_{2}+3a_{3}$ [/hide]"
}
{
  "Tag": [],
  "Problem": "I am sorry for asking a very basic question about the Math Main Exam of\r\nIIT-JEE. In main exam, there are too many questions for 120 minutes.\r\nI cannot believe that we should sove as many as 15 to 20 questions \r\nin only 120 minutes.\r\n\r\n  I wonder how many questions in the IIT-JEE Math Main exam we should \r\nsolve in this very limited 120 minites.",
  "Solution_1": "Did you come out through a time machine from the stone age?\r\n\r\nThere are no mains no more. :wink:",
  "Solution_2": "I am ashamed of asking a silly question. Actually, I am not Indian but Japanese\r\n(I come from not a stone age but from abroad)\r\nand I bought a book on IIT-JEE (from 1992-2005) during my stay in India, \r\nand knew about the IIT-JEE. (until 2005, I remeber that there WAS a Main Exam).\r\n\r\nComparing with the Japanese math entrance exam (for example,\r\nin Tokyo University, there are 6 questions within 150 minutes)\r\nand INMO (6 or 7 questions within 3 hours?),\r\nI was surprised at the vast amount of questions in IIT-JEE.\r\nAnd I wonder whether the candidates were expected to solve so many\r\nquestions when there was a main exam.\r\n(or was it practically selective in that it was enough to solve 7 or 8 questions out of 15 to 20).",
  "Solution_3": "as far as I remember the papers were always solvable within the time (or at least most of it, even without rigorous time training, something I'm dead against).",
  "Solution_4": "Thank you for your reply. I learned that the IIT-JEE is different\r\nfrom INMO or Japanese entrance exams of mathematics.\r\n(A question of Tokyo University of typical level is, for example,\r\nFor -1<x<1, x \\neq 0, prove (1-x)^{1-1/x} < (1+x)^{1/x}. \r\nHence or otherwise, prove 0.9999^{101}<0.99<0.9999^{100},\r\nwhich we are expected to solve within 150/6=25 minutes.)",
  "Solution_5": "That involves a lot of \"pen work\", besides \"brain work\". The JEE however, is more of \"brain work\" and [b]less[/b] of \"pen work\", in [b]most[/b] questions.\r\n\r\n(those words are not made bold for the fun of it, I really mean to stress on those)"
}
{
  "Tag": [
    "geometry",
    "analytic geometry",
    "graphing lines",
    "slope",
    "trigonometry",
    "rectangle"
  ],
  "Problem": "$\\triangle ABC$ is right-angled at $B$. Point $M$ is located on side $AB$ so that $AM=BC$, and point $N$ is on side $BC$ so that $CN = MB$. Find the angle between $AN$ and $CM$.",
  "Solution_1": "When two lines intersect there are two different (supplementary) angles you can measure, which one are you referring to?",
  "Solution_2": "Any of them  :)",
  "Solution_3": "Loads of analytical geometry says 45 degrees.",
  "Solution_4": "if you do the slopes (make one of them a, and the other b), and the tangent subtracting formula, you get 45",
  "Solution_5": "Well , I tried for a while for a geometrical proof but I failed. I send an analytical proof.\r\n\r\n\r\nWe put $NA = x$ , $BN = y$ , so $BM = x$   and $MC = x + y$ .Consider a Cartesian system at  $B$ with axis $BC , BA$. Then the slope of  $CN$  is  $- \\frac{y }{2x+y}$ and the slope of $AM$  is $- \\frac{x + y }{x}$ .\r\n\r\n If $\\phi$ is the desired angle ,  we find that $\\tan\\phi$ = 1 and the angle is 45 .\r\n\r\n [i]Now I see that this is a hint of Altheman , too![/i]\r\n\r\nBabis",
  "Solution_6": "We consider the rectangle $ABCD$ .Take a point $F$ on $CD$ with  $CF = AM$. Then $\\triangle FNA$ is an isosceles and right triangle. But  $CM // FA$and $\\angle FAN = 45$.\r\n\r\n see also the solution of M@re  in\r\n\r\n http://www.mathlinks.ro/Forum/viewtopic.php?t=57468\r\n\r\n and my observations in the topic.\r\n\r\n [i]Babis[/i]"
}
{
  "Tag": [],
  "Problem": "prove: \r\n\r\n$ 1/1^2 \\plus{} 1/2^2 \\plus{} ... \\plus{} 1/n^2 \\equal{} 1^2 \\plus{} 2^2 \\plus{} ... \\plus{} n^2$  $ mod(n \\plus{} 1)$\r\n\r\nsorry if this is too easy for this section, i was not sure..\r\n\r\nedit: n+1 is prime",
  "Solution_1": "The divisions on the LHS are clearly not all valid unless $ n \\plus{} 1$ is prime.",
  "Solution_2": "In the book i am reading it says that $ 1/1 \\plus{} 1/2 \\plus{} ... \\plus{} 1/(p \\minus{} 1) \\equal{} 1 \\plus{} 2 \\plus{} ... \\plus{} p \\minus{} 1$ $ mod(p)$ where p is prime\r\nthe proof is that each non-zero residue mod(p) is equal to the reciprocal of one and only one non-zero residue mod(p)\r\n\r\nit then states the equality given in the OP. in this equality we can see that the numbers on the RHS do not cover all the non-zero residues of mod(p) so how do we know the LHS is a permutation of exactly those residues on the RHS? this is probably real simple but i can't see it atm so please show me?",
  "Solution_3": "Let $ a_1, a_2 \\ldots a_{p \\minus{} 1}$ be a permutation of $ 1, 2, \\ldots p \\minus{} 1$ so that:\r\n\r\n$ 1 \\cdot a_1 \\equiv 1$\r\n$ 2 \\cdot a_2 \\equiv 1$\r\n$ 3 \\cdot a_3 \\equiv 1$\r\n$ \\ldots$\r\n$ (p \\minus{} 1)a_{p \\minus{} 1} \\equiv 1$\r\n\r\nWe have $ m \\equiv \\frac {1}{a_m}$ from here, which was what was used to solve your example proof. We can just square this to get $ m^2 \\equiv \\frac {1}{a_m ^2}$\r\n\r\nIn other words, the same permutation that solves your example question will solve your original problem.",
  "Solution_4": "ok thanks, i suck.\r\n\r\nbtw can you tell me what a generator of z/pz means?"
}
{
  "Tag": [
    "exterior angle"
  ],
  "Problem": "Prove that the angles of a triangle add up to 180 degrees.",
  "Solution_1": "This is easy.\r\n\r\nDraw a triangle ABC.\r\n\r\nThen, draw a line, XY, containing A that is parallel to BC with points Y and C to the right of A and the other two points to the left.\r\nThen, extend BC segment into a line.\r\n\r\nThen, using corresponding angles, <YAC = <ACB and <XAB = <ABC.\r\n\r\nSince <YAC+<XAB+<BAC = 180,\r\n\r\nthe three angles of ABC also add up to 180.",
  "Solution_2": "[hide]Exterior angles are the sums of two remote angles, so $\\angle 1=\\angle B+\\angle C;\\,\\angle 2=\\angle A+\\angle C;\\,\\angle 3=\\angle A+\\angle B.$\n\nWe can combine that into $\\angle 1+\\angle 2+\\angle 3=2\\angle A+2\\angle B+2\\angle C.$\n\nThe sum of the exterior angles of a polygon are 360, so $\\angle 1+\\angle 2+\\angle 3=360.$\n\nSubstituting, we get $2\\angle A+2\\angle B+2\\angle C=360.$\n\nSimplifying gives us $\\angle A+\\angle B+\\angle C=180.$\n\n[img]http://img308.imageshack.us/img308/4405/trianglenf8.jpg[/img][/hide]",
  "Solution_3": "[quote=\"i_like_pie\"]The sum of the exterior angles of a polygon are 360[/quote] That seems like a more advanced theorem than what we are trying to prove.  :wink:",
  "Solution_4": "[quote=\"lotrgreengrapes7926\"][quote=\"i_like_pie\"]The sum of the exterior angles of a polygon are 360[/quote] That seems like a more advanced theorem than what we are trying to prove.  :wink:[/quote]\r\nwell don't we get that fact from what we're trying to prove\r\nso its a bit circular reasoning",
  "Solution_5": "[quote=\"junggi\"][quote=\"lotrgreengrapes7926\"][quote=\"i_like_pie\"]The sum of the exterior angles of a polygon are 360[/quote] That seems like a more advanced theorem than what we are trying to prove.  :wink:[/quote]\nwell don't we get that fact from what we're trying to prove\nso its a bit circular reasoning[/quote]\r\n\r\nconsider a walk around a polygon, at each vertex, you turn precisely the exterior angle, once you return to your starting vertex, you have spun around 360 degrees",
  "Solution_6": "[quote=\"Altheman\"][/quote][quote=\"junggi\"][quote=\"lotrgreengrapes7926\"][quote=\"i_like_pie\"]The sum of the exterior angles of a polygon are 360[/quote] That seems like a more advanced theorem than what we are trying to prove.  :wink:[/quote]\nwell don't we get that fact from what we're trying to prove\nso its a bit circular reasoning[/quote][quote=\"Altheman\"]\n\nconsider a walk around a polygon, at each vertex, you turn precisely the exterior angle, once you return to your starting vertex, you have spun around 360 degrees[/quote]\r\nok that's cool"
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "number theory unsolved"
  ],
  "Problem": "let:\r\n \\[ \\frac{1\\plus{}xy}{x\\minus{}y}\\equal{}k, \\frac{1\\plus{}xz}{z\\minus{}x})\\equal{}m,\\frac{1\\plus{}yz}{y\\minus{}z})\\equal{}l\\]\r\nprove that :\r\n\\[ gcd(m.l)\\equal{}gcd(m,k)\\equal{}gcd(l,k)\\equal{}1\\]\r\n($ x,y,z \\in R$)\r\n($ m,l,k \\in Z$)",
  "Solution_1": "$ km\\plus{}ml\\plus{}lk\\equal{}1$ $ \\Longrightarrow gcd(m,l)\\equal{}gcd(l,k)\\equal{}gcd(k,m)\\equal{}1$"
}
{
  "Tag": [
    "geometry",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Some $1\\times 2$ dominoes, each covering two adjacent unit squares, are placed on a board of size $n\\times n$ such that no two of them touch (not even at a corner). Given that the total area covered by the dominoes is $2008$, find the least possible value of $n$.",
  "Solution_1": "Consider a set of dominoes placed on an $ n \\times n$ board, no touching allowed. Extend the board to $ (n\\plus{}1) \\times (n\\plus{}1)$ by adding an empty column on the right and an empty row at the top. For each domino consider it's \"shadow\" which are all the squares lying directly above, directly to the right, or directly above and to the right of a cell covered by the domino.\r\n\r\nIn the diagram below, the dominoes are the XX cells, and the shadows are designated by the dots.\r\n\r\n[code]\n...\nXX.\n\n..\nX.\nX.\n[/code]\r\n\r\nIt is clear that each domino has precisely $ 4$ shadow cells, and that different dominoes have disjoint shadows. The point of extending the board is that now even dominoes that were placed along the original right or top borders of the board have their full shadows. \r\n\r\nIf there are $ 1004$ dominoes, then they cover $ 6 \\times 1004$ cells (including their shadows), so the extended $ (n\\plus{}1) \\times (n\\plus{}1)$ board must have area at least $ 6024 > 77^2$. Therefore $ n \\geq 77$.\r\n\r\nIf $ n\\equal{}77$, we can achieve the required result by placing all the dominoes horizontally on the odd-numbered rows, having a single cell separating each domino from the next one. We can pack $ 26$ dominoes per row since $ 26 \\times 3 \\minus{} 1 \\equal{} 77$, and they lie in $ 39$ rows, so overall we have $ 39 \\times 26 \\equal{} 1014$ dominoes, a bit more than necessary."
}
{
  "Tag": [
    "geometry",
    "radical axis",
    "power of a point",
    "geometry solved"
  ],
  "Problem": "Given a triangle $ABC$ and $AA_0,BB_0,CC_0$ are converge at M.\r\nThe circle with diameter $BC$ intersect the circle with diameter $AA_0$ at $A_1$ and $A_2$. Similar we get $B_1,B_2$ and $C_1,C_2$. Prove that six points $A_1,A_2,B_1,B_2,C_1,C_2$ are lie on a circle",
  "Solution_1": "This was posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=3535 , but there is no correct solution yet.\r\n\r\n  darij",
  "Solution_2": "Let $A', B', C'$ be the midpoints of the sides $BC, CA, AB$ and $D, E, F$ the feet of altitudes $AD, BE, CF$ from the vertices $A, B, C$. The circles $(A'), (B'), (C')$ with diameters $BC, CA, AB$ obviously pairwise meet at the vertices $A = (B') \\cap (C')$, $B = (C') \\cap (A')$, $C = (A') \\cap (B')$ and also at the feet of the altitudes $D = (B') \\cap (C')$, $E = (C') \\cap (A')$, $F = (A') \\cap (B')$. Hence, the altitudes $AD, BE, CF$ are the pairwise radical axes of the circle pairs $(B') - (C')$, $(C') - (A')$, $(A') - (B')$ and the orthocenter $H$ is their radical center. Let $P, Q, R$ be the midpoints of the cevians $AA_0, BB_0, CC_0$. Similar claim can be made for the circles $(P), (Q), (R)$ with diameters $AA_0, BB_0, CC_0$ - they pass through the feet of the altitudes and their radical center is the triangle orthocenter $H$. Actually, the 3 cevians $AA_0, BB_0, CC_0$ do not even have to be concurrent at all. For example, the circle $(P)$ intersects the side $BC$ at the point $A_0$ and at one other point $D'$. Since $AA_0$ is the circle diameter, the angle $\\measuredangle AD'A_0 = 90^o$ is right and consequently, $AD'$ is the altitude from the vertex $A$ and the points $D' \\equiv D$ are identical. Similarly, the circles $(Q), (R)$ with diameters $BB_0, CC_0$ pass through the feet $E, F$ of the altitudes $BE, CF$. Consequently the power of the ortocenter to the circles $(P), (Q), (R)$ is $HA \\cdot HD = HB \\cdot HE = HC \\cdot HF$, respectively, is the same and the orthocenter $H$ therefore must be the radical center of these 3 circles. Of course, the power of the orthocenter to the circles $(A'), (B'), (C')$ is exactly the same as for the circles $(P), (Q), (R)$ and the orthocenter is thus the common radical center of all 6 circles $(A'), (B'), (C'), (P), (Q), (R)$, regardless of whether the cevians concur at a point $M$ or not. But then the radical axis of any pair from these 6 circles passes through the orthocenter and the power of the orthocenter to all 6 circles is exactly the same, in particular,\r\n\r\n$HA_1 \\cdot HA_2 = HB_1 \\cdot HB_2 = HC_1 \\cdot HC_2$\r\n\r\nAs a result, [color=blue]the 6 points $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a single circle, regardles of whether the cevians $AA_0, BB_0, CC_0$ are concurrent or not[/color]. In the original thread [url=http://www.mathlinks.ro/Forum/topic-3535.html]Geo4 [the points A1, A2, B1, B2, C1, C2 lie on a circle][/url], grobber arrived to the same conclusion. Gotcha, Darij.  :D",
  "Solution_3": "[quote=\"yetti\"]in particular,\n\n$HA_1 \\cdot HA_2 = HB_1 \\cdot HB_2 = HC_1 \\cdot HC_2$\n\nAs a result, [color=blue]the 6 points $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a single circle, regardles of whether the cevians $AA_0, BB_0, CC_0$ are concurrent or not[/color].[/quote]\r\n\r\nSorry, Yetti - same error as Grobber did...\r\n\r\n  darij",
  "Solution_4": "[quote=\"darij grinberg\"]\nSorry, Yetti - same error as Grobber did...\n\n  darij[/quote]\r\n\r\nAll right, I see that I am wrong. Darn.\r\n\r\nYetti",
  "Solution_5": "Phew!!! Thanks everyone , I think I finally got it.  :D \r\nLet $X,Y,Z$ and $U,V,T$ be the midpoint of $BC,CA,AB$ and $AA_0,BB_0,CC_0$ respectively. \r\nSo we easily find that $XU,YV,ZT$ are concurent at point K.\r\nHence $K$ is also the center of $(A_1A_2B_1B_2)$ and $(B_1B_2C_1C_2)$ and $(C_1C_2A_1A_2)$ and we get the result",
  "Solution_6": "[quote=\"Stun\"]Phew!!! Thanks everyone , I think I finally got it.  :D... [/quote]\r\n\r\nAfter a good sleep, I got it as well. The centers of the circles $(P), (Q), (R)$ with the diameters $AA_0, BB_0, CC_0$, are the midpoints of these cevians, hence, they lie on the sides $B'C', C'A', A'B'$ of the medial triangle $\\triangle A'B'C'$. From similarity of the triangle $\\triangle ABC$ and its medial triangle,\r\n\r\n$\\frac{C'P}{BA_0} = \\frac{B'P}{CA_0},\\ \\frac{A'Q}{CB_0} = \\frac{C'Q}{AB_0},\\ \\frac{B'R}{AC_0} = \\frac{A'R}{BC_0}$\r\n\r\nConsequently,\r\n\r\n$\\frac{C'P}{B'P} \\cdot  \\frac{A'Q}{C'Q} \\cdot \\frac{B'R}{A'R} = \\frac{BA_0}{CA_0} \\cdot  \\frac{CB_0}{AB_0} \\cdot \\frac{AC_0}{BC_0}$\r\n\r\ni.e., the cevians $A'P, B'Q, C'R$ of the medial triangle $\\triangle A'B'C'$ are concurrent iff the cevians $AA_0, BB_0, CC_0$ of the original triangle $\\triangle ABC$ are concurrent. Due to\r\n\r\n$HA_1 \\cdot HA_2 = HB_1 \\cdot HB_2 = HC_1 \\cdot HC_2$\r\n\r\nthe point quadruples $(B_1, B_2, C_1, C_2)$, $(C_1, C_2, A_1, A_2)$, $(A_1, A_2, B_1, B_2)$ are concyclic, and the centers of these 3 circles are $A_3 \\equiv B'Q \\cap C'R$, $B_3 \\equiv C'R \\cap A'P$, $C_3 \\equiv A'P \\cap B'Q$, respectively. Iff the cevians $A'P, B'Q, C'R$ concur, the centers $A_3, B_3, C_3$ are identical and because the 3 circles pairwise intersect in at least 2 points, they cannot be concentric, hence, they are identical."
}
{
  "Tag": [
    "induction",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "The statement is here http://eridian.xhost.ro/autom.jpg\r\nWhat do the hints mean, exactly?\r\nIs it possible to apply induction in order to solve the problem?\r\nThanks!",
  "Solution_1": "[quote=\"tomsofts\"]The statement is here ...[/quote]\r\n\r\nNo, it isn't. owk",
  "Solution_2": "[img]http://eridian.xhost.ro/autom.jpg[/img]\r\nHope it works now"
}
{
  "Tag": [
    "inequalities",
    "search",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "graph theory",
    "inequalities theorems"
  ],
  "Problem": "Hello,\r\n\r\nI heard that one has to prove Muirhead on a competition, and you can always site AM-GM and other trivial inequalities. So could some one post here which inequalities can be just sited on a contest, and which have to be proved?\r\n\r\nThank you.",
  "Solution_1": "[quote=\"rem\"]Hello,\n\nI heard that one has to prove Muirhead on a competition, and you can always site AM-GM and other trivial inequalities. So could some one post here which inequalities can be just sited on a contest, and which have to be proved?\n\nThank you.[/quote]\r\n\r\nHey rem, I think it mostly depends on the contest.For example I think in the IMO you can easily use Muirhead and other rather famous ineq's without any proof. But if you use that kind of inequalities in let's say a regional olympiad  you  will probably have hard time with jury :lol: .",
  "Solution_2": "It does depend in the contest, but I think that if you can attribute a theorem to a particular person (i.e. [b]Schur's[/b] inequality, [b]Holder[/b] or [b]Muirhead[/b]) then you should always be able to quote them. I see no reason why a jury in a regional olympiad would have a problem with someone quoting Muirhead- it's not a particularly obscure theorem.",
  "Solution_3": "[quote=\"tim1234133\"]I see no reason why a jury in a regional olympiad would have a problem with someone quoting Muirhead- it's not a particularly obscure theorem.[/quote]\r\n\r\nOne reason is that if the jury don't have enough qualification you can easily get missunderstood and lose important points. Also the problems in that kind of olympiads are usualy easy and you can avoid using complex theorems and formuls. Believe me that I am telling you this form my own experience that it is better to solve problems as simple as possible especialy in the regional or other similar olympiads. :wink:",
  "Solution_4": "I recently read this somewhere (a couple days ago?) that a person in the 2002 IMO was describing their experiences as part of the committee and later grader, and a student sited an obscure graph theory theorem that trivialized (maybe not trivialized but helped a lot) on a problem. No one had heard of it, and for days the committee tried to prove it, and no one could construct a counterexample. Finally, a counterexample was found and a 0 was given on the problem; however, according to the writer, he would have recieved a 7 if someone could prove it. I think this was on an IMO compendium or something, really sorry I can't remember where I found this but try a google search :wink:",
  "Solution_5": "Half-way down the (very long) first post of this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=15623\r\n\r\nI think Tiks' point is that in some smaller olympiads, (not IMO) there will be much less leeway as to what is considered 'well-known'.",
  "Solution_6": "Does anybody have specific information about Muirhead (especially on the USAMO)? Some people have told me it's okay, some people that you have to prove it. Since the proof is NASTY and it is extremely well known, I would like some more info.",
  "Solution_7": "[quote=\"jb05\"]Does anybody have specific information about Muirhead (especially on the USAMO)? Some people have told me it's okay, some people that you have to prove it. Since the proof is NASTY and it is extremely well known, I would like some more info.[/quote]\r\n\r\nI am sure you don't need to prove it in IMO.Although,I am not sure about USAMO(never participated :blush: ), but I think that you don't need to prove it there, too."
}
{
  "Tag": [
    "function",
    "search",
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "let $ X$ be a compact metric space and $ f: X\\to X$ be a function  that \r\n\r\n$ d(f(x),f(y))\\geq d(x,y)$  for all $ x\\neq y$ , show that $ f$ is isometric.",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1962216639&t=18269"
}
{
  "Tag": [],
  "Problem": "Place the integers $ 1,2,3,...,n^2$ (without duplication) in any order onto an $ n$ x $ n$ chessboard, with one integer per square. Show that there exist two adjacent entries whose difference is at least $ n\\plus{}1$. Adjacent means horizontally or vertically or diagonally adjacent.",
  "Solution_1": "I think it's just because 1, 1+n, 1+2n, ... and 1+(n-1)n < n^2 for n > 2.  \r\n\r\nSo starting from 1, you need the shortest path (counting diagonals, horizontal, vertical) from 1 to n^2 to be at least n, but in n by n board, the shortest distance from a square block to another is at most n-1.  \r\n\r\nAnd for n = 2, it's clear.",
  "Solution_2": "[hide=\"Proof\"]A path can be traced from square $ 1$ to square $ n^2$ in such a way that the minimal amount of squares is covered along this path. The length of this path, in squares, is at most $ n\\minus{}2$. This means the number of differences between square $ 1$ and square $ n^2$ is at most $ n\\minus{}1$.\n\nThe difference between $ n^2$ and $ 1$ is $ n^2\\minus{}1$, or $ (n\\plus{}1)(n\\minus{}1)$.\n\nOn average, the difference between any two adjacent squares on this path is $ \\frac{(n\\plus{}1)(n\\minus{}1)}{n\\minus{}1}\\equal{}n\\plus{}1$.\n\nTherefore, at least one pair of adjacent squares exists such that their difference is $ n\\plus{}1$ or greater.[/hide]"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "projective geometry",
    "conics",
    "hyperbola",
    "geometry proposed"
  ],
  "Problem": "Let H and O be the orthocenter and the circumcenter of a triangle ABC, respectively, and let D, E, F be the orthocenters of triangles AHO, BHO, CHO. Prove that the midpoints of the segments AD, BE, CF are collinear.\r\n\r\nAh yes, and I've almost forgotten to say that the line through these midpoints touches the nine-point circle of triangle ABC at the point X(115) (the center of the Kiepert hyperbola). This wasn't part of the olympiad problem, but it helps in the solution.\r\n\r\n  Darij",
  "Solution_1": "I proved the following LemmaRL041104 from which a more general results of that claimed follows.\r\nActually O no needs to be the circumcenters and H be the orthocenter and again the midpoints of the segments AD, BE, CF remain collinear. Where D, E and F need to be redefined in a little bit more general way.\r\nI don't have time now to translate in english my items: I'll do it later and will post here.\r\n\r\nLemmaRL041104\r\n\r\nLet O be a oint on the plane of triangle  ABC and let a,b and c the parallel lines from O to BC, CA and AB respectively. If d is a line through A wich intersect BC at D and l is a parallel to d from B, let R=b^l, T=a^l and S=c^l, then ST/TR=CD/DB.\r\n\r\nProof.\r\n\r\nAs RTO~ADC then RT/OT = AD/CD. As OTS~BDA then OT/TS = BD/DA. From this the thesis.\r\n\r\nNow one can prove the following LemmaRL041104a\r\n\r\nLet O be a oint on the plane of triangle  ABC and let a,b and c the parallel lines from O to BC, CA and AB respectively and let a', b' and c' three line through A,B and C which are parallel eachother. Let a'^a=A1, a'^b=A2 and a'^c=A3, and Bi and Ci are defined similarly (see fig. 1 attached).\r\nIf X=AB^A1C3 and Y=AC^A1B2 then XY//CC3.\r\n\r\n\r\nProof\r\nIf C'3=c'^B2A1 then C'3C=B3B. Indeed B2B1/C1C'3 = B1A1/A1C1 = B2B1/B1B3 which leads to C1C3'=B1B3 from which, as CC1=BB1, C'3C=B3B=A3A=C3C' where C'=c'^AB.\r\nNow, from various similarity, it results XC3/C3A1 = C'C3/A3A1 = AA3/A3A1 = YC'3/C'3A1 which, by Thales implies the thesis.\r\n\r\n\r\nFrom the latter lemma, appropiately appling Desargues, one can deduce  the fact that midpoints of AA1, BB2, CC3 are collinear.\r\n\r\nI will give later the details.",
  "Solution_2": "The points referred are those you can see in the fig. 2 attacched. I renamed the point to match as much is possible the original problem.\r\n\r\n\r\n\r\nLets have a look at the triabgles BRE and FSC. From lemma, it holds that BE^FC = oo, BR^FS = X ans CS^ER = Y are collinear. Then, by Desargues, BF, EC and RS are concurrent, lets say, at T (*).\r\nLet, now, have a look at the triangles ABc and DEF. Since AD//BE//FC are concurrents then, by Desargues, R = AB^DE, S = AC^DF and U = BC^EF are collinear.\r\nIt is a well known fact that UT bisects BE and all chords of triangle UBE parallel to BE, in particular UT bisects FC. As ASd~CSF, UT bisects  AD  too.\r\n\r\nNow what is missing is to prove that the original problem is a corollary of this fact, but this is quite easy.\r\n\r\n\r\n(*) This fact, by Brianchon, iplies that B,C,D,E,T,F stay on a conic. Perhaps this is the hyperbola mentioned by Darij  :maybe: .",
  "Solution_3": "Sorry for bumping a thirteen year old topic, but this problem seems nice.\n\nNote that the midpoint of $AD$ is the antipode of the midpoint of $OH$ in the nine point circle of $AOH$, so all we need to show is that the nine point centers of $AOH,BOH,COH$ are collinear. This in turn is equivalent to showing that they are coaxial, since they share the midpoint of $OH$ asa common point. But the center of the rectangular circumhyperbola through $O$ is on the nine point circle of all the three triangles. (Since $H$ is always on this hyperbola). So they are coaxial and we are done. \n\nJust for fun, here's a solution using Cartesian coordinates. \nSuppose that hyperbola is $xy=1$ and the point $X$ on it has a parameter $t$ when $X=(t,\\frac 1t)$. Suppose the parameters of the points are denoted by the smaller letter. Then $habc=-1$ (well known). We also have $daho=-1$, so the midpoint of $AD$ is $\\frac{1}{2} \\left( a + \\frac{bc}{o}, \\frac 1a + \\frac{o}{bc} \\right)$. So the slope of the line joining the midpoints of $AD, BE$ is $\\frac{o}{abc}$. Thus we are done.  \n\nNote that both the proofs show the fact that $O$ doesn't need to be the circumcenter of $ABC$ and the problem is true for any point replacing $O$."
}
{
  "Tag": [],
  "Problem": "Prove that the square of any odd multiple of $ 3$ is the difference of two triangular numbers.",
  "Solution_1": "The difference between the nth and n+1th triangular numbers is n+1, so every number can be expressed as the difference of 2 triangular numbers."
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "There are 2n+1 pirates.The distance between each one of them is different.\r\nEveryone has only one bullet left in their guns.Now each one is shooting bullet to that pirate who is standing nearest to him.\r\n\r\nProve that there would be no one who will be shot by more than 5 pirate's bullets.",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?t=13356",
  "Solution_2": "THANX A LOT DUDE.\r\nI GOT THE ANSWER."
}
{
  "Tag": [
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "Hello!\r\n\r\nI know different constructions of real numbers, one with Dedekind's cuts, another one with Cauchy's sequences...\r\n\r\nI understand them and they are \"pretty\", BUT these constructions are not natural for me, I mean they are not intuitive...\r\n\r\nI have heard that a construction of real numbers by decimal expansions is possible, and it seems very intuitive then...\r\n\r\nBut the problem is that arithmetical operations (+,x) are difficult to define formally, and show that R is a field too...\r\n\r\nCould everyone give me links or complete proofs with construction of real numbers by decimal expansions please?\r\n\r\n\r\nThank you in advance, this subject is wonderful and thinking about constructions of \"basic\" sets is very interesting :)\r\n\r\nBest regards.\r\n\r\nPS: I am french, and my english is poor!",
  "Solution_1": "Try Spivak's book Calculus. There's a chapter at the end which constructs the real numbers (by Dedekind cuts) and then has a pair of long exercises leading you through alternate constructions (Cauchy sequences and decimal expansions).\r\n\r\nIn any case, the important theorem is that there is only one complete ordered field up to isomorphism. The real numbers are unique any way we construct them, and the construction merely proves existence.\r\n\r\nWe don't normally use the decimal expansion construction because it's harder in every way. The ambiguity of decimal representations complicates everything, and arithmetic is hard because we can't just inherit it from the rationals.",
  "Solution_2": "I know all this, but what I am looking for is a way to construct real numbers that is intuitive and natural!\r\n\r\nAnd this is because construct them with decimal expansions is a difficult job that I asked this question here: how define arithmetical operations (+,x) on decimal expansions and how prove that one gets a field...\r\n\r\nThe great advantage of this method is that with decimal expansions, the fact that every nonempty set of real numbers with an upper bound has a least upper bound is \"obvious\" !\r\n\r\nThanks for the link, I will try to find the book\r\n\r\nBut I still need your help if someone want to help me :)",
  "Solution_3": "If someone can confirm or scan the exercice about construction of real numbers by decimal expansions in Spivak book, it would be so great! Thanks !",
  "Solution_4": "Intuitively, what makes us believe that every decimal expansion corresponds to a real number?  Doesn't this depend on already having accepted (on intuitive grounds) the completeness axiom?  We have an idea of what real numbers are and how they behave before we know anything about decimal expansions.  I don't find it especially natural to start with decimal expansions as a foundation for understanding what the real numbers are.\r\n\r\nIn my mind, I start with an intuitive understanding of what real numbers are supposed to be... an intuitive understanding which is captured by the axioms for the real numbers.  In my mind, the only reason to construct the real numbers at all, using Dedekind cuts or any other method, is just to prove that the axioms for the real numbers are consistent.  (Of course, we expected they would be, but it doesn't hurt to check.)\r\n\r\nKind of philosophical and vague, but that's my two cents.",
  "Solution_5": "Is the axiom of upper bound not actually an axiom? By axiom of upper bound I mean that every non-empty set of real numbers bounded upwards has a least upper bound. I always learned this as an axiom in school. Is it not true?",
  "Solution_6": "There is always one completeness axiom. There are many equivalent choices for this axiom- least upper bounds are simply the most common choice (and generally easy to work with)."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "vector",
    "abstract algebra",
    "trigonometry",
    "rectangle"
  ],
  "Problem": "Ok, this is a classical idea and you probably know it, but here it is anyway:\r\n\r\nA sphere of radius $R$ is charged with the surface charge density given by $\\sigma = \\vec a \\cdot \\hat r$, where $\\vec a$ is a constant vector, and $\\hat r$ is the unit radius-vector at the given point. Find the intensity of the electric field in the center of the sphere.",
  "Solution_1": "I'm not sure that I understood the problem well :( .But according to what I understood,I got the answer $E=\\frac{aR}{e_{0}}$\r\nWhat is the answer?",
  "Solution_2": "is the sphere empty in the middle?",
  "Solution_3": "I think it mustn't be empty, because $sigma$ is changing with radius $\\sigma=ar$",
  "Solution_4": "The sphere is empty, yes. And it's not changing with the radius, but with the direction of the radius vector ($\\hat r$ is a unit vector, and $\\vec a\\cdot \\hat r$ is its dot product with the constant $\\vec a$).",
  "Solution_5": "If r with a hat is r vector/r than is the answer E=4a/3epsilon ?  a is the the module of the vector",
  "Solution_6": "thats the right answer :D .now for extra points find it for a solid sphere. also show that the resulting charge distribution foe a hollow sphere can be represented as the sum of two solid charged spheres with a small distance between their centres.",
  "Solution_7": "You've mentioned the neat trick :) But the answer isn't correct, it doesn't have the 4...",
  "Solution_8": "If it has the three and not the four then could you post the sollution?",
  "Solution_9": "By examining the dot product we can notice that the charge density is $\\sigma (\\phi) = a \\cos\\phi$, so the sphere can be considered to be made up of rings of various radii and charges. We have: \\[E(\\phi) = \\frac{kq(\\phi) z}{R^{3}},\\] which is the well-known formula for the el. field of a charged ring; a ring corresponding to the angle $\\phi$ has a charge $q(\\phi) = 2a\\pi R^{2}\\sin\\phi\\cos\\phi\\text d\\phi$, and is at height $z = R\\cos\\phi$. \r\n\r\nNow, plugging these back into $E(\\phi)$ and integrating over $(-\\pi/2, \\pi/2)$ while taking care of the changing of the sign of $\\sigma$, we get \\[E = \\frac{a}{3\\varepsilon_{0}}.\\]",
  "Solution_10": "isn't q(phi)=2a pi R sin(phi)cos(phi) ?",
  "Solution_11": "No, the dimensions would be wrong then (for a start...)",
  "Solution_12": "than the surface on which q(phi) is is 2 pi sqr(R ) sin phi d(phi). how come?",
  "Solution_13": "The surface is a rectangle; one side is $R \\text d\\phi$, and the other one is $2r\\pi$, where $r = R\\sin\\phi$.",
  "Solution_14": "i understand now",
  "Solution_15": "I hope so :) Personally, I think I've explained horribly... but if you have any questions do ask!"
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "Just out of curiosity, are we allowed to use pen in the AMC 10?",
  "Solution_1": "Well the forms are scantrons, so I would think, no.",
  "Solution_2": "[url=http://xkcd.com/499/]This[/url] will happen.",
  "Solution_3": "I'm sure that you would be allowed to use a pen to solve the problems, however, you must use a #2 pencil to bubble in your answers."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c\\geq 0$, no two of which are zero. Prove that:\r\n$ 8\\sum{\\frac {a^2 \\plus{} b^2}{a \\plus{} b}} \\plus{} 64\\sum_{cyc}{\\frac {ab}{a \\plus{} 2b \\plus{} c}}\\leq 25(a \\plus{} b \\plus{} c)$",
  "Solution_1": "We have\r\n$ 8\\sum{\\frac{a^2\\plus{}b^2}{a\\plus{}b}}\\plus{}16\\sum_{cyc}{\\frac{ab}{a\\plus{}2b\\plus{}c}}\\leq 16(a\\plus{}b\\plus{}c)$\r\nIt is equivalent to\r\n$ \\sum{[3a^3bc\\plus{}21a^2bc(b\\plus{}c)\\plus{}77ab^2c^2\\plus{}2b^2c^2(b\\plus{}c)](a\\minus{}b)(a\\minus{}c)}$\r\n$ \\plus{}\\sum{a^2b^2(a\\plus{}b\\plus{}3c)(b\\minus{}c)^2}$\r\n$ \\plus{}128(abc)^2(a\\plus{}b\\plus{}c)$\r\n$ \\geq0$.",
  "Solution_2": "Sorry for mistakes, I have edited.",
  "Solution_3": "$ 8\\sum{\\frac {a^2 \\plus{} b^2}{a \\plus{} b}} \\plus{} 64\\sum_{cyc}{\\frac {ab}{a \\plus{} 2b \\plus{} c}}\\leq 25(a \\plus{} b \\plus{} c) (*)$\r\n\r\nis equivalent to \r\n\r\n      $ \\sum{[18a^4(b\\plus{}c)\\plus{}180a^3bc\\plus{}210a^2bc(b\\plus{}c)\\plus{}572ab^2c^2\\plus{}20b^2c^2(b\\plus{}c)](a\\minus{}b)(a\\minus{}c)}$\r\n\r\n     $ \\plus{}21(a\\plus{}b\\plus{}c)(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2\\plus{}64\\sum{a^2b^2(a\\plus{}b\\plus{}3c)(b\\minus{}c)^2}\\geq0$,\r\n\r\nthe (*) holds!",
  "Solution_4": "Could you write a full detailed solution,\r\nsuch that everyone would understand all your steps?\r\nThank you very much.",
  "Solution_5": "Can take to the equal sign?\n",
  "Solution_6": "After expanding using software, we can see this is not a symmetric polynomial(it is cyclic, but not symmetric),this is\nquite surprising, so can't be solved by uvw method. BTW,seems that change RHS \u201825\u2019to '24', still holds.\n[quote=Inequalities Master]Let $ a,b,c\\geq 0$, no two of which are zero. Prove that:\n$ 8\\sum{\\frac {a^2 \\plus{} b^2}{a \\plus{} b}} \\plus{} 64\\sum_{cyc}{\\frac {ab}{a \\plus{} 2b \\plus{} c}}\\leq 25(a \\plus{} b \\plus{} c)$[/quote]",
  "Solution_7": "this solution is wrong, after full expanding ,they are not equivalent inequality.\n[quote=fjwxcsl]$ 8\\sum{\\frac {a^2 \\plus{} b^2}{a \\plus{} b}} \\plus{} 64\\sum_{cyc}{\\frac {ab}{a \\plus{} 2b \\plus{} c}}\\leq 25(a \\plus{} b \\plus{} c) (*)$\n\nis equivalent to \n\n      $ \\sum{[18a^4(b\\plus{}c)\\plus{}180a^3bc\\plus{}210a^2bc(b\\plus{}c)\\plus{}572ab^2c^2\\plus{}20b^2c^2(b\\plus{}c)](a\\minus{}b)(a\\minus{}c)}$\n\n     $ \\plus{}21(a\\plus{}b\\plus{}c)(a\\minus{}b)^2(b\\minus{}c)^2(c\\minus{}a)^2\\plus{}64\\sum{a^2b^2(a\\plus{}b\\plus{}3c)(b\\minus{}c)^2}\\geq0$,\n\nthe (*) holds![/quote]\n\n"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "if [size=150][b]f'(X)[/b] [/size] [b]=[size=150] f(X) + integral 0 to 1 of f(X) dx then integral f(X) dx = [/size] :?: [/b]\r\n\r\nplease give me a solution..",
  "Solution_1": "The trivial $ f(x)\\equal{}0$ works.",
  "Solution_2": "[quote=\"vijaymenon\"]if [size=150][b]f'(X)[/b] [/size] [b]=[size=150] f(X) + integral 0 to 1 of f(X) dx then integral f(X) dx = [/size] :?: [/b]\n\nplease give me a solution..[/quote]\r\n\r\nThe question is $ f'(x)\\equal{}f(x)\\plus{}\\int_{0}^{1} f(x) dx.....................(1)$ then $ \\int f(x) dx\\equal{}?$\r\n\r\n[b]Solution:[/b]\r\nLet $ k\\equal{}\\int_{0}^{1} f(x) dx$ as a constant.\r\n\r\nDifferentiate both sides of [b](1)[/b] with respect to $ x$ , then we get $ f''(x)\\equal{}f'(x)$\r\n\r\n$ \\frac{d}{dx}(e^{\\minus{}x}f'(x))\\equal{}e^{\\minus{}x}(f''(x)\\minus{}f'(x))\\equal{}0$\r\n\r\nSo $ f'(x)\\equal{}ce^{x}$, where $ c$ is any constant, which gives $ f(x)\\equal{}ce^{x}\\minus{}k$ from $ (1)$\r\n\r\nNow $ k\\equal{}\\int_{0}^{1} f(x) dx\\equal{}\\int_{0}^{1} (ce^{x}\\minus{}k) dx\\equal{}ce\\minus{}k\\minus{}c$\r\n\r\n$ \\Rightarrow k\\equal{}\\frac{c(e\\minus{}1)}{2}$\r\n$ \\Rightarrow f(x)\\equal{}ce^{x}\\minus{}\\frac{c(e\\minus{}1)}{2}$\r\n\r\nSo $ \\int f(x) dx\\equal{}ce^x\\minus{}\\frac{c(e\\minus{}1)}{2}x\\plus{}C$, where $ c$ and $ C$ are any constants.",
  "Solution_3": "gr8 solution but how did u get the idea of differentiating [img]http://alt2.mathlinks.ro/latexrender/pictures/e/3/8/e3829e5a2157817b62803e68c845c0f974c7a119.gif[/img] this.please help",
  "Solution_4": "[quote=\"vijaymenon\"]gr8 solution but how did u get the idea of differentiating [img]http://alt2.mathlinks.ro/latexrender/pictures/e/3/8/e3829e5a2157817b62803e68c845c0f974c7a119.gif[/img] this.please help[/quote]\r\n\r\nthats  something  which we come across quite often .",
  "Solution_5": "Hi :) !!!\r\n\r\ni think that the result of [b]stephencheng[/b] is not enough and is correct if $ f$ is the class $ C^2$ ( because we just know that $ f$ are one time derivable) and thanks !!\r\n\r\nmy answer :\r\n\r\n$ f(x)\\equal{}0$ is one solution obvious...\r\n \r\nthe other:\r\n\r\n$ f'(x) \\equal{} f(x) \\plus{} a$ with $ a\\equal{} \\int_0^1 f(x)dx$\r\n\r\nthe soulution of differentielle equation : $ y'\\minus{}y \\equal{} a$ are the fuctions: $ x \\rightarrow Ae^x \\minus{} a$ with $ A$ is a real constante.\r\n\r\nthan $ f(x) \\equal{} Ae^x \\minus{} a$\r\n\r\nand $ a\\equal{}  \\int_0^1 f(x) dx \\equal{} Ae \\minus{} a \\minus{}A \\Rightarrow  a \\equal{} \\frac{A(e\\minus{}1)}{2}$\r\n\r\nthan $ f(x) \\equal{} Ae^x \\minus{} \\frac{A(e\\minus{}1)}{2}$ \r\n\r\nthe rest are easy ...\r\n\r\nPS: i didn't talk by the second derivate ....",
  "Solution_6": "anyways thanks for the solution"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "the sequence (an) is difined  by a1>0 and a_{n+1}=c.an\u00b2+an where c is a costant.prove that:\r\na) an >=sqrt(c^{n-1}.n^n.a1^{n+1})\r\nb)a1+a2......+an >n(n.a1-1/c)   ,where n is a naturel number",
  "Solution_1": "no body? :blush:"
}
{
  "Tag": [
    "modular arithmetic",
    "quadratics",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Prove that the following propositions (i) and (ii) are equivalent\r\n\r\n$ (i)$ $ char\\mathbb{F}_q\\equal{}2$ , there exist a prime number $ p$ and a primitive $ p\\minus{}$ root of unity say $ \\zeta$ such that $ \\deg Irr\\left(\\zeta\\plus{}\\zeta^{\\minus{}1},\\mathbb{F}_q\\right)\\equal{}\\frac{p\\minus{}1}{2}$ and $ \\mathbb{F}_{q^n}\\equal{}\\mathbb{F}_q\\left(\\zeta\\plus{}\\zeta^{\\minus{}1}\\right)$\r\n\r\n$ (ii)$ There exist a prime $ p$ such that $ n\\equal{}\\frac{p\\minus{}1}{2}$ and at least one of the following holds:\r\n\r\n   $ (a)$ 2 is primitive element of $ \\mathbb{Z}_{2n\\plus{}1}^{\\star}$ or\r\n\r\n    $ (b)$ $ 2n\\plus{}1\\equiv 3 \\pmod{4}$ and $ 2$ is a generator of the quadratic residues of $ \\mathbb{Z}_{2n\\plus{}1}$ .\r\n\r\n\r\nPlease give all the details of the proof.",
  "Solution_1": "Can anyone post a proof for this proposition? It seems not very difficult.\r\n\r\nThanks a lot,\r\n\r\nAlexandros"
}
{
  "Tag": [
    "function",
    "\\/closed"
  ],
  "Problem": "Hmm...is it just me, or does the \"Log me on automatically each visit\" function not work on AoPS right now?",
  "Solution_1": "its probably just you, try deleting the artofproblemsolving cookie and that should fix your problem.",
  "Solution_2": "Yep, good call, everything works now. Thanks.",
  "Solution_3": "Actually, that problem is happening to me too. I'd get logged out randomly. Let me check something...\r\n\r\nI found the cookie, but it said nothing conclusive. I can venture a guess that somehow the AOPS forum cookie expiration setting was made too low - so it'd expire after half a day or so, and you have to relogin - but this is only a guess.",
  "Solution_4": "No, since I checked the forum right after I had clicked on the option and then closed the browser.\r\n\r\nBut everything works now. :)"
}
{
  "Tag": [
    "search",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all $n\\in N$ such that\r\n\r\n$x+y+z+w = n\\sqrt{xyzw}$ \r\n\r\nhas a solution $x,y,z,w\\in N$",
  "Solution_1": "Search the forum for Harazi's solution (I think).",
  "Solution_2": "I am sure it use the same idea as I have written in http://www.mathlinks.ro/Forum/viewtopic.php?t=113734"
}
{
  "Tag": [
    "complex numbers",
    "calculus",
    "calculus computations"
  ],
  "Problem": "$ \\sqrt{5\\plus{}\\sqrt{13\\plus{}\\sqrt{5\\plus{}\\sqrt{13\\plus{}....\\plus{}\\sqrt{5\\plus{}\\sqrt{13}}}}}}$ :D",
  "Solution_1": "or $ x_{n}\\equal{}\\sqrt{5 \\plus{}\\sqrt{13\\plus{}x_{n\\minus{}1}}}$ and $ x_{0}\\equal{}0$\r\nat first you need to prove that there exist limit. \r\nand after you suppose that $ |x_{n}\\minus{}x_{n\\minus{}1}| \\to 0$ as $ n \\to \\infty$\r\nand your last step will be to solve this equation:\r\n$ x\\equal{}\\sqrt{5 \\plus{}\\sqrt{13\\plus{}x}}$\r\n[hide=\"answer\"]\n3\n[/hide]",
  "Solution_2": "I make $ x\\equal{}\\sqrt{5\\plus{}\\sqrt{13\\plus{}x}}$\r\n$ (x^2\\minus{}5)^2\\minus{}13\\minus{}x\\equal{}0$\r\n$ x^4\\minus{}10x^2\\minus{}x\\plus{}12\\equal{}0$\r\n$ Xn<1\\plus{}\\sqrt{5\\plus{}\\sqrt{13}}$\r\n$ Xn\\plus{}1\\equal{}\\sqrt{5\\plus{}\\sqrt{13\\plus{}Xn}}<1\\plus{}\\sqrt{5\\plus{}\\sqrt{13}}$\r\nI am stuck :roll:",
  "Solution_3": "Well, Mathematica gives some ugly solutions. One is integer and equals 3, two are suspiciously complex (one with negative real part and one with positive but smaller than $ \\sqrt {5 \\plus{} \\sqrt {13}}$) and one is real but negative. And by \"suspiciously complex\" I mean that numerically, their imaginary parts are of the order $ 10^{\\minus{}6}$. Therefore, the answer should be $ x \\equal{} 3$ (and that is backed up with some numerical approximation)."
}
{
  "Tag": [
    "trigonometry",
    "Online Math Open"
  ],
  "Problem": "\u0391\u03a3\u039a\u0397\u03a3\u0397 1\r\n\u039d\u03b1 \u03b2\u03c1\u03b5\u03b8\u03bf\u03cd\u03bd \u03cc\u03bb\u03b1 \u03c4\u03b1 \u03c0\u03bf\u03bb\u03c5\u03cc\u03bd\u03c5\u03bd\u03b1 \u03bc\u03b5 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03cd\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03bb\u03b5\u03c3\u03c4\u03ad\u03c2 \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03b9\u03ba\u03b1\u03bd\u03bf\u03c0\u03bf\u03b9\u03bf\u03cd\u03bd \u03c4\u03b7\u03bd \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03c3\u03c7\u03ad\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03cc\u03bb\u03b1 \u03c4\u03b1 $x \\in R$.\r\n$P(2x^2-1)=\\frac{(P(x))^2}{2} -1 $.",
  "Solution_1": "\u039d\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b1\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03ac\u03bb\u03c5\u03c4\u03b7 \u03c0\u03ac\u03bd\u03c9 \u03b1\u03c0\u03cc \u03bc\u03b9\u03b1 \u03b2\u03b4\u03bf\u03bc\u03ac\u03b4\u03b1. \u0398\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b5\u03c2 \u03bd\u03b1 \u03b4\u03ce\u03c3\u03b5\u03b9\u03c2 \u03ad\u03bd\u03b1 hint \u03a3\u03c4\u03ad\u03c1\u03b3\u03b9\u03bf?",
  "Solution_2": "Geia sas, tora mporesa na mpo sto internet proti fora. Signomi gia tin kathisterisi alla den briskomai spiti.\r\n\r\nTha doso tin apantisi kai prospathiste ti lisi...\r\n\r\nTo p(x) einai stathero...",
  "Solution_3": "ok, tha doso kai ligi akoma boitheia.... to p(1) einai ritos i arritos ??",
  "Solution_4": "Afu to $P(x)$ einai polionimo eukola pairnume apo ti dosmeni sxesi oti to $P$ eite einai artio eite einai peritto. An einai peritto, dld $P(x)=-P(-x)$ tote $P(0)=0$. \r\n$x=0$ stin arxiki $\\rightarrow P(-1)=-1$\r\n$x=-1$ stin arxiki $\\rightarrow P(1)=-\\dfrac{1}{2}$ \r\npu erxetai se antithesi me tin \"peritotita\" tu $P$.\r\nAra $P$ artio kai $P(1)=P(-1)$.\r\n$x=-1$ stin arxiki $\\rightarrow P(1)=1+\\sqrt{3}=P(-1)$ h $P(1)=1-\\sqrt{3}=P(-1)$.\r\nTheoro tin proti periptosi pu einai akrivos idia me ti deuteri.\r\n$x=0$ stin arxiki $\\rightarrow (P(0))^2=(1+\\sqrt{3})^2$\r\nParagogizo tin arxiki sxesi kai pairno $P'(2x^2-1)4x=P(x)P'(x)$(1), ap' opu $P'(0)=0$. Theoro akoluthia $a_n$ \r\nme $a_1=0$ kai $a_{n+1}=\\sqrt{\\dfrac{1+a_n}{2}}$. Einai apli epagogi na deiksei kaneis oti $a_n=\\cos(\\dfrac{\\pi}{2^n})$ . Epomenos i $a_n$ einai gnisios monotoni kai diafori tu 0. Akoma apo tin (1) exume:\r\n$P'(a_n)4a_{n+1}=P(a_{n+1})P'(a_{n+1})$. An deiksume oti $P(a_{n+1})$ diaforo tu 0 exume teleiosei afu tote exume $P'(a_n)=0$ gia ola ta $n$, pu simainei oti $P'(x)=0$ dld $P(x)$ stathero.\r\nEsto oti $P(a_n)=0$ gia kapoio $n$. Apo tin arxiki sxesi exume oti $P(a_n)=\\dfrac{(P(a_{n+1}))^2}{2}-1$. Tote $P(a_{n-1})=-1$, $P(a_{n-2})=-\\dfrac{1}{2}$ kai profanos oso proxorame pros ta piso tha pairnume ritus, pragma pu simainei oti $P(a_1)=P(0)$ ritos,atopo afu $(P(0))^2=(1+\\sqrt{3})^2$\r\n\r\nTelika $P$ stathero kai $P(x)=1+\\sqrt{3}$ h $P(x)=1-\\sqrt{3}$",
  "Solution_5": "[quote=\"Anto\"]Geia sas, tora mporesa na mpo sto internet proti fora. Signomi gia tin kathisterisi alla den briskomai spiti.\n\nTha doso tin apantisi kai prospathiste ti lisi...\n\nTo p(x) einai stathero...[/quote]\r\n\r\nme prolaves prin kano post tin lisi!",
  "Solution_6": "Oraia lisi  ;)  ksero alles 2 alla kai i dio diaforetikes, sas grafo tin diki mou.\r\nEsto oti den einai stathero.  :o \r\nOs polionimo to P einai paragogisimo. 'Esto oti $P'(x)=(x-1)^r \\cdot q(x)$, opou r fisikos mporei kai miden,alla q(1) pote miden. Tote paragogizontas kai antikathistontas, exo oti : $4x(2x^2-2)^r q(2x^2-1)= P(x)(x-1)^r q(x)$. Alla apo tin ipothesi to q(x) den einai to mideniko kai epeidi milame gia polionima kai i pano sxesi isxuei gia apeira x. Pairno oti $2^{r+2}x(x+1)^r q(2x^2 -1)=P(x)q(x)$ gia ola ta x. Tote omos to P(1) einai akeraios arithmos (apla balte stin teleutaia sxesi apou x to 1).\r\nAntifasi dioti stin arxi sxesi an baloume opou x to 1, pairnoume oti to x einai arritos.",
  "Solution_7": "opos sinithizeis teleutaia, pio apli lisi apo mena!\r\nparomoio stil :\r\n\r\n$P$ polionimo tetoio oste $P(x^2-1)=(P(x))^2-1$. Na vrethei to $P$.",
  "Solution_8": "\u03a4\u03ad\u03b8\u03b7\u03ba\u03b5 \u03c3\u03c4\u03b7 \u03a3\u03b5\u03c1\u03b2\u03af\u03b1 \u03c3\u03c4\u03bf TST \u03c6\u03ad\u03c4\u03bf\u03c2. \u03a0\u03bf\u03bb\u03cd \u03ba\u03b1\u03bb\u03ae \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 . \u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf \u03c6\u03cc\u03c1\u03bf\u03c5\u03bc \u03b1\u03bb\u03bb\u03ac \u03ba\u03b1\u03bb\u03cc \u03b8\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b4\u03bf\u03cd\u03bc\u03b5 \u03b5\u03b4\u03ce \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ad\u03c2.",
  "Solution_9": "tin allaksa ligo tora!",
  "Solution_10": "[quote=\"galeos\"]tin allaksa ligo tora![/quote]\r\n\r\n\u0398\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03bf\u03cd\u03c3\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b1\u03b6\u03b5\u03c2. \u039d\u03b1 \u03b2\u03bb\u03ad\u03c0\u03b1\u03bc\u03b5 \u03bc\u03b9\u03b1 \u0395\u039b\u039b\u0397\u039d\u0399\u039a\u0397 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf \u03b1\u03c4\u03cc\u03c6\u03b9\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1",
  "Solution_11": "[quote=\"silouan\"]\n\n\u0398\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03bf\u03cd\u03c3\u03b1 \u03bd\u03b1 \u03bc\u03b7\u03bd \u03c4\u03b7\u03bd \u03ac\u03bb\u03bb\u03b1\u03b6\u03b5\u03c2. \u039d\u03b1 \u03b2\u03bb\u03ad\u03c0\u03b1\u03bc\u03b5 \u03bc\u03b9\u03b1 \u0395\u039b\u039b\u0397\u039d\u0399\u039a\u0397 \u03bb\u03cd\u03c3\u03b7 \u03c3\u03c4\u03bf \u03b1\u03c4\u03cc\u03c6\u03b9\u03bf \u03c0\u03c1\u03cc\u03b2\u03bb\u03b7\u03bc\u03b1[/quote]\r\nto arxiko provlima  periorize tin idia sxesi se polionima peritu vathmu. k efoson exei idi lithei sto forum to ekana na exei ligo perisotero endiaferon(oso endiaferon mporei na exei meta ti lisi tis proigumenis ;) :roll:  )",
  "Solution_12": "Geia sas, \r\n(den exo poli xrono na parakoloutho to forum, giauto na me sinxoreite pou kathistero)\r\nLoipon, oso afora to problima pou these o galeos. Den to brika katholou eukolo kai den to exo lisi.\r\nGrafo parakato ti proodo pou exo kanei :\r\nAn o bathmos tou P einai perritos tote einai \"sxetika\" eukolo na deiksoume oti $P(x) =x$.(to afino auto os askisi gia ta ipolipa meli)\r\nAn o bathmos tou P einai artios arithmos, tote an einai 0 exoume statheres liseis.\r\nAn einai dinami tou 2, tote mporoume na deiksoume oti iparxei akrivos mia lisi, i $f(f(...f(x)...))$ opou $f(x)=x^2-1$.\r\nKai genika mporoume na deiksoume oti gia kathe bathmo iparxei to poli mia lisi. \r\nMia sigkrisi sinteleston se 6 bathmou polinomio, odigei se antifasi. Pragma pou me odigise na ipotheso (eikaso) oti an o bathmos tou P einai artios tote prepei na einai dinami tou 2. \r\n\r\nSizitisa to problima auto kai me allous lites kai den mporesan oute auti na to lisoun.\r\n\r\nAntrea, exeis lisi se auto to problima ? i to egrapses apo periergia ?\r\n\r\nXereto, \r\nStergios",
  "Solution_13": "[quote=\"Anto\"]\nAntrea, exeis lisi se auto to problima ? i to egrapses apo periergia ?\n[/quote]\r\ndistixos to deutero  :yinyang: Stin arxi nomiza oti den eixe tpt idiaitero, alla argotera \"eida\" oti mporume na parume apeires liseis k sineiditopoihsa oti den einai kai toso trivial. :blush:",
  "Solution_14": "Loipon, istera apo ola auta pou eixa grapsei stamatisa na to skeftomai...\r\nalla simera to proi, enas filos edo, nomizo oti einai grammenos sto site me to nickname Fiachra, brike mia lisi sto problima. Kai pragmati i eikasia mou itan sosti. An to n einai artio , kai oxi miden, tote einai dinami tou 2. Kai an perittos tote P(x)=x. Einai oles i liseis. I lisi tou, pou einai teleios steixeiodeis, apoteleitai apo ta parakato bimata, prota deixnei oti i sintelestes tou P(x) einai akeraioi arithmoi. Meta xrisimopoiei mia morfi epagogis kai apodeikniei oti an $2^a | n$ kai $n\\geq 2^{a+1} +1$ tote $2^{a+1} | n$. Entiposiastika poli dioti i lisi tou , otan tin diavazeis, fainete poli apli na tin skefteis kai poli profaneis... anyway an thelete olokliri tin lisi mporo na tin grapso, alla den to brisko katholou endiaferon, apla ergapsa tin idea. Opoios endiaferete parapano tha mporouse na to apodeiksi apo monos tou, dioti apotelei mia poli kali askisi.\r\n\r\nStergios"
}
{
  "Tag": [
    "geometry solved",
    "geometry"
  ],
  "Problem": "let XOY be a fixed angle .and (C) is a fixed circle in the interior of it.\r\n\r\ndraw a circle wich is tangent to OX,OY,and (C).",
  "Solution_1": "You've posted this before, I think.",
  "Solution_2": "oh... yes dear Arne.but that was because of seeing my solution.\r\n\r\nnow im waiting for u nice solutions. ;)",
  "Solution_3": "Is this problem open ? If yes we shall cry for humanity for not being able to solve this problem ,but i can assure you that this problem is not Open ,therefore it should not be in this section .:\r\n\r\nAs for the problem ,this is one of the  problems which ,i have personally disccused it with you lots of times ,so are you playing with these guys here man , time is valuable .\r\n\r\nYou will find nice conversations here ,mate:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/topic-46945.html",
  "Solution_4": "shhhhhhhhhhhhhhh,\r\n\r\ni dont need u informations :mad:"
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "A certain kind of tree increases in height by 25 percent each year. It was 2.5 feet tall when it was planted in January 2000. Approximately how tall will the tree be in December 2008?",
  "Solution_1": "[hide=\" click to reveal hidden content\"]\n\n$ 2.5 \\cdot(1.25)^{8} \\equal{} 14.90116119...$, or qapproximately $ 14.9$ metres.\n\n[/hide]",
  "Solution_2": "[quote=\"AndrewTom\"]\n\n$ 2.5 \\cdot(1.25)^{8} \\equal{} 14.90116119...$, or qapproximately $ 14.9$ metres.\n\n[/quote]\r\n\r\nCloser to $ 9$ years pass, so shouldn't we have $ 2.5 \\cdot(1.25)^{9}$.",
  "Solution_3": "True, but it is not yet $ 9$ years, and the question does not specify on sub-annual growth amounts. It could suggest continuous growth, but I doubt that we're dealing with tropical trees here, as that isn't the tree the normal person, or at least the normal math-question writer, would think of.",
  "Solution_4": "[quote=\"PhireKaLk6781\"]True, but it is not yet $ 9$ years, and the question does not specify on sub-annual growth amounts. It could suggest continuous growth, but I doubt that we're dealing with tropical trees here, as that isn't the tree the normal person, or at least the normal math-question writer, would think of.[/quote]\r\n\r\nThe question asks for an approximation."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "quadratics",
    "function"
  ],
  "Problem": "Let $ P(x) \\equal{} (x \\minus{} 1)(x \\minus{} 2)(x \\minus{} 3)$. For how many polynomials $ Q(x)$ does there exist a polynomial $ R(x)$ of degree 3 such that $ P(Q(x)) \\equal{} P(x) \\cdot R(x)$?\r\n\r\n$ \\textbf{(A)}\\ 19\\qquad\r\n\\textbf{(B)}\\ 22\\qquad\r\n\\textbf{(C)}\\ 24\\qquad\r\n\\textbf{(D)}\\ 27\\qquad\r\n\\textbf{(E)}\\ 32$",
  "Solution_1": "[hide=\"partial answer\"]$P(Q(x)) = (Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3)\\cdot R(x)$\n\n$(Q(x))^3 - 6(Q(x))^2 + 11(Q(x)) - 6 = R(x)x^3 - 6R(x)x^2 + 11R(x)x - 6$\n\nNow since $R(x)$ is of degree 3, we know it's in the form of $ax^3 + bx^2 + cx + d$ so...\n\n$(Q(x))^3 - 6(Q(x))^2 + 11(Q(x)) - 6 = ax^6 + bx^5 + cx^4 + dx^3 - 6ax^5 - 6bx^4 - 6cx^3 - 6dx^2 + 11ax^4 + 11bx^3 + 11cx^2 + 11dx - 6ax^3 + 6bx^2 + 6cx + 6d = ax^6 + (b-6a) + (c - 6b + 11a)x^4 + (d-6c+11b-6a)x^3 +(-6d +11c-5b)x^2 + (11d - 6c)x + 6d$\n\nNow, since $Q(x)$ is a polynomial and the greatest power of $(Q(x))^3$ is 6, the degree of $Q(x)$ is 2. So it is of the form $ex^2 + fx + g$\n\nso:\n\n\\[ (ex^2 + fx + g)^3 - 6(ex^2 + fx + g)^2 + 11(ex^2 + fx + g) - 6  \\]\n\\[ (ex^2 + fx + g)(ex^4 + 2efx^3 + (2eg + f^2)x^2 + 2fgx + g^2)  \\]\n\\[ e^3x^6 + 3e^2fx^5 + (3ef^2+3e^2g)x^4 + (f^3 + 6efg)x^3 +(3eg^2 + 3f^2g)x^2 + 3fg^2x + g^3  \\]\n\n$e^3x^6 + 3e^2fx^5 + (3ef^2+3e^2g)x^4 + (f^3 + 6efg)x^3 +(3eg^2 + 3f^2g)x^2 + 3fg^2x + g^3 = ax^6 + (b-6a) + (c - 6b + 11a)x^4 + (d-6c+11b-6a)x^3 +(-6d +11c-5b)x^2 + (11d - 6c)x + 6d$\nso we have the system of 6 equations in 6 variables:\n\n$e^3 = a$\n$3e^2f = b-6a$\n$3ef^2+3e^2g = c-6b+11a$\n$f^3 + 6efg = d-6c+11b-6a$\n$3eg^2+3f^2g = -6d+11c-5b$\n$3fg^2 = 11d-6c$\n$g^3 = 6d$\n\nWould anyone like to solve those equations? I have no idea how to solve a system this complicated. (the number of solutions will be the answer)[/hide]",
  "Solution_2": "Considering this is a AMC test question, Hamster1800 perhaps you should try another method...",
  "Solution_3": "Check out the AMC 12A math jam from last year, it goes over the last five problems on the AMC 12, so this one is obvious included.",
  "Solution_4": "[quote=\"10000th User\"]Considering this is a AMC test question, Hamster1800 perhaps you should try another method...[/quote]\r\n\r\nI would....but I don't see the other method......\r\n\r\n[hide=\"trying something else\"]\nI'm going to put the polynomials in the form of (x-a)(x-b) and (x-c)(x-d)(x-e) in hopes of getting an easier system.\nso....\n\n$((x-a)(x-b)-1)((x-a)(x-b)-2)((x-a)(x-b)-3) = (x-1)(x-2)(x-3)(x-c)(x-d)(x-e)$\n\nOH!!! dur!!!\n\nok...say $y=Q(x)$\n\n$(y-1)(y-2)(y-3) = (x-1)(x-2)(x-3)(x-c)(x-d)(x-e)$\nok. Now each factor on the left is the product of two on the right.\n\nI think I'm still missing something....The answer that I got is 90....but that's way to high. Am I at least on the right path now though?\n\nOh...ok I see something else.\n\nIt only works when the the two a values for the two (x-a) factors add up to the same number. Then the products have to be 3 consecutive numbers.\n\nSo....I know how to do it but I don't know how to quickly just eliminate all the ones that won't work.\n[/hide]\n[/hide]",
  "Solution_5": "[hide]\nFirst we notice that $Q(x)$ is degree two . Then we know that $Q(1),Q(2),Q(3)$ can only be $1,2,3$ since it must satisfy $P(Q(1))=P(Q(2))=P(Q(3))=0$ and $P(x)$ is degree 3 . So now , if $Q(1)=1$ , then $Q(2)=Q(3)=2$ or $=3$ because $Q(x)$ is quadratic function and the most is only two same roots. So this happens to be one of the way $Q(x)$ can be formed.Similarly we can work out such permutation and get $3\\times 3P2=18$ . But we also have another case which is all $Q(1),Q(2),Q(3)$ is different which has $3!$ method . So total is $\\boxed{24}$ ways  ;) \n[/hide]",
  "Solution_6": "The answer is actually B) 22.\r\n[hide]Since $R(x)$ has degree three, then $P(x)\\cdot R(x)$ has degree six. Thus, $P(Q(x))$ has degree six, so $Q(x)$ must have degree two, since $P(x)$ has degree three. \\[ P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\\cdot R(1)=0, \\] \\[ P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\\cdot R(2)=0, \\] \\[ P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\\cdot R(3)=0. \\] Hence, we conclude $Q(1)$, $Q(2)$, and $Q(3)$ must each be $1$, $2$, or $3$. Since a quadratic is uniquely determined by three points, there can be $3*3*3=27$ different quadratics $Q(x)$ after each of the values of $Q(1)$, $Q(2)$, and $Q(3)$ are chosen.\n\nHowever, we have included $Q(x)$ which are not quadratics. Namely, \\[ Q(1)=Q(2)=Q(3)=1 \\Rightarrow Q(x)=1, \\] \\[ Q(1)=Q(2)=Q(3)=2 \\Rightarrow Q(x)=2, \\] \\[ Q(1)=Q(2)=Q(3)=3 \\Rightarrow Q(x)=3, \\] \\[ Q(1)=1, Q(2)=2, Q(3)=3 \\Rightarrow Q(x)=x, \\] \\[ Q(1)=3, Q(2)=2, Q(3)=1 \\Rightarrow Q(x)=4-x. \\] Clearly, we could not have included any other constant functions. For any linear function, we have $2\\cdot Q(2)=Q(1)+Q(3)$. Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is $27-5=\\boxed{22}$.[/hide]",
  "Solution_7": "[quote=\"towersfreak2006\"]\n[hide]Since $ R(x)$ has degree three, then $ P(x)\\cdot R(x)$ has degree six. Thus, $ P(Q(x))$ has degree six, so $ Q(x)$ must have degree two, since $ P(x)$ has degree three.\n\\[ P(Q(1)) \\equal{} (Q(1) \\minus{} 1)(Q(1) \\minus{} 2)(Q(1) \\minus{} 3) \\equal{} P(1)\\cdot R(1) \\equal{} 0,\n\\]\n\n\\[ P(Q(2)) \\equal{} (Q(2) \\minus{} 1)(Q(2) \\minus{} 2)(Q(2) \\minus{} 3) \\equal{} P(2)\\cdot R(2) \\equal{} 0,\n\\]\n\n\\[ P(Q(3)) \\equal{} (Q(3) \\minus{} 1)(Q(3) \\minus{} 2)(Q(3) \\minus{} 3) \\equal{} P(3)\\cdot R(3) \\equal{} 0.\n\\]\nHence, we conclude $ Q(1)$, $ Q(2)$, and $ Q(3)$ must each be $ 1$, $ 2$, or $ 3$.[/hide][/quote]\r\n\r\nWhy must Q(x) be 1,2,3?",
  "Solution_8": "[quote=\"Smartguy\"][quote=\"towersfreak2006\"]\n[hide]Since $ R(x)$ has degree three, then $ P(x)\\cdot R(x)$ has degree six. Thus, $ P(Q(x))$ has degree six, so $ Q(x)$ must have degree two, since $ P(x)$ has degree three.\n\\[ P(Q(1)) \\equal{} (Q(1) \\minus{} 1)(Q(1) \\minus{} 2)(Q(1) \\minus{} 3) \\equal{} P(1)\\cdot R(1) \\equal{} 0,\n\\]\n\n\\[ P(Q(2)) \\equal{} (Q(2) \\minus{} 1)(Q(2) \\minus{} 2)(Q(2) \\minus{} 3) \\equal{} P(2)\\cdot R(2) \\equal{} 0,\n\\]\n\n\\[ P(Q(3)) \\equal{} (Q(3) \\minus{} 1)(Q(3) \\minus{} 2)(Q(3) \\minus{} 3) \\equal{} P(3)\\cdot R(3) \\equal{} 0.\n\\]\nHence, we conclude $ Q(1)$, $ Q(2)$, and $ Q(3)$ must each be $ 1$, $ 2$, or $ 3$.[/hide][/quote]\n\nWhy must Q(x) be 1,2,3?[/quote]\r\n\r\nThey don't have to be, that's just one of the possible options which doesn't give you a quadratic polynomial for $ Q(x)$. Can you specify which part of his solution you are unclear about? There are multiple parts of the solution you're question can be asking about.",
  "Solution_9": "[quote=\"sunehra\"]\nThere are multiple parts of the solution you're question can be asking about.[/quote]\r\n\r\nthe part i quoted\r\n\r\nWhy he used P(Q(1)), and P(Q(2)) etc.",
  "Solution_10": "OK, why did he plug in x=1,2,3??\r\nIf you do Q(1,2,3) what will happen?",
  "Solution_11": "[quote=\"Smartguy\"]OK, why did he plug in x=1,2,3??\nIf you do Q(1,2,3) what will happen?[/quote]If you read carefully, $ x\\equal{}1,2,3$ are zeroes of $ P(x)$. Your objective is to look for polynomials $ Q(x)$ satisfying $ P(Q(x))\\equal{}P(x)\\cdot R(x)$, and now do you see why he plugged in $ x\\equal{}1,2,3$? (Obvious hint: They are also zeroes of $ P(Q(x))$!!!  :ninja:",
  "Solution_12": "thanks, i get it now. I got a little confused over $ 3^3$ but i see how Q(1), Q(2), Q(3) each can be 1, 2 or 3.\r\n\r\nThanks everyone",
  "Solution_13": "$P(Q(x)) = P(x) \\cdot R(x) \\implies $ $(Q(x)-1)(Q(x)-2)(Q(x)-3)=$ $(x-1)(x-2)(x-3)R(x)$. From here, we can conclude that $Q(1)$ is either $1,2,3$, $Q(2)$ is either $1,2,3$, and $Q(3)$ is either $1,2,3$, yielding $3^3=27$ possibilities but there's a catch! Because the degree of $R$ is $3$, the degree of $1$ must be $2$, so if $Q(1)=Q(2)=Q(3)=1$, $2$, or $3$, then $Q$ is not a quadratic, but a horizontal line. Similarly, if $Q(1)=1,Q(2)=2,$ and $Q(3)=3$, then we have the line $y=x$ or if $Q(1)=3,Q(2)=2,Q(3)=1$, we have the line $y=-x+4$. In essence, so long as $Q$ satisfies that $Q(1),Q(2),Q(3) \\in \\{ 1,2,3 \\}$ and not one of the five cases highlighted above, then it satisfies the given equation. Hence, there are $27-5 = \\boxed{22}$ possibilities for $Q$.",
  "Solution_14": "[hide=another sol?]Let $Q(x)=ax^2+bx+c$ as it has a degree of at most $2$ since $P(x) \\cdot R(x)$ has a degree of $6$. Thus, we have $(ax^2+bx+c-1)(ax^2+bx+c-2)(ax^2+bx+c-3)=(x-1)(x-2)(x-3)(R(x))$. Plugging in $x=1,2,3$, we get the equations $(a+b+c-1)(a+b+c-2)(a+b+c-3)=0, (4a+2b+c-1)(4a+2b+c-2)(4a+2b+c-3)=0, (9a+3b+c-1)(9a+3b+c-2)(9a+3b+c-3)=0$. Choosing one condition from each, we get that there are $27$ possible conditions. However, we cannot have $a=0$, because this will force the degree of $R(x)$ to not be $3$. The cases where this happens is with $a+b+c=4a+2b+c=9a+3b+c=1, a+b+c=4a+2b+c=9a+3b+c=2, a+b+c=4a+2b+c=9a+3b+c=3, a+b+c=1; 4a+2b+c=2; 9a+3b+c=3$, and $a+b+c=3; 4a+2b+c=2; 9a+3b+c=3$. To get these cases, we do: [hide]Basically, let $a+b+c=t_1, 4a+2b+c=t_2, 9a+3b+c=t_3$. With the first two equations, we have $3a+b=t_2-t_1$ and with the last two equations, we have $5a+b=t_3-t_2$. Note that if $t_2-t_1=t_3-t_2$, then this will force $a$ to be $0$. The only cases where this happens is $(t_1,t_2,t_3)=(1,1,1), (2,2,2), (3,3,3), (1,2,3), (3,2,1)$.[/hide]These make up $5$ cases, so we have $\\boxed{22}$. I have a feeling that this solution might be flawed in some way. Apologies if I missed something or messed something up.[/hide]",
  "Solution_15": "Clearly $Q(x)$ is a quadratic polynomial we have $P(Q(1))=P(Q(2))=P(Q(3))=0$ so $Q(1), Q(2), Q(3) \\in \\{1,2,3\\}$ but also $Q(1) =Q(2)=Q(3)$ is not permissible since $Q(x)$ is a quadratic polynomial.\\\\\n\nA quadratic is uniquely determined by three points so $Q(1),Q(2),Q(3)$ can be from  $\\{1,2,3\\}$ only now $3 \\times 3 \\times 3=27$ ways are there to place numbers\n\n\nBut exclude cases where all are same that is $3$\\\\\n\nAnd there are just two cases where it is linear\\\\\n\n$Q(1)=1,Q(2)=2,Q(3)=3 ,Q(x)=x$ and $Q(1)=3, Q(2)=2 ,Q(3)=1 , Q(x)=4-x$\n\nSo $27-3-2=\\boxed{22}$ such polynomials are there"
}
{
  "Tag": [],
  "Problem": "given dat,if a+b+c=0,then a3+b3+c3=3abc.show dat (2+rt.5)^1/3  +(2-rt.5)^1/3 is a rational number.",
  "Solution_1": "[quote=\"williams\"]given dat,if a+b+c=0,then a3+b3+c3=3abc.show dat (2+rt.5)^{1/3 } +(2-rt.5)^{1/3} is a rational number.[/quote]\r\n\r\n\r\nThe question is that:\r\n\r\ngiven that,if $ a\\plus{}b\\plus{}c\\equal{}0,$ then $ a^3\\plus{}b^3\\plus{}c^3\\equal{}3abc$.show that $ (2\\plus{}\\sqrt 5)^{1/3}  \\plus{}(2\\minus{}\\sqrt 5)^{1/3}$ is a rational number.\r\n\r\nIs it right?\r\n\r\n[hide=\"The first one\"]\n$ a\\plus{}b\\plus{}c\\equal{}0 \\implies a\\plus{}b\\equal{}\\minus{}c \\implies (a\\plus{}b)^3\\equal{}\\minus{}c^3$\n$ \\implies a^3\\plus{}b^3\\plus{}c^3\\equal{}\\minus{}3ab(a\\plus{}b)\\equal{}3abc$ \nsince $ a\\plus{}b\\equal{}\\minus{}c$\n[/hide]",
  "Solution_2": "[quote=\"vinskman\"][quote=\"williams\"]given dat,if a+b+c=0,then a3+b3+c3=3abc.show dat (2+rt.5)^{1/3 } +(2-rt.5)^{1/3} is a rational number.[/quote][/quote]\r\n\r\n\r\nThe question is that:\r\n\r\ngiven that,if $ a \\plus{} b \\plus{} c \\equal{} 0,$ then $ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$.show that $ (2 \\plus{} \\sqrt 5)^{1/3} \\plus{} (2 \\minus{} \\sqrt 5)^{1/3}$ is a rational number.\r\n\r\nIs it right?\r\n\r\n[hide=\"The first one\"]\n$ a \\plus{} b \\plus{} c \\equal{} 0 \\implies a \\plus{} b \\equal{} \\minus{} c \\implies (a \\plus{} b)^3 \\equal{} \\minus{} c^3$\n$ \\implies a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} \\minus{} 3ab(a \\plus{} b) \\equal{} 3abc$ \nsince $ a \\plus{} b \\equal{} \\minus{} c$\n[/hide]",
  "Solution_3": "show that $ (2 \\plus{} \\sqrt 5)^{1/3} \\plus{} (2 \\minus{} \\sqrt 5)^{1/3}$ is a rational number.\r\n\r\n[hide=\" The second one \"]\n\nLet $ a\\equal{}(2 \\plus{} \\sqrt 5)^{1/3},b\\equal{}(2 \\minus{} \\sqrt 5)^{1/3}$\n\nSince\n$ a^3\\plus{}b^3\\equal{} 2 \\plus{} \\sqrt 5\\plus{}2 \\minus{} \\sqrt 5\\equal{}4$\n\n$ ab\\equal{}(2 \\plus{} \\sqrt 5)^{1/3} (2 \\minus{} \\sqrt 5)^{1/3}\\equal{}(\\minus{}1)^{1/3}\\equal{}\\minus{}1$\n\nThen $ (a\\plus{}b)^3\\equal{} a^3\\plus{}b^3\\plus{}3ab(a\\plus{}b)\\equal{}4\\minus{}3(a\\plus{}b)$\n\nSuppose that $ x\\equal{}a\\plus{}b$\n\n$ \\therefore x^3\\plus{}3x\\minus{}4\\equal{}0$\n\n$ \\implies (x\\minus{}1)(x^2\\plus{}x\\plus{}4)\\equal{}0$\n\n$ \\delta_x\\equal{}1^2\\minus{}4\\cdot1\\cdot4<0 \\implies x^2\\plus{}x\\plus{}4\\equal{}0$ no real roos.\n\n$ \\therefore x\\equal{}1$ only.\n\n$ \\implies a\\plus{}b\\equal{}1 \\quad$  so that the sum of two cubic roots is rational.\n\n[/hide]",
  "Solution_4": "[hide=\"Alt solution\"]\n\nfor the first one, simplification gives\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a\\plus{}b\\plus{}c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ac)$\n\n$ a \\plus{} b \\plus{} c \\equal{} 0 \\ \\Rightarrow \\ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$\n\n\n\nnext part, take $ a \\equal{} (2\\plus{}\\sqrt{5})^{1/3} , b \\equal{} (2\\minus{}\\sqrt{5})^{1/3}, c \\equal{} \\minus{}1$\n\nObserve that $ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$\n\nand $ a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} ac \\minus{} bc > 0 \\ if \\ a \\neq b \\neq c$ (from A.M > G.M)\n\nso a + b = -c = rational number.\n\n[b]Proved.[/b]\n\n[/hide]",
  "Solution_5": "[quote=\"Nora.91\"][hide=\"Alt solution\"]\n\nfor the first one, simplification gives\n\n$ a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} 3abc \\equal{} (a \\plus{} b \\plus{} c)(a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} bc \\minus{} ac)$\n\n$ a \\plus{} b \\plus{} c \\equal{} 0 \\ \\Rightarrow \\ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$\n\n\n\nnext part, take $ a \\equal{} (2 \\plus{} \\sqrt {5})^{1/3} , b \\equal{} (2 \\minus{} \\sqrt {5})^{1/3}, c \\equal{} \\minus{} 1$\n\nObserve that $ a^3 \\plus{} b^3 \\plus{} c^3 \\equal{} 3abc$\n\nand $ a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} ab \\minus{} ac \\minus{} bc > 0 \\ if \\ a \\neq b \\neq c$ (from A.M > G.M)\n\nso a + b = -c = rational number.\n\n[b]Proved.[/b]\n\n[/hide][/quote]\r\n\r\n\r\ngood, very nice solution :P",
  "Solution_6": "[color=green][b]Thank You vinskman[/b][/color]  :D",
  "Solution_7": "Let $ x \\equal{} \\sqrt [3]{2 \\plus{} \\sqrt {5}} \\plus{} \\sqrt [3]{2 \\minus{} \\sqrt {5}}\\Longrightarrow x^3 \\minus{} (2 \\plus{} \\sqrt {5}) \\minus{} (2 \\minus{} \\sqrt {5}) \\equal{} 3x\\sqrt [3]{2 \\plus{} \\sqrt {5}}\\cdot \\sqrt [3]{2 \\minus{} \\sqrt {5}}$\r\n\r\n$ \\therefore x^3 \\plus{} 3x \\minus{} 4 \\equal{} 0\\Longleftrightarrow (x \\minus{} 1)(x^2 \\plus{} x \\plus{} 4) \\equal{} 0$, yielding $ x\\in\\mathbb{R},\\ x \\equal{} 1.$",
  "Solution_8": "[quote=\"kunny\"]Let $ x \\equal{} \\sqrt [3]{2 \\plus{} \\sqrt {5}} \\plus{} \\sqrt [3]{2 \\minus{} \\sqrt {5}}\\Longrightarrow x^3 \\minus{} (2 \\plus{} \\sqrt {5}) \\minus{} (2 \\minus{} \\sqrt {5}) \\equal{} 3x\\sqrt [3]{2 \\plus{} \\sqrt {5}}\\cdot \\sqrt [3]{2 \\minus{} \\sqrt {5}}$\n\n$ \\therefore x^3 \\plus{} 3x \\minus{} 4 \\equal{} 0\\Longleftrightarrow (x \\minus{} 1)(x^2 \\plus{} x \\plus{} 4) \\equal{} 0$, yielding $ x\\in\\mathbb{R},\\ x \\equal{} 1.$[/quote]\r\n\r\n\r\nhaha , very funny \r\n\r\nyours is just as the same as mine. :P  :lol:",
  "Solution_9": "No. My solution is $ x \\equal{} a \\plus{} b\\Longleftrightarrow x \\plus{} ( \\minus{} a) \\plus{} ( \\minus{} b) \\equal{} 0\\Longrightarrow x^3 \\plus{} ( \\minus{} a)^3 \\plus{} ( \\minus{} b)^3 \\equal{} 3x( \\minus{} a)( \\minus{} b).$\r\n\r\nYour solution is $ x \\equal{} a \\plus{} b\\Longleftrightarrow x^3 \\equal{} (a \\plus{} b)^3\\Longleftrightarrow x^3 \\equal{} a^3 \\plus{} b^3 \\plus{} 3ab(a \\plus{} b).$",
  "Solution_10": "[quote=\"kunny\"]No. My solution is $ x \\equal{} a \\plus{} b\\Longleftrightarrow x \\plus{} ( \\minus{} a) \\plus{} ( \\minus{} b) \\equal{} 0\\Longrightarrow x^3 \\plus{} ( \\minus{} a)^3 \\plus{} ( \\minus{} b)^3 \\equal{} 3x( \\minus{} a)( \\minus{} b).$\n\nYour solution is $ x \\equal{} a \\plus{} b\\Longleftrightarrow x^3 \\equal{} (a \\plus{} b)^3\\Longleftrightarrow x^3 \\equal{} a^3 \\plus{} b^3 \\plus{} 3ab(a \\plus{} b).$[/quote]\r\n\r\n\r\nSorry , mistaken  :D",
  "Solution_11": "No problem, but both of us finally obtained $ x^3\\plus{}3x\\minus{}4\\equal{}0.$ :lol:",
  "Solution_12": "$ a^3\\plus{}b^3\\plus{}c^3\\minus{}3abc\\equal{}(a\\plus{}b\\plus{}c)(a^2\\plus{}b^2\\plus{}c^2\\minus{}ab\\minus{}ac\\minus{}bc)\\equal{}(a\\plus{}b\\plus{}c)((a\\minus{}b)^2\\plus{}(b\\minus{}c)^2\\plus{}(c\\minus{}a)^2)\\frac{1}{2}$. This is useful to know.",
  "Solution_13": "A simple way for number 2:\r\n\r\n$ \\left(\\dfrac{1\\plus{}\\sqrt{5}}{2}\\right)^3 \\equal{} 2 \\plus{} \\sqrt{5}$\r\n\r\n$ \\left(\\dfrac{1\\minus{}\\sqrt{5}}{2}\\right)^3 \\equal{} 2 \\minus{} \\sqrt{5}$\r\n\r\nThen the question becomes: is $ \\dfrac{1\\plus{}\\sqrt{5}}{2}\\plus{}\\dfrac{1\\minus{}\\sqrt{5}}{2}$ a rational number? \r\n\r\n$ \\dfrac{1\\plus{}\\sqrt{5}}{2}\\plus{}\\dfrac{1\\minus{}\\sqrt{5}}{2}\\equal{}\\dfrac{2}{2}\\equal{}1$ so the answer is yes.",
  "Solution_14": "[quote=\"tonypr\"]A simple way for number 2:\n\n$ \\left(\\dfrac{1 \\plus{} \\sqrt {5}}{2}\\right)^3 \\equal{} 2 \\plus{} \\sqrt {5}$\n\n$ \\left(\\dfrac{1 \\minus{} \\sqrt {5}}{2}\\right)^3 \\equal{} 2 \\minus{} \\sqrt {5}$\n\nThen the question becomes: is $ \\dfrac{1 \\plus{} \\sqrt {5}}{2} \\plus{} \\dfrac{1 \\minus{} \\sqrt {5}}{2}$ a rational number? \n\n$ \\dfrac{1 \\plus{} \\sqrt {5}}{2} \\plus{} \\dfrac{1 \\minus{} \\sqrt {5}}{2} \\equal{} \\dfrac{2}{2} \\equal{} 1$ so the answer is yes.[/quote]\r\n\r\n\r\n[b] Good One !!  [/b]   :coolspeak:",
  "Solution_15": "Hmm is there a fast way to simplify $ \\sqrt[3]{2\\plus{}\\sqrt5}$ to $ \\frac{1\\plus{}\\sqrt5}2$?\r\nOf course it's easier to verify than to derive..."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Do the items in the online school Archive ever expire?  If so, when do they expire?",
  "Solution_1": "We don't anticipate removing them, but we're reluctant to ever say forever.",
  "Solution_2": "Is there a way to save every question with its solution using a script as a PDF file that we can save on our computers"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $ p$ be a prime number from $ 4\\mathbb{Z}\\plus{}3$\r\n\r\nthen prove $ x^2 \\equal{}\\minus{}1 [p]$ impossible\r\n\r\nthanks",
  "Solution_1": "Use Euler's criterion http://en.wikipedia.org/wiki/Euler's_criterion\r\nOr if you dont want to quote it, learn its proof and apply to this situation!",
  "Solution_2": "[quote=\"FOURRIER\"]Let $ p$ be a prime number from $ 4\\mathbb{Z} \\plus{} 3$\n\nthen prove $ x^2 \\equal{} \\minus{} 1 [p]$ impossible\n\nthanks[/quote]\r\n\r\n$ 1\\equiv x^{p\\minus{}1}\\equiv (x^2)^{\\frac{p\\minus{}1}{2}}\\equiv (\\minus{}1)^{\\frac{p\\minus{}1}{2}}\\equiv \\minus{}1$ impossible"
}
{
  "Tag": [
    "topology"
  ],
  "Problem": "Prove that every point in the Cantor Set is a cluster point of its complement.",
  "Solution_1": "the cantor set is the set of numbers in $ [0,1]$ that don't require the digit $ 1$ for their base-3 representations. let $ x$ be a member of the cantor set. we need to show that there are numbers arbitrarily close to $ x$ which require $ 1$'s in their base 3 representation. by definition, we may assume $ x\\equal{}0.d_1d_2d_3...$, where each $ d_i$ is 0 or 2. if infinitely many $ d$'s are 0, then the set of numbers obtained by replacing exactly one of those 0's by 1 is a set of numbers all requiring 1 in their base 3 rep's and containing elements arbitrarily close to $ x$. if only finitely many digits are 0, the digits are all 2 from some point on. this representation of $ x$ may then be replaced with one having 0's from some point on. then, as in the first case, the set of numbers obtained by replacing exactly one of these 0's by a 1 does the job."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "$x_{1},x_{2},...,x_{n}$ are real numbers.and $\\sum_{i=1}^{n}x_{i}=1$.prove:\r\n$\\sum_{i=1}^{n}\\frac{x_{i}x_{i+1}}{x_{i-1}x_{i}+x_{i}x_{i+1}+x_{i-1}x_{i+1}}>\\sum_{i=1}^{n}\\frac{x_{i-1}^{2}(x_{i}+x_{i+1})+x_{i-1}x_{i}x_{i+1}}{x_{i-1}x_{i}+x_{i}x_{i+1}+x_{i-1}x_{i+1}}$",
  "Solution_1": "i was solving a problem,and then i found this inequality.i think it is nice but i don't know it is hard or not,because i didn't solve it yet. :D",
  "Solution_2": "[quote=\"jensen\"]$x_{1},x_{2},...,x_{n}$ are real numbers.and $\\sum_{i=1}^{n}x_{i}=1$.prove:\n$\\sum_{i=1}^{n}\\frac{x_{i}x_{i+1}}{x_{i-1}x_{i}+x_{i}x_{i+1}+x_{i-1}x_{i+1}}>\\sum_{i=1}^{n}\\frac{x_{i-1}^{2}(x_{i}+x_{i+1})+x_{i-1}x_{i}x_{i+1}}{x_{i-1}x_{i}+x_{i}x_{i+1}+x_{i-1}x_{i+1}}$[/quote]\r\n\r\nI think $\\sum_{i=1}^{n}\\frac{x_{i-1}^{2}(x_{i}+x_{i+1})+x_{i-1}x_{i}x_{i+1}}{x_{i-1}x_{i}+x_{i}x_{i+1}+x_{i-1}x_{i+1}}$ is just $\\sum_{i=1}^{n}x_{i}=1$. Is it a typo?\r\n\r\nIf it is intended so, then let $y_{i}=\\frac{1}{x_{i}}$. Then $\\sum_{i=1}^{n}\\frac{x_{i}x_{i+1}}{x_{i-1}x_{i}+x_{i}x_{i+1}+x_{i-1}x_{i+1}}$\r\n$=\\sum_{i=1}^{n}\\frac{y_{i-1}}{y_{i-1}+y_{i}+y_{i+1}}$\r\n$>\\sum_{i=1}^{n}\\frac{y_{i-1}}{\\sum_{i=1}^{n}y_{i}}$\r\n$=1$."
}
{
  "Tag": [
    "factorial",
    "symmetry"
  ],
  "Problem": "Find the only integer n>32 that satisfies                               (n-23)!(23)!=(n-32)!(32)!",
  "Solution_1": "[quote=\"mdk\"]Find the only integer n>32 that satisfies                               (n-23)!(23)!=(n-32)!(32)![/quote]\r\n\r\n[hide]\nDivide both sides by $23!$ to get, $(n-23)!=(n-32)!(32*31*30*...*25*24)$. Divide both sides by $(n-32)!$ to get, $(n-23)(n-24)(n-25)...(n-30)(n-31)=32*31*30*...*25*24$ So, since both of these are products of 9 consecutive integers, the terms have to be equivilent, so $n-23=32, n-24=31,...,n-31=24$ So, $n=55$[/hide]",
  "Solution_2": "nicely done.",
  "Solution_3": "Right from Math League #2, huh.\r\n\r\nAnd yeah, that is the best way of approaching it, or notice the symmetry involved, and there you have it.",
  "Solution_4": "[quote=\"mathgeniuse^ln(x)\"]Right from Math League #2, huh.\n\nAnd yeah, that is the best way of approaching it, or notice the symmetry involved, and there you have it.[/quote]\r\n\r\nyep",
  "Solution_5": "take reciprocals, and multiply by n!, the equation is then $\\binom{n}{23}=\\binom{n}{32}$, which has the obvious connection to $\\binom{n}{k}=\\binom{n}{n-k}$, giving $(n-k)+k=23+32=55$",
  "Solution_6": "[quote=\"Altheman\"]take reciprocals, and multiply by n!, the equation is then $\\binom{n}{23}=\\binom{n}{32}$, which has the obvious connection to $\\binom{n}{k}=\\binom{n}{n-k}$, giving $(n-k)+k=23+32=55$[/quote]\r\n\r\neven prettier than the first response.",
  "Solution_7": "prettiest:\r\n\r\nn-23=32\r\n\r\nn-32=23\r\n\r\nBoth give n=55",
  "Solution_8": "[quote=\"1=2\"]prettiest:\n\nn-23=32\n\nn-32=23\n\nBoth give n=55[/quote]\r\n\r\nnice observation. Your answer is nice."
}
{
  "Tag": [],
  "Problem": "Prove that if $ p$ is a prime number greater than $ 3$, then $ 24$ divides $ p^{2}-1$.",
  "Solution_1": "[hide]\nSince all primes greater than three are in the the form $ 6k \\pm 1$, then\n$ (6k \\pm 1 )^{2}-1 = 36k \\pm 12k+1-1 = 48k , 24k \\Rightarrow \\boxed{24 | 48k \\ , \\ 24 | 24 k }\\ \\ \\ \\mathbb{Q.E.D.}$\n[/hide]",
  "Solution_2": "[hide=\"alternate solution\"] \n$ p^{2}-1 = (p+1)(p-1)$, and p is prime, so p mod 4 = 1 mod 4 or 3 mod 4. When p = 3mod4, then p+1 = 0 mod 4, and p-1= 2mod4, so the product is divisible by 8. When p=1 mod 4, then (p+1)(p-1) = 2mod4 * 0 mod 4, so it is also divisible by eight. Finally, it must also be divisible by 3, to have 24 divide it, and p must be 1 mod 3 or 2 mod 3. By same process as for 4, you get that it is divisible by 3. [/hide]\r\n\r\nbtw, how did you know that all primes after 3 are $ 6k\\pm1$? is there a proof or something? :huh:",
  "Solution_3": "[hide=\"proof for the prime form\"]\nlooking at $ \\mathbb{Z}$ in mod 6\n\n$ 0 \\mod{6}$ is divisible by 6\n$ 2 \\mod{6}$ is divisible by 2\n$ 3 \\mod{6}$ is divisible by 3\n$ 4 \\mod{6}$ is divisible by 2\n\nwhich leaves us with $ 1,5 \\equiv \\pm{1}\\mod{6}$ or $ 6k\\pm1$\n[/hide]",
  "Solution_4": "oh, haha, that was simple  :blush:",
  "Solution_5": "So how do you prove?",
  "Solution_6": "What do you mean? Prove what? Both of the proofs posted above are valid proofs.",
  "Solution_7": "[quote=\"The great one\"]So how do you prove?[/quote]\r\nLook at AstroPhys's proof (first reply)."
}
{
  "Tag": [
    "calculus",
    "MIT",
    "college",
    "videos",
    "vector",
    "calculus computations"
  ],
  "Problem": "I recently finished Calculus 2 and I'm looking forward to setting out a journey to Multivariable Calculus. Can anyone recommend me any good resources or books to study so that I can acquire the most out of it?\r\nI'm also planning on brushing up my skills on calc 1 a little bit more at the same time. That will be really nice if anyone knows how to study them at once.",
  "Solution_1": "Are you doing this alone or do you have a teacher? What book did you use for Calculus 2? Are you interested in science and/or engineering?",
  "Solution_2": "I'm very interested in science and engineering. I would like to learn chemistry more in depth and hopefully astronomy and all other sorts of fields. My teacher for calc 2 used Stewart single variable calc book which I didn't think so self-contained and also Robin Gottlieb's calc textbook, which in my opinion, better than Stewart, but somewhat lacked in explanation. I'm going to learn Multivariable on my own, so I will be happy if the textbook or other kinds of resources could explain really well.",
  "Solution_3": "A good resource for you might be the MIT OpenCourseWare site for [url=http://ocw.mit.edu/OcwWeb/Mathematics/18-02Fall-2007/CourseHome/]18.02 Multivariable Calculus[/url]. They have video lectures, (sparse) notes, problem sets, and exams.\r\n\r\nYou also might find some of the resources from http://teachers.sduhsd.k12.ca.us/abrown/classes.htm (under Calculus D) and http://multivariablecalculus.wordpress.com/ useful. There are lots of homepages for multivariable calculus courses out there with course notes and practice tests.\r\n\r\nAs for books, one of my favorite multivariable calculus books (especially for aspiring physicists/engineers) is [u]Vector Calculus[/u] by Marsden and Tromba. It has good application problems, but it is not an easy book without a teacher. It's not easy to read on your own and, though most of the problems are computational, they are generally harder than the ones in Stewart.\r\n\r\nOn the other hand, a book with good explanations and with pretty decent problems and exercises is [u]Calculus Early Transcendentals[/u] by Varberg and Purcell. It isn't very advanced though. Just bare bones multivariable calculus computation skills.\r\n\r\nFor a book that may be in the middle - try [u]Thomas' Calculus[/u] (10th ed or later) by Thomas, Weir, Hass, and Giordano. It comes in various formats, but they're all basically the same. Of the \"basic\" calculus books that cover both single and multivariable calculus, I think it has the best exercises for people interested in science/engineering. It's very similar to Stewart in style and difficulty, though Stewart has more concept questions and less application questions. There's also a solutions manual for the 11th edition floating around on the internet that's not hard to find."
}
{
  "Tag": [
    "function",
    "puzzles"
  ],
  "Problem": "Suppose f:[a,b]--> is bounded and f^2 is Riemann integrable on [a,b]. Must f also be Riemann integrable on [a, b]? Either prove that the answer is \"Yes\" or give an example showing that it is \"No\".\r\n\r\n\r\nGood Luck :)",
  "Solution_1": "This is very much not the correct forum for this question.  On the other hand, I do not think it is too hard.  Take a non-integrable function $g: [a, b] \\to \\{1, -1\\}$ and any integrable function $h: [a, b] \\to \\bf R$ which is within our bound.  Then $f = g\\cdot h$ satisfies $f^2 = h^2$ is integrable while $f$ is not.  (If the bound is of a more general nature than $|f| < R$ for some $R > 0$, this method obviously doesn't work.)"
}
{
  "Tag": [
    "Support",
    "algebra",
    "function",
    "domain",
    "\\/closed"
  ],
  "Problem": "Is it just me, or is this site becoming progressively slower by the day since the merge with mathlinks? It used to be rather fast (pages loaded within a second or two), but now it's taking in excess of 1 min 10 sec (I just stopwatched it) for pages to fully load, which is rather painful. I have a 2 megabit connection and other pages have been loading at their normal speed, so I have a feeling that the problem is not on my end.",
  "Solution_1": "Weird - I'm not seeing this at all, but a few students in class had this problem today.  How widespread is this?",
  "Solution_2": "I noticed it also but I thought it was because I had spyware or something.",
  "Solution_3": "Is this just today, or over the last week, or what?",
  "Solution_4": "It's been happening to me too...for about a week or so.",
  "Solution_5": "By the way, shouldn't this thread be in Support rather than in Questions and Suggestions? I just tried deleting my browser (Mozilla) cache, closing my browser, and starting over. The whole site, including images, loaded in ten seconds with a cable modem connection. I logged in instantly, and then I got here right away. No noticeable slow-down here. I get cable modem pool latency problems once in a while, when a lot of other subscribers are online at the same time as I, but don't notice any particular slowness to this AoPS site on a consistent basis.",
  "Solution_6": "Now it takes my computer 4-5 seconds instead of 2-3.  On my computer there isn't much of a difference.",
  "Solution_7": "Is it the whole site that is a problem, or just the message board?\r\n\r\nAre there others besides me and tokenadult who are *not* seeing any difference?",
  "Solution_8": "Also, for those of you experiencing slowness - what browser/OS are you using?",
  "Solution_9": "I am using IE 6 and I have Windows XP.  Most of the time the site is fine but every once in a while it will take a large amount of time to load something.",
  "Solution_10": "I've been having the same problems, along many pages not loading and tex not working sporadically.",
  "Solution_11": "Please note how long you've been having the problems, as well as your internet connection type and browser you are using - this will help us diagnose the problem, as we've made several server changes over the last couple weeks.  Knowing exactly when the problems happen will help us zero on which change caused them.",
  "Solution_12": "[quote=\"rrusczyk\"]Please note how long you've been having the problems, as well as your internet connection type and browser...[/quote]\r\n\r\nI'm not complaining.  But since you are asking...  The site has been noticeably slower for me, since the merge.  I can't honestly quantify the amount.  It just feels slower.  [Maybe half as fast, on average, if I had to make a guess.]\r\n\r\nSometimes -- perhaps two or three times a week -- pages will consistently take a minute or more to load.  I think I've always seen this behavior in the wee hours of the morning (USA).  The site is essentially unusable when this occurs.  Most times it seems to clear up fairly quickly (under ten minutes?)  A few times I've just given up for the night.  This never occured prior to the merge.  I think I was at least as active on this web site prior to the merge, so I think this is a new behavior.\r\n\r\nI see both of these behaviors with IE5, IE6, and NS4.7x.",
  "Solution_13": "[quote=\"rrusczyk\"]Please note how long you've been having the problems, as well as your internet connection type and browser you are using - this will help us diagnose the problem, as we've made several server changes over the last couple weeks.  Knowing exactly when the problems happen will help us zero on which change caused them.[/quote]\r\nThe speed change coincided precisely with the merge. Some days it has been a bit better, and some a bit worse, but it seems to be going slowly downhill on average since then.\r\n\r\nMy computer, etc:\r\nWin XP Home\r\nMozilla 1.6\r\n2 megabit down / 128 kilobit up cable connection through Charter Communications in Traverse City, MI\r\nMore than capable hardware to run the computer acceptably",
  "Solution_14": "Windows XP Pentium 4, AOL or Microsoft Internet Explorer.\r\n\r\nTex images aren't loading and whenever I click on basically anything, it says \"Bad Request.\"  I can get to the forums in two ways: either going to the homepage and clicking the forum button, or hitting the browser's back button.  I couldn't click the button now.",
  "Solution_15": "[quote=\"confuted\"]The speed change coincided precisely with the merge. Some days it has been a bit better, and some a bit worse, but it seems to be going slowly downhill on average since then.[/quote] Please report any abonormality in the functionality of the site as soon as it appears. The merge took place a couple of weeks ago. By reporting quickly we can operate faster in order to solve the problem. \r\n\r\nOn the other hand, it is strange, as only some people are affected by this behaviour, which seems to indicate that it is not a server problem. However you all have this since the merge and this is clue pointing that it is related closely to the server ...  :?",
  "Solution_16": "[quote=\"Valentin Vornicu\"]Please report any abonormality in the functionality of the site as soon as it appears. The merge took place a couple of weeks ago. By reporting quickly we can operate faster in order to solve the problem.[/quote]\nI actually complained about it the day of the merge to Mr. Crawford, who reassured me:\n\n[quote=\"Mathew Crawford\"]The site will get faster. We're adding RAM on the server end and much of the slowness is due to all the little quirks of combining so many things at once. Ultimately the site will probably get faster soon over the next couple of years because the influx of members means a greater influx of capital with which to make improvements.\n\nI hope you enjoy the new international base of students.[/quote]\r\n\r\nSo I held off on further complaints for a while.",
  "Solution_17": "[quote=\"rrusczyk\"]Is it the whole site that is a problem, or just the message board?[/quote]\r\nExcuse the double post, but it's the entire site, not just the message board.",
  "Solution_18": "Has it always been the entire site?  To the degree I see a slow-down, I only see it on the forum, and the merge didn't touch anything else on the site (though we have made server changes since doing the merge that might affect the whole site).",
  "Solution_19": "[quote=\"rrusczyk\"]Has it always been the entire site?  To the degree I see a slow-down, I only see it on the forum, and the merge didn't touch anything else on the site (though we have made server changes since doing the merge that might affect the whole site).[/quote]\r\nI don't know; I don't often visit the rest of the site. I come directly to the message board when I visit, without going through the \"main page.\"",
  "Solution_20": "To me the whole site is slow, whatever browser or computer I use, but I live on the east coast, maybe that has something to do with it. Interestingly, to me the mathlinks.ro website is much faster.",
  "Solution_21": "RC-7th: How do you connect to the internet?",
  "Solution_22": "Through a network, either home or work. I use cable or DSL",
  "Solution_23": "Are all pages always slow?  Sometimes slow?  Have you tried deleting your cookies? \r\n\r\n(As for the east coast - our server is on the east coast, so I don't think geography is the problem)\r\n\r\nAlso, what are you calling slow?  5 seconds?  10 seconds?  They're loading in around 1 second on cable for me, 5 seconds on dialup.",
  "Solution_24": "[quote=\"RC-7th\"]To me the whole site is slow, whatever browser or computer I use, but I live on the east coast, maybe that has something to do with it. Interestingly, to me the mathlinks.ro website is much faster.[/quote] the only real difference between the two domains is the DNS addressing. Other than that, which should not affect this, both sites should have the same speeds.",
  "Solution_25": "Did you try changing templates? Do you have the problem on the SubSilver template?"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $ G$ be a simple graph with vertex set $ V\\equal{}\\{0,1,2,3,\\cdots ,n\\plus{}1\\}$ .$ j$and$ j\\plus{}1$ are connected by an edge for $ 0\\le j\\le n$. Let $ A$ be a subset of $ V$ and $ G(A)$ be the induced subgraph associated with $ A$. Let $ O(G(A))$ be number of components of $ G(A)$ having an odd number of vertices.\r\nLet\r\n$ T(p,r)\\equal{}\\{A\\subset V \\mid 0.n\\plus{}1 \\notin A,|A|\\equal{}p,O(G(A))\\equal{}2r\\}$ for $ r\\le p \\le 2r$.\r\nProve That $ |T(p,r)|\\equal{}{n\\minus{}r \\choose{p\\minus{}r}}{n\\minus{}p\\plus{}1 \\choose{2r\\minus{}p}}$.",
  "Solution_1": "[hide=\"hint\"]\nuse lattice path. start from 0,0. at any step if next no. is taken go right, else go up.\nnext use two specific operations on the path:\n1.delete 1 right step from every odd lengthed horizontal strip to make it even lengthed.\n2.all consequtive right steps are halved,and other half is discarded.\n now find the inverse process of above operations.\n[/hide]",
  "Solution_2": "Sorry ith_power, could you please elaborate your hint ?",
  "Solution_3": "[hide]the first term is no of paths possible when both operations are done. i.e. without any restriction other that $ 0,n\\plus{}1 \\notin A$.\nthe second term is no of paths $ P$ from which we can arive at same path after first operation.[/hide]"
}
{
  "Tag": [
    "induction",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There are $ n\\geq 5$ people in a party. Assume that among any three of them some two know each other. Show that one can select at least $ \\frac{n}{2}$ people and arrange them at a round table so that each person sits between two of his/her acquaintances.",
  "Solution_1": "Use induction on $ n$ :\r\n\r\nBasis cases are straightforward.\r\n\r\nAssume the result holds for $ n$ and let's consider a collection of $ n\\plus{}2$ people.\r\nIf any two persons are acquainted the result holds trivially.\r\nAssume that people $ A$ and $ B$ are not acquainted, and using the induction hypothesis to the group of the $ n$ other people, we may choose at least $ \\frac n 2$ of them to be disposed as desired. Let $ M$ be any of these people sitting around the table, then $ M$ is acquainted to $ A$ or $ B$, so that $ A$ or $ B$, say $ A$, is acquainted to at least half of the people around the table. If to of the people around the table, sau $ C$ and $ D$ are acquainted with $ A$, then just insert $ A$ between $ C$ and $ D$.\r\nIf $ A$ is not acquainted to two neighbours around the table, then $ \\frac n 2$ is even and the people around the table are $ M_1P_1M_2P_2 \\cdots M_kP_k$ such that $ A$ is acquainted to each of the $ M_i$'s, and not to any of the $ P_i$'s thus $ B$ is acquainted to each of the $ P_i$'s. Note that $ n \\geq 5$ implies that $ k \\geq 2$.\r\nIf $ k\\equal{}2$ then we may use the arrangement $ M_1AM_2P_1BP_2M_1$.\r\nIf $ k \\geq 3$ then, some two among $ P_1P_2P_3$ are acquainted, say $ P_1$ and $ P_2$. Then, instead of $ M_1P_1M_2P_2$ then use $ M_1AM_2P_1P_2$ which gives an arrangement of $ 2k\\plus{}1$ persons, as desired.\r\n\r\nPierre.",
  "Solution_2": "if we mention the friends of A( one person in n people) group B and the others group C then because one of group C is not friends with A, each two people in C are friends and if the number of people in C is more than n/2 it's solved.\nso we mention that each person has more than n/2 friends so if we start from A and choose one of its friends and so on then after on less n/2 times (if get cycle we have to come more than n/2 times) we get to a person if it was a friend of A than it's solved if not it's from the group C and because it has more than n/2 friends it has on less one friend which is also friend with A."
}
{
  "Tag": [
    "algorithm",
    "number theory",
    "Euclidean algorithm"
  ],
  "Problem": "hey ppl ...help me with dis\r\nthis is the given problem\r\n\r\nGiven two vessels, one of which can accommodate a litres of water and the other - b litres of water, determine the minimum number of steps required to obtain exactly c litres of water in one of the vessels.\r\n\r\nAt the beginning both vessels are empty. The following operations are counted as 'steps':\r\n\r\n    * emptying a vessel,\r\n    * filling a vessel,\r\n    * pouring water from one vessel to the other, without spilling, until one of the vessels is either full or empty. \r\n\r\nInput\r\n\r\nAn integer t, 1<=t<=100, denoting the number of testcases, followed by t sets of input data, each consisting of three positive integers a, b, c, not larger than 40000, given in separate lines.\r\nOutput\r\n\r\nFor each set of input data, output the minimum number of steps required to obtain c litres, or -1 if this is impossible.\r\n\r\nin c++\r\ni have done  a bfs on the possible states ...and i am storing the distance in a map corresponding to every state.. checking if the state has occured previously in the map ..if not then updating distance\r\n\r\nnow this approach times out ..time limit 1 sec..\r\nwhat optimisation can i make\r\n\r\nbest regards\r\nvasu",
  "Solution_1": "As far as I can see, theres only two possibilities (this is a well-known problem, after all). \r\nEither:\r\nfill A, pour A->B, empty B, pour A->B. Repeat.\r\nOr the same, but starting with B.\r\n\r\nJust run them both, and see which one is shorter.",
  "Solution_2": "[quote=\"TripleM\"]As far as I can see, theres only two possibilities (this is a well-known problem, after all). \nEither:\nfill A, pour A->B, empty B, pour A->B. Repeat.\nOr the same, but starting with B.\n\nJust run them both, and see which one is shorter.[/quote]\r\nThat's an NP algorithm. There is a better way. I can transfer water from one container to another. I can also compute $x-(Y-y)$ where $Y$ is the capacity of a container, and $y$ the amount it holds. These are the operations the Extended Euclidean Algorithm uses. So now we have to translate back to pours. This is pretty easy.",
  "Solution_3": "Yeah, that would be the faster (algorithmically) way to do it, but with such small bounds on the sizes of the two vessels, either way works fine. (my way got AC in 0.33 seconds, and thats in java!)",
  "Solution_4": "i just got a TLE trying to run the same algorithm as stephen's....probably i am doing the some stupid mistake\r\njust tel me if i have got it right(ur algo that is)\r\nwe run thru the set of states starting with(a,0) and pouring from 1st to 2nd always..and emptying 2nd if it is full..filling 1st if its empty\r\n\r\nand then run thru set of states starting with(0,b) and pouring from 2nd to 1st always..emptying 1st when its full...and filling 2nd when its empty\r\n\r\nam i right??\r\nand one more thing ..just chekd spoj status..does the test data really require the huge amt of memory ur program uses..or is it because of java...(this might sound like a dumb question:) ...anyways)\r\n\r\nthx",
  "Solution_5": "Yup, thats all I did.\r\nAnd yeah, java uses a huge amount of memory for all sorts of weird stuff, its not really the program.",
  "Solution_6": "Can you elaborate more on that answer?\r\n\r\nI deeply apologize for my incompetence, but I don\u00b4t understand what you meant, starting by the meaning of the x."
}
{
  "Tag": [],
  "Problem": "Prije par dana sam obrisao sa na\u0161eg podforuma ovakvu temu\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=248100\r\n\r\nu dogovoru sa ostalim moderatorima (direktiva s vrha je da se ovakvi crankovi bri\u0161u). Ko \u017eeli da pogleda rad, na gore navedenom threadu se nalazi link. Za\u0161to ovo pi\u0161em:\r\n\r\nNisam znao kakvoj nau\u010dnoj veli\u010dini se zamjerim! http://uncyclopedia.wikia.com/wiki/Tsolkas  :rotfl:",
  "Solution_1": "Zanimljivo...\r\nAli, 2. link svako moze da edituje i da stavi svoj tekst, pa ne dokazuje tacnost, a 1. link treba detaljno procitati i onda se moze zakljuciti da li ima greske, ili je dokaz tacan.",
  "Solution_2": "2. link je poprili\u010dno duhovit (mada Uncyclopedia ima mnogo smje\u0161nijih \u010dlanaka, dakako), dok je prvi prevelik \u010ditala\u010dki zalogaj za mene. Naime, pro\u010ditao sam rad, ali nisam na\u0161ao previ\u0161e smisla (priznajem, nisam se ni trudio ba\u0161 previ\u0161e da ga na\u0111em). Na svakom nau\u010dnom forumu postoje ljudi koji su sigurni da su dokazali da je Einstein pogrije\u0161io, da imaju dokaz FLT na pola strane, dokaz Riemannove hipoteze na dvije strane i sl (dru\u0161tvo sa elite securityja ili math.e foruma \u0107e se sjetiti Spre\u010de). A ovaj Grk je dodatno nanervirao moderatore postaju\u0107i svoj link po cijelom forumu, te ulaze\u0107i u sva\u0111u s pojedinim moderatorima... Ovaj thread sam postavio tek toliko da se ne osjetite zakinutim u odnosu na ostale nacionalne podforume na kojim thread (jo\u0161 uvijek) nije obrisan.",
  "Solution_3": "Vi vjerojatno niste upoznati s Gezom... autorom, pogledao sam, 3417 postova na forum.hr. E, uglavnom, taj je (na\u017ealost, nai\u0161ao sam na njegove postove tek nakon \u0161to je prestao objavljivati) imao sve mogu\u0107e teorije... to trebate pro\u010ditati. No, posebnu je zainteresiranost pokazivao za broj 4 za koji je zaklju\u010dio da upravlja prirodom, \u0161to li.\r\n\r\nNaravno, volio je tetraedre (napisao je i knjigu o njihovim svojstvima) i, po njemu, postoje \u010detiri spola \u010dovjeka.\r\n\r\nOd svoje daljnje djelatnosti, tvrdio je da je 1:2 oznaka za dvije polovine i da je to strogo ve\u0107e od 1. Mislim da je imao i ideju da matrice zapravo u sebi nu\u017eno imaju neko dublje zna\u010denje (i barem \u010detiri dimenzije). Volio je velike rije\u010di i pisao ovakve izjave: \"[i]... DOVOLJAN JE MOJ PRIMJER. \u0160TO GOD SAM PAMETNO NAPISAO KAO PAMET SVJETSKE ZNANOSTI POPLJUVANO JE OD REDUKCIONISTA I NEZNALICA \u0160TO JE ZNANOST. \nTEK KADA ZNANOST POSTANE SLOBODNA UMNOST TEK TADA JE MOGU\u0106 POVRATAKA ZNANOSTI. SVAKO HERMETIZIRANJE, SHEMATIZIRANJE I TVR\u0110ENJE DA POSTOJI ZNANSTVENA ISTINA DO KOJE SE DOLAZI SAMO EKSPERIMENTOM JE TU\u017dALJKA ZNANOSTI.[/i]\"\r\n\r\nJo\u0161 \u0107u vas ostaviti s ovom njegovom misli: (na pitanje kakvi su ljudi kao on) \"[i]Visoki, elegantni, kao mu\u0161karci lijepi, nao\u010diti, obrazovani, samoobrazovani, dru\u017eeljubivi, pokretljivi, dobrotvori, plemeniti, po prirodi istra\u017eiva\u010dii i kreatori, u zvanju i karijeri uspje\u0161ni, imaju veliki radni i kretaivni kapacitet, dobri su ronioci, ribari, skiperi, plesa\u0107i, ljubavnici, pijanci, vlo dobro kuhaju, razumiju se prakti\u010dki i sve vje\u0161tine ...., jedina im je velika mana \u0161to nemaju sluha.[/i]\"\r\n\r\nUostalom, upi\u0161ite u Google geza site:forum.hr pa se smijte. Ja se smijem ve\u0107 dvije godine.\r\n\r\nSve gornje izjave mogu se prona\u0107i na gorespomenutom sajtu."
}
{
  "Tag": [
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "For $  f:[0,1]\\to [0,1] $ continuous on [0,1] let    $ J_n(f) : = \\int\\limits_{0}^1 |f(x)|^n \\;dx \\; ,\\; \\; \\; (n\\ge 1) ,   J_0(f):=\\frac{1}{2} , $ and suppose $ J_1(f) \\ne 0  . $ \r\nBy supposing that   $ n\\cdot J_n(f)=(n-1)\\cdot J_{n-2}(f) $  ,$ n=2,3,... , $[b] to find [/b]  $ \\int\\limits_{0}^1 f(x)\\;  dx\\;    . $",
  "Solution_1": "[quote=\"flip2004\"]  $ J_n(f)  = \\int\\limits_{0}^1 |f(x)|^n dx$\n$  J_0(f):=\\frac{1}{2}$ [/quote]\r\n\r\nWith the definition of $J_n(f)$ we have $J_0(f)=\\int\\limits_{0}^1 1dx=1$",
  "Solution_2": "$J_n$ has the same property of Wallis integral \r\n$J_{n+1}\\leq J_n$,  $J_{2n}=\\frac{(2n)!}{(2^n.n!)^2}J_0$, $J_{2n+1}=\\frac{(2^n.n!)^2}{(2n+1)!}J_1$, $(n+1)/(n+2)=\\frac{J_{n+2}}{J_n}\\leq \\frac{J_{n+1}}{J_n}\\leq1$ we get $J_{2n+1}/J_{2n}$ tends to $1$ when n tends \r\n$+\\infty$ with Stirling formulae we get $J_1$",
  "Solution_3": "The definiton of $ J_n(f) $ as an integral is valid  for $ n\\ge 1 $ . However, thanks for remark.Alex"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Can any of you find f(x,y,z) that the following ineq is right where x,y,z\u22650:\r\n$ (x^2\\plus{}xy\\plus{}y^2)(y^2\\plus{}yz\\plus{}z^2)(z^2\\plus{}zx\\plus{}x^2)$\u2265f(x,y,z)\r\nI found that f(x,y,z)=$ \\frac{(x\\plus{}y\\plus{}z)^2(xy\\plus{}yz\\plus{}zx)^2}{3}$\r\nI am needing another f(x,y,z) that is stronger than this one.Let me know your idea",
  "Solution_1": "It is rather hard to understand what exactly you are looking for. Why, for example, not to take $ f(x,y,z)$ to be just the left hand side? :?",
  "Solution_2": "$ f(x, y, z)\\equal{}\\frac{27}{64}(x\\plus{}y)^2(y\\plus{}z)^2(z\\plus{}x)^2$  :maybe:"
}
{
  "Tag": [
    "search",
    "puzzles"
  ],
  "Problem": "Think of one word for each of the following properties.\r\n\r\n1) This 8-letter word has five vowels in a row.\r\n2) This 9-letter word has exactly one vowel (this includes y)\r\n3) This 10-letter word can be typed on the top row of a standard QWERTY typewriter.\r\n4) This word has all of the vowels in order (a,e,i,o,u).\r\n5) This word has all of the vowels out of order (a,e,i,o,u), but is only 7 letters long.\r\n6) This ten letter word has 1 of one letter, 2 of another letter, 3 of another letter, and 4 of another letter.",
  "Solution_1": "*** Spoilers Ahead ***\r\n.......partial list of answers\r\n>\r\n>\r\n>\r\n>\r\n.\r\n.\r\n.\r\n.\r\n.>\r\n>\r\n>\r\n>\r\n>\r\n1) queueing\r\n\r\n2) strengths\r\n\r\n3) typewriter\r\n\r\n4) facetious\r\n\r\n5) \r\n\r\n6)\r\n>",
  "Solution_2": "I'm sure queueing isn't a word",
  "Solution_3": "On Today, at 10:31 am ; junggi wrote:\r\n[quote]I'm sure queueing isn't a word[/quote]\r\n\r\nTrue. According to dictionary.com queueing is not a proper word. \r\n\r\nMy attempt at solving the problem is given below.\r\n\r\n[hide]PARTIAL ANSWERS:\n\n1) Not found a solution. Solution with 7 letters: queuing.\n\n2) strengths \n\n3) typewriter, proprietor, repertoire\n\n4) abstemious,  abstemiously,  abstentious,  acheilous,  acheirous,  acleistous,  affectious,  aleikoum,  anemious,  annelidous,  arsenious,  arteriosus,  arterious,  bacterious,   facetious,  facetiously,  fracedinous,  majestious,  parecious,  pareciously,  tragedious \n\n5) Sequoia, Oiseaux\n\nNote:  Till date, I have not been able to solve (6)[/hide]",
  "Solution_4": "[quote=\"junggi\"]I'm sure queueing isn't a word[/quote]\r\nHowever, [url=http://www.google.com/search?hl=en&q=queueing&btnG=Google+Search]Google[/url] and [url=http://en.wikipedia.org/wiki/Queuing_theory]wikipedia[/url] don't seem to have much of a problem with it. In fact, wikipedia even redirects \"queuing theory\" to \"queueing theory\".\r\n\r\n\r\n(6) is fun.",
  "Solution_5": "randomdragoon wrote:\r\n\r\n[quote](6) is fun[/quote]\r\n\r\nSolution to (6):\r\n\r\n[hide]SLEEVELESS -----> This word contains precisely one 'V', two 'L's, three 'S's and four 'E's.[/hide]",
  "Solution_6": "The instructions at the top of the first post include:\r\n\r\n\"Think of one word...\"  \r\n\r\nFor any entry of 1) - 6) you should have THAT disqualified from counting for NOT following directions.  I don't care if more than one word fits the other criteria; it's a puzzle or game, and there are rules to be upheld.\r\n\r\nK. Sengupta, for each numbered entry where you entered more than one word, that numbered entry should not count.\r\n\r\nFollowing directions is part of puzzle solving.",
  "Solution_7": "box of rocks, are you [b]joking[/b]? \r\n\r\nAnother fun but slightly easier one:\r\nFind a word (any length) with 6 consecutive consonants. (where y counts as a vowel).",
  "Solution_8": "On #3, \"typewriter\" is not a coincidence- it's marketing. The QWERTY keyboard was originally designed to slow down typists so that keys wouldn't get stuck. It's impractical with modern equipment, but persists because everyone's familiar with it.\r\n\r\nTo \"box of rocks\": Don't be so obnoxious. In this context, there's nothing wrong with finding multiple answers.\r\n\r\nNot a real answer to #6: \"these teeth\"",
  "Solution_9": "tim1234133,\r\n\r\nno, I'm not joking.  The user who calls my correct answer wrong will learn about following directions.\r\n------------------------------------------------------------\r\n\r\njmerry,\r\n\r\nyou have an insecurity about following instructions too?\r\n\r\nThere is absolutely nothing obnoxious about what I've done.  It's called reading the directions and adhering to them.  You're \r\nthe person having trouble with that.  There is no \"in this context.\"  You give one word or you don't.\r\n\r\nWhen I make up a puzzle, I don't want multiple answers given with \"one answer.\"  That's plain.  If I want multiple answers, THEN I'll type that.\r\n\r\nDon't affix the name \"obnoxious\" to a user; it shows you are being a name-caller yourself, which is rude.",
  "Solution_10": "Please stop accusing and actually get back to the problem:\r\n\r\n[quote=\"tim1234133\"]Find a word (any length) with 6 consecutive consonants. (where y counts as a vowel).[/quote]\r\n\r\n[hide]Rhythms, dysrhythmic, dysthymia...[/hide]",
  "Solution_11": "[quote=\"mathnerd314\"]Please stop accusing and actually get back to the problem:\n\n[quote=\"tim1234133\"]Find a word (any length) with 6 consecutive consonants. (where y counts as a vowel).[/quote]\n\n[hide]Rhythms, dysrhythmic, dysthymia...[/hide][/quote]y counts as a vowel?",
  "Solution_12": "I can't get the six consecutive consonants... but here is another one\r\n\r\n\r\nFind a 7-letter word that has exactly one vowel (where y counts a vowel), but also has no \"s\".",
  "Solution_13": "box of rocks wrote:\r\n[quote]K. Sengupta, for each numbered entry where you entered more than one word, that numbered entry should not count. [/quote]\r\n\r\nMy opinion is at direct variance with the foregoing observation.\r\nIt would be observed that  if one is asked to write just one number satisfying x^2-5x+6 = 0; then x=2,3 is as valid as a response as x=2 or x=3. Accordingly,for any given query possessing multiple answers, it is NOT improper to write more than one answer.\r\n\r\nConsequiently, I do not agree at all with the foregoing observation since my response is fully in conformity with the provisions governing this sub-forum.",
  "Solution_14": "Also, remember that if you write three words, you have also written one word. I did not use the word \"exactly\".",
  "Solution_15": "[quote=\"randomdragoon\"]I can't get the six consecutive consonants... but here is another one\n\n\nFind a 7-letter word that has exactly one vowel (where y counts a vowel), but also has no \"s\".[/quote]\r\n\r\nscratch",
  "Solution_16": "Scratch has an \"s\" in it. Try again."
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "circumcircle",
    "inequalities proposed"
  ],
  "Problem": "Prove that in any triangle $ ABC$ exists the inequality\r\n\r\n$ \\boxed {\\sqrt {\\frac {R \\plus{} 2r}{R}}\\le\\frac {OH}{OI}\\le\\sqrt {\\frac {3(3R \\minus{} 2r)}{R}}}\\ .$",
  "Solution_1": "[quote=\"Virgil Nicula\"]Prove that in any triangle $ ABC$ exists the inequality $ \\boxed {\\frac {OH}{OI}\\ge \\sqrt {\\frac {R + 2r}{R}}}\\ .$[hide=\"Hint.\"]\n${ \\frac {OH}{OI}\\ge \\sqrt {\\frac {R + 2r}{R}}}\\Longleftrightarrow R^2\\le OH^2 + 4r^2\\Longleftrightarrow p^2\\le 4R^2 + 4Rr + 3r^2\\ .$ This means that\n\nthe power of the orthocenter $ H$ w.r.t. the circumcircle of $ \\triangle ABC$ is at least equal to $ - 4r^2\\ .$[/hide][/quote]\r\nI think it's Gerretsen inequality, we can prove it if uses fomula: $ IH^2=(4R^2+3r^2+4Rr)-p^2 \\ge 0$",
  "Solution_2": "[quote=\"Virgil Nicula\"] [color=darkred]Prove that in any triangle $ ABC$ exists the inequality $ \\boxed {\\sqrt {\\frac {R + 2r}{R}}\\le\\frac {OH}{OI}\\le\\sqrt {\\frac {3(3R - 2r)}{R}}}\\ .$[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof.[/u][/b] The relations $ \\|\\begin{array}{c} a^2 + b^2 + c^2 = p^2 + r^2 + 4Rr \\\\\n \\\\\nOI^2 = R^2 - 2Rr \\\\\n \\\\\nOH^2 = 9R^2 - (a^2 + b^2 + c^2) \\\\\n \\\\\nIH^2 = 4R^2 + 4Rr + 3r^2 - p^2 \\\\\n \\\\\n9\\cdot IG^2 = p^2 + 5r^2 - 16Rr\\end{array}$ are well-known. Thus, \n\n${ \\frac {OH}{OI}\\ge \\sqrt {\\frac {R + 2r}{R}}}\\Longleftrightarrow R^2\\le OH^2 + 4r^2\\Longleftrightarrow p^2\\le 4R^2 + 4Rr + 3r^2\\ .$\n\nThe power of $ H$ w.r.t. the circumcircle of $ \\triangle ABC$ is at least equal to $ - 4r^2\\ .$\n\n[b][u]Remark.[/u][/b] Prove easily that $ \\boxed {R\\cdot OH^2 = 2R\\cdot IH^2 + (R + 2r)\\cdot OI^2}\\ .$\n\nNow the proposed left inequality is evidently !\n\nApply the [b]Stewart's theorem[/b] to the cevian $ [IG]$ in $ \\triangle HIO$ :\n\n$ 3\\cdot IH^2 + 6\\cdot OI^2 = 9\\cdot IG^2 + 2\\cdot OH^2\\ .$  Using and the upper identity eliminate :\n\n$ \\|\\begin{array}{ccc} OH\\ \\nearrow & \\implies & 4\\cdot (R - r)\\cdot OI^2 = 9R\\cdot IG^2 + R\\cdot IH^2 \\\\\n \\\\\nIH\\ \\nearrow & \\implies & 3(3R - 2r)\\cdot OI^2 = 18R\\cdot IG^2 + R\\cdot OH^2 \\\\\n \\\\\nOI\\ \\nearrow & \\implies & 4(R - r)\\cdot OH^2 = 3(3R - 2r)\\cdot IH^2 + 9(R + 2r)\\cdot IG^2\\end{array}$\n\nThus, $ \\|\\begin{array}{c} \\frac {IH}{IO}\\le2\\cdot\\sqrt {1 - \\frac rR} \\\\\n \\\\\n\\frac {IG}{IO}\\le\\frac 23\\cdot\\sqrt {1 - \\frac rR} \\\\\n \\\\\n\\frac {HI}{HO}\\le \\sqrt \\frac {4(R - r)}{3(3R - 2r)}\\end{array}$ and $ \\sqrt {\\frac {R + 2r}{R}}\\le\\frac {OH}{OI}\\le\\sqrt {\\frac {3(3R - 2r)}{R}}\\ .$\n\n[b]Remark.[/b] Prove easily that $ 9\\cdot(IG^2+IH^2-GH^2)=-12r(R-2r)\\le 0$,\n\ni.e. $ \\boxed {m(\\widehat {GIH})\\ge 90^{\\circ}}$ and $ \\frac {2\\cdot IO^2+IH^2}{2\\cdot GO^2+GI^2}=3\\ .$[/color]"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "If $x_1 ,x_2, \\ldots ,x_n$  are the roots of the equation \\[x^n\\plus{}p_1 x^{n\\minus{}1}\\plus{}p_2 x^{n\\minus{}2}\\plus{} \\cdots \\plus{}p_{n\\minus{}1}x\\plus{}p_n\\equal{}0,\\] where $p_1,p_2, \\ldots, p_n \\in \\mathbb R$. Prove that \\[ (1\\plus{}x_1 ^2)(1\\plus{}x_2 ^2)\\cdots (1\\plus{}x_n ^2)\\equal{}(1\\minus{}p_2\\plus{}p_4\\minus{}p_6\\plus{}\\cdots)^2\\plus{}(p_1\\minus{}p_3\\plus{}p_5\\minus{}p_7\\plus{}\\cdots)^2.\\]",
  "Solution_1": "[quote=\"jackie-chan01\"]$ If x_1 ,x_2,...,x_n \\ are \\ nth \\ roots \\ of \\ equation \\ x^n \\plus{} p_1 x^{n \\minus{} 1} \\plus{} p_2 x^{n \\minus{} 2} \\plus{} ... \\plus{} p_{n \\minus{} 1}x \\plus{} p_n \\equal{} 0 \\ where \\ p_1,p_2...p_n \\in R$ .\nProve that $ (1 \\plus{} x_1 ^2)(1 \\plus{} x_2 ^2)..(1 \\plus{} x_n ^2) \\equal{} (1 \\minus{} p_2 \\plus{} p_4 \\minus{} p_6 \\plus{} ...)^2 \\plus{} (p_1 \\minus{} p_3 \\plus{} p_5 \\minus{} p_7 \\plus{} ..)^2$[/quote]\r\n\r\nLet $ P(x)\\equal{}\\prod_{k\\equal{}1}^n(x\\minus{}x_k)\\equal{}x^n\\plus{}p_1x^{n\\minus{}1}\\plus{}p_2x^{n\\minus{}2}\\plus{}...\\plus{}p_n$\r\n\r\n$ P(i)\\equal{}i^n(1\\minus{}ip_1\\minus{}p_2\\plus{}ip_3\\plus{}p_4\\minus{}ip_5 ...)$\r\n$ P(\\minus{}i)\\equal{}(\\minus{}1)^ni^n(1\\plus{}ip_1\\minus{}p_2\\minus{}ip_3\\plus{}p_4\\plus{}ip_5...)$\r\n\r\nSo $ 1\\minus{}p_2\\plus{}p_4\\minus{}p_6...\\equal{}\\frac{P(i)\\plus{}(\\minus{}1)^nP(\\minus{}i)}{2i^n}$ and $ p_1\\minus{}p_3\\plus{}p_5\\minus{}p_7...\\equal{}\\frac{P(i)\\minus{}(\\minus{}1)^nP(\\minus{}i)}{(\\minus{}2i)i^n}$\r\n\r\nSo $ (1\\minus{}p_2\\plus{}p_4\\minus{}p_6..)^2\\plus{}(p_1\\minus{}p_3\\plus{}p_5\\minus{}p_7...)^2$ $ \\equal{}\\frac{(P(i)\\plus{}(\\minus{}1)^nP(\\minus{}i))^2}{4(\\minus{}1)^n}$ - $ \\frac{(P(i)\\minus{}(\\minus{}1)^nP(\\minus{}i))^2}{4(\\minus{}1)^n}$\r\n\r\nSo $ (1\\minus{}p_2\\plus{}p_4\\minus{}p_6..)^2\\plus{}(p_1\\minus{}p_3\\plus{}p_5\\minus{}p_7...)^2$ $ \\equal{}P(i)P(\\minus{}i)$ $ \\equal{}\\prod_{k\\equal{}1}^n(i\\minus{}x_k)\\prod_{k\\equal{}1}^n(\\minus{}i\\minus{}x_k)$ $ \\equal{}\\prod_{k\\equal{}1}^n(1\\plus{}x_k^2)$\r\n\r\nQ.E.D.",
  "Solution_2": "thank you  :)"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "perpendicular bisector",
    "geometry proposed"
  ],
  "Problem": "According to the estimated number of participants who gave a correct solution, this was the hardest (!) problem from today's paper. So here is this great German killer - be warned!\r\n\r\nGiven a circle k and three pairwisely distinct points A, B, C on this circle. Let h and g be the perpendiculars to the line BC at the points B and C. The perpendicular bisector of the segment AB meets the line h at a point F; the perpendicular bisector of the segment AC meets the line g at a point G.\r\n\r\nProve that the product $BF\\cdot CG$ is independent from the position of the point A, as long as the points B and C stay fixed.\r\n\r\nThe actual problem behind the problem: Why on hell should the points B and C stay fixed?\r\n\r\n  Darij",
  "Solution_1": "Wow, the German olympiads are SO tough these days :D\r\n[That's called sarcasm, for those who didn't realize that ;)]\r\n\r\nIt is really obvious that BF = cR/b with R the radius of the circumcircle, and CG = bR/c, so that the product equals R<sup>2</sup>  (for all possible positions of A, B, C)."
}
{
  "Tag": [
    "function",
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "for every closed set $ E$  on real line, is there a differentiable function $ f$ which satisfies $ f^ {\\minus{} 1}\\{0\\} \\equal{} E$?",
  "Solution_1": "0\r\n\r\n :clap2:",
  "Solution_2": "thanks\r\nbut i don't know what you mean",
  "Solution_3": "I think SouthStar interpreted the question to be \"Is there a differentiable function $ f: E\\to \\mathbb{R}$, ...\". Then $ f\\equiv 0$ works. I'm assuming the question is actually $ f: \\mathbb{R}\\to\\mathbb{R}$.",
  "Solution_4": "Ooooh. Couldn't have been that easy. It was 2 am. :)\r\n\r\nI have seen some results for infinitely differentiable functions but not for this general case.",
  "Solution_5": "Define $ \\varphi(x)\\equal{}\\begin{cases}x^2(1\\minus{}x)^2& x\\in(0,1)\\\\0&\\text{otherwise}\\end{cases}.$\r\n\r\nNote that $ \\varphi$ is differentiable and that $ \\varphi(x)\\ne0$ precisely on $ (0,1).$\r\n\r\nAny open set in $ \\mathbb{R}$ can be written as a countable disjoint union of open intervals, so let $ \\mathbb{R}\\setminus E\\equal{}\\bigcup_{k\\equal{}1}^{\\infty}I_k.$\r\n\r\nIf there are any infinite intervals there, pull them aside to be treated separately. Write $ I_k\\equal{}(a_k,b_k).$\r\n\r\nDefine $ f(x)\\equal{}\\sum_{k\\equal{}1}^{\\infty}2^{\\minus{}k}(b_k\\minus{}a_k) \\varphi\\left(a_k\\plus{}\\frac{x\\minus{}a_k}{b_k\\minus{}a_k}\\right)$\r\n\r\nIf we have an interval of the form $ (a,\\infty),$ then add $ (x\\minus{}a)^2$ to $ f(x)$ for $ x>a.$\r\n\r\nIf we have an interval of the form $ (\\minus{}\\infty,b),$ then add $ (b\\minus{}x)^2$ to $ f(x)$ for $ x<b.$\r\n\r\nThe coefficients $ 2^{\\minus{}k}(b_k\\minus{}a_k)$ were added to assure that the series for $ f'(x)$ would converge uniformly so that $ f$ would be differentiable. We have a differentiable function $ f$ which is zero on $ E$ and strictly positive on the complement of $ E.$\r\n\r\nWe can see how to adjust this proof to get a $ C^{k}$ function - all that requires is a modification of the definition of $ \\varphi.$ With just a few more adjustments, both to $ \\varphi$ and to the coefficients of the sum, we can make a $ C^{\\infty}$ function.\r\n\r\nBut this proof doesn't generalize so easily to $ \\mathbb{R}^{n}$ for $ n>1,$ since I used a specifically $ \\mathbb{R}^1$ characterization of open sets.",
  "Solution_6": "[quote=\"Kent Merryfield\"]\nBut this proof doesn't generalize so easily to $ \\mathbb{R}^{n}$ for $ n > 1,$ since I used a specifically $ \\mathbb{R}^1$ characterization of open sets.[/quote]\r\nActually, it does: of course, it has been used that every open set is a [i]countable union[/i] of open intervals, but the fact that those intervals are [i]disjoint[/i] hasn't been really used anywhere ;)."
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $S=\\{1,2,\\cdots,40\\}$. Find the minimum value of $n$ such that in any $n$ element subset $P$ of $S$ there exists distinct $a,b,c,d$ in $P$ with $a+b=c+d$.\r\n\r\ni discovered a general formula but i'm not sure if its correct. because if we take 20 instead of 40 my formula gives n=9, but i think that it should be n=8. if you want i can post my formula.",
  "Solution_1": ":blush: sorry my formula had some mistakes. now it gives n=8 in case of 20. why hasn't anyone posted a solution yet? is it too easy or too hard? will i post my solution? say something..."
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Calculate the following integrals: (in the first dx is omited)",
  "Solution_1": "$ \\int\\frac{\\ln{e^{x\\plus{}1}}}{e^{2x}}\\ \\mathrm{d}x\\equal{}\\minus{}\\frac{e^{\\minus{}2x}}{4}(2\\ln{e^{x\\plus{}1}}\\plus{}1)$.",
  "Solution_2": "what about the second? I'm not sure if in the first integral it is not: $ ln(e^{x}\\plus{}1)$ instead of $ ln(e^{x\\plus{}1})$. I found these integrals in Internet and I show them here, because I think they are interesting.",
  "Solution_3": "[hide=\"Number 2\"]Integrating by parts,\n\n$ \\int \\frac{\\ln(1 \\plus{} \\sqrt{x})}{(1\\plus{}x)^2}\\,dx \\equal{} \\underbrace{\\minus{} \\frac{\\ln(1 \\plus{} \\sqrt{x})}{1\\plus{}x}}_{\\equal{} A} \\plus{} \\underbrace{\\frac{1}{2} \\int\\frac{dx}{\\sqrt{x}(1 \\plus{} \\sqrt{x})(1\\plus{}x)}}_{make\\,\\,x \\equal{} t^2} \\equal{}$\n\n$ \\equal{} A \\plus{} \\int\\frac{dt}{(1\\plus{}t)(1\\plus{}t^2)} \\equal{} A \\plus{} \\frac{1}{2} \\int \\left(\\frac{1}{1\\plus{}t} \\minus{} \\frac{t}{1\\plus{}t^2} \\plus{} \\frac{1}{1\\plus{}t^2}\\right)\\,dt \\equal{}$\n\n$ \\equal{} A \\plus{} \\frac{1}{2} \\ln(1\\plus{}t) \\minus{} \\frac{1}{4} \\ln(1\\plus{}t^2) \\plus{} \\frac{1}{2} \\arctan t \\plus{} C$.\n\nYou can finish from here.[/hide]",
  "Solution_4": "Thank you guys, excellent solutions!"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "For each positive integer number n let A be set of all positive integers x such that gcd(x,n)>1\r\nn is called good if for any x,y in A then x+y also in A. Find all good number.\r\n       ( Sorry,my English is bad  :| )",
  "Solution_1": "[quote=\"Nbach\"]For each positive integer number $n$, let $A_{n}$ be the set of all positive integers $x$ such that $\\gcd(x,n)>1$: $n$ is called good if, for any $x,y \\in A_{n}$, then also $x+y \\in A_{n}$. Find all good number.[/quote]\r\nObviously $n = 1$ isn't good. So suppose $n \\ge 2$. If $n = p^{k}$, where $p \\in \\mathbb{N}$ is prime and $k$ is any positive integer, then $\\gcd(x,n) > 1$ means that $p \\mid x$, whenever $x \\in \\mathbb{N}$. Hence $p \\mid (x+y)$, if $x, y \\in A_{n}$, proving that any prime power is good. Now suppose that $n$ has at least two distinct prime factors $p, q \\in \\mathbb{N}$. Then there exists a minimal $m \\in \\mathbb{N}^{+}$ such that $\\gcd(mq+p,n) = 1$, which eventually implies that $n$ isn't good."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "$ a_1,a_2,...,a_n$ is a sequence of real numbers. $ a_{i \\plus{} j} \\le a_i \\plus{} a_j$ for every natural $ i,j; i \\plus{} j \\le n$  .prove that  $ a_1 \\plus{} \\frac {a_2}{2} \\plus{} \\frac {a_3}{3} \\plus{} ... \\plus{} \\frac {a_k}{k} \\ge a_k$ for every natural $ k \\le n$",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?p=456607#456607"
}
{
  "Tag": [],
  "Problem": "How do you think Armaggedon will be?\r\nI'm looking for different scenarios so don't be afraid to post your ideas.",
  "Solution_1": "It will be terrifying with a nice scenario, loud music and many people crying  :( \r\nHiiiiiiiii, Haaaaaaa.\r\n\r\n\r\nBUT !! Don't worry man, USA president, Schwarzie, will save all of us with his big muscles  :D",
  "Solution_2": "I understand you foretell a sudden, but expected, end for mankind. Something like an asteroid or comet.",
  "Solution_3": "hey the movie is quite good (armaggedon :D).",
  "Solution_4": "The movie was good, the end was good, but Bruce Willis(with his big muscles, and he is not even president  :D ) could not have missed the chance to save the world.",
  "Solution_5": "Would you? :)",
  "Solution_6": "Of course not, but I don't uderstand why couldn't Ben Affleck have saved the day.",
  "Solution_7": "Three world superpowers will rise (Asia(Asian Empire), Russia+Germany(Russian Empire), and Europe+America(Allied Forces)). They will fight a huge nuclear war(World War 3) which will destroy near all the planet. The Russians have the best tacticians in the world(chess players like Karpov and Kasparov) and the best machines from German car companies. The Russians will win. Iceland will be searched, when Bobby Fischer is found he will be executed and the Russians will be the Best at Chess Forever. The Russians will create a huge superpower spanning over the entire world, everyone will be schooled, live, and die Russian. The Russians will have a Communist Government. Now that the Caribbean, Taiwan, Japan, and all the Islands north of Australia sank, South America is no more, most of China is a feast for scavengers, and whats left of America and Europe are the two biggest trophies, Russia will have full dominance. This will leave three occupied continents: Russia(the habitable part of Asia), Europe, and North America. Then they will split and have another war(World War 4) which will destroy all but Russia. Then the sun will explode and the world will be lost forever....",
  "Solution_8": "Here is a good (and hilarious) scenario of how the world will end:\r\nhttp://www.albinoblacksheep.com/flash/end.php",
  "Solution_9": "i haven't seen any chuck norris bs yet\r\n\r\nWHAT IS WRONG WITH YOU PEOPLE",
  "Solution_10": "[hide=\"my prediction\"]A chinese Hitler will gain control of China, start shipping out poisonous stuff, like toothpaste, or Water guns, and Chaos will ensue. Then China will just go out there, and capture all of the world, and then Communism will ensue, And communism=bad. Then some random person will realize \"OMG!!!!!!!!!\" Then the 7th year of the Tribulation will ensue.[/hide]",
  "Solution_11": "Has anyone else noticed that this thread is about 4 years old?",
  "Solution_12": "lol I was about to wonder how guests can post  :D",
  "Solution_13": "lol apprently its not really called \"armageddon\" in scientific terms.  its actually global killer (or something like that).  i took a science pre test for some disasters unit and i put armageddon as my answer and it was wrong..."
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $n \\geq 1$ be a natural number. And let $a_n$ be the number of different possibilities to fill a  $2 \\times 2 \\times n$ cuboid by means of $1 \\times 1 \\times 2$ cuboids which are positioned in a axis-parallel way. Determine $a_n$.",
  "Solution_1": "Isn't the recursion straightforward?\r\n\r\n$a_{n+2}=2a_{n+1}+5a_n$, I think (there might be something I'm missing, of course :)).",
  "Solution_2": "[quote=\"grobber\"]Isn't the recursion straightforward?\n\n$a_{n+2}=2a_{n+1}+5a_n$, I think (there might be something I'm missing, of course :)).[/quote]\r\n\r\nI think you got wrong. Please rethink.  :)  And be so kind and find an explicit formula of $a_n$.",
  "Solution_3": "Yup, I realized what I did wrong. Sorry about that, I'll rethink :).",
  "Solution_4": "How's this: $a_{n+2}=2a_{n+1}+a_n+4\\sum_{i=0}^n a_i$, where we define $a_0=1$. Still wrong?  :blush:\r\n\r\n[Edit: actually, from that recurrence formula, we get the nicer one $a_{n+3}=3a_{n+2}+3a_{n+1}-a_n,\\ \\forall n\\ge 0$ by subtracting $a_{n+2}$ from $a_{n+3}$]",
  "Solution_5": "I give you another try.  ;)  A hint the recursive formula is something like:\r\n\r\n\\[a_n = u \\cdot a_{n-1} + v \\cdot a_{n-2} + w \\cdot a_{n-3}\\] with $u,v,w \\in \\mathbb{Z}$. This formula is valid for most cases. Two special case considerations are necessary.  :)",
  "Solution_6": "I edited my post before seeing the hint :D. Besides, orl, the fact that a recurrence formula is different from the one you had in mind when you initiated the topic, doesn't necessarily make that particular recurrence formula incorrect.  ;)",
  "Solution_7": "You are right. I must say: I am impressed. 4000 postings and most of them really on mathematics. That is really an :alien: performance. !!! Keep up the good work.  :)",
  "Solution_8": "Thank you, that's very kind of you :).\r\n\r\nIf I am not mistaken (again :D), the closed form should be something like $a_n=\\frac{2(-1)^n+(2-\\sqrt 3)^{n+1}+(2+\\sqrt 3)^{n+1}}6$. Even if this is not exactly accurate due to computational mistakes, it can be obtained in the usual manner: $a_n=\\sum_{i=1}^3 \\alpha_i\\cdot r_i^n$, where $r_1,r_2,r_3$ are the three roots of $x^3-3x^2-3x+1$, the characteristic equation of the recurrence. From the first three terms, we compute $\\alpha_i$, and I found them to be $\\frac 13,\\frac{2-\\sqrt 3}6,\\frac{2+\\sqrt 3}6$ for $r_1=-1,r_2=2-\\sqrt 3,r_3=2+\\sqrt 3$.",
  "Solution_9": "Your expicit formula is perfectly right. Then we just would like to know how you obtained the recurrence formula.  :lol:",
  "Solution_10": "How any recurrence formula is obtained :). Just consider how the first few pieces \"at the bottom\" of the box can be arranged. You'll obtained the first one, that ugly one with $\\sum_{i=0}^n$. Then subtract two consecutive such expressions to get the short and nice one :)."
}
{
  "Tag": [],
  "Problem": "hi,\r\ni cant seem to get the answer when i try to find a discount from an original price tag or find the original price when discount has already been giving.\r\n\r\neg:\r\n\r\nwhat does  ring originally sell for if it has been given 20% discount and is on sale for 149.99\r\n\r\ndont you find the discount price of 149.99 using 20% then add it to 149.99?",
  "Solution_1": "No, that would be finding 20% of the sale price, and then adding it, which doesn't give you anything.  20% off is the same as 80% of the original.  So\r\n$ .8x\\equal{}149.99\\Rightarrow x\\equal{}187.49$",
  "Solution_2": "The thing to be careful of here is of what you are taking the percentage. You can think from the shopkeeper's perspective - how would she have changed the price?",
  "Solution_3": "Or instead of thinking in terms or the discount, think in terms of the sale price.\r\n\r\nSo if there's a 20% discount, think of it as 80% of the normal price.\r\n\r\nIf there's a 6% tax, then think of it as 106% (1.06) of the original price."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "AMC 10",
    "USA(J)MO",
    "USAMO",
    "percent",
    "AMC 12"
  ],
  "Problem": "Does anybody think the floor score could possibly be a $ 7$? I'm a sophomore, and I think I got a 7 maximum (unless I guessed right...). I thought this time's AIME was a bit easier.",
  "Solution_1": "Maybe. I'm a sophomore too, and I got an 8 or 9, depending on 12. It might be a 7. Or an 8, if you're really unlucky.\r\n\r\nLast year it was a 6. Maybe something like that'll happen again.",
  "Solution_2": "yea im a sophomore and i got a 8\r\n\r\nbut anyways,\r\n\r\ni would compare this test as the same difficulty as the 2006 AIME\r\n\r\nif you look at the first 10, they are about the same difficulty of  the first 10 that year.\r\n\r\nthat year's cutoff was 8 and that was limited expansion.\r\n\r\nim betting this year, with a full expansion, is going to be 7 (someones gonna get like 144/7), and they will probably make the cutoff like with a 193 index...",
  "Solution_3": "WHOO! Following the weird pattern here, I'm a sophomore and got an 8, a huge improvement over a 4 from last year.  This is crazy, because i barely made the AIME, getting a 117 on the AMC 10A",
  "Solution_4": "I feel pathetic in terms of improvement...I've had a 7-7-9 the last three years (including this year).",
  "Solution_5": "If the floor is an 8 this year, I'll be so mad, because I made stupid mistakes.",
  "Solution_6": "I think it is certainly possible that the floor will be $ 7$, but I think the high AMC/low AIME profile won't be as prevalent this year as in the past.  I would guess $ 8$ until I see a counterexample.  Sorry about that (I got $ 7$ on the 2006 AIME, which had an $ 8$ floor, so I feel your pain).",
  "Solution_7": "[quote=\"E^(pi*i)=-1\"]I think it is certainly possible that the floor will be $ 7$, but I think the high AMC/low AIME profile won't be as prevalent this year as in the past.  I would guess $ 8$ until I see a counterexample.  Sorry about that (I got $ 7$ on the 2006 AIME, which had an $ 8$ floor, so I feel your pain).[/quote]\r\n\r\ni know someone that got a 144/7 (well he thinks he got) if he didnt guess one right",
  "Solution_8": "Was that $ 144$ on the AMC 12 or AMC 10?  If it was on the 10, it doesn't affect floor score placement, as far as I know.",
  "Solution_9": "compare this contest to the difficulty of the 2006 AIME where the floor was 8",
  "Solution_10": "[quote=\"the future\"]i know someone that got a 144/7 (well he thinks he got) if he didnt guess one right[/quote]\r\nDid he take the AMC 10 or the 12?",
  "Solution_11": "i think 12...\r\nbut i dont know if 1 person can decide it..\r\n\r\nif he does then it will be problaby with a really high index\r\n\r\nso \r\n\r\neither 8 floor/no index required\r\n\r\nor 7 floor/200 index\r\n\r\nwhich means some people with 117/8 or 118.5/8 will be excluded",
  "Solution_12": "im not sure about anything else, but my brother did get a 145.5/ 8 on the amc 12..\r\nwhich im pretty sure passes him.\r\n\r\nI wrote down my amc 10 score, will they count that for my index?",
  "Solution_13": "[quote=\"miller4math\"]im not sure about anything else, but my brother did get a 145.5/ 8 on the amc 12..\nwhich im pretty sure passes him.\n\nI wrote down my amc 10 score, will they count that for my index?[/quote]\r\n\r\nthat wasnt suppose to be for anything but identification...\r\n\r\nso it doenst really matter if you wrote down 12 or 10, they should already have it in the database, the AMC score you wrote down for the AIME should only be for identifying you...i think",
  "Solution_14": "The score you write down has no bearing on what they use for qualification. It's just an identity check.\r\n\r\nI would guess 8, personally, as this AIME seemed easier than the past ones. Of course, having gotten a 9, I'm alright with that, but it would suck to be someone who got a 7.\r\n\r\nAnd in terms of improvement: 1-5-9.  :D",
  "Solution_15": "[quote=\"unimpossible\"]james4l did you take AIME I or II?[/quote]\r\nI took AIME II. I also took AMC10. My score isn't too good, so I am on the edge of being bumped off  :(",
  "Solution_16": "Uhh unimpossible, he took the AMC 10/AIME (hes only in 6th grade) so the 144/7 is irrelevant (but i do know someone without an AoPS account that got a 144/7 on the 12/AIME).\r\n\r\nBut seriosuly, you guys should stop worrying.\r\n\r\nTheres no way the cutoff will be greater than 7.  And theres no way the index will be greater  than 196.",
  "Solution_17": "[quote=\"the future\"]Uhh unimpossible, he took the AMC 10/AIME (hes only in 6th grade) so the 144/7 is irrelevant (but i do know someone without an AoPS account that got a 144/7 on the 12/AIME).\n\nBut seriosuly, you guys should stop worrying.\n\nTheres no way the cutoff will be greater than 7.  And theres no way the index will be greater  than 196.[/quote]\r\n\r\nI hope you're right  :)",
  "Solution_18": "On a different note, when do they announce whos made USAMO.\r\n\r\nIve read somewhere that the AMC Director said that the AMC would be sending out on April 11, 2008 the scores for the AIME, the solutions for the AIME, and a list of all the USAMO qualifiers to each school.\r\n\r\nHowever, I have spring break all next week.  So do they also email every person on whos made USAMO (i know they did last year), and would that also be on April 11, or would that be on an earlier date?\r\n\r\nThanks.",
  "Solution_19": "I pretty sure that it says on the teachers manual that invitations for USAMO would be sent out on April 6-11. I think the solutions and results for the AIME I have been sent out already, but I'm not sure about the AIME II...",
  "Solution_20": "[quote=\"AMCDirector\"]AIME results for AIME I and AIME II will be mailed out starting approximately April 11.  This is because we include a list of USAMO qualifiers along with the results, the answer pamphlet and the certificates of participation.  See the AIME Teachers' Manual, Section V, page 5.\n\nSteve Dunbar\nMAA Director, American Mathematics Competitions[/quote]\r\n\r\nThis is what the AMC Director wrote but thats when they are mailed out.  So I was just wondering when do we acutally find out by email (if we do).\r\n\r\nIt seems from what he wrote, the AIME I and AIME II results are mailed out at the same time.  So I dont think anyones gotten their I results back yet...",
  "Solution_21": "Well, the AIME I results were supposed to come out about 10 days after the test, and there's already been a thread asking about what people's official scores were. I took the AIME I, but I don't think my school's received the scores yet...",
  "Solution_22": "Yeah I created that thread when I thought that the results were already being sent out.  The AMC director corrected me, saying that the results won't be mailed out until the 11th.  We can just revive the thread around the 11th.",
  "Solution_23": "Hmm... How about this?\r\n\r\nI am an 8th grader...\r\n\r\n[b]AMC 10: 150.0/150.0\n\nAMC 12: Didn't take.\n\nAIME II : 8/10[/b]\r\n\r\nUSAMO: ?/9+\r\n\r\nOMG I wrote 502 for #8....\r\n\r\nOMG I wrote 120 for #10....\r\n\r\nOMG I could've guessed that 8 moves would make some sort of a pattern for #9....\r\n\r\nBut I dind't.... :(\r\n\r\nSilly, silly mistakes.....I shouldn't make any in next year's AIME II!!!\r\n[b]\nAnyway, my index is 230. A good possibility of taking the USAMO?[/b]",
  "Solution_24": "[quote=\"physisnu\"]Hmm... How about this?\n\nI am an 8th grader...\n\n[b]AMC 10: 150.0/150.0\n\nAMC 12: Didn't take.\n\nAIME II : 8/10[/b]\n\nUSAMO: ?/9+\n\nOMG I wrote 502 for #8....\n\nOMG I wrote 120 for #10....\n\nOMG I could've guessed that 8 moves would make some sort of a pattern for #9....\n\nBut I dind't.... :(\n\nSilly, silly mistakes.....I shouldn't make any in next year's AIME II!!!\n[b]\nAnyway, my index is 230. A good possibility of taking the USAMO?[/b][/quote]\r\n\r\nNo question, you will make it--don't stress it. Btw, AIME is out of 15.\r\n\r\nI missed #1 (along with two other mistakes) on the AIME last year, which got me a 5, missing the floor by 1 (which was low enough; I shouldn't complain).\r\nI missed #1 again this year, which got me a 6. If it misses a floor of 7, I'm seriously going to grind my head into the wall  :wallbash_red:",
  "Solution_25": "[quote=\"mathcrazed\"]Just kidding =P...  Honestly, the first 7 or 8 questions of the AIME II were pretty simple.  #8 was slightly harder, but other than that... Probably just me though.[/quote]2 was very hard if you're talking percentage of people who answered correctly.",
  "Solution_26": "[quote=\"worthawholebean\"][quote=\"mathcrazed\"]Just kidding =P...  Honestly, the first 7 or 8 questions of the AIME II were pretty simple.  #8 was slightly harder, but other than that... Probably just me though.[/quote]2 was very hard if you're talking percentage of people who answered correctly.[/quote]\r\n\r\nIs it just me, or do the AMC people just like messing with people on #2s?",
  "Solution_27": "I think people just don't pay enough attention to #2's. They're like, okay, I better not miss #1 because that would look stupid, but then focus their attention on the later problems.",
  "Solution_28": "Like me.  I got an 11 on AIME last year as a sophomore, and this year I missed #1 and #2 but still scraped out an 8.  With an almost decent AMC12B score I'm right on the bubble.  It's sad because, excluding 8th grade, my USAMO index has steadily dropped since freshman year.b",
  "Solution_29": "Actually, I would say that #2 on last year's AIME I and this year's AIME II were just downright tricky. Especially last year."
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "Putnam",
    "algebra",
    "partial fractions",
    "calculus computations"
  ],
  "Problem": "Evaluate\r\n\r\n$\\sum_{n=1}^{\\infty}\\sum_{m=1}^{\\infty}\\frac{1}{m^{2}n+mn^{2}+2mn}$\r\n\r\n[hide]\nso far the only ideas i have are, partial fractions and finding some clever way to turn it into a double integral. neither of which have gotten me very far.\n[/hide]",
  "Solution_1": "I got, if my calculations are correct, that the value of your sum is $\\frac{7}{4}$.",
  "Solution_2": "could you please provide a proof for your answer.",
  "Solution_3": "I would try the integral representation\r\n\\[\\frac{1}{m+n+2}= \\int_{0}^{1}x^{m+n+1}\\, dx \\, . \\]",
  "Solution_4": "The sum equals \r\n$\\int_{0}^{1}x(\\ln(1-x))^{2}dx=\\frac{7}{4}$ by Ravi`s idea, however I have another method for calculating this sum based on a, let`s say , long computational method.\r\nDidi",
  "Solution_5": "[quote=\"didilica\"]The sum equals \n$\\int_{0}^{1}x(\\ln(1-x))^{2}dx=\\frac{7}{4}$ by Ravi`s idea, however I have another method for calculating this sum based on a, let`s say , long computational method.\nDidi[/quote]\r\n\r\nhow did you get that integral. i follow what Ravi did but i dont see how they are related.",
  "Solution_6": "Write the sum as $\\sum_{n=1}^\\infty \\sum_{m=1}^\\infty \\int_{0}^{1}\\frac{x^{m}}{m}\\cdot \\frac{x^{n}}{n}\\cdot x \\, dx$.\r\nNow interchange the sums with the integral and use the Taylor expansion of $\\log$ around $1$.\r\n\r\nPS: I'm not sure what steps to do in order to justify that interchange... But this is the idea, I think.",
  "Solution_7": "to interchange i think it just has to be a sequence of only positive terms. but i could be wrong.",
  "Solution_8": "Here is a nice generalization:\r\n\r\nLet $k\\geq 2$ be a natural number.\r\n\r\nCalculate,\r\n\r\n$\\sum_{n=1}^{\\infty}\\sum_{m=1}^{\\infty}\\frac{1}{nm^{2}+n^{2}m+kmn}$.",
  "Solution_9": "This problem has been discussed in detail in the Putnam section, if I am not wrong.",
  "Solution_10": "i am pretty sure that it was. however i was unable to find it.",
  "Solution_11": "solution involving partial fractions is quite ugly:\r\n\r\n$S=\\sum\\frac{1}{n}\\sum\\frac{1}{m(m+n+2)}$\r\n$=\\sum\\frac{1}{n(n+2)}[1+\\frac{1}{2}+...+\\frac{1}{n+2}]$\r\n\r\ndo another partial fractions, then combine terms.\r\n\r\nanswer is indeed 7/4."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "I think it's very well-known , but i just can't find it out . \r\n\r\n$a , b , c > 0$ with $abc = 1$ , prove that \r\n\r\n$a^{2}+b^{2}+c^{2}+3 \\ge 2(ab+ac+bc)$",
  "Solution_1": "actually this problem if i am not mistaken was discovered by darij and it was without the condition of $abc=1$:\r\n\r\n$a^{2}+b^{2}+c^{2}+2abc+1 \\geq  2 (ab+bc+ca)$ , for all positive reals.\r\n\r\nthere are a lot of ways to solve it, one of them is with schur, $2abc+1 \\geq  \\frac{9abc}{a+b+c}\\geq  2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})$.\r\n\r\nactually the \"stronger\" one holds, and it is equivalent with the one with the condition $abc=1$ : $a^{2}+b^{2}+c^{2}+3 \\sqrt[3]{a^{2}b^{2}c^{2}}\\geq  2 (ab+bc+ca)$, and it can be solved in the same way, or directly by popoviciu for $f(x)=e^{2x}$",
  "Solution_2": "As for the initial problem i think it could be solved by mixing variables."
}
{
  "Tag": [
    "geometry",
    "rectangle"
  ],
  "Problem": "29 seems to be a pretty straightforward. Let x be the weight of the tank when empty and y be the weight of the fuel needed to completely fill up the tank. Then you can express both statements in the problem in x and y, so you just have to solve the simultaneous equations.\r\n\r\nFierytycoon",
  "Solution_1": "The numbers given in problem 29 make it especially nice. This will happen often in math contests. Note that the difference between 25% and 75% is 50%, so half a tank of fuel (without the container) is 1200 tons. Thus 25% is 600. Thus the weight of the empty tanker is 2500-600=1900 tons.\r\n\r\nI don't understand question 30. Is the original rectangle ABCD?",
  "Solution_2": "Well I have run out of paper within reaching distance of this seat so I'll just start this and get someone else to finish  :D \r\nFirst of all you know that AB = BC because thats how we've folded it. So AB=5 and B to the other top corner is 3. Now that right angled triangle is a 3,4,5 triangle so C is 4 from the top. Now we got a heap of right angled triangles.. the line connecting A to C must have length :sqrt:80, dividing that in half gives :sqrt:20 somewhere, and you can finish it off from there, I think it comes out to 5:sqrt:5 or something?",
  "Solution_3": "Diagram for #30. It was included in the problem. \n\n[asy]\nunitsize(20);\ndraw((0,0)--(0,8)--(5,8)--(5,0)--cycle);\nlabel(\"$C$\",(0,5),W);\ndot((0,5));\ndot((5,8));\nlabel(\"$A$\",(5,8),E);\nlabel(\"$8\\\"$\",(2.5,0),S);\ndraw((10,0)--(10,8)--(10+15/8,8)--(10,5)--(15,3)--(15,0)--cycle);\ndraw((10+15/8,8)--(15,3),dashed);\nlabel(\"$C$\",(10,5),W);\nlabel(\"$B$\",(10+15/8,8),N);\nlabel(\"$8\\\"$\",(12.5,0),S);\nlabel(\"$L$\",(13.4375,6),E);\ndot((10,5));\ndot((10+15/8,8));\n[/asy]",
  "Solution_4": "No need to revive the thread."
}
{
  "Tag": [],
  "Problem": "caculate:\r\n$ \\frac{2^3\\minus{}1}{2^3\\plus{}1} \\plus{}\\frac{2^3\\minus{}1}{2^3\\plus{}1}\\plus{}...\\plus{}\\frac{2009^3\\minus{}1}{2009^3\\plus{}1}$",
  "Solution_1": "Are you sure it is sum, and not product?\r\n\r\nBecause $ \\prod_{k\\equal{}2}^n \\frac {k^3\\minus{}1} {k^3\\plus{}1} \\equal{} \\left ( \\prod_{k\\equal{}2}^n \\frac {k\\minus{}1} {k\\plus{}1} \\right ) \\left ( \\prod_{k\\equal{}2}^n \\frac {k^2 \\plus{} k \\plus{} 1} {k^2 \\minus{}k \\plus{} 1} \\right ) \\equal{}$ $ \\frac {2} {n(n\\plus{}1)}\\cdot \\frac {n^2\\plus{}n\\plus{}1} {3}$, while the sum does not seem to telescope that well.",
  "Solution_2": "sorry problem this:\r\ncaculate:\r\n$ \\frac {2^3 \\minus{} 1}{2^3 \\plus{} 1} . \\frac {2^3 \\minus{} 1}{2^3 \\plus{} 1}  ... \\frac {2009^3 \\minus{} 1}{2009^3 \\plus{} 1}$",
  "Solution_3": "[quote=\"wall e\"]sorry problem this:\ncaculate:\n$ \\frac {2^3 \\minus{} 1}{2^3 \\plus{} 1} . \\frac {2^3 \\minus{} 1}{2^3 \\plus{} 1} ... \\frac {2009^3 \\minus{} 1}{2009^3 \\plus{} 1}$[/quote]\r\n\r\nI assume you meant $ \\frac{2^{3}\\minus{}1}{2^{3}\\plus{}1}.\\frac{3^{3}\\minus{}1}{3^{3}\\plus{}1}...\\frac{2009^{3}\\minus{}1}{2009^{3}\\plus{}1}$\r\nmavropnevma already showed how it telescopes because you get $ \\frac{1}{3}.\\frac{2}{4}.\\frac{3}{5}...\\frac{2007}{2009}.\\frac{2008}{2010}\\equal{}\\frac{2}{2009.2010}$ for the first part of the product and $ \\frac{7}{3}.\\frac{13}{7}.\\frac{21}{13}...\\frac{2008^2\\plus{}2008\\plus{}1}{2008^2\\minus{}2008\\minus{}1}.\\frac{2009^2\\plus{}2009\\plus{}1}{2009^2\\minus{}2009\\plus{}1}\\equal{}\\frac{2009^2\\plus{}2009\\plus{}1}{3}$ for the second part of the product. This gives us a final answer of $ \\frac{(2)(2009^2\\plus{}2009\\plus{}1)}{(3)(2009)(2010)}$"
}
{
  "Tag": [
    "USAMTS"
  ],
  "Problem": "I would like to participate in USAMTS this year. I am a resident of the United States but I am not a citizen of the United States. Would I be able to participate in USAMTS?",
  "Solution_1": "I think so, because I'm only a resident too.",
  "Solution_2": "Yes, as long as you are a resident (in other words, have a US mailing address), you may participate."
}
{
  "Tag": [
    "search",
    "blogs"
  ],
  "Problem": "N\u0103m nay K\u1ef3 thi SMO do H\u1ed9i to\u00e1n h\u1ecdc Singapore t\u1ed5 ch\u1ee9c, d\u00e0nh cho h\u1ecdc sinh c\u00e1c n\u01b0\u1edbc trong khu v\u1ef1c s\u1ebd \u0111\u01b0\u1ee3c t\u1ed5 ch\u1ee9c v\u00e0o ng\u00e0y 29 th\u00e1ng 5 n\u0103m 2007. D\u01b0\u1edbi s\u1ef1 \u0111i\u1ec1u ph\u1ed1i c\u1ee7a h\u1ed9i to\u00e1n h\u1ecdc H\u00e0 N\u1ed9i, t\u00ednh \u0111\u1ebfn nay k\u1ef3 thi n\u00e0y \u0111\u00e3 \u0111\u01b0\u1ee3c t\u1ed5 ch\u1ee9c hai l\u1ea7n t\u1ea1i Vi\u1ec7t Nam. \r\n\r\nN\u0103m nay c\u00f3 s\u1ef1 tham gia r\u1ed9ng r\u00e3i c\u1ee7a nhi\u1ec1u t\u1ec9nh h\u01a1n nh\u1eefng n\u0103m tr\u01b0\u1edbc. \r\n\r\nC\u00f3 em n\u00e0o \u0111\u00e3 \u0111\u01b0\u1ee3c l\u1ecdt v\u00e0o v\u00f2ng lo\u1ea1i th\u00ec l\u00ean ti\u1ebfng nh\u00e9.",
  "Solution_1": "[quote=\"pvthuan\"]N\u0103m nay K\u1ef3 thi SMO do H\u1ed9i to\u00e1n h\u1ecdc Singapore t\u1ed5 ch\u1ee9c, d\u00e0nh cho h\u1ecdc sinh c\u00e1c n\u01b0\u1edbc trong khu v\u1ef1c s\u1ebd \u0111\u01b0\u1ee3c t\u1ed5 ch\u1ee9c v\u00e0o ng\u00e0y 29 th\u00e1ng 5 n\u0103m 2007. D\u01b0\u1edbi s\u1ef1 \u0111i\u1ec1u ph\u1ed1i c\u1ee7a h\u1ed9i to\u00e1n h\u1ecdc H\u00e0 N\u1ed9i, t\u00ednh \u0111\u1ebfn nay k\u1ef3 thi n\u00e0y \u0111\u00e3 \u0111\u01b0\u1ee3c t\u1ed5 ch\u1ee9c hai l\u1ea7n t\u1ea1i Vi\u1ec7t Nam. \n\nN\u0103m nay c\u00f3 s\u1ef1 tham gia r\u1ed9ng r\u00e3i c\u1ee7a nhi\u1ec1u t\u1ec9nh h\u01a1n nh\u1eefng n\u0103m tr\u01b0\u1edbc. \n\nC\u00f3 em n\u00e0o \u0111\u00e3 \u0111\u01b0\u1ee3c l\u1ecdt v\u00e0o v\u00f2ng lo\u1ea1i th\u00ec l\u00ean ti\u1ebfng nh\u00e9.[/quote]\r\n\r\nC\u00f3 em :)\r\nB\u00e2y gi\u1edd em m\u1edbi c\u00f3 \u0111\u1ec1 c\u00e1c n\u0103m 2004, 2005 v\u00e0 2006 n\u1ebfu ai c\u00f3 \u0111\u1ec1 n\u0103m n\u00e0o kh\u00e1c th\u00ec cho em v\u1edbi :D",
  "Solution_2": "Ch\u00fa t\u00ecm th\u00eam [hide=\"\u1edf \u0111\u00e2y\"] http://sms.math.nus.edu.sg/Competitions/CompetitionHomePage.aspx[/hide] xem sao? C\u1ee9 search tr\u00ean Google l\u00e0 ra c\u1ea3 \u0111\u1ed1ng \u1ea5y m\u00e0 :D",
  "Solution_3": "N\u0103m nay kh\u1ed1i c\u1ea5p 2 th\u00ec Hanoi Ams \u0111\u01b0\u1ee3c 5 em, Nguy\u1ec5n Tr\u01b0\u1eddng T\u1ed9 \u0111\u01b0\u1ee3c 1 em, H\u1ea3i D\u01b0\u01a1ng c\u00f3 4 em. Ch\u00fac m\u1eebng nh\u00e9. V\u1eady c\u00f2n 10 v\u1ecb tr\u00ed n\u1eefa s\u1ebd thu\u1ed9c v\u1ec1 c\u00e1c t\u1ec9nh V\u0129nh Ph\u00fac, Ph\u00fa Th\u1ecd, ....",
  "Solution_4": "[quote=\"pvthuan\"]N\u0103m nay kh\u1ed1i c\u1ea5p 2 th\u00ec Hanoi Ams \u0111\u01b0\u1ee3c 5 em, Nguy\u1ec5n Tr\u01b0\u1eddng T\u1ed9 \u0111\u01b0\u1ee3c 1 em, H\u1ea3i D\u01b0\u01a1ng c\u00f3 4 em. Ch\u00fac m\u1eebng nh\u00e9. V\u1eady c\u00f2n 10 v\u1ecb tr\u00ed n\u1eefa s\u1ebd thu\u1ed9c v\u1ec1 c\u00e1c t\u1ec9nh V\u0129nh Ph\u00fac, Ph\u00fa Th\u1ecd, ....[/quote]\r\nTh\u1ea7y c\u00f3 danh s\u00e1ch kh\u00f4ng \u1ea1, share cho em v\u1edbi :)",
  "Solution_5": "Sao \u1edf mi\u1ec1n trung thi\u1ec7t th\u00f2i th\u1ebf nh\u1ec9? Ch\u1eb3ng bao gi\u1edd \u0111\u01b0\u1ee3c tham gia m\u1ea5y c\u00e1i nh\u01b0 th\u1ebf n\u00e0y!",
  "Solution_6": "Ch\u1eafc ng\u01b0\u1eddi ta ngh\u0129 c\u00e1c b\u1ea1n kh\u00f4ng gi\u1ecfi English \u0111\u1ea5y th\u00f4i :D",
  "Solution_7": "Kh\u1ed1i Senior H\u1ea3i D\u01b0\u01a1ng \u0111\u01b0\u1ee3c 4 b\u1ea1n  :D C\u00f2n Junior c\u0169ng \u0111\u01b0\u1ee3c 4 h\u1ea3 th\u1ea7y Thu\u1eadn? \u01a0 sao \u0111\u1ec1u th\u1ebf nh\u1ec9  :D \r\n\r\nCh\u00e1n qu\u00e1 c\u00e1i \u0111\u1ec1 v\u00f2ng lo\u1ea1i c\u1ee7a Senior b\u00e0i BDT c\u00f2n b\u1ecb sai \u0111\u1ec1  :( l\u00e0m em m\u1ea5t bao th\u1eddi gian, t\u01b0\u1edfng t\u00ednh sai (ng\u1ed1  :blush: ) M\u00e0 sao thi lo\u1ea1i v\u1eabn \u0111\u01b0\u1ee3c d\u00f9ng m\u00e1y t\u00ednh \u1ea1? Thi th\u1eadt c\u00f3 \u0111\u01b0\u1ee3c kh\u00f4ng \u1ea1  :D (th\u1ea5y trong \u0111\u1ec1 ghi l\u00e0 not allowed) M\u00e0 em th\u1ea5y c\u1ea7n ph\u1ea3i cho gi\u1ea5y nh\u00e1p tho\u1ea3i m\u00e1i m\u1ed9t ch\u00fat  :D h\u00f4m thi th\u1ea7y Th\u1ee7y b\u1ea3o l\u00e0 thi th\u1eadt nh\u00e1p lu\u00f4n c\u1ea3 v\u00e0o b\u00e0i l\u00e0m cho ng\u01b0\u1eddi ta ch\u1ea5m  :D",
  "Solution_8": "[quote=\"N.T.TUAN\"]Ch\u00fa t\u00ecm th\u00eam [hide=\"\u1edf \u0111\u00e2y\"] http://sms.math.nus.edu.sg/Competitions/CompetitionHomePage.aspx[/hide] xem sao? C\u1ee9 search tr\u00ean Google l\u00e0 ra c\u1ea3 \u0111\u1ed1ng \u1ea5y m\u00e0 :D[/quote]\r\n\r\nL\u00e0m g\u00ec c\u00f3 \u0111\u1ec1 \u1edf \u0111\u00f3 anh \u01a1i! Ch\u1ec9 c\u00f3 2 c\u00e1i \u0111\u1ec1 Senior in tr\u00ean THTT 2 n\u0103m v\u1eeba r\u1ed3i thui  :lol:",
  "Solution_9": "[quote=\"hieuchuoi@\"]Kh\u1ed1i Senior H\u1ea3i D\u01b0\u01a1ng \u0111\u01b0\u1ee3c 4 b\u1ea1n  :D C\u00f2n Junior c\u0169ng \u0111\u01b0\u1ee3c 4 h\u1ea3 th\u1ea7y Thu\u1eadn? \u01a0 sao \u0111\u1ec1u th\u1ebf nh\u1ec9  :D \n\nCh\u00e1n qu\u00e1 c\u00e1i \u0111\u1ec1 v\u00f2ng lo\u1ea1i c\u1ee7a Senior b\u00e0i BDT c\u00f2n b\u1ecb sai \u0111\u1ec1  :( l\u00e0m em m\u1ea5t bao th\u1eddi gian, t\u01b0\u1edfng t\u00ednh sai (ng\u1ed1  :blush: ) M\u00e0 sao thi lo\u1ea1i v\u1eabn \u0111\u01b0\u1ee3c d\u00f9ng m\u00e1y t\u00ednh \u1ea1? Thi th\u1eadt c\u00f3 \u0111\u01b0\u1ee3c kh\u00f4ng \u1ea1  :D (th\u1ea5y trong \u0111\u1ec1 ghi l\u00e0 not allowed) M\u00e0 em th\u1ea5y c\u1ea7n ph\u1ea3i cho gi\u1ea5y nh\u00e1p tho\u1ea3i m\u00e1i m\u1ed9t ch\u00fat  :D h\u00f4m thi th\u1ea7y Th\u1ee7y b\u1ea3o l\u00e0 thi th\u1eadt nh\u00e1p lu\u00f4n c\u1ea3 v\u00e0o b\u00e0i l\u00e0m cho ng\u01b0\u1eddi ta ch\u1ea5m  :D[/quote]\r\n\u0110i thi kh\u00f4ng \u0111\u01b0\u1ee3c d\u00f9ng m\u00e1y t\u00ednh \u0111\u00e2u anh, c\u00f2n \u0111\u1ec1 thi th\u1eadt \u0111\u1ebfn 2,3 trang (t\u1ee9c 4-6 m\u1eb7t), nh\u00e1p v\u00f4 t\u01b0 m\u00e0 anh :D",
  "Solution_10": "[quote=\"tunganh\"][quote=\"hieuchuoi@\"]Kh\u1ed1i Senior H\u1ea3i D\u01b0\u01a1ng \u0111\u01b0\u1ee3c 4 b\u1ea1n  :D C\u00f2n Junior c\u0169ng \u0111\u01b0\u1ee3c 4 h\u1ea3 th\u1ea7y Thu\u1eadn? \u01a0 sao \u0111\u1ec1u th\u1ebf nh\u1ec9  :D \n\nCh\u00e1n qu\u00e1 c\u00e1i \u0111\u1ec1 v\u00f2ng lo\u1ea1i c\u1ee7a Senior b\u00e0i BDT c\u00f2n b\u1ecb sai \u0111\u1ec1  :( l\u00e0m em m\u1ea5t bao th\u1eddi gian, t\u01b0\u1edfng t\u00ednh sai (ng\u1ed1  :blush: ) M\u00e0 sao thi lo\u1ea1i v\u1eabn \u0111\u01b0\u1ee3c d\u00f9ng m\u00e1y t\u00ednh \u1ea1? Thi th\u1eadt c\u00f3 \u0111\u01b0\u1ee3c kh\u00f4ng \u1ea1  :D (th\u1ea5y trong \u0111\u1ec1 ghi l\u00e0 not allowed) M\u00e0 em th\u1ea5y c\u1ea7n ph\u1ea3i cho gi\u1ea5y nh\u00e1p tho\u1ea3i m\u00e1i m\u1ed9t ch\u00fat  :D h\u00f4m thi th\u1ea7y Th\u1ee7y b\u1ea3o l\u00e0 thi th\u1eadt nh\u00e1p lu\u00f4n c\u1ea3 v\u00e0o b\u00e0i l\u00e0m cho ng\u01b0\u1eddi ta ch\u1ea5m  :D[/quote]\n\u0110i thi kh\u00f4ng \u0111\u01b0\u1ee3c d\u00f9ng m\u00e1y t\u00ednh \u0111\u00e2u anh, c\u00f2n \u0111\u1ec1 thi th\u1eadt \u0111\u1ebfn 2,3 trang (t\u1ee9c 4-6 m\u1eb7t), nh\u00e1p v\u00f4 t\u01b0 m\u00e0 anh :D[/quote]\r\nEm th\u1ea5y xin th\u00eam nh\u00e1p v\u1eabn \u0111\u01b0\u1ee3c m\u00e0, ch\u1eb3ng qua ng\u01b0\u1eddi ta mu\u1ed1n \u0111\u00e1nh gi\u00e1 k\u1ef9 n\u0103ng c\u1ee7a m\u00ecnh qua c\u00e1ch nh\u00e1p \u0111\u00f3. V\u1eady ai c\u00f3 \u0111\u1ec1 Junior c\u00e1c cu\u1ed9c thi kh\u00e1c c\u0169ng \u0111\u01b0\u1ee3c share cho em v\u1edbi",
  "Solution_11": "Theo y\u00eau c\u1ea7u c\u1ee7a m\u1ed9t s\u1ed1 b\u1ea1n kh\u00f4ng c\u00f3 \u0111i\u1ec1u ki\u1ec7n v\u1ec1 H\u00e0 N\u1ed9i, t\u00f4i tr\u00edch d\u01b0\u1edbi \u0111\u00e2y m\u1ed9t s\u1ed1 b\u00e0i to\u00e1n t\u1eadp hu\u1ea5n \u0111\u00e3 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng.",
  "Solution_12": "Ui thay` \u01a1i ,cho em h\u1ecfi th\u1ebf \u0111\u00e1 c\u00f3 danh s\u00e1ch ru\u00ec \u1ea1 ,th\u1ea7y c\u00f3 th\u1ec3 post danh s\u00e1ch l\u00ean \u0111c ko \u1ea1 ?\u0111\u1ebfn gi\u1edd m\u00e0 ch\u01b0a \u0111c th\u00f4ng b\u00e1o ngh\u0129a l\u00e0... :(",
  "Solution_13": "Kh\u1ed1i Senior \u1edf H\u00e0 N\u1ed9i ko \u0111c ai h\u1ea3 th\u1ea7y?N\u1ebfu c\u00f3 danh s\u00e1ch th\u1ea7y post qua cho b\u1ecdn em coi  th\u1ea7y nh\u00e1",
  "Solution_14": "Theo t\u00f4i bi\u1ebft th\u00ec \u1edf Hanoi c\u00f3 4 em sau \u0111\u00e2y\r\n\r\n1) Pham Thuy Duong\r\n2) Dang Dung Ha\r\n3) Truong Ba Thang\r\n4) Ha Khuong Duy\r\n\r\nB\u1ed1n em n\u00e0y \u0111\u1ec1u l\u00e0 h\u1ecdc sinh c\u1ee7a Kh\u1ed1i chuy\u00ean KHTH Hanoi. S\u1ed1 c\u00f2n l\u1ea1i th\u00ec t\u00f4i kh\u00f4ng bi\u1ebft.",
  "Solution_15": "Vang r\u1ed9i nh\u1ea5t v\u1eabn l\u00e0 n\u0103m 2005. T\u00f4i nh\u1edb n\u0103m \u0111\u00f3 l\u00e0 \u0111\u1ed9 V\u0129nh ph\u00fac c\u00f3 20 em d\u1ef1 thi th\u00ec c\u00f3 18 HCV, 2 HCB. N\u0103m \u0111\u00f3 V\u0129nh ph\u00fac t\u1eadp trung \u0111\u1ed9i tuy\u1ec3n kh\u00e1 s\u1edbm (tr\u01b0\u1edbc 1 th\u00e1ng) v\u00e0 t\u1eadp hu\u1ea5n kh\u00e1 nhi\u1ec1u (ri\u00eang Mr Thu\u1eadn \u0111\u00e3 kho\u1ea3ng 10 bu\u1ed5i). N\u0103m nay c\u00f3 Hanoi Ams c\u0169ng r\u1ea5t m\u1ea1nh, ra qu\u00e2n 5 th\u00ec \u0111\u1ec1u \u0111\u01b0\u1ee3c 5 HCV.",
  "Solution_16": "[quote=\"quangpbc\"]Kh\u00f4ng bi\u1ebft qua c\u00e1c n\u0103m th\u00ec n\u0103m nay th\u00e0nh t\u00edch \u0111\u01b0\u1ee3c \u0111\u00e1nh gi\u00e1 l\u00e0 t\u0103ng hay gi\u1ea3m anh Tu\u00e2n nh\u1ec9  :([/quote]\r\nGi\u1ea3m so v\u1edbi c\u00e1c n\u0103m tr\u01b0\u1edbc. N\u00f3i chung m\u1ea5y c\u00e1i tr\u00f2 n\u00e0y anh ch\u1ea3 quan t\u00e2m.  :lol:",
  "Solution_17": "H\u00ea!V\u00f4 l\u00fd.L\u00e0m g\u00ec c\u00f3 chuy\u1ec7n kh\u00f4ng cho d\u00f9ng m\u00e1y t\u00ednh,n\u1ebfu g\u1eb7p s\u1ed1 to th\u00ec l\u00e0m th\u1ebf n\u00e0o?! :wink:  :(  :huh:",
  "Solution_18": "[quote=\"Tr\u1ecbnh L\u01b0u Ly\"]H\u00ea!V\u00f4 l\u00fd.L\u00e0m g\u00ec c\u00f3 chuy\u1ec7n kh\u00f4ng cho d\u00f9ng m\u00e1y t\u00ednh,n\u1ebfu g\u1eb7p s\u1ed1 to th\u00ec l\u00e0m th\u1ebf n\u00e0o?! :wink:  :(  :huh:[/quote]\r\nKh\u00f4ng \u0111\u01b0\u1ee3c d\u00f9ng m\u00e1y t\u00ednh m\u00e0 b\u1ea1n! Th\u1eb1ng b\u1ea1n m\u00ecnh \u0111i thi v\u1ec1 b\u1ea3o v\u1eady, \u0111\u00e2y l\u00e0 thi ki\u1ebfn th\u1ee9c, ch\u1ee9 c\u00f3 ph\u1ea3i thi m\u00e1y t\u00ednh \u0111\u00e2u! M\u00e0 th\u1ea5y n\u00f3 b\u1ea3o, l\u00fac thi 4 \u0111\u1ee9a TH \u0111\u01b0\u1ee3c ng\u1ed3i c\u1ea1nh nhau, ko bi\u1ebft t\u1eeb TA n\u00e0o c\u00f3 th\u1ec3 h\u1ecfi nhau \u0111\u01b0\u1ee3c, th\u1ebf m\u1edbi hay!  :wink:",
  "Solution_19": "[quote=\"NguyenDungTN\"] M\u00e0 th\u1ea5y n\u00f3 b\u1ea3o, l\u00fac thi 4 \u0111\u1ee9a TH \u0111\u01b0\u1ee3c ng\u1ed3i c\u1ea1nh nhau, ko bi\u1ebft t\u1eeb TA n\u00e0o c\u00f3 th\u1ec3 h\u1ecfi nhau \u0111\u01b0\u1ee3c, th\u1ebf m\u1edbi hay!  :wink:[/quote]\r\nC\u00e1i n\u00e0y \u0111\u1eebng n\u00f3i tr\u00ean m\u1ea1ng nhe em . :P",
  "Solution_20": "[quote=\"pvthuan\"]Vang[b] r\u1ed9i[/b] .[/quote] \r\n\r\n\r\n :D Sai roi` tha^y` MrThuan oi .\r\n\r\nMa`sao ko cho cac tinh? khac' thi anh TUAN nhi?. Ba^t' co^ng qua'  :(",
  "Solution_21": "Theo nh\u01b0 m\u1ed9t th\u00e0y n\u00f3i th\u00ec do l\u00e0 c\u00e1c t\u1ec9nh kh\u00e1c kh\u00f4ng ch\u1ecbu c\u1ea5p kinh ph\u00ed , th\u1ebf th\u00f4i. Ch\u1ee9 thi c\u00e1i n\u00e0y ai ch\u1eb3ng thi \u0111\u01b0\u1ee3c. M\u00e0 blog c\u1ee7a anh ngon r\u1ed3i \u0111\u1ea5y, v\u00e0o comment mau. :P",
  "Solution_22": "Comment r\u1ed3i \u00f4ng anh \u1ea1. M\u00e0 \u0111i n\u00f3i chuy\u1ec7n n\u00e0y \u1edf \u0111\u00e2y m\u1edbi kh\u1ed5 ch\u1ee9 :wacko:  :wallbash:  :whistling:",
  "Solution_23": "C\u00f3 kinh ph\u00ed l\u00e0 \u0111\u01b0\u1ee3c tham gia thi ha anh",
  "Solution_24": "[quote=\"pvthuan\"]Vang r\u1ed9i nh\u1ea5t v\u1eabn l\u00e0 n\u0103m 2005. T\u00f4i nh\u1edb n\u0103m \u0111\u00f3 l\u00e0 \u0111\u1ed9 V\u0129nh ph\u00fac c\u00f3 20 em d\u1ef1 thi th\u00ec c\u00f3 18 HCV, 2 HCB. N\u0103m \u0111\u00f3 V\u0129nh ph\u00fac t\u1eadp trung \u0111\u1ed9i tuy\u1ec3n kh\u00e1 s\u1edbm (tr\u01b0\u1edbc 1 th\u00e1ng) v\u00e0 t\u1eadp hu\u1ea5n kh\u00e1 nhi\u1ec1u (ri\u00eang Mr Thu\u1eadn \u0111\u00e3 kho\u1ea3ng 10 bu\u1ed5i). N\u0103m nay c\u00f3 Hanoi Ams c\u0169ng r\u1ea5t m\u1ea1nh, ra qu\u00e2n 5 th\u00ec \u0111\u1ec1u \u0111\u01b0\u1ee3c 5 HCV.[/quote]\r\nN\u00f3i chung hai n\u0103m nay Hanoi ams n\u0103m n\u00e0o c\u0169ng h\u1ea7u nh\u01b0 to\u00e0n v\u00e0ng (\u0111\u1ea5y l\u00e0 THCS, c\u00f2n THPT n\u0103m ngo\u00e1i thi th\u00ec ch\u1ec9 c\u00f3 b\u1ea1c th\u00f4i)",
  "Solution_25": "cho em hoi the bao gio`,giay chung nhan gui ve` truong` a,em muon biet vi` can co' 1 chut viec :)",
  "Solution_26": "Th\u01b0\u1eddng th\u00ec H\u1ed9i to\u00e1n h\u1ecdc H\u00e0 N\u1ed9i s\u1ebd t\u1ed5ng k\u1ebft v\u00e0 trao ch\u1ee9ng nh\u1eadn cho c\u00e1c em tham gia v\u00e0o d\u1ecbp \u0111\u1ea7u n\u0103m h\u1ecdc m\u1edbi ho\u1eb7c th\u00e1ng 11. Em n\u00e0o tham gia th\u00ec tr\u1ef1c ti\u1ebfp h\u1ecfi hi\u1ec7u tr\u01b0\u1edfng c\u1ee7a tr\u01b0\u1eddng \u0111\u00f3 v\u1ec1 vi\u1ec7c n\u00e0y.",
  "Solution_27": "Ng\u00e0y 28, 29 th\u00e1ng 11, n\u0103m 2007, nh\u00e2n H\u1ed9i th\u1ea3o To\u00e1n Gi\u1ea3i t\u00edch v\u00e0 \u1ee8ng d\u1ee5ng, H\u1ed9i To\u00e1n H\u1ecdc H\u00e0 N\u1ed9i s\u1ebd trao gi\u1ea3i c\u1ee7a cu\u1ed9c thi SMO 2007. C\u00e1c em tham gia, \u0111o\u1ea1t gi\u1ea3i, \u0111\u00e3 bi\u1ebft th\u00f4ng tin n\u00e0y ch\u01b0a?\r\n \r\n\r\n(s\u1ebd c\u00f3 \u00edt nh\u1ea5t l\u00e0 3 em kh\u00f4ng ch\u1ee9ng ki\u1ebfn gi\u00e2y ph\u00fat n\u00e0y v\u00ec \u0111ang \u1edf Singapore r\u00f9i)  :)",
  "Solution_28": "T__T sao bon e ko dc bao' a?\r\nMuon nhan giai thuong thi` lay o dau a? hicc",
  "Solution_29": "Ch\u00e0o c\u00e1c anh ch\u1ecb em. H\u00ec, h m\u1edbi bi\u1ebft c\u00f3 ch\u1ed7 \u0111\u1ec3 m\u00ecnh vi\u1ebft Ti\u1ebfng Vi\u1ec7t tr\u00ean Mathlinks. Mong m\u1ecdi ng\u01b0\u1eddi gi\u00fap \u0111\u1ee1."
}
{
  "Tag": [],
  "Problem": "The $64$ squares of an $8 \\times 8$ chessboard are filled with positive integers in such a way that each integer is the average of the integers on the neighbouring squares. Show that in fact all the $64$ entries are equal.",
  "Solution_1": "where do you get these problems?(you know,you send a lot of problem) :D",
  "Solution_2": "Take the smallest integer among the ones written on the chessboard. Then all neighbouring integers must be equal to this one. Continue repeating this process and you easily get that they are all equal. Too easy, and too well-known.",
  "Solution_3": "Proof: by contradiction. Suppose that every square has a different integer value. Then, without loss of generality, we will pick an arbituary square and say this has the lowest value. This square, though, is equal to the arithmetic mean of its neighbours. This implies that one of its neighbours has a lower value than it. But we specified the square to have  the lowest valu. Contradiction! Thus, all squares have equal values.\r\n\r\nQ.E.D",
  "Solution_4": "we can also introduce a maximal element..\n(the proof is still the same as above)",
  "Solution_5": "Choose the square (any one) which is filled up with the smallest of all the positive integers filled in the 64 squares. It any of the neighbors is filled with positive integer greater then the smallest positive integer, then the average of all the integers in the squares occupying the neighbor\u2019s positions will be greater than the said smallest integer. This is contradiction !Hence all the neighboring squares are filled with same smallest integer continuing is this way till all the squares are filled in finite number of steps",
  "Solution_6": "If you know $Geogebra$ and $\\LaTeX$$,$ please use them neelanjan. Your solution will look good. :)\nBTW, your solution is completely wrong - from the beginning to the end.",
  "Solution_7": "atfirst , i do not know asymptote . you will see only my solutions ! , see it and tell me where i am wrong . :mad: ",
  "Solution_8": "[quote=eshan]If you know $Asymptote$ and $\\LaTeX$$,$ please use them neelanjan. Your solution will look good. :)[/quote]\n\n[quote=neelanjan]atfirst , i do not know asymptote . you will see only my solutions ! , see it and tell me where i am wrong . :mad:[/quote]\n\ni think neelanjan is right, it is the depth of the solution you need to consider, not the way it is presented. Everybody is not supposed to know everything!"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "Rational Root Theorem"
  ],
  "Problem": "Please, help me to solve this problem: \r\n\"Define the number of positive roots of the equation:\r\n12x^5 + 6x^4 \u2013 4x^3 \u2013 x \u2013 34 = 0\"",
  "Solution_1": "The solutions are going to be of the form $ \\frac{a}{b}$, where $ a|12$ and $ b|34$.\r\nWhat are the divisors of 12 and 34?\r\n$ a = {\\pm1, \\pm 2, \\pm 3, \\pm 4, \\pm 6, \\pm 12}$\r\n$ b = { \\pm 1, \\pm 2, \\pm 17, \\pm 34}$\r\n\r\nSince we are looking for positive roots, $ a$ and $ b$ must have the same sign. \r\nThus our answers fall in this list:\r\n$ \\frac{a}{b} = {1, \\frac{1}{2}, \\frac{1}{17}, \\frac{1}{34}, 2, \\frac{2}{17}, 3, \\frac{3}{2}, \\frac{3}{17}, \\frac{3}{34}, 4, \\frac{4}{17}, 6, \\frac{6}{17}, 12, \\frac{12}{17}}$\r\n\r\nNote that it is not all of numbers that are going to satisfy your equation. But you know where to look now for your answers and that was from a theorem known as Rational Root Theorem which you should study.",
  "Solution_2": "[quote=\"akech\"]The solutions are going to be of the form $ \\frac {a}{b}$, where $ a|12$ and $ b|34$.\nWhat are the divisors of 12 and 34?\n$ a = {\\pm1, \\pm 2, \\pm 3, \\pm 4, \\pm 6, \\pm 12}$\n$ b = {\\pm 1, \\pm 2, \\pm 17, \\pm 34}$\n\nSince we are looking for positive roots, $ a$ and $ b$ must have the same sign. \nThus our answers fall in this list:\n$ \\frac {a}{b} = {1, \\frac {1}{2}, \\frac {1}{17}, \\frac {1}{34}, 2, \\frac {2}{17}, 3, \\frac {3}{2}, \\frac {3}{17}, \\frac {3}{34}, 4, \\frac {4}{17}, 6, \\frac {6}{17}, 12, \\frac {12}{17}}$\n\nNote that it is not all of numbers that are going to satisfy your equation. But you know where to look now for your answers and that was from a theorem known as Rational Root Theorem which you should study.[/quote]\r\nYou'd still need to look for irrational positive roots, also.\r\n\r\nYou could use [url=http://www.artofproblemsolving.com/Wiki/index.php/Polynomial#Descartes.27_Law_of_Signs]Descartes' Rule of Signs[/url] to find the possible numbers of positive roots."
}
{
  "Tag": [],
  "Problem": "How many ways can you arrange 7 green bottles and 8 blue bottles in a row, if exactly $1$ pair of green bottles is side by side?",
  "Solution_1": "[hide]We have six blocks of green bottles, one containing two bottles and five containing one bottle each.  There are 6 ways to order these blocks.  Then let $x_{n}$ be the number of blue bottles between the $(n-1)$th and $n$th blocks of green bottles.  We have $x_{1}+x_{2}+\\cdots+x_{7}=8$.  Here $x_{1}$ and $x_{7}$ are nonnegative integers, and the rest are positive.  Let $x_{1}=y_{1}-1$ and $x_{7}=y_{7}-1$.  Substituting, we obtain $y_{1}+x_{2}+x_{3}+\\cdots+x_{6}+y_{7}=10$, where all variables are positive integers.  There are $\\binom{9}{6}$ solutions to this equation, so the answer is $6\\binom{9}{6}$.[/hide]",
  "Solution_2": "Are the bottles distinct?",
  "Solution_3": "[quote=\"calculuslover800\"]Are the bottles distinct?[/quote]\r\nNo, they generally aren't for these types of problems.\r\n\r\nAlthough, you could solve it so that they are distinct.  :D"
}
{
  "Tag": [
    "function",
    "search",
    "algebra",
    "functional equation",
    "algebra proposed"
  ],
  "Problem": "Find all functions $f: \\mathbb N\\cup\\{0\\}\\rightarrow \\mathbb N\\cup\\{0\\}$ that \\[f(f(n))+f(n)=2n+3k\\] which $k\\in\\mathbb Z$",
  "Solution_1": "Omid,\r\n\r\nedit out the word \"easy.\"\r\nI told you I would come back to check up on this.\r\nIt's not mathematical AT ALL.",
  "Solution_2": "I apologize for spamming.\r\n[i]To A cute angle :[/i]\r\nIf you search the topics that has the word \"easy\" you will see a great list.\r\n\r\nPlease finish it and don't tell me what to do :mad:",
  "Solution_3": "Stating something is easy (or hard) is wrong.  I expect it out of elementary school children, but not serious people who call themselves mathematicians.",
  "Solution_4": "1) You still decided not to make a change in the word.\r\n\r\n2) An apology for spamming from you is EMPTY; it is actions insofar as NOT typing out said words in the future that matter.\r\n\r\n3) Argumentation/Logic -- Don't change the subject on me.  We are talking about you and your words.  I am focusing my attention on one person (user) at a time.  Do not bring in other users for an attempted comparison.",
  "Solution_5": "The word \"easy\" in titles is indeed not the best, especially when not anything else is given.\r\nReason: it's impossible to know what the thread is about before opening -> searching is hard for it, and finding even more.\r\nThis changes if there is more information given, e.g. that it is a functional equation; but it would be even better to give \"parts\" of the problem in the title, like the equation itself.\r\n\r\nBut more on what Pleaseread(/A cute angle/sohcohtoa/all the other names he uses) refers to:\r\nI agree that \"easy\" is subjective; but there are easier and harder problems (here \"easy\" is defined on how hard the average mathematicians solving it considers it), and when the author considered this one to be easy, then it is at least not one of the hardest problems he knows (otherwise he wouldn't say so). These kind of statements are for example usefull if you are looking for not-to-hard problems; you just have to remember who stated this (e.g. an \"easy\" from vess is normally not what most people call easy ;) ), but then there is no problem in using this terms.\r\n\r\n\r\nAnd it is much more childish to use a couple of accounts instead of one; but I think you will never stop this...\r\n\r\n\r\nConclusion:\r\n@all: try not to use \"easy\" in titles, especially without more specific terms on the problems.\r\nBut telling that one thinks it is easy is indeed ok.\r\n@Pleaseread: stop doing this finally!\r\n@moderators: delete this in some days please.",
  "Solution_6": "hallo,\r\ni think the problem is nice.\r\n$f(n)=n+k$ is this right ?",
  "Solution_7": "Yes, that's the only solution. But post your solution."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "modular arithmetic",
    "pigeonhole principle"
  ],
  "Problem": "(1) Given $ n$ integers. Prove that there are some integers which sum is divisible by $ n$.\r\n(2) Put 10 points in a circle with radius of 5. Prove that there are 2 points which distance is less than 2.",
  "Solution_1": "Not sure if I have misunderstood this, but for \"given n integers\" we will always have n 1s or n 2s or n 3s or more generally n ms. Summing them you will have nm, which is divisible by n.\r\n\r\nFor the second one, you could evenly spread out all the points on the sphere, even when you do that, the points will always be just over 3 apart.",
  "Solution_2": "One valid set of $ n$ integers is $ 1, 2, 3, \\ldots, n\\minus{}1, n\\plus{}1$, in which case it is not as trivial as you have made it out to be. Obviously in this case $ 1 \\plus{} (n\\minus{}1)$ is one sum that works, but proving there is always such a sum is slightly trickier.\r\n\r\n[hide=\"1\"]Let the numbers be $ k_1, k_2, \\ldots k_n$. Consider each of the following sums modulo $ n$: $ k_1$, $ k_1 \\plus{} k_2$, $ \\ldots$, $ k_1 \\plus{} k_2 \\plus{} \\ldots \\plus{} k_n$. If any of them are $ 0 \\pmod{n}$, then that is the desired sum. If none of them are $ 0 \\pmod{n}$, then they are one of the following $ \\mod n$: $ 1, 2, 3, \\ldots n\\minus{}1$. By the pigeonhole principle, at least two of the sums have the same remainder ($ n$ sums and $ n\\minus{}1$ possible remainders). Then their positive difference is $ 0 \\pmod{n}$, giving the desired sum.[/hide]"
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Find the dimension of the subspace of those linear operators $\\varphi$ on $M_{n}(\\mathbb{R})$ for which the identity $\\varphi(A^{T})=(\\varphi(A))^{T}$ holds for every matrix $A\\in M_{n}(\\mathbb{R}).$ \r\n(A. Oliynyk)",
  "Solution_1": "sorry, I think it should be $\\frac{n^{2}(n^{2}+1)}{2}$. May I know if i'm right?\r\n\r\nDefine $A_{ij}$ to be matrix that is 0 except the ij-th position, where it is 1.\r\n\r\nThese $n^{2}$ matrices constitutes a basis for $Mn(R)$.\r\n\r\nThus we only need to consider $\\phi$ on these matrices.\r\n\r\nFor $A_{ii}$, $\\phi(A)$ must be symmetric.\r\nFor $A_{ij}$, $i < j$, $\\phi(A)$ is arbitrary.\r\n\r\nThus the answer is $n \\frac{n(n+1)}{2}+\\frac{n(n-1)}{2}n^{2}= \\frac{n^{4}+n^{2}}{2}$",
  "Solution_2": "seems more acceptable\r\nWould you post a complete proof ?\r\nthank's",
  "Solution_3": "[quote=\"Soarer\"]Thus the answer is $n \\frac{n(n+1)}{2}+\\frac{n(n-1)}{2}n^{2}= \\frac{n^{4}+n^{2}}{2}$[/quote]\r\n\r\ncheck this ?"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra unsolved"
  ],
  "Problem": "Given a natural number $n,$ find all polynomials $P(x)$ of degree less than $n$ satisfying the following condition \\[ \\sum^n_{i=0} P(i) \\cdot (-1)^i \\cdot \\binom{n}{i} = 0. \\]",
  "Solution_1": "Just use Lagrange Interpolatin formula for P(x)."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "x,y are rational, solve this equation:\r\nx^n+y^n=x^m+y^m\r\n\r\n**m,n are natural",
  "Solution_1": "?\r\n\r\nyou sure you didnt make any mistake in the qn?"
}
{
  "Tag": [
    "pigeonhole principle"
  ],
  "Problem": "Let $ a_1, a_2, \\dots , a_{99}$ be positive integers all less than 100 (but not necessarily distinct). If the sum of any two or more of them is never a multiple of 100, show that they all must be equal.\r\n\r\n\r\nYeah, I fail at pigeonhole.",
  "Solution_1": "Should I move this to HSI?",
  "Solution_2": "[hide=\"Part of a solution\"]\nSo define $ s_n$ to be the cumulative sums of $ a_n$, i.e., $ s_n \\equal{} a_1 \\plus{} a_2 \\plus{} \\ldots \\plus{} a_n$.\nDefine $ s_0 \\equal{} 0$.\nThen if $ s_i \\equiv s_j \\bmod{100}$ and $ i \\neq j$, we have $ s_{i\\plus{}1} \\plus{} s_{i\\plus{}2} \\plus{} \\ldots \\plus{} s_{j} \\equiv 0 \\bmod{100}$.\n(Since no $ a_n$ can be 0 or 100, there is no way the above sum could consist of just one number.)\nSo, assume that $ s_0, s_1, s_2, \\ldots, s_{99}$ are all distinct.\nThe above holds if you permute the elements of $ a_n$ in any way.\n[/hide]\r\n\r\nI keep feeling that the last step I'm missing is either obvious or trivial, and possibly both."
}
{
  "Tag": [
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Does there exist a continuous function which is not monotonic on any interval?",
  "Solution_1": "$ f(x) \\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {\\cos(3^{n}x)}{2^{n}}$ has the property stated.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nEdited: $ x$ is installed back where it was missing upon a request by an acute reader",
  "Solution_2": "Another approach: for any $ 0\\le a<b\\le 1$ the set of all functions $ f\\in C[0,1]$ that are monotone on $ [a,b]$ is closed and has no interior points. Taking the union over intervals with rational endpoints, we conclude that the set of functions that are monotone on some interval is of first category in $ C[0,1]$.",
  "Solution_3": "The classical Weierstrass function $ f(x)\\equal{}\\sum_k 10^{\\minus{}k}\\cos(100^k x)$ is the most famous example. To see that it works, just consider $ f(\\pi(n\\plus{}1)/100^k)\\minus{}f(\\pi n/100^k)$ with odd and even $ n$. What is interesting is that you can even find a differentiable (though, of course, not continuously differentiable) function with this property ;)\r\n\r\nEdit: mlok and akech were faster :)",
  "Solution_4": "[quote=\"akech\"]$ f(x) \\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {\\cos(3^{n})}{2^{n}}$ has the property stated.[/quote]\r\n\r\nthere is no $ x$ in the expression...  :oops:",
  "Solution_5": "[quote=\"bookworm_vn\"][quote=\"akech\"]$ f(x) \\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {\\cos(3^{n})}{2^{n}}$ has the property stated.[/quote]\n\nthere is no $ x$ in the expression...  :oops:[/quote]\r\n\r\nThanks, it was too obvious to include;-"
}
{
  "Tag": [
    "geometry",
    "area of a triangle"
  ],
  "Problem": "I need help on this area problem, please. You have a regular hexagon with area 36. Find the side length.",
  "Solution_1": "[quote=\"Arvind_sn\"]I need help on this area problem, please. You have a regular hexagon with area 36. Find the side length.[/quote]\r\n[hide]The area of a hexagon is $\\frac{3x\\sqrt{3}}{2}$. Solving the equation $\\frac{3x\\sqrt{3}}{2}=36$, $x\\sqrt{3}=24$ and $\\boxed{x=8\\sqrt{3}}$. [/hide]",
  "Solution_2": "I think area of regular hexagon is ((3)pow(1/2)*3)/2*side*side",
  "Solution_3": "[quote=\"merajatartofprobsolving\"]I think area of regular hexagon is ((3)pow(1/2)*3)/2*side*side[/quote]\r\n\r\nIt is, so chinesechess, you should solve for x in this:$\\frac{3x^2\\sqrt3}{2}=36$.",
  "Solution_4": "Oops. Yeah, you're right...that was weird. [hide]$x=2\\sqrt[4]{12}$ i think. [/hide]",
  "Solution_5": "Or you could divide the hexagon into 6 congruent equilateral triangles from the centre... If the area is 36, the area of a triangle is 6, and therefore each side is 2....",
  "Solution_6": "the answer should be 2 the power of 3 over 4",
  "Solution_7": "sorry i was wrong",
  "Solution_8": "$[ABCDEF]=6\\frac{a^2\\sqrt{3}}{4}$\r\n\r\nwe have: $[ABCDEF]=36$, that's why $a^2=8\\sqrt{3}\\iff a=\\sqrt{8}\\sqrt[4]{3}=2\\sqrt{2}\\sqrt[4]{3}$  :)\r\n\r\nEDIT: We can generalize it to regular $n$ polygon and compute with letters, not with $36$.. :P",
  "Solution_9": "I think you divide that 36 by 6 but dont take my word 4 it k? get it checked afterwards k?",
  "Solution_10": "[quote] I think you divide that 36 by 6 but dont take my word 4 it k? get it checked afterwards k? [/quote]\r\n\r\nI don't think it works that way...in doing that you split the hexagon into 6 pieces, each with area 6, but that doesn't get you much closer to finding the side length",
  "Solution_11": "Here is my approach:\r\n[hide]Since a hexagon consists of 6 equilateral triangles, each triangle has an area of 6. Area of an equilateral triangle is $\\frac{(s^2)\\sqrt{3}}{4}$. So 24=$s^2\\sqrt{3}$, so $s^2=8\\sqrt{3}$. Therefore $s= 2\\sqrt[4]{12}$.[/hide]\r\n\r\nCan a mod please insert the symbols since it is kind of messy like this and I don't know how to insert them? Thank you.\r\n[size=75][color=green]\nMod note: Latexed as much as I could [/color][/size] :)",
  "Solution_12": "[quote=\"life=tennis+math\"]Here is my approach:\n[hide]Since a hexagon consists of 6 equilateral triangles, each triangle has an area of 6. Area of an equilateral triangle is $\\frac{((s^2)\\sqrt{3})}{4}$. So $24=(s^2)\\sqrt{3}$, so $s^2=8\\sqrt{3}$. Therefore $s= 2\\sqrt[4]{12}$.[/hide]\n\nCan a mod please insert the symbols since it is kind of messy like this and I don't know how to insert them? Thank you.[/quote]\r\nFixed :)",
  "Solution_13": "the main approach is that when the opp. vertices of a hexagon are joined the diagonals are concurrent. thus 6 equilateral triangles are formed with each side being each side of the hexagon.",
  "Solution_14": "Dude... this thread is 6-7 years old.... don't revive it...."
}
{
  "Tag": [
    "geometry",
    "LaTeX",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "number theory",
    "prime numbers"
  ],
  "Problem": "The problems I'll be posting in this topic are just some problems from previous Canadian contests.  Anyone is welcome to try them.\r\n\r\n@mods : If these questions do not suit the level of Getting Started, please move it to the forum of your desire.\r\n\r\n1. A square whose area is [tex]100cm^2[/tex] is inscribed in a semicircular region.  The radius of the semicircle, in cm, is...\r\n\r\na)[tex]  \\sqrt{50}[/tex]  b) [tex]\\sqrt{75}[/tex] c)[tex]  \\sqrt{125}[/tex] d) [tex] \\sqrt{200}[/tex]  e)[tex]  \\sqrt{500}[/tex]\r\n\r\n2.  If [tex]\\sqrt{x} + \\sqrt{y}=\\sqrt{135}[/tex], where x and y are positive integers, then the value of x+y is...\r\n\r\na) 135 b) 75 c) 45 d) 30 e) 15\r\n\r\n3.  [tex]\\frac{2^{1990}-2^{1989}}{2^{1990}+{2^{1989}}}[/tex], equals\r\n\r\na) 0 b) [tex]2^{-3978}[/tex] c) [tex]\\frac{1}{3}[/tex] d) [tex]\\frac{1}{2}[/tex] e) [tex]2^{3978}[/tex]\r\n\r\n4  If [tex]a,b,$ and c[/tex] are prime numbers with [tex]a+b=c[/tex] and [tex]1<a<b<c<30[/tex], then the number of possible sets [tex]{a,b,c}[/tex] is...\r\n\r\na) 0 b) 2 c) 3 d) 4 e) 5\r\n\r\n5.  If 630 is expressed as the product of three positive integers with a sum of 26, then the smallest of these integers is...\r\n\r\na) 2 b) 3 c) 5 d) 6 e)7\r\n\r\nI will post more once these are completed.",
  "Solution_1": "1.[hide]The diagnol of the square is the radius of the smeicirle.\n\n\n\nThe side is 10, so the diagnol is  :sqrt: 10 * :sqrt:2 = :sqrt: 200 or d[/hide]\n\n\n\n5.[hide]The prime factorization of 630 is 2*3<sup>2</sup>*5*7.\n\n\n\nJust by looking at the possible way there can be 3 numbers, we have 10, 9, 7.  The smallest number in this set is 7, or 7 [/hide]\n\n\n\nAnd btw, is this representative of what would be on the first exam for the CMO?  Like if I got 4/5 on this one, then I would get 4 problems out of every 5 problems on the exam the CMO has?",
  "Solution_2": "Oh no.  These are some problems from the Grade 10 Math Contest in Canada.  It's called Cayley and I'll be taking this this year =D.  So I'm preparing for it as well.\r\n\r\nThe CMO is different.  There are around 25 problems, similar to the AMC.  I don't have much information about the CMO as I have not yet taken it yet but I will be taking it this year.",
  "Solution_3": "(3) [hide]\n\n\n\n(2^1990 - 2^1989)/(2^1990+2^1989)\n\n\n\n2^1990 = 2^1989*2^1\n\n\n\n[tex]\\frac{2^{1989}(2-1)}{2^{1989}(2+1)}=\\frac13[/tex]\n\n\n\n[/hide]",
  "Solution_4": "(2)\r\n$\\sqrt{145} = 3\\sqrt{15}$\r\nthus $\\sqrt{x} = \\sqrt{15}$ and $\\sqrt{y} = 2\\sqrt{15}$\r\n$x = 15, y=4\\cdot 15 = 60$ hence $x+y=75$",
  "Solution_5": "(4)\r\nSince c>b>a, and all must be prime, the least c is (2+3) or 5. Hence, every value of c is odd since the only even prime is 2. An odd can only be written as a sum of an odd and an even, so we know a must always be 2. So, the number of ways such that a + b = c where all are primes between 1 and 30 is simplythe number of twin primes between 1 and 30. So, there are four twins, (3,5), (5,7), (11,13), and (17,19). ;)",
  "Solution_6": "Good job everyone.  I don't have the answers but they all look correct to me since they do correspond to one of the multiple choice.\r\n\r\nNext set of questions.\r\n\r\n6.  The distance from the point (-1,-2) to the midpoint of the line segment joining points (2,3) and (-4,3) is...\r\n\r\na) [tex]\\sqrt{20}[/tex] b) [tex]\\sqrt{24}[/tex] c) 1 d) 4 e) 5\r\n\r\n7. If [tex]x^2 + y^2 = -2xy[/tex], then [tex]x^3 + 2y^3[/tex] equals...\r\n\r\na) [tex]y^3[/tex] b) [tex]y^6[/tex] c) [tex]3y^3[/tex]  d) [tex]7y^3[/tex]  e) [tex]9y^3[/tex] \r\n\r\n8. The base of a triangle is increased by 10%.  For the area to remain unchanged, the altitude must be decreased by...\r\n\r\na) 9% b) 9/1/11% c) 10% d) 11%\r\ne) 11/1/9%\r\n\r\nNote : Does anyone know how to work percentage in Latex\r\n\r\n9. The sum of four numbers is 64.  If 3 is added to the first number, 3 is subtracted from the second, the third is multiplied by 3, and the fourth is divided by 3, then all four results will be equal.  The difference between the largest and the smallest of the original numbers is...\r\n\r\na) 27 b) 32 c) [tex]\\frac{128}{3}[/tex] d) [tex]\\frac{130}{3}[/tex] e) 52\r\n\r\n10.  If [tex](x-a)(x-10) + 3 = (x+b)(x+c)[/tex] is true for all [tex]x[/tex] and if [tex]a,b,$ and c[/tex] are integers, then the number of possible values of [tex]a+b[/tex] is\r\n\r\na) 0 b) 2 c) 3 d) 4 e) 6",
  "Solution_7": "(6) is very very straightful...\r\nthe midpoint of (-4,3) and (2,3) is (-1, 3).\r\nthe distance from (-1,2) to (-1,3) is 5.\r\n\r\n(7) $x^2+y^2=-2xy \\Longrightarrow x^2+y^2+2xy=0 \\Longrightarrow (x+y)^2=0$ therefore $x=-y$, hence $x^3+2y^3 = (-y)^3+2y^3 = y^3$\r\n\r\n(8) suppose it decreases by $x$. and the original base and altitude are $a$ and $h$.\r\n$\\frac{1}{2}ah = \\frac{1}{2}\\cdot 1.1a \\cdot h(1-x)$\r\nwe get $x=1-1/1.1 = 1/11$.\r\n\r\nuse \\% to implement $\\%$.",
  "Solution_8": "9.[hide]\n\n{w+x+y+z}={64}\n\n\n\n [/hide]\n\n\n\nI know I went wrong somwhere, since [hide]256[/hide] isn't one of the answer choices..",
  "Solution_9": "(9)\r\nw+x+y+z=64\r\nw+3=x-3=3y=z/3\r\nx=w+6, y=$\\frac{w+3}{3}$, z = 3w+9\r\nw + w + 6 + $\\frac{w+3}{3}$ + 3w + 9 = 64\r\n5w + w/3 = 48\r\nw = 9, (w,x,y,z) = (9,15,4,36)\r\n36 - 4 = 32(B)",
  "Solution_10": "Oops I think I messed up\n\n9.\n\n[hide] We have a+3=b-3=3c=d/3\n\na+b+c+d=64, so dividing by 4 we have each equation must equal 16.  Solving gives a=19, b=13, c=48, d=16/3.  c is the largest and d is the smallest, so 144/3-16/3=128/3 C[/hide]",
  "Solution_11": "that's the same thing I did at first too",
  "Solution_12": "9. [hide](x-3) + (x+3) + (3x) + (x/3) = 64\n\nx-3+x+3+3x+1/3x = 64\n\n5_1/3x = 64\n\n64/5_1/3 = x\n\nx=12\n\n\n\nplug it in, and the highest number is 36, and the lowest is 4. 36-4 = 32.[/hide]",
  "Solution_13": "[quote=\"h_s_potter2002\"]\nAnd btw, is this representative of what would be on the first exam for the CMO?  Like if I got 4/5 on this one, then I would get 4 problems out of every 5 problems on the exam the CMO has?[/quote]\r\n\r\nErm.... correct me if I'm not mistaken, but the only CMO I know of is the Canadian Mathematical Olympiad... which is the Canadian equivalent of the USAMO and is FAR harder than these questions. It is also 5 questions long, however, and people qualify through the COMC.\r\n\r\nUnless there is a different CMO I don't know of....",
  "Solution_14": "Could someone explain number 3? I can't seem to solve it.",
  "Solution_15": "Ah ok, got it now. Thanks.",
  "Solution_16": "On #1 isnt the radius the diagonal of half the square?",
  "Solution_17": "I believe so\r\n\r\nIn which case, the answer would be root 125",
  "Solution_18": "[quote=\"Lucky707\"][quote=\"h_s_potter2002\"]\nAnd btw, is this representative of what would be on the first exam for the CMO?  Like if I got 4/5 on this one, then I would get 4 problems out of every 5 problems on the exam the CMO has?[/quote]\n\nErm.... correct me if I'm not mistaken, but the only CMO I know of is the Canadian Mathematical Olympiad... which is the Canadian equivalent of the USAMO and is FAR harder than these questions. It is also 5 questions long, however, and people qualify through the COMC.[/quote]\r\n\r\nI agree; the COMC is, to my knowledge, the 25-question multiple-choice test that kool_dudy was referring to.  The CMO is definitely olympiad-style; I believe it's 5 questions long.\r\n\r\nHowever, these questions aren't from either test - they're from one of {Fermat, Cayley, Hypatia, etc.}.  (I can't keep track of all of them...)  And there are many other CMOs - China, for one.  But the Chinese olympiad problems are nowhere near as hard as these.   :D  (I jest.)",
  "Solution_19": "(2^1990 - 2^1989) = (2^1989)(2-1) = 2^1989\r\n(2^1990 + 2^1989) = (2^1989)(2+1) = 3*2^1989\r\n\r\n[2^(1990) - 2^(1989)]/[2^(1990) + 2^(1989)]\r\n\r\n= (2^1989)/(3*2^1989)\r\n= 1/3\r\n\r\nedit - oops too late.. didn't see the second page"
}
{
  "Tag": [
    "vector",
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "So far, we are in the second week of class and last class we talked about linear independence and linear dependence..\r\n\r\nwithout further ado, here is a problem assigned to us by the prof.:\r\n\r\nLet V be a vector space over the reals\r\nLet u and v be distinct vectors in V. \r\nProve that {u,v} is linearly independent if and only if {u+v,u-v} is linearly independent.\r\n\r\nSo... basically, how do you approach this problem? All I have written down is:\r\nAssume {u,v} is linearly independent.\r\nI'm trying to prove the if first, and then the only if second... and yet i'm still completely stuck.\r\n\r\nHints would be greatly appreciated.. thanks.",
  "Solution_1": "$a_{1}\\mathbf{u}+a_{2}\\mathbf{v}=\\left(\\frac{a_{1}+a_{2}}2\\right)(\\mathbf{u}+\\mathbf{v})+\\left(\\frac{a_{1}-a_{2}}2\\right)(\\mathbf{u}-\\mathbf{v})$\r\n\r\nNote: $\\mathsf{V}$ is over a field with characteristic other than 2.",
  "Solution_2": "Generalization: Suppose $A$ is an $n\\times m$ matrix whose rows are independent vectors in $F^{m}$ and suppose $\\{u_{1}\\,u_{2},\\dots u_{m}\\}$ is an independent set in $V,$ which is a vector space over $F.$ For $i=1,2,\\dots,n$ let $v_{i}=\\sum_{j=1}^{m}a_{ij}u_{j}.$ Then $\\{v_{1},v_{2},\\dots,v_{n}\\}$ is independent.\r\n\r\nAs a corollary, if $A$ is an $n\\times n$ matrix with independent columns (that is, an invertible matrix), then $\\{u_{1}\\,u_{2},\\dots u_{n}\\}$ is independent if and only if $\\{v_{1},v_{2},\\dots,v_{n}\\}$ is independent.\r\n\r\n10000th user's insistence that the field not have characteristic $2$ is tantamount to insisting that $\\begin{bmatrix}1&1\\\\ 1&-1\\end{bmatrix}$ be invertible. In his post, he has in effect used $A^{-1}$ and the $2$ that appears in the denominator of his fractions is simply $-\\det(A).$\r\n\r\nProof of the claim: Suppose $\\sum_{i=1}^{n}\\alpha_{i}v_{i}=0.$\r\n\r\nThen $\\sum_{i=1}^{n}\\alpha_{i}v_{i}=\\sum_{i=1}^{n}\\left(\\sum_{j=1}^{m}a_{ij}u_{j}\\right)= \\sum_{j=1}^{m}\\left(\\sum_{i=1}^{n}\\alpha_{i}a_{ij}\\right)u_{j}=0.$\r\n\r\nSince the $u_{j}$ are independent, this means that for each $j,$ $\\sum_{i=1}^{n}\\alpha_{i}a_{ij}=0.$\r\n\r\nBut that then means that a linear combination of the rows of $A$ is the zero vector in $F^{m}.$ Since the rows of $A$ are independent, that means that $\\alpha_{i}=0$ for each $i.$\r\n\r\nHence we have proved that the $v_{i}$ are independent.",
  "Solution_3": "[quote=\"10000th User\"]$a_{1}\\mathbf{u}+a_{2}\\mathbf{v}=\\left(\\frac{a_{1}+a_{2}}2\\right)(\\mathbf{u}+\\mathbf{v})+\\left(\\frac{a_{1}-a_{2}}2\\right)(\\mathbf{u}-\\mathbf{v})$\n\nNote: $\\mathsf{V}$ is over a field with characteristic other than 2.[/quote]\r\n\r\nI was wondering, how did you come up with that solution? I know it works out  algebraically, but I'm still trying to arrive at that solution",
  "Solution_4": "Gilbert0000, what is the inverse of $\\begin{bmatrix}1&1\\\\1&-1\\end{bmatrix}\\ ?$",
  "Solution_5": "[quote=\"Kent Merryfield\"]Gilbert0000, what is the inverse of $\\begin{bmatrix}1&1\\\\1&-1\\end{bmatrix}\\ ?$[/quote]\r\n\r\n.5 .5\r\n.5 -.5\r\n\r\n\r\nstill.. im not exactly following how i'm supposed to use this in my proof..\r\n\r\n :(",
  "Solution_6": "[hide=\"Spoiler inside\"]\nWrite $a_{1}\\mathbf{u}+a_{2}\\mathbf{v}=b_{1}(\\mathbf{u}+\\mathbf{v})+b_{2}(\\mathbf{u}-\\mathbf{v})$\n\n$a_{1}\\mathbf{u}+a_{2}\\mathbf{v}=(b_{1}+b_{2})\\mathbf{u}+(b_{1}-b_{2})\\mathbf{v}$\n\n$\\begin{pmatrix}a_{1}\\\\a_{2}\\end{pmatrix}=\\begin{bmatrix}1&1\\\\1&-1\\end{bmatrix}\\begin{pmatrix}b_{1}\\\\b_{2}\\end{pmatrix}$\n\nSolve for $b_{1},b_{2}$\n[/hide]\r\nNice generalisation Prof. Merryfield :) Is this true?--> $A$ is invertible iff the vectors are linearly independent",
  "Solution_7": "if you mean ALL the row vectors ( or equivalently column vectors), yes. otherwise no"
}
{
  "Tag": [
    "inequalities",
    "Euler",
    "algebra",
    "polynomial",
    "number theory",
    "prime numbers",
    "absolute value"
  ],
  "Problem": "I saw the following problem in Komal and I don't have any idea of a solution. Can someone help me, please? \r\n Does there exist an integer n such that every prime factor of (2^n)-1 is at most (2^(n/1993))-1?",
  "Solution_1": "What is Komal?",
  "Solution_2": "is the Hungarian equivalent of AMM. :)",
  "Solution_3": "And what is AMM?\r\n\r\nLet n<sub>1</sub>,...,n<sub>s</sub> are pairwise coprime numbers. It is well known (2<sup>ni</sup>-1,2<sup>nj</sup>-1)=1.\r\n\r\nConsider n=n<sub>1</sub>...n<sub>s</sub>. We have 2<sup>ni</sup>-1 | 2<sup>n</sup>-1 ==> (2<sup>n</sup>-1) /  (2<sup>n1</sup>-1)...(2<sup>ns</sup>-1) = m is an integer number and m < 2<sup>s</sup>.\r\n\r\nSo our task to find s and n<sub>i</sub> (i=1..s) such that n<sub>i</sub> < n<sub>1</sub>...n<sub>s</sub>/1993 and s < n<sub>1</sub>...n<sub>s</sub>/1993. \r\nFor example, s=1993, n<sub>i</sub>=p<sub>1</sub>...p<sub>1993</sub>+p<sub>i</sub> where p<sub>i</sub> are successive prime numbers 2,3,5,7 and so on.",
  "Solution_4": "Kmal informatation can be found at:\r\n\r\nhttp://komal.elte.hu/info/bemutatkozas.e.shtml\r\n\r\nAMM (American Mathematical Monthly): \r\n\r\nhttp://gort.ucsd.edu/newjour/a/msg02973.html",
  "Solution_5": "It were grateful if you could (please) be more explicitely in your solution.",
  "Solution_6": "My solution is rigorous. Do you not understand some details?",
  "Solution_7": "To be sincerely, I don't understand how you proved that every prime factor of 2^n-1 is smaller than 2^(n/1993)-1.",
  "Solution_8": "We obtained 2<sup>n</sup>-1 = m(2<sup>n1</sup>-1)...(2<sup>ns</sup>-1) and each factor is less than 2<sup>n/1993</sup>-1 ==> each prime divisor is less than 2<sup>n/1993</sup>-1.\r\n\r\nPlease, write private message next time.",
  "Solution_9": "I thought I saw an error. Your last statement is false. Indeed, I will prove that the factor m is not smaller than 2^(n/19993)-1. It is obvious that prod (2^n_i-1) is strictly smaller than prod 2^n_i, which is 2^(n_1+...+n_s). Let us denote for simplicity 2^(n/1993)=a. Then it is clear that 2^n-1 is strictly greater than a^1992*(a-1). So proving that m>a-1 reduces to proving that a^1992>2^(n1+...+n_k). This is equivalent to proving that 1992 prod n_i>1993 sum n_i. Since n_i are different, it is clear that sum n_i/ prod n_i is smaller than (1+2+...+s)/s!, which is obviously smaller than 1992/1993. So, I believe that the whole argument of the solution is incorrect.",
  "Solution_10": "You are wrong.\r\n\r\n[quote]So proving that m>a-1 reduces to proving that a^1992>2^(n1+...+n_k). This is equivalent to proving that 1992 prod n_i>1993 sum n_i.[/quote]\r\nIt is a gross error: m>a-1 reduces to prooving 1992 \\sum n<sub>i</sub> > 1993 \\sum n<sub>i</sub> !!!\r\n\r\nI use obvious inequality 2<sup>ni</sup>-1 \\geq 2<sup>ni-1</sup> ==> m=(2<sup>n1+...+ns</sup>-1) / (2<sup>n1</sup>-1)...(2<sup>ns</sup>-1) \\leq 2<sup>n1+...+ns</sup> / 2<sup>n1-1</sup>...2<sup>ns-1</sup> = 2<sup>s</sup>.\r\nAll numbers n<sub>i</sub> are greater than 1993 ==> n/1993 = (n<sub>1</sub>+...+n<sub>s</sub>)/1993 > 1993 = s ==> m \\leq 2<sup>s</sup>-1 < 2<sup>n/1993</sup>-1.",
  "Solution_11": "Well, I'm a bit confused by your argument. You have said first that you take n=n_1*n_2*...n_s and noe you tell me that n=n_1+n_2+...+n_s. What should I believe? If you take n=n_1*n_2*...*n_k, then my argument ( which is good-try to look very well) shows that m>2^(n/1993)-1, while if you take n=n_1+...+n_k, why is it neccesarily that 2^n_i-1|2^n-1. And I wonder why doesn't any other meber interfere and say anything?",
  "Solution_12": "Sorry for my poor english, but I think that it's time someone different from myth and harazi read the proofs given by myth and harzi and completely close this useless dialog.",
  "Solution_13": "Noone wants to help me! Too bad!",
  "Solution_14": "So I should understand that everyone believes Myth's solution is corect? Ok then, I won't insist anymore. But I must say that I'm not satisfied with that solution and that, in my opinion, the problem is not solved yet.",
  "Solution_15": "Do you read private messages, harazi?\r\n\r\nI wrote to you that I really was wrong and my solution is incorrect. Execuse me for my self-confidence.",
  "Solution_16": "Then I must apologise. Please, excuse me.",
  "Solution_17": "Let Q_k(x) denote the k th cyclotomic polynomial. We have deg(Q_k(x))= phi(k), where phi is the Euler totient function.\r\nWe have (2^n)-1= Product Q_k(2) where k takes divisors of n. Since Q_k(2) is an integer for all k, we have for all prime divisor of (2^n)-1 is a divisor of Q_k(2) for some divisor k of n. Thus we have p<(n^2)*2^phi(n)\r\n(because all the coefficent of P_k(x) is smaller than n). Now we only have to choose n such that (n^2)*(2^phi(n)) < 2^(n/1993). This can easily done for n^2< a^n for any a>1 and n sufficently large and phi(n) <n/c for any given c>0 if we choose n=p(1)*p(2)*...*p(l) for sufficent large ( here p(i) denotes the i th prime number.).\r\nIn fact by choosing n=p(1)*p(2)*...*p(m) for sufficent large m, and using more sophisticated method to estimate phi(n), we can choose n such that for all prime divisors p of (2^n)-1 we have p<n^d for some constant d.",
  "Solution_18": "I don't know cyclotomic polinomial's theory. Can anyone verify this slution?",
  "Solution_19": "I think the solution Ihvietbao posted is corect. There is something I don't understand though:\r\n[quote](because all the coefficent of P_k(x) is smaller than n)[/quote]\r\n\r\nI think that a proper form of this should be: because all the coefficient of Q_k(x) are smaller that n (or k). \r\nCould you, Ihvietbao, please explain it to me why this is true?",
  "Solution_20": "I think that a proper form of this should be: because all the coefficient of Q_k(x) are smaller that n (or k). \r\nCould you, Ihvietbao, please explain it to me why this is true?\r\n\r\nWell, I can't prove that statement, so I have a correction:\r\nWe don't need to estimate the coefficent of P_k(x) that \"sharp\".\r\nWe only have to use a weaker estimation, which is almost obvious:\r\nthe absolute value of the coefficent of x^j in P_k(x) doesn't exceed jCphi(k) (0<=j<=k) (uCv is the binomial coefficent), because all the roots of P_k(x) has modul 1.\r\nThus we have abs(P_k(2))<= 2^k*(\\sum jCphi(k)) = 4^phi(k) <= 4^phi(n) (because k is a divisor of n). (in the above sum j runs from 0 to phi(k))\r\n\r\nSo for each prime divisor of (2^n)-1 we have p<=4^phi(n), and now the proof continues as posted and we are done.\r\n(At first I thought that all the coefficents of cyclotomic polynomials are either -1,1 or 0, but in fact, P_105(x) has a coefficent -2!)"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove this inequality for [tex] a,b,c >0 [/tex]\r\n[tex] \\sqrt{\\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}}+\\sqrt{\\frac{2(ab+bc+ca)}{a^2+b^2+c^2}} \\geq 1+\\sqrt{2} [/tex]  :coool:",
  "Solution_1": "I think I have a very funny solution.\r\n\r\nIf $\\sqrt{a},\\sqrt{b},\\sqrt{c}$ are sidelengths of a triangle, then the inequality is trivial, because $2(ab+bc+ca) \\ge a^2+b^2+c^2$\r\n\r\nIf not, let me use a lemma,\r\nAssume x,y,p,q are positive, then if $x \\ge p \\ge y$, $q \\ge y$, $xy \\ge pq$, then $x+y \\ge p+q$\r\nProof:\r\nAssume $xy=kpq$ where k is a constant > 1. Then, $x=\\frac{kpq}{y}$.\r\n$x+y \\ge p+q$\r\niff $kpq+y^2 \\ge y(p+q)$\r\niff $(k-1)pq + (p-y)(q-y) \\ge 0$\r\nSince $k-1 >0$ and $q-y \\ge 0$, $p-y \\ge 0$, the inequality is true.\r\n\r\nNow, let $x = \\sqrt{\\frac{\\sum a^4}{\\sum a^2b^2}}$, $y = \\sqrt{\\frac{2(\\sum ab)}{\\sum a^2}}$, $p=1$, $q=\\sqrt{2}$\r\nIt's easy to see that $x \\ge p$, $q \\ge y$.\r\n$p \\ge y$ because $\\sqrt{a},\\sqrt{b},\\sqrt{c}$ does not form a triangle.\r\nTo prove $xy \\ge pq$, we need to prove \r\n$(\\sum a^4)(\\sum ab) \\ge (\\sum a^2b^2)(\\sum a^2)$\r\nThis resolves to $(5,1,0) \\succ (4,2,0)$ and $(4,1,1) \\succ (2,2,2)$ and muirhead. \r\nSo we can apply our lemma and the question is done.",
  "Solution_2": "Very nice approach  :) , but I still don't understand why it is trivial when [tex] \\sqrt{a},\\sqrt{b},\\sqrt{c} [/tex] are sidelenghs of a triangle  :( .",
  "Solution_3": "[quote=\"Anh Cuong\"]Very nice approach  :) , but I still don't understand why it is trivial when [tex] \\sqrt{a},\\sqrt{b},\\sqrt{c} [/tex] are sidelenghs of a triangle  :( .[/quote]\r\n\r\noops, I think I must have something wrong in my mind... let me think again..  :blush:",
  "Solution_4": "I tried this inequality but haven't finished yet\r\nMy approach is to prove two inequalities:\r\n$\\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}\\ge\\frac{a^2+b^2+c^2}{ab+bc+ca}$\r\n[tex] \\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\\frac{2(ab+bc+ca)}{a^2+b^2+c^2}} \\geq 3 [/tex]",
  "Solution_5": "Yes,nice!",
  "Solution_6": "[quote=\"keira_khtn\"]Yes,nice![/quote]\r\nDo you have a aolution for the ideas of Minhkoa?",
  "Solution_7": "My solution was not completed, can anyone give me a hand?",
  "Solution_8": "[quote=\"silouan\"][quote=\"keira_khtn\"]Yes,nice![/quote]\nDo you have a aolution for the ideas of Minhkoa?[/quote]\r\nYes,\r\n1/my proof for first lemma:\r\nAfter expandind we get the equivalent ineq:\r\n$(a^4bc+b^4ca+c^4ab-3a^2b^2c^2)+\\sum(a^5b+b^5a-a^2b^4-a^4b^2)\\geq 0$\r\nwhich is obviously true by Cauchy and Muirhead\r\n2/Solution of second lemma:Using Cauchy ineq for 3 expression:\r\n$LHS\\geq 3 \\sqrt[3]{\\frac{(a^4+b^4+c^4)(ab+bc+ca)^2}{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)^2}}$\r\nIt remains to prove:\r\n$(a^4+b^4+c^4)(ab+bc+ca)^2 \\geq (a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)^2$\r\nWhich is equivalent to:\r\n$abc^2(a-b)(a^3-b^3)+ab^2c(c-a)(c^3-a^3)+a^2bc(b-c)(b^3-c^3)+abc[a^5+b^5+c^5-abc(a^2+b^2+c^2)] \\geq 0$\r\n********\r\nReturn the given ineq,Multiplying both two hands we get:\r\n$\\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\\frac{2(ab+bc+ca)}{a^2+b^2+c^2}+2\\sqrt{\\frac{2(a^4+b^4+c^4)(ab+bc+ca)}{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)}}\\geq 3+2\\sqrt2$\r\nSince Minhkhoa's two lemma and it follows"
}
{
  "Tag": [
    "function",
    "inequalities",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Determine, with proof, all functions $f: %Error. \"matbbl\" is a bad command.\n{R}\\longrightarrow %Error. \"matbbl\" is a bad command.\n{R}$ that map every closed interval onto a closed interval of the same length.\r\n\r\n\r\n[hide]\ni dont have a proof but i think that the answer is $f=x+b$ for some constant $b$.\n[/hide]",
  "Solution_1": "Since $[x,y]$ is sent to an interval of length $x-y$, $|f(x)-f(y)|\\le |x-y|$ for all $x,y$. This gives continuity. Moreover, $|f(a)-f(b)|=|x-y|\\ge |a-b|$ for some pair $a,b\\in [x,y]$. We must have $\\{a,b\\}=\\{x,y\\}$ for this pair of inequalities to hold, and $|f(x)-f(y)|=|x-y|$ for all $x,y$. For each $x$, either $f(x)=f(0)+x$ or $f(x)=f(0)-x$. By continuity, the same choice must be made for all positive $x$ and the same choice must be made for all negative $x$. A simple check ($f(x)\\neq f(-x)$) shows that either $f(x)=f(0)+x$ for all $x$ and $f(x)=f(0)-x$. Making the choice $f(0)=c$, both choices work."
}
{
  "Tag": [
    "AwesomeMath",
    "summer program",
    "HCSSiM",
    "Mathcamp",
    "Ross Mathematics Program"
  ],
  "Problem": "Sorry if this has been asked before - I did not spend an hour searching on the forums for this question.\r\n\r\nCan you really apply to more than one math summer camp? (i.e. I am refering to ones like Awesomemath, HCSSiM, mathcamp, ross program, etc) All these camps don't mind if you are accepted and then choose to not go?\r\n\r\nThanks.",
  "Solution_1": "Of course.  You can apply to more than one college, can't you?",
  "Solution_2": "are the apps to any of these programs open yet?",
  "Solution_3": "I am trying to decide whether I should plan to set aside time to do some of the awesome math early registration problems, but they won't be up for another 3 weeks.\r\n\r\nanother question: if I am accepted to more than one camp, about when would I need to decide which one I want to go to? I realize this would probably vary from camp to camp. Also, if I apply early to AwesomeMath, do I have to go?\r\n\r\nThanks."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "Calculate these:\r\n$ \\int\\frac{x^4\\plus{}1}{x^6\\plus{}1}dx$\r\n$ \\int ln(lnx)dx$",
  "Solution_1": "[quote=\"dulldrake\"]\n$ \\int\\frac {x^4 \\plus{} 1}{x^6 \\plus{} 1}dx$\n\n[/quote]\r\n\r\nTry:\r\n\r\n$ (\\cos x)^{\\frac{1}{4}} \\equal{} t$\r\n\r\n\r\n\r\n\r\n.",
  "Solution_2": "hello, to your first integral, note that\r\n$ \\frac{x^4\\plus{}1}{x^6\\plus{}1}\\equal{}\\frac{1}{3}\\frac{x^2\\plus{}1}{x^4\\minus{}x^2\\plus{}1}\\plus{}\\frac{2}{3(x^2\\plus{}1)}$.\r\nSonnhard.",
  "Solution_3": "hello, to your second integral, set $ \\ln(x)\\equal{}t$, then we have to solve $ \\int e^t \\cdot \\ln(t)\\,dt$.\r\nSonnhard.",
  "Solution_4": "[hide=\"1\"]\n$ \\frac{x^4\\plus{}1}{x^6\\plus{}1}\\equal{}\\frac{x^2\\plus{}1}{3(x^4\\minus{}x^2\\plus{}1)} \\plus{} \\frac{2}{3(x^2\\plus{}1)}$\nso ur integral \n$ \\int \\frac{x^4\\plus{}1}{x^6\\plus{}1}\\equal{}\\int \\frac{x^2\\plus{}1}{3(x^4\\minus{}x^2\\plus{}1)}dx \\plus{}\\int \\frac{2}{3(x^2\\plus{}1)}dx$\n$ \\equal{}\\int \\frac {1 \\plus{}\\frac{1}{x^2}}{3(x^2\\minus{}1\\plus{}\\frac{1}{x^2})} \\plus{} \\frac{2}{3}tan^{\\minus{}1}(x)$\n=$ \\int \\frac {1 \\plus{}\\frac{1}{x^2}}{3((x\\minus{}\\frac{1}{x})^2\\plus{}1)}\\plus{} \\frac{2}{3}tan^{\\minus{}1}(x)$\n=$ \\frac{1}{3}tan^{\\minus{}1}(x\\minus{}\\frac{1}{x}) \\plus{} \\frac{2}{3}tan^{\\minus{}1}(x)$\n[/hide]",
  "Solution_5": "hello, this is ok, Maple 12 gives us the following answer\r\n$ \\int\\ln(\\ln(x))\\,dx\\equal{}\\ln  \\left( \\ln  \\left( x \\right)  \\right) x\\plus{}{\\it Ei} \\left( 1,\\minus{}\\ln \r\n \\left( x \\right)  \\right)$\r\nSonnhard."
}
{
  "Tag": [
    "Pascal\\u0027s Triangle",
    "Pythagorean Theorem",
    "geometry"
  ],
  "Problem": "im rather confused on how there are so many results attributed to china, arabia, and india, \r\n(such as, they attribute pascal's triangle, the bernolli numbers, the pythagorean theorem, the base 10 system, etc) but the people who invented/discovered these are never named in the books i read asides from ramanujan, but perhaps that is because he went under the wing of a western european mathematician.Al kwarizi is only known (did i spell that rihgt?) because we coined algebra after his name. perhaps it is because i live in america, so the resources around me are completely biased for west european mathematicians\r\n\r\ndoes anyone have any names?",
  "Solution_1": "The reason (in my personal opinion) is that we are from the west, and for centuries the west misjudge as poor and non-important the science in the east. It is just rescently (not more than two centuries) that the west realize that there are as many geniuses as in the West. How could a genius not to be European, or descendent of europeans in the case of America. Just a personal opinion. Al Kwarizmi is just an exception to my argument, but, well, all non-mathematical rule has its exceptions.",
  "Solution_2": "So, just for the record, Al Kwarizmi is the name from which we derive the word \"algorithm.\"  The word algebra comes from the Arabic [i]al-jabr[/i], used in the title of one of his works.  As for the actual content of your question, we also don't know many particular ancient Babylonian, Egyptian, or Mayan mathematicians, but they certainly must have existed.  Very early mathematics was largely not about actually studying mathematics but instead about such practical tasks as calculating boundaries between lands (surveying) and astronomical calculations.  As such, many of the people doing mathematics weren't doing it in the way that gets your name attached to a theorem."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ a$ be a real number. Prove that if the equation $ x^2\\minus{}ax\\plus{}a\\equal{}0$ has two real roots $ x_1$ and $ x_2$, then: $ x_1^2\\plus{}x_2^2 \\ge 2(x_1\\plus{}x_2).$",
  "Solution_1": "$ a^2\\minus{}4a\\geq 0$  $ x_{1}^{2}\\plus{}x_{2}^{2}\\equal{}( x_{1}\\plus{}x_{2})^2\\minus{}2x_1.x_2\\equal{}a^2\\minus{}2a$\r\n$ a^2\\minus{}2a\\geq 2a$",
  "Solution_2": "$\\color{blue} \\boxed{\\textbf{SOLUTION}}$\n\nBy $\\textbf{Vieta's Formula,}$\n$$x_1+x_2=a, x_1x_2=a $$\n\nSo, ${x_1}^2+{x_2}^2=a^2 -2a$\nAnd, $2(x_1+x_2)=2a$\n\nWe need to show, \n${x_1}^2+{x_2}^2 \\ge 2(x_1+x_2) \\implies a^2-2a \\ge 2a \\implies a \\ge 4$\n\nWhich is true because $\\triangle \\ge 0 \\implies a^2-4a \\ge 0 \\blacksquare$",
  "Solution_3": "[quote=lian_the_noob12]Here is a mistake in the Question, It should be $a \\ge 4$\n\n$\\color{blue} \\boxed{\\textbf{SOLUTION}}$\n\nBy $\\textbf{Vieta's Formula,}$\n$$x_1+x_2=a, x_1x_2=a $$\n\nSo, ${x_1}^2+{x_2}^2=a^2 -2a$\nAnd, $2(x_1+x_2)=2a$\n\nWe need to show, \n${x_1}^2+{x_2}^2 \\ge 2(x_1+x_2) \\implies a^2-2a \\ge 2a \\implies a \\ge 4$ as desired $\\blacksquare$[/quote]\n\nNo problem w/ the question. As you know the problem is to prove that $a^2-4a\\geq0$, which is true since $x_1,x_2$ being real leads to $a\\leq0$ or $a\\geq4$.\n",
  "Solution_4": "[quote=qwedsazxc][quote=lian_the_noob12]Here is a mistake in the Question, It should be $a \\ge 4$\n\n$\\color{blue} \\boxed{\\textbf{SOLUTION}}$\n\nBy $\\textbf{Vieta's Formula,}$\n$$x_1+x_2=a, x_1x_2=a $$\n\nSo, ${x_1}^2+{x_2}^2=a^2 -2a$\nAnd, $2(x_1+x_2)=2a$\n\nWe need to show, \n${x_1}^2+{x_2}^2 \\ge 2(x_1+x_2) \\implies a^2-2a \\ge 2a \\implies a \\ge 4$ as desired $\\blacksquare$[/quote]\n\nNo problem w/ the question. As you know the problem is to prove that $a^2-4a\\geq0$, which is true since $x_1,x_2$ being real leads to $a\\leq0$ or $a\\geq4$.[/quote]\n\nOh Yes, Sorry I didn't notice this, So for real roots\n$$\\triangle \\ge 0 \\implies a^2-4a \\ge 0$$",
  "Solution_5": "[quote=moldovan]Let $ a$ be a real number. Prove that if the equation $ x^2\\minus{}ax\\plus{}a\\equal{}0$ has two real roots $ x_1$ and $ x_2$, then: $ x_1^2\\plus{}x_2^2 \\ge 2(x_1\\plus{}x_2).$[/quote]\n\nNote that $x_1+x_2=x_1x_2=a$, so the inequality is equivalent to $a^2-4a\\geq 0$ which is true because this is the discriminant. "
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "ratio",
    "inequalities",
    "geometry proposed"
  ],
  "Problem": "Let $ ABC$ be an acute triangle. Points $ D$, $ E$, and $ F$ lie on segments $ BC$, $ CA$, and $ AB$, respectively, and each of the three segments $ AD$, $ BE$, and $ CF$ contains the circumcenter of $ ABC$. Prove that if any two of the ratios $ \\frac{BD}{DC}$, $ \\frac{CE}{EA}$, $ \\frac{AF}{FB}$, $ \\frac{BF}{FA}$, $ \\frac{AE}{EC}$, $ \\frac{CD}{DB}$ are integers, then triangle $ ABC$ is isosceles.",
  "Solution_1": "No solution for this problem  :o  ?",
  "Solution_2": "If i have understood the problem, $AD,BE,CF$ are concurrent and the point of concurrency is the circumcenter of \n$\\triangle ABC.$\nIf we consider four following cases then we are done.(Other cases are symmetric)\n$\\boxed {case (i):-}$\n$\\frac {BD}{CD}$ and $\\frac {CD}{BD}$ are both integers then $BD=CD \\implies AB=AC$\n$\\boxed {case (ii):-}$\n$\\frac {AF}{BF}$ and $\\frac {AE}{CE}$ are both integers.\n$\\implies \\frac {b^2(c^2+a^2-b^2)}{a^2(c^2+b^2-a^2)}$ and $\\frac {c^2(b^2+a^2-c^2)}{a^2(c^2+b^2-a^2)}$ are integers.\n$\\implies \\frac {b^2(c^2+a^2-b^2)}{a^2(c^2+b^2-a^2)}- \\frac {c^2(b^2+a^2-c^2)}{a^2(c^2+b^2-a^2)}=\\frac {b^2-c^2}{a^2}$ is an integer.\nSince $\\triangle ABC$ is acute, So $-1 < \\frac {b^2-c^2}{a^2} < 1$\nhence $AB=AC$\n$\\boxed {case (iii):-}$\n$\\frac {AF}{BF}$ and $\\frac {CE}{AE}$ are integers.\n$\\implies \\frac {CD}{BD}$ is an integer.\nNow similarly as previous we get $CA=CB$\n$\\boxed {case (iv):-}$\n$\\frac {BF}{AF}$ and $\\frac{CE}{AE}$ are integers.\n[hide]Someone for this case please ??? [/hide]",
  "Solution_3": "Yeah, I can help with the last case ($\\frac {a^2(c^2+b^2-a^2)}{b^2(c^2+a^2-b^2)}$ and $\\frac {a^2(c^2+b^2-a^2)}{c^2(b^2+a^2-c^2)}$ are integers). Denote $a^2=x,b^2=y, c^2=z$ and $x,y,z$ are the sidelenghts of a triangle, so we can write $x=v+w, y=u+w, z=u+v$. Now, we know that $\\frac {a^2(c^2+b^2-a^2)}{b^2(c^2+a^2-b^2)} \\ge 2$ and $\\frac {a^2(c^2+b^2-a^2)}{c^2(b^2+a^2-c^2)} \\ge 2$ (in fact, if we have both of them $2$, we are done since $D$ is the midpoint of $BC$, case already treated). If you are patient enough to multiply the paranthesis, you will get $xz+2y^2 \\ge x^2+xy+2yz, xy+2z^2 \\ge x^2+2yz+xz$. We have 3 cases:\n1. $x \\ge y \\ge z$ , when $xy+2z^2 <x^2+2yz+xz$, impossible (the equations are symmetric in $y$ and $z$).\n2. $y \\ge z \\ge x$ which implies $ u \\ge w \\ge v$ and the two inequlities become $uw \\ge uv+2vw ; uv \\ge uw+2vw$, obviously impossible.\n3. $y \\ge x \\ge z$ which implies $ w \\ge u \\ge v$ and the two inequlities become $uw \\ge uv+2vw ; uv \\ge uw+2vw$, obviously impossible.\n\nSo, at least one of the ratios is $1$ and we already solved this case."
}
{
  "Tag": [
    "complex numbers"
  ],
  "Problem": "From what I have learned of complex numbers, it seems to me that any number's nth root yields $n$ complex roots.\r\n\r\nSo for example, $8$ must have $3$ 3rd roots.\r\n\r\nWhich should be:\r\n\r\n$2 cis 0$\r\n$2 cis 120$\r\n$2 cis 240$\r\n\r\n\r\nSo say your solving the simple equation: $x^3 = 8$\r\n\r\nShould one include all 3 roots in their answer or just the real roots? I was taught that if  the equation is $x^2 = 4$, then the answer must be $\\pm2$",
  "Solution_1": "ask your teacher what he/she wants, on contests give all solutions unless it says real roots, etc",
  "Solution_2": "And sometimes (in lower classes, Algebra 1 and such) they only ask for the single real positive root. If you're learning De Moivre's theorem, then the assignment would probably to provide all roots.",
  "Solution_3": "[quote=\"breez\"]it seems to me that any number's nth root yields $n$ complex roots.[/quote]\r\n\r\nThis depends strongly on exactly what you mean.  The equation $x^n = a$ has $n$ complex roots - however, strictly speaking, $x = \\sqrt[n]{a}, a \\in \\mathbb{R}$ is defined to produce only the principal value - the single positive real root.  \r\n\r\nSo if you wanted to find the solutions to $x^3 = 8$, you would include all the roots.  If you wanted to evaluate $\\sqrt[3]{8}$, you would include only $2$.  Strictly speaking."
}
{
  "Tag": [],
  "Problem": "The sum of three different positive unit fractions is $ \\frac78$. What is the least number that can be the sum of the denominators of these fractions?",
  "Solution_1": "Rather obviously, it's $ \\frac{1}{2}\\plus{}\\frac{1}{4}\\plus{}\\frac{1}{8}$, where the sum of the denominators is $ 2\\plus{}4\\plus{}8\\equal{}\\boxed{14}$.",
  "Solution_2": "It's relatively easy to see that $ frac12\\plus{}\\frac14\\plus{}\\frac18\\equal{}\\frac78$, and since no smaller values work, $ 2\\plus{}4\\plus{}8\\equal{}\\boxed{14}$.",
  "Solution_3": "What if there was an extension to this problem, like if 3 unit fractions add to 37/81?\n\nIs there a way to solve this systematically?"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "Bag A contains four yellow marbles, two green marbles and three blue marbles.  \tBag B contains four blue marbles and four green marbles.  If you choose one marble from each bag, what is the probability that you end up with one blue marble and one yellow marble?  Express your answer as a common fraction.",
  "Solution_1": "There are no yellow marbles in Bag B, so they have to come from Bag A. Thus our answer is $ \\frac{4}{9}\\cdot\\frac{4}{8}\\equal{}\\boxed{\\frac{2}{9}}$."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "\\/closed"
  ],
  "Problem": "What is the significance of the Art of Problem Solving logo?",
  "Solution_1": "It's an optical illusion. There are different amounts of little cubes depending on how you look at it.",
  "Solution_2": "That's not the only significance.  Imagine that the colors were not highlighted and the tiling was arbitrary within the same hexagon.  Can you think of a math problem that might be asked about the figure?",
  "Solution_3": "I don't know if I see it correctly, but it looks like it could be an area/volume question. Clarification?",
  "Solution_4": "See http://www.mathlinks.ro/Forum/viewtopic.php?t=21705",
  "Solution_5": "Thanks I got it now.",
  "Solution_6": "Indeed, that problem is the inspiration for the logo."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Show that 26 is the only number between a square and a cube.",
  "Solution_1": "Posted before."
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "Putnam",
    "calculus computations"
  ],
  "Problem": "find $\\int_{0}^{1}\\ln{(1-x)}\\cdot\\ln{(1+x)}dx$",
  "Solution_1": "The integral equals: $2-\\frac{\\pi^{2}}{6}-2\\ln 2+(\\ln 2)^{2}$.\r\nTo see this use the following expansions:\r\n\r\n$\\ln(1-x)=-\\sum_{m=1}^{\\infty}\\frac{x^{m}}{m}$, $-1\\leq x<1$,\r\n\r\n$\\ln(1+x)=-\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}x^{n}}{n}$, $-1<x\\leq 1$\r\nand \r\n$\\frac{1}{2}(\\ln(1-z))^{2}=\\sum_{n=1}^{\\infty}\\frac{H_{n}}{n+1}z^{n+1}$, $|z|<1$, wher $H_{n}$ is the $n^{th}$ harmonic number\r\n\r\nand \r\n\r\n$\\sum_{k=1}^{\\infty}\\frac{(-1)^{k+1}}{k^{2}}=\\frac{\\pi^{2}}{12}$\r\nLet $I$ be the value of your integral, then \r\n\r\n$I=\\sum_{n,m=1}^{\\infty}\\frac{(-1)^{n}}{nm(n+m+1)}=\\sum_{n=1}^{\\infty}\\left(\\frac{(-1)^{n}H_{n+1}}{n}-\\frac{(-1)^{n}H_{n+1}}{n+1}\\right)$.\r\nFrom now on, using some simple *rearrangements*, the calculations are straightforward.",
  "Solution_2": "[quote=\"didilica\"]\nFrom now on, using some simple *rearrangements*, the calculations are straightforward.[/quote]\r\n\r\ni dont see it :wink: \r\nplease elaborate.",
  "Solution_3": "Here is the detailed work:\r\n\r\n$I=\\sum_{n=1}^{\\infty}\\left(\\frac{(-1)^{n}H_{n}}{n}+\\frac{(-1)^{n}}{n(n+1)}\\right)-\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}}{n+1}\\left(H_{n}+\\frac{1}{n+1}\\right)$.\r\n\r\n$=\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}H_{n}}{n}+\\sum_{n=1}^{\\infty}(-1)^{n}\\frac{1}{n}-\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}}{n+1}-\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}H_{n}}{n+1}-\\sum_{n=1}^{\\infty}\\frac{(-1)^{n}}{(n+1)^{2}}$\r\n\r\n$=-1+\\sum_{n=2}^{\\infty}\\frac{(-1)^{n}H_{n-1}}{n}+\\sum_{n=2}^{\\infty}\\frac{(-1)^{n}}{n^{2}}-\\ln 2-\\ln 2+1+\\frac{(\\ln 2)^{2}}{2}+\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{(n+1)^{2}}$\r\n\r\n$=-1+\\frac{(\\ln 2)^{2}}{2}-\\frac{\\pi^{2}}{12}+1-2\\ln 2+1+\\frac{(\\ln 2)^{2}}{2}+\\frac{-\\pi^{2}}{12}+1=2+\\ln^{2}2-\\frac{\\pi^{2}}{6}-2\\ln 2$.\r\nI hope that there are no computational mistakes. Anyway this is my approach for solving this problem.",
  "Solution_4": "haha, this problem is suddenly popping up in so many places.\r\n\r\ni was given the exact problem in my putnam meetings the other night.\r\nand the same problem was used in our university's challenge problem.\r\n\r\nmy solution is almost identical to the didilica's, so i agree with the answer.\r\n\r\ni wonder if there's a shorter and/or more elegant way to do this!"
}
{
  "Tag": [
    "number theory",
    "prime factorization"
  ],
  "Problem": "1.\tHow many positive integers are divisors of 444,444?\r\n[hide]96[/hide]",
  "Solution_1": "i believe the answer to this problem is:\r\n\r\n[hide]96.\n\n[/hide]hi roland",
  "Solution_2": "[quote=\"helga7032Z\"]i believe the answer to this problem is:\n\n[hide]96.\n\n[/hide]hi roland[/quote]\r\neh.. next time can you explain?\r\n\r\nI believe prime factorization of $ 444444$ is $ 2^2\\cdot3\\cdot7\\cdot11\\cdot13\\cdot37$\r\n\r\nSo (exponent+1)$ \\rightarrow 3\\cdot2\\cdot2\\cdot2\\cdot2\\cdot2$ and multiply to get $ \\boxed{96}$",
  "Solution_3": "[quote=\"helga7032Z\"]i believe the answer to this problem is:\n\n[hide]96.\n\n[/hide]hi roland[/quote]\r\nI believe that you should post a solution with your answers?\r\nBecause he told us the answers, and wanted a solution."
}
{
  "Tag": [
    "function",
    "geometry",
    "3D geometry",
    "sphere"
  ],
  "Problem": "Prove that the following equation has exactly one root in $ (\\frac{1}{2},1)$:\r\n$ \\frac{e^x}{(x\\plus{}1)^2}\\equal{}x$",
  "Solution_1": "Let $ f(x) \\equal{} e^x \\minus{} (x)(x \\plus{} 1)^2$.\r\n\r\nWe can use crude approximations to show that:\r\n\r\n$ f(1/2) > 0 > f(1)$\r\n\r\nHence there is at least one root due to the intermediate value theorem. (since obviously f is continuous).\r\n\r\nThen $ f'(x) \\equal{} e^x \\minus{} (x \\plus{} 1)(3x \\plus{} 1)$\r\n\r\nagain, we can use crude approximations to show that\r\n\r\n$ f'(x) < 0$ for $ x\\in[0.5,1]$ [that is a very weak inequality...]\r\n\r\nIn particular, $ e^x\\le e^{1} < 3$\r\n\r\nand $ \\minus{} (x \\plus{} 1)(3x \\plus{} 1) \\le \\minus{} 3.75 < \\minus{} 3$\r\n\r\nSumming those gives us the result...\r\n\r\nHence $ f(x)$ is strictly decreasing in that interval and it has only one root.",
  "Solution_2": "This is not a formal proof but, \r\nConsider the graphs of $ y \\equal{} e^x$ and $ y \\equal{} f(x) \\equal{} x(x \\plus{} 1)^2$in the interval $ (\\frac {1}{2},1)$ (at this point it is almost evident)\r\nNote that $ f$ and $ e^x$ both are increasing continuous functions, for $ x > 0$\r\nSince $ f(\\frac {1}{2}) < \\sqrt {e}$ and $ f(1) > e^1$, So(Here it comes the key of this problem):\r\n\"The line joining $ f(\\frac {1}{2})$ and $ f(1)$ must intercept the graph $ y \\equal{} e^x$ somewhere, when $ x\\in(\\frac {1}{2},1)$, it cannot intercept it twice since (reread the above conditions) for $ x \\equal{} 1$ $ f$ is still inside the region bounded by $ x \\equal{} 0$ and $ y \\equal{} e^x.$\"\r\n\r\n\r\n :)"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "[b]1.2[/b] [i] A coin is flipped 1001 times.  What is the probablity that at least 3 heads will appear consecutively in the set of heads and tials?[/i]\r\n\r\n[i]Masoud Zargar[/i]",
  "Solution_1": "I was just about to delete this post...",
  "Solution_2": "[quote=\"draperp\"]There are $1001-2=999$ ways that the three flips could be consecutive.  Therefore, the probability is $\\frac{999}{2^{3}}=\\frac{999}{8}$.[/quote]\r\n\r\nUmm, probability is never greater than 1, and what if I flipped an alternating pattern of tails and heads? That means that the probability of this is lower than 1.",
  "Solution_3": "Your avatar is a game of Go.  It looks like the players are not very aggressive.  :)\r\n\r\nEdit:  A recursion is possible, but solving it would not be very nice as it would include the number of ways that at least two consecutive heads can be flipped.  Two recursions, really.   :|",
  "Solution_4": "If you find the recursion, you can relate it to a simpler sequence.",
  "Solution_5": "[hide=\"this is what I got\"]\nlet $p_{n}$ be the total number of $n$ term sequences that satisfy the given condition.  then we wish to calculate $p_{n+1}$.\nIf we stick a head or a tail to the end of one of the $p_{n}$ sequence, it'll still work.  so $2p_{n}$ sequences work.\nOtherwise, if we stick a head to the end of a sequence, and it contains 2 consecutive heads at the end, and no consecutive 3 heads block before, we got a new satisfying sequence.\nIn the latter case, the last two must be heads, and the third one from the last must be tails.  All the terms before can be anything as long as it does not have a consecutive 3 block.\nSo we get:\n\n$p_{n+1}= 2p_{n}+(2^{n-3}-p_{n-3})$.\n\nWith $p_{0}= p_{1}= p_{2}= 0$, and $p_{3}= 1$.  And to get the probability, divide $p_{n}$ by $2^{n}$.\n[/hide]\r\n\r\nSomeone can figure out the close form from now on...",
  "Solution_6": "It is very, very close to $1$."
}
{
  "Tag": [],
  "Problem": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188613[/url]\r\n\r\n\r\n\r\nI guess this is a reminder....",
  "Solution_1": "Yes, it is",
  "Solution_2": "Let's break a record people!",
  "Solution_3": "Woohoo! Haha, this will be fun  :D"
}
{
  "Tag": [
    "function",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "I just want to make sure I did this right.\r\n\r\nSuppose the group G acts on the set X.  Let x be in X.  Find a bijection from the set of right cosets of the stabilizer to the G-orbit of x.  Prove that the function is indeed a bijection.\r\n\r\nWe need to prove a bijection between the right cosets of the stabilizer and the orbit.\r\n\r\nDefining the function as f: H--> Orb(x)\r\nH=gG_x\r\n\r\nOne-to-one\r\nf(g_1H)=f(g_2H)\r\ng_1x=g_2x\r\nx=(g_1)(g_2)^{-1}x  This stabilizes x, through moving stuff around we know that g_1 and g_2 are in H.  This satisfies 1-1.\r\n\r\nOnto\r\n\r\nGiven some element b in Orb(x)\r\nb=g*x=f(gh)",
  "Solution_1": "A cleaner proof I suppose:\r\n\r\nf(gGx)=Orb(x)\r\n\r\n1-1\r\n\r\nf(g1Gx)=f(g2Gx)\r\n\r\nI'm going to use ' to denote inverse\r\n\r\ng2'g1x=x, this shows that g2'g1 is in the left coset.  g2'g1x=x shows that g1x=g2z.  So we are done\r\n\r\nOnto\r\n\r\nGiven any gx=y with y  in Orb(x) we know that f(gGx)=y,this shows that it is onto."
}
{
  "Tag": [
    "calculus",
    "LaTeX",
    "symmetry",
    "calculus computations"
  ],
  "Problem": "I need help with the following two questions.\r\n\r\n1)  If f(x) = sqrt{x} times g(x), where g(4) = 8 and g ' (4) = 7, find f '(4).\r\n\r\n2)  If f(3) = 4, g(3) = 2, f '(3) = -6 and g '(3) = 5, find the following numbers a through d:\r\n\r\na) (f + g) ' (3)\r\n\r\nb) (fg) ' (3)\r\n\r\nc) (f/g) ' (3)\r\n\r\nd) (f/(f - g)) ' (3)",
  "Solution_1": "1) Post this in the right section (Computations and tutorials).\r\n\r\n2) Learn how to use LaTeX.",
  "Solution_2": "3) We're not going to do your homework for you.",
  "Solution_3": "[hide=\"#1\"]$f(x) = g(x)\\sqrt{x} \\Rightarrow f'(x) = g'(x)\\sqrt{x}+\\frac{g(x)}{2\\sqrt{x}} \\Rightarrow f'(4) = 16$[/hide]\n\n[hide=\"#2\"]For the first three, use $[f(x)+g(x)]' = f'(x) + g'(x)$, the product rule, and the quotient rule, and for the forth:\n\n$\\left[\\frac{f(x)}{f(x)-g(x)}\\right]' = \\frac{f'(x)\\left[f(x)-g(x)\\right]-f(x)\\left[f'(x)-g'(x)\\right]}{\\left[f(x)-g(x)\\right]^2}$[/hide]\r\n\r\nJmerry has a good point, if you don't understand a concept we'd be glad to help, but don't post a list of homework problems.",
  "Solution_4": "For the millionth time, I am NOT, NOT, NOT, posting homework.\r\nI am preparing for a state test that I need to pass.  I am NOT, NOT, NOT in high school.\r\n\r\nTo Jimmy:\r\n\r\nThank you for your help.  I can now go back into the test prep book and solve similar questions.",
  "Solution_5": "I know, that's why I didn't post that, but at least try to learn LaTeX and post in the right section! You just keep ignoring that!",
  "Solution_6": "I will look into Latex this week.",
  "Solution_7": "[quote=\"symmetry\"]I need help with the following two questions.\n\n1)  If f(x) = sqrt{x} times g(x), where g(4) = 8 and g ' (4) = 7, find f '(4).\n\n2)  If f(3) = 4, g(3) = 2, f '(3) = -6 and g '(3) = 5, find the following numbers a through d:\n\na) (f + g) ' (3)\n\nb) (fg) ' (3)\n\nc) (f/g) ' (3)\n\nd) (f/(f - g)) ' (3)[/quote]symmetry,\r\n\r\nKey to both problems: Differentiate the general expression first, then plug in specific values of x.\r\n\r\nFor example, 1) $ f(x)g(x) \\Rightarrow \\frac{\\mathrm{d}fg}{\\mathrm{d}x} = \\frac{\\mathrm{d}f}{\\mathrm{d}x}g(x) + f(x)\\frac{\\mathrm{d}g}{\\mathrm{d}x} = \\frac{1}{2\\sqrt{x}}g(x) + \\sqrt{x}\\cdot g'(x) $"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Reading a proof of \"R\\Q is not a countable union of closed subsets of R, (Q is such a union as [b]Q= union {q}[/b])\r\n\r\nSuppose by contradiction, that [b]R\\Q [/b]=[b] countable union {Fn}[/b], Fn is equals to it's closure. Then for every n, Fn is nowhere dense,\r\nsince\r\n[b] interior(Fn closure)[/b] =[b] int(Fn)[/b] is contained in [b]int(R\\Q) [/b]= [b]empty set[/b]. (Since Q is dense in R). Hence,\r\n[b]R[/b]=[b](R/Q) union Q[/b] =[b] [countable union Fn] union [union{q}] [/b]\r\nis a countable union of nowhere dense sets.  :? \r\nBut by Baire's Category Theorem, [b]int {[countable union Fn] union [union{q}] } [/b]= empty set does not equals to R = int R...so a contradiction...\r\n\r\non the sentence marked with  :? , why it can be said that =[b] [countable union Fn] union [union{q}] [/b] is a countable union ofnowhere dense sets??  [b]Q= union {q}[/b] is nowhere dense?? why?",
  "Solution_1": "Use the single element sets as your other nowhere dense sets, not their union.",
  "Solution_2": "so does it means every singleton {q} is nowhere dense.....so is it's union?"
}
{
  "Tag": [
    "symmetry",
    "Vieta",
    "algebra",
    "polynomial"
  ],
  "Problem": "I did it with long way.Is there any short solutions? :blush: \r\n[hide][img]http://img180.imageshack.us/img180/9038/17688677se5.gif[/img][/hide]",
  "Solution_1": "What's the long way?\r\n\r\n[hide=\"maybe\"]\nFind the second symmetry $ab+ac+bc$. Then use $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)$ to find $abc$. Then use vieta's, create the polynomial with roots a, b, c, then find $s_{4}$\n[/hide]",
  "Solution_2": "[hide]\n$(a+b+c).(a^{2}+b^{2}+c^{2})=a^{3}+ab^{2}+ac^{2}+ba^{2}+b^{3}+bc^{2}+ca^{2}+cb^{2}+c^{3}=(a^{3}+b^{3}+c^{3})+(ab^{2}+ac^{2}+ba^{2}+bc^{2}+ca^{2}+cb^{2})=2$ $(ab^{2}+ac^{2}+ba^{2}+bc^{2}+ca^{2}+cb^{2})=-1$  ${(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(ab^{2}+ac^{2}+ba^{2}+bc^{2}+ca^{2}+cb^{2})+6abc]}$ ${1=3-3+6abc abc=1/6}$  ${(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+ac+bc)}$ ${1=2+2(ab+ac+bc) ab+ac+bc=-1/2}$  ${(ab+ac+bc)^{2}=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+2(a^{2}bc+ab^{2}c+abc^{2})}$  ${1/4=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+2[a(abc)+b(abc)+c(abc)]}$  ${1/4=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+1/3}$  ${a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}=-1/12}$  \n${(a^{2}+b^{2}+c^{2})^{2}=a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})}$  ${4=a^{4}+b^{4}+c^{4}-1/6 a^{4}+b^{4}+c^{4}=4+1/6}$ \n${a^{4}+b^{4}+c^{4}=25/6 }$ [/hide]",
  "Solution_3": "That's monstrous.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=131617",
  "Solution_4": "i don't understand vieta's theory completely. :blush:"
}
{
  "Tag": [
    "function",
    "limit",
    "integration",
    "calculus",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Suppose $ \\sum_{0}^{\\infty}|a_n|^2$ converges.  Show that $ f(z)\\equal{}\\sum_{0}^{\\infty}a_nz^n$ is anlytic for $ |z|<1$.\r\n\r\nMore over, prove $ \\lim_{r\\rightarrow 1}\\int_{0}^{2\\pi}|f(re^{it})|^2dt$ exists and find its value.",
  "Solution_1": "First question is definition of radius of convergence. Second is Parseval's formula.",
  "Solution_2": "i understand the first question, but i stilldont know how to get the second one!\r\nany help :maybe: ?",
  "Solution_3": "Do you know Parseval's formula? Here is a derivation\r\n\r\n$ |f(re^{it})|^2 \\equal{} \\left(\\sum_{n \\equal{} 0}^\\infty a_n r^n e^{int} \\right)\\left(\\sum_{n \\equal{} 0}^\\infty \\overline{a_n} r^n e^{ \\minus{} int} \\right) \\equal{} \\sum_{m,n \\equal{} 0}^\\infty a_n\\overline{a_m}r^{m \\plus{} n}e^{i(n \\minus{} m)t}$\r\nIt's trivial to justify interchanging the sum and integral so we get\r\n$ \\int_0^{2\\pi}|f(re^{it})|^2 \\frac {dt}{2\\pi} \\equal{} \\sum_{m,n \\equal{} 0}^\\infty a_n\\overline{a_m}r^{m \\plus{} n}\\int_0^{2\\pi}e^{i(n \\minus{} m)t} \\frac {dt}{2\\pi} \\equal{} \\sum_{n \\equal{} 0}^\\infty |a_n|^2r^{2n}$"
}
{
  "Tag": [],
  "Problem": "For a world record, George Adrian picked 15,832 pounds of\napples in 8 hours. Assuming he maintained a constant rate of\npicking, how many pounds of apples did he pick in 3 hours?",
  "Solution_1": "To figure this out, let's find out how many apples he picked per hour, and then multiply by 3.\r\n\r\n$ \\frac{15832}{8} \\equal{} 1979$\r\n\r\n$ 1979 \\cdot 3 \\equal{} 5937$\r\n\r\nSo our answer is $ \\boxed{5937}$.",
  "Solution_2": "Hmmmmmmmm. I have 8 hours of free time today! Why not go pick over 7 tons of apples!!! :rotfl:"
}
{
  "Tag": [],
  "Problem": "A bookstore sells used books at a 40$ \\%$ discount off the regular price of the book when it was new. The price of a used book is $ \\$18$. What was the regular price of that book when it was new?",
  "Solution_1": "since there is $ 40\\%$, that means the book costs $ 60\\%$ of it's original price, so we just divide\r\n$ 60\\%$ is equal to $ 0.6$, so we get...\r\n\r\n$ \\frac {18} {0.6} \\equal{} 30$"
}
{
  "Tag": [
    "floor function",
    "Divisibility Theory"
  ],
  "Problem": "Let $ a$ and $ b$ be natural numbers such that\r\n\\[ \\frac{a}{b}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots-\\frac{1}{1318}+\\frac{1}{1319}. \\]\r\nProve that $ a$ is divisible by $ 1979$.",
  "Solution_1": "I will assume paragraphs 1. and 2. of http://www.mathlinks.ro/Forum/viewtopic.php?t=150391 (or http://www.problem-solving.be/pen/viewtopic.php?t=25 ) post #4 as known.\r\n\r\nNote that $ 1979$ is a prime (I am wondering whether the IMO contestants had to check that by hand or they were supposed to know that). We are going to prove $ v_{1979}\\left(\\frac {a}{b}\\right)\\geq 1$, because this will yield $ v_{1979}\\left(a\\right) \\equal{} v_{1979}\\left(\\frac {a}{b}\\cdot b\\right) \\equal{} \\underbrace{v_{1979}\\left(\\frac {a}{b}\\right)}_{\\geq 1} \\plus{} \\underbrace{v_{1979}\\left(b\\right)}_{\\geq 0\\text{, since }b\\text{ is an integer}}\\geq 1$, so that $ 1979\\mid a$, and thus the problem will be solved.\r\n\r\nIn order to prove that $ v_{1979}\\left(\\frac {a}{b}\\right)\\geq 1$, we note that $ \\frac {a}{b} \\equal{} 1 \\minus{} \\frac {1}{2} \\plus{} \\frac {1}{3} \\minus{} \\frac {1}{4} \\plus{} \\cdots \\minus{} \\frac {1}{1318} \\plus{} \\frac {1}{1319} \\equal{} \\sum_{i \\equal{} 1}^{\\left\\lfloor 2\\cdot 1979/3\\right\\rfloor}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i}$, so we have to show that $ v_{1979}\\left(\\sum_{i \\equal{} 1}^{\\left\\lfloor 2\\cdot 1979/3\\right\\rfloor}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i}\\right)\\geq 1$. This is a particular case ($ p \\equal{} 1979$) of the following theorem:\r\n\r\n[color=blue][b]Theorem 1.[/b] Let $ p$ be a prime such that $ p > 3$. Then, $ v_{p}\\left(\\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i}\\right)\\geq 1$.[/color]\r\n\r\n[i]Proof of Theorem 1.[/i] First we prove that\r\n\r\n[color=green][b](1)[/b][/color] $ p \\minus{} \\left\\lfloor \\left\\lfloor 2p/3\\right\\rfloor /2\\right\\rfloor \\equal{} \\left\\lfloor 2p/3\\right\\rfloor \\plus{} 1$.\r\n\r\nIn fact, since $ p$ is a prime and $ p > 3$, we have $ 3\\nmid p$. Thus, either $ p\\equiv 1\\mod 3$ or $ p\\equiv 2\\mod 3$.\r\n\r\nIf $ p\\equiv 1\\mod 3$, then $ \\frac {p \\minus{} 1}{3}\\in\\mathbb{Z}$, so that $ \\left\\lfloor\\frac {2p}{3}\\right\\rfloor \\equal{} 2\\cdot\\frac {p \\minus{} 1}{3}$ (since $ 2\\cdot\\frac {p \\minus{} 1}{3}\\in\\mathbb{Z}$ (because $ \\frac {p \\minus{} 1}{3}\\in\\mathbb{Z}$) and $ 2\\cdot\\frac {p \\minus{} 1}{3}\\leq\\frac {2p}{3} < 2\\cdot\\frac {p \\minus{} 1}{3} \\plus{} 1$) and therefore $ \\left\\lfloor\\left\\lfloor\\frac {2p}{3}\\right\\rfloor /2\\right\\rfloor \\equal{} \\frac {p \\minus{} 1}{3}$ (since $ \\left\\lfloor\\frac {2p}{3}\\right\\rfloor /2 \\equal{} \\left(2\\cdot\\frac {p \\minus{} 1}{3}\\right)/2 \\equal{} \\frac {p \\minus{} 1}{3}$ and $ \\frac {p \\minus{} 1}{3}\\in\\mathbb{Z}$), so that\r\n\r\n$ p \\minus{} \\left\\lfloor\\left\\lfloor\\frac {2p}{3}\\right\\rfloor /2\\right\\rfloor \\equal{} p \\minus{} \\frac {p \\minus{} 1}{3} \\equal{} 2\\cdot\\frac {p \\minus{} 1}{3} \\plus{} 1 \\equal{} \\left\\lfloor\\frac {2p}{3}\\right\\rfloor \\plus{} 1$,\r\n\r\nand therefore [color=green][b](1)[/b][/color] is proven for the case $ p\\equiv 1\\mod 3$.\r\n\r\nIf $ p\\equiv 2\\mod 3$, then $ \\frac {p \\minus{} 2}{3}\\in\\mathbb{Z}$, so that $ \\left\\lfloor\\frac {2p}{3}\\right\\rfloor \\equal{} 2\\cdot\\frac {p \\minus{} 2}{3} \\plus{} 1$ (since $ 2\\cdot\\frac {p \\minus{} 2}{3} \\plus{} 1\\in\\mathbb{Z}$ (because $ \\frac {p \\minus{} 2}{3}\\in\\mathbb{Z}$) and $ 2\\cdot\\frac {p \\minus{} 2}{3} \\plus{} 1\\leq\\frac {2p}{3} < \\left(2\\cdot\\frac {p \\minus{} 2}{3} \\plus{} 1\\right) \\plus{} 1$) and therefore $ \\left\\lfloor\\left\\lfloor\\frac {2p}{3}\\right\\rfloor /2\\right\\rfloor \\equal{} \\frac {p \\minus{} 2}{3}$ (since $ \\left\\lfloor\\frac {2p}{3}\\right\\rfloor /2 \\equal{} \\left(2\\cdot\\frac {p \\minus{} 2}{3} \\plus{} 1\\right)/2 \\equal{} \\frac {p \\minus{} 2}{3} \\plus{} \\frac {1}{2}$ and $ \\frac {p \\minus{} 2}{3}\\in\\mathbb{Z}$), so that\r\n\r\n$ p \\minus{} \\left\\lfloor\\left\\lfloor\\frac {2p}{3}\\right\\rfloor /2\\right\\rfloor \\equal{} p \\minus{} \\frac {p \\minus{} 2}{3} \\equal{} \\left(2\\cdot\\frac {p \\minus{} 2}{3} \\plus{} 1\\right) \\plus{} 1 \\equal{} \\left\\lfloor\\frac {2p}{3}\\right\\rfloor \\plus{} 1$,\r\n\r\nand therefore [color=green][b](1)[/b][/color] is proven for the case $ p\\equiv 2\\mod 3$.\r\n\r\nHence, [color=green][b](1)[/b][/color] is proven in both possible cases, and we can step to the actual proof of Theorem 1.\r\n\r\nWe work with fractions modulo $ p$, noting that we can divide by every $ i\\in\\left\\{1;\\;2;\\;...;\\;p \\minus{} 1\\right\\}$ modulo $ p$. Obviously,\r\n\r\n$ \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i} \\equal{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is odd}}}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i} \\plus{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is even}}}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i}$\r\n$ \\equal{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is odd}}}\\frac {1}{i} \\plus{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is even}}}\\frac { \\minus{} 1}{i} \\equal{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is odd}}}\\frac {1}{i} \\minus{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is even}}}\\frac {1}{i}$\r\n$ \\equal{} \\left(\\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is odd}}}\\frac {1}{i} \\plus{} \\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is even}}}\\frac {1}{i}\\right) \\minus{} 2\\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is even}}}\\frac {1}{i} \\equal{} \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {1}{i} \\minus{} 2\\sum_{\\substack{1\\leq i\\leq \\left\\lfloor 2p/3\\right\\rfloor ; \\\\\r\ni\\text{ is even}}}\\frac {1}{i}$\r\n$ \\equal{} \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {1}{i} \\minus{} 2\\sum_{j \\equal{} 1}^{\\left\\lfloor \\left\\lfloor 2p/3\\right\\rfloor /2\\right\\rfloor}\\frac {1}{2j} \\equal{} \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {1}{i} \\plus{} \\sum_{j \\equal{} 1}^{\\left\\lfloor \\left\\lfloor 2p/3\\right\\rfloor /2\\right\\rfloor}\\frac {1}{ \\minus{} j}$.\r\n$ \\equiv\\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {1}{i} \\plus{} \\sum_{j \\equal{} 1}^{\\left\\lfloor \\left\\lfloor 2p/3\\right\\rfloor /2\\right\\rfloor}\\frac {1}{p \\minus{} j}$\r\n$ \\equal{} \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {1}{i} \\plus{} \\sum_{i \\equal{} p \\minus{} \\left\\lfloor \\left\\lfloor 2p/3\\right\\rfloor /2\\right\\rfloor}^{p \\minus{} 1}\\frac {1}{i}$      (here we substituted $ j$ by $ p \\minus{} i$ in the second sum)\r\n$ \\equal{} \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {1}{i} \\plus{} \\sum_{i \\equal{} \\left\\lfloor 2p/3\\right\\rfloor \\plus{} 1}^{p \\minus{} 1}\\frac {1}{i}$      (by [color=green][b](1)[/b][/color])\r\n$ \\equal{} \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i}\\mod p$.\r\n\r\nNow, $ \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i}\\equiv 0\\mod p$, as can be shown in different ways - e. g. as follows:\r\n\r\n$ \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i} \\equal{} \\frac12\\cdot\\left(\\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i} \\plus{} \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i}\\right)$\r\n$ \\equal{} \\frac12\\cdot\\left(\\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i} \\plus{} \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{p \\minus{} i}\\right)$             (here we substituted $ i$ by $ p \\minus{} i$ in the second sum)\r\n$ \\equal{} \\frac12\\cdot\\sum_{i \\equal{} 1}^{p \\minus{} 1}\\left(\\frac {1}{i} \\plus{} \\frac {1}{p \\minus{} i}\\right)\\equiv\\frac12\\cdot\\sum_{i \\equal{} 1}^{p \\minus{} 1}\\left(\\frac {1}{i} \\plus{} \\frac {1}{ \\minus{} i}\\right) \\equal{} \\frac12\\cdot\\sum_{i \\equal{} 1}^{p \\minus{} 1}0 \\equal{} 0\\mod p$;\r\n\r\nhereby, we were allowed to divide by $ 2$ because $ 2\\not\\equiv 0\\mod p$ (since $ p > 3$). Thus,\r\n\r\n$ \\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i}\\equiv \\sum_{i \\equal{} 1}^{p \\minus{} 1}\\frac {1}{i}\\equiv 0\\mod p$,\r\n\r\nso that $ v_{p}\\left(\\sum_{i \\equal{} 1}^{\\left\\lfloor 2p/3\\right\\rfloor}\\frac {\\left( \\minus{} 1\\right)^{i \\minus{} 1}}{i}\\right)\\geq 1$, and Theorem 1 is proven.\r\n\r\n  Darij",
  "Solution_2": "1979 is prime.\r\n$ 1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots-\\frac{1}{1318}+\\frac{1}{1319}\\#=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{1318}+\\frac{1}{1319}-2(\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{1318}) \\#=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{1318}+\\frac{1}{1319}-(1+\\frac{1}{2}+\\cdots+\\frac{1}{659})\\#=\\frac{1}{660}+\\frac{1}{661}+\\cdots+\\frac{1}{1319}\\#=(\\frac{1}{660}+\\frac{1}{1319})+(\\frac{1}{661}+\\frac{1}{1318}) \\cdots+( \\frac{1}{989}+\\frac{1}{990}) =1979 \\frac{d}{c}$\r\n(where $ c, d$ are relative prime.)\r\n\r\nBecause $ c$ is not multiple of 1979, $ 1979 \\mid a$",
  "Solution_3": "The generalisation of the above is : Let $ p$ be a prime greater than $ 3$, and $ q\\equal{}[\\frac{2p}{3}]$ where $ [x]$ is the greatest integer function.Let $ m$ and $ n$ be positive integers such that$ \\frac{m}{n}\\equal{}1\\minus{}\\frac{1}{2}\\plus{}\\frac{1}{3}\\minus{}\\frac{1}{4}.....\\plus{}(\\minus{}1)^{q\\minus{}1}\\frac{1}{q}$. then $ m$ is divisible by $ p$.",
  "Solution_4": "Just an observation: if I'm not mistaken, this problem (and manjil's generalization) is a direct consequence [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150392]A 24[/url]. Why? \r\n\r\nFirst, I claim that when $ 0 \\leq k < p$, $ {p \\minus{} 1 \\choose k} \\equiv (\\minus{}1)^k \\mbox{ mod p}$. If $ k \\equal{} 0$, this is obvious. Otherwise, this is a consequence of the fact that $ \\binom{p \\minus{} 1}{k} \\equal{} \\frac{(p\\minus{}1) \\cdot (p\\minus{}2) \\cdot \\ldots \\cdot (p\\minus{}(k\\plus{}1)}{k \\cdot (k\\minus{}1) \\cdot \\ldots \\cdot 1}$; taking the numerator mod $ p$ and cancelling wtih the denominator, our result follows easily. \r\n\r\nNext, observe that $ p \\binom{p \\minus{} 1}{k\\minus{}1} \\equal{} k\\binom{p}{k}$; this is a trivial consequence of the definition of $ \\binom{n}{k}$. Rearranging, we see that $ \\frac{\\binom{p\\minus{}1}{k\\minus{}1}}{k} \\equal{} \\frac{\\binom{p}{k}}{p}$. \r\n\r\nNow, let $ p$ be a prime and $ q \\equal{} \\lfloor \\frac{2p}{3} \\rfloor$. We wish to show that $ 1 \\minus{} \\frac{1}{2} \\plus{} \\frac{1}{3} \\ldots \\plus{} (\\minus{}1)^{q\\minus{}1} \\frac{1}{q}$ is 0 mod $ p$, where $ \\frac{1}{n}$ is taken to be the inverse of $ n$ modulo $ p$. Our sum is\r\n\r\n$ \\sum_{k \\equal{} 0}^q (\\minus{}1)^k \\frac{1}{k} \\equiv \\sum_{k \\equal{} 0}^q \\frac{\\binom{p\\minus{}1}{k\\minus{}1}}{k} \\equiv \\sum_{k\\equal{}0}^q \\frac{{n \\choose k}}{p} \\mbox{ mod } p$. However, by problem A 24, $ \\sum_{k\\equal{}0}^q {n \\choose k} \\equiv 0 \\mbox{ mod } p^2$, so $ \\sum_{k\\equal{}0}^q \\frac{{n \\choose k}}{p} \\equiv 0 \\mbox{ mod } p$, as desired. \r\n\r\n[Clearly, we may also go in reverse and use this generalization to solve problem A 24.]",
  "Solution_5": "We don't need generalizations :D\n\n[url]https://artofproblemsolving.com/community/c6h61006p367332[/url]"
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "If would appreciate it if you could tell me how to solve these 2 problems \r\n\r\n1st Problem\r\n\r\nIf Bob can decorate a room for a party in 8 hours and Jim can do it in 12 hours. If Jen helps, all three can do the job in 3 hours working together. How long would it take Jen if she works alone? \r\n\r\n2nd Problem \r\n\r\n2/3 of men are married to 3/4 of the women (assume each woman is married to exactly one man and conversely). What is the ratio of number of men to the number of women",
  "Solution_1": "Here's the first one:\r\n\r\n[hide]\nIf Bob can decorate the room in 8 hours, he can do 1/8 of a room per hour. So Jim can do 1/12 per hour. If Jen helps, they can do 1/3 per hour. 1/8 + 1/12 = 5/24, and 1/3 = 8/24. So Jen can do 8/24 - 5/24 = 3/24 = 1/8 per hour, or one room in eight hours.[/hide]",
  "Solution_2": "[hide=\"hint for 2\"]\n\nLet x be the total number of couples.\n\nThen there are $\\frac{3}{2}x$ men and $\\frac{4}{3}x$ women.\n\n[/hide]",
  "Solution_3": "THANKS FOR 1ST one ... now so im guessing for the 2nd one i would do 2/3(x) / 3/4(x) and get 8:9 ?",
  "Solution_4": "[quote=\"fero4u123\"]If would appreciate it if you could tell me how to solve these 2 problems \n\n1st Problem\n\nIf Bob can decorate a room for a party in 8 hours and Jim can do it in 12 hours. If Jen helps, all three can do the job in 3 hours working together. How long would it take lois if she works alone? \n\n2nd Problem \n\n2/3 of men are married to 3/4 of the women (assume each woman is married to exactly one man and conversely). What is the ratio of number of men to the number of women[/quote]\r\n\r\n\r\n\r\n[hide]\n$\\frac{1}{8}+\\frac{1}{12}+\\frac{1}{x}=\\frac{1}{3}$\n\n$3+2+\\frac{24}{x}=8$\n\n$5x+24=8x$\n\n$x=\\boxed{8}$\n[/hide]\r\n\r\nwhat's it mean by conversely?",
  "Solution_5": "[quote=\"xxxxx\"][quote=\"fero4u123\"]If would appreciate it if you could tell me how to solve these 2 problems \n\n1st Problem\n\nIf Bob can decorate a room for a party in 8 hours and Jim can do it in 12 hours. If Jen helps, all three can do the job in 3 hours working together. How long would it take lois if she works alone? \n\n2nd Problem \n\n2/3 of men are married to 3/4 of the women (assume each woman is married to exactly one man and conversely). What is the ratio of number of men to the number of women[/quote]\n\n\n\n[hide]\n$\\frac{1}{8}+\\frac{1}{12}+\\frac{1}{x}=\\frac{1}{3}$\n\n$3+2+\\frac{24}{x}=8$\n\n$5x+24=8x$\n\n$x=\\boxed{8}$\n[/hide]\n\nwhat's it mean by conversely?[/quote]\r\nBasically meaning that also assume that one man is married to a woman as well.",
  "Solution_6": "conversly means \"also reverse in order\" so that would mean to say each woman is married to exactly one man / each man is married to exactly one woman",
  "Solution_7": "ok thx. \r\n\r\nnumber 2: (is this right?)\r\n[hide]\n \n$\\frac{2}{3}m=\\frac{3}{4}w$\n\n$m=\\frac{9}{8}w$\n\n$\\boxed{8: 9}$[/hide]",
  "Solution_8": "For Question 1, who is Lois?",
  "Solution_9": "yeah it makes no sense if lois just appears\r\n\r\n[hide=\"xxxxxhowevermany xs\"]\n9:8\nis what i get\nso there were 9 men\n6=3/4x\n8=x\nx is number of women\n9:8[/hide]",
  "Solution_10": "*im sorry that lois was suppose to say Jen * i will change and 9:8 does seem right because it makes the 1:1 ratio but so does (8/9):1 ? im not sure of exactly how u are suppose to work the problem",
  "Solution_11": "Well, since 2/3 of the men is equal to 3/4 of the women, there must be more men than women.",
  "Solution_12": "[quote=\"fero4u123\"]*im sorry that lois was suppose to say Jen * i will change and 9:8 does seem right because it makes the 1:1 ratio but so does (8/9):1 ? im not sure of exactly how u are suppose to work the problem[/quote]\r\n\r\nwell i mean it is sort of gross if you are going to start cutting women in ninths....go ahead if you want to"
}
{
  "Tag": [
    "induction",
    "number theory solved",
    "number theory"
  ],
  "Problem": "Let p,q be natural numbers, and n=2^q(2^p+1).Prove that the binomial number Cn,k( combinations) is odd if and only if k is in the set {0,n,2^q,2^(p+q)}",
  "Solution_1": "The sufficience:\r\n\r\nIf k is 0 or n the corresponding binomial coefficient is 1, so it's done. For the other 2 we write the binomial coefficient:\r\n\r\n[2^q+2^(p+q)]!/[2^q]!*[2^(p+q)]!. To prove that it's odd all we have to do is prove that the exponent of 2 in the numerator=the exponent of 2 in the denominator:\r\n\r\n \\sum {k=1->00} [2^q(2^p+1)/2^k] = \r\n \\sum {k=1->00} ([2^q/2^k]+[2^(p+q)/2^k]), but it's easy to prove that for each k>=1 we have [2^q(2^p+1)/2^k]=[2^q/2^k]+[2^(p+q)/2^k], where [x]=integer part of x.\r\n\r\nI think the necessity has a similar idea, but I haven't given it any thought yet.",
  "Solution_2": "This a special case of a more general (and very nice :D)  result :\r\n\r\nWrite n = sum 2^ai (n in base 2)\r\n\r\nC(n,k) is odd iff k = sum 2^bj where bj  {ai}, i.e. the bj are chosen among the binary digits of n\r\n\r\n\r\nIt's fun to show using modular polynomials.\r\n\r\n. (1+a)^2 = 1+a (mod 2)\r\n. by induction (1+a)^(2^n) = 1 + a^(2^n) (mod 2) : it means C(2^n,k) are even for k != 0,2^n\r\n. (1+a)^(2^n + 2^p) = 1 + a^(2^n) + a^(2^p) + a^(2^(n+p)) (mod 2)\r\n. .....\r\n. by induction :\r\n(1+a)^(sum 2^ai) = 1 + sum a^(something) (mod 2) where something is a sum of some of the ai.\r\nHence the result.\r\n\r\n\r\nSo :\r\n. C(2^n, k) is odd for k != 0,2^n\r\n. C(2^m+2^n, k) is odd for k != 0,2^m,2^n,2^(m+n)\r\n. C(2^l+2^m2^n,k) is odd for k != 0,2^l,2^m,2^n,2^(l+n), .., 2^(l+m+n).\r\n. .....\r\n\r\nNice isn't it  :)"
}
{
  "Tag": [
    "function",
    "calculus",
    "quadratics",
    "algebra"
  ],
  "Problem": "Find the local mimimum of $f(x)=x^4+2x^3+3x^2+5x+4$.\r\n\r\nMasoud Zargar",
  "Solution_1": "Do you mean local maximum? \r\nYour function is unbounded above.\r\nBesides, your function has no local maximums.\r\n$f'(x) = 4x^3+6x^2+6x+5=0$ if and only if x = -1.143955(approx.)\r\n$f''(x)=12x^2+12x+6=7.976135>0=$ when x =-1.143955.\r\nTherefore, f(x) has only one local minimum but it has no maximums.",
  "Solution_2": "yes, local maximum.",
  "Solution_3": "lol, what a foolish problem...very well then; find the local minimum.",
  "Solution_4": "So in other words, find the root of $4x^3+6x^2+6x+5=0$. (Just to get rid of the calculus in it.)\r\nCan this root even be found exactly (without ugly cubic formula)?",
  "Solution_5": "absolutely...I made the problem, so yes. :D",
  "Solution_6": "are you saying that is supposed to factor nicely?",
  "Solution_7": "Yes, it is a foolish problem, for beginners ! [color=darkred]But we must make it.[/color]\r\n$\\left(\\forall\\right) x\\in R\\ ,\\ f''(x)>0$ $\\Longrightarrow$  $f'\\nearrow$ $\\Longrightarrow$ $\\left(\\dot {\\exists}\\right)\\ \\alpha\\in (-2,-1)$ so that\r\n$f'(x)\\ .s.s.\\ (x-\\alpha )$ $\\Longleftrightarrow$ $\\min_{x\\in R} f(x)=f(\\alpha)=\\frac 38(2\\alpha^2+8\\alpha +9)\\ .$\r\n[u]Notations.[/u] $x\\ .s.s.\\ y\\Longleftrightarrow x=y=0\\ \\vee\\ xy>0\\ ;$\r\n$\\left(\\dot {\\exists }\\right) \\Longleftrightarrow$ exists and is uniquely .",
  "Solution_8": "[quote=\"Virgil Nicula\"]\n$\\left(\\dot {\\exists }\\right) \\Longleftrightarrow$ exists and is uniquely .[/quote]\r\nIsn't there an international symbol for that? Namely : $\\exists !$?",
  "Solution_9": "Yes, but I use often this symbol which is more suggestive, in my opinion !",
  "Solution_10": "Well, my solution goes as follows; I am currently busy with other math.\r\n\r\n1. Take the derivative.\r\n2. Let $x=y-\\frac{b}{3}$, where $b$ is in the second coifficient.\r\n3. Simplify and set $y=z+\\frac{p}{3z}$, where $p$ is the second coifficient of the new equation.\r\n4. Solve the quadratic equation and solve  for $x$.\r\n\r\nMasoud Zargar"
}
{
  "Tag": [
    "inequalities",
    "algebra"
  ],
  "Problem": "Fie $a,b,c>0$. Aratati ca: \\[\\frac{1}{a+3b}+\\frac{1}{b+3c}+\\frac{1}{c+3a}\\geq\\frac{1}{a+2b+c}+\\frac{1}{b+2c+a}+\\frac{1}{c+2a+b},\\ \\ \\ \\ \\ \\ \\textbf{(1)}\\]  \\[\\frac{3}{a}+\\frac{4}{b}+\\frac{5}{c}\\geq\\frac{9}{a+2b}+\\frac{18}{b+2c}+\\frac{9}{c+2a},\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\textbf{(2)}\\]  \\[\\frac{a^{2}}{1+b^{2}}+\\frac{b^{2}}{1+a^{2}}+\\frac{1}{a^{2}+b^{2}}\\geq\\frac{1}{a+b}+\\frac{a}{1+b}+\\frac{b}{a+1}.\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\textbf{(3)}\\]",
  "Solution_1": "Inegalitatea $\\ (1)$\r\nFolosind spargerile cu medii obtinem inegalitatea $\\frac{\\frac{4}{7}}{3a+b}+\\frac{\\frac{1}{7}}{3b+c}+\\frac{\\frac{2}{7}}{3c+a}\\geq\\frac{1}{2b+c+a}$.Adunand si analoagele, obtinem $\\ (1)$.\r\nInegalitatea $\\ (2)$\r\nTot cu spargerile cu medii avem inegalitatea $\\frac{\\frac{a}{3}}{3}+\\frac{\\frac{8b}{9}}{4}\\geq\\frac{9}{a+2b}$ si iese inegalitatea $\\ (2)$.\r\nInegalitatea $\\ (3)$\r\n..nu mai scriu ca oricum e la fel, tot cu spargeri cu medii. :)",
  "Solution_2": "$(2)$ Cautam o spargere convenabila:\r\n$\\frac{x}{a}+\\frac{y}{b}\\geq \\frac{9}{a+2b}$\r\n$\\frac{x^{2}}{ax}+\\frac{y^{2}}{by}\\geq \\frac{(x+y)^{2}}{ax+by}\\Rightarrow\\ x=1,\\ y=2$\r\nRezulta:\r\n$\\frac{1}{a}+\\frac{2}{b}\\geq \\frac{9}{a+2b}$\r\nsi\r\n$\\frac{1}{c}+\\frac{2}{a}\\geq \\frac{9}{a+2b}$\r\nRamane sa demonstram ca\r\n$\\frac{2}{b}+\\frac{4}{c}\\geq \\frac{18}{b+2c}\\Leftrightarrow \\frac{1}{b}+\\frac{2}{c}\\geq \\frac{9}{a+2b}$, care este evident.",
  "Solution_3": "Let $a,b,c>0$. Prove that$$\\frac{3}{a}+\\frac{4}{b}+\\frac{5}{c}\\geq\\frac{9}{a+2b}+\\frac{18}{b+2c}+\\frac{9}{c+2a}$$$$\\frac{a^{2}}{1+b^{2}}+\\frac{b^{2}}{1+a^{2}}+\\frac{1}{a^{2}+b^{2}}\\geq\\frac{1}{a+b}+\\frac{a}{1+b}+\\frac{b}{a+1}$$\n\nVery nice."
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Solve the equation:\r\n\r\n\\[ 2^{x\\plus{}1}\\equal{}(x\\minus{}1)(x\\minus{}2)\\]",
  "Solution_1": "[quote=\"moldovan\"]Solve the equation:\n\\[ 2^{x \\plus{} 1} \\equal{} (x \\minus{} 1)(x \\minus{} 2)\\]\n[/quote]\r\n\r\nThe function $ 2^{x \\plus{} 1} \\minus{} (x \\minus{} 1)(x \\minus{} 2)$ is strictly increasing. We have just one solution $ x\\equal{}0$"
}
{
  "Tag": [
    "ARML",
    "MATHCOUNTS",
    "geometry",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "trigonometry"
  ],
  "Problem": "P.K,tell me the thoughts crossing your minds,even the randomest.",
  "Solution_1": "why do people seem to get confused with what they feel about themselves??!!!",
  "Solution_2": "ARML in 19 days!",
  "Solution_3": "gah you know what I really need to go to the bathroom right now and why did mandelbrot send me a shirt that was so small that's sort of weird heh when I was thinking of it I accidentally thought of mathcounts but we all know that's not true as THEY EXIST ONLY IN THE MINDS OF HORRIBLE PEOPLE THAT oh man I think I'll go to the bathroom now it'll be pretty awesome maybe there'll be like an orchestra y'know like in heaven with the angels and they're all like whoa if you mistype angels it looks like angles angles angles like geometry yea geometry is awesome except for solid geometry because that's like oh no I haven't learned that yet I should though that sounds sort of like fun but then again arrgh",
  "Solution_4": "MIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\nMIMC DUE TOMMORROW\r\n\r\n\r\nGET IT IN YOUR HEAD PEOPLE",
  "Solution_5": "does that mean i can submit it tomorrow and not be late or do i have to submit before tomorrow?\r\n\r\nmy thoughts...\r\n[size=200]$7\\frac{1}{2}$ more days of school!![/size]",
  "Solution_6": "why do we drive on a parkway...and park on a driveway??",
  "Solution_7": "Would you stone me if I dropped out of MIMC?  I totally forgot about it, and at this point, I have too much work...\r\n\r\nEdit: 10 more posts...",
  "Solution_8": "well...schools almost over!!!!!\r\n12 more days!",
  "Solution_9": "man MIMC that makes no sense at all oh man oh man should I do it no I think not HAHA LIKE DESCARTES EH GET IT GET IT this thread is such a good stress reliever wait no it isn't I wasn't even stressed in the first place YOU LIE",
  "Solution_10": "[quote=\"fireemblem13\"][size=200]$7\\frac{1}{2}$ more days of school!![/size][/quote]\n[quote=\"math92\"]12 more days![/quote]\r\n\r\nI have like around a month left.[img]http://smilies.sofrayt.com/fsc/sobbing.gif[/img]",
  "Solution_11": "what's the point??",
  "Solution_12": "a bear came to me bearing a bear...i can bear a bear but not a bear's bear.........get it???!!!",
  "Solution_13": "I went shopping for me yesterday and the bill was like over 5000 rupees,so i thiought the bags would be goddarn heavy and when i carried them,they were so light,i'm so disappointed....\r\n\r\nou know some typos which can be really misunderstood,like when you write 'busy',you accidentally type the 't' next to y... :o \r\n\r\n\r\nacidpop,did you mean you can't be the mother of the bear's child bear?That's totally possible ...",
  "Solution_14": "I have no chance of qualifying for IMO. Got 13 out of 40 in Practcie Test I and wrote a horrible selection test I some hours ago. Dash it!",
  "Solution_15": "i just saw my chem teacher from last year- and in his classroom, there's this clock the has element symbols instead of numbers- like where 1 should be, it says H, then He, according to atomic number.",
  "Solution_16": "[quote=\"MysticTerminator\"]man MIMC that makes no sense at all oh man oh man should I do it no I think not HAHA LIKE DESCARTES EH GET IT GET IT this thread is such a good stress reliever wait no it isn't I wasn't even stressed in the first place YOU LIE[/quote]\r\n\r\nYOU BETTER DO IT\r\n\r\nIT'S DUE TODAY\r\n\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY\r\nMIMC DUE TODAY",
  "Solution_17": "wondering if that is all you think about lately\r\n\r\n(by the way, i just sent in my mimc answers)",
  "Solution_18": "I'm so sad i didn't make usamo this year...\r\nI can solve ~3 problems per practice, sometimes some partial stuff on other ones...\r\nI might have done well...",
  "Solution_19": "[quote=\"perfectnumber628\"]wondering if that is all you think about lately\n\n(by the way, i just sent in my mimc answers)[/quote]\r\n\r\nthis is the only way I can get people to remember...by spamming about this. We only have around 30 people that sent in their answers out of 120 that signed up! :o , so either people forgot or don't care. *hoping that it was the former*, I'm trying to remind people.",
  "Solution_20": "[quote=\"Iversonfan2005\"][quote=\"perfectnumber628\"]wondering if that is all you think about lately\n\n(by the way, i just sent in my mimc answers)[/quote]\n\nthis is the only way I can get people to remember...by spamming about this. We only have around 30 people that sent in their answers out of 120 that signed up! :o , so either people forgot or don't care. *hoping that it was the former*, I'm trying to remind people.[/quote]\r\n\r\nmaybe it's because most of them (like me?) didn't sign up  :o",
  "Solution_21": "whoops,didn't submit my answers,i shiuld sttill get a 50 though,according to the rules,right?",
  "Solution_22": "currently thinking about writing \"the greatest english paper you've ever written\" because that's what my teacher said to do... :(  :oops:",
  "Solution_23": "When was the last time i had a bath......",
  "Solution_24": "these last 3 or 4 days of school are gonna be awesome. :lol:",
  "Solution_25": "my scar is literally shiny",
  "Solution_26": "today i realized i don't have school this friday!  i'll probably forget, and then remember on thursday, and it'll make my whole day  :roll:",
  "Solution_27": "i wish aops didn't have censors\r\n\r\ndon't eat the fish",
  "Solution_28": "You should put that on the wish list...I'll back it\r\n\r\nOh mann..,..a huge chemistry test and REALLY HUGE physics test this week...arrrrrrrghhhhhh",
  "Solution_29": "[quote=\"fireemblem13\"]these last 3 or 4 days of school are gonna be awesome. :lol:[/quote]\r\n\r\ni need to edit that. yesterday(monday), we had our annual class trip to schlitterbahn and i lost my glasses. :blush:",
  "Solution_30": "i should probably be working on my english homework right now.  odd how that works out.",
  "Solution_31": "[quote=\"perfectnumber628\"]i should probably be working on my english homework right now.  odd how that works out.[/quote]\r\n:huh: \r\n\r\nit was 2:48 when you wrote that. are you already out of school?\r\n\r\nEDIT: oh you're in NY. that means it's 4:48. it's still kind of early to start on homework. don't you think?",
  "Solution_32": "actually, school ends at 2:13 ish, but if i stay after (like today) i get home at like, 3:15.\r\n\r\nand it is ALWAYS too early to do homework.",
  "Solution_33": "[quote=\"perfectnumber628\"]actually, school ends at 2:13 ish, but if i stay after (like today) i get home at like, 3:15.\n\nand it is ALWAYS too early to do homework.[/quote]\r\n\r\ndo u mean\r\n\r\nit is NEVER too early to do homework.\r\n\r\ncuz if it's ALWAYS too early then if i procrastinate, it's still early?",
  "Solution_34": "procrastination is an important skill to have.\r\n\r\neven more important: when in doubt, use law of cosines!  :lol:",
  "Solution_35": "i'm out of school before all of you. i got out last thursday!\r\n :rotfl:      :rotfl:   :rotfl: \r\ni'm already enjoying the benefits of sleeping in and being lazy and getting more stuff done than when i was in school and stuff!\r\n\r\n\r\n\r\n[size=200][color=red]I'M FREE!!!!!!!!!!!!!!![/color][/size]",
  "Solution_36": "I may not even be free for a month ;) .\r\n\r\nBIRTHDAY TOMORROW!",
  "Solution_37": "random comment: no school tomorrow!  actually, we have a 4-day weekend: tomorrow is off because we didn't have snow days, and monday is memorial day.",
  "Solution_38": "[size=200]SCHOOL IS OUT!!!!!!![/size] (well, yesterday, but i couldn't get on to the internet) [size=200]WOOOOOHOOOOOO!!!!!!!!!![/size]",
  "Solution_39": "wait a minute... I didn't order any iced tea!  :o \r\n\r\n.......................RANDOM EXPLOSION.....................",
  "Solution_40": "I'm so exited! My birthday was yesterday!!!!! .......wait. D'oh!",
  "Solution_41": "arml in some smallish number of days!!!  :clap:  :trampoline:",
  "Solution_42": "All math competitions gone! NOOOOOOOOOOOOOOOOOoooooooooooooooooooooooooooo!!!!!!!!!!!!!!!!!!"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "$x,y,z \\in R+$.Prove that:\r\n$(x+y)\\sqrt \\frac{xy}{(x+z)(y+z)}+(y+z)\\sqrt \\frac{yz}{(x+y)(x+z)}+(z+x)\\sqrt \\frac{xz}{(y+z)(y+x)}$\r\n$\\geq x+y+z$ :?:",
  "Solution_1": "[quote=\"Denbox\"]$x,y,z \\in R+$.Prove that:\n$(x+y)\\sqrt \\frac{xy}{(x+z)(y+z)}+(y+z)\\sqrt \\frac{yz}{(x+y)(x+z)}+(z+x)\\sqrt \\frac{xz}{(y+z)(y+x)}$\n$\\geq x+y+z$ :?:[/quote]\r\nTake a look at \r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=462855#p462855]Moldova TST 2006 Test II problem 3[/url] :wink:",
  "Solution_2": "I have a nice solution.\r\nLet $a= y+z, b= z+x, c= x+y$ then $a, b, c$ are side lenghts of a triangle $ABC$.\r\nAnd now, we have to prove that\r\n$\\sum_{}asin\\frac{A}{2}\\geq  p$\r\nFirst, note that $cos\\frac{B-C}{2}= \\frac{b+c}{a}sin\\frac{A}{2}$ therefor $\\sum_{}asin\\frac{A}{2}\\geq  \\sum_{}(b+c)sin^{2}\\frac{A}{2}= \\frac{1}{2}\\sum_{}(b+c)(1-cosA)= 2p-\\frac{1}{2}\\sum_{}(b+c)cosA$\r\nOn the other hand, $a= bcosC+ccosB$. Applying this equality, the inequality is clearly true. The equality holds iff $A= B= C$ or $x= y= z$."
}
{
  "Tag": [
    "induction",
    "greatest common divisor"
  ],
  "Problem": "I m given these rules for division:\r\n\r\na)$ t|t$\r\nb)$ 1|t$\r\nc)$ t|0$\r\nd)$ t|s\\wedge s|r \\rightarrow t|r$\r\ne)$ s \\neq 0 \\wedge t|s \\rightarrow t \\le s$\r\nf)$ t|s \\wedge s|t \\rightarrow s\\equal{}t$\r\ng)$ t|s\\rightarrow t|(r\\cdot s)$\r\nh)$ t|s \\wedge t|r \\rightarrow t|(s\\plus{}r)$\r\n\r\nnote: $ \\wedge$ = \"and\" , $ \\rightarrow$ = implies\r\n\r\n\r\nhow would I show $ (t|s\\wedge t|s')\\rightarrow t\\equal{}1$?\r\n\r\n\r\n\r\nBasically $ s'$ is the successor of $ s$ like 4 is the successor of 3. Thus the statement above is saying if $ t$ divides $ s$and $ t$ divides the successor of $ s$, then $ t$ is $ 1$.\r\n\r\nWhat would be a rigorous way to show this?",
  "Solution_1": "So are we assuming $ t$ is a positive integer then?  If these divisibility rules are allowed to extend to the integers, we use h.\r\nYou have $ t|s \\implies t|\\minus{}s$.  So because $ t|s'$, we have $ t|s'\\minus{}s$, so $ t|1$.  If $ t$ is given to be a positive integer, then $ 1|t, t|1 \\implies t\\equal{}1$.",
  "Solution_2": "these rules are only for the natural numbers. Though, I wish we were in the integers.",
  "Solution_3": "There are many arithmetic progressions of the form ks-1 (k>1), ks-2 (k>2)... ks-n (k>n).\r\ntry to use that, even though I ain't exactly getting anywhere",
  "Solution_4": "I don't think it's possible just from those rules.  If you had the rule $ \\left(t | s \\right)\\wedge\\left( t | (r \\plus{} s)\\right) \\to t | r$ then you would be all set.\r\n\r\nThe reason it's not possible:\r\nRule (e) is clearly irrelevant.  For rules (d), (g) and (h), we always end up with $ t$ dividing something larger than what we started with.  However, our conclusion can only come from rule (f), which requires that we prove $ t | 1$.  Thus, it's only possible if one of $ s, s'$ is smaller than or equal to 1, which is not true in general.",
  "Solution_5": "what about, call s'=s+1 and s=s then for the statement to be correct we must have t|(1+s), but t|s, hence by h) t|1, or t=1",
  "Solution_6": "It is not possible to conclude from the facts $ t | s$, $ t | s \\plus{} 1$ and (h) that $ t | 1$.  This was exactly the point of my comment.",
  "Solution_7": "Another way to see that this cannot work:\r\nInstead of using the normal definition of $ a|b$, we write $ a|b$ iff $ b\\equal{}0$ or $ 0<a \\leq b$ (thus we see $ 0$ as some infinite number and $ |$ as standard order relation)\r\nIt's easy to check that all the mentioned statements are fulfilled, but the conclusion is not.",
  "Solution_8": "What if we are also given that $ a|b \\Leftrightarrow \\exists c, a\\cdot c\\equal{}b$? What conditions do we need to show the requested result?",
  "Solution_9": "Then it works as this is the standard definition of $ |$, thus all the \"known\" properties hold.",
  "Solution_10": "Hmm..I was asking which of the properties listed above are necessary to prove the result? Some of them are given by the definition, but not all of them. For example, what if these weren't integers, and therefore not given e? Is e necessary?",
  "Solution_11": "yes Hamster1800. It's stupid of me not to include that we're given the definition $ a|b \\Leftrightarrow \\exists c, a\\cdot c\\equal{}b$. I just thought it was assumed. my fault.",
  "Solution_12": "[quote=\"mstrip\"]yes Hamster1800. It's stupid of me not to include that we're given the definition $ a|b \\Leftrightarrow \\exists c, a\\cdot c \\equal{} b$. I just thought it was assumed. my fault.[/quote]\r\n\r\nIf we have this, we can reason by contradiction.",
  "Solution_13": "Prove$ (t|s\\wedge t|s')\\rightarrow t\\equal{}$\r\n\r\nNeed to show for two cases $ s\\equal{}0$ and $ s\\neq0$\r\n\r\n$ s\\equal{}0$\r\n\r\nwe simply have the case $ t|1\\rightarrow t\\equal{}1$\r\n\r\nby hypothesis, we have $ t|1$   \r\nby proposition b) we have $ 1|t$    \r\nthen by prop. f) we have $ t|1 \\wedge 1|t \\rightarrow t\\equal{}1$   \r\n\r\n$ s\\neq0$\r\n\r\nlet $ s'\\equal{}s\\plus{}1$\r\nwe'll try to prove a contradiction\r\nby hypothesis we have$ t|s$ and $ t|s'$ and assume $ t\\neq1$\r\nthen by prop e)$ s \\neq 0 \\wedge t|s \\rightarrow t \\le s$\r\nthus we have two cases: i) $ t\\equal{}s$ and ii)$ t<s$\r\ncase i)\r\nif $ t\\equal{}s$ then we know that $ t|s\\equal{}1$ but $ t\\nmid s'\\equal{}s\\plus{}1$.\r\nhow do we know this? proceed by induction.\r\nbase case:( since $ t\\neq 1$) we start with t=2\r\nclearly, $ 2|2$ and for $ n \\in\\mathbb{N} , 2|2n$ but it's clear that $ 2\\nmid 2n\\plus{}1$\r\nstep:Assume true for $ k$ where $ k\\equal{}mn$ for $ m,n\\in\\mathbb{N}$\r\n\r\nthis is where I'm stuck. I just dont see how i can show definitively for $ k\\plus{}1$\r\n\r\nthe idea of this proof is to show that for $ t\\equal{}s$, we can't have $ t\\equal{}s$ since $ t\\nmid s'$ and thus doesn't satisfy our hypothesis. and for $ t<s$, show that becuase $ t<s$ and $ t<s'$ for some arbitrary $ s$, then work backwards inductively ( method of infinte descent?) such that we get $ t<1$ which contradicts the fact that $ t\\in\\mathbb{N}$. So I sorry that I m so bad at both cases.\r\n\r\nany help is very much appreciated.\r\n\r\nThanks",
  "Solution_14": "I'd like to point out that in no way is property (e) given by the standard definition of divisibility. It happens to be true in the naturals, but it is not true in general. My question is: Is property (e) necessary?",
  "Solution_15": "well this is a proof in the naturals ( at least attempting a proof). I don't know if it's \"necessary\" but if you have a different idea of how to start,  it's probably better than my approach.",
  "Solution_16": "Let $ (a,b)$ denote the greatest common divisor of $ a$ and $ b$. This is defined as being the number $ g$ such that $ g|a$, $ g|b$, and $ d|a, d|b \\implies d|g$. Now consider $ g' \\equal{} (b\\minus{}a,b)$. Suppose $ d|a$ and $ d|b$. Then $ d|b\\minus{}a$ and $ d|b$. Thus, $ g' | g$. Then suppose $ d|b\\minus{}a$, $ d|b$. Then $ d|b\\minus{}(b\\minus{}a) \\equal{} a$, so $ g | g'$. Therefore, $ (a,b) \\equal{} (b\\minus{}a,b)$, and the result follows.\r\n\r\nNotice that there was no way in which the ordering was required.",
  "Solution_17": "I hate to be a jerk, but the reason I didn't use subtraction was because I m not given subtraction. If i was, as you showed, this problem would be a little bit better.\r\n\r\nam I at least on the right track( in terms of my proof), or can you suggest improvements?\r\n\r\nthanks",
  "Solution_18": "You are actually given subtraction, and here's why:\r\n\r\n$ t|a, t|b \\implies t|a, t|\\minus{}1\\cdot b \\implies t|a\\plus{}\\minus{}1\\cdot b \\implies t|a\\minus{}b$",
  "Solution_19": "I appreciate the help",
  "Solution_20": "As far as I'm aware, Peano arithmetic deals with $ \\mathbb{N}_0$, so the use of $ \\minus{}1$ is not valid.",
  "Solution_21": "That is correct.  Hamster's proof is bogus for Peano arithmetic.\r\n\r\nUnrelatedly, a notational point: the problem with the notation $ \\bf N$ is that it is unenlightening -- there's no way to know whether that's meant to include 0 or not, unless you've already established context.  The problem with the notation $ \\mathbb{N}_0$ is that it's equally unenlightening.  I strongly advocate the use of notations that no one can possibly be confused by, especially things like $ {\\bf Z}_{\\geq 0}$.  (No one uses $ \\bf Z$ for anything other than the integers, so not great likelyhood of confusion there.)",
  "Solution_22": "I haven't worked with peano arithmetic directly before, so I'll trust you that my proof doesn't work. How would you go about the proof then?",
  "Solution_23": "don't be shy, someone please help.\r\n\r\nAlso, I was wondering if my attempt to prove the question in my previous post was a good start, or should I have attempted another direction?\r\n\r\nthanks",
  "Solution_24": "Let's say we're given induction for this problem $ (t|s\\wedge t|s')\\rightarrow t\\equal{}1$\r\n\r\nif we applied induction:\r\n\r\nbase: $ s\\equal{}1$\r\n\r\n$ t|1$ and $ t|2$. Since we're working in $ \\mathbb{N}$ with peano arithmetic, then $ t\\equal{}1$\r\n\r\nstep: assume true for $ s \\equal{}k$, then $ (t|k\\wedge t|k')\\rightarrow t\\equal{}1$\r\nwe will note that proposition (h) $ \\rightarrow t|2k\\plus{}1$, in other words t divides odd numbers by our induction step\r\nassume $ k' \\equal{} k\\plus{}1$\r\n\r\nattempt to show for $ s\\equal{}k\\plus{}1$ in which we have $ t|k\\plus{}1 \\wedge t|k\\plus{}2 \\rightarrow t|2k \\plus{}3$ , in other words t divides an odd number.\r\n\r\nDoes this prove our assumption then that $ t|k\\plus{}1 \\wedge t|k\\plus{}2 \\rightarrow t\\equal{}1$?\r\n\r\nAlso, what would be wrong if I took $ k\\plus{}1 \\equal{} m$ and then showed that $ t|m \\wedge t|m\\plus{}1$ is in the form of our induction step, thus proving $ t\\equal{}1$ for $ s\\equal{}k\\plus{}1$?\r\n\r\nthanks for the help",
  "Solution_25": "[quote=\"mstrip\"]$ s \\equal{} k \\plus{} 1$ in which we have $ t|k \\plus{} 1 \\wedge t|k \\plus{} 2 \\rightarrow t|2k \\plus{} 3$ , in other words t divides an odd number.\n\nDoes this prove our assumption then that $ t|k \\plus{} 1 \\wedge t|k \\plus{} 2 \\rightarrow t \\equal{} 1$? [/quote]\n\nEr, no.  This doesn't do anything to show that $ t \\equal{} 1$ in this case.  \n\n[quote=\"mstrip\"]Also, what would be wrong if I took $ k \\plus{} 1 \\equal{} m$ and then showed that $ t|m \\wedge t|m \\plus{} 1$ is in the form of our induction step[/quote]\r\n\r\nYou seem to be misunderstanding induction.  What we are trying to prove is the statement\r\n\r\n$ P(n) \\equal{} \\text{If t divides n and t divides n \\plus{} 1 then t \\equal{} 1}$\r\n\r\nfor all positive integers $ n$.  The inductive hypothesis is that we assume $ P(k)$ for some [b]fixed[/b] $ k$ and look at \r\n\r\n$ P(k \\plus{} 1) \\equal{} \\text{If t divides k \\plus{} 1 and t divides k \\plus{} 2 then t \\equal{} 1}$\r\n\r\nUnfortunately, the set of $ t$ such that $ t | k, t | k \\plus{} 1$ is by no means necessarily related to the set of $ t$ such that $ t | k \\plus{} 1, t | k \\plus{} 2$, so in fact induction is a very poor choice of strategy here.",
  "Solution_26": "$ (s|s\\wedge s|s')\\rightarrow s \\equal{} 1$\r\nYou can prove that by induction.\r\n\r\nAlso $ t \\leq s$ but $ s\\equal{}1$ so $ t\\equal{}1$.",
  "Solution_27": "[quote=\"Fraenkel\"]$ (s|s\\wedge s|s')\\rightarrow s \\equal{} 1$\nYou can prove that by induction.[/quote]\r\n\r\n$ s | s$ is given.  Moreover, [b]there is nothing on which to induct.[/b]  This is not a statement about all natural numbers $ n$.",
  "Solution_28": "fraenkel, so then how do you do it by induction? it seems doable, but then t0rajir0u says induction wouldn't be a good idea. So who's right? and how would you prove it( as in I need someone to walk me through it) please?\r\n\r\n\r\nthanks!",
  "Solution_29": "[b]Given subtraction:[/b] \r\n\r\n$ t | k, t | k\\plus{}1 \\implies t | (k\\plus{}1) \\minus{} k \\implies t | 1 \\implies t \\equal{} 1$\r\n\r\n[b]Given the standard definition of divisibility:[/b]\r\n\r\n$ t | k \\Leftrightarrow \\exists a : k \\equal{} at$\r\n$ t | k\\plus{}1 \\Leftrightarrow \\exists b : k\\plus{}1 \\equal{} bt$\r\n$ (k\\plus{}1) \\minus{} k \\equal{} 1 \\equal{} bt \\minus{} at \\equal{} (b \\minus{} a)t \\implies$\r\n$ t | 1 \\implies$\r\n$ t \\equal{} 1$",
  "Solution_30": "if we did not have subtraction then we can't do it?",
  "Solution_31": "If we had neither subtraction nor the standard definition of divisibility, then [b]no.[/b]",
  "Solution_32": "in my previous post i said that in order for the statement to be true t|1, now by b) and f) t|1 \"and\" 1|t implies that 1=t. If there's anything wrong with that please pint it out coz i don't get it :!:",
  "Solution_33": "[quote=\"ggbg90\"]in my previous post i said that in order for the statement to be true t|1[/quote]\r\n\r\nYou can't conclude this without either subtraction or the standard definition of divisibility.",
  "Solution_34": "... or the variation on rule (h) that I suggested back in my first post on the thread.",
  "Solution_35": "Your proof with the standard definition of divisibility relies on the fact that multiplication distributes over subtraction, which is sufficient to fix my proof from before. Why is this true?",
  "Solution_36": "[quote=\"Hamster1800\"]Your proof with the standard definition of divisibility relies on the fact that multiplication distributes over subtraction, which is sufficient to fix my proof from before. Why is this true?[/quote]\r\n\r\n[b]Definition:[/b]  We say that $ a \\minus{} b \\equal{} c$ when $ a \\equal{} b \\plus{} c$.\r\n\r\n[b]Given distributivity over addition[/b] (which is not an axiom in Peano arithmetic):  \r\n\r\n$ a \\minus{} b \\equal{} c \\Leftrightarrow$\r\n$ a \\equal{} b \\plus{} c \\implies$\r\n$ da \\equal{} d(b \\plus{} c) \\equal{} db \\plus{} dc \\Leftrightarrow$\r\n$ da \\minus{} db \\equal{} dc \\equal{} d(a \\minus{} b)$\r\n\r\nAlternately, prove that $ a \\minus{} b \\equal{} a \\plus{} (\\minus{}1)b$.",
  "Solution_37": "So, I know I'm just making a bother of myself now, but I would like to suggest that any solution using subtraction is \"inferior\" from the point of view of Peano arithmetic to any solution not using subtraction (i.e., solutions either using my proposed additional axiom or solutions that use a quantified definition of divides) because Peano arithmetic doesn't have subtraction and negative numbers -- you can sort-of define subtraction, but you have to decide what to do when you subtract a larger number from a smaller number.  So, there is a lot more hidden behind any proof that has a \"$ \\minus{}$\" in it than there is in proofs that can avoid it.",
  "Solution_38": "but we are assuming that t|s and t|s+1, so in h) we can put t|s \"and\" t|1 implies t|s+1. We know that t|s and t|s+1 by our assumption, so if t doesn't divide 1 then condition h) fails",
  "Solution_39": "[quote=\"ggbg90\"]but we are assuming that t|s and t|s+1, so in h) we can put t|s \"and\" t|1 implies t|s+1. [/quote]\r\n\r\nh) tells us that if $ t | s$ and $ t | 1$ then $ t | s \\plus{} 1$.  It gives us no information about what happens if $ t | s$ and $ t | s \\plus{} 1$.  \r\n\r\nHere's another example:  Say that h) was the statement \"if I light a match and it is not raining then I will start a fire.\"  You are attempting to conclude from this that \"if I have started a fire and it is not raining then I lit a match.\"  But there are many other ways to start a fire (for example, a flamethrower).\r\n\r\nPut another way:  you are assuming that $ t | 1$, and from the premise $ t | s$ you can conclude that $ t | s \\plus{} 1$.  All this tells you is that it's [b]possible[/b] that $ t | 1$, not that it [b]must be true.[/b]"
}
{
  "Tag": [
    "articles"
  ],
  "Problem": "[hide]1)     x = y \n2)     x^2 = xy      (multiply both sides by x)\n3)     x^2 - y^2 = xy - y^2  (subtract y^2 from each side)\n4)     (x + y)(x - y) = y(x - y)     (factor)\n5)     x + y = y     (divide out (x - y))\n6)     2y = y      (substitute y for x from #1)\n7)     2 = 1      (What do you think of that?!)\n[/hide]\r\n\r\n\r\n :D  I copied this of a site!....Cool, huh?\r\n\r\n\r\n[color=white]BTW:we cannot divide by (x-y) so don't annoy me too much[/color]",
  "Solution_1": "Actually, one step doesn't work.  You can't divide out (x-y) since that's equal to 0 and you can't divide by 0.",
  "Solution_2": "way to ruin our fun",
  "Solution_3": "If I'd get a Euro every time I see that proof  :( ... Let me say it this way:\r\n[img]http://users.telenet.be/frederic_bel6/Fun/everytimeyouprovethat2improved.jpg[/img]",
  "Solution_4": "It's a perfectly legal answer. Since that's not true, we know $x\\not=y$.\r\nEither that, or it's an inequality. :P",
  "Solution_5": "[quote=\"fredbel6\"]If I'd get a Euro every time I see that proof  :( ... Let me say it this way:\n[img]http://users.telenet.be/frederic_bel6/Fun/everytimeyouprovethat2improved.jpg[/img][/quote]\r\n\r\n :lol: nice response",
  "Solution_6": "hahahahaaha",
  "Solution_7": "Yes, nice reply! (reminds me of another forum, when someone was spamming and then some other person gave a link to a pic with a 5yo kid praying: \"Please God, make it stop!\")\r\n\r\nBut what about those who don't like kittens? You know, the kind of people that have been terorized in their childhood by kittens and want to get even with them...",
  "Solution_8": "For an explanation of the picture, please see this Wikipedia article\r\n\r\n[url]http://en.wikipedia.org/wiki/Every_time_you_masturbate%E2%80%A6_God_kills_a_kitten[/url]",
  "Solution_9": "yea, I've definitely heard the phrase several times\r\n\r\nso I was reading the wiki article and then linked to dysphemism and then to Gresham's Law where I found\r\n\r\n\"In democracy, a less intelligent majority may be able to out-vote an exceptionally intelligent minority, effectively driving out the most intelligent voters by guaranteeing that their votes will be defeated. See also: IQ\"",
  "Solution_10": "[quote=\"fredbel6\"]If I'd get a Euro every time I see that proof  :( ... Let me say it this way:\n[img]http://users.telenet.be/frederic_bel6/Fun/everytimeyouprovethat2improved.jpg[/img][/quote]\r\n\r\n\r\nI hate kittens!! :diablo: \r\n\r\n[color=white]and mystic is right,\"less-inteligent people rock\"[/color]"
}
{
  "Tag": [
    "limit",
    "function",
    "calculus",
    "Asymptote",
    "calculus computations"
  ],
  "Problem": "$ N$ is a positive integer. Show that the equation $ \\sqrt{2x\\plus{}3} \\equal{} \\frac{N}{x^{3}}$ has exactly one real root.",
  "Solution_1": "Let $ f(x) \\equal{} \\sqrt {2x \\plus{} 3} \\minus{} N/x^3\\colon [ \\minus{} 3/2, \\plus{} \\infty)\\to {\\bf R}.$  [b]1)[/b] $ \\lim_{x\\to( \\minus{} 3/2)^ \\plus{} } f(x) \\equal{} 8N/27 > 0,$  [b]2)[/b]  \r\n$ \\lim_{x\\to0^ \\minus{} } f(x) \\equal{} \\plus{} \\infty,$ [b]3)[/b] $ \\lim_{x\\to0^ \\plus{} } f(x) \\equal{} \\minus{} \\infty,$  [b]4)[/b] $ f'(x) \\equal{} {1\\over \\sqrt {2x \\plus{} 3}} \\plus{} {3N\\over x^4} > 0$ for all $ x > \\minus{} 3/2.$ [b]1)--4)[/b] togheter with Weierstrass theorem on continuous functions imply the existence of a single \r\nreal numbers such that $ f(x) \\equal{} 0$  \r\n\r\n[b] Edit [/b] Add the fifth condition $ \\lim_{x\\to\\plus{}\\infty} f(x)\\equal{}\\plus{}\\infty$\r\n\r\nP.S. may be you thought of a simpler proof (differential calculus free?)[/u]",
  "Solution_2": "Thanks, Arrigo-Sacchi. By the way, what does Weierstrass's theorem state exactly?\r\n\r\nSketching the graphs. But perhaps there's a better way.",
  "Solution_3": "Let $ K$ be a compact subset of $ {\\bf R}^n$ and let $ f\\colon K\\to{\\bf R}$ be a continuous function on $ K.$ For instance $ n \\equal{} 1$ and $ K$ a closed and bounded interval $ [a,b]$ . Then $ f$ is bounded in $ [a,b],$ $ m \\le f(x)\\le M,$ and moreover there exist $ x_1,$ $ x_2$ such that $ f(x_1) \\equal{} m,$ $ f(x_2) \\equal{} M.$ \r\n\r\nIn your problem it sufficies to find two real positive values  $ y_1$ and $ y_2$ such that $ f(y_1) < 0$ and $ f(y_2) > 0.$ This is not difficult as the behaviors at $ x\\to0^ \\plus{}$ and $ x\\to\\ \\plus{} \\infty$ show (by the way in my post I forgot to write $ \\lim_{x\\to \\plus{} \\infty} f(x) \\equal{} \\plus{} \\infty$ becaquse otherwise the function could be negative having $ y \\equal{} 0$ as horizontal aymptote"
}
{
  "Tag": [
    "geometry",
    "\\/closed"
  ],
  "Problem": "When I see some users' mini-profiles on the left of their post, I see two national flags below their rating.  Is this new?  How can I turn it off or on?\r\nEDIT:  Ok, I see this can be edited in the profile options under Contact information.  The first flag represents the country the user was originally from, and the second represents te country the user is currently residing in.\r\n\r\nWhy does mine come above my rating?",
  "Solution_1": "Out of interest, can we request new flags (my country isn't there :()?",
  "Solution_2": "[quote=\"SimonM\"]Out of interest, can we request new flags (my country isn't there :()?[/quote]What country would that be? We're using the ISO 3661 standards btw.",
  "Solution_3": "Guernsey (one of the Channel Islands between England and France): https://www.cia.gov/library/publications/the-world-factbook/geos/gk.html\r\n\r\nTotal area, 78 square km, or about half the size of Washington, D.C.",
  "Solution_4": "I always thought the flags were a nice little touch, and I'm glad they're back. And having two flags allows for a little more nuance in the personal preferences.\r\n\r\nI'd like to see more individual users turn this preference on.",
  "Solution_5": "[quote=\"JBL\"]Guernsey (one of the Channel Islands between England and France): https://www.cia.gov/library/publications/the-world-factbook/geos/gk.html\n\nTotal area, 78 square km, or about half the size of Washington, D.C.[/quote]\r\n\r\nYeah, JBL's got it :) You're not the only site which doesn't have it. I had to request it on ProjectEuler (and  on a few other sites)\r\n\r\nSarnia Cherie!\r\n\r\nBTW, that CIA page could do with some improvements though.\r\n\r\nOur suffrage is 16. I have the right to vote too!",
  "Solution_6": "Ok it's there now.",
  "Solution_7": "Thanks :)\r\n\r\n*Goes to add it",
  "Solution_8": "Cool! I like the feature too.",
  "Solution_9": "One more time: how can we make a flag appear? Why hasn't some administrator explained this? Is that a secret?",
  "Solution_10": "It's in the first post- you can indicate your country in your profile. It's also an old feature reenabled; I haven't done anything, since I was already set up for it.",
  "Solution_11": "[quote=\"Carcul\"]One more time: how can we make a flag appear? Why hasn't some administrator explained this? Is that a secret?[/quote]\r\n\r\nMore specifically, go to your profile, click \"profile\", then click \"personal\". You will see two drop down menus where you can choose your flags.",
  "Solution_12": "[quote=\"jmerry\"]It's in the first post- you can indicate your country in your profile. It's also an old feature reenabled; I haven't done anything, since I was already set up for it.[/quote]\r\n\r\nWhy was it disabled before?",
  "Solution_13": "[quote=\"BanishedTraitor\"]Why was it disabled before?[/quote]We made some upgrades to the site (major ones) a while ago, and I didn't have time to upgrade the flag display up until recently.",
  "Solution_14": "Thank you very much Jmerry and Buzzer11."
}
{
  "Tag": [
    "LaTeX",
    "\\/closed"
  ],
  "Problem": "When will the WOOT message boards and classroom be open, and how would one navigate to those pages? \r\n\r\nThanks.  :lol:\r\n\r\n--200th post!--",
  "Solution_1": "We'll likely activate 2008-09 WOOT Message Boards and classroom towards the end of the month.  There won't be any classes until September, but we'll start some message board discussion when we open the board, by posting occasional problems to challenge enrolled students.",
  "Solution_2": "i have a question that is somewhat related to this topic and i figured it would be useless to make another topic, but just for conformation, students can type latex in the WOOT classroom? because i remember attending some math jams and only the instructor could use latex",
  "Solution_3": "[quote=\"cognos599\"]i have a question that is somewhat related to this topic and i figured it would be useless to make another topic, but just for conformation, students can type latex in the WOOT classroom? because i remember attending some math jams and only the instructor could use latex[/quote]\r\n\r\nStudents can use LaTeX in the classroom, even in the Math Jams (we sometimes don't mention that at the start of Math Jams, because, particularly in large Math Jams, it leads to some confusion, as some students think we mean that they have to know LaTeX, etc).",
  "Solution_4": "I (my parents) purchased the WOOT class for me today, but I still cannot find the corresponding message board. Is it under the Online School forum? I looked there, but I can't find it.\r\nedit: Yay, thanks!",
  "Solution_5": "You should see it now."
}
{
  "Tag": [
    "geometry",
    "trigonometry"
  ],
  "Problem": "Using a balance scale, it was determined that an algebra book\nand a geometry book together balance a trigonometry book,\nand two trigonometry books balance three geometry books.\nHow many geometry books are needed to balance four algebra\nbooks?",
  "Solution_1": "Let $ a$= the weight of an algebra book, $ g$= the weight of a geometry book, and $ t$= the weight of a trigonmetry book. We have the following: \r\n\r\n$ a \\plus{} g \\equal{} t$\r\n$ 3g \\equal{} 2t$\r\n\r\nTherefore, $ 2a \\plus{} 2g \\equal{} 2t \\implies 2a \\plus{} 2g \\equal{} 3g \\implies 2a \\equal{} g \\implies 4a \\equal{} 2g$. \r\n\r\nSo, $ \\boxed{2}$ geometry books balance $ 4$ algebra books."
}
{
  "Tag": [
    "search",
    "function"
  ],
  "Problem": "Can the number obtained by writing numbers from 1 to n be a palindrome?",
  "Solution_1": "[hide=\"this is what i have done\"]\nwe have to see if the number - $ 12345....(n\\minus{}2)(n\\minus{}1)(n)$ where n is replaced by its digits form a palindrome.\nWe see that $ n$ has to end with $ 321$ to make the number palindrome.\nso $ (n\\minus{}1)$ ends in 0.\nconsider - $ 123456789101112131415161718192021222324252627282930313234.....$\nfor the whole number to be a palindrome we chose n such that the digit after n in this sequence is 0 and n in the sequence starts from the left hand most and keeps on going until when we choose.\nso $ n$ is of $ 1987654321$ or of the form $ X9(X\\minus{}1)8(X\\minus{}1)7(X\\minus{}1)6(X\\minus{}1).....$ where $ (X\\minus{}1)$ is replaced by its digits.\nthis means that $ (n\\minus{}1)$ ends $ 7X6X5X4X3X2X1X0$. BUT $ (n\\minus{}1)$'s last digits are $ 54320$ so even if we set X=2 this is not possible so the number cant be a palindrome . However i dont think we are done as we also have to consider the case when n has digits $ 3201$.[/hide]\r\n i think it is a beautiful question!",
  "Solution_2": "c'mon mathlinkers! no one interested in this problem?",
  "Solution_3": "1) This is not a personal help line.  There is no reason to go bumping posts a mere 12 hours after they were posted while they still remain at the top of the forum.\r\n\r\n2) This question is very easy to search for: putting \"Russia 1996\" or \"Russia palindrome\" into the search function leads to threads like \r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=19872\r\nwith very little effort."
}
{
  "Tag": [],
  "Problem": "What is the value of the expression $ 5^3 \\minus{} 3 \\times 5^2 \\plus{} 3 \\times 5 \\minus{} 1$?",
  "Solution_1": "Whenever you see a question in this form ($ ax^3 \\minus{} 3x^2 \\plus{} 3x \\minus{} 1$), the answer is simply $ (x \\minus{} 1)^3$, so in this case, the answer is $ \\boxed{64}$.",
  "Solution_2": "hello, it is $ 125\\minus{}75\\plus{}15\\minus{}1\\equal{}64$.\r\nSonnhard.",
  "Solution_3": "The order of operations goes as follows:\r\n\r\nP: Parentheses\r\nE: Exponents\r\nM/D: Multiplication/Division(in the order from left to right)\r\nA/S: Addition/Subtraction(in the order from left to right)\r\n\r\nThere aren't any parentheses, so we do exponents first.\r\n\r\n$ 5^3 \\equal{} 125$\r\n\r\nand\r\n\r\n$ 5^2 \\equal{} 25$\r\n\r\n$ 125 \\minus{} 3 \\times 25 \\plus{} 3 \\times 5 \\minus{} 1$\r\n\r\nNow multiplication.\r\n\r\n$ 3 \\times 25 \\equal{} 75$\r\n\r\n$ 3 \\times 5 \\equal{} 15$\r\n\r\n$ 125 \\minus{} 75 \\plus{} 15 \\minus{} 1$\r\n\r\nFinally addition and subtraction.\r\n\r\n$ 125 \\minus{} 75 \\equal{} 50$\r\n\r\n$ 50 \\plus{} 15 \\equal{} 65$\r\n\r\n$ 65 \\minus{} 1 \\equal{} \\boxed{64}$\r\n\r\nEDIT: Beaten to it."
}
{
  "Tag": [],
  "Problem": "\u5728[url]http://en.wikipedia.org/wiki/The_High_School_Affiliated_to_Renmin_University_of_China[/url] \u6709\u4e00\u53e5\uff1a2 Gold Medalists in the International Mathematics Olympiad \r\n\r\nOut of the 6 Chinese Competitors, who is from \u4eba\u5927\u9644\u4e2d\uff1f   \uff08\u5bf9\u4e0d\u8d77\u3002\u6211\u7684\u6c49\u8bed\u4e0d\u597d....\uff09 \r\n\r\n\u6211\u5728\u97e9\u56fd\u7684\u5927\u5143\u5916\u8bed\u9ad8\u4e2d\u4e0a\u5b66\uff0c\u6211\u7684\u5b66\u6821\u548c\u4eba\u5927\u9644\u4e2d are sister schools\u3002\u4e0a\u4e2a\u6691\u5047\u6211\u5230\u5317\u4eac\u53bb\u8bbf\u95ee\u3002Just curious....[/url]",
  "Solution_1": "According to the listing on most Chinese websites, there was only one team member from \u4eba\u5927\u9644\u4e2d in 2008, Ruixiang Zhang (\u5f20\u745e\u7965), who earned a gold medal."
}
{
  "Tag": [
    "function",
    "logarithms",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Let $ h: \\mathbb{R} \\minus{} > \\mathbb{R}$ a function that satisfies:\r\ni) $ h(xy) \\equal{} h(x) \\plus{} h(y)$\r\nii)$ h(max(\\frac {m}{2^p3^q},\\frac {n}{2^p}) \\plus{} min(\\frac {n}{2^p3^q},\\frac {m}{2^p}))\\equal{}\\log_2 (q\\plus{}1)  \\forall m, n \\in \\mathbb{Z} , \\forall p,q \\in \\mathbb{N}$\r\nShow that if $ x \\in \\mathbb{Q}$ then $ h(x) \\equal{} 0$.",
  "Solution_1": "You haven't written the second condition well.\r\n\r\n- What you've said is basically the equivalent of saying $ h(z)$, which means nothing. What's the other side of the equation?\r\n- What are $ m,n,p,q$? Integers? Positive integers?",
  "Solution_2": "[quote=\"MellowMelon\"]You haven't written the second condition well.\n\n- What you've said is basically the equivalent of saying $ h(z)$, which means nothing. What's the other side of the equation?\n- What are $ m,n,p,q$? Integers? Positive integers?[/quote]\r\n\r\nNow I guess it's ok.",
  "Solution_3": "From the first condition: $ h(x) \\equal{} c\\log_a x , a > 0$,  $ a \\neq 1, c \\neq 0$ or $ h(x) \\equal{} 0$. I was trying to prove that $ h(x) \\equal{} c\\log_a x$ cannot be satisfied. But it's not working.",
  "Solution_4": "[quote=\"danilorj\"]Let $ h: \\mathbb{R} \\minus{} > \\mathbb{R}$ a function that satisfies:\ni) $ h(xy) \\equal{} h(x) \\plus{} h(y)$\nii)$ h(max(\\frac {m}{2^p3^q},\\frac {n}{2^p}) \\plus{} min(\\frac {n}{2^p3^q},\\frac {m}{2^p})) \\equal{} \\log_2 (q \\plus{} 1) \\forall m, n \\in \\mathbb{Z} , \\forall p,q \\in \\mathbb{N}$\nShow that if $ x \\in \\mathbb{Q}$ then $ h(x) \\equal{} 0$.[/quote]\r\n\r\nIt seems obviously wrong.\r\n\r\nTake property ii with $ p\\equal{}q\\equal{}m\\equal{}n\\equal{}1$ and you have : $ h(max(\\frac {1}{6},\\frac {1}{2}) \\plus{} min(\\frac {1}{6},\\frac {1}{2})) \\equal{} \\log_2 (2)$ And so :\r\n\r\n$ h(\\frac{2}{3})\\equal{}1\\neq 0$",
  "Solution_5": "ii) $ h( \\max ( \\frac{m}{2^{p}(q\\pi \\plus{} 1 )}, \\frac{n}{2^{p}}) \\plus{}  min (\\frac{m}{2^{p}}, \\frac{n}{2^{p}(q\\pi \\plus{} 1)})) \\equal{} log _{2} (q \\plus{} 1)$\r\n\r\nNow the second condition is correct.",
  "Solution_6": "Still doesn't look right. You need to clarify the first condition too (sorry for not noticing in first reply). If it's for all reals $ x,y$, then $ x \\equal{} 0$ gives $ h(y) \\equal{} 0$ for all reals $ y$, trivializing the problem.\r\n\r\nPerhaps the function is $ \\mathbb{R}^\\plus{} \\rightarrow \\mathbb{R}$?",
  "Solution_7": "[quote=\"danilorj\"]ii) $ h( \\max ( \\frac {m}{2^{p}(q\\pi \\plus{} 1 )}, \\frac {n}{2^{p}}) \\plus{} min (\\frac {m}{2^{p}}, \\frac {n}{2^{p}(q\\pi \\plus{} 1)})) \\equal{} log _{2} (q \\plus{} 1)$\n\nNow the second condition is correct.[/quote]\r\nLet $ n>0,m<0$, then we get $ h(\\frac{n\\plus{}m}{2^p})\\equal{}log_2(q\\plus{}1)\\not \\equal{}0.$",
  "Solution_8": "I received an e-mail that this question is totally incorrect. So forget about it."
}
{
  "Tag": [
    "calculus",
    "integration",
    "algebra",
    "function",
    "domain",
    "abstract algebra",
    "analytic geometry"
  ],
  "Problem": "Must $ A$ be an integral domain if $ A_p$ is an integral domain for every prime ideal $ p$?",
  "Solution_1": "in same spirit:\r\n\r\nLet $ A$ be a ring such that each local ring $ A_p$ is Noetherian. Must $ A$ itself be Noetherian?",
  "Solution_2": "@ integral domain:\r\n\r\n[quote=\"Jacobson\"]Must $ A$ be an integral domain if $ A_p$ is an integral domain for every prime ideal $ p$?[/quote]\r\n\r\nWhat about $ A \\equal{} \\mathbb{Z}\\slash 6\\mathbb{Z}$ ? This is hardly an integral domain, but the localizations at the prime ideals $ \\left(2\\right)$ and $ \\left(3\\right)$ are (if I am not mistaken, they are $ \\mathbb{Z}\\slash 2\\mathbb{Z}$ and $ \\mathbb{Z}\\slash 3\\mathbb{Z}$, up to isomorphism).\r\n\r\n  darij",
  "Solution_3": "@Noetherian:\r\n\r\n[quote=\"Jacobson\"]in same spirit:\n\nLet $ A$ be a ring such that each local ring $ A_p$ is Noetherian. Must $ A$ itself be Noetherian?[/quote]\r\n\r\nhttp://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2005;task=show_msg;msg=2110.0001 gives a nice counterexample, which would be even more nice if I would understand it. Here is a modification which I can prove: Let $ A \\equal{} \\oplus_{i\\in I}F$, where $ F$ is a field and $ I$ is an infinite index set. Then, for every prime ideal $ \\mathfrak{p}$ of $ A$, the ring $ A_{\\mathfrak{p}}$ is a field and therefore Noetherian, but $ A$ itself is not Noetherian. How to show that $ A_{\\mathfrak{p}}$ is a field? For every $ i\\in I$, if $ \\mathfrak{p}$ contains an element with nonzero $ i$-th coordinate (say $ \\left(x_1,x_2,...,x_{i \\minus{} 1},x_i,x_{i \\plus{} 1},...\\right)$ with $ x_i\\neq 0$), then $ \\mathfrak{p}$ contains the element $ \\left(0,0,...,0,1,0,0,...\\right)$, where the $ 1$ stands on the $ i$-th place (because $ \\left(0,0,...,0,1,0,0,...\\right) \\equal{} \\left(0,0,...,0,\\frac {1}{x_i},0,0,...\\right)\\cdot\\left(x_1,x_2,...,x_{i \\minus{} 1},x_i,x_{i \\plus{} 1},...\\right)$). Conclude from this that there exists a subset $ J\\subseteq I$ such that $ \\mathfrak{p} \\equal{} \\oplus_{i\\in I}\\left\\{\\begin{array}{c}F,\\text{ if }i\\in J; \\\\\r\n0,\\text{ if }i\\notin J\\end{array}\\right.$; in other words, $ \\mathfrak{p}$ is the set of all elements of $ I$ whose $ i$-th coordinates are $ 0$ for $ i\\notin J$, but all other coordinates can be arbitrary. Now show that $ \\left|I\\setminus J\\right| \\equal{} 1$ because $ \\mathfrak{p}$ is a prime ideal (in fact, if $ I\\setminus J$ would contain two distinct elements $ u$ and $ v$, then the product of $ \\left(0,0,...,0,1,0,...\\right)$ with $ 1$ on the $ u$-th place and $ \\left(0,0,...,0,1,0,...\\right)$ with $ 1$ on the $ v$-th place would be $ 0$ and thus lie in $ \\mathfrak{p}$, but none of the two factors lie in $ \\mathfrak{p}$). Conclude from this that $ A_{\\mathfrak{p}}$ is a field, namely $ F$.\r\n\r\nI hope I didn't make any stupid mistakes here.\r\n\r\n  darij",
  "Solution_4": "thanks darij\r\n\r\nCan someone expand my list?\r\n\r\n[b]Local property[/b]\r\n-flatness\r\n-injectivity\r\n-surjectivity\r\n\r\n[b]Non-local property[/b]\r\n-being an integral domain\r\n-being noetherian",
  "Solution_5": "a geometric motivation for darij's example: due to stone's representation theorem, the question, if noetherian is a local property, for boolean rings is if there is a totally disconnected hausdorff space, which is not artinian for closed subsets. well, $ \\mathbb{N}$ is an example.",
  "Solution_6": "further local properties for rings: reduced, normal (given integral domains), regular\r\nthere are more for schemes.",
  "Solution_7": "[quote=\"Jacobson\"]Must $ A$ be an integral domain if $ A_p$ is an integral domain for every prime ideal $ p$?[/quote]\r\n\r\nIf I recall correctly, for $ A$ to be an integral domain, $ Spec(A)$ must be connected. If that's the case, it's quite easy to construct such an example: let $ F$ be a field and $ A \\equal{} F \\times F$. Then $ A$ has two prime ideals $ 0 \\times F$ and $ F \\times 0$. Their corresponding localisations are isomorphic to F and hence integral domains - on the other hand $ A$ clearly is not.\r\n\r\nGeometrically, we're just taking the union of two points.",
  "Solution_8": "another local property: being torsion-free"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Here is a well used inequality and I think it needs a new topic to put nice solutions generalisations and maybe some stronger ones.\r\nHere is the problem\r\n$a,b,c\\geq 0$ . Prove that  \r\n$(a+b+c)^6\\geq 64(a^2+bc)(b^2+ca)(c^2+ab)$",
  "Solution_1": "[quote=\"silouan\"]\n$a,b,c\\geq 0$ . Prove that  \n$(a+b+c)^6\\geq 64(a^2+bc)(b^2+ca)(c^2+ab)$[/quote]\r\nsilouan,it follows from a wonderful hungkhtn's inequality:\r\n$a+b+c\\geq\\frac{2}{3}\\left(\\sqrt{a^2+bc}+\\sqrt{b^2+ac}+\\sqrt{c^2+ab}\\right).$ ;)",
  "Solution_2": "It is equivalent to\r\n\r\n$360(3,2,1)+3(6,0,0)+36(5,1,0)+90(4,2,0)\\geq 102(4,1,1)+132(3,3,0)$\r\n\r\nThen you can use Muirhead to solve it anyway.",
  "Solution_3": "[quote=\"libra_gold\"]It is equivalent to\n\n$360(3,2,1)+3(6,0,0)+36(5,1,0)+90(4,2,0)\\geq 102(4,1,1)+132(3,3,0)$\n\nThen you can use Muirhead to solve it anyway.[/quote]\r\n\r\nHow you arrive at such expression ? And is this the expansion of first or second inequality ? (I have checked and the expansion of first inequality doesnt look such way,and I doubt the second inequality can be \" expanded\" ? :?: )",
  "Solution_4": "[quote]Here is a well used inequality and I think it needs a new topic to put nice solutions generalisations and maybe some stronger ones. \nHere is the problem \n\nProve that \n\\[ (a+b+c)^6\\ge 64(a^2+bc)(b^2+ca)(c^2+ab) \\][/quote]\r\n\r\nMy direct solution for it is using $(a-b)^2(b-c)^2(c-a)^2 \\ge 0$ and SOS method, but it's truly hard, \r\n\r\nAlso, Manlio has asked me this question one time, in topic [b]Nice[/b].",
  "Solution_5": "I have a nice solution to this inequality in Algebraic Inequality. \r\nI finished the last correction of the book.  :("
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "Let $A(x)$ and $B(x)$ polynomials with degree greater than 1 and assume that exists polynomials $C(x)$ and $D(x)$ such that:\r\n\\[ A(x)\\cdot C(x)+B(x)\\cdot D(x)=1,~\\forall\\, x\\,\\in\\,\\mathbb{R}. \\]\r\nProve that $A(x)$ isn't divisible by $B(x)$.",
  "Solution_1": "[quote=\"LoboSolitario\"]Let $A(x)$ and $B(x)$ polynomials with degree greater than 1 and assume that exists polynomials $C(x)$ and $D(x)$ such that:\n\\[ A(x)\\cdot C(x)+B(x)\\cdot D(x)=1,~\\forall\\, x\\,\\in\\,\\mathbb{R}.  \\]\nProve that $A(x)$ isn't divisible by $B(x)$.[/quote]\r\n\r\nAssume for the sake of contradiction that $B(x)|A(x)$. Then $A(x) = Q(x)B(x)$ so $A(x)C(x)+B(x)D(x) = B(x)[Q(x)C(x)+D(x)] = 1 \\Rightarrow B(x)|1$. But since $B$ has degree greater than $1$, it obviously cannot divide something of degree $0$. Contradiction."
}
{
  "Tag": [
    "geometry",
    "trigonometry",
    "similar triangles",
    "power of a point",
    "trig identities",
    "Law of Cosines"
  ],
  "Problem": "Prove the Power of a Pt. Thm. for the case where the pt. of intersection lies within the circle. There is a simple proof, different from the more complete one in AOPS vol. II.\r\n\r\nCouldn't get a good picture up. I'm still learning how to do that kind of thing  :roll:",
  "Solution_1": "Well, while we're on the subject, I know two good geometry programs :\r\n[url=http://www.matf.bg.ac.yu/~janicic/gclc/]GCLC[/url] (Ugh, I don't like that new thing about proving theorems at all, though)\r\n[url=http://www.mit.edu/~ibaran/kseg.html]KSEG[/url]\r\n\r\nDo you mean AoPS I?!  I'm sure they don't do Power of a Point in Volume II\u2014it seems too elementary.  Someone else is borrowing my Vol. I right now, so how do they prove it there?  [hide]Do they use similar triangles?[/hide]",
  "Solution_2": "Oh, I'm sorry. I accidentally claimed that they proved the pwr. of a pt. thm. in AOPS II. Instead, I was thinking of the proof of the Law of Cosines, where the power of a pt. thm. was used throughout most of the proof.\r\n\r\nSimilar triangles is the right track. Someone should give a clear solution though."
}
{
  "Tag": [
    "MATHCOUNTS",
    "modular arithmetic"
  ],
  "Problem": "if 3^1993 = 100[i]k[/i]+[i]n[/i], where [i]n[/i]<100 is a nonnegative integer, what is [i]n[/i]?",
  "Solution_1": "[hide=\"um\"]\n23?\n\nThe solution is a little above mathcounts\n[/hide]",
  "Solution_2": "no it is not.\r\n\r\nnotice 9^10 ends in 01",
  "Solution_3": "[quote=\"unimpossible\"]no it is not.\n\nnotice 9^10 ends in 01[/quote]\r\n\r\nokay.\r\n\r\nThats exactly the point, so if 9^10 is 1 mod 100, then 3^20 is and so is 3^1980.\r\n\r\nSo we want to find just 3^13 mod 100, which is 23.",
  "Solution_4": "[quote=\"andrewliu\"]if 3^1993 = <br> 100[i]k[/i]+[i]n[/i] <br>, where [i]n[/i]<100 is a nonnegative integer, what is [i]n[/i]?[/quote]\r\n\r\nThat part gives away that it is (mod 100).",
  "Solution_5": "My idea is that u can find a pattern in the 1's digits. 3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243 and so on. If u look at the ones digit, it always is 3, 9, 7, 1, 3, 9, ... To find the units digit of 3^1933, just divide 1933 by 4 and find the remainder. Since every 4th power's units digit is 1, and u get a remainder of 1 which tells u to go 1 further than the 4th possible digit, then the units digit is 3. Since anything multiplyed by 100 has a units digit of 0 the answer is 3. Hope it helps  :D  :lol:  :)  :oops:  :wink:",
  "Solution_6": "it doesn't ask for units digit :wink:",
  "Solution_7": "[quote=\"501316\"]My idea is that u can find a pattern in the 1's digits. 3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243 and so on. If u look at the ones digit, it always is 3, 9, 7, 1, 3, 9, ... To find the units digit of 3^1933, just divide 1933 by 4 and find the remainder. Since every 4th power's units digit is 1, and u get a remainder of 1 which tells u to go 1 further than the 4th possible digit, then the units digit is 3. Since anything multiplyed by 100 has a units digit of 0 the answer is 3. Hope it helps  :D  :lol:  :)  :oops:  :wink:[/quote]\r\n\r\nNot asking for units digit, asking for modular residue in modulo 100, or in simpler words, the first 2 digits. Although there indeed is a pattern for the hundred's digit, this solution is not very \"elegant\" and takes a while. \r\n\r\nYou first want to write it as $ 3^{1993}\\equiv{x}\\pmod{100}$. Based on unimpossible's observation (you can confirm it by entering it in to your calculator), we can rewrite is as $ 3^{20} * 3^{20}... 3^{13}$. Since $ 3^{20}\\equiv{1}\\pmod{100}$, they cancel out, leaving out $ 3^{13}\\equiv{x}\\pmod{100}$. This number is small enough for your calculator to handle; you get\r\n\r\n$ \\boxed{n \\equal{} 23}$"
}
{
  "Tag": [
    "geometry",
    "ratio",
    "geometry proposed"
  ],
  "Problem": "Let $M$ be the midpoint of side $BC$ of a triangle $ABC$. Let the median $AM$ intersect the incircle of $ABC$ at $K$ and $L$, $K$ being nearer to $A$ than $L$. If $AK=KL=LM$, prove that the sides of triangle $ABC$ are in the ratio $5: 10: 13$ in some order.\r\n :wink:",
  "Solution_1": "You will find a solution in this forum very easy , because it is posted again recently. I'm not at home this moment , so I can not search.I hope , somebody will send you the link. Ah! I think Nicula has recently sent a nice solution. \r\n   I had proposed the same problem in the past too !!!\r\n[u]Babis[/u]\r\n\r\nWell , see\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=incircle&t=49419"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "This has appeared a ton of times on mathcounts cd rounds when they say like leaps = some lops = some loops and then they want you to find the leaps for something. I always get intimidated by this and screw up, does anyone have any tips for this?",
  "Solution_1": "Lets say that $ \\frac{a}{b}logs\\equal{}\\frac{c}{d}fraps$ and $ \\frac{e}{f}fraps\\equal{}\\frac{g}{h}$ slaps.\r\n\r\nIf we put in numbers for the variables and think it through, we can find a pattern. Lets say we want to go from slaps to fraps. If we have $ x$ slaps, we have $ x*\\frac{h}{g}*\\frac{e}{f}$ fraps. Can you see a pattern of multiplying certain reciprocals by certain fractions?\r\n\r\nNow if you wanted to from slaps to logs, just convert slaps to fraps and then convert fraps to logs using similar logic."
}
{
  "Tag": [
    "function",
    "limit",
    "logarithms",
    "integration",
    "calculus",
    "derivative",
    "calculus computations"
  ],
  "Problem": "If $f(x)$ be a differentaible function such that $ \\lim_{x \\longrightarrow \\infty} f(x) = a \\in \\mathbb{R} $. Suppose that $ \\lim_{x \\longrightarrow \\infty} x f'(x) $ exists. Find this limit.",
  "Solution_1": "The limit $\\lim_{x\\to\\infty} xf'(x)$ must be zero. If not, WLOG, suppose it is a positive real. Then there exists $X>1$ such that $xf'(x)>a>0$ for $x>X$. $f'(x)>a/x$ for $X>0$. Choose $b>X$.\r\n\r\nLet us prove that $f(x)>f(b)+a\\ln x-a\\ln b$ for $x>b$.\r\nTo see this, let $g(x) = f(x)-f(b)-a\\ln x+a\\ln b$. In fact, $g'(x) = f'(x) - \\frac{a}{x} > 0$ for $x>b$ and $g(b)=0$. Thus $g(x)>0$ for $x>b$.\r\n\r\n$f(x)>f(b)+a\\ln x-a\\ln b$ implies that $f$ is unbounded. Contradiction.",
  "Solution_2": "thanks!!! Good proof!",
  "Solution_3": "A small comment: A more natural thought is use $\\int_b^x f'(x)dx > \\int_b^x \\frac{a}{x}dx$ to get the contradiction. But is $f'(x)$ integrable? We need some words to argue. So I use derivatives instead of integration in the proof."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Hello.\r\n\r\nDoes anybody know if there's a way to get two lines of conditions below a sum in LaTeX? i.e. sum over d<x and Y_1 < T < Y_2, with the \"d<x\" and \" Y_1 < T < Y_2 \" on two separate lines under the sum? (because it looks ugly with one rediculously long line)  I tried using double subscripts and putting in a \\\\ but to no avail.\r\n\r\nThanks,\r\n\r\nian",
  "Solution_1": "$\\sum_{\\stackrel{d<x}{Y_{1}<T<Y_{2}}}$\r\n\r\nI don't know how to make them equally sized (the second line has bigger chars). :(",
  "Solution_2": "So what's the code?  How did you do it?",
  "Solution_3": "\\sum_{\\stackrel{d<x}{Y_{1}<T<Y_{2}}}\r\n\r\nYou can view the code of a $\\LaTeX$ picture by putting your mouse over it or viewing the picture's properties.",
  "Solution_4": "awesome.  thanks.\r\n\r\nian",
  "Solution_5": "Use the \\atop command as follows: $\\sum_{ d<x \\atop Y_{1}<T<Y_{2}}$.",
  "Solution_6": "The problem with \\stackrel and \\atop is that the size of the text may be too small. So amsmath has \\substack which also allows as many lines as you like. To compare:\r\n\\substack: $\\sum_{\\substack{d<x\\\\Y_{1}<T<Y_{2}}}$ \\stackrel: $\\sum_{\\stackrel{d<x}{Y_{1}<T<Y_{2}}}$  \\atop: $\\sum_{d<x \\atop Y_{1}<T<Y_{2}}$\r\nThe substack version looks nice and uses \\sum_{\\substack{d<x\\\\Y_{1}<T<Y_{2}}. It can be extended to say for example: $\\sum_{\\substack{d<x\\\\Y_{1}<T<Y_{2}\\\\a>2\\\\b\\neq0}}$ with each line being separated by \\\\."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "if $ {S_a}\\plus{}{S_b}\\plus{}{S_c}\\geq 0$ and $ {S_a}{S_b}\\plus{}{S_b}{S_c}\\plus{}{S_c}{S_a}\\geq 0$\r\ncan we prove that:$ {S_a}{(b\\minus{}c)^2}\\plus{}{S_b}{(a\\minus{}c)^2}\\plus{}{S_c}{(a\\minus{}b)^2}\\geq 0$\r\nthank you for your help~",
  "Solution_1": "[quote=\"xiaohui\"]if $ {S_a} \\plus{} {S_b} \\plus{} {S_c}\\geq 0$ and $ {S_a}{S_b} \\plus{} {S_b}{S_c} \\plus{} {S_c}{S_a}\\geq 0$\ncan we prove that:$ {S_a}{(b \\minus{} c)^2} \\plus{} {S_b}{(a \\minus{} c)^2} \\plus{} {S_c}{(a \\minus{} b)^2}\\geq 0$\nthank you for your help~[/quote]\r\nLet $ S_a\\plus{}S_b\\geq0.$\r\nHence, $ {S_a}{(b \\minus{} c)^2} \\plus{} {S_b}{(a \\minus{} c)^2} \\plus{} {S_c}{(a \\minus{} b)^2}\\equal{}$\r\n$ \\equal{}{S_a}{(b \\minus{} c)^2} \\plus{} {S_b}{(a \\minus{} b\\plus{}b\\minus{}c)^2} \\plus{} {S_c}{(a \\minus{} b)^2}\\equal{}$\r\n$ \\equal{}(S_a\\plus{}S_b)(b \\minus{} c)^2 \\plus{} 2S_b(a \\minus{} b)(b\\minus{}c) \\plus{} (S_c\\plus{}S_b)(a \\minus{} b)^2\\geq0$\r\nsince, $ S_b^2\\minus{}(S_a\\plus{}S_b)(S_c\\plus{}S_b)\\leq0.$ :wink:",
  "Solution_2": "$ S_{b}^{2}\\minus{}(S_{a}\\plus{}S_{b})(S_{c}\\plus{}S_{b})\\leq0.$\r\nyes,it is the key to the problem,thank you~"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let $ a,b,c\\geq0$.Prove that:\r\n$ (a^2\\minus{}bc)\\sqrt{b\\plus{}c}\\plus{}(b^2\\minus{}ca)\\sqrt{c\\plus{}a}\\plus{}(c^2\\minus{}ab)\\sqrt{a\\plus{}b}\\geq0$",
  "Solution_1": "I think SOS might be useful and this: [b]2(a^2 -bc)= (a-b)(a+c)+(a-c)(a+b)[/b]",
  "Solution_2": "Let $ x \\equal{} \\sqrt {a \\plus{} b}, y \\equal{} \\sqrt {b \\plus{} c}, z \\equal{} \\sqrt {c \\plus{} a}$\r\n\r\nthen this inequality is equivalent to $ \\sum_{cyc} x(y^4 \\plus{} z^4 \\minus{} x^2y^2 \\minus{} x^2z^2) \\geq 0$,\r\n\r\nwhich is true by Muirhead's inequality. ($ \\sum_{sym} x^4y \\geq \\sum_{sym} x^3y^2$)",
  "Solution_3": "[quote=\"M.A\"]I think SOS might be useful and this: [b]2(a^2 -bc)= (a-b)(a+c)+(a-c)(a+b)[/b][/quote]\r\nI think you're right , M.A . This inequality is equivalent to :\r\n$ \\sum (a \\minus{} b)^2.[\\frac {1}{2}\\frac {\\sqrt {a \\plus{} c}\\sqrt{b \\plus{} c}}{\\sqrt {a \\plus{} c} \\plus{} \\sqrt {b \\plus{} c}} ]\\geq 0$",
  "Solution_4": "The following inequality is also true\n[img]http://s3.sinaimg.cn/middle/0018zOAxgy70KPcMrfA92&690[/img]",
  "Solution_5": "[quote=luofangxiang]The following inequality is also true\n[img]http://s3.sinaimg.cn/middle/0018zOAxgy70KPcMrfA92&690[/img][/quote]\n$\\sum_{cyc}(a^2-bc)(b+c)^{\\lambda}=\\frac{1}{2}\\sum_{cyc}((a-b)(a+c)-(c-a)(a+b))(b+c)^{\\lambda}=$\n$=\\frac{1}{2}\\sum_{cyc}((a-b)\\left((a+c)(b+c)^{\\lambda}-(b+c)(a+c)^{\\lambda}\\right)=$\n$=\\frac{1}{2}\\sum_{cyc}(a+c)^{\\lambda}(b+c)^{\\lambda}(a-b)\\left((a+c)^{1-\\lambda}-(b+c)^{1-\\lambda}\\right)\\geq0$.\n\n",
  "Solution_6": "Nice arqady"
}
{
  "Tag": [],
  "Problem": "For what real number value of $n$ does the system of equations\r\n\r\n$nx+y=1$\r\n$ny+z=1$\r\n$nz+x=1$\r\n\r\nhave no solution?",
  "Solution_1": "[quote=\"13375P34K43V312\"]For what real number value of $n$ does the system of equations\n\n$nx+y=1$\n$ny+z=1$\n$nz+x=1$\n\nhave no solution?[/quote]\r\n[hide]Adding them we get $x(n+1)+y(n+1)+z(n+1)=3$\n$(n+1)(x+y+z)=3$\nSo, if n=-1, then there is no solution...[/hide]",
  "Solution_2": "Yes, nice solution.",
  "Solution_3": "[quote=\"13375P34K43V312\"]Yes, nice solution.[/quote]\r\nThis is a great problem...\r\nThis and that one other one that I said was great..."
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "circumcircle",
    "number theory",
    "prime numbers",
    "national olympiad"
  ],
  "Problem": "5. http://www.mathlinks.ro/Forum/viewtopic.php?p=910666#910666\r\nFor all positive reals $ a$, $ b$, and $ c$, what is the value of positive constant $ k$ satisfies the following inequality?\r\n$ \\frac{a}{c\\plus{}kb}\\plus{}\\frac{b}{a\\plus{}kc}\\plus{}\\frac{c}{b\\plus{}ka}\\geq\\frac{1}{2007}$\r\n\r\n6. http://www.mathlinks.ro/Forum/viewtopic.php?p=910667#910667\r\n$ ABC$ is a triangle which is not isosceles. Let the circumcenter and orthocenter of $ ABC$ be $ O$, $ H$, respectively,\r\nand the altitudes of $ ABC$ be $ AD$, $ BC$, $ CF$. Let $ K\\neq A$ be the intersection of $ AD$ and circumcircle of triangle $ ABC$, $ L$ be the intersection of $ OK$ and $ BC$, $ M$ be the midpoint of $ BC$, $ P$ be the intersection of $ AM$ and the line that passes $ L$ and perpendicular to $ BC$, $ Q$ be the intersection of $ AD$ and the line that passes $ P$ and parallel to $ MH$, $ R$ be the intersection of line $ EQ$ and $ AB$, $ S$ be the intersection of $ FD$ and $ BE$.\r\nIf $ OL \\equal{} KL$, then prove that two lines $ OH$ and $ RS$ are orthogonal.\r\n\r\n7. http://www.mathlinks.ro/Forum/viewtopic.php?p=910669#910669\r\nIn each $ 2007^{2}$ unit squares on chess board whose size is $ 2007\\times 2007$,\r\nthere lies one coin each square such that their \"heads\" face upward.\r\nConsider the process that flips four consecutive coins on the same row, or flips four consecutive coins on the same column.\r\nDoing this process finite times, we want to make the \"tails\" of all of coins face upward, except one that lies in the $ i$th row and $ j$th column.\r\nShow that this is possible if and only if both of $ i$ and $ j$ are divisible by $ 4$.\r\n\r\n8. http://www.mathlinks.ro/Forum/viewtopic.php?p=910671#910671\r\nFor all positive integer $ n\\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$.",
  "Solution_1": "Thank you very much!  :wink: I posted help you all link.  :)",
  "Solution_2": "When will the final round take place?",
  "Solution_3": "[quote=\"modal_soul\"]8. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=910671#910671\nFor all positive integer $ n\\geq 2$, prove that product of all prime numbers less or equal than $ n$ is smaller than $ 4^{n}$.[/quote]Anyone know why this one was deleted? I reposted it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=364002]here[/url]."
}
{
  "Tag": [
    "trigonometry",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Let $ K$ denote the intersection of the two diagonals of a quadrilateral $ ABCD$ where $ \\angle ABC\\equal{}\\angle ADC\\equal{}90^o$ and $ AC\\equal{}AB\\plus{}AD$.Prove that the radius of inscribed of triangle $ ABK$ and $ ADK$ are equal",
  "Solution_1": "Let D'=>triangle ADC and triangle D'BC is congruent.\r\nAB+AD=AB+AD'=AD'=AC\r\nso, triangle AD'C is isoceles triangle. Let, angle AD'C =a\r\nangle DAC=a, and angle DCA=angle BCD'=90-a\r\nangle D'CA=a, so, angle BCA=2a-90 ,and angle BAC is 2a\r\nLook at triangle AD'C, you can fing 4a=180, a=45..\r\nso, quadrilateral ABCD is square. so, triangle ABK, ADK is congruent.\r\nso, the radius is same.",
  "Solution_2": "[quote=\"woofunky\"]Let D'=>triangle ADC and triangle D'BC is congruent.\nAB+AD=AB+AD'=AD'=AC\nso, triangle AD'C is isoceles triangle. Let, angle AD'C =a\nangle DAC=a, and angle DCA=angle BCD'=90-a\nangle D'CA=a, so, angle BCA=2a-90 ,and angle BAC is 2a\nLook at triangle AD'C, you can fing 4a=180, a=45..\nso, quadrilateral ABCD is square. so, triangle ABK, ADK is congruent.\nso, the radius is same.[/quote]\r\nSorry\uff0cbut you are wrong",
  "Solution_3": "That \\[ \\Leftrightarrow \\frac{{S_{\\vartriangle ABK} }}\r\n{{S_{\\vartriangle ADK} }} \\equal{} \\frac{{BK}}\r\n{{KD}} \\equal{} \\frac{{AB \\plus{} BK \\plus{} KA}}\r\n{{AD \\plus{} DK \\plus{} KA}} \\equal{} \\frac{{AB \\plus{} KA}}\r\n{{AD \\plus{} KA}}\\]",
  "Solution_4": "Let $ \\angle BAK \\equal{} \\theta$ ,$ \\angle DAK \\equal{} \\phi$ and $ \\angle BKA \\equal{} \\psi$.Then use the Law of Sine: $ BK: KA: AB \\equal{} \\sin \\theta : \\cos \\phi : \\sin \\psi$ ,$ AK: KD: DA \\equal{} \\cos \\theta : \\sin \\phi : \\sin \\psi$.Then $ \\frac{{BK}}{{KD}} \\equal{} \\frac{{BK}}{{KA}} \\cdot \\frac{{KA}}{{KD}} \\equal{} \\frac{{\\sin \\theta \\cos \\theta }}{{\\sin \\phi \\cos \\phi }}$ ,$ \\frac{{AB \\plus{} AK}}{{AD \\plus{} AK}} \\equal{} \\frac{{1 \\plus{} {\\textstyle{{AB} \\over {AK}}}}}{{1 \\plus{} {\\textstyle{{AD} \\over {AK}}}}} \\equal{} \\frac{{\\cos \\theta \\left( {\\cos \\phi  \\plus{} \\sin \\psi } \\right)}}{{\\cos \\phi \\left( {\\cos \\theta  \\plus{} \\sin \\psi } \\right)}}$.\r\nSo $ \\sin \\theta \\cos \\theta  \\plus{} \\sin \\theta \\sin \\psi  \\equal{} \\sin \\phi \\cos \\phi  \\plus{} \\sin \\phi \\sin \\psi$.\r\nAs $ \\frac{\\pi }{2} \\equal{} \\theta  \\plus{} \\psi  \\minus{} \\phi$ ,everything left is so easy.",
  "Solution_5": "Stop this topic!\r\nThis problem is in a problem of a continous contest in \"Math and Youth Magazine\", so I think the topic should be locked in order to ensure the justice for this contest.\r\nLast note: In fact, this problem has many synthetic proofs and excellent extensions. However, we will refer to them after the contest of M&Y finishes. \r\n[hide=\"To Mathlinkers of Vietnam\"]Mong c\u00e1c b\u1ea1n \u0111\u1ebfn t\u1eeb Vi\u1ec7t Nam \u0111\u1eebng n\u00ean post nh\u1eefng b\u00e0i t\u1eeb THTT s\u1ed1 m\u1edbi nh\u1ea5t, \u0111\u00f3 ko ph\u1ea3i l\u00e0 b\u00e0i t\u1eadp b\u1eaft bu\u1ed9c n\u00ean h\u00e3y t\u1ef1 suy ngh\u0129 \u0111\u1ec3 gi\u1ea3i quy\u1ebft ch\u00fang.[/hide]"
}
{
  "Tag": [],
  "Problem": "Arrange these Phosphorous Trihalides in the order of Bond Angle\r\n\r\n$ PF3, PCl3, PBr3, PI3$.\r\n\r\nNot meant for chemrock, he has already been told the ans",
  "Solution_1": "[hide=\"Answer\"]Bond angle increases from left to right.[/hide]",
  "Solution_2": "the answer i think is \r\n\r\n[hide]$ {PCl3} < {PBr3} < {PF3} < {PI3}$[/hide]",
  "Solution_3": "Valerium reconsider ur answer a little. :D",
  "Solution_4": "forget about whatever the answer is if the answer by Carcul(the logical one) is not right then how are we supposed to know them without any data or experiment....not to sound bad what's the use of such a problem :(",
  "Solution_5": "well u all the posted ur answers but has anyone given an explanation for it? :huh: \r\n\r\n@pardesi: Once u reason out then obviously u will find the answer :)",
  "Solution_6": "i guess it will be impossible to answer the reason without reasoning the answer...and that too insuffcient",
  "Solution_7": "The size of the halogen increases.",
  "Solution_8": "[quote=\"pardesi\"]i guess it will be impossible to answer the reason without reasoning the answer...and that too insuffcient[/quote]\r\n\r\nHo Ho Mokka Da! \r\n\r\n@Carcul: One small thing I think you are missing.",
  "Solution_9": "I am not missing anything. The halogen size increases, that's a fact. I have said anything else.",
  "Solution_10": "see i can put all the things i know into pen and paper but can never predict until i am given the data or ofc the answer what's the order\r\n1)the electronegativity difference is decreasing down the group and that of iodine and phosphorous are nearly the same so logically the bond angle shud be decreasing taking this into account as the bond pairs are attracte dless towards the central atom as we move down so repulsion decreases\r\n\r\n2)the sizes as Carcul said is increasing so in that way the bond pair-bond pair repulsion is increasing so looking at it this way the bond angle shud increase as we move down...\r\nso the thing happening depends on what factor outweighs what....which is IMPOSSIBLE to predict unless u r given the data or u know the answer :)",
  "Solution_11": "the bond angle is PI3>PBr3>PCl3. but i wont say anything about where pf3 should be placed, as i think there is an anomaly in case of pf3 as the hybridisation of pf3 is sp3, but in case of the others, as the size of halogen is great, the overlapping of  orbitals is reduced ,and therefore greater deviation of bond angle from 109 deg.\r\npersonally i agree with pardesi, without experimental evidence we are in no position to answer the question, but of course we mite get such questions in our exams. :wink:",
  "Solution_12": "[quote=\"pardesi\"]see i can put all the things i know into pen and paper but can never predict until i am given the data or ofc the answer what's the order\n1)the electronegativity difference is decreasing down the group and that of iodine and phosphorous are nearly the same so logically the bond angle shud be decreasing taking this into account as the bond pairs are attracte dless towards the central atom as we move down so repulsion decreases\n\n2)the sizes as Carcul said is increasing so in that way the bond pair-bond pair repulsion is increasing so looking at it this way the bond angle shud increase as we move down...\nso the thing happening depends on what factor outweighs what....which is IMPOSSIBLE to predict unless u r given the data or u know the answer :)[/quote]\r\nconsidering your first point pardesi, as the electroneg diff, decreases down the group, bond angle shud increase not decrease.",
  "Solution_13": "Varun you are right!! Its illogical to compare PCl3 with PI3.. But definitely you can logically compare all adjacent groups???",
  "Solution_14": "[quote=\"pardesi\"]the bond angle is PI3>PBr3>PCl3. but i wont say anything about where pf3 should be placed, as i think there is an anomaly in case of pf3 as the hybridisation of pf3 is sp3, but in case of the others, as the size of halogen is great, the overlapping of orbitals is reduced ,and therefore greater deviation of bond angle from 109 deg. [/quote]\r\n\r\nEDIT: never mind.",
  "Solution_15": "[quote=\"VARUNARASU\"]the bond angle is PI3>PBr3>PCl3. but i wont say anything about where pf3 should be placed, as i think there is an anomaly in case of pf3 as the hybridisation of pf3 is sp3, but in case of the others, as the size of halogen is great, the overlapping of  orbitals is reduced ,and therefore greater deviation of bond angle from 109 deg.\npersonally i agree with pardesi, without experimental evidence we are in no position to answer the question, but of course we mite get such questions in our exams. :wink:[/quote]\r\nur right. PF3 cannot be placed without factual evidence but for the rest it is correct. :)",
  "Solution_16": "[quote=\"VARUNARASU\"][quote=\"pardesi\"]see i can put all the things i know into pen and paper but can never predict until i am given the data or ofc the answer what's the order\n1)the electronegativity difference is decreasing down the group and that of iodine and phosphorous are nearly the same so logically the bond angle shud be decreasing taking this into account as the bond pairs are attracte dless towards the central atom as we move down so repulsion decreases\n\n2)the sizes as Carcul said is increasing so in that way the bond pair-bond pair repulsion is increasing so looking at it this way the bond angle shud increase as we move down...\nso the thing happening depends on what factor outweighs what....which is IMPOSSIBLE to predict unless u r given the data or u know the answer :)[/quote]\nconsidering your first point pardesi, as the electroneg diff, decreases down the group, bond angle shud increase not decrease.[/quote]\r\n\r\nwhy as the elevtronegativity of the central atom decreases so the eletrons tend to be relatively more away from the central atom hence they are more away than they would have been if they were near the angular points(wher the repulsion is max)  :)",
  "Solution_17": "Answer\r\n\r\n$ PI_3 > PBr_3 > PCl_3 < PF_3$\r\n\r\nCare to explain??",
  "Solution_18": "[quote=\"pardesi\"]\nwhy as the elevtronegativity of the central atom decreases so the eletrons tend to be relatively more away from the central atom hence they are more away than they would have been if they were near the angular points(wher the repulsion is max)  :)[/quote]\r\n\r\nDei only the surrounding atom changes and not the central atom :)",
  "Solution_19": "[b]Cyber-Shreyas[/b], your answer is wrong.",
  "Solution_20": "[quote=\"Carcul\"][b]Cyber-Shreyas[/b], your answer is wrong.[/quote]\r\n\r\nI verified values. Wiki says http://en.wikipedia.org/wiki/Phosphorus_chlorides I am wrong. \r\n\r\nHowever a book of mine says PF3 > PCl3 because there is back bonding which makes the BP - BP repulsion slightly significant increasing the P-X-P bond. :("
}
{
  "Tag": [
    "number theory",
    "prime numbers"
  ],
  "Problem": "For how many prime numbers between 10 and 100 is the last digit also a prime number?",
  "Solution_1": "the prime numbers are 2, 3, 5, and 7\r\n\r\nhowever, prime numbers ending in 2 and 5 are not prime, so the only digits that the prime numbers can end in are 3 and 7\r\n\r\nlet's list the numbers out\r\n13\r\n17\r\n23\r\n37\r\n43\r\n47\r\n53\r\n67\r\n73\r\n83\r\n97\r\n\r\nwe have 11 numbers\r\n\r\ncan anyone confirm?\r\ni feel like i made a mistake somewhere....",
  "Solution_2": "since 2 and 5 are prime number we can conclude that 15 12 25 22 are prime # hence as they say that you can include any number as long as they have ANY PRIME # \nSO THE LIST WILL BE WHAT VALLON22 WROTE \n\n\n\n\nIf this is a straight line [showing his audience a straight line drawn by a ruler], then it necessarily ensues that the sum of the angles of the triangle is equal to two right angles, and conversely, if the sum is not equal to two right angles, then neither is the triangle rectilinear. \nAristotle (384-322 B. C. E)"
}
{
  "Tag": [
    "inequalities",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Given are triangle $ABC$ with any point $M$ on plane prove that\r\n$\\sqrt{bc}MA^{2}+\\sqrt{ca}MB^{2}+\\sqrt{ab}MC^{2}\\ge abc$\r\nsimilar like\r\n$aMA^{2}+bMB^{2}+cMC^{2}\\ge abc$\r\n :D",
  "Solution_1": "We have:\r\n$\\left(\\sum\\sqrt{bc}.\\overrightarrow{MA}\\right)^{2}\\ge 0\\Longrightarrow\\sum bc.MA^{2}+2\\sum a\\sqrt{bc}.\\overrightarrow{MB}.\\overrightarrow{MC}\\ge 0\\Longrightarrow\\sum bc.MA^{2}+\\sum a\\sqrt{bc}.(MB^{2}+MC^{2}-a^{2})\\ge 0\\Longrightarrow\\left(\\sum\\sqrt{bc}\\right)\\cdot\\left(\\sum{\\sqrt{bc}.MA^{2}}\\right)\\ge\\sum a^{3}\\sqrt{bc}\\Longrightarrow\\sum\\sqrt{bc}.MA^{2}\\ge\\frac{\\sum a^{3}\\sqrt{bc}}{\\sum\\sqrt{bc}}$\r\nWe have to prove that:\r\n$\\frac{\\sum a^{3}\\sqrt{bc}}{\\sum\\sqrt{bc}}\\ge\\prod a\\Longleftrightarrow\\sum\\frac{a^{3}\\sqrt{bc}}{\\prod a}\\ge\\sum\\sqrt{bc}\\Longleftrightarrow\\sum\\frac{a^{2}}{\\sqrt{bc}}\\ge\\sum\\sqrt{bc}$\r\nBut we easy see that:\r\n $\\sum\\frac{a^{2}}{\\sqrt{bc}}\\ge\\frac{(\\sum a)^{2}}{\\sum\\sqrt{bc}}\\ge\\frac{\\left(\\sum\\sqrt{bc}\\right)^{2}}{\\sum\\sqrt{bc}}=\\sum\\sqrt{bc}\\; \\mathbb{GED}$"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Let be given a natural number $n\\ge 2$.Find all real numbers $\\alpha$ such that for all sets of positive real numbers $a_1\\ge a_2\\ge a_3\\ge ...\\ge a_n$ we have:\r\n$\\frac{a_1}{a_n}+\\frac{a_2}{a_{n-1}}+....+\\frac{a_{n-1}}{a_2}+\\frac{a_n}{a_1}\\le n+\\frac{(a_1-a_n)^2+(a_2-a_n)^2+...+(a_{n-1}-a_n)^2+(a_n-a_n)^2}{a_n^\\alpha}$",
  "Solution_1": "If $\\alpha>2:$\r\nFor $a_2=a_3=\\ldots=a_n=r>2^{\\frac{1}{\\alpha-2}}$\r\n$a_1=2r$\r\nThe term of the left is equal to $n+\\frac{1}{2}$\r\nand the term of the right is equal to $n+\\frac{r^2}{r^\\alpha}$\r\nBut $r^{\\alpha-2}>2 \\Rightarrow \\frac{1}{2}>r^{2-\\alpha}$ (contradiction)\r\nIf $\\alpha<2:$\r\nanalogously\r\nIf  $\\alpha=2:$\r\nadding extrems It prove the inequality"
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "How many odd numbers with third digit 5 are there between 20000 and 69999 inclusive?",
  "Solution_1": "i think this should be moved to high school... it gave me a headache... T-T",
  "Solution_2": "This problem isn't hard... but I'll move it to mathcounts \r\n\r\n[hide=\"hint\"]think about how many digits are possible for each place. [/hide]",
  "Solution_3": "[hide][quote=\"4everwise\"]How many odd numbers with third digit 5 are there between 20000 and 69999 inclusive?[/quote]\n\n5*10*1*10*5=2500 numbers[/hide]"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "For a triple $ (m,n,r)$ of integers with $ 0 \\leq r \\leq n \\leq m \\minus{} 2$, define\r\n$ p(m,n,r) \\equal{} \\sum_{k \\equal{} 0}^{r}{m \\plus{} n \\minus{} 2(k \\plus{} 1) \\choose n}{r \\choose k}$\r\nProve that $ p(m,n,r)$ is positive and that\r\n$ \\sum_{r \\equal{} 0}^{n}{p(m,n,r} \\equal{} {m \\plus{} n \\choose n}$",
  "Solution_1": "It looks like the factor $ (\\minus{}1)^k$ is missing in the definition of $ p(m,n,r)$. \r\nIf so, first notice that\r\n\\[ \\sum_{r \\equal{} 0}^{n} p(m,n,r) \\equal{} \\sum_{r \\equal{} 0}^{n} \\sum_{k \\equal{} 0}^{r}(\\minus{}1)^k {m \\plus{} n \\minus{} 2(k \\plus{} 1) \\choose n}{r \\choose k} \\equal{} \\sum_{k \\equal{} 0}^{n} (\\minus{}1)^k {m \\plus{} n \\minus{} 2(k \\plus{} 1) \\choose n} \\sum_{r \\equal{} k}^{n} {r \\choose k} \\equal{} \\sum_{k \\equal{} 0}^{n} (\\minus{}1)^k {m \\plus{} n \\minus{} 2(k \\plus{} 1) \\choose n} {n\\plus{}1\\choose k\\plus{}1} \\equal{} {m\\plus{}n\\choose n} \\minus{} \\sum_{k \\equal{} \\minus{}1}^{n}  {m \\plus{} n \\minus{} 2(k \\plus{} 1) \\choose n} \\cdot (\\minus{}1)^{k\\plus{}1} {n\\plus{}1\\choose k\\plus{}1}\\]\r\nThe first factor of the summand in the last sum is the coefficient of $ x^{m \\plus{} n \\minus{} 2(k \\plus{} 1)}$ in $ x^n (1\\plus{}x)^{\\minus{}(n\\plus{}1)}$ while the second factor is the coefficient of $ x^{k\\plus{}1}$ in $ (1\\minus{}x)^{n\\plus{}1}$, so that the sum equals the coefficient of $ x^{m \\plus{} n}$ in\r\n\\[ x^n (1\\plus{}x)^{\\minus{}(n\\plus{}1)} (1\\minus{}x^2)^{n\\plus{}1} \\equal{} x^n (1\\minus{}x)^{n\\plus{}1}\\]\r\ni.e., the coefficient of $ x^m$ in $ (1\\minus{}x)^{n\\plus{}1}$ which is zero since $ m>n\\plus{}1$.\r\nTherefore,\r\n\\[ \\sum_{r \\equal{} 0}^{n} p(m,n,r) \\equal{} {m\\plus{}n\\choose n}.\\]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Find: $ \\int \\frac{xarctgx}{\\sqrt{x^{2}\\minus{}1}} \\,dx$\r\n\r\nP.S. Do not use integration software :-)",
  "Solution_1": "Integrate by parts and you're left with $ \\int \\frac{\\sqrt{x^2 \\minus{} 1}}{x^2 \\plus{} 1} \\, dx$.\r\n\r\nWrite that as $ \\int \\frac1{\\sqrt{x^2 \\minus{} 1}} \\cdot \\frac{x^2 \\minus{} 1}{x^2 \\plus{} 1} \\, dx$.\r\n\r\nHence, it suffices to compute $ \\int \\frac1{\\left( x^2 \\plus{} 1 \\right) \\sqrt{x^2 \\minus{} 1}} \\, dx$.\r\n\r\nMake the substitution $ t \\equal{} \\frac1{x^2} \\, \\Longrightarrow \\, \\int \\frac1{\\left( 1 \\plus{} t \\right) \\sqrt{1\\minus{}t}} \\, dt$.\r\n\r\nFinal substitution $ u^2 \\equal{} 1\\minus{}t \\, \\Longrightarrow \\, \\int \\frac1{u^2 \\minus{} 2} \\, du$.\r\n\r\nFor final answer use integration software.",
  "Solution_2": "What do you think? Was this integral interesting and/or difficult?",
  "Solution_3": "[color=darkblue][b][u]The Abel's substitution.[/u][/b] $ \\boxed {t \\equal{} \\frac {2ax \\plus{} b}{\\sqrt {ax^2 \\plus{} bx \\plus{} c}}}$\n\nfor $ \\int\\frac {1}{\\left(ax^2 \\plus{} bx \\plus{} c\\right)\\cdot\\sqrt [n] {ax^2 \\plus{} bx \\plus{} c}}\\ \\mathrm {dx}\\ .$\n\n\nAlthough $ a\\ne b$, you can try similarly and here : \n\n$ I(a,b)\\equiv\\int\\frac {1}{(x^2 \\plus{} a)\\sqrt {x^2 \\plus{} b}}\\ \\mathrm {dx} \\equal{} F\\left(\\frac {x}{\\sqrt {x^2 \\plus{} b}}\\right) \\plus{} \\mathcal C$,\n\nwhere $ F\\equiv\\int\\frac {1}{(b \\minus{} a)t^2 \\plus{} a}\\ \\mathrm {dt}\\ .$[/color]"
}
{
  "Tag": [
    "pigeonhole principle"
  ],
  "Problem": "Let a finite sequence consist of $N$ positive integers. There are $n$ distinct positive integers in this sequence with the condition that $N\\geq 2^{n}$. \r\nShow that we can find a block of consecutive integers whose product is equal to a perfect square. \r\n\r\nSo far I considered the following cases\r\n-------------------------------------\r\n1)Let $S=\\{a_{1}<a_{2}<...<a_{n}\\}$ be these integers. And let the holes be the proper subsets of $S$ and therefore a total of $2^{n}-1$. And we have $N>2^{n}-1$ but that doesnt seem to work.\r\n\r\n2)Let the pigeons be the different possible consecutive blocks. Which is $(N-1)+(N-2)+...+2+1 = \\frac{N(N-1)}{2}\\geq 2^{n-1}(2^{n}-1)$. But I have no idea what the holes I have to take.\r\n\r\n\r\nI got a seperate question. Do the Pigeonhole skill eventually get good and fast?",
  "Solution_1": "It is certainly not easy. Pigeonhole is really ingenuity; you can memorize many of the constructions, but hard work can only get you so far. \r\n\r\n[hide=\"Solution\"]\n\nConsider the sequences \n$a_{1}\\\\ a_{1}a_{2}\\\\ ... \\\\ a_{1}a_{2}...a_{n}$\n\nEach of the integers can be written as $a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{n}^{k_{n}}$. If any of the above have $k_{1}...k_{n}$ all even, we are done. Otherwise, there are $2^{n}-1$ different possibilities for the parities of the $k_{i}$. By pigeonhole, some two are equal. Their difference is a string of the form $a_{i}...a_{j}$ and all the $k_{i}$ are even. [/hide]\r\n\r\nNote: this is a rather brutal estimate.",
  "Solution_2": "[quote=\"Phelpedo\"]It is certainly not easy. Pigeonhole is really ingenuity; you can memorize many of the constructions, but hard work can only get you so far. \n\n[hide=\"Solution\"]\n\nConsider the sequences \n$a_{1}\\\\ a_{1}a_{2}\\\\ ... \\\\ a_{1}a_{2}...a_{n}$\n\nEach of the integers can be written as $a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{n}^{k_{n}}$. If any of the above have $k_{1}...k_{n}$ all even, we are done. Otherwise, there are $2^{n}-1$ different possibilities for the parities of the $k_{i}$. By pigeonhole, some two are equal. Their difference is a string of the form $a_{i}...a_{j}$ and all the $k_{i}$ are even. [/hide]\n[/quote]\r\nCan you be more explanatory. I am not good with this pigeons as you are. Thanks.",
  "Solution_3": "Let me give a similar problem that may seem more natural:\r\n\r\nProve that any 16-digit number contains a subsequence of digits whose product is a perfect square\r\n\r\n[hide]\nLet $a_{i}$ be the $i$th digit. Consider the products\n$a_{1}$\n\n$a_{1}\\cdot a_{2}$\n\n$a_{1}\\cdot a_{2}... a_{16}$\n\nNote that each of these numbers can be factorized as $2^{w}3^{x}5^{y}7^{z}$. If any of these $16$ products has $w,x,y,z$ all even, then we have a perfect square. Otherwise, there are $2^{4}-1=15$ possible parity combination. By pigeonhole, some two are equal. Divide them to get a product $a_{i}...a_{j}$ with all even exponents. [/hide]",
  "Solution_4": "[quote=\"Phelpedo\"]\n\nConsider the sequences \n$a_{1}\\\\ a_{1}a_{2}\\\\ ... \\\\ a_{1}a_{2}...a_{n}$\n\nEach of the integers can be written as $a_{1}^{k_{1}}a_{2}^{k_{2}}...a_{n}^{k_{n}}$. If any of the above have $k_{1}...k_{n}$ all even, we are done. Otherwise, there are $2^{n}-1$ different possibilities for the parities of the $k_{i}$. By pigeonhole, some two are equal. Their difference is a string of the form $a_{i}...a_{j}$ and all the $k_{i}$ are even. [/quote]\r\n\r\nI don't see the pigeonhole though -- there are $2^{n}-1$ different parities, but only $n$ different sequences.",
  "Solution_5": "Ok, I can't edit the post but the $n$ in the sequences should be a capital $N$, which by problem statement is greater than $2^{n}-1$.",
  "Solution_6": "Thank you very much, Phelpedo! That really is a tough pigeonhole argument and tought to explain but you did a good job :). I only have one complaint I think you should have used $b_{1}, b_{1}b_{2},...,b_{1}b_{2}...b_{N}$ because the way you wrote it it was confusing it seems the $a_{i}$ where distint but with these $b_{j}$ you can say they arent necessarily distinct."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "homothety",
    "rotation",
    "circumcircle",
    "rectangle",
    "geometry proposed"
  ],
  "Problem": "ABC is a triangle and M, N, P are pts on (AC), (AB), (BC) respectively s.t. MP, PN, NM are perpendicular to BC, AB, CA respectively. Show that if the centroid of MNP is the Lemoine point of ABC then ABC is equilateral.",
  "Solution_1": "This is interesting. If I remember correctly (I am writing from memory now), triangle MNP is the pedal triangle of one of the two Brocard points of triangle ABC, and triangle MNP is the image of triangle ABC in a spiral homothety with the center of rotation at a Brocard point. Maybe, this helps?\r\n\r\n  Darij",
  "Solution_2": "I don't think MNP is the pedal triangle of a Brocard point, but the second part (the one about the similarity or spiral homothety) is, indeed, true. The triangles ABC and MNP share one of the Brocard points of ABC. \r\n\r\nIt didn't help me, though :). I'd love to see a solution which uses that.",
  "Solution_3": "[quote=\"grobber\"]I don't think MNP is the pedal triangle of a Brocard point,[/quote]\r\n\r\nYes, sorry, I was wrong. As I've said, I was writing from memory. Actually, the points M, N, P are the intersections of the lines CA, AB, BC with the antiparallels to AB, BC, CA through the symmedian point of triangle ABC. Hence, these points M, N, P lie on the second Lemoine circle of triangle ABC. Therefore, the circumcenter of triangle MNP is the center of this second Lemoine circle, i. e. the symmedian point of triangle ABC. Now, since we assumed that this symmedian point coincides with the centroid of triangle MNP, we conclude that the circumcenter of triangle MNP coincides with the centroid of triangle MNP. Hence, triangle MNP is equilateral, and triangle ABC, being similar to triangle MNP, must be equilateral, too. Theorem proven.\r\n\r\nAh yes, and how I showed that the points M, N, P are the intersections of the lines CA, AB, BC with the antiparallels to AB, BC, CA through the symmedian point of triangle ABC ?\r\n\r\nWell, let K' be the circumcenter of triangle MNP, and let L be the point diametrically opposite to M on the circle MNP. Then, < MPL = 90, but we know that < MPB = 90, hence L lies on the line BC. It is very easy to show that the segment LM is antiparallel to AB; now, the point K' is the midpoint of this segment (since L and M are diametrically opposite points on the circle MNP, and K' is the center of the circle), and hence, K' lies on the C-symmedian of triangle ABC. Similarly, K' lies on the A-symmedian and on the B-symmedian, and consequently, K' is the symmedian point of triangle ABC. This not only entails that the point M is the intersection of the line CA with the antiparallel to AB through the symmedian point of triangle ABC (and similarly for N and P), but this also immediately implies that the circumcenter of triangle MNP is the symmedian point of triangle ABC, the only fact we really needed in our proof.\r\n\r\n  Darij",
  "Solution_4": "Wow! It's so cool! :). I had a completely different approach:\r\n\r\nI tok the point X to be the common point (X=K(ABC)=G(MNP)). X is projected in M', P' on NP, NM respectively, and it's projected in B', C' on AC, AB respectively. Triangles ABC and NPM are directly similar, X is K(ABC) and X=G(MNP)=the isogonal conjugate of K(MNP), so XP'/XM'=XC'/XB'=>XP'*XB'=XM'*XC', and since XB'MP' and XC'NM' are rectangles, it means that S(XMP')=S(XNM') (*). Remember that X is the centroid of MNP and P', M' are its projections on the sides. If we write all the relations analogous to (*) it's easy to derive the fact that MNP is equilateral."
}
{
  "Tag": [
    "geometry",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find, with proof, all triples of integers $ (a,b,c)$ such that $ a, b$ and $ c$ are the lengths of the sides of a right angled triangle whose area is $ a \\plus{} b \\plus{} c$",
  "Solution_1": "W.L.O.G. $ a\\leq b < c$, by Pythagorean's theorem $ a^2 \\plus{} b^2 \\equal{} c^2$, we can set $ a \\equal{} m^2 \\minus{} n^2, \\ b \\equal{} 2mn,\\ c \\equal{} m^2 \\plus{} n^2$ for positive integers $ m,\\ n\\ (m > n)$.\r\nThen the area : $ \\frac {1}{2}ab \\equal{} a \\plus{} b \\plus{} c\\Longleftrightarrow \\frac {1}{2}(m^2 \\minus{} n^2)\\cdot 2mn \\equal{} (m^2 \\minus{} n^2) \\plus{} 2mn \\plus{} (m^2 \\plus{} n^2)$\r\n\r\n$ \\Longleftrightarrow m(m \\plus{} n)(m \\minus{} n) \\cdot n \\equal{} 2m(m \\plus{} n) > 0$\r\n\r\n$ \\therefore (m \\minus{} n)n \\equal{} 2\\Longleftrightarrow (m \\minus{} n,\\ n) \\equal{} (1,\\ 2),\\ (2,\\ 1)$, yielding $ (m,\\ n) \\equal{} (3,\\ 2), (3,\\ 1)$.\r\n\r\nAns. $ (a,\\ b,\\ c) \\equal{} (5,\\ 12, \\ 13),\\ (6,\\ 8,\\ 10),\\ (8,\\ 6,\\ 10),\\ (12,\\ 5,\\ 13)$.",
  "Solution_2": "$ K \\equal{} rs \\equal{} r \\cdot \\frac{a \\plus{} b \\plus{} c}{2} \\equal{} a \\plus{} b \\plus{} c \\Leftrightarrow r \\equal{} 2$.  Kunny is using the identity $ r \\equal{} n(m\\minus{}n)$ (a rather particular identity for right triangles, but one that I find useful for problems of this type anyway).  Using this technique you can easily calculate the Pythagorean triangles of any given inradius.",
  "Solution_3": "That's right. Especially for a right angled triangle with side lengths $ a,\\ b,\\ c\\ (a\\leq b < c)\\ \\boxed{r \\equal{} \\frac {a \\plus{} b \\minus{} c}{2}}$ holds."
}
{
  "Tag": [
    "floor function",
    "trigonometry",
    "linear algebra",
    "matrix",
    "algebra",
    "polynomial",
    "calculus"
  ],
  "Problem": "Prove \r\n\r\n\r\n$ \\prod_{k\\equal{}1}^{\\lfloor \\frac{n\\minus{}1}{2}\\rfloor} \\tan\\left(\\frac{k\\pi}{n}\\right)\\equal{}\\left\\{ \\begin{matrix} \\sqrt{n} & \\text{for} & n \\;\\; \\text{odd} \\\\ \\\\ 1 & \\text{for} & n\\;\\; \\text{even} \\end{matrix} \\right.$",
  "Solution_1": "consider the polynomial $ p(x) \\equal{} \\frac {x^n \\minus{} 1}{x \\minus{} 1} \\equal{} \\prod_{k \\equal{} 1;n \\minus{} 1}(x \\minus{} e^{i2k\\pi/n})$ ,and evaluate $ p(1)$ and $ p(\\minus{}1)$",
  "Solution_2": "another way:\r\n$ \\sin n\\alpha \\plus{}i\\cos n\\alpha \\equal{} (\\sin \\alpha \\plus{}\\cos \\alpha)^{n}\\equal{} \\cos^{n}\\alpha(\\tan \\alpha \\plus{}i)^{n}\\equal{}\\cos^{n}\\alpha (\\sum_{k\\equal{}0}^{n}C_{n}^{k}\\tan^{n\\minus{}k}i^{k})$\r\nhence you have :\r\n$ \\sin n\\alpha\\equal{}\\cos ^{n}\\alpha P_{m}(\\tan \\alpha)$ where you already know all coeficient of polynom $ P_{m}(x)$\r\nhence  $ \\alpha \\equal{} \\frac{k\\pi}{n}$ will be root \r\nor for polynom you have :\r\n$ x_{k}\\equal{}\\tan \\frac{k\\pi}{n}$ and final use Viete theorem  you are done"
}
{
  "Tag": [
    "function",
    "floor function",
    "inequalities",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Consider the function $f$ defined by\r\n\r\n$f(n)=\\frac{1}{n}{\\sum^{n}_{i=1}}{\\lfloor \\frac{n}{i}\\rfloor}$\r\n\r\nfor all positive integers $n$. Prove that:\r\n(a)  $f(n+1)>f(n)$ for infinitely many $n$\r\n(b)  $f(n+1)<f(n)$ for infinitely many $n$",
  "Solution_1": "For each positive integer $n$, denote $S_{n}= \\sum_{i=1}^{n}\\left[\\frac{n}{i}\\right]$, then $f(n)=\\frac{1}{n}S_{n}$. \r\n\r\nAssume that $f(k) > f(k+1), i.e., S_{k+1}<(1+1/k)S_{k}$ for only finite many positive integer k, then there is an positive integer $n$ such that for all positive integer $m$, we have \\[S_{n+m+1}\\geq \\left(1+\\frac{1}{n+m}\\right)S_{n+m}\\geq\\cdots\\geq \\left(1+\\frac{1}{n+m}\\right)\\cdots \\left(1+\\frac{1}{n}\\right) S_{n}.\\] Let $a=1+\\frac{1}{n}>1$, then $S_{n+m+1}\\geq a^{m+1}S_{n}\\geq a^{m+1}\\quad (*)$. However for each positive integer k, $S_{k}\\geq k+(k-1)+\\cdots+1=k(k+1)/2$, it implies that $S_{n+m+1}/a^{m+1}$ reaches to $0$ when $m$ tends to $\\infty$. A contradiction with (*).\r\n\r\nAssume that $f(k) < f(k+1)$ for only finite many positive integer k. then there is an positive integer $n$ such that $f(m)\\leq f(n)$ for all $m\\geq n$. Let $b=f(n)$, then $S_{m}\\leq bm\\quad(**)$.  Put $j=[m/2]$, we have \\[S_{m}= \\left(\\left[\\frac{m}{1}\\right]+\\cdots+\\left[\\frac{m}{j}\\right]\\right)+\\left(\\left[\\frac{m}{j+1}\\right]+\\cdots+\\left[\\frac{m}{m}\\right]\\right)\\\\ \\geq m\\left(\\frac{1}{1}+\\cdots+\\frac{1}{j}+\\right)-j+(m-j) \\geq m\\left(\\frac{1}{1}+\\cdots+\\frac{1}{j}\\right).\\] For each $x\\geq 0, ln(1+x)\\geq x$, then the above inequality implies \r\n$S_{m}/m \\geq ln(\\frac{2}{1})+ln(\\frac{3}{2})+\\cdots ln(\\frac{j+1}{j})=ln(j+1) =ln([m/2]+1)$. Hence $S_{m}/m$ reaches to $\\infty$ if $m$ tends to $\\infty$. A contradiction with (**).",
  "Solution_2": "For each positive integer $n$, denote $S_{n}= \\sum_{i=1}^{n}\\left[\\frac{n}{i}\\right]$, then $f(n)=\\frac{1}{n}S_{n}$. \r\n\r\nAssume that $f(k) < f(k+1)$ for only finite many positive integer k. then there is an positive integer $n$ such that $f(m)\\leq f(n)$ for all $m\\geq n$. Let $b=f(n)$, then $S_{m}\\leq bm\\quad(*)$.  Put $j=[m/2]$, we have \\[S_{m}= \\left(\\left[\\frac{m}{1}\\right]+\\cdots+\\left[\\frac{m}{j}\\right]\\right)+\\left(\\left[\\frac{m}{j+1}\\right]+\\cdots+\\left[\\frac{m}{m}\\right]\\right)\\\\ \\geq m\\left(\\frac{1}{1}+\\cdots+\\frac{1}{j}+\\right)-j+(m-j) \\geq m\\left(\\frac{1}{1}+\\cdots+\\frac{1}{j}\\right)\\quad (**).\\] For each $x\\geq 0, ln(1+x)\\leq x$, then the above inequality implies \r\n$S_{m}/m \\geq ln(\\frac{2}{1})+ln(\\frac{3}{2})+\\cdots ln(\\frac{j+1}{j})=ln(j+1) =ln([m/2]+1)$. Hence $S_{m}/m$ reaches to $\\infty$ if $m$ tends to $\\infty$. A contradiction with (*).\r\n\r\nNow we prove that there are infinitely many $k$ such that $f(k) > f(k+1)$. For each prime number $p> 8$, from (**), we get $S_{p}>2p$. Note that $\\left[\\frac{p-1}{i}\\right]=\\left[\\frac{p}{i}\\right]$ for all $i=2,\\cdots,p-1$. It implies that $S_{p}=S_{p-1}+2$. Hence \\[f(p)-f(p-1)=\\frac{S_{p}}{p}-\\frac{S_{p-1}}{p-1}=\\frac{S_{p}}{p}-\\frac{S_{p}-2}{p-1}=-\\frac{S_{p}-2p}{p(p-1)}<0.\\] Because there are infinite primes, thus the assertion holds true. QED"
}
{
  "Tag": [
    "calculus",
    "integration"
  ],
  "Problem": "Let $ z$ and $ z\\plus{}1$ be $ n$th roots of unity, where $ n$ is a positive integral multiple of 7. Compute the smallest possible numerical value for $ n\\minus{}z^3.$",
  "Solution_1": "[quote=\"pi^2/6\"]Let $ z$ and $ z \\plus{} 1$ be $ n$th roots of unity, where $ n$ is a positive integral multiple of 7. Compute the smallest possible numerical value for $ n \\minus{} z^3.$[/quote]\r\n[hide=\"Is this right?\"]\nIf z and z+1 are roots of unity, they must be either cis60 and cis120, or cis -60 and cis-120. For n to be an integral multiple of 7, while have roots at the 60 and 120 degree positions, n must be >=42. 42 is the minimum, so we take that to be n.\nz^3= either cis(120x3) or cis(-120x3), both of which are equal to 1, so z^3=1.\nn-z^3=41[/hide]",
  "Solution_2": "[quote=\"serialk11r\"][hide=\"Is this right?\"]\nIf z and z+1 are roots of unity, they must be either cis60 and cis120, or cis -60 and cis-120. For n to be an integral multiple of 7, while have roots at the 60 and 120 degree positions, n must be >=42. 42 is the minimum, so we take that to be n.\nz^3= either cis(120x3) or cis(-120x3), both of which are equal to 1, so z^3=1.\nn-z^3=41[/hide][/quote]\n[hide=\"Almost\"]z=cis 60 or cis-60, so z^3=-1, and the total answer is 43.[/hide]",
  "Solution_3": "What? How is z cis60? If z=cis60=1/2+root3/2, then z+1=3/2+root3/2 which is not a root of unity..."
}
{
  "Tag": [
    "function",
    "limit",
    "real analysis",
    "topology",
    "absolute value",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $f_{n} : \\mathbb{R} \\to  \\mathbb{R}$ be a sequence of continous functions that converge at every point. Prove that there exist an interval $I \\subset \\mathbb{R}$  and a number $M$ such that $|f_{n}(x)| \\leq M$  for every $x \\in I$ and every $n$  \r\n\r\nAny suggestion?\r\nThanks.",
  "Solution_1": "Do you know Baire's theorem ? If so, it is an easy consequence of that theorem.",
  "Solution_2": "Hello! Well, I don't know is it required to know Baire theorem (since I when I saw this question I did not know in which context it is included). Anyway, I know that Baire theorem states that if a metric space is complete it is nonmeager in intself...how you would work out this solution using this theorem?\r\nThanks!",
  "Solution_3": "[quote=\"niski\"]Hello! Well, I don't know is it required to know Baire theorem (since I when I saw this question I did not know in which context it is included). Anyway, I know that Baire theorem states that if a metric space is complete it is nonmeager in intself...how you would work out this solution using this theorem?\nThanks![/quote]\r\n\r\nNote, that since $\\lim_{n \\to \\infty}f_n(x)$ exists for every real $x$, for each fixed $x$, the sequence $\\{f_n(x) \\}_n$ - Cauchy and,thus, for every $x$ and every $n$ there exists $p$ such that for all $q \\geq p$ we have:\r\n$|f_q(x) - f_p(x)| < \\frac{1}{n}$, which means that $x \\in S_{n,p} = \\bigcap \\{ y, |f_q(y) - f_p(y)| < \\frac{1}{n} \\forall q \\geq p \\}$ , so for all $n$ $\\bigcup_{p} S_{n,p} = \\mathbb{R}$ and each of $S_{n,p}$ is closed. Now, you may apply Baire Category Theorem and the left is for you ...",
  "Solution_4": "$F_M=\\{x : |f_n(x)| \\leq M, \\forall n > 0\\}$ is closed. $\\bigcup_{M \\in \\mathbb{N}} F_M = \\mathbb{R}$ so ..",
  "Solution_5": "[quote=\"alekk\"]$F_M=\\{x : |f_n(x)| \\leq M, \\forall n > 0\\}$ is closed. $\\bigcup_{M \\in \\mathbb{N}} F_M = \\mathbb{R}$ so ..[/quote]\r\nOh, stupid me, we can do it easier.\r\nNiski, follow Alekk - his idea is much more appropriate in this problem.",
  "Solution_6": "[quote=\"eugene\"][quote=\"alekk\"]$F_M=\\{x : |f_n(x)| \\leq M, \\forall n > 0\\}$ is closed. $\\bigcup_{M \\in \\mathbb{N}} F_M = \\mathbb{R}$ so ..[/quote]\nOh, stupid me, we can do it easier.\nNilski, follow Alekk - his idea is much more appropriate in this problem.[/quote]\r\n\r\nI saw the Baire theorem statment in Rudin Book and it said that\r\n\"If $X$ is a complete metric space, the intersection of every countable collection of dense open subsets of $X$ is dense in $X$\"\r\n\r\nSo if we are working in $\\mathbb{R}$, the theorem means that if we pick any countable collection of dense open subsets of $\\mathbb{R}$ and consider its intersection, then  this intersection will be again dense (and open) in $\\mathbb{R}$.\r\n\r\nSo in this case, why you guys are considering $\\mathbb{R}$ as union of closed subsets?",
  "Solution_7": "Ok. Using the other statement of the theorem, then I know that at least one \r\n$F_{M}$ contains a non empty open subset. This $F_{M}$ is the candidate for $I$ ? why  $F_{M}$ must be an interval?\r\nThanks",
  "Solution_8": "in open sets you can always find intervals !",
  "Solution_9": "Rigth! So the only thing left to prove is that $F_{M}$ is closed. Let's try\r\nFix $M$ and let $(x_{m})$ be a convergent sequence of elements of $F_{M}$  with $\\bar{x}$ as limit. We have to prove that $\\bar{x} \\in F_{M}$. \r\nNow we have that $|f_{n}(x_{m})| \\leq M, \\forall n > 0$ and $\\forall m > 0$ so, because the absolute value is continous and $f_{n}$ is contiunous for every $n > 0$ then\r\n$|f_{n}(\\bar{x})| = \\lim_{m \\to  \\infty} |f_{n}(x_{m})| \\leq M$..\r\nbut I think something must be wrong since I did not used the hypothesis that $f_{n}$ is a sequence of continuous functions [b]that converges at every point[/b]...\r\nany suggestions?\r\n\r\nThanks again!",
  "Solution_10": "I think it is correct.\r\nOr you write $F_M = \\bigcap_n f_n^{-1}([-M ; M])$ and because of the continuity of $f_n$, $f_n^{-1}([-M ; M])$ is closed."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "When I have\r\n\r\nProblem Set: \\printproblemset  Problem Number: \\theqnumber  Page: \\thepage\r\n\r\nHow do I get spaces between \"\\printproblemset\" and \"Problem\"?  Because right now, I'm getting something like (if problemset = 1)\r\n\r\nProblem Set: 1Problem [more stuff]",
  "Solution_1": "Try \\ followed by a space so you get\r\nProblem Set: \\printproblemset \\ Problem Number: \\theqnumber Page: \\thepage"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $ABC$ be an acute triangle and let $D,\\,E,$ and $F$ be the feet of the altitudes from $A,\\,B,$ and $C$ to sides $BC,\\,CA,$ and $AB$ respectively. Let $P,\\,Q,$ and $R$ be the feet of the  perpendiculars from $A,\\,B,$ and $C$ to $EF,\\, FD,$ and $DE$ respectively. Prove that\r\n\\[2(PQ+QR+RP) \\ge DE+EF+FD\\]",
  "Solution_1": "Hint: Triangle inequality w/ Ptolemy's inequality.",
  "Solution_2": "[quote=\"April\"]Let $ABC$ be an acute triangle and let $D,\\,E,$ and $F$ be the feet of the altitudes from $A,\\,B,$ and $C$ to sides $BC,\\,CA,$ and $AB$ respectively. Let $P,\\,Q,$ and $R$ be the feet of the  perpendiculars from $A,\\,B,$ and $C$ to $EF,\\, FD,$ and $DE$ respectively. Prove that\n\\[2(PQ+QR+RP) \\ge DE+EF+FD \\]\n[/quote]\r\nI wonder if there exists a triangle $\\triangle XYZ$ such that $X,Y,Z$ lie on $BC,CA,AB$ respectively,and $\\triangle XYZ \\sim \\triangle PQR$ s.t $\\frac{XY}{PQ}=2$... :maybe: \r\nthen the problem will follow immediatly..."
}
{
  "Tag": [
    "LaTeX",
    "MATHCOUNTS"
  ],
  "Problem": "i will need 9 people to join\r\nthere will be 2 teams\r\nthe war begins when one person from one team insults another person from another team (dont take these insults seriously, its just a game). Then the other team has a choice: either the insulted person can insult back with a counter-insult, or a teammate of the insulted person can compliment the insulted person to prove the insult wrong. Once a team has no good move to make (cannot form an insult or compliment), the person currently insulted is \"dead\". Then the other team starts off the next round. when a team loses all its members, it loses the war. Signups are first come first serve, need nine people!\r\nEdit: when you sign up, you must specify if you are on my team or the other team. 4 members can join my team, while 5 members can join the team opposing me. my team gets to strike first, but dont worry, this war will be done twice where the other team strikes first. if each team wins one war, the final result is a tie. if one team wins both wars, that team is the final winner.",
  "Solution_1": "yeah why not ill join the team without mathelete...",
  "Solution_2": "Yeah, I'll join too... As an insult, I won't join mathelete's team either.  :)",
  "Solution_3": "same thing here.",
  "Solution_4": "okay:\r\n\r\nteam 1:\r\nmathelete\r\nteam 2:\r\nchompy\r\nlatex\r\nspaceguy524\r\n\r\nstill need 4 members on team 1 and 2 members on team 2",
  "Solution_5": "join mathelete",
  "Solution_6": "okay:\r\nTeam 1:\r\nmathelete\r\ngogators\r\nTeam 2:\r\nchompy\r\nlatex\r\nspaceguy524",
  "Solution_7": "I'll join Team 2.",
  "Solution_8": "mathelete, gogators\r\n\r\nchompy, latex, spaceguy524, westiepaw\r\n\r\njust need one more on team 2 and that team will be complete.",
  "Solution_9": "join mathlete",
  "Solution_10": "join team 2. so that completes team 2",
  "Solution_11": "Join the available team. (Mathelete's team)",
  "Solution_12": "Hm...I'll give it a shot :wink:",
  "Solution_13": "Join Mathlete's team",
  "Solution_14": "[quote=\"abcak\"]Join Mathlete's team[/quote]\r\n\r\nI think you're a bit too late... The teams are\r\n\r\nmathelete, gogators, EggyLv.999, meewhee009, and KING 1\r\n\r\nhjoon0125, chompy, latex, spaceguy524, westiepaw",
  "Solution_15": "I'm ready when everyone else is :wink:",
  "Solution_16": "Dang, i checked the Games Forum too late :wink: \r\n\r\nMy sis could've help me with this :lol:",
  "Solution_17": "Okay:\r\nmathelete, gogators, EggyLv.999, mwehee009, KING 1\r\n\r\nhjoon0125, chompy, latex, spaceguy524, westiepaw\r\n\r\nwar will start this friday, cause i am too busy. i will let mwehee009 (chosen by random.org) fire the first insult.\r\nremember: it is just a game, dont take the insults seriously. also, dont use profanity in the \"insults\".",
  "Solution_18": "I SO wish i could join.\r\n\r\n\r\nCould you just please, please, PLEASE have 5 people on each team?",
  "Solution_19": "[quote=\"polkadotman1\"]\nCould you just please, please, PLEASE have 5 people on each team?[/quote]\r\n\r\nThere are 5 people on each team.",
  "Solution_20": "kk srry 6 :blush:",
  "Solution_21": "Go try to PM mathlete. Maybe you can convince him. :)",
  "Solution_22": "Oh YES, I fire first insult. :evilgrin:\r\nNoting from the rules;\r\nwhoever I insult... do not, I repeat DO NOT take it seriously, I'm nice but when told to, I can be mean. So yes... early sorry!",
  "Solution_23": "Just insult someone already.",
  "Solution_24": "@spaceguy524\r\n[quote=\"mathelete\"]Okay:\n[b]war will start this friday[/b], cause i am too busy. i will let mwehee009 (chosen by random.org) fire the first insult.\nremember: it is just a game, dont take the insults seriously. also, dont use profanity in the \"insults\".[/quote]",
  "Solution_25": "i'm not in this game but am anxious to see how fail you bunch of losers are at coming up with insults(don't take this seriously either)",
  "Solution_26": "Can we start?\r\nOkay, I will.\r\n\r\n\r\n\r\nHeyyy... Latex...I need you...........I want you............To get out of my face.\r\n\r\n\r\n\r\n(IDMI. (guess what that is))",
  "Solution_27": "Lol...nice one meewhee09! High Five!",
  "Solution_28": "@LaTeX: You are very smart. :) :) :)",
  "Solution_29": "[quote=\"spaceguy524\"]@LaTeX: You are very smart. :) :) :)[/quote]\r\nreally? well, you cannot tell if somebody is smart unless you are also smart, which you are not!!!",
  "Solution_30": "*sniff*, *sniff*\r\n\r\n\r\nu left me out!!!!!!!!!!!!!!!!\r\n\r\n\r\nmathelete, you don't deserve to be an AoPSer.\r\n\r\n\r\n(again, don't take insults seriously)",
  "Solution_31": "Uh, I don't think you are playing.",
  "Solution_32": "I know.\r\n\r\n\r\ni don't care (yes i do).",
  "Solution_33": "[quote=\"mathelete\"][quote=\"spaceguy524\"]@LaTeX: You are very smart. :) :) :)[/quote]\nreally? well, you cannot tell if somebody is smart unless you are also smart, which you are not!!![/quote]\r\n\r\nSpaceguy not smart? Are you kidding!?  :o  Spaceguy made Mathcounts, which is more than you could do!",
  "Solution_34": "meewhee is so forgetful, she... um... what was I talking about?",
  "Solution_35": "xD\r\n\r\n\r\nI hope meewhee forgets.\r\n\r\nthen i can joinz :evilgrin:  :evilgrin:  :evilgrin:",
  "Solution_36": "oh crap sry i thought this game didn't start yet. Silly me.",
  "Solution_37": "Since everyone forgot about this, then am i automatically in?",
  "Solution_38": "This is slow....and yes most are forgetting.\r\npolkadot can take over for me :D",
  "Solution_39": "that is, if he is able to do so!!! can you now???\r\ni did not think so.",
  "Solution_40": "*sigh*\r\n\r\n\r\ni hav 1 more post than you mathlete!!!! yay!!!!!\r\n\r\nand i hav the right to take over chompy if he lets me and he does.",
  "Solution_41": "dead topic.\r\n\r\nmaybe too inappropriate.\r\n\r\nlock this topic up mods",
  "Solution_42": "too late, you should have called the mods sooner. i guess you are really slow. and yay i have one less post than you, take that!",
  "Solution_43": "You're one to talk. You've been losing the game for quite some time. You couldn't win the game if my life depended on it.\r\nAlso, when was the last time you cooked cookies? You haven't cooked cookies since your ma helped you. I can do it [i]all by myself![/b]",
  "Solution_44": "and because you do it all by yourself, they turn out bad. while mine turn out really good since my mother helps me. also, the first insult does not count since you can only do one."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Every finite group is isomorphic to a subgroup of some $A_n$.",
  "Solution_1": "Let $G$ act on two copies of itself. Each element acts as a permutation, and each such permutation is obviously even.\r\n\r\nTherefore, if $G$ has $k$ elements, $G$ is isomorphic to a subgroup of $A_{2k}$.",
  "Solution_2": "Or:\r\nevery group of order $n$ is subgroup of $S_n$ (Cayley's theorem)\r\nConsider this subgroup as subgroup of $S_{n+2}$ and 'adjungate' the cycle $(n+1 , n+2)$ to all odd permutations, then you get it as subgroup of $A_{n+2}$",
  "Solution_3": "[quote=\"jmerry\"]Let $ G$ act on two copies of itself. Each element acts as a permutation, and each such permutation is obviously even.\n\nTherefore, if $ G$ has $ k$ elements, $ G$ is isomorphic to a subgroup of $ A_{2k}$.[/quote]\r\n\r\nwhat do u mean act on two copies of itself , Could you clarify the action only ?\r\n\r\nDid u mean $ G$ is $ G \\minus{} Set$ via $ g.x \\equal{} g. \\minus{}(x)$ , where is $ g.\\minus{}$ is a permutation on $ G$, if yes , how do u sure this is even permutation ?\r\n\r\n :)",
  "Solution_4": "$ G$ acts fauthfully on $ G$ by left multiplication, i.e. we have a monomorphism $ G \\to Sym(G)$. this yields a monomorphism $ G \\to Sym(G) \\times Sym(G) \\to Sym(G \\dot{\\cup} G)$. the images have the form $ \\sigma \\circ \\sigma'$, where $ \\sigma,\\sigma \\in Sym(G)$ are actually the same but regarded on different copies of $ G$. in particular, they have the same sign and thus the sign of their product is $ 1$."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "geometry",
    "geometric transformation",
    "homothety",
    "rectangle",
    "parameterization"
  ],
  "Problem": "The solutions by this Web Site's instructors are posted in the Math Jams transcripts.  The solutions are, of course, very good!\r\n\r\nHow about posting novel solutions to see some different methods?\r\n\r\nI'll start the ball with a different solution to #10.",
  "Solution_1": "Nicely done.",
  "Solution_2": "In our Olympiad Geometry class we teach homothecy.  Students asked if it could be used on the AIME.  I answered 'rarely, though we often wrote Mandelbrot problems for homothecy.'\r\n\r\nThen you give this cool solution, in which the homothecy of the 'desired' area to the half-rectangle solves the problem quickly.\r\n\r\n8-)",
  "Solution_3": "2004 AIME-I #15\r\n\r\nI do not consider the following to be elegant.  But it is different.\r\n\r\nI am hopeful someone will have a truly elegant solution they will be willing to share.  This problem is so out of the ordinary for AIME, there *must* be an elegant solution that is within the norm for AIME problems.\r\n\r\nAs MCrawford suggested during the Math Jam, false starts are a pitfall for this problem.  I considered trying to count the cases directly.  I considered taking the 9's complement of the decimal input parameter (and adapting the function).  I considered making a tree.  I considered making a 20-stage finite state machine.  I considered making a single-stage finite state machine.  While thinking about this, I thought about Turing's tape.  This eventually led to the following.  [You will be thinking \"only from the mind of a software engineer\".]  Cheers!  ...\r\n\r\n\r\nAs the sequence x1, x2, x3, \u2026, xn is generated, let\u2019s create a string to represent which of the three operations the function performs with each step.  Let c(x) be a string representing the operation(s) of function f(x) in generating d(x).\r\n\r\nWe\u2019ll use \u201cI\u201d to indicate the function incremented the parameter.  f(x)=x+1\r\nWe\u2019ll use \u201c/\u201d to indicate the function shifted the parameter by a factor of 10.  f(x)=x/10\r\nAnd we\u2019ll use \u201cZ\u201d to indicate the function has found the terminal result.  f(x)=1\r\n\r\nEvery pattern ends with \"Z\".  Except for the coding of x=1, every pattern ends with \u201c/Z\u201d.\r\n\r\nIf the input parameter, x1, contains trailing zeros, the first steps to be performed by the function are to shift the parameter by factors of 10.  We insert a shift symbol (\u201c/\u201d) at the beginning of the output string for each trailing zero in the input parameter.\r\n\r\nLeading and trailing zeros have now been stripped from our decimal input parameter.  If what remains is 1, we're done, and the coding is some consecutive sequence of zero or more \u201c/\u201d symbols, followed by a \u201cZ\u201d.\r\n\r\nOtherwise, there are digits to be \u201ccoded\u201d.  We append an initial \u201cI\u201d to the output string.  Then consider each digit of the input parameter, from right-to-left, and append the output string with the appropriate code from the following table:\r\n[code]\n\t0\tIIIIIIIII/\n\t1\tIIIIIIII/\n\t2\tIIIIIII/\n\t3\tIIIIII/\n\t4\tIIIII/\n\t5\tIIII/\n\t6\tIII/\n\t7\tII/\n\t8\tI/\n\t9\t/\n[/code]\n\nThe value of d(x) is the length of the string, c(x) we have just determined.\n\nTry some examples, and convince yourself the coding correctly represents the operation of function, f.\n[code]\n        c(1)        Z                    d(1)=1\n        c(10)       /Z                   d(10)=2\n        c(100)      //Z                  d(100)=3\n        c(9)        I/Z                  d(9)=3\n        c(90)       /I/Z                 d(90)=4\n        c(87)       III/I/Z              d(87)=7\n        c(349)      I/IIIII/IIIIII/Z     d(349)=16\n        c(34900)    //I/IIIII/IIIIII/Z   d(34900)=18\n[/code]\n\nOur goal is to determine how many valid strings have a length of 20.  All strings are of the following form, where \u201cx\u201d can represent a \u201c/\u201d symbol or a \u201cI\u201d symbol.\n\n[code]\n\tc(x) = \u201cxxxxxxxxxxxxxxxxxx/Z\u201d\n[/code]\n\nThe problem is now a combinatorics problem\u2026  \u201cHow many ways can we place \u201cI\u201d and \u201c/\u201d symbols in a string of length 18, without violating certain conditions?\u201d\n\nI identify the exceptions we must consider:[list=1]\n[*]A low-order zero digit is never coded as \u201cIIIIIIIII/\u201d.  [The function would use shifts to code low-order zeros.]\n[*]A high-order zero digit is never coded as \u201cIIIIIIIII/\u201d.  [The function does not code high-order zero digits.  It \"stops\".]\n[*]A stand-alone one digit is never coded as \u201cIIIIIIII/\u201d.  [The function never performs in increment if the input parameter is a power of 10.]\n[*]The longest run of \"I\"s is produced by the code for zero.  No longer run of \"I\"s can be produced by the function.[/list]\n\nNext I consider how these exceptions translate into conditions for the combinatorics problem:[list]\n[*]We can never allow a run of 10 or more consecutive \u201cI\u201d symbols in the string.  This one condition deals with two issues.  First, there is no digit that generates a code of 10 or more consecutive \u201cI\u201d symbols.  Second, the low-order digit in the coded string can never be \u201c0\u201d.  Since the code for \u201c0\u201d is nine consecutive \u201cI\u201ds, and since we always begin the coding of digits with an \u201cI\u201d, the prohibition against 10 consecutive \u201cI\u201d symbols solves both problems.\n\n[*]The high-order digit in the coded string can never be \u201c0\u201d.  Testing for this one condition also deals with the other special case\u2026\n\nThe string cannot contain the code for \u201c1\u201d, unless it also contains the code for at least one other digit.  The only case this affects is \u201c/////////IIIIIIIII/Z\u201d, which is a false coding for \u201c1,000,000,000\u201d.[/list]\n\nI enumerate the possibilities.  The left column shows various patterns of the strings of length 20.  The right column describes why they are a special case.  Each \u201cx\u201d in the pattern strings indicates a position that can be filled by an increment (\u201cI\u201d) or a shift (\u201c/\u201d).  The total number of choices within each pattern is just 2^n, where n is the number of \u201cx\u201d symbols in the pattern.  I have carefully organize the patterns so there is no duplication between the impossible choices.\n\n[code]\n        xxxxxxxxxxxxxxxxxx/Z    2^18 total choices.  But some of these are impossible\u2026\n\n         xxxxxxxx/IIIIIIIII/Z    256 impossible choices.  x cannot code a high-order zero-digit, and x=1,000,000,000 cannot be coded in this manner.\n\n        IIIIIIIIIIxxxxxxxx/Z    256 impossible choices.  No code has a run of 10 \u201cI\u201ds.\n        /IIIIIIIIIIxxxxxxx/Z    128 impossible choices.  Ditto.\n        x/IIIIIIIIIIxxxxxx/Z    128 impossible choices.  Ditto.\n        xx/IIIIIIIIIIxxxxx/Z    128 impossible choices.  Ditto.\n        xxx/IIIIIIIIIIxxxx/Z    128 impossible choices.  Ditto.\n        xxxx/IIIIIIIIIIxxx/Z    128 impossible choices.  Ditto.\n        xxxxx/IIIIIIIIIIxx/Z    128 impossible choices.  Ditto.\n        xxxxxx/IIIIIIIIIIx/Z    128 impossible choices.  Ditto.\n        xxxxxxx/IIIIIIIIII/Z    128 impossible choices.  Ditto.\n[/code]\r\n\r\nI have 2^18=262144 total choices.  Subtract 1536 impossible choices.  We are left with 260608 valid choices.\r\n\r\nPrime factor:  260608 = 2*2*2*2*2*2*2*2*2*509.\r\n\r\nSum the distinct prime factors:  2+509=511.",
  "Solution_4": "WOW! That's amazing. I probably will never come up something like this(I probably will never solve this problem)",
  "Solution_5": "(Worked w/ Fergie to get this)\r\n\r\nSo yeah the basic idea is to work backwards. Start at 1 then multiply by 10 (denoted by B) or subtract 1 (denoted by A) at each step. Then you get a 19 letter word of A's and B's, with the following conditions.\r\n\r\n  \r\n1. The word starts with a B (you must go to 10 first) \r\n2. The word does not start with BAAAAAAAAA (B, 9 A's) (otherwise you hit 1. But 20 is the first spot where you hit 1, so no good).\r\n\r\n3. The word does not have AAAAAAAAAA (10 A's), Since you cant do 100 -> 99 -> ... -> 91 -> 90, as 90 can only go to 9, not 91.\r\n\r\n  \r\nThis is now a simpler counting problem. \r\n  \r\nConsider only the remainder of the word after the B (18 letters). Then there are 2^18 possibilities. But some do not work. They fall into two nonoverlapping categories.\r\n\r\n  \r\n1. Starts with 9 A's but has no 10 consecutive A's. \r\n2. Has 10 consecutive A's. \r\n  \r\n1) The word is of form AAAAAAAAABXXXXXXXX, where each X can be anything without violating any conditions. This makes 2^8 words of this form.\r\n\r\n  \r\n2) Either the word starts with 10 A's (2^8 words), or has a BAAAAAAAAAA (B, 10A's) in the word. There are 8 ways to choose where this goes in the word (pick the number of letters before it from 0 to 7), and 2^7 ways to pick the other letters (7 spots remain), making 8 * 2^7 = 2 ^ 10 words.\r\n\r\n  \r\nTherefore, there are 2^18 - 2 * 2^8 - 2^10 = 2^18 - 3 * 2^9 = 509 * 2^9 words. The sum of its prime divisors is 509 + 2 = 511 as desired.",
  "Solution_6": "#14.\r\n\r\nTake an overhead view and use power of a point to find the distance (on the ground) between tangent point and the unicorn is 4 :sqrt: 5. Then unwind the rope into a right triangle ABC, with right angle C, and BC = 4, AB = 20. Mark point D on AC s.t. DC = 4:sqrt: 5 and draw a perpendicular through D to AC letting this intersect AB at E. Then find the length AE and you have your answer.",
  "Solution_7": "#6 -- multiply by 1+0x and pair stuff up:\r\n\r\n(1+0x)(1-x) = 1 - x - 0*1*x^2\r\n(1+2x)(1-3x) = 1 - x - 2*3*x^2\r\n(1+4x)(1-5x) = 1 - x - 4*5*x^2\r\n.\r\n.\r\n.\r\n(1+14x)(1-15x) = 1 - x - 14*15*x^2\r\n\r\nThe the only way to get an x^2 term is either to take 2 -x terms and all the rest 1's -- there are 8 C 2 = 28 ways to do that, so that contributes 28 to the product, or to take a 2k(2k+1)x^2 term from one factor and all the rest 1, which contributes\r\n-0*1 + 2*3 + 4*5 + ... + 14*15 = -616 (add it all up or derive a formula) to make 28-616 = -588.\r\n\r\nI had an alternate solution to #15 that wasn't particularly slick but was a different approach.  So, instead of looking at x, you look at the tens complement of x, that is, 10^n - x, where 10^n is the least power of 10 >= x (or n is one less than the number of digits of x), and where leading zeros are added such that the tens complement of x has the same number of digits as x.  Given leading zeros, the tens complement of x uniquely defines x, and the tens complement of x can be any nonnegative integer (with leading zeros) that does not begin with 9 (this is the part I missed on the test).\r\n\r\nThen if you look at what f does to the tens complement of x, either it decreases the last digit by 1, or if the last digit is 0, it removes the last digit.  Hence d(x) is equal to the sum over all the digits in d(x) (including leading 0's) of one plus the digit -- this is a sum of a sequence of numbers between 1 and 10 such that the first number is not 10 (because the first digit of a tens complement is not 9).\r\n\r\nSo what we want to count is the number of sequences of numbers between 1 and 10 with the first number not 10 that sum to 19.  This is now fairly doable.  If we didn't have the restrictions, the answer would be 2^18 -- if there are 19 stars in a row, there are 18 spots between them in which to put a bar or not put a bar, and then the number of stars between successive bars gives such a sequence.  After that, it's just subtracting stuff off, which is fairly routine and not so bad.",
  "Solution_8": "For the coefficient problem...\r\n\r\nAll I did was find all possible pairwise products between -1, 2, -3 ..., -15.\r\n\r\nI did this by (-1 + 2 - 3 + 4... - 15)^2 -  :Sigma: x = 1 to 15, x^2, and divided the whole thing by 2. I got -588, so |-588| = 588.",
  "Solution_9": "[quote=\"Nukular\"]#15.  So yeah the basic idea is to work backwards.[/quote]\r\n\r\nVery nice!  Combinatorically, this is equivalent to my solution.  But your solution provides two major advantages:\r\n\r\nWith your solution, it wasn't necessary to go through the intermediate \"coding\" step.\r\n\r\nSecond, by working backwards, you had only two special cases to deal with.  I had four special cases to deal with.  Until I noticed I could combine my special cases into just two, I had several dozen exception patterns.\r\n\r\n\r\nI also like your other alternate solution and those by Alison and masamune!",
  "Solution_10": "Not elegant solutions, just a little different thinking ...\r\n\r\n#7. \r\nLet \r\nP(x)=(1-x)(1+2x)...(1+14x)(1-15x) = 1 - 8x + C x^2 + ...\r\nreplace x by -x, we have\r\nP(-x)=(1+x)(1-2x)...(1-14x)(1+15x) = 1 + 8x + C x^2 + ...\r\nSo\r\nP(x)P(-x) = 1 + (2C-8^2) x^2 + ...\r\nOn the other hand,\r\nP(x)P(-x)=(1-1^2x^2)(1-2^2x^2)...(1-15^2x^2)=1 - (1^2+2^2+...+15^2) x^2 + ...\r\nComparing the coefficients of x^2, we get\r\n2C - 8^2 = -(1^2+2^2+...+15^2) \r\nTherefore\r\nC= [8^2 - (1^2+2^2+...+15^2)] / 2\r\n\r\n#13.\r\nThe expression 1+x+...+x^17 in P(x) makes you think to multiply it by x-1\r\nSo\r\n(x-1)^2 * P(x) = (x^18 - 1)^2 - (x-1)^2 * x^17\r\n= (x^36 - 2x^18 + 1) - (x^2 - 2x + 1) * x^17\r\n= x^36 - x^19 - x^17 + 1\r\n=(x^19 - 1)(x^17 - 1)\r\nThe rest is easy.",
  "Solution_11": "Nukular, how do you know fergie?",
  "Solution_12": "#9\r\n U_2 is a 3-4-5 triangle. The segment has to be drawn from the vertex of the rectangle to one of the opposite side. The longer leg of U_2 has to be 6 or 7.  The greater angle of U_2 is  :cong: to one of the angle of V_2.\r\n\r\nU_1 is a 3-4-5 triangle. It must have the greater angle  :cong: one of the angle of V_1. So the segment must be drawn from the hypotenuse to the longer leg that's parallel to the shorter leg. \r\n\r\nSuppose longer leg iof U_2is 7, then the shorter leg has to be 21/4, so shorter base of V_2 has to be 3/4 and the longer base is 6. (3/4)/6=x/3 where x is the shorter leg of U_1. x=3/8 \r\nSuppose it'6, then the shorter leg is 9/2, so the shorter base is 5/2. (5/2)/7=x/3, so x=15/14.\r\nWe are going to have smaller area if x is smaller, so x=3/8\r\n\r\nThe longer leg is 1/2, 1/2*(3/8)*1/2=3/32. 3+32=35",
  "Solution_13": "Nukular knows me from many places, most notably PROMYS '03, since he was my roommate.",
  "Solution_14": "For AIME problem #3 --\n\n\n\n   I'm not sure how elegant this is, but .....\n\n\n\n   In each of the 12 quadrilaterals, draw a single diagonal line segment. By assuming a little bending on these segments, you now have a convex polyhedron with 48 triangular faces, but the same 26 vertices. The effect of these auxilary line segments is the creation of an addition 12 space diagonals. In the original quadrilaterals, the vertices not connected by the auxilary form space diagonals in the new convex polyhedron.\n\n\n\n   So, at this point, it's just of matter of computing the number of space diagonals in a convex polyhedron (48 triangles, 26 vertices) then subtracting 12 for the additional ones you created.\n\n\n\nSpoiler:\n\n[hide] 3*48 = 144 vertices, with redundancies.\n\nIf k triangles meet at a single vertex, then the number of space diagonals that can be formed from that vertex is (25 - k).\n\nAssume for a moment, that k can be either 5 or 6, and x is the number of vertices with k=5.  The redundancies handled by:\n\n    5x + 6 (26-x) = 144     which yields  x = 12\n\n\n\n   So, 12 vertices with 5, and 14 vertices with 6.\n\n   This yields    12 * (25 -5) + 14 * (25 - 6) = 506.\n\n    Dividing by 2 (since each space diagonal counted twice), yields 253. \n\n\n\n   Subtracting 12 (created by splitting quadrilaterals) and you get 241.\n\n\n\n  Note:\n\n  Other combinations like k either 5 or 7 produces\n\n      19 5-type vertices and 7 7-type which yield..\n\n      19 * (25 -5) + 7 * (25 - 7) = 506\n\n\n\n   k either 4 or 6 produces\n\n      6 4-type vertices and 20 6-type which yield..\n\n      6 * (25 -4) + 20 * (25 - 6) = 506\n\n\n\n   \n\n[/hide]",
  "Solution_15": "For #3, what I did was\r\n\r\nthe total number of segments joined by any vertices is 26C2\r\n\r\nthat include the 60 edges, which we don't want\r\n\r\nit also include all the diagonals of Quads that we don't want\r\nthere are 2 diagonals for each quad, so total 12*2=24\r\n26C2-60-24=241",
  "Solution_16": "beta's way is by far the best I have seen (it's the way I did it; it surprised me how easy the solution was to find.)\r\n\r\nAnd rcv, wow. That's just really, REALLY amazing. The odd thing being that I started thinking about Turing too during that problem, but dismissed it as nothing more than my ADD. (It was really just a shadow of an idea; wondering if I could use computers to solve it.)",
  "Solution_17": "Yep. Same way as beta.",
  "Solution_18": "[quote=\"masamune2387\"]For the coefficient problem...\n\nAll I did was find all possible pairwise products between -1, 2, -3 ..., -15.\n\nI did this by (-1 + 2 - 3 + 4... - 15)^2 -  :Sigma: x = 1 to 15, x^2, and divided the whole thing by 2. I got -588, so |-588| = 588.[/quote]\r\n\r\nThat's amazing.  Nice job.",
  "Solution_19": "Thanks Magnara :D\r\n\r\nGotta love working with generating functions.",
  "Solution_20": "#4,\r\nnote that the midpt. of the hypotenuse is equidistant from all vertices of a right triangle, so the locus of points that's 1 any of the vertices is 4 quarter of a circle with radius 1. The enclosed region by the locus is 4- :pi:  :approx: 86.",
  "Solution_21": "Beta-\r\n  Awesome solutions for 3 and 4.  I came up with the all triangles approach for #3, because I was having trouble visualizing the quads and triangles together.  Yours is so much slicker.",
  "Solution_22": "For #8, it's like stepping around a clock with a thousand numbers, without hitting the same spot twice, before coming back to you starting point (and you can't step by 1). Stepping by more than 500, is like stepping by less than 500 in the opposite direction.\r\n \r\n So, it's all the integers > 1 and < 500 that are relatively prime to 1000 = (2*5)^3\r\n\r\n  So, (499) -  (250 multiples of 2) - (50 odd multiples of 5) = 199",
  "Solution_23": "Yep. For that star problem:\r\n\r\nphi(1000) = 4*25*4\r\n\r\nthere are 400 relatively prime numbers. Get rid of 1 and 999, so there are 398, and they pair up, so divide by two. The answer is then 199.",
  "Solution_24": "[quote=\"beta\"]#4. note that the midpt. of the hypotenuse is equidistant from all vertices of a right triangle[/quote]\r\nNice, simple explanation, Beta.  For those who are visual, I've attached a picture."
}
{
  "Tag": [
    "calculus",
    "logarithms",
    "probability",
    "number theory",
    "least common multiple",
    "ratio",
    "function"
  ],
  "Problem": "Prove that \\[ x^a y^b \\leq ax + yb \\] whenever $x,y > 0$ and $a+b=1$. [edit]  of course, $a,b > 0$ [/edit]\r\n\r\nCan anybody show me a proof without calculus?\r\n\r\n(My proof would go as follows: we prove it first for $a,b \\in \\mathbb{Q}$ using normal AM-GM; after that, using a continuity argument - :(, we prove it for $a,b \\in \\mathbb{R} - \\mathbb{Q}$)\r\n\r\nI'm really, really, really interested in a more elementary proof ...",
  "Solution_1": "Just A Hint: Prove rational situation first. \r\n                 Show the irrational in continued fraction.",
  "Solution_2": "That is exactly what perfect_radio wrote, Skyward_sea.\r\n\r\nI don't have a more elementary prove but I do have a very short one -\r\n\r\nsince ln is concave we have $\\ln{\\left(x^ay^b\\right)} = a \\ln{x} + b \\ln{y} \\leq \\ln{\\left(ax + by\\right)}$ :)",
  "Solution_3": "[quote=\"Arne\"]That is exactly what perfect_radio wrote, Skyward_sea.\n\nI don't have a more elementary prove but I do have a very short one -\n\nsince ln is concave we have $\\ln{\\left(x^ay^b\\right)} = a \\ln{x} + b \\ln{y} \\leq \\ln{\\left(ax + by\\right)}$ :)[/quote]\r\n\r\nof course, your proof is correct. but showing that ln is concave implies some calculus :(, so, indeed, it isn't what i'm looking for",
  "Solution_4": "[quote=\"Arne\"]That is exactly what perfect_radio wrote, Skyward_sea.\n\nI don't have a more elementary prove but I do have a very short one -\n\nsince ln is concave we have $\\ln{\\left(x^ay^b\\right)} = a \\ln{x} + b \\ln{y} \\leq \\ln{\\left(ax + by\\right)}$ :)[/quote]\r\n\r\nSorry I said \"continued fraction\"(like $\\sqrt2=1+\\frac{1}{2+\\frac{1}{2+...}}$), I thought it was different from \"a continuity argument \"",
  "Solution_5": "Ummmm...\r\n\r\nx=1\r\ny=1/2\r\n\r\na=-1\r\nb=2\r\n\r\nThat would mean that \r\n1/4 < 0, which is false.\r\nAre we assuming a,b > 0?",
  "Solution_6": "Yes, $a$ and $b$ (or more generally, $x_1,x_2,...,x_n$ with sum $1$) define a probability distribution, so they must be positive.",
  "Solution_7": "FWIW - Here is an interesting  proof when  $a$ and $b$ are  rational and positive,  using only elementary math.\r\n\r\nFirst since  $(\\sqrt x_1-\\sqrt x_2)^2  \\ge 0$ gives\r\n$\\frac {x_1+x_2}  2 \\ge  (x_1 x_2)^{(1/2)}$\r\nSimilarly  $\\frac {x_3+x_4}  2 \\ge  (x_3 x_4)^{(1/2)}$\r\nwhich gives:\r\n$\\frac {x_1+x_2+x_3+x_4}  4 \\ge \\frac {(x_1 x_2)^{(1/2)}+ (x_3 x_4)^{(1/2)}} 2 \\ge (x_1 x_2 x_3 x_4)^{(1/4)}$\r\n\r\nKeep doing it for $k$ times and if  $n=2^k$ then: (Eq 1): \r\n$\\frac {x_1+x_2+x_3+x_4+ \\dots +x_n}  n   \\ge (x_1 x_2 x_3 x_4 \\dots x_n)^{(1/n)}$\r\n\r\nNow if $a$ and $b$ are rational and $m$ is LCM of the denominators  then one can write:\r\n$a = p/m $ and $b = q/m$ where $p$ and $q$ are integers and $p+q=m$ \r\n\r\nNow select  $k$ large  enough so that $n=2^k \\ge  m$ and select:\r\n\r\nFirst $p$ ....   $x_1 x_2 \\dots  x_p = x$\r\nNext $q$ ....  $x_{p+1} x_{p+2} \\dots  x_{p+q} = y $ \r\nand the rest $(n - p - q)=n-m $ ...  $x_{p+q+1} \\dots x_n =  (ax+by) $ \r\n\r\nNow eq 1 gives:\r\n\r\n${\\frac  {p x + q y + (n -m) (ax+by)} n } \\ge  (x^p y^q (ax+by)^{n-m})^{(1/n)}$\r\nwhich gives: $ax + by \\ge (x^a y^b)$\r\n\r\nIf $a$ and $b$ are not rational, one can use the  method suggested (approximate as continued fraction).\r\n step 1 show that it is true for x=y\r\nstep 2 show that if x is not equal to y  the difference between LHS and RHS is posiitive, and  use the limit for LHS and RHS... etc..",
  "Solution_8": "how can you expect a calculus theorem to be proven properly without calculus? :?\r\n\r\nI mean, the reason they ever started with calculus was to avoid long basic proofs like that...",
  "Solution_9": "Peter, did you follow the proof at all?\r\n\r\nthe majority of it is just a proof of AM-GM, by appropriate choice of the numbers, we can prove it for rationals, and by continuity it holds for all reals.  Of course there is a faster proof using calculus, but you need to know that $f''(x)\\geq 0$ implies convexity, which is not particularly simple.\r\n\r\nHere is the continuity part.\r\n\r\nSay that $G(\\lambda)=\\lambda a+(1-\\lambda)b-a^\\lambda b^{1-\\lambda}\\geq 0$ for all $\\lambda\\in\\mathbb{Q}\\cap (0,1)$.  Then if there is some $t\\in\\mathbb{R\\setminus Q}\\cap (0,1)$ for which $G(t)<0$, by continuity (since half of the equation is linear in $\\lambda$ and the other part is continuous), there exists an interval $(t-\\epsilon,t+\\epsilon)$ in which $G(t)<0$.  But any such interval contains rationals, a contradiction.",
  "Solution_10": "$x^ay^b\\leq ax+by\\ (a,b,x,y>0,a+b=1)$\r\n\r\n$\\Longleftrightarrow \\frac{a\\log x+b\\log y}{b+a}\\leq \\log \\left(\\frac{ax+by}{b+a}\\right)$     \r\n\r\nYou can imagine [color=blue]Internal Ratio[/color]. :) \r\n\r\nkunny",
  "Solution_11": "Well in my opinion showing the graph of ln() is enough to deduce convexity if you don't know the details... but of course proofs without calculus are \"possible\", as shown above. It just seems a little pointless to me. (but then again, me = a person knowing calculus)",
  "Solution_12": "Me,too. :)",
  "Solution_13": "Well, why don't you prove that a function with $f''(x)>0$ is convex then?",
  "Solution_14": "Of course I know, blahblahblah. I am just lazy, rather than , I would say the very Seeing is Believing!\r\n\r\nI wish I could draw a graph with Tex."
}
{
  "Tag": [],
  "Problem": "A man of weight $w$ is in an elevator of weight $w$.The elevator accelerates vertically up at a rate $a$ and at a certain instant has a speed $V$.\r\n\r\n(a) What is the  apparent weight of the man?\r\n\r\n(b) The man climbs a vertical ladder within the elevator at a speed $v$ relative to the elevator.What is the man's rate of expenditure of energy?",
  "Solution_1": "a\r\n$m|(g-2a)|$\r\n\r\nfor b, i'm not sure what is does the word expenditure means but if it means change, then \r\n$\\frac{w}{2g}(v)^2$",
  "Solution_2": "Your answer is incorrect, amirhtulsa :(\r\n\r\nThe rate of expenditure of energy means power output.\r\n\r\nkunny",
  "Solution_3": "for the first one, in terms of $w$\r\n$w(1-\\frac{a}{g})$??",
  "Solution_4": "No,it isn't.",
  "Solution_5": "[hide=\"a\"]$W = w(1+\\frac{a}{g})$[/hide]\n\n[hide=\"b\"]$P = Wv = vw(1+\\frac{a}{g})$\nThe velocity V is doomed to be irrelevant, I think.[/hide]",
  "Solution_6": "The first part.\r\n[hide]\n$N-mg=ma$\nSo\n$N=ma+mg$[/hide]\n\nFor the second one, I still don't quite understand the meaning. But if it means power, then i think that:\n[hide]$P=(v+V)(ma+mg)$[/hide]",
  "Solution_7": "Shobber's a) part is the same as mine, but in b) I still think that V should NOT play any role in any of those problems.\r\n\r\nRecall the Einstein's (or whoever it was) law of relativity of two systems moving with constant speed against each other. How can you even FIND OUT how fast is the elevator moving? (I definitely think it shouldn't be possible by measuring your power climbing a ladder.)\r\n\r\nDo you really think that the stewardess in an airplane, who is pushing the drink&meals cart with velocity 0.3 m/s = 1 ft/s has to exert 1000 times larger power when the plane is in the air at cruising speed 300 m/s, than when it is sitting on the ground? (Not talking of a stewardess pushing the cart in opposite direction;-)",
  "Solution_8": "Do you have the answers to this question Kunny?",
  "Solution_9": "Yes. Here are the answers.\r\n\r\n(a) $w\\left(1+\\frac{a}{g}\\right)$\r\n\r\n(b) $w\\left(1+\\frac{a}{g}\\right)(V+v)$",
  "Solution_10": "I think your B) answer is not right. Maybe I'm wrong, but still, look at my previous post in this forum.\r\n\r\nFrom your formula for (b), when the man climbs with a zero speed and the acceleration is zero and the elevator is moving at a high speed, the man has a big power output. Which doesn't seem right to me at all."
}
{
  "Tag": [
    "probability",
    "geometry",
    "perpendicular bisector"
  ],
  "Problem": "Ok, so you are given the following diagram and asked to find the length of two segments one connecting the center of the circle to the midpoint of the triangle's base and the other connecting a perpendicular bisector on the base to the circle . [geogebra]19c921643b05462da5a71f324ff7a16724611df9[/geogebra] This is what I did: I made the segment connecting the center of the circle to the midpoint of the base of the triangle = 2root3. Then I subtracted it from the radius: root16 - root12=root4. This means that the remaining segment is 2. Subtracting that from the radius gave me 2. Which means root12=2. What I am I doing wrong? Please help.",
  "Solution_1": "I haven't seen the applet yet, but $ \\sqrt{16} \\minus{} \\sqrt{12} \\neq \\sqrt{16\\minus{}12}$.",
  "Solution_2": "oops  :blush: this is worse than when I was a poker probability did not work out because 4^2 is not eight. Thank you."
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope"
  ],
  "Problem": "What is the slope of any line parallel to the graph of the line determined by the equation $ \\frac{2}{3} x \\plus{} \\frac{2}{5} y \\equal{}2$?",
  "Solution_1": "$ 2/3(x)\\plus{}2/5(y)\\equal{}2$\r\n$ 2/5(y)\\equal{}\\minus{}2/3(x)\\plus{}2$\r\n$ y\\equal{}((\\minus{}2/3)/(2/5))(x)\\plus{}2/(2/5)$\r\n$ y\\equal{}\\minus{}5/3(x)\\plus{}5$",
  "Solution_2": "But then they ask us for the slope, so the answer is -5/3."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "1/ Let $ a,b,c>0$ such that $ abc\\equal{}1$. Prove that \r\n$ a\\plus{}b\\plus{}c \\ge \\frac2{a\\plus{}b}\\plus{}\\frac2{b\\plus{}c}\\plus{}\\frac2{c\\plus{}a}$\r\n\r\n2/ Let $ a,b,c,d>0$ such that $ abcd\\equal{}1$. Prove that \r\n$ a\\plus{}b\\plus{}c\\plus{}d \\ge \\frac3{a\\plus{}b\\plus{}c}\\plus{}\\frac3{b\\plus{}c\\plus{}d}\\plus{}\\frac3{c\\plus{}d\\plus{}a}\\plus{}\\frac3{d\\plus{}a\\plus{}b}$\r\n\r\n\r\n3/ Let $ a_i>0$ such that $ \\Pi a_i\\equal{}1$. Prove that\r\n$ a_1\\plus{}a_2\\plus{}...\\plus{}a_n\\ge \\frac{n\\minus{}1}{a_1\\plus{}a_2\\plus{}...\\plus{}a_{n\\minus{}1}} \\plus{}...\\plus{}\\frac{n\\minus{}1}{a_n\\plus{}a_1\\plus{}...\\plus{}a_{n\\minus{}2}}$",
  "Solution_1": "[quote=\"akai\"]1/ Let $ a,b,c > 0$ such that $ abc \\equal{} 1$. Prove that \n$ a \\plus{} b \\plus{} c \\ge \\frac2{a \\plus{} b} \\plus{} \\frac2{b \\plus{} c} \\plus{} \\frac2{c \\plus{} a}$\n[/quote]\r\n$ a\\plus{}b\\plus{}c\\geq\\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c\\geq \\frac2{a \\plus{} b} \\plus{} \\frac2{b \\plus{} c} \\plus{} \\frac2{c \\plus{} a}.$",
  "Solution_2": "[quote=\"akai\"]\n2/ Let $ a,b,c,d > 0$ such that $ abcd \\equal{} 1$. Prove that \n$ a \\plus{} b \\plus{} c \\plus{} d \\ge \\frac3{a \\plus{} b \\plus{} c} \\plus{} \\frac3{b \\plus{} c \\plus{} d} \\plus{} \\frac3{c \\plus{} d \\plus{} a} \\plus{} \\frac3{d \\plus{} a \\plus{} b}$\n[/quote]\n\nRinse repeating in a similar way to arqady\n\n$ \\frac {a \\plus{} b \\plus{} c}{3} \\ge \\sqrt [3]{abc} \\Longleftrightarrow \\frac {3}{a \\plus{} b \\plus{} c} \\leq \\frac {1}{\\sqrt [3]{abc}} \\equal{} \\sqrt [3]{\\frac {1}{abc}} \\leq d$\n\n[quote]3/ Let $ a_i > 0$ such that $ \\Pi a_i \\equal{} 1$. Prove that\n$ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n\\ge \\frac {n \\minus{} 1}{a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_{n \\minus{} 1}} \\plus{} ... \\plus{} \\frac {n \\minus{} 1}{a_n \\plus{} a_1 \\plus{} ... \\plus{} a_{n \\minus{} 2}}$[/quote]\r\n\r\nFor the more general one, we may also rinse repeat\r\n\r\n$ \\frac {{\\displaystyle \\sum_{k \\in S \\minus{} \\{a_m\\}}k}}{n \\minus{} 1} \\geq \\sqrt [n \\minus{} 1]{\\prod_{k \\in S \\minus{} \\{a_m\\}}} \\Longleftrightarrow \\frac {n \\minus{} 1}{{\\displaystyle \\sum_{k \\in S \\minus{} \\{a_m\\}}k}} \\leq \\sqrt [n \\minus{} 1]{a_m} \\leq a_m$\r\n\r\nWhere $ S \\equal{} \\{ a_1,a_2,a_3, \\ldots, a_n\\}$",
  "Solution_3": "Pease can you explain more arqady? :|",
  "Solution_4": "rachid, of cause! \r\n$ \\sqrt a\\equal{}\\frac{1}{\\sqrt{bc}}\\geq\\frac{2}{b\\plus{}c}$ and $ a\\plus{}b\\plus{}c\\geq\\sqrt a\\plus{}\\sqrt b\\plus{}\\sqrt c\\Leftrightarrow\\sum_{cyc}(x^6\\minus{}x^4yz)\\geq0,$ which is true by Muirhead, where $ a\\equal{}x^6,$ $ b\\equal{}y^6$ and $ c\\equal{}z^6,$ $ \\{x,y,z\\}\\subset\\mathbb R^\\plus{}.$",
  "Solution_5": "thank you arqady",
  "Solution_6": "[quote=\"arqady\"]rachid, of cause! \n$ \\sqrt a \\equal{} \\frac {1}{\\sqrt {bc}}\\geq\\frac {2}{b \\plus{} c}$ and $ a \\plus{} b \\plus{} c\\geq\\sqrt a \\plus{} \\sqrt b \\plus{} \\sqrt c\\Leftrightarrow\\sum_{cyc}(x^6 \\minus{} x^4yz)\\geq0,$ which is true by Muirhead, where $ a \\equal{} x^6,$ $ b \\equal{} y^6$ and $ c \\equal{} z^6,$ $ \\{x,y,z\\}\\subset\\mathbb R^ \\plus{} .$[/quote]\r\n $ a \\plus{} 1    \\geq 2\\sqrt{a }$ \r\n  hence    we have  : \r\n $ a \\plus{} b\\plus{}c  \\plus{}3  \\geq 2(\\sqrt{a}  \\plus{}\\sqrt{b} \\plus{}\\sqrt{c})  \\equal{} (\\sqrt{a}  \\plus{}\\sqrt{b} \\plus{}\\sqrt{c})\\plus{}(\\sqrt{a}  \\plus{}\\sqrt{b} \\plus{}\\sqrt{c})  \\geq  (\\sqrt{a}  \\plus{}\\sqrt{b} \\plus{}\\sqrt{c})  \\plus{} 3\\sqrt[6]{abc}$\r\n    or  $ a\\plus{}b\\plus{}c \\geq  \\sqrt{a}  \\plus{}\\sqrt{b} \\plus{}\\sqrt{c}$\r\n i think  this solution is better",
  "Solution_7": "I predict that the following result also true: \r\n\r\n$ a\\plus{}b\\plus{}c\\plus{}6 \\ge \\frac6{a\\plus{}b}\\plus{}\\frac6{b\\plus{}c}\\plus{}\\frac6{c\\plus{}a}$\r\n\r\nwhere $ a,b,c>0$ and $ abc\\equal{}1$.",
  "Solution_8": "Thank for nice solution and nine problem.",
  "Solution_9": "[quote=\"dduclam\"]I predict that the following result also true: \n\n$ a \\plus{} b \\plus{} c \\plus{} 6 \\ge \\frac6{a \\plus{} b} \\plus{} \\frac6{b \\plus{} c} \\plus{} \\frac6{c \\plus{} a}$\n\nwhere $ a,b,c > 0$ and $ abc \\equal{} 1$.[/quote]\r\n\r\nNo body have idea for it? :wink: \r\n\r\n\r\n[b]*[/b]A my new result: \r\n\r\n$ a \\plus{} b \\plus{} c  \\ge \\frac2{a \\plus{} b} \\plus{} \\frac2{b \\plus{} c} \\plus{} \\frac2{c \\plus{} a}$\r\n\r\nwhere $ a,b,c \\ge 0$ and $ ab\\plus{}bc\\plus{}ca \\equal{} 3$\r\n\r\nIt's nice but easy  :)",
  "Solution_10": "[quote=\"dduclam\"]\n\n$ a \\plus{} b \\plus{} c \\ge \\frac2{a \\plus{} b} \\plus{} \\frac2{b \\plus{} c} \\plus{} \\frac2{c \\plus{} a}$\n\nwhere $ a,b,c \\ge 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 3$\n\n[/quote]\nIt is equivalent to $ \\sum_{cyc}(a^3b \\plus{} a^3c \\minus{} 2a^2bc)\\geq0.$\n[quote=\"dduclam\"]I predict that the following result also true: \n\n$ a \\plus{} b \\plus{} c \\plus{} 6 \\ge \\frac6{a \\plus{} b} \\plus{} \\frac6{b \\plus{} c} \\plus{} \\frac6{c \\plus{} a}$\n\nwhere $ a,b,c > 0$ and $ abc \\equal{} 1$.[/quote]\r\nYes, it is true. The Mixing Variables method kills it very easy.",
  "Solution_11": "[quote=\"dduclam\"] \n\n\n[b]*[/b]A my new result: \n\n$ a \\plus{} b \\plus{} c \\ge \\frac2{a \\plus{} b} \\plus{} \\frac2{b \\plus{} c} \\plus{} \\frac2{c \\plus{} a}$\n\nwhere $ a,b,c \\ge 0$ and $ ab \\plus{} bc \\plus{} ca \\equal{} 3$\n\nIt's nice but easy  :)[/quote]\r\n\r\nMy solution:\r\n\r\n$ a \\plus{} b \\plus{} c \\ge \\frac2{a \\plus{} b} \\plus{} \\frac2{b \\plus{} c} \\plus{} \\frac2{c \\plus{} a}$\r\n\r\n<=> $ 3(a\\plus{}b\\plus{}c)\\ge \\frac{2(ab \\plus{} (a\\plus{}b)c)}{a\\plus{}b} \\plus{}\\frac{2(bc \\plus{} (b\\plus{}c)a)}{b\\plus{}c}\\plus{}\\frac{2(ca \\plus{} (c\\plus{}a)b)}{c\\plus{}a}$\r\n\r\n<=> $ a\\plus{}b\\plus{}c\\ge \\sum\\frac{2ab}{a\\plus{}b}$ also true because  $ (a\\plus{}b)^2\\ge 4ab$\r\n :)"
}
{
  "Tag": [
    "real analysis",
    "function",
    "real analysis unsolved"
  ],
  "Problem": "Show that if $ (f_n)_n$ is a sequence in $ L^1[0, 1]$ is such that $ \\sum_{n\\equal{}1}^{\\infty}  \\parallel{} f_n \\parallel{}_1 < \\infty$ then $ \\sum_{n\\equal{}1}^{\\infty} | f_n(s) | < \\infty$ for almost every $ s \\in [0, 1]$.\r\n\r\n\r\nThis looks simple to prove.  However, I do not see how to prove this right now.  Any hints on how to proceed would be appreciated.",
  "Solution_1": "Hint: is the function $ g(x)\\equal{}\\sum_{n\\equal{}1}^{\\infty}|f_n(x)|$ integrable?"
}
{
  "Tag": [
    "Support",
    "\\/closed"
  ],
  "Problem": "Is it possible for you guys to post PDFs of Transcripts instead of having to copy and paste into text editors?  I think it would help [b][i]Lots[/i][/b] of people.  Just a suggestion. :D",
  "Solution_1": "If you really want to use a PDF copy, then just copy and paste to MS word and then use [url=http://www.primopdf.com/]PrimoPDF[/url].\r\n\r\nI support HTML copies, so ccy's transcript player can be used.",
  "Solution_2": "PrimoPDF doesn't work with some systems. http://cutepdf.com/ is a better choice.\r\n\r\nThere should be a php that echoes the transcripts based on a url var.",
  "Solution_3": "Guys! The whole point of this is that converting to PDF takes too LONG!!! See, waiting for the transcript to load completely (without any squares with x's inside them) takes like 10 tries. Next, pasting into word takes 4 minutes, and finally converting takes 2 minutes. So, basically could you give us the pdf's ready to download in the \"my classes\" page?",
  "Solution_4": "[quote=\"Nexmus\"]Guys! The whole point of this is that converting to PDF takes too LONG!!! See, waiting for the transcript to load completely (without any squares with x's inside them) takes like 10 tries. Next, pasting into word takes 4 minutes, and finally converting takes 2 minutes. So, basically could you give us the pdf's ready to download in the \"my classes\" page?[/quote]\r\n\r\nThose times are exaggerated.\r\nOn a standard connection, in Firefox, loading takes about 30 seconds.\r\n\r\nOn a normal Windows XP Dell, pasting takes maybe 10 seconds.\r\nConverting takes maybe a minute, probably less.",
  "Solution_5": "Just in case you guys didn't notice, there's a new \"Print\" feature now, so it makes it printer friendly and all. While downloading a 1.7 GB file, I loaded a transcript (1.5 hrs) in about 10 secs. (SDSL here)\r\n\r\nIt doesn't take too long to convert after that. Maybe 40 secs?",
  "Solution_6": "Sorry guys, maybe it's just me. But, I'm in a town in the middle of nowhere with a connection sucks !@#$ and am about 1000 miles from the nearest server.",
  "Solution_7": "Note #1: ad above.\r\n\r\nNote two: I support this too, but I believe there are problems ya'll aren't seeing. First, PDFs are easier to transfer between people. This means one dishonest person can pay for WOOT and pass *all* the class materials around quite easily. Second, it's not straightforward to convert it to PDF on the server side. BBCode is not perfectly easily translatable. Third, in a browser, the images may be cached so that multiple loads of one transcript only cause, say, 1.5 full page loads worth of traffic. In a PDF, someone clicks it 8 times, it get downloaded 8 times: guess what, you've tripled the load on the server on  the transcripts."
}
{
  "Tag": [
    "AMC",
    "AIME"
  ],
  "Problem": "So I've compiled all the answers from my own test and from the various threads on here. I've got what I believe to be the definitive list, but I would like some confirmation on 15.\r\n\r\n1. 372\r\n2. 200\r\n3. 578\r\n4. 450\r\n5. 888\r\n6. 640\r\n7. 553\r\n8. 896\r\n9. 259\r\n10. 710\r\n11. 179\r\n12. 91\r\n13. 640\r\n14. 676\r\n15. 389 (pretty sure)",
  "Solution_1": "Number 15 is correct.",
  "Solution_2": "im pretty sure the first 13 are correct",
  "Solution_3": "How sure are you guys about 11? Because I was pretty sure I was right with 227.\r\nI got 128pi + 96sqrt3",
  "Solution_4": "[quote=\"Arnizzle\"]How sure are you guys about 11? Because I was pretty sure I was right with 227.\nI got 128pi + 96sqrt3[/quote]\r\n\r\n100%.",
  "Solution_5": "Yes, #11 is confirmed as 179.",
  "Solution_6": "I got 389 for #15.",
  "Solution_7": "All of the above must be right.",
  "Solution_8": "All of the above ARE correct, according to the Math Jam."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "prism"
  ],
  "Problem": "The length and width of a right rectangular prism are each doubled and the height remains the same. By what factor is the volume of the original prism increased?\n\n[asy]draw((0,0)--(10,0)--(10,7)--(0,7)--cycle);\ndraw((0,7)--(5,12)--(15,12)--(10,7));\ndraw((15,12)--(15,5)--(10,0));\ndraw((18,0)--(38,0)--(38,7)--(18,7)--cycle);\ndraw((18,7)--(28,17)--(48,17)--(38,7));\ndraw((48,17)--(48,10)--(38,0));[/asy]",
  "Solution_1": "The volume of the new shape is $ 2\\cdot2\\cdot1\\equal{}\\boxed{4}$ times the volume of the original shape."
}
{
  "Tag": [
    "ARML"
  ],
  "Problem": "I wrote a tournament test and I discovered that it was incredibly hard, much harder than I intended. In fact, people who got a 7 at ARML last year got a 4 on my ARMLish ciphering test. I've posted it at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=932033#932033 if you'd like to take a look at it, answers will be posted on Tuesday.",
  "Solution_1": "Where is the solution thread to the problems (if any)? Thanks!"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove the existence of $ f,g: \\mathbb{R}\\rightarrow\\mathbb{R}$ measurable such that $ f\\circ g$ is not measurable.",
  "Solution_1": "I would use two Cantor-type sets $ C_1$ (with positive measure) and $ C_2$ (with zero measure). They can be arranged so that there is a continuous (even Lipschitz) bijection $ g\\colon C_1\\to C_2$ that can be extended to $ \\mathbb R$. Choose a nonmeasurable set $ E\\subset C_1$ and let $ f$ be the characteristic function of $ g(E)$."
}
{
  "Tag": [],
  "Problem": "Given that $ n$ is an integer and $ 0 &lt; 4n &lt;30$, what is the sum of all possible integer values of $ n$?",
  "Solution_1": "$ 0 &lt; 4n &lt; 30$ simplifies to $ 0 &lt; n &lt; 7 \\frac{1}{2}$. Thus, positive integers $ 7$ and less fit the restrictions on $ n$. Adding these $ 7$ positive integers, $ 1\\plus{}2\\plus{}3\\plus{}4\\plus{}5\\plus{}6\\plus{}7\\equal{}\\frac{7(7\\plus{}1)}{2}\\equal{}7(4)\\equal{}\\fbox{28}$."
}
{
  "Tag": [],
  "Problem": "Propose a synthesis of p-aminoacetophenone from aniline.",
  "Solution_1": "Quite very simple\r\n1. Protect the NH2 by converting it into NHCOCH3 ucing acetic anhydride\r\n2. acylate in the para posn using acetic anhydride\r\n3. use NaOH to convert NHCOCH3 back into NH2",
  "Solution_2": "[quote=\"Madness\"]Quite very simple[/quote]\n\nIs that so? Humm, let's see:\n\n[quote=\"Madness\"]2. acylate in the para posn using acetic anhydride[/quote]\r\n\r\nAnd what are the reaction conditions to do that? Aren't you forgeting something?"
}
{
  "Tag": [
    "function",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Find a bijection between the real line R and the unbounded interval (0, inf) = {x in R: x>0}.\r\n\r\nCan the bijection be f: x-->x+1?",
  "Solution_1": "A function $ f: X\\to Y$ is [i]surjective[/i] if for each $ y\\in Y$ there is at least one $ x\\in X$ such that $ f(x) \\equal{} y$.  A function is [i]injective[/i] if for each $ y\\in Y$ there is at most one $ x\\in X$ such that $ f(x) \\equal{} y$.  A function is a bijection if it is both injective and surjective.\r\n\r\nYour function is a bijection from $ \\mathbb{R}\\to \\mathbb{R}$, but its range is not $ (0,\\infty)$, so it is not the bijection you're looking for.\r\n\r\nTry $ f(x) \\equal{} e^x$."
}
{
  "Tag": [],
  "Problem": "Let $R = gs - 4$.  What $S = 8$, $R = 16$.  What $S = 10$, $R$ is equal to:\r\n\\[ \\text{(A)}\\ 11 \\qquad \\text{(B)}\\ 14 \\qquad \\text{(C)}\\ 20 \\qquad \\text{(D)}\\ 21 \\qquad \\text{(E) none of these} \\]",
  "Solution_1": "[hide][color=green]the answer is 21, or D.[/color][/hide]",
  "Solution_2": "[hide]The first condition gives us that when S is 8,  R is 16, so when we substitute the values into the equation, we get\n16 = 8g - 4,\ng = 20/8 = 5/2, solving for g.\nNow when S is 10, R is:\n(5/2)(10) - 4,\nor 21, which is [color=brown]D[/color][/hide]",
  "Solution_3": "[hide]D[/hide]",
  "Solution_4": "[hide]R=21\nD[/hide]"
}
{
  "Tag": [
    "inequalities",
    "algorithm",
    "inequalities proposed"
  ],
  "Problem": "Given four nonnegative reals $a$, $b$, $c$, $d$.\r\nProve that:\r\n$a^{4}+b^{4}+c^{4}+d^{4}+2abcd \\geq a^{2}b^{2}+b^{2}c^{2}+c^{2}d^{2}+d^{2}a^{2}+a^{2}c^{2}+b^{2}d^{2}$.",
  "Solution_1": "[i]\"If it is dumb and works, then it is not dumb.\"[/i] (Apocryphal Murphy law)\r\n\r\nThere seems to be a general method for killing inequalities like this one. I will demonstrate it at the example of your nice inequality (which surely has a much simpler proof).\r\n\r\nDenote\r\n\r\n$ f\\left(a, b, c, d\\right)$\r\n$ \\equal{} \\left(a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} d^4 \\plus{} 2abcd\\right) \\minus{} \\left(a^2 b^2 \\plus{} b^2 c^2 \\plus{} c^2 d^2 \\plus{} d^2 a^2 \\plus{} a^2 c^2 b^2 d^2\\right)$.\r\n\r\nWe must show that $ f\\left(a, b, c, d\\right) \\geq 0$ for all positive a, b, c, d. Now, since $ f\\left(a, b, c, d\\right)$ is obviously symmetric in a, b, c, d, we may WLOG assume that\r\n\r\n  $ a\\geq b\\geq c\\geq d$.\r\n\r\nAfter some algebra,\r\n\r\n  $ f\\left(a, b, c, d\\right) \\minus{} f\\left(a, b, c, c\\right)$\r\n$ \\equal{} \\left(c \\minus{} d\\right)\\left(c \\left(a^2 \\plus{} b^2 \\minus{} 2ab\\right) \\plus{} d\\left(a^2 \\plus{} b^2 \\minus{} cd \\minus{} d^2\\right)\\right)$.\r\n\r\nNow, a well-known inequality states $ a^2 \\plus{} b^2 \\minus{} 2ab \\geq 0$; also, from $ a\\geq b\\geq c\\geq d$, we have $ c \\minus{} d \\geq 0$ and $ a^2 \\plus{} b^2 \\minus{} cd \\minus{} d^2 \\geq 0$. Thus,\r\n\r\n  $ f\\left(a, b, c, d\\right) \\minus{} f\\left(a, b, c, c\\right) \\geq 0$,\r\n\r\nand $ f\\left(a, b, c, d\\right) \\geq f\\left(a, b, c, c\\right)$.\r\n\r\nAnother calculation gives\r\n\r\n  $ f\\left(a, b, c, c\\right) \\minus{} f\\left(a, b, b, b\\right)$\r\n$ \\equal{} \\left(b \\plus{} c\\right)\\left(b \\minus{} c\\right)\\left(2a^2 \\plus{} b^2 \\minus{} 2ab \\minus{} c^2\\right)$.\r\n\r\nAgain, $ b \\minus{} c \\geq 0$ and $ 2a^2 \\plus{} b^2 \\minus{} 2ab \\minus{} c^2 \\geq 0$ because of $ a\\geq b\\geq c\\geq d$. Therefore,\r\n\r\n  $ f\\left(a, b, c, c\\right) \\minus{} f\\left(a, b, b, b\\right) \\geq 0$,\r\n\r\nand $ f\\left(a, b, c, c\\right) \\geq f\\left(a, b, b, b\\right)$.\r\n\r\nFinally,\r\n\r\n  $ f\\left(a, b, b, b\\right) \\equal{} a \\left(a \\minus{} b\\right)^2 \\left(a \\plus{} 2b\\right)$,\r\n\r\nwhat is obviously $ \\geq 0$. Hence,\r\n\r\n  $ f\\left(a, b, c, d\\right) \\geq f\\left(a, b, c, c\\right) \\geq f\\left(a, b, b, b\\right) \\geq 0$,\r\n\r\nqed..\r\n\r\nSome comments about this solution:\r\n\r\nSince $ f\\left(a, a, a, a\\right) \\equal{} 0$, we could also write\r\n\r\n  $ f\\left(a, b, c, d\\right) \\geq f\\left(a, b, c, c\\right) \\geq f\\left(a, b, b, b\\right)\\geq f\\left(a, a, a, a\\right)$\r\n\r\ninstead of\r\n\r\n  $ f\\left(a, b, c, d\\right) \\geq f\\left(a, b, c, c\\right) \\geq f\\left(a, b, b, b\\right)\\geq 0$.\r\n\r\nThis would show the algorithm more clearly.\r\n\r\nOne thing more: If I were at the IMO, I wouldn't have written the full argument I wrote above, but I would just prove by calculation the identity\r\n\r\n  $ f\\left(a, b, c, d\\right) \\equal{} \\left(c \\minus{} d\\right)\\left(c \\left(a^2 \\plus{} b^2 \\minus{} 2ab\\right) \\plus{} d\\left(a^2 \\plus{} b^2 \\minus{} cd \\minus{} d^2\\right)\\right)$\r\n$ \\plus{} \\left(b \\plus{} c\\right)\\left(b \\minus{} c\\right)\\left(2a^2 \\plus{} b^2 \\minus{} 2ab \\minus{} c^2\\right) \\plus{} a \\left(a \\minus{} b\\right)^2 \\left(a \\plus{} 2b\\right)$,\r\n\r\nwithout giving any hints how I found this formula, just as if I had guessed it. Subsequently I would conclude that $ f\\left(a, b, c, d\\right) \\geq 0$. The corrector surely would be fascinated about my ingeniousity and I'd get a special prize for an elegant solution. ;-)\r\n\r\n[b]EDIT:[/b] The method of showing that $ f\\left(a, b, c, d\\right)\\geq 0$ by subsequently passing from $ \\left(a,b,c,d\\right)$ to other quadruples such that the value of $ f$ becomes smaller and smaller on each step is the so-called [b]mixing variables[/b] technique of solving inequalities. Hence, in the above proof, I more or less discovered this method for myself. I am obviously not the first one to discover it; I could bet that the famous special prize solution of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55327]IMO 1983 problem 6[/url] was also found by this method.\r\n\r\n  Darij",
  "Solution_2": "Dear Anh Cuong,\r\n\r\nThis is Turkevici's inequality... ;)",
  "Solution_3": "Let [tex]S=a^4+b^4+c^4+d^4+2abcd-(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)[/tex].\r\n\r\nThen S is symmetric, so we can suppose that the biggest two numbers among [tex]a,b,c,d[/tex] is [tex]a,b[/tex].\r\n\r\nTherefore\r\n\r\n[tex]S=(ab+cd-c^2-d^2)^2+2cd(c-d)^2+(a-b)^2((a+b)^2-(c^2+d^2))\\geq 0[/tex].",
  "Solution_4": "The above problem was used in KMO winter program, and In-seok Seo (2002 IMO) proposed the above solution :)"
}
{
  "Tag": [
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "I have seen the term \"subset ordering of sets\" at http://en.wikipedia.org/wiki/Order_theory\r\n\r\nWhat I can understand now is it is something related to the ordering of sets.\r\n\r\nBut I can't understand literally what \"subset ordering of sets\" means. \r\nWhat is the subset, what are the sets, and how they relate to each other?",
  "Solution_1": "The page offers a definition: [quote]If a set A contains all the elements of a set B, then B is said to be smaller than or equal to A. [/quote]\r\nThat is, we consider $B$ to be smaller than (or equal to) $A$ precisely when $B$ is a subset of $A$. It is a rather natural ordering of sets.",
  "Solution_2": "Thanks, but I still not really understand the whole picture.\r\nCan you please literally explain what is \"subset ordering of sets\"?\r\nVery thanks=)",
  "Solution_3": "The concept can be broken down as follows.\r\n1. Ordering\r\n2. Ordering of sets\r\n3. Subset ordering of sets. \r\n\r\n1. Ordering is a relation between some (possibly quite abstract) objects that is [b]reflexive, antisymmetric, and transitive[/b] (these terms are explained by Wikipedia). We usually write $a\\le b$ to indicate this kind of relation between objects $a$ and $b$. This need not have anything to do with usual inequalities.\r\n\r\n2. A special case of part 1 arises when the objects that we are trying to order are subsets of the same set. For example, they could be intervals on the real axis, in which case we could order them by declaring that $I\\le J$ when the length of interval $I$ is less than or equal to the length of interval $J$. According to this ordering, $[0,2]\\le [1,6]$.\r\n\r\n3. This is a special case of part 2. Again the objects to be ordered are subsets of the same set. But this time we consider a very special ordering: $A\\le B$ if $A$ is contained in $B$. According to this ordering, $[0,2]\\le [-1,4]$, but it is not true that $[0,2]\\le [1,6]$.",
  "Solution_4": "Thanks for the wonderful explanation!!!"
}
{
  "Tag": [
    "geometry",
    "trapezoid",
    "similar triangles",
    "geometry unsolved"
  ],
  "Problem": "Dear Mathlinkers,\r\nlet ABC be a triangle, P a point, P* the isogonal of P wrt ABC, Pb the perpendicular to AB at A, Pc the perpendicular to AC at A, \r\nB\" the meet point of Pb and BP, C\" the meet point of Pc and CP \r\nand A' the meet point of the perpendicular to BC through P* with BC.\r\nProve that AA' is perpendicular to B\"C\".\r\nSincerely\r\nJean-Louis",
  "Solution_1": "Hello!\r\n\r\nConstruct $ B\"M\\parallel{}AC$, $ M\\in BC$ and $ \\left\\{S\\right\\}\\equal{}AB\"\\cap BC$.\r\n$ \\frac{CM}{CS}\\equal{}\\frac{AB\"}{AS}\\equal{}\\frac{AB\\cdot tan\\left(\\angle ABP\\right)}{AB\\cdot tan\\left(\\angle B\\right)}\\equal{}\\frac{tan\\left(\\angle A'BP^{*}\\right)}{tan\\left(\\angle B\\right)}\\equal{}\\frac{A'P^{*}}{A'B\\cdot tan\\left(\\angle B\\right)}\\Rightarrow$\r\n\r\n$ \\Rightarrow CM\\cdot A'B\\equal{}\\frac{A'P^{*}}{tan\\left(\\angle B\\right)}\\cdot CS\\equal{}\\frac{A'P^{*}}{tan\\left(\\angle B\\right)}\\cdot\\frac{AC\\cdot sin\\left(90^{\\circ}\\minus{}\\angle A\\right)}{sin\\left(90^{\\circ}\\minus{}\\angle B\\right)}\\equal{}$\r\n\r\n$ \\equal{}\\frac{A'P^{*}}{tan\\left(\\angle B\\right)}\\cdot\\frac{AC\\cdot cos\\left(\\angle A\\right)}{cos\\left(\\angle B\\right)}\\Rightarrow CM\\cdot A'B\\equal{}A'P^{*}\\cdot\\frac{AC\\cdot cos\\left(\\angle A\\right)}{sin\\left(\\angle B\\right)}$     (1)\r\n\r\nSimilarly construct $ C\"N\\parallel{}AB$, $ N\\in BC$ and $ \\left\\{T\\right\\}\\equal{}AC\"\\cap BC$\r\n$ \\Rightarrow BN\\cdot A'C\\equal{}A'P^{*}\\cdot\\frac{AB\\cdot cos\\left(\\angle A\\right)}{sin\\left(\\angle C\\right)}$     (2)\r\n\r\n(1), (2) $ \\Rightarrow CM\\cdot A'B\\equal{}BN\\cdot A'C\\Rightarrow\\frac{A'B}{BN}\\equal{}\\frac{A'C}{CM}$     (3)\r\n$ \\left\\{V'\\right\\}\\equal{}B\"M\\cap AA'$ and $ \\left\\{V\"\\right\\}\\equal{}C\"N\\cap AA'$\r\n$ \\frac{A'A}{AV\"}\\equal{}\\frac{A'B}{BN}\\equal{}\\frac{A'C}{CM}\\equal{}\\frac{A'A}{AV'}\\Rightarrow V'\\equal{}V\"\\equal{}V$.\r\n$ C\"V\\parallel{}AB$ and $ AB\\bot AB\"\\Rightarrow C\"V\\bot AB\"$ (4)\r\n$ B\"V\\parallel{}AC$ and $ AC\\bot AC\"\\Rightarrow B\"V\\bot AC\"$ (5)\r\n(4), (5) $ \\Rightarrow$ the point $ V$ is the orthocenter of the triangle $ AB\"C\"\\Rightarrow VA\\bot B\"C\"\\Rightarrow AA'\\bot B\"C\"$\r\n\r\nBest regards, Petrisor Neagoe :)",
  "Solution_2": "Let $ X,\\ Z,\\ Y$ be, the orthogonal projections of $ A,\\ B'',\\ C''$ respectively, on the side-segment $ BC,$ of the given triangle $ \\bigtriangleup ABC.$\r\n\r\nAlso, le $ P'$ be ( instead of $ P^{*}$ ), the isogonal conjugate of $ P,$ wrt $ \\bigtriangleup ABC.$\r\n\r\nThe lines through the points $ X,\\ Y,\\ Z$ and perpendicular to $ B''C'',\\ AB'',\\ AC''$ respectively, are concurrent at one point so be it $ Q,$ as well ( [b][size=100]Carnot theorem[/size][/b] ).\r\n\r\nSo, because of the similar triangles $ \\bigtriangleup ABC,\\ \\bigtriangleup QYZ,$ with $ AB\\parallel QY$ and $ AC\\parallel QZ,$ it is enough to prove that $ AA'\\parallel XX',$ which is equivalent to prove that $ \\frac {A'B}{A'C} \\equal{} \\frac {XY}{XZ}.$\r\n\r\n$ \\bullet$ From the similar right triangles $ \\bigtriangleup A'BP',\\ \\bigtriangleup ABB'',$ we have that $ \\frac {A'B}{AB} \\equal{} \\frac {A'P'}{AB''}$ $ \\Longrightarrow$ $ \\frac {A'B}{A'P'} \\equal{} \\frac {AB}{AB''}\\ \\ ,(1)$\r\n\r\nSimilarly, from the similar right triangles $ \\bigtriangleup A'CP',\\ \\bigtriangleup ACC'',$ we have that $ \\frac {A'C}{A'P'} \\equal{} \\frac {AC}{AC''}\\ \\ ,(2)$\r\n\r\nFrom $ (1),\\ (2)$ $ \\Longrightarrow$ $ \\frac {A'B}{A'C} \\equal{} \\frac {AB}{AC}\\cdot \\frac {AC''}{AB''}\\ \\ ,(3)$ \r\n\r\n$ \\bullet$ Based on the below [b][size=100]Lemma[/size][/b], we have $ \\frac {AB}{AC}\\cdot \\frac {AC''}{AB''} \\equal{} \\frac {RC''}{RB''} \\equal{} \\frac {XY}{XZ}\\ \\ ,(4)$ $ ($ because of the trapezium $ YZB''C''$ $ ).$\r\n\r\nHence, from $ (3),\\ (4)$ $ \\Longrightarrow$ $ \\frac {XY}{XZ} \\equal{} \\frac {A'B}{A'C}\\ \\ ,(5)$\r\n\r\nFrom $ (5)$ and because of the similarity of the triangles $ \\bigtriangleup ABC,\\ \\bigtriangleup QYZ,$ with $ AB\\parallel QY$ and $ AC\\parallel QZ,$ we conclude that $ AA'\\parallel XX'$ $ \\Longrightarrow$ $ AA'\\perp B''C''$ and the proof is completed.\r\n\r\n[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \\bigtriangleup ABC$ is given and let $ AD$ be, its $ A$-altitude. Let $ P$ be, an arbitrary point on $ AD,$ between $ A,\\ D$ and we draw an arbitrary line through $ P,$ which intersects the lines through vertex $ A$ and perpendicular to $ AC,\\ AB,$ at points $ E,\\ F,$ respectively. Prove that $ \\frac {PE}{PF} \\equal{} \\frac {AB}{AC}\\cdot \\frac {AE}{AF}.$[/color][/size][/b]\r\n\r\n$ \\bullet$ Because of the proof of the above [b][size=100]Lemma[/size][/b] is very easy, I post only the schema and I think it is enough.\r\n\r\nKostas Vittas.",
  "Solution_3": "Dear Kostas and Mathlinkers,\r\nI found your approach (orthopole) very interesting. I ask me always the same question: can we obtain the result with a minimum of calculation?\r\nSincerely\r\nJean-Louis",
  "Solution_4": "I liked also the argument of [b][size=100]orthopole[/size][/b] dear [b][size=100]Jean Louis[/size][/b], but after the reading of the solution by [b][size=100]Petrisor[/size][/b], I think we can have a more easier proof of your proposed problem, without this argument and also without trigonometric relations.\r\n\r\nSo, we denote the points $ N'\\equiv AC\\cap NC'',\\ M\\equiv AB\\cap MB''$ and applying the [b][size=100]Thales theorem[/size][/b] we have :\r\n\r\nFrom $ AB\\parallel NN'$ $ \\Longrightarrow$ $ \\frac {BN}{BC} \\equal{} \\frac {AN'}{AC}$ $ ,(1)$\r\n\r\nFrom $ AC\\parallel MM'$ $ \\Longrightarrow$ $ \\frac {CM}{CB} \\equal{} \\frac {AM'}{AB}$ $ ,(2)$\r\n\r\nFrom $ (1),\\ (2)$ $ \\Longrightarrow$ $ \\frac {BN}{CM} \\equal{} \\frac {AB}{AC}\\cdot \\frac {AN'}{AM'}$ $ ,(3)$\r\n\r\nFrom the similar right triangles $ \\bigtriangleup AN'C'',\\ \\bigtriangleup AM'B''$ we have that $ \\frac {AN'}{AM'} \\equal{} \\frac {AC''}{AB''}$ $ ,(4)$\r\n\r\nFrom $ (3),\\ (4)$ $ \\Longrightarrow$ $ \\frac {BN}{CM} \\equal{} \\frac {AB}{AC}\\cdot \\frac {AC''}{AB''}$ $ ,(5)$\r\n\r\n$ \\bullet$ From the similarities of the triangles $ \\bigtriangleup A'BP'\\approx \\bigtriangleup ABB''$ and $ \\bigtriangleup A'CP'\\approx \\bigtriangleup ACC'',$ we have $ \\frac {AB}{AC}\\cdot \\frac {AC''}{AB''} \\equal{} \\frac {A'B}{A'C}$ $ ,(6)$ ( please see the [b][size=100]post[/size][/b] #3# ).\r\n\r\nHence, from $ (5),\\ (6)$ $ \\Longrightarrow$ $ \\frac {BN}{CM} \\equal{} \\frac {A'B}{A'C}$ $ \\Longrightarrow$ $ \\frac {A'B}{BN} \\equal{} \\frac {A'C}{CM}$ $ ,(7)$\r\n\r\nFrom $ (7),$ based on the [b][size=100]Thales theorem[/size][/b], we conclude that the line segment $ AA'$ passes through the point $ V\\equiv NC''\\cap MB''.$\r\n\r\nBecause of now, the point $ A$ is the orthocenter of the triangle $ \\bigtriangleup VB''C'',$ we conclude that $ VA\\equiv AA'\\perp B''C''$ and the proof is completed.\r\n\r\nKostas Vittas.",
  "Solution_5": "Dear Mathlinkers,\r\nafter toying with my programm,\r\nlet A* be the meetpoint of BC\" and CB\".\r\nProve that A*, A and P* are collinear.\r\nSincerely\r\nJean-Louis",
  "Solution_6": "[quote=\"jayme\"]Dear Mathlinkers,\nafter toying with my programm,\nlet A* be the meetpoint of BC\" and CB\".\nProve that A*, A and P* are collinear.\nSincerely\nJean-Louis[/quote]\r\nWe can prove this result dear [b][size=100]Jean Louis[/size][/b], as a direct application of the [b][size=100]Isogonic theorem[/size][/b].  :wink: \r\n\r\nKostas Vittas.",
  "Solution_7": "Dear Mathlinkers,\nan article concerning the Vecten's figure and its developpement can be seeing on my site\nhttp://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 133\nSincerely\nJean-Louis"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "tetrahedron",
    "geometry proposed"
  ],
  "Problem": "Let a tetrahedron $ ABCD$. Find $ M,N$ in $ AB,CD$ so that all planes over $ M,N$ cut $ ABCD$ become $ 2$ constant volume parts.",
  "Solution_1": "If $ M$ midpoint of $ [AB]$ and $ N$ midpoint of $ [CD]$, then the 2 volume is =, for all planes $ \\pi; M,N\\in\\pi.$"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $x_{1}, x_{2},\\ldots, x_{n}>0$ such that \r\n\r\n                \\[ \\frac{1}{1+x_{1}}+\\frac{1}{1+x_{2}}+\\ldots+\\frac{1}{1+x_{n}}\\geq n-1.  \\]\r\n\r\n Prove that:\r\n                               \\[{ \\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\ldots+\\frac{1}{x_n}}\\geq n(n-1).  \\]",
  "Solution_1": "I'm really not sure how looks the correct statement of a problem, because the inequality is not true for example for $n=3$, $a=b=c=1/3$. But I think that if Cezar meant $n(n-1)$ instead of $n(n+1)$ I got it. Here is my proof:\r\n\r\nDenote $a_i=\\frac{1}{x_i+1}$. Then $a_i \\leq 1$, $\\sum a_i \\geq n-1$ and $x_i=\\frac{1-a_i}{a_i}$.\r\n\r\nSo we have to prove that: \r\n$\\sum \\frac{a_i}{1-a_i} \\geq n(n-1)$ \r\nBut using the Cauchy-Shwarz, the inequality $\\sum a_i^2 \\geq \\frac{(\\sum a_i)^2}{n}$ and the condition gives us:\r\n$\\sum \\frac{a_i}{1-a_i} = \\sum \\frac{a_i^2}{a_i-a_i^2} \\geq \\frac{(\\sum a_i)^2}{\\sum a_i - \\sum a_i^2} \\geq \\frac{(\\sum a_i)^2}{\\sum a_i - \\frac{(\\sum a_i)^2}{n}} = \\frac{\\sum a_i}{1-\\frac{\\sum a_i}{n}} \\geq \\frac{n-1}{1-\\frac{n-1}{n}}=n(n-1)$\r\nQED.",
  "Solution_2": "Nice! :) !",
  "Solution_3": "nice solution TomciO ,(it was the same that mine :lol: )"
}
{
  "Tag": [
    "algebra",
    "difference of squares",
    "special factorizations",
    "AMC"
  ],
  "Problem": "The fraction \\[\\frac {(3^{2008})^2 - (3^{2006})^2}{(3^{2007})^2 - (3^{2005})^2}\\] simplifies to which of the following?\n\n$ \\textbf{(A)}\\ 1 \\qquad \\textbf{(B)}\\ \\frac {9}{4} \\qquad \\textbf{(C)}\\ 3 \\qquad \\textbf{(D)}\\ \\frac {9}{2} \\qquad \\textbf{(E)}\\ 9$",
  "Solution_1": "[hide=\"Solution\"]The given expression equals\n$ \\frac{9^{2008}\\minus{}9^{2006}}{9^{2007}\\minus{}9^{2005}}$\nFactoring, we have\n$ \\frac{9^{2006}}{9^{2005}}\\cdot \\frac{9^2\\minus{}1}{9^2\\minus{}1}\\equal{}9 \\Rightarrow \\boxed{E}$[/hide]",
  "Solution_2": "[hide=\"Alternative Solution\"]\n\\[ \\begin {align*} \\dfrac {(3^{2008})^2 - (3^{2006})^2}{(3^{2007})^2 - (3^{2005})^2} &= \\dfrac {3^{4016} - 3^{4012}}{3^{4014} - 3^{4010}} \\\\&= \\dfrac {3^{4010} \\cdot (3^{6} - 3^{2})}{3^{4010} \\cdot (3^{4} - 3^{0})} \\\\&= \\dfrac {3^6 - 3^2}{3^4 - 3^0} \\\\&= \\dfrac {3^2 \\cdot (3^4 - 3^0)}{3^0 \\cdot (3^4 - 3^0)} \\\\&= \\dfrac {3^2}{3^0} \\\\&= \\boxed {\\textbf {(E) } 9} \\]\n[/hide]\n[hide=\"Unnecessarily long Solution\"]\nDifference of squares: \\[ \\begin {align*} \\dfrac {(3^{2008})^{2}-(3^{2006})^2}{(3^{2007})^{2}-(3^{2005})^{2}} &= \\dfrac {(3^{2008} + 3^{2006})(3^{2008} - 3^{2006})}{(3^{2007} + 3^{2005})(3^{2007} - 3^{2005})} \\\\&= \\dfrac {3^{2006} \\cdot (3^{2} + 3^{0}) \\cdot 3^{2006} \\cdot (3^{2} - 3^{0})}{3^{2005} \\cdot (3^2 + 1) \\cdot 3^{2005} (3^2 - 1)} \\\\&= 3^1 \\cdot 3^1 = 3^2 \\\\&= \\boxed {\\textbf {(E) } 9} \\]\n[/hide]",
  "Solution_3": "\\begin{align*}\n\\frac{(3^{2008})^2-(3^{2006})^2}{(3^{2007})^2-(3^{2005})^2} &= \\frac{(3^{2006})^2}{(3^{2005})^2} \\\\\n&= \\boxed{9}\n\\end{align*}"
}
{
  "Tag": [],
  "Problem": "The world would definately be a different place... my question, for worse? Or better?",
  "Solution_1": "Is this even a question?\r\n\r\nedit: is this meant to imitate the below \"If America didn't go to Iraq....\" post?",
  "Solution_2": "Of course it is a spoof of that.\r\nBy the way, the US did not join the war out of pity for the jews - they did it because there submarines were being attacked. So it was not really a choice, you see.",
  "Solution_3": "Was it the US's subs or their merchant vessels that were being attacked?",
  "Solution_4": "[quote=\"solafidefarms\"]Was it the US's subs or their merchant vessels that were being attacked?[/quote]\r\n\r\nMerchant vessels? That was in WWI... USA entered the WWII after the Japanese attack on the Pearl Harbor, if they thought me good in my history classses...",
  "Solution_5": "[quote=\"bubka\"]Of course it is a spoof of that.\nBy the way, the US did not join the war out of pity for the jews - they did it because there submarines were being attacked. So it was not really a choice, you see.[/quote]\r\n\r\nIndeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.\r\n\r\nNo country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war \r\n\r\nUrsula",
  "Solution_6": "[quote=\"ursula\"][quote=\"bubka\"]Of course it is a spoof of that.\nBy the way, the US did not join the war out of pity for the jews - they did it because there submarines were being attacked. So it was not really a choice, you see.[/quote]\n\nIndeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.\n\nNo country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war \n\nUrsula[/quote]\r\nWell the holocaust was by large unknown until the advance into germany started happening.",
  "Solution_7": "[quote=\"ursula\"][quote=\"bubka\"]Of course it is a spoof of that.\nBy the way, the US did not join the war out of pity for the jews - they did it because there submarines were being attacked. So it was not really a choice, you see.[/quote]\n\nIndeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.\n\nNo country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war \n\nUrsula[/quote]\r\n\r\nAmendment of the Neutrality Act, Lend-Lease, \"Four Freedoms\", Atlantic Charter, [i]then[/i] Pearl Harbor.",
  "Solution_8": "[quote=\"Piggypi314\"]\nAmendment of the Neutrality Act, Lend-Lease, \"Four Freedoms\", Atlantic Charter, [i]then[/i] Pearl Harbor.[/quote]\r\n\r\nI don't understand. Is this supposed to be an argument against the claim that the US entered WWII after Pearl Harbor? Because I don't get it, of the four things you mention, only the Lend Lease program influenced directly and significantly in the course of the war - it's actual significance might be a bit exaggerated but nonetheless -. And with a world scale war going on, surely you are not telling me that the US approving to loan stuff to the allies marks its entrance in the war. Four Freedoms, do you mean that stating some basic freedoms you think people should have is anyhow historically as important as the Pearl Harbor attack and the events that broke out? The US politics those days was of course, shaped by the international conflict but I think the statement remains, the US joins WWII after Pearl Harbor.\r\n\r\nUrsula",
  "Solution_9": "hi there but people, FDR was against Germany even before WWII had started.\r\n\r\nAnd...\r\n\r\nGreat Britain headed the war against the U-Boats, and decrypted various Enigma messages throughout world war II and was able to read coded messages within hours of transmission which is very good\r\n\r\nalso, the enigma could have 15 billion+ ways to scramble a single letter so...",
  "Solution_10": "The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.",
  "Solution_11": "[quote=\"ursula\"]\n\nIndeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.\n\nNo country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war \n\nUrsula[/quote]\r\n\r\nussr wouldn't be able to win a victory if usa hadn't helped them (remember who told the soviets that the japan has no intentions in attacking ussr so stalin sent his elite divisions to europe) and don't forget all the armory that was sent to moscow\r\n\r\nussr hasn't won 70% of the war\r\n\r\nbtw, remember that hitler and stalin made an agreement about poland and borders in eastern europe",
  "Solution_12": "[quote=\"deimos\"]\nussr wouldn't be able to win a victory if usa hadn't helped them (remember who told the soviets that the japan has no intentions in attacking ussr so stalin sent his elite divisions to europe) and don't forget all the armory that was sent to moscow[/quote]\n\nA lot of weapons were sent to the USSR, however three times more were sent to Britain. I don't think you are correct in thinking the USSR wouldn't be able to win a victory, without the US. Sorry but the men in the front were the soviet ones, and they were the ones using those weapons you mention which I don't think represent a major fraction of the weapons the USSR was using. Anyway, this is Lend Lease and I reckon the importance of it, that pretty much makes the 30% I'm conceding to the other allies. I don't know according to you who told the soviets japan had no intentions in attacking the ussr, to my knowledge this information was provided by Richard Sorge, prominent spy of the soviet secret services.\n\n[quote=\"deimos\"]btw, remember that hitler and stalin made an agreement about poland and borders in eastern europe[/quote]\r\n\r\nI remember deimos, however one thing has nothing to do with the other. I also said above that no country behaved acceptably morally speaking. I'm not saying Stalin did the best things of his life during WWII. I'm saying the USSR won the war, I'm saying that the D-Day the war was won and the allies knew better than not coming and keeping a portion of the western world. If they hadn't they would have learned russian in Paris.\r\n\r\nUrsula\r\n\r\nEdited: I was kind kind of pissed of at the world last night and was a bit rude.",
  "Solution_13": "I think it's an interesting question whether the United States would have entered the war in Europe, had Hitler not declared war after the U.S. declared war on Japan.  \r\n\r\nWhile I'm always up for a little left-wing revisionist history, the idea that the U.S. entered the war in Europe to stop the Soviet Union, only a week after the Germans gave up their assault on Moscow, is a strained interpretation at best.  I think a better case can be made that stopping the Soviet Union was one of the reasons we nuked Japan.\r\n\r\nWhat appears to be a good timeline of the war is available [url=http://www.historyplace.com/worldwar2/timeline/ww2time.htm]here[/url].",
  "Solution_14": "[quote=\"ursula\"][quote=\"Piggypi314\"]\nAmendment of the Neutrality Act, Lend-Lease, \"Four Freedoms\", Atlantic Charter, [i]then[/i] Pearl Harbor.[/quote]\n\nI don't understand. Is this supposed to be an argument against the claim that the US entered WWII after Pearl Harbor? Because I don't get it, of the four things you mention, only the Lend Lease program influenced directly and significantly in the course of the war - it's actual significance might be a bit exaggerated but nonetheless -. And with a world scale war going on, surely you are not telling me that the US approving to loan stuff to the allies marks its entrance in the war. Four Freedoms, do you mean that stating some basic freedoms you think people should have is anyhow historically as important as the Pearl Harbor attack and the events that broke out? The US politics those days was of course, shaped by the international conflict but I think the statement remains, the US joins WWII after Pearl Harbor.\n\nUrsula[/quote]\r\n\r\nHave you read the actual speech?  The whole thing is like \"Hello.  These are four wonderful freedoms.  WE MUST GO TO WAR to protect these wonderful freedoms.  WE WILL GO TO WAR AND WE WILL WIN.\"\r\n\r\nOf course, the US officially joined after Pearl Harbor, meaning Congress voted and yadda yadda.  But you should read up on undeclared wars, for fun if not for anything else.",
  "Solution_15": "[quote=\"JBL\"]\nWhile I'm always up for a little left-wing revisionist history, the idea that the U.S. entered the war in Europe to stop the Soviet Union, only a week after the Germans gave up their assault on Moscow, is a strained interpretation at best.  [/quote]\n\nI didn't knew the  germans abandoned the assault on Moscow on june, 1944 and I don't think so. Are you sure about this? Because, as it appears in the link you gave us, on january 1944, the ussr already was advancing into poland, out of their borders. The front line was very far from Moscow by then, the siege of Leningrad was over and there was also the ukrainian front who took up Ukraine, all this happened before Normandy. I'd say that the soviets in june 1944 were knocking at the doors of eastern europe with inexorable drive. \n\nHowever, I understand you can't make a good case of my opinion and it can be bashed because of the lack of supporting evidence. But the fact that the allies, with the US being essential - without the US there is no D day - enter Europe a bit en retard I think is clear. And although I'm always up for a little left-wing revisionist history myself, I think I'm not trying to do so right now, in the sense I'm not praising left-wing side actions, well may be so, but my intention rather than giving credit to left-wing as opposed to right-wing is giving credit to the other pole that existed as opposed to the only pole that survived. I think history in our schools is taught to minimize the role of the USSR in WWII. The landing on Normandy is presented as one of the greatest victories of humanity, which I think it is, but there are three or four victories by the soviets at least comparable to Normandy of which you are told very little, ok, maybe it didn't happened in your school. If you don't set up to find out yourself, you learn almost nothing about Kursk or Stalingrad or Moscow or the siege of Leningrad or even Berlin.\n\nTotally unrelatedly, this also happens with physics. The physics made in the USSR was truly amazing, however most of it remained unknown in the west until someone rediscovered it. Then they received all the credit and many times this is what we know today. Names like Feynman, Gell-Mann, Wilson, Weinberg, I could go on, are all well known even to the broad public. However, russians like Landau, Zel'dovich, Bogoliubov, Ginzburg, Kapitsa, etc are much less well known although their contributions are truly magnificent aswell, in many ocasions, as is the case with Wilson and Bogoliubov some contributions were very related. I had a professor who told me about a conference he assisted where the speaker was explaining his result and there was this russian guy who stood up to say that all he was saying actually was known 40 years ago in goo ol' UUSR and was discovered by Landau. Of course, you had to go to some russian journal and read it in russian to be aware. Of these soviet physicists you got to know what was so outstanding that the world [i]had to know[/i] but no more.\n\n[quote=\"Piggypi314\"]\nHave you read the actual speech? The whole thing is like \"Hello. These are four wonderful freedoms. WE MUST GO TO WAR to protect these wonderful freedoms. WE WILL GO TO WAR AND WE WILL WIN.\"\nOf course, the US officially joined after Pearl Harbor, meaning Congress voted and yadda yadda. But you should read up on undeclared wars, for fun if not for anything else.\n[/quote]\r\n\r\nI've read it diagonally, I think I've read enough and Hello, you keep missing the point. Roosevelt saying 'we will go to war and we will win' [i]is not[/i] going to war and winning. Others really [i]were[/i] at war and had not been for Pearl Harbor one could ask when the US would have actually entered the war. And above all, why are you telling me this? I say US doesnn't enters the war until Pearl Harbor and I also recognize the importance of Lend Lease, are you saying there is something absolutely inaccurate in this statement? What undeclared war are you speaking of? What the US was doing at least cualitatively is known and was somewhat known at the time as Mr Roosevelt comes to remind us, and I'm saying that it certainly wasn't fighting a war. Right until Pearl Harbor.\r\n\r\nUrsula",
  "Solution_16": "[quote=\"ursula\"]I didn't knew the  germans abandoned the assault on Moscow on june, 1944 and I don't think so. Are you sure about this? Because, as it appears in the link you gave us, on january 1944, the ussr already was advancing into poland, out of their borders. The front line was very far from Moscow by then, the siege of Leningrad was over and there was also the ukrainian front who took up Ukraine, all this happened before Normandy. I'd say that the soviets in june 1944 were knocking at the doors of eastern europe with inexorable drive. [/quote]\nWhoa, whoa, 194[b]4[/b]?  The United States entered the war in Europe shortly after Germany declared war at the end of 194[i]1[/i].  By the time of the D-Day landings and Normandy in 1944, there were American troops in Rome, among other places.  From my earlier link: [quote=\"[url=http://www.historyplace.com/worldwar2/timeline/ww2time.htm#1941]historyplace.com[/url]\"]Dec 5, 1941 - German attack on Moscow is abandoned.\nDec 6, 1941 - Soviet Army launches a major counter-offensive around Moscow.\nDec 7, 1941 - Japanese bomb Pearl Harbor; Hitler issues the Night and Fog decree.\nDec 8, 1941 - United States and Britain declare war on Japan.\nDec 11, 1941 - Germany declares war on the United States.\n...\nJan 26, 1942 - First American forces arrive in Great Britain.\n[/quote]",
  "Solution_17": "[quote=\"JBL\"]\nWhoa, whoa, 194[b]4[/b]?  The United States entered the war in Europe shortly after Germany declared war at the end of 194[i]1[/i].  By the time of the D-Day landings and Normandy in 1944, there were American troops in Rome, among other places.  From my earlier link: \n[/quote]\r\n\r\nOK sorry, I misunderstood you, I already mentioned the presence of American troops before D-Day in Europe, I just thought you were referring to Normandy since I still think represents the turning point in the other allies side - the turning point for the soviets had ocurred long ago as we seem to agree now. However what I thought was clearly permeated by my ignorance on the subject, I just checked some numbers and I was wrong, I thought the campaigns in the Mediterraneum (italy and north africa) were largely conducted by the british and that's not true, the US contribution was at least comparable to the british one. So  I was wrong and will have to study my history again :) So bad you were right so quickly, :lol: it removes the fun and I stop learning\r\nWith no intention to keep the argument because it was incorrect I might nevertheless say that anyhow the greatest enemy by far of the war was Germany. Italians weren't serious about the war and was comparably easy to overpower them. And I still think the war with Germany was essentially won by the USSR.\r\n\r\nUrsula",
  "Solution_18": "[quote=\"ursula\"]However, it was the USSR who turned the tide and I'd say won 70% of the war \n\nUrsula[/quote]Really? Who stormed the beaches of Normandy? Who (I really hate to say this) dropped the nukes on Hiroshima and Nagasaki? The Soviets really didn't have that great of a strategy.",
  "Solution_19": "[quote=\"kstan013\"][quote=\"ursula\"]However, it was the USSR who turned the tide and I'd say won 70% of the war \n\nUrsula[/quote]Really? Who stormed the beaches of Normandy? Who (I really hate to say this) dropped the nukes on Hiroshima and Nagasaki? The Soviets really didn't have that great of a strategy.[/quote]\r\n\r\nNukes and strategy? Oh, please... Don't be silly. If Soviets were in the position in which USA was, the nukes would never been used, you can be sure of that! \r\nAnd remember that every history book states that the crucial turn in WWII was the break of Stalingrad siege!\r\n\r\nBTW, did you know that the US army losses in WWII were smaller than the losses of Yugoslav army? \r\n\r\nRemember one thing: Soviets lost 10,000,000 soldiers and 12,000,000 civilians in WWII. The greatest war toll in history.",
  "Solution_20": "[quote=\"hsiljak\"][quote=\"solafidefarms\"]Was it the US's subs or their merchant vessels that were being attacked?[/quote]\n\nMerchant vessels? That was in WWI... USA entered the WWII after the Japanese attack on the Pearl Harbor, if they thought me good in my history classses...[/quote]\r\nAnd the Black Tom explosion! :ninja:",
  "Solution_21": "[quote=\"hsiljak\"][quote=\"kstan013\"][quote=\"ursula\"]However, it was the USSR who turned the tide and I'd say won 70% of the war \n\nUrsula[/quote]Really? Who stormed the beaches of Normandy? Who (I really hate to say this) dropped the nukes on Hiroshima and Nagasaki? The Soviets really didn't have that great of a strategy.[/quote]\n \nAnd remember that every history book states that the crucial turn in WWII was the break of Stalingrad siege!\n\nBTW, did you know that the US army losses in WWII were smaller than the losses of Yugoslav army? \n\nRemember one thing: Soviets lost 10,000,000 soldiers and 12,000,000 civilians in WWII. The greatest war toll in history.[/quote]1) [i]crucial[/i] turn? yes, definitely a turn, but I don't know if it was the crucial hit.\r\n2) Maybe they fought better and were less prone to be killed\r\nAnother thing is who *ended* the war. The Soviets take more credit than they deserve after completely taking control of Eastern Europe in the 1950's to the 1980's, that was a complete act of ruthlessness.",
  "Solution_22": "If you are going to say the nukes or the invasion of normandy was the crucial turn then you defiantely have no idea what you are talking about\r\n\r\nAlso America is the country that takes more credit then it deserves the war would be over if they hadn't joined it would have just taken a while longer\r\n\r\nps the bombs were not needed, the fact was the Soviets were freeing China faster then France surrendered, and the generals of Japan said they would have continued fighting after the bombs and did until they saw that there was no hope in China\r\n\r\nAnd about fighting better yeah everyone fights better when they can just bomb from across the ocean, they did not even have to send troops unlike yugoslaiva which had to fight an invasion\r\n\r\nOnly thing in Europe that I can say America actually did well and before the war was determined was Africa and Italy but in the first case it was mostly Britian still and in Italy the Italians were the ones who actually overthrew Mussolini, America was mostly the supplier,\r\n\r\nAlso your last point, America shot, bombed, and well attacked Soviet troops to delay them in Germany so the United states could take over the lands, if that is not ruthless then I have no clue what is\r\n\r\nFinally the dividing of Germany was America's idea the soviets said that it should be kept whole and did not care for it at all\r\n\r\nFinally about strategy America had none what so ever it was kill everthing on your path to Berlin and had no way of freeing China\r\n\r\nThey also told the Soviets to free Korea but then said no don't and came back there\r\n\r\nIf you have anything else about strategy to talk about go ahead but I believe you have no clue what you are saying\r\n\r\nThis of Course is not to commend Stalin's actions after the war in any way",
  "Solution_23": "That's the power of money right there. Where did the money come from? A great army. Where did the great army come from? Money. It's a never-ending cycle.",
  "Solution_24": "Yeah if America was in Europe I doubt you would be talking about the money cycle now\r\n\r\nAmerica did not have a great Army in comparison to the one Germany raised in its depression\r\n\r\nThe only Reason America had money was because it used its isolation to declare nuetrality and sell things\r\n\r\nYes now I agree its army is great but of course there is the fact it is all in Iraq\r\n\r\nAt the time its army was worse then almost all other powers armies\r\n\r\nThe money came from isolation and industry nothing to do with the army\r\n\r\nIt was not the power of money but of isolation\r\n\r\nand you still have not answered any of my points",
  "Solution_25": "[quote=\"ursula\"]I don't know according to you who told the soviets japan had no intentions in attacking the ussr, to my knowledge this information was provided by Richard Sorge, prominent spy of the soviet secret services.[/quote]\nhe was working for soviets?! :o sorry, i didn't know that :blush: \n\n[quote=\"hsiljak\"]And remember that every history book states that the crucial turn in WWII was the break of Stalingrad siege! \n[/quote]\n[quote=\"hsiljak\"]Remember one thing: Soviets lost 10,000,000 soldiers and 12,000,000 civilians in WWII.[/quote]\r\n\r\nsoviets had a great lost of soldiers in stalingrad battle because stalin wanted a glorious victory, and he didn't care about soldiers life.\r\ndo you know what soviet general zukov [as far as i have heard] have said about that battle (concerning stalin and hitler)",
  "Solution_26": "[quote=\"deimos\"]do you know what soviet general zukov [as far as i have heard] have said about that battle (concerning stalin and hitler)[/quote]\r\n\r\nNo, what?\r\n\r\nBTW, Russians lost 800,000 men (both soldiers and civilians) in Stalingrad battle. That's less than 5% of all their losses.",
  "Solution_27": "[quote=\"bubka\"]The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.[/quote]\r\n\r\nOKAY. SO IF SOME DUDE GOES AND DOES SOME GREAT FEAT, THEN HE IS EXEMPT FROM ANY BAD STUFF HE DOES?\r\n\r\nAlso, Turing was not the only person behind it? True that he was the one who designed the Turing Bombe, but Welchman and Winterbotham were also very involved in it. :wink:\r\n\\[\\ast\\qquad\\ast\\qquad\\ast\\qquad \\]\r\nAlso, the US won a predominant part of WWII, especially in Japan. The man who cracked Purple did not even know how purple was like and stuff.\r\n\r\nAlso, the US was helping UK for a long time before it entered the war.\r\n\r\nFurthermore, would anyone in their right mind seriously join someone who had just bombed them in some place killing 3403 people?\r\n\r\nAlso, as of 1942, the US read a lot of enigma messages using the US Navy Cryptanalytic Bombe by Joe Desch. :wink:\r\n\r\nNext Up: Who dropped the atomic bomb?",
  "Solution_28": "[quote=\"anirudh\"][quote=\"bubka\"]The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.[/quote]\n\nOKAY. SO IF SOME DUDE GOES AND DOES SOME GREAT FEAT, THEN HE IS EXEMPT FROM ANY BAD STUFF HE DOES?\n\nAlso, Turing was not the only person behind it? True that he was the one who designed the Turing Bombe, but Welchman and Winterbotham were also very involved in it. :wink:\n\\[\\ast\\qquad\\ast\\qquad\\ast\\qquad \\]\nAlso, the US won a predominant part of WWII, especially in Japan. The man who cracked Purple did not even know how purple was like and stuff.\n\nAlso, the US was helping UK for a long time before it entered the war.\n\nFurthermore, would anyone in their right mind seriously join someone who had just bombed them in some place killing 3403 people?\n\nAlso, as of 1942, the US read a lot of enigma messages using the US Navy Cryptanalytic Bombe by Joe Desch. :wink:\n\nNext Up: Who dropped the atomic bomb?[/quote]\r\nThe atomic bomb drop is regarded more of as a bad thing rather than a good one (okay, so granted maybe it stopped the Soviet Union from invading more territory, but that's not really that important). Japan was already losing the fight against the American pacific force, and if France and England would direct all their forces to French Indochina and India, in addition with the Soviet war machine turning east, then the war would've swiftly been won anyways. The Atomic bomb killed thousands of innocent civilians; its an example of terrorism on an enormous scale.\r\n\r\nAnd do you consider homosexuality a crime punishable by death? What did Turing's sexual orientation ever do to anyone?\r\n\r\nIn any case, America did do something, of course. For example, if 1,000 pounds of food need to be outputted by a factory, and I outputted 100 while someone else outputted 400 and another outputted 550, then I could claim that we wouldn't have met the quota without me, but it doesn't make me a predominant player.",
  "Solution_29": "[quote=\"anirudh\"][quote=\"bubka\"]The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.[/quote]\n\nOKAY. SO IF SOME DUDE GOES AND DOES SOME GREAT FEAT, THEN HE IS EXEMPT FROM ANY BAD STUFF HE DOES?\n\nAlso, Turing was not the only person behind it? True that he was the one who designed the Turing Bombe, but Welchman and Winterbotham were also very involved in it. :wink:\n[/quote]\r\nActually, turing designed the collosus mark I which cracked the enigma code. That was his major accomplishment.",
  "Solution_30": "[quote=\"bpms\"][quote=\"anirudh\"][quote=\"bubka\"]The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.[/quote]\n\nOKAY. SO IF SOME DUDE GOES AND DOES SOME GREAT FEAT, THEN HE IS EXEMPT FROM ANY BAD STUFF HE DOES?\n\nAlso, Turing was not the only person behind it? True that he was the one who designed the Turing Bombe, but Welchman and Winterbotham were also very involved in it. :wink:\n[/quote]\nActually, turing designed the collosus mark I which cracked the enigma code. That was his major accomplishment.[/quote]\r\n\r\nthe Colossus was to recover key settings for the Lorenz :wink: \r\n\r\nThe Lorenz was used by Hitler and the German High Command\r\n\r\nthe bombe was what helped crack enigma :wink:"
}
{
  "Tag": [
    "number theory open",
    "number theory"
  ],
  "Problem": "Is anybody aware of a definition of generalized Carmichael numbers using a modified Korselt criterion\r\n\r\nn is called r-Carmichael number,\r\nif n is squarefree and if for all primes p with $ n \\equiv r \\mod(p)$ then $ n \\minus{} 1 \\equiv r \\mod(p \\minus{} 1)$\r\n\r\nThe normal Carmichael numbers would be 0-Carmichael numbers in the above sense.\r\nAre there any composite r-Carmichael numbers for $ a\\not \\equal{} 0$ ?",
  "Solution_1": "Is anybody aware of a definition of generalized Carmichael numbers using a modified Korselt criterion \r\n\r\nn is called r-Carmichael number, \r\nif n is squarefree, n-r composite and if for all primes p with  $ n\\equiv r\\mod(p)$ then $ n\\minus{}1\\equiv r\\mod(p\\minus{}1)$ then  \r\n\r\nThe normal Carmichael numbers would be 0-Carmichael numbers in the above sense. \r\nAre there any composite r-Carmichael numbers for  $ a\\not\\equal{}0$"
}
{
  "Tag": [
    "function",
    "linear algebra",
    "matrix",
    "logarithms",
    "calculus",
    "integration"
  ],
  "Problem": "For any positive integer $ n$, let \\[f(n) \\equal{} \\begin{cases} \\log_8{n}, & \\text{if }\\log_8{n}\\text{ is rational,} \\\\\n0, & \\text{otherwise.} \\end{cases}\\] What is $ \\sum_{n \\equal{} 1}^{1997}{f(n)}$?\n\n$ \\textbf{(A)}\\ \\log_8{2047}\\qquad \\textbf{(B)}\\ 6\\qquad \\textbf{(C)}\\ \\frac {55}{3}\\qquad \\textbf{(D)}\\ \\frac {58}{3}\\qquad \\textbf{(E)}\\ 585$",
  "Solution_1": "[hide=\"Click for solution\"]\nSince $ \\log_8n\\equal{}\\frac{1}{3}(\\log_2n)$, it follows that $ \\log_8n$ is rational iff $ \\log_2n$ is rational.  The nonzero numbers in the sum will therefore be all numbers of the form $ \\log_8n$, where $ N$ is an integral power of $ 2$.  The highest power of $ 2$ that does not exceed $ 1997$ is $ 2^{10}$, so the sum is: \n\n$ \\log_81\\plus{}\\log_82\\plus{}\\log_82^2\\plus{}\\log_82^3\\plus{}\\ldots\\plus{}\\log_82^{10}\\equal{}0\\plus{}\\frac{1}{3}\\plus{}\\frac{2}{3}\\plus{}\\frac{3}{3}\\plus{}\\ldots\\plus{}\\frac{10}{3}\\equal{}\\frac{55}{3}$, or $ \\boxed{\\textbf{(C)}}$.\n[/hide]"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "$x+y=2, x,y \\geq 0$\r\nprove that $x^{2}y^{2}(x^{2}+y^{2}) \\leq 2$.\r\n :P",
  "Solution_1": "Let $xy=a$. We then have $x^{2}+y^{2}=(x+y)^{2}-2a=4-2a$\r\nTherefore:\r\n$x^{2}y^{2}(x^{2}+y^{2})\\le 2$\r\n$\\Leftrightarrow a^{2}(4-2a)\\le 2$\r\n$\\Leftrightarrow a^{2}(2-a)\\le 1$\r\n$\\Leftrightarrow a^{3}-2a^{2}+1\\ge 0$\r\n$\\Leftrightarrow (a^{3}-a^{2})-(a^{2}-1)\\ge 0$\r\n$\\Leftrightarrow (a-1)(a^{2}-a-1)\\ge 0$\r\nbecause $a=xy\\le \\frac{(x+y)^{2}}{4}=1$ we have $a\\le 1$ so $a-1 \\le 0$ and $a^{2}-a-1=a(a-1)-1<0$ so we have $(a-1)(a^{2}-a-1)\\ge 0$ and the result follows.",
  "Solution_2": "my solution\r\n\r\nBecause \\[x^{2}y^{2}(x^{2}+y^{2})=2x^{2}y^{2}(2-xy) \\leq 2 \\Longleftrightarrow (xy)^{3}-2(xy)^{2}+1 \\leq 0\\]  \\[\\Longleftrightarrow xy(xy-1)^{2}+(1-xy) \\geq 0\\] Since $x+y=2,~x+y \\geq 2\\sqrt{xy}\\leftrightarrow 1-xy \\geq 0$, therefore, $x^{2}y^{2}(x^{2}+y^{2}) \\leq 2$.",
  "Solution_3": "Moreover, if $x_{1}+x_{2}+\\cdots+x_{n}=n$ we have \\[x_{1}^{r}x_{2}^{r}\\cdots x_{n}^{r}(x_{1}^{s}+x_{2}^{s}+\\cdots+x_{n}^{s})\\leq n\\] for all $r,s\\geq1.$"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Given are the natural number $ n\\ge 2$ and the real positive numbers $ x,y,a,b$ . Prove that \r\n\r\n$ \\boxed {\\ \\frac {a^nx^{n \\plus{} 1} \\plus{} b^ny^{n \\plus{} 1}}{(x \\plus{} y)^{n \\plus{} 1}}\\ge\\left(\\frac {ab}{a \\plus{} b}\\right)^n\\ }$ . Particularly, $ \\boxed {\\ \\left\\{\\begin{array}{c} x \\plus{} y \\equal{} 1 \\\\\r\n\\ a \\plus{} b \\equal{} 1\\end{array}\\right|\\ \\Longrightarrow\\ \\frac {x^{n \\plus{} 1}}{a^n} \\plus{} \\frac {y^{n \\plus{} 1}}{b^n}\\ge 1\\ }$ .\r\n\r\nSee [url=http://www.mathlinks.ro/viewtopic.php?t=193348]here[/url]",
  "Solution_1": "[quote=\"Virgil Nicula\"]Given are the natural number $ n\\ge 2$ and the real positive numbers $ x,y,a,b$ . Prove that \n\n$ \\boxed {\\ \\frac {a^nx^{n \\plus{} 1} \\plus{} b^ny^{n \\plus{} 1}}{(x \\plus{} y)^{n \\plus{} 1}}\\ge\\left(\\frac {ab}{a \\plus{} b}\\right)^n\\ }$ . Particularly, $ \\boxed {\\ \\left\\{\\begin{array}{c} x \\plus{} y \\equal{} 1 \\\\\n\\ a \\plus{} b \\equal{} 1\\end{array}\\right|\\ \\Longrightarrow\\ \\frac {x^{n \\plus{} 1}}{a^n} \\plus{} \\frac {y^{n \\plus{} 1}}{b^n}\\ge 1\\ }$ .\n\nSee [url=http://www.mathlinks.ro/viewtopic.php?t=193348]here[/url][/quote]\r\nBy Holder inequality, we have:\r\n$ \\left(\\frac{1}{a}\\plus{}\\frac{1}{b}\\right)^n\\left(a^nx^{n \\plus{} 1} \\plus{} b^ny^{n \\plus{} 1}\\right) \\ge \\left(x\\plus{}y\\right)^{n\\plus{}1}$\r\nSo we are done!"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra unsolved"
  ],
  "Problem": "Is there any easy solution without expanding to show :\r\n$ \\minus{}\\sum \\frac{a^3}{(a\\minus{}b)(c\\minus{}a)}\\equal{}a\\plus{}b\\plus{}c$",
  "Solution_1": "\\begin{eqnarray*}\r\n-\\sum_{cyc}\\frac{a^3}{(a-b)(c-a)}\r\n& = & -\\sum_{cyc}\\frac{a^3(b-c)}{(a-b)(b-c)(c-a)} \\\\\r\n& = & -\\frac{1}{(a-b)(b-c)(c-a)}\\sum_{cyc}a^3(b-c) \\\\\r\n& = & -\\frac{1}{(a-b)(b-c)(c-a)} \\times(-(a-b)(b-c)(c-a)(a+b+c)) \\\\\r\n& = & a+b+c\r\n\\end{eqnarray*}\r\n\r\nBecause,\r\nLet $ P = a^3(b - c) + b^3(c - a) + c^3(a - b)$.\r\n$ P$ is quartic alternating polynomial of $ a$, $ b$, $ c$.\r\nSo, $ (a - b)(b - c)(c - a)(a+b+c)|P$.\r\n$ \\therefore P = k(a - b)(b - c)(c - a)(a+b+c)$  ($ k$:const)\r\n$ P_{(a, b, c) = (0, 1, 2)} = -6$, $ k(a - b)(b - c)(c - a)(a+b+c)|_{(a, b, c) = (0, 1, 2)} = 6k$\r\n$ \\therefore k = -1$\r\n\r\nThus, $ \\sum_{cyc}a^3(b-c) = a^3(b - c) + b^3(c - a) + c^3(a - b) = -(a - b)(b - c)(c - a)(a+b+c)$."
}
{
  "Tag": [
    "combinatorics solved",
    "combinatorics"
  ],
  "Problem": "Let k and n be positive integers:$\\frac{n}{2}<k\\leq\\frac{2n}{3}$Find the least number m such that we can place each of m pawns on a squares of an n*n chessboard so that no column and row contains a block of k adjacent unoccupied squares",
  "Solution_1": "This kind of problem is well-known.For example,see IMO3-1999",
  "Solution_2": "Can you post the solution?"
}
{
  "Tag": [],
  "Problem": "HADAFE SHOMA AZ IN ZENDEGI CHIYE? :?:",
  "Solution_1": "[quote=\"mohamadhosein\"]HADAFE SHOMA AZ IN ZENDEGI CHIYE? :?:[/quote]\r\nmishe yek kami bishtar tozih bedi?",
  "Solution_2": "[quote=\"jensen\"][quote=\"mohamadhosein\"]HADAFE SHOMA AZ IN ZENDEGI CHIYE? :?:[/quote]\nmishe yek kami bishtar tozih bedi?[/quote]\r\n\r\nmanzooreshon ineh keh, har ensan e salem ye hadafi too zendegish dareh.\r\n\r\nmasalan hadafe to gold medal of imo e, amma omran be in arezoo beresi. ;)  :D",
  "Solution_3": "fekr konam soalam be andazeye kafi vazeh bashe.hammon chizi ke ashegh gofte.\r\nbaraye che hadafi darin talash mikonin?"
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "show:\r\n\r\n$\\textbf{(i)}\\;\\;\\;\\prod_{k=2}^{\\infty}\\;\\left( 1+\\frac{2}{k}\\right)^{(-1)^{k}k}\\;=\\;\\boxed{\\frac{6}{\\pi e}}$\r\n\r\n\r\n$\\textbf{and\\;\\;\\;(ii)}\\;\\;\\;\\prod_{k=1}^{\\infty}\\;\\left( 1+\\frac{2}{k}\\right)^{(-1)^{k+1}k}\\;=\\;\\boxed{\\frac{\\pi}{2 e}}$\r\n\r\n\r\n :D",
  "Solution_1": "[quote=\"misan\"]show:\n\n$\\textbf{(i)}\\;\\;\\;\\prod_{k=2}^{\\infty}\\;\\left( 1+\\frac{2}{k}\\right)^{(-1)^{k}k}\\;=\\;\\boxed{\\frac{6}{\\pi e}}$\n\n\n$\\textbf{and\\;\\;\\;(ii)}\\;\\;\\;\\prod_{k=1}^{\\infty}\\;\\left( 1+\\frac{2}{k}\\right)^{(-1)^{k+1}k}\\;=\\;\\boxed{\\frac{\\pi}{2 e}}$\n\n\n :D[/quote]\r\nsome idea:\r\nStirling forever :\r\nby stirling we also can find asymptotic of this expression:\r\n$(2n)!!$~...\r\nand \r\n$(2n+1)!!$~...\r\ntry to find it and your problem will be solved\r\nit's easy"
}
{
  "Tag": [
    "calculus",
    "integration",
    "MATHCOUNTS",
    "modular arithmetic",
    "search",
    "function"
  ],
  "Problem": "How do you find the smallest integral x for simultaneous congruences?\r\n(i.e. x congruent to 2 (mod 3) and also congruent to 3 (mod 4) I'm trying to find a systematic way.\r\n\r\nI came upon this problem in problem 9 of Warm-Up 10 (2005-2006) MATHCOUNTS School Handbook:\r\n\r\nA college class has exactly enough students to from eight equal rows. On Monday a student is absent, and the professor is able to seat the students into five equal rows. On Tuesday two students are absent, and the professor can seat the students into nine equal rows. What is the least possible number of students in the class?\r\n\r\nI have:\r\nx congruent to 0 (mod 8)\r\nx-1 congruent to 0 (mod 5)\r\nx-2 congruent to 0 (mod 9)",
  "Solution_1": "You can use certain techniques to limit it, but you can't actually find it without some testing (most of the time). For example, you know that it's a multiple of 8. Next, you know that it's one more than a multiple of 5. If this was an even multiple of 5, then the number would end in a 1, which is impossible, so you know that it's a multiple of 8 that ends in a 6. Finally, you know that it is two more than a multiple of 9. For all multiples of 9, the sum of the digits is a multiple of 9, so this must add to 11. (20 and higher are too high, because you need a three-digit number for that.)\r\n\r\nSo, in this case, you can actually get the answer right off. It ends in a 6, and the digits sum to 11, so the answer is [hide]56[/hide].",
  "Solution_2": "[quote=\"hillarryous\"]How do you find the smallest integral x for simultaneous congruences?\n(i.e. x congruent to 2 (mod 3) and also congruent to 3 (mod 4) I'm trying to find a systematic way.\n\n[/quote]\r\n\r\nYou have \r\n\r\n$x\\equiv2\\pmod3$ \r\n\r\n$x\\equiv3\\pmod4$. \r\n\r\nYou add one to both sides to get \r\n\r\n$x+1\\equiv0\\pmod3$\r\n\r\n$x+1\\equiv0\\pmod4$. \r\n\r\nThat must mean that $x+1\\equiv0\\pmod{12}$, $x\\equiv11\\pmod{12}$.",
  "Solution_3": "In general modular congruences can be solved with the Chinese Remainder Theorem.  I think it's a bit too advanced to put in this forum (and I'm too lazy to), but if you use the search function you can probably find some stuff about it."
}
{
  "Tag": [
    "logarithms",
    "function"
  ],
  "Problem": "Find root of this equation: $ 7^{x \\minus{} 1} \\equal{} 6.log_{7}(6x \\minus{} 5) \\plus{} 1$",
  "Solution_1": "[hide]\nLet $ z \\equal{} x \\minus{} 1$, and the equation becomes $ \\frac {7^z \\minus{} 1}{6} \\equal{} \\log_7 (6z \\plus{} 1)$. But the LHS and RHS are inverse functions (if $ f(z) \\equal{} \\frac {7^z \\minus{} 1}{6}$, then $ f^{ \\minus{} 1}(z) \\equal{} \\log_7 (6z \\plus{} 1)$), so equality comes when $ f(z) \\equal{} \\frac {7^z \\minus{} 1}{6} \\equal{} z \\Longrightarrow z \\equal{} 0, 1$, or $ x \\equal{} \\boxed{1,2}$.\n[/hide]",
  "Solution_2": "[quote=\"azjps\"][hide]\nLet $ z \\equal{} x \\minus{} 1$, and the equation becomes $ \\frac {7^z \\minus{} 1}{6} \\equal{} \\log_7 (6z \\plus{} 1)$. But the LHS and RHS are inverse functions (if $ f(z) \\equal{} \\frac {7^z \\minus{} 1}{6}$, then $ f^{ \\minus{} 1}(z) \\equal{} \\log_7 (6z \\plus{} 1)$), so equality comes when $ f(z) \\equal{} \\frac {7^z \\minus{} 1}{6} \\equal{} z \\Longrightarrow z \\equal{} 0, 1$, or $ x \\equal{} \\boxed{1,2}$.\n[/hide][/quote]\n\n[hide]Oh ! It's nice. Thank you. :P [/hide]"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "invariant",
    "incenter",
    "circumcircle",
    "perpendicular bisector"
  ],
  "Problem": "[color=green]I tried to invent a transformation like isogonal and isotomic transformations  :blush: . I thought that if Isogonal keep invariant incenter and \nisotomic keep invariant baricenter then there's a tranformation like \nthose that keep invariant the circuncenter (I was searching some result \nlike the concurrence). So I got this results  :( .\n\n   Given $ \\triangle ABC$ and given a point $ P$ in its plane, let be $ O$ the \ncircumcenter of $ \\triangle ABC$ and $ P_{a}$, $ P_{b}$, $ P_{c}$ the simetrics of $ P$ respect the \nperpendicular bisector of sides $ \\overline{BC}$, $ \\overline{AC}$, $ \\overline{AB}$ respectly. Then\n\n   $ \\blacksquare$ $ O$ is the circumcenter of the triangle $ P_{a}P_{b}P_{c}$.\n   $ \\blacksquare$ $ P$ lies on circumcicle of $ P_{a}P_{b}P_{c}$.\n   $ \\blacksquare$ $ \\triangle P_{a}P_{b}P_{c}\\sim \\triangle ABC$.\n   $ \\blacksquare$ Let $ M, N, P$ the midpoints of sides $ \\overline{BC}$, $ \\overline{AC}$, $ \\overline{AB}$ and let be $ M'$, $ N'$, $ P'$ \nthe midpoints of $ \\overline{P_{b}P_{c}}$, $ \\overline{P_{a}P_{c}}$, $ \\overline{P_{a}P_{b}}$, then ,obviusly $ \\triangle MNP$ is similar to \n$ \\triangle M'N'P'$. Also $ \\triangle MNP$ and $ \\triangle M'N'P'$ are in perspective.\n\n   I also got some others results but I'd like to anybody helps me to \nfind out a very interesting result there. \n\n   Thanks a lot[/color]\r\n\r\n\r\n",
  "Solution_1": "All of your results are true, and I found particularly the last one very interesting. I have given proofs of these results now in my paper \"Isogonal conjugation with respect to a triangle\", which can be found as an attachment at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=569871#569871]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 post #6[/url] and on [url=http://de.geocities.com/darij_grinberg/]my website[/url] (Theorems 10 and 11).\r\n\r\n  Darij",
  "Solution_2": "[quote=\"darij grinberg\"]All of your results are true, and I found particularly the last one very interesting. I have given proofs of these results now in my paper \"Isogonal conjugation with respect to a triangle\", which can be found as an attachment at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=569871#569871]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 post #6[/url] and on [url=http://de.geocities.com/darij_grinberg/]my website[/url] (Theorems 10 and 11).\n\n  Darij[/quote]\r\n\r\nYour proof is very nice, my proof is longer than yours  :blush:",
  "Solution_3": "Here is another result\r\n\r\nGiven $A, B, C \\in l$ and a point $P \\not \\in l$. Let $A'$, $B'$, $C'$ the circumcenters of $\\triangle PBC$, $\\triangle PCA$, $\\triangle PAB$ respectively. Then:\r\n\r\n$\\bigstar$ The orthogonal proyections of $P$ on the sides of $\\triangle A'B'C'$ are collinear.\r\n$\\bigstar$ $P$ lies on the circumcircle of $\\triangle A'B'C'$.\r\n\r\nSould anybody post anothers??",
  "Solution_4": "[quote=\"hucht\"]Here is another result\n\nGiven $A, B, C \\in l$ and a point $P \\not \\in l$. Let $A'$, $B'$, $C'$ the circumcenters of $\\triangle PBC$, $\\triangle PCA$, $\\triangle PAB$ respectively. Then:\n\n$\\bigstar$ The orthogonal proyections of $P$ on the sides of $\\triangle A'B'C'$ are collinear.\n$\\bigstar$ $P$ lies on the circumcircle of $\\triangle A'B'C'$.\n\nSould anybody post anothers??[/quote]\r\n\r\nWhy did you post this in this topic? Actually, it's a different problem, I don't see a direct relation... It's ok if I move it into a new topic?\r\n\r\n  darij"
}
{
  "Tag": [
    "trigonometry",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let $ ABC$ be a triangle for which $ b\\ne c$ . Consider the points  $ \\{M,D\\}\\subset BC$ so that \r\n\r\n$ \\left\\|\\begin{array}{c} MB \\equal{} MC \\\\\r\n \\\\\r\n\\widehat {DAB}\\equiv\\widehat {DAC} \\\\\r\n \\\\\r\nm(\\widehat {DAM}) \\equal{} \\phi\\end{array}\\right\\|$ . Show that $ \\boxed {\\ \\tan\\phi \\equal{} \\frac {|b \\minus{} c|}{b \\plus{} c}\\cdot\\tan\\frac A2 \\equal{} \\tan^2\\frac A2\\cdot\\tan\\frac {|B \\minus{} C|}{2}\\ }$ .\r\n\r\nDenote the points $ \\left\\|\\begin{array}{c} P\\in AB\\ ,\\ PD\\perp AM \\\\\r\n\\ R\\in AD\\ ,\\ RP\\perp AB\\end{array}\\right\\|$ . Prove that $ \\boxed {\\ RM\\perp BC\\ \\Longleftrightarrow\\ B \\equal{} 90^{\\circ}\\ \\vee\\ C \\equal{} 90^{\\circ}\\ }$ .",
  "Solution_1": "I corrected \"two letters\". Sorry !",
  "Solution_2": "i think the first one must be $ \\tan\\phi\\equal{}\\tan^2\\dfrac{A}{2}\\cdot\\tan\\dfrac{|B\\minus{}C|}{2}$...\r\n\r\nsuppose wlog that $ B\\geq C$. then the order of the points is $ B,D,M,C$, and $ DM\\equal{}\\dfrac{a(b\\minus{}c)}{2(b\\plus{}c)}$...\r\n\r\nwe have that \r\n\r\n$ \\dfrac{\\sin\\left(\\frac{A}{2}\\minus{}\\phi\\right)}{\\sin\\phi}\\equal{}\\dfrac{CM\\cdot AD}{DM\\cdot AC}\\equal{}\\dfrac{2c\\cos\\frac{A}{2}}{b\\minus{}c}$\r\n\r\nthis implies that \r\n\r\n$ \\sin\\frac{A}{2}\\cot\\phi\\equal{}\\cos\\frac{A}{2}\\plus{}\\dfrac{2c\\cos\\frac{A}{2}}{b\\minus{}c}\\equal{}\\dfrac{(b\\plus{}c)\\cos\\frac{A}{2}}{b\\minus{}c}$,\r\n\r\nfrom where we conclude that \r\n\r\n$ \\tan \\phi\\equal{}\\tan\\frac{A}{2}\\cdot\\dfrac{b\\minus{}c}{b\\plus{}c}\\equal{}\\tan^2\\frac{A}{2}\\tan\\dfrac{B\\minus{}C}{2}$.",
  "Solution_3": "Thank you, [b]Campos[/b] ! I corrected and added ...  I am interested by the second part of it.",
  "Solution_4": "for the second part I assume that $ B\\geq C$ still holds, i don't have checked the other case but it must follow in the same way... \r\n\r\nnote that \r\n$ DR\\equal{}\\dfrac{PD\\sin\\left(\\frac{A}{2}\\plus{}\\phi\\right)}{\\sin\\left(\\frac{B\\plus{}C}{2}\\right)}\\cdot\\dfrac{AD\\sin\\frac{A}{2}}{PD\\cos\\left(\\frac{A}{2}\\plus{}\\phi\\right)}\\equal{}AD\\tan\\frac{A}{2}\\tan\\left(\\frac{A}{2}\\plus{}\\phi\\right)\\equal{}\\dfrac{2bc\\sin\\frac{A}{2}\\tan\\left(\\frac{A}{2}\\plus{}\\phi\\right)}{b\\plus{}c}$\r\n\r\nnote also that $ RM\\perp BC$ iff $ PBMR$ is cyclic, which holds iff $ \\angle DRM\\equal{}\\frac{B\\minus{}C}{2}$. then, \r\n\r\n$ RM\\perp BC \\Leftrightarrow DR\\equal{}\\dfrac{DM}{\\sin\\left(\\frac{B\\minus{}C}{2}\\right)}\\equal{}\\dfrac{a(b\\minus{}c)}{2(b\\plus{}c)\\sin\\left(\\frac{B\\minus{}C}{2}\\right)}\\equal{}\\dfrac{a^2}{2(b\\plus{}c)\\cos\\frac{A}{2}}$\r\n\r\nthen $ RM\\perp BC$ iff $ \\dfrac{\\sin A}{2\\sin B\\sin C}\\equal{}\\tan\\left(\\frac{A}{2}\\plus{}\\phi\\right)$\r\n\r\nusing the formula $ \\tan(x\\plus{}y)\\equal{}\\dfrac{\\tan x\\plus{}\\tan y}{1\\minus{}\\tan x\\tan y}$ it follows that the last equation holds iff\r\n\r\n$ \\dfrac{\\sin A}{2\\sin B\\sin C}\\minus{}\\tan\\frac{A}{2}\\equal{}\\tan \\phi\\left(1\\plus{}\\dfrac{\\tan\\frac{A}{2}\\sin A}{2\\sin B\\sin C}\\right)$\r\n\r\n\r\nnote that $ \\dfrac{\\sin A}{2\\sin B\\sin C}\\minus{}\\tan\\frac{A}{2}\\equal{}\\dfrac{\\sin\\frac{A}{2}\\sin^2\\left(\\frac{B\\minus{}C}{2}\\right)}{\\cos\\frac{A}{2}\\sin B\\sin C}$ \r\n\r\nand $ 1\\plus{}\\dfrac{\\tan\\frac{A}{2}\\sin A}{2\\sin B\\sin C}\\equal{}\\dfrac{\\cos^2\\left(\\frac{B\\minus{}C}{2}\\right)}{\\sin B\\sin C}$\r\n\r\ntherefore, $ RM\\perp BC$ iff $ \\tan\\phi\\equal{}\\tan\\frac{A}{2}\\tan^2\\frac{B\\minus{}C}{2}$, which holds iff $ \\tan\\frac{A}{2}\\equal{}\\tan\\frac{|B\\minus{}C|}{2}$..\r\n\r\nsince we assumed that $ B\\geq C$ then we must have $ \\frac{A}{2}\\equal{}\\frac{B\\minus{}C}{2}$, from where we conclude that this holds iff $ B\\equal{}90^{\\circ}$, and we're done.  :D"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Triangle ABC has area 84 and is such that AB=13, AC=15. Some point D is chosen in the same plane of ABC such that BD=DC=7. E is the midpoint of AC and F is the midpoint of AB. Let G be the centroid of triangle DEF, and let H be the intersection of FD and BE. Find the area of triangle FGH.\r\n\r\n[hide=\"hint\"]Where/What must D be?[/hide]",
  "Solution_1": "[quote=\"13375P34K43V312\"]Triangle ABC has area 84 and is such that AB=13, AC=15. Some point D is chosen in the same plane of ABC such that BD=DC=7. E is the midpoint of AC and F is the midpoint of AB. Let G be the centroid of triangle DEF, and let H be the intersection of FD and BE. Find the area of triangle FGH.\n\n[hide=\"hint\"]Where/What must D be?[/hide][/quote]\n\n[hide=\"Hint\"]$ D$ is the midpoint of $ BC$.[/hide]\n[hide=\"Solution\"]Since $ BC \\equal{} 14$, $ D$ is the midpoint of $ BC$. So, $ \\triangle DEF$ is formed by connecting the midpoints of $ \\triangle ABC$.\n\nIf you draw $ BE$, you'll see that it actually contains one of the medians of $ \\triangle DEF$. So, $ \\triangle FGH$ is just one-sixth of the area of $ \\triangle DEF$, which is $ 84\\cdot 1/4\\cdot 1/6 \\equal{}\\boxed{7/2}$[/hide]",
  "Solution_2": "Was that whole ordeal with point D really necessary? :O\r\n\r\n\r\nAnd what is it with those contest writers, they really like 13-14-15 triangles for some reason",
  "Solution_3": "[quote=\"randomdragoon\"]Was that whole ordeal with point D really necessary? :O\n\n\nAnd what is it with those contest writers, they really like 13-14-15 triangles for some reason[/quote]\r\n\r\nThats because a 13-14-15 triangle has an altitude of [b]12[/b] xDDD\r\n\r\nwell from the side that is length 15 atleast...",
  "Solution_4": "I mean seriously it can't be the only non-right triangle with integer side lengths AND integer area ...\r\n\r\n.. can it?",
  "Solution_5": "Try $ 51,52,53$\r\n\r\nBy the way, memorize this triple. Contests love it :)",
  "Solution_6": "[hide]I mean seriously it can't be the only non-right triangle with integer side lengths AND integer area ... \n\n.. can it?[/hide]\r\n\r\nWe had this problem in int nt..."
}
{
  "Tag": [
    "function",
    "calculus",
    "derivative",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ f : [0,1] \\longrightarrow \\mathbb{R}$ be continuous and $ f'(x) > 0$ for all but countably many points. Prove that $ f$ is increasing. Would the result still be true if we had $ f'(x) > 0$ almost everywhere?",
  "Solution_1": "Here's a thought: The fact that $ f'(x)>0$ [b]at all but countably many points[/b] means that $ f'(x)\\leq 0$ only at isolated points, not an interval, because any interval contains uncountably many points.",
  "Solution_2": "You're not going to get anywhere with the interval idea. Here's a concrete example of a continuous function with positive derivative almost everywhere, which is not increasing:\r\nLet $ f$ be the Cantor devil's staircase function on $ [0,1]$, and let $ g(x)\\equal{}\\frac12x\\minus{}f(x)$. With a different function with zero derivative a.e., I could make it nonicreasing on every interval."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "A positive number $x$ satisfies the inequality $\\sqrt{x} < 2x$ if and only if\r\n$\\text{(A)} \\ x > \\frac{1}{4} \\qquad \\text{(B)} \\ x > 2 \\qquad \\text{(C)} x > 4 \\qquad \\text{(D)} \\ x < \\frac{1}{4}\\qquad \\text{(E)} x < 4$",
  "Solution_1": "[quote=\"4everwise\"]A positive number $x$ satisfies the inequality $\\sqrt{x} < 2x$ if and only if\n$\\text{(A)} \\ x > \\frac{1}{4} \\qquad \\text{(B)} \\ x > 2 \\qquad \\text{(C)} x > 4 \\qquad \\text{(D)} \\ x < \\frac{1}{4}\\qquad \\text{(E)} x < 4$[/quote]\r\n[hide]\n$\\sqrt{x} < 2x$\n$x < 4x^2$\n$0 < x(4x-1)$\n$0 < 4x-1$\n$1 < 4x$\n$x > \\frac{1}{4}$\n(A)[/hide]",
  "Solution_2": "[hide]$\\text{Answer: (A)}$\n\nThe best way to solve this seems to be by squaring and regrouping. \n\n$\\sqrt{x}<2x$\n$x<4x^2$\n$0<x(4x-1)$\n$\\frac{1}{4}<x$[/hide]",
  "Solution_3": "[hide]You get two inequalities (see above messages), x>0, and x>1/4. The x>1/4 overshadows, so to say, the x>0...I think...anyone want to confirm? [/hide]",
  "Solution_4": "[quote=\"imber.matutinis\"][hide]You get two inequalities (see above messages), x>0, and x>1/4. The x>1/4 overshadows, so to say, the x>0...I think...anyone want to confirm? [/hide][/quote]\r\nIf x>1/4, then x must be greater than 0 as well. :)",
  "Solution_5": "[quote=\"catcurio\"][quote=\"imber.matutinis\"][hide]You get two inequalities (see above messages), x>0, and x>1/4. The x>1/4 overshadows, so to say, the x>0...I think...anyone want to confirm? [/hide][/quote]\nIf x>1/4, then x must be greater than 0 as well. :)[/quote]\r\n\r\nRight, that makes sense. But, what if you have an equation that yields two inequalities that don't work out that way--i.e. you would have to put the word [i]and[/i] or [i]or[/i]. Can't really think of an example...but are there specific instances/guidelines where you choose either [i]and[/i] or [i]or[/i]--how can you tell which word you're supposed to use?",
  "Solution_6": "[hide=\"Solution\"]\n$\\\\\\sqrt{x}<2x\\\\ x<4x^2\\\\ \\frac{1}{x}<4\\\\ x>\\frac{1}{4}\\\\ \\boxed{A}$\n[/hide]",
  "Solution_7": "[quote=\"boxedexe\"][hide=\"Solution\"]\n$\\\\\\sqrt{x}<2x\\\\ x<4x^2\\\\ \\frac{1}{x}<4\\\\ x>\\frac{1}{4}\\\\ \\boxed{A}$\n[/hide]\n\nMasoud Zargar[/quote]\r\n\r\nI thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?",
  "Solution_8": "[quote=\"imber.matutinis\"][quote=\"boxedexe\"][hide=\"Solution\"]\n$\\\\\\sqrt{x}<2x\\\\ x<4x^2\\\\ \\frac{1}{x}<4\\\\ x>\\frac{1}{4}\\\\ \\boxed{A}$\n[/hide]\n\nMasoud Zargar[/quote]\n\nI thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]\r\n\r\nthat's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D",
  "Solution_9": "[quote=\"ch1n353ch3s54a1l\"][quote=\"imber.matutinis\"][quote=\"boxedexe\"][hide=\"Solution\"]\n$\\\\\\sqrt{x}<2x\\\\ x<4x^2\\\\ \\frac{1}{x}<4\\\\ x>\\frac{1}{4}\\\\ \\boxed{A}$\n[/hide]\n\nMasoud Zargar[/quote]\n\nI thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]\n\nthat's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D[/quote]\r\n\r\nBut there's an x<1/4 in the answer choices. Unless you don't count that? Why don't you count that? \r\n\r\nBut x CAN be zero! In the inequality-- x>0 and x>1/4.",
  "Solution_10": "[quote=\"imber.matutinis\"][/quote][quote=\"ch1n353ch3s54a1l\"][quote=\"imber.matutinis\"][quote=\"boxedexe\"][hide=\"Solution\"]\n$\\\\\\sqrt{x}<2x\\\\ x<4x^2\\\\ \\frac{1}{x}<4\\\\ x>\\frac{1}{4}\\\\ \\boxed{A}$\n[/hide]\n\nMasoud Zargar[/quote]\n\nI thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]\n\nthat's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D[/quote][quote=\"imber.matutinis\"]\n\nBut there's an x<1/4 in the answer choices. Unless you don't count that? Why don't you count that? \n\nBut x CAN be zero! In the inequality-- x>0 and x>1/4.[/quote]\r\n\r\noh, i didn't see that\r\nwell according to the equation x CAN'T be zero  :P so i guess that means u can cancel out the Xs because it won't get rid of any solutions",
  "Solution_11": "[quote=\"ch1n353ch3s54a1l\"][/quote][quote=\"imber.matutinis\"][/quote][quote=\"ch1n353ch3s54a1l\"][quote=\"imber.matutinis\"][quote=\"boxedexe\"][hide=\"Solution\"]\n$\\\\\\sqrt{x}<2x\\\\ x<4x^2\\\\ \\frac{1}{x}<4\\\\ x>\\frac{1}{4}\\\\ \\boxed{A}$\n[/hide]\n\nMasoud Zargar[/quote]\n\nI thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]\n\nthat's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D[/quote][quote=\"imber.matutinis\"]\n\nBut there's an x<1/4 in the answer choices. Unless you don't count that? Why don't you count that? \n\nBut x CAN be zero! In the inequality-- x>0 and x>1/4.[/quote][quote=\"ch1n353ch3s54a1l\"]\n\noh, i didn't see that\nwell according to the equation x CAN'T be zero  :P so i guess that means u can cancel out the Xs because it won't get rid of any solutions[/quote]\r\n\r\nWait, really? You can't take the square root of zero? I thought that was just for negative numbers...",
  "Solution_12": "You can take the square root of 0. It equals 0. But if we want $\\sqrt{x}<2x$, then x cannot be 0. :)",
  "Solution_13": "[quote=\"catcurio\"]You can take the square root of 0. It equals 0. But if we want $\\sqrt{x}<2x$, then x cannot be 0. :)[/quote]\r\n\r\nOh...haha. That makes sense. 0 is not less than 0. Oookaay. \r\n\r\nDoes anyone know anything about the [i]and[/i] and [i]or[/i] circumstances? Not that it's really all that relevant..."
}
{
  "Tag": [
    "function",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Prove that for each positive integer n and each number t in the range (1/2, 1) there exist numbers a,b in the range (1999, 2000) such that (a^n)/2 + (b^n)/2 < (ta + (1-t)b)^n",
  "Solution_1": "We can take $b=0$ and $a$ lies in $[\\frac{1}{t},t]$ with $t=\\frac{2000}{1999}$\r\n\r\n$f(a)=a^n+1-2(ta+1-t)^n$\r\n\r\n$f'(a)=na^{n-1}-2nt(ta+1-t)^{n-1}$\r\n\r\nbecause $f(1)=n(1-2t) < 0$\r\n\r\nThen exist $u$,$1 \\le u \\le t$ and $f'(x) < 0$  for all $x \\in [1,u]$\r\n\r\n(Because $f'(a)$ is continuous function)\r\n\r\nThen $f(u) \\le f(1)=0$,let $u=a$ and we are done."
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "How do you think this year's AMC 12 A compares to 2001 AMC 12 (arguably the hardest AMC 12). Harder, easier?",
  "Solution_1": "2005 was much, much easier than last year (which probably explains my 34.5 point increase in score).  Of course, I started doing AoPS stuff after last year's AMC.  I took the 2001 about a week ago, and I got 125.5, and that was probably because I had done a few problems before."
}
{
  "Tag": [],
  "Problem": "Calculate: $ (243)^{\\frac35}$",
  "Solution_1": "Recall the identity $ \\left(a^b\\right)^{c}\\equal{}a^{bc}.$ \r\n$ 243\\equal{}3^5$ and applying our identity, we have $ 243^{3/5}\\equal{}\\left(3^5\\right)^{3/5}\\equal{}3^3\\equal{}\\boxed{27}.$",
  "Solution_2": "So basically we have to convert 243 = 3 to the 5th so we can multiply it out ans get 3 to the third. Is there any way to do so without converting 243 to a power?\n"
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "(i) Show that, for $ m > 0$,\r\n\r\n$ \\int_{\\frac{1}{m}}^{m}\\, \\frac{x^{2}}{x\\plus{}1}\\,dx \\equal{} \\frac{(m\\minus{}1)^{3}(m\\plus{}1)}{2m^{2}} \\plus{} \\ln m$.\r\n\r\n(ii) Show by means of a substitution that\r\n\r\n$ \\int_{\\frac{1}{m}}^{m}\\, \\frac{1}{x^{n}(x\\plus{}1)}\\,dx \\equal{} \\int_{\\frac{1}{m}}^{m}\\, \\frac{u^{n\\minus{}1}}{u\\plus{}1}\\,du$.\r\n\r\n(iii) Evaluate \r\n\r\n(a) $ \\int_{\\frac{1}{2}}^{2}\\, \\frac{x^{5} \\plus{} 3}{x^{3}(x\\plus{}1)}\\, dx$\r\n\r\n(b) $ \\int_{1}^{2}\\, \\frac{x^{5}\\plus{}x^{3}\\plus{}1}{x^{3}(x\\plus{}1)}\\, dx$.",
  "Solution_1": "[hide=\"(i)\"]\nNote that $ \\frac{x^2}{x+1} = \\frac{(x+1)^2 - 2 (x + 1) + 1}{x+1} = (x+1) - 2 + \\frac1{x+1} = x - 1 + \\frac1{x+1}$.\n\nThus we get:\n\n$ \\int\\limits_{\\frac1m}^m \\frac{x^2}{x+1}dx = \\int\\limits_{\\frac1m}^m \\left[x - 1 + \\frac1{x+1}\\right] dx$\n\n$ = \\left[\\frac12 x^2 - x + \\ln(x + 1)\\right|_{\\frac1m}^m = \\frac12 m^2 - m + \\ln(m + 1) - \\frac1{2 m^2} + \\frac1m - \\ln(1 + \\frac1m)$.\n\nNow note that ${ \\ln(m + 1) - \\ln(1 + \\frac1m) = \\ln( \\frac{1+m}{1 + \\frac1m}}) = \\ln(m \\frac{1 + m}{m + 1}) = \\ln(m)$.\n\nAnd that $ \\frac12 m^2 - m - \\frac1{2 m^2} + \\frac1m = \\frac12 \\left( \\frac{m^4 - 2 m^3 - 1 + 2m}{m^2}\\right) = \\frac{m^4 - 2 m^3 + 2m - 1}{2m^2}$.\n\nFinally, after expanding\n\n$ (m-1)^3(m+1) = (m^2 -1)(m-1)^2 = (m^2 - 1)(m^2 - 2m + 1)$\n$ = m^4 - 2 m^3 + m^2 - m^2 + 2m -1 = m^4 - 2 m^3 + 2m - 1$\n\nwe get the result.\n[/hide]\n\n[hide=\"(ii)\"]\nLet $ u = \\frac1x$, then $ du = - \\frac1{x^2} dx$.\n\nWe get:\n\n${ \\int\\limits_{\\frac1m}^m \\frac1{x^n(x+1)} dx = \\int\\limits_m^{\\frac1m}} \\frac{- u^{n-2}}{1 + \\frac1u} du = \\int\\limits_{\\frac1m}^m \\frac{u^{n-1}}{u + 1} du$.\n[/hide]\r\n\r\nI leave (iii) for someone else."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "probability",
    "geometry",
    "circumcircle",
    "incenter",
    "LaTeX"
  ],
  "Problem": "So iniquitus and I (Jerry and Oliver), think that math is fun.  We want to post problems for others (mostly from GA) to work on.  These should be at about an AIME level.  Who's interested?\r\n\r\nWe'll have about 35 questions.  Sadly, we probably won't give prizes.",
  "Solution_1": "Fun.\r\n\r\nThe message is too small. Please make the message longer before submitting.",
  "Solution_2": "I'll go for it.",
  "Solution_3": "Ok, I'll swear my death oath.  :P",
  "Solution_4": "May I participate as well, although I'm not from Georgia?  :P",
  "Solution_5": "@funcia- the math marathon won't be THAT hard.\r\n\r\n@xpmath- Sure.",
  "Solution_6": "Bring it. :ninja:",
  "Solution_7": "Sitan, you're Chinese.  Thus, you are not a ninja.\r\n\r\nTo keep this post on-topic, the last seven questions might be more proof-based than the first 28, which are straight up answers.",
  "Solution_8": "I'm in. :cleaning: \r\n\r\nI've always wanted to use that smiley. -points-",
  "Solution_9": "Hi ! i would like to participate in the marathon too. :lol:",
  "Solution_10": "Wait, how are we doing this? All 35 at once or like in rounds or like the first one to put Sitan in a trash can wins?\r\n\r\nIm confuzzled.  :huh:",
  "Solution_11": "I am too pro for all of you. Sitan, my name is copyrighted from now on.",
  "Solution_12": "Frisbeeballer- are you Heng?  I've never seen you before.\r\n\r\nArvind- Seriously?  Of course it's the first one to put Sitan in the trash can.\r\n\r\nThe problems will probably be given out at a rate of seven per week.  Thus, it's almost like a problem of the day.\r\n\r\nAnd they should also be published REALLY soon...we're making the final edits to the first round.",
  "Solution_13": "Yes, it's HENG TONG (there, I said it, wut are you going to do?).  He told me.\r\n\r\nAnd you can tell by the username.",
  "Solution_14": "Waaaaaaait.\r\n\r\nSo like 6 normal and 1 proof every week?\r\n\r\nor one week w/ 7 proofs? that's hardcore.",
  "Solution_15": "[hide=\"hints\"]1. conditional probability\n\n2. (failed) seems bruteforce-able with coordinates + shoelace...anyone knows of a geometry solution?\n\n3. let all three logs equal x\n\n4. newton's sums\n\n5. consider 200th roots of unity\n\n6. recursion\n\n7. split into prime power moduli and solve\n[/hide]",
  "Solution_16": "[hide]Actually, If you used coordinates on #2, you didn't need the shoelace theorem since one side was parallel to the 14-side. So, you could use the (1/2)base*height formula[/hide]",
  "Solution_17": "The answers for set 1 are:\r\n\r\n1. $ 144$\r\n2. $ 244$\r\n3. $ \\sqrt{2}\\minus{}1$\r\n4. $ 528$\r\n5. $ 60501$\r\n6. $ 133$\r\n7. $ 373$",
  "Solution_18": "Feel free to post solutions if you want to; if I don't see any being posted up, I will post up mine 1 by 1, in no particular order. The second round will probably start next Monday.\r\n\r\n\r\n[hide=\"Solution to Question 6\"]\n\nLet the 200-gon be $ A_1 A_2...A_{200}$.\n\n\nLemma: \n\n$ |\\overline{A_1 A_2}| \\times |\\overline{A_1 A_3}| \\times... \\times |\\overline{A_1 A_{200}}| = 200$\n\nProof:\n\n$ x^{200}-1=(x-1)(x-w)(x-w^2)...(x-w^{199})$, where $ w=e^{\\frac{2\\pi i}{200}}$\n\nDividing $ x-1$, we get\n$ x^{199}+x^{198}+...+x^0 = (x-w)(x-w^2)...(x-w^{199})$\n\nPlug in $ x=1$ to get\n$ 200 = (1-w)(1-w^2)...(x-w^{199})$\n\nNoting that $ 1,w,w^2,...,w^{199}$ are the vertices of a regular 200-gon in the complex plane, take the magnitude of both sides to achieve the desired result.\n\n\nThe number we want to find, by the lemma, is $ \\sqrt{200^{200}} = 200^{100}=2^{300}5^{200}$\nThis number has $ 60501$ factors.\n[/hide]",
  "Solution_19": "[hide=\"Solution_to_Question_3\"]Let $ \\log_9 (a) \\equal{} \\log_{15} (b) \\equal{} \\log_{25} {(a\\plus{}2b)} \\equal{} x.$ Then $ 9^x\\equal{}a, 15^x\\equal{}b, 25^x\\equal{}a\\plus{}2b.$ You can see that \\[ 9^x*25^x\\equal{}(15^x)^2\\] \\[ a(a\\plus{}2b)\\equal{}b^2\\] \\[ a^2\\plus{}2ab\\plus{}b^2\\equal{}(a\\plus{}b)^2\\equal{}2b^2\\] \\[ a\\plus{}b\\equal{}\\sqrt 2 b\\] \\[ \\frac{a}{b}\\equal{}\\boxed{\\sqrt 2 \\minus{} 1}\\] Note that it can't be $ a\\plus{}b\\equal{}\\minus{}\\sqrt 2 b$ because then it implies the ratio is negative, which is impossible looking at the original equations.[/hide]",
  "Solution_20": "[hide=\"A MEDIOCRE SOLUTION TO QUESTION 4\"]Let $ x=r_n$ for any arbitrary value $ n$ of the equation such that $ r_n$ is a root of the equation and such that $ n$ is any positive integral value between $ 1$ and $ 16$, inclusive.  Rearranging the equation to find the values of $ r^{16}_n$ and $ r^{15}_n$ gives us the values of $ 16r_n-18$ and $ 16-\\frac {18} {r_n}$.  Subtracting the second value from the first gives us:\n\n\\[ \\displaystyle\\sum_{n=1}^{16}{r^{16}_n-r^{15}_n}=\\displaystyle\\sum_{n=1}^{16}{16r_n-34+\\frac {18} {r_n}}\\].\n\nSeparating the sum into the three parts $ 16r_n$, $ -34$, and $ \\frac {18} {r_n}$ and finding these values individually, we know using Vieta's that the first value is equal to $ 0$, the second is $ -34\\cdot16=-544$, and the third can be expressed as $ 18$ multiplied by the fifteenth(I think) symmetric sum of the roots and divided by the product of the roots, or $ 16$ as it turns out.\n\nHence, the final value is $ |16-544|=\\boxed {528}$. $ \\blacksquare$[/hide]\r\n\r\nApologies for the ugliness of language, lack of adequate explanation, and overall failure. ;\\",
  "Solution_21": "[quote]1. Jerry has $ 3$ bags of aerodynamic marbles. Each marble is either tan or beige. Bag A has $ 40$ marbles, Bag B has $ 50$ marbles, and Bag C has $ 60$ marbles. Jerry knows that if he takes Bag A and Bag B and draws a tan marble from them, the probability that it came from Bag A is $ 25/77$. If he takes Bag B and Bag C and draws a tan marble from them, the probability it came from Bag B is $ 13/38$. If he takes Bag C and Bag A and draws a tan marble from them, the probability it came from Bag C is $ 4/5$. What is the sum of the possible total numbers of tan marbles that Jerry has in his bags? \n[/quote]\r\n\r\nLet the number of tan marbles in bag $ A$ be $ x$\r\n\r\nLet the number of tan marbles in bag $ B$ be $ y$\r\n\r\nLet the number of tan marbles in bag $ C$ be $ z$\r\n\r\nGiven $ (x/40)/(x/40) \\plus{} (y/50)$ = $ 25/77$\r\n\r\nImplies $ 13x/10$ = $ y/2$ \r\n\r\nImplies $ y \\equal{} 13x/5$\r\n\r\nFrom the second statement we have $ (y/50)/(y/50) \\plus{} (z/60)$ = $ 13/38$\r\n\r\nWe have $ y/2$ = $ 13z/60$\r\n\r\nImplies $ y \\equal{} 13z/30$\r\n\r\nThe third equation i felt is extraneous.\r\n\r\nSo since the number of marbles is integral we see z is a multiple of 30.\r\n\r\nCase 1: z is 30 \r\n\r\nthen y is 13.\r\n\r\nand x is 5.\r\n\r\nCase 2: z is 60.\r\n\r\nthen y is 26.\r\n\r\nand x is 10.\r\n\r\nSumming all the possibilities the sum of  number of possible marbles is 144. :D\r\n\r\nP.S. come on guys post solutions for some problems.",
  "Solution_22": "[quote=\"iniquitus\"]Feel free to post solutions if you want to; if I don't see any being posted up, I will post up mine 1 by 1, in no particular order. The second round will probably start next Monday.\n\n\n[hide=\"Solution to Question 6\"]\n\nLet the 200-gon be $ A_1 A_2...A_{200}$.\n\n\nLemma: \n\n$ |\\overline{A_1 A_2}| \\times |\\overline{A_1 A_3}| \\times... \\times |\\overline{A_1 A_{200}}| = 200$\n\nProof:\n\n$ x^{200} - 1 = (x - 1)(x - w)(x - w^2)...(x - w^{199})$, where $ w = e^{\\frac {2\\pi i}{200}}$\n\nDividing $ x - 1$, we get\n$ x^{199} + x^{198} + ... + x^0 = (x - w)(x - w^2)...(x - w^{199})$\n\nPlug in $ x = 1$ to get\n$ 200 = (1 - w)(1 - w^2)...(x - w^{199})$\n\nNoting that $ 1,w,w^2,...,w^{199}$ are the vertices of a regular 200-gon in the complex plane, take the magnitude of both sides to achieve the desired result.\n\n\nThe number we want to find, by the lemma, is $ \\sqrt {200^{200}} = 200^{100} = 2^{300}5^{200}$\nThis number has $ 60501$ factors.\n[/hide][/quote]\r\n\r\nUh, Oliver, don't you mean solution to question 5?",
  "Solution_23": "Yep, so here's 6:\r\n\r\n[hide=\" Question 6\"]\nThe number of ways to put n stamps on, or $ a_n$, satisfies the recurrence relation $ a_n \\equal{} a_{n\\minus{}1} \\plus{} 6a_{n\\minus{}2}$.\n\nBy the characteristic equation $ x^2 \\minus{} x \\minus{} 6$, we have\n\n$ a_n \\equal{} A(3)^n \\plus{} B (\\minus{}2)^n$\n\nWe know that $ a_0 \\equal{} 1$ and $ a_1 \\equal{} 1$.\nPlugging these in, we get\n\n$ 1 \\equal{} A\\plus{}B$, and\n$ 1 \\equal{} 3A \\minus{} 2B$.\n\nThis gives A = 3/5, B = 2/5\n\nso $ a_{1337} \\equal{} \\frac{3}{5} 3^{1337} \\plus{} \\frac{2}{5} (\\minus{}2)^{1337} \\equal{} \\frac{1}{5}(3^{1338} \\minus{} 2^{1338})$\n\nConveniently, 223 is prime, so $ n^{222} \\equiv 1 \\;\\; (\\text{mod} 223)$\n\nTherefore, $ a_{1337} \\equiv \\frac{1}{5}( 3^6 \\minus{} 2^6) \\equiv \\frac{1}{5}( 3^3 \\minus{} 2^3)(3^3\\plus{}2^3) \\equiv (19)(7) \\equiv 133 \\;\\; (\\text{mod} 223)$[/hide]",
  "Solution_24": "The solution I have to number 2 is not very \"elegant,\" but it works.\r\n\r\n[hide]A $ 13 \\minus{} 14 \\minus{} 15$ triangle can be broken into a $ 9 \\minus{} 12 \\minus{} 15$ and a $ 5 \\minus{} 12 \\minus{} 13$ triangle, which makes it very easy to use coordinates for the triangle.  Thus, place the vertexes of the triangle at $ (0,0), (5,12), (14,0)$.\n\nThe coordinates of the centroid are given as the average of the x and y values in the coordinate plane.  So the centroid is at $ (\\frac {19}{3},4)$.\n\nThe incenter of the triangle can be found from the radius of the incircle.  This is given as $ A \\equal{} rs$, where $ A$ is the area of the triangle, $ r$ is given as the radius, and $ s$ is the semiperimeter.  $ A \\equal{} 84$, $ s \\equal{} 21$, so $ r \\equal{} 4$.  Now we need to find where the incircle meets the triangle in the coordinate system.  Use the two tangent theorem to discover that, WLOG, $ a \\plus{} b \\equal{} 13, b \\plus{} c \\equal{} 14, a \\plus{} c \\equal{} 15$.  Solving, we discover that the x coordinate of the incenter is $ 6$.  Since it is perpendicular to the triangle, the incenter is at $ (6,4)$.\n\nFinally, to solve for the circumcenter, we will place the three vertexes of the triangle into the equation of a circle.  $ (x \\minus{} h)^2 \\plus{} (y \\minus{} k)^2 \\equal{} r^2$.  Substitute $ (0,0), (14,0)$ to get that $ x \\equal{} 7$,  Then, substitute in $ (5,12)$ and solve to get that $ y \\equal{} \\frac {33}{8}$.\n\nWith the three coordinates, we can solve for the area.  $ A \\equal{} \\frac {1}{2}bh$.  $ b \\equal{} \\frac {1}{3}, h \\equal{} \\frac {1}{8}$.  $ 4*1 \\plus{} 5*48 \\equal{} 244$.[/hide]",
  "Solution_25": "[hide=\"Solution to Question 7\"]We first split 1323 into $ 3^3$ and $ 7^2$, and then use CRT to work our way back up.\n\n1)\n\n$ 13x^7 \\minus{}42x \\plus{} 674 \\equiv 0 \\;\\; (\\text{mod} 9)$, and it's obvious that x=0 is not a solution.\n$ 4x^7 \\plus{} 3x \\plus{}8 \\equiv 0 \\;\\; (\\text{mod} 9)$\n$ 4x \\plus{} 3x \\plus{} 8 \\equiv 0 \\;\\; (\\text{mod} 9)$ by Euler's Theorem\n$ 7x\\plus{}8 \\equiv 0 \\;\\; (\\text{mod} 9)$\n$ x \\equiv 4 \\;\\; (\\text{mod} 9)$.\n\nThen, by Hansel's lemma, we get the solution\n\n$ x \\equiv 4 \\plus{} 2\\times 9 \\equal{} 22 \\;\\; (\\text{mod} 27)$\n\n2)\n\n$ 13x^7 \\minus{}42x \\plus{} 674 \\equiv 0 \\;\\; (\\text{mod} 7)$, and it's obvious that x=0 is not a solution.\n$ \\minus{}x^7 \\plus{}2 \\equiv 0 \\;\\; (\\text{mod} 7)$\n$ \\minus{}x \\plus{} 2 \\equiv 0 \\;\\; (\\text{mod} 7)$\n$ x \\equiv 2 \\;\\; (\\text{mod} 7)$\n\nBy Hansel's lemma, since the derivative is a multiple of 7 and the function of 2 is a multiple of 49, the solution is\n\n$ x \\equiv 2 \\;\\; (\\text{mod} 7)$\n\n\nCombining these with CRT, we have:\n\n$ x(7\\plus{}27) \\equiv (\\minus{}5)(7) \\plus{} (2)(27) \\equiv 19 \\equiv 19 \\minus{} 189 \\equiv \\minus{}170 \\;\\; (\\text{mod} 189)$\n$ x \\equiv \\minus{}5 \\;\\; (\\text{mod} 189)$, so the desired answer is $ 2 \\times 189 \\minus{} 5 \\equal{} 373$\n\n\n(In both cases, you actually only had to check 2 values, so if you didn't know Hansel's lemma, you could've used either mod manipulation or just checked 4, -5 in part (1) and 2, -5 in part (2))\n[/hide]",
  "Solution_26": "Once Again:\r\n\r\n\r\n\r\nSome rules:\r\n\r\nPlease do not look questions up on the internet.\r\nPlease do not use a calculator.\r\nThis is not competitive so don't waste your time brute forcing.\r\nPM your answers to me (no proofs or justifications required, just answers)\r\n\r\nSET 2\r\n\r\nYou have until Sunday (January 11) 11:59:59\r\n\r\n[hide=\"GOOD LUCK!\"]\n1. This is how Oliver does a long division problem.\n\n$ .\\;\\;\\;\\;\\;\\;322$\n$ 31 \\big)\\hspace{ \\minus{} 0.3em}\\overline{\\hspace{0.3em} 9982}$\n$ . \\;\\;\\,\\: \\underline{\\hspace{0.3em}93}$\n$ . \\;\\;\\;\\;\\;\\;68$\n$ . \\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}62}$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;62$\n$ . \\;\\;\\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}62}$\n\nOliver has also done a longer long division problem, but he has accidentally erased all the digits except for seven sevens (He could\u2019ve erased some 7\u2019s as well, so some asterisks may be 7\u2019s). The missing digits are marked below as asterisks and as the two letters $ A$ and $ B$. What is the ordered pair $ (A,B)$?\n\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\: **\\: 7AB$\n$ ****7* \\big)\\hspace{ \\minus{} 0.3em}\\overline{\\hspace{0.3em} **7******\\: *\\ }$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}*****\\: *}$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;*****7\\: *$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}*******}$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\,*7****$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}*\\: 7****}$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\,******\\: *$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}****7**}$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\,******$\n$ . \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\underline{\\hspace{0.3em}*****\\: *}$\n\n2. Find a function $ f(x)$ that satisfies $ f^{2009}(x) \\equal{} \\frac {x}{x \\plus{} 1}$ \n(Remember that $ f^n (x) \\equal{} f(f(f(..._{\\text{n times}} ...(x))...)$)\n\n3. Find the number of ways to make a 12 letter \u2018word\u2019 from 3 A\u2019s, 3 B\u2019s, 3 C\u2019s, 3 D\u2019s, and 3 E\u2019s. How many factors does this number have?\n\n4. Let $ x, y, z$ be real numbers such that $ x \\plus{} y \\plus{} z \\equal{} 6$ and $ xy \\plus{} yz \\plus{} zx \\equal{} a$ for some real parameter $ a$. What is the value of $ a$ such that the difference between the maximum and minimum possible values of $ x$ is $ 12$?\n\n5. Convex nondegenerate Pentagon $ ABCDE$ is inscribed in a circle $ O$ with radius of $ 1/2$. The angles of the pentagon have the relations \n$ \\sin A \\plus{} \\sin B \\equal{} 2 \\sin (C/2)\\cos(C/8)$,\n$ \\sin D \\plus{} \\sin E \\equal{} 2(\\sin [C \\plus{} 30])(\\cos C/8)$,\n$ \\sin B \\plus{} \\sin E \\equal{} 2 \\sin C \\cos A/2$. \nFurthermore, $ A < B < C < D < E$. \n\nIf the sum of all the products of the lengths of 2 line segments (formed from the five vertices) that do not share a common vertex (For example, $ AB*CD$ is counted, but $ AB*BC$ is not) can be expressed as $ \\frac {a \\sqrt {b} \\plus{} c\\sqrt {d} \\plus{} e\\sqrt {f} \\minus{} g}{h}$, where $ h$ is the smallest denominator such that a single fraction can express the sum, find $ a \\plus{} b \\plus{} c \\plus{} d \\plus{} e \\plus{} f \\plus{} g \\plus{} h$\n\n6. Jerry is hosting an ultimate Frisbee tournament, but only fifteen people show up. Fortunately, the field is so small that Jerry knows a fair ultimate Frisbee game can be played as long as a team has five people. As he is about to divide teams, he realizes that he forgot to distribute Frisbee jerseys for the team. He has four colors of jerseys and only five of each. How many possible combinations of team players with jerseys are there? Assume all team members are distinct.\n\n7. $ P(x)$ is a monic (leading coefficient is 1) polynomial that satisfies $ (x^3 \\plus{} 3x^2 \\plus{} 3x \\plus{} 2)P(x \\minus{} 1) \\equal{} (x^3 \\minus{} 3x^2 \\plus{} 3x \\minus{} 2)P(x)$. Find $ P(3)$.[/hide]",
  "Solution_27": "Question for #6: How many teams are there, how many people are in each team, and do all team members have to have the same colored jersey? Or do I totally misunderstand the problem...",
  "Solution_28": "Sorry about the confusion.\r\n\r\nThere are only two teams.\r\n\r\nA team does not have to have the same colored jersey (in fact, it can't, since there's only four of one color, and a team must have at least five people).",
  "Solution_29": "Hey guys, sorry if you've already submitted your answers.\r\n\r\nFor the team and jersey problem, it involves a lot of computation that is ugly without a calculator.\r\n\r\nThe following is a QUESTION CHANGE!\r\n\r\nInstead of four colors of jerseys, suppose he only has two colors of jerseys.  Also suppose he only has three of each type of jersey.  THERE ARE NOW ONLY SIX JERSEYS!\r\n\r\nAGAIN, I apologize for the poorly created question.  It's doable, but very very very ugly. :oops_sign:  :oops_sign:"
}
{
  "Tag": [],
  "Problem": "The product of two consecutive even integers is 32 less than the square of the greatest integer. Find the smaller integer.",
  "Solution_1": "$ n(n\\plus{}2)\\equal{}(n\\plus{}2)^2\\minus{}32$\r\n$ n^2\\plus{}2n\\equal{}n^2\\plus{}4n\\minus{}28$\r\n$ 2n\\equal{}28$\r\n$ n\\equal{}14$."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "LEt $ p>3$ be a prime number. Prove that each of the divisors (bigger than $ 1$) of $ \\frac{2^{p}+1}{3}$  are in the form $ 2pt+1$ where $ t$ is a positive integer. :)",
  "Solution_1": "It is enough to prove that for prime divisors. Let $ q$ be one of them. Then obviously $ q \\not = 2$ and $ q \\not = 3$ since if $ 9|2^{a}+1$ then $ 3|a$. We have $ 2^{p}\\equiv-1 \\mod{q}$ and $ 2^{2p}\\equiv 1 \\mod{q}$. Let $ k$ be the order of $ 2$ $ \\mod{q}$. Then $ k|2p$ and $ k \\not | p$, so $ k=2$ or $ k=2p$ since $ p$ is prime. But $ k \\not = 2$ since we would have that $ 2 \\equiv-1 \\mod{q}$, so $ q=3$. It means that $ k=2p$ and $ 2p|q-1$, conclusion follows."
}
{
  "Tag": [
    "inequalities",
    "triangle inequality",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "slove this system of equation: $ \\begin {cases} x^2\\plus{}y^2\\equal{}\\frac{82}{9}\\\\\\displaystyle \\left | y\\plus{}\\frac{1}{y}\\right |\\plus{}\\displaystyle \\left |\\frac{10}{3}\\minus{}x\\plus{}y \\right |\\equal{}\\frac{10}{3}\\plus{}y\\plus{}\\frac{1}{y}\\end {cases}$",
  "Solution_1": "(First post here!)\r\nEDIT EDIT: I've gotten rid of the old solution. I've got to stop making careless mistakes.\r\n[hide=\"Solution\"]Notice that \n\n$ \\displaystyle \\left | y \\plus{} \\frac {1}{y} \\right | \\ge y \\plus{} \\frac {1}{y}$\n\n$ \\displaystyle \\left | \\frac {10}{3} \\minus{} x \\plus{} y \\right | \\ge \\frac {10}{3}$\n\nClearly $ \\displaystyle \\left | y \\plus{} \\frac {1}{y}\\right | \\plus{} \\displaystyle \\left |\\frac {10}{3} \\minus{} x \\plus{} y \\right | \\equal{} \\frac {10}{3} \\plus{} y \\plus{} \\frac {1}{y}$ iff equality holds in both inequalities. So $ y \\plus{} \\frac {1}{y} \\ge 0$ (implying $ y > 0$) and \n\n$ \\displaystyle \\left | \\frac {10}{3} \\minus{} x \\plus{} y \\right | \\equal{} \\frac {10}{3} \\implies \\minus{} x \\plus{} y \\equal{} 0 \\text{ OR } \\minus{}x \\plus{} y \\equal{} \\frac{\\minus{}20}{3}\\implies x \\equal{} y \\text{ OR } x \\minus{} y \\equal{} \\frac{20}{3}$\n\nThe $ x \\equal{} y$ case is easy. \n\nIf $ x \\minus{} y \\equal{} 20/3$, then note that the max for $ x$ is $ \\frac{\\sqrt{82}}{3}$, and the min for $ \\minus{}y$ is $ \\minus{}\\frac{\\sqrt{82}}{3}$. So the max for $ x \\minus{} y$ is $ 2\\sqrt{82}/3$, but this less than $ 20/3$.\n\nWe're not supposed to consider imaginary numbers, right?[/hide]",
  "Solution_2": "[quote=\"vishalarul\"]\n\n\nIf $ | \\minus{} x \\plus{} y| > 0$ then this would mean that:\n\\[ \\displaystyle \\left |\\frac {10}{3} \\minus{} x \\plus{} y \\right | > \\frac {10}{3} \\plus{} | \\minus{} x \\plus{} y| > \\frac {10}{3}\n\\]\n\n\n[/quote]\r\nI think there is a very small mistake in this step... :wink:",
  "Solution_3": "[quote=\"ali666\"][quote=\"vishalarul\"]\n\n\nIf $ | \\minus{} x \\plus{} y| > 0$ then this would mean that:\n\\[ \\displaystyle \\left |\\frac {10}{3} \\minus{} x \\plus{} y \\right | > \\frac {10}{3} \\plus{} | \\minus{} x \\plus{} y| > \\frac {10}{3}\n\\]\n[/quote]\nI think there is a very small mistake in this step... :wink:[/quote]\r\n? :P",
  "Solution_4": "vishalarul means $ \\minus{} x \\plus{} y > 0$.  And in that case the first inequality is an equality.  (The triangle inequality goes in the other direction.)  Unfortunately, nothing is said about the case that $ \\minus{} x \\plus{} y < 0$.",
  "Solution_5": "[quote=\"thanhnam2902\"]slove this system of equation: $ \\begin {cases} x^2 \\plus{} y^2 \\equal{} \\frac {82}{9} \\\\\n\\displaystyle \\left | y \\plus{} \\frac {1}{y}\\right | \\plus{} \\displaystyle \\left |\\frac {10}{3} \\minus{} x \\plus{} y \\right | \\equal{} \\frac {10}{3} \\plus{} y \\plus{} \\frac {1}{y}\\end {cases}$[/quote]\r\nHope it is correct.\r\n[hide]Case 1: $ y > 0$\n\nWe have then $ |\\frac {10}{3} \\minus{} x \\plus{} y| \\equal{} \\frac {10}{3}$\n\n 1.1: suppose that $ \\frac {10}{3} \\minus{} x \\plus{} y > 0$, then we get $ x \\equal{} y$ \n\n 1.2: else we get $ x \\minus{} y \\equal{} 20/3$\n\nCase 2: $ y < 0$\n \n Our result is then $ |\\frac {10}{3} \\minus{} x \\plus{} y| \\equal{} \\frac {10}{3} \\plus{} 2(y \\plus{} \\frac {1}{y})$, we get something absurd here because\n\\[ 2(y \\plus{} \\frac {1}{y})\\leq \\minus{} 4\n\\]\n[/hide]",
  "Solution_6": "[quote=\"Amazigh\"][quote=\"thanhnam2902\"]slove this system of equation: $ \\begin {cases} x^2 \\plus{} y^2 \\equal{} \\frac {82}{9} \\\\\n\\displaystyle \\left | y \\plus{} \\frac {1}{y}\\right | \\plus{} \\displaystyle \\left |\\frac {10}{3} \\minus{} x \\plus{} y \\right | \\equal{} \\frac {10}{3} \\plus{} y \\plus{} \\frac {1}{y}\\end {cases}$[/quote]\nHope it is correct.\n[hide]Case 1: $ y > 0$\n\nWe have then $ |\\frac {10}{3} \\minus{} x \\plus{} y| \\equal{} \\frac {10}{3}$\n\n 1.1: suppose that $ \\frac {10}{3} \\minus{} x \\plus{} y > 0$, then we get $ x \\equal{} y$ \n\n 1.2: else we get $ x \\minus{} y \\equal{} 20/3$\n\nCase 2: $ y < 0$\n \n Our result is then $ |\\frac {10}{3} \\minus{} x \\plus{} y| \\equal{} \\frac {10}{3} \\plus{} 2(y \\plus{} \\frac {1}{y})$, we get something absurd here because\n\\[ 2(y \\plus{} \\frac {1}{y})\\leq \\minus{} 4\n\\]\n[/hide][/quote]\r\n\r\nTHANK YOU VERY MUCH!"
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "let $f(x)=sinx+sin(x\\sqrt{2})$\r\nprove$f$ is not period  function .",
  "Solution_1": "We haved see that: If $f(x)=asinux+bsinvx$ is period function iff $\\frac{u}{v}$ is rational number.\r\nHence applied the event to this problem too clearly"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "AMC"
  ],
  "Problem": "Let $ M_{a}$ be the $ a$ by $ a$ matrix for which every entry below the main diagonal is $ 0$ and the rest of the entries are 1's. If $ N=(M_{4})^{0}+(M_{4})^{1}+(M_{4})^{2}+...+(M_{4})^{2398}$, and the sum of the entries of $ N$ is $ A$, then find the remainder when $ A$ is divided by $ 1000$.",
  "Solution_1": "[hide=\"hint\"]If an $ a \\times a$ matrix $ M$ represents an a-state system in which the number of ways to get from state $ i$ to state $ j$ in one step is the element in the $ i$th row and $ j$th column, then the $ a \\times a$ matrix $ M^{k}$ represents the set of numbers of ways to get from one state to another in $ k$ steps.\n[/hide]\n\n[hide=\"solution\"]\nThe matrix $ M_{a}$ represents an $ a$-state system in which any step taken goes from a state to one strictly preceding it.  Then, because no paths of length greater than or equal to $ a$ can be taken, we have that $ M_{a}^{k}$ is a zero matrix for $ k \\ge a$.  Therefore, all we need to calculate is $ M_{4}^{0}+M_{4}+M_{4}^{2}+M_{4}^{3}$.  The sume of the entries in the sum of these is the sum of the sum of entries in each, because there are no negative entries.  The sum of the entries of $ M_{4}^{0}$ is 4, because it is the identity matrix.  The next one gives $ 6$, the next gives $ 3$, and the last $ 1$, for a total of $ \\boxed{14}$.\n\nIn general, the sum of the entries of $ M_{a}^{k}$ for $ 0<k < a$ is the $ a-k$th triangular number.[/hide]",
  "Solution_2": "Err...I don't think that is correct. The main diagonal has ones. Hence you can stay at the same place. \r\n\r\nI was pretty proud of this problem. It is correct to interpret this as a graph. There is a very nice bijection.\r\n\r\nEDIT:\r\n\r\n[I flipped the matrix since that makes the meaning more clear; it obviously doesn't change the sum.]\r\n\r\n1 0 0 0\r\n1 1 0 0\r\n1 1 1 0\r\n1 1 1 1 \r\n\r\nIf we consider this as an adjacency matrix of a directed graph, then we have the following edges.\r\n\r\n1->1,2,3,4\r\n2->2,3,4\r\n3->3,4\r\n4->4\r\n\r\nWe are creating a bijection. The first number in the sequence is where you start [a number from 1 to 4]. Each turn can be represented by a 0,1,2,3,4, representing how many spaces that you jumped. The sum of all these entries must be less than or equal to 4. Consider another dummy spot at the end that can take up the rest of the 1's. So we have: \r\n[1 to 4] {[0 to 4] [0 to 4]....[0 to 4] } {a moves} [0 to 4]\r\n\r\nIn otherwords, we are placing 3 items in a+2 places, so this is a simple counting problem. There are $ \\binom{a+2+3-1}{3}$.\r\n\r\nWe have to sum this over $ a=0,1,...,2398$. I remember that there was a slick way to circumvent this, but anyway, we can just sum them with basic binomial stuff, we get \r\n\r\n$ \\binom{2398+4}{4}=2402*2401*100*2399\\equiv 200\\mod 1000$.",
  "Solution_3": "[quote=\"Altheman\"]Err...I don't think that is correct. The main diagonal has ones. Hence you can stay at the same place. \n[/quote]\r\n\r\nThat's a good point.  I switched the 1's and 0's (I thought it was everything below the main daigonal is a 1, and everything else is a 0).  The argument gets a little more complicated but is not that much more difficult.  You can calculate the number of paths of length $ k$ which take $ m$ backwards steps: it's just $ \\binom{k}{m}$.  Then considering there are $ a-m$ places you can start, we find that the sum of the entries of $ M_{a}^{k}$ is $ \\sum_{m=0}^{a-1}(a-m)\\binom{k}{m}$.",
  "Solution_4": "[quote=\"K81o7\"][quote=\"Altheman\"]Err...I don't think that is correct. The main diagonal has ones. Hence you can stay at the same place. \n[/quote]\n\nThat's a good point.  I switched the 1's and 0's (I thought it was everything below the main daigonal is a 1, and everything else is a 0).  The argument gets a little more complicated but is not that much more difficult.  You can calculate the number of paths of length $ k$ which take $ m$ backwards steps: it's just $ \\binom{k}{m}$.  Then considering there are $ a-m$ places you can start, we find that the sum of the entries of $ M_{a}^{k}$ is $ \\sum_{m=0}^{a-1}(a-m)\\binom{k}{m}$.[/quote]\r\n\r\nWell, right you can go through that whole mess to evaluate all the sums. I wish I could remember how to do the argument by one bijection. Maybe ask JSteinhardt, I think that he helped me check this problem when I made it. \r\n\r\n\r\nOn a slightly different note, I totally forgot about that mock test. I don't think I ever put up results because not enough people took it. It still had a lot of nice problems though. It is nice to see some of them show up on the forums again.",
  "Solution_5": "Wait isn't the answer $ \\binom{2403}{4}-1$?\r\n\r\n[hide]\nWe want the number of nondecreasing sequences consisting of the numbers 1-4 and having at most 2399 terms (each sequence corresponds to the vertices of a path in the graph Altheman described). If we let a be the number of 1's, b be the number of 2's, etc., then we want the number of nonnegative integer solutions to $ 1 \\le a+b+c+d \\le 2399$. If we define the nonnegative integer $ e=2399-(a+b+c+d)$, we can turn this into the equality $ a+b+c+d+e=2399$. The number of solutions to this is $ \\binom{2399+4}{4}$, but we have to throw out $ (0,0,0,0,2399)$ because $ a+b+c+d \\ge 1$. So the final answer is $ \\binom{2403}{4}-1$, which I think is 599 mod 1000.\n[/hide]",
  "Solution_6": "[quote=\"Altheman\"]\nIf we consider this as an adjacency matrix of a directed graph, then we have the following edges.\n\n1->1,2,3,4\n2->2,3,4\n3->3,4\n4->4\n\nWe are creating a bijection. The first number in the sequence is where you start [a number from 1 to 4]. Each turn can be represented by a 0,1,2,3,4, representing how many spaces that you jumped. The sum of all these entries must be less than or equal to 4. Consider another dummy spot at the end that can take up the rest of the 1's. So we have: \n[1 to 4] {[0 to 4] [0 to 4]....[0 to 4] } {a moves} [0 to 4]\n\n[/quote]\r\n\r\nWell, actually I don't quite get what you're doing. How come the adjacency matrix (what is it?) and the edges?  :blush: \r\n\r\nCan you help explain it to me?\r\n\r\nEDIT: OK, I understand now  :D",
  "Solution_7": "[quote=\"Gaku Liu\"]Wait isn't the answer $ \\binom{2403}{4}-1$?\n\n[hide]\nWe want the number of nondecreasing sequences consisting of the numbers 1-4 and having at most 2399 terms (each sequence corresponds to the vertices of a path in the graph Altheman described). If we let a be the number of 1's, b be the number of 2's, etc., then we want the number of nonnegative integer solutions to $ 1 \\le a+b+c+d \\le 2399$. If we define the nonnegative integer $ e=2399-(a+b+c+d)$, we can turn this into the equality $ a+b+c+d+e=2399$. The number of solutions to this is $ \\binom{2399+4}{4}$, but we have to throw out $ (0,0,0,0,2399)$ because $ a+b+c+d \\ge 1$. So the final answer is $ \\binom{2403}{4}-1$, which I think is 599 mod 1000.\n[/hide][/quote]\r\n\r\nI think that the M^0 term includes the 'no terms in the sequence'.",
  "Solution_8": "[quote=\"Altheman\"]I think that the M^0 term includes the 'no terms in the sequence'.\n[/quote]\r\nWell, my sequence was listing the vertices encountered in a walk, not the edges. So M^n would correspond to sequences with n+1 terms. M^0 would correspond to walks without an [i]edge[/i]; but they would still have a starting vertex and therefore the sequence would have 1 term. Besides, the sum of the elements of M^0 is 4, which is the number of sequences with 1 term."
}
{
  "Tag": [
    "FTW",
    "MATHCOUNTS",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "algorithm"
  ],
  "Problem": "So, this is a tournament. In the tournament, there willl be 8 problems, 2 school/chapter difficulty, 3 state-national difficulty, 3 national and beyond. You will be each given 2 problems at the beginning. After solving a problem, pm me the answers and I will pm the next problem if you are correct. You will also gain a point. If you are incorrect, you will lose a point and will not get the next problem. (But you can try to answer the problem again, if you are correct this time you get the next problem: Otherwise you are done with the round)\r\n\r\nWhen you finish all 8 problems, you will get 2 extra points.\r\n\r\nYou have a week to solve the problems. After the week, the people with the lowest score will play a tiebrake if lowest score was less than 6. (If the lowest score is 6 or above, then nobody is eliminated) The tiebraker will consist of 5 state-national level problems. Everyone with the lowest score on that will be eliminated.\r\n\r\nWhen there are 4 people left, then they are matched up against each other and have an elimination tournament:\r\n\r\nThe format will be the same, except that if both people tie on the tiebraker, then they have a FTW countdown match.\r\n\r\nThe tournament will start on August 20 or when we get 20 people.\r\n\r\nTo signup, post here or PM me.\r\n\r\nSignups:\r\n1. shentang\r\n2. Poincare",
  "Solution_1": "I WANNA PLAY!!!  :D",
  "Solution_2": "ill join  :D",
  "Solution_3": "On second thought, I'll just post all 8 problems at once.\r\n\r\nSignups:\r\n1. shentang\r\n2. Poincare\r\n3. FantasyLover\r\n4. qwertythecucumber\r\n5. 5849206328x",
  "Solution_4": "dude when pythag says mathcounts he means IMO #1 or 4 (or since he doesn't want proofs, AIME #15)\r\nif you can get 1 point, I'll be impressed\r\n@pythag: I'm watching you (not really, but i know who you are)",
  "Solution_5": "I wanna join!",
  "Solution_6": "me want to join",
  "Solution_7": "i don't mind USAMO or AIME problems but IMO might be getting out of hand!",
  "Solution_8": "There will be no IMO problems except for really really easy ones. Have you seeen the first IMO problem?\r\n\r\nIs goes like:\r\n\r\nProve that gcd(21n + 4, 14n +3) = 1, which is just standard euclidean algorithm on that.\r\n\r\ngcd(21n + 4, 14n + 3) = gcd(7n +1, 14n+3) = gcd(7n+1, 1) = 1.",
  "Solution_9": "WOAH THAT IS PRETTY INSANELY EASY\r\n\r\nLIKE\r\n\r\nI CAN ACTUALLY UNDERSTAND THE PROBLEM\r\n\r\nAND SOLVE IT\r\n\r\n<7 HRS\r\n\r\nTHAT IS A BAD PROBLEM",
  "Solution_10": "BTW, IMO difficulty is about the same as USAMO difficulty.\r\n\r\nIn fact, this year's IMO is probably much easier than this year's USAMO.",
  "Solution_11": "which is generally the case",
  "Solution_12": "Whether easy or not, proofs are beyond MathCounts level. I suggest sticking to MathCounts.  :wink:",
  "Solution_13": "I'll sign up.",
  "Solution_14": "/in, but I'm not really good at math, so I will probably fail :(",
  "Solution_15": "actually I sent the answer to number one to the host.",
  "Solution_16": "neutrus, Poincare, Eulcild_great and 5849206328x are eliminated.\r\n\r\nRound 2! (last 3 Problems with proof)\r\n\r\nPlease answer. Deadline next Sunday September 14th midnight.\r\n\r\nThe Problems:\r\n\r\n(Note. Next round has been purposely made pretty hard.)\r\n\r\n1.\ta + b = 5 and ab = 10. If a and b are complex numbers, find $ a^4 \\plus{} b^4$.\r\n2.\tA set of positive integers has the triangle property if it has 3 numbers in it that could be the sides of a non-degenerate triangle. Consider the set {4, 5, 6, $ \\ldots$ n}. If every 10-element subset of this set has the triangle property, what is the maximum possible value of n?\r\n3.\tLet\u2019s play a game on an n x n board. You are in the center square (or one of the center squares.) Every turn, you can move to an adjacent square(but not diagonally). When it is my turn, I can signify a square to make impassable (i.e. you can never move to that square.) If you reach the edge of the board, you win. If, at some point, you cannot move, you lose. Find the largest n such that you have a winning strategy.\r\n4.\tLet S be a subset of $ {1, 2, 3, \\ldots, 2000}$ such that no two members of S have a positive difference of 4 or 7. Which is maximum number of elements in S?\r\n5.\tThere is an acute triangle with side lengths BC = a, AC = b, and AB = c. There are points D, E,F, and G with  D on side AC, E on side AB, and F and G on side BC such that DEFG is a square. Find the length of this square in terms of a, b, and c.\r\n6.\tWe have $ 3\\plus{}x^6 \\plus{} y^6 \\plus{} z^6 \\equal{} 2x^2y \\plus{} 2y^2z \\plus{} 2z^2x.$ If x, y, and z are positive real numbers, find all possible values of xyz. If your answer is something like, all real numbers greater than 1/2, then use standard set-builder notation (i.e. {x| x > 1/2}. (Proof!)\r\n7.\tIn a soccer tournament, every team plays every other team once. A team receives 4 points for a win, 2 points for a draw, and 1 point for a loss. After the tournament, the tournament director notes that the team who had the highest number of points at the end also won the fewest number of games out of all the teams. Find the smallest possible number of teams. (Proof!)\r\n8.\tFind all the pairs of positive integers (x, p) such that p is a prime, $ x \\le 2p$ and $ x^{p\\minus{}1}$ is a divisor of $ (p\\minus{}1)^x \\plus{} 1$. (Proof!)\r\n\r\nPM me for clarifications.\r\n\r\nAlso: The sources for the Round 1 Problems:\r\n\r\n5. An old tounament\r\n6. 2007-2008 WOOT (You are allowed to use individula problems, just not a lot of problems at one time...this will be the only problem from 2007-2008 WOOT I will use.)\r\n7. IMO Shorlist\r\n8. Tournament of Towns.\r\n\r\nSee? IMO problems can be eaSY. \r\n\r\nStill participating\r\n\r\nbudi713\r\nshentang\r\nFantasyLover\r\nchenhsi\r\nTextangle\r\nWickedstjr\r\nxpmath\r\nalingaling\r\n\r\nProblems and Answers:\r\n\r\n1.\tHow many two-digit integers are multiples of 4 but not 5? (18)\r\n2.\tA triangle ABC has side lengths AB = 3, AC = 4, BC = 5. Find the length of the altitude from A. Express your answer as a fraction in simplest form? (12/5)\r\n3.\tBruce plays a game. He rolls 2 dice. If the sum of the numbers on the two dice is 7, he wins. Otherwise, he rolls again. If it is 6 or 8, he wins. Otherwise, he loses. What is the probability Bruce wins? (43/108)\r\n4.\tThere are four people Albert, Bruce, Casey, and Dale who want to split up 26 apples among themselves. If Dale must have the same number of apples as Casey and Bruce combined, and Albert must get an even number of apples, how many ways are there to split up the apples if everyone must get at least one apple? (66)\r\n5.\tHow many numbers less than or equal to 1000 can be expressed as the difference of two perfect squares? (749 or 750)\r\n6.\tA floor can be covered with n identical square tiles of a certain size. If a certain smaller square tile is used, then n + 76 tiles are required. Find n. (324)\r\n7.\tOn a 5 x 5 board, two players alternately mark numbers on empty cells. The first player marks 1\u2019s, the second 0\u2019s. One number is marked per turn, until the board is filled. For each of the nine 3 x 3 squares the sum of the nine numbers on its cells is computed. Denote by A the maximum of these sums. How large can the first player make A, regardless of the responses of the second player? (6)\r\n8.\tYou have n colors and a 10x10 board. You must color every square of the board, and every color must be used at least once. If a square is colored color A, then at least 2 squares adjacent to it must be colored A. What is the maximum possible value of n? (25)",
  "Solution_17": "Yes, I've never been in the top eight in any competition ever. I'll get working on these problems as soon as possible.",
  "Solution_18": "Okay when are we going to continue this tournament. I think we shoud move on to the next round.",
  "Solution_19": "...you are teh only person who sent me answers  :mad:  :rotfl:",
  "Solution_20": "You should send them reminders. That's what I do, and most of the time they end up answering the questions immediately after that. \r\n\r\nOR \r\n\r\n\r\n\r\nYou could make me the champion of pythag011'stournament.  :)",
  "Solution_21": "[quote=\"Wickedestjr\"]You should send them reminders. That's what I do, and most of the time they end up answering the questions immediately after that. \n\nOR \n\n\n\nYou could make me the champion of pythag011'stournament.  :)[/quote]\r\n\r\nFine. Solve problem 8 and I will.",
  "Solution_22": "I can't solve it, but I am obviously going to be the only active competitor in this competition. budi713 who gets on like every other day of the week hasn't been submitting answers, so therefore, he probably doesn't want to compete anymore.",
  "Solution_23": "Just declare me champion already!!!",
  "Solution_24": "Well, fine. I have a competitor who has solved all 8 problems already. So yoou are 2nd place.",
  "Solution_25": "Who got first place?  :mad:",
  "Solution_26": "Obviously pythag011.\r\n\r\nDuh :|",
  "Solution_27": "/w Wickedesjr I don't think he liked your nagging....",
  "Solution_28": "Brut3Forc3.\r\n\r\nI made him do it because its sad when theres only 1 person competiting...\r\n\r\nAnywyas, he got all 7. Excpet 8. So if you solve all 8, you might get first! \r\n\r\nBTW, #8 is an IMO level problem.\r\n\r\nFor now:\r\n :winner_first:  Brut3Forc3\r\n :winner_second: Wickedstrj\r\n :winner_third: spammer123",
  "Solution_29": "Yay!!! Even though you misspelled my name, I am glad that I got second place."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "How do you make a bar in a LaTeX document?\r\n\r\nFor example:\r\n\r\n________________________________________________________________________________\r\nThanks.",
  "Solution_1": "\\hrulefill may be what you're looking for.  (if that doesn't work, try \\hrule)",
  "Solution_2": "Yep, that's it. :)"
}
{
  "Tag": [],
  "Problem": "I don't know if \"period\" is the right name for how often something repeats itself. For example, $ 0.\\overline{19}$ has a \"period\" of $ 2$ digits. Please let me know if that's the correct term.\r\n\r\nAnyways, calculate how many digits are in the period of $ 0.\\overline{19}\\cdot0.\\overline{199}$",
  "Solution_1": "We have that $ \\overline{.19}\\equal{} \\frac {19}{99}$, and $ \\overline{.199}\\equal{}\\frac{199}{999}$.  Multiplied, these fractions\r\nequal $ \\frac{3781}{98901}$, which in decimal form is equal to about\r\n$ 0.0382301493$.  When I plugged it into my calculator, I got that \r\nthe period was $ \\boxed {54}$ digits long.",
  "Solution_2": "Thanks [b]Z-coli[/b], but the idea is to figure out how long the period is without using calculators.\r\n\r\nHere is an attempt I had of organizing the number, but I couldn't get anything useful from it:\r\n\r\n[hide=\"Attempt\"]In fractions:\n\n$ \\frac{19}{99}\\cdot\\frac{199}{999}\\equal{}\\frac{(20\\minus{}1)(200\\minus{}1)}{(100\\minus{}1)(1000\\minus{}1)}\\equal{}\\frac{4000\\minus{}220\\plus{}1}{100000\\minus{}1100\\plus{}1}\\equal{}\\frac{3781}{98901}$\n\nLet $ y$ be our number:\n\n$ 100000y\\minus{}1000y\\minus{}100y\\plus{}y\\equal{}3781$[/hide]",
  "Solution_3": "In general, to find the period of the reduced fraction $ \\frac pq$, we note that factors of $ 2$ and $ 5$ in the denominator won't affect the period. If, after we take out these factors, the denominator is $ q'$, then the period is the smallest $ k$ such that $ q'|10^k \\minus{} 1$. This is because if we can express $ \\frac p{q'}$ in the form $ \\frac {np}{\\underbrace{999\\ldots999}_k}$ such that $ q'n\\equal{}\\underbrace{999\\ldots999}_k$ for some integer $ n$ and $ np \\equal{} \\overline{p_1p_2p_3\\ldots p_k}$, then we can easily see that the decimal form of this is $ 0.\\overline{p_1p_2p_3\\ldots p_k}$, and the period is $ k$.\r\n\r\nNow, for this problem, we know that $ 98901 \\equal{} 3^5\\cdot11\\cdot37 \\equal{} 3^4\\cdot11\\cdot111$. This means that the number of $ 9$'s in the denominator must be a multiple of $ 2$ to be divisible by $ 11$ and a multiple of $ 3$ to be divisible by $ 111$. For $ \\underbrace{999\\ldots999}_m$ to be divisible by $ 3^4$, it is sufficient to find the smallest $ m$ such that $ \\underbrace{111\\ldots111}_m$ is divisible by $ 3^2 \\equal{} 9$. Since the sum of the digits of $ \\underbrace{111\\ldots111}_m$ is equivalent to its residue modulo $ 9$, $ m \\equal{} 9$ is the smallest possible value. Finally, we see that $ k \\equal{} 18i$ for some positive integer $ i$ (I don't think we can just conclude that the period is $ 2\\cdot3\\cdot9 \\equal{} 54$, even though it gives the right answer), from which it is easier to see that $ \\boxed{54}$ is the period.",
  "Solution_4": "I understood the idea, but there is one thing I didn't catch on. \r\n\r\nYou said that $ k \\equal{} 18n$ , for some integer $ n$, and since you found $ n \\equal{} 3$ , then:\r\n\r\n$ \\frac {np}{q'} \\equal{} \\frac {p}{q}\\Leftrightarrow q' \\equal{} nq$\r\n\r\nSince $ 3\\cdot 98901 \\equal{} 296703$ which is very diferent from what is expected. Are the $ n$'s you used different or is there something wrong with my thinking?",
  "Solution_5": "Sorry, they are different. :blush: I'll edit it. The bottom paragraph is just executing the idea in the top paragraph. Again, I'm very sorry!\r\n\r\nAnd for further clarification: $ q'n\\equal{}10^k\\minus{}1$, in case something wasn't clear about that.",
  "Solution_6": "Ah ok, just a small detail hehehe...\r\n\r\nHey, don't worry about it, no need to apologize for helping me with the problem :)\r\n\r\nThanks for the solution."
}
{
  "Tag": [
    "Putnam",
    "calculus",
    "linear algebra",
    "\\/closed"
  ],
  "Problem": "There will be a Putnam Competition Math Jam on Wednesday 15 September at 7:30ET / 4:30PT / 2330 GMT.\r\n\r\nThis is the first of three scheduled Math Jams for the Putnam:\r\nSep 15: Calculus & Analysis Problems\r\nOct 20: Linear Algebra & Abstract Algebra Problems\r\nNov 9: Miscellaneous Problems\r\n\r\nFor more info on the Putnam Competition, visit \r\n[url]http://math.scu.edu/putnam/[/url] and \r\n[url]http://www.unl.edu/amc/a-activities/a7-problems/putnam/[/url]\r\n\r\nEven if you do not fit the Putnam criteria (for instance, if you're not a college student in the US or Canada), please nonetheless feel free to attend the Math Jam.  The only prerequisite is a working knowledge of basic calculus.\r\n\r\nAlso, if anyone has favorite Putnam problems that they'd like to see worked through in one of the Math Jams, please do tell.",
  "Solution_1": "I saw this problem in the Art and Craft of Problem Solving that looks pretty cool (but I can't solve it of course :D ):\r\n\r\nA hole puncher is placed on a point on a plane.  When operated, the hole puncher removes every point whose distance from the hole puncher is irrational.  How many times must the hole puncher be operated to remove every point?\r\n\r\nSadly, I can't make it to the math jam :(",
  "Solution_2": "[quote=\"nr1337\"]I saw this problem in the Art and Craft of Problem Solving that looks pretty cool (but I can't solve it of course :D ):\n\nA hole puncher is placed on a point on a plane.  When operated, the hole puncher removes every point whose distance from the hole puncher is irrational.  How many times must the hole puncher be operated to remove every point?\n[/quote]\r\nThis problem was on the 1990 Putnam (it was Problem A4).  It happens to be my all-time favorite Putnam problem!  We're going to do it in the \"Miscellaneous Problem\" Jam on November 9th.  Wednesday's Jam will be all calculus problems.",
  "Solution_3": "Oh.  I tried the problem once and got the answer 2, but then noticed a fatal error in my reasoning.  Well, I still have 7 more years before I can actually be in Putnam, so I have a while to train ;)",
  "Solution_4": "Doesnt that make you in 6th grade (if u mean that you are only going to take it in college)?",
  "Solution_5": "Seventh.  Oh wait... only six more years (13-7).",
  "Solution_6": "[quote=\"nr1337\"]I saw this problem in the Art and Craft of Problem Solving that looks pretty cool (but I can't solve it of course :D ):\n\nA hole puncher is placed on a point on a plane.  When operated, the hole puncher removes every point whose distance from the hole puncher is irrational.  How many times must the hole puncher be operated to remove every point?\n[/quote]\r\n\r\nJust take the points $(0,0),(1,0),(\\sqrt 2,0)$. Now take a point $(x,y)$ in the plane. If $\\sqrt{x^2+y^2}$ is irrational, then we are done. Otherwise, $x^2+y^2$ is also rational, so there are two possibilities: either $x$ is irrational, in which case $(x-1)^2+y^2$ is irrational, and then so is $\\sqrt{(x-1)^2+y^2}$, or $x$ is rational, in which case $(x-\\sqrt 2)^2+y^2$ is irrational, so $\\sqrt{(x-\\sqrt 2)^2+y^2}$ is irrational as well. On the other hand, it's easy to show that two points do not suffice, because there are two circles of rational radii, one centered at each one of the two points, which intersect. Their points of intersection will not be removed."
}
{
  "Tag": [],
  "Problem": "Do any of you play USCF chess tournaments??? \r\n\r\nIf you do I'd like to know your rating.\r\n\r\nme, im only a 1046  :blush: ...i got careless in a couple tournaments but managed a first place in one  :first:",
  "Solution_1": "I do... and I'm rated 1708  :D",
  "Solution_2": "me.... 1325... 8th grader rite now hasn't played since last year and has only played in 7 tournaments.\r\n\r\nI went mostly because of the school team back in 5th grade. I have only gone to 2 or 3 tournaments since.",
  "Solution_3": "wow...all higher than me...grr! lol how long have you guys actually started chess? i'm in eighth grade and i started in first  :lol: ...",
  "Solution_4": "In my active playing chess heydays,1979-80,i got official\r\nbeginners rating at 1680. Though,not much higher,i appreciate\r\nsuch given rating.I played couple of 'systematic blindfold games'with my brother.I found it very exercising,imaginatively.Have you ever try playing 'BLINDFOLD CHESS?'\r\n\r\n                                                     Respectfully,\r\n                                                    A.Eddington",
  "Solution_5": "yeah its fun\r\n\r\nbut i dont have a coach so i quit last summer since it got hard to improve.\r\n\r\nmath is more challenging now",
  "Solution_6": "uuhhh started in like 5th or 6th grade... haven't really played since like 8th or 9th grade... rating's like 1000 something :blush:",
  "Solution_7": "[quote=\"arthur eddington\"]Have you ever try playing 'BLINDFOLD CHESS?'\n\n                                                     Respectfully,\n                                                    A.Eddington[/quote]\r\n\r\nYea except we don't use blindfolds. Instead, (like on our carpools to tournaments), we sometimes just say the notation out. Is your version of BLINDFOLD CHESS like that or do you actually use chess boards and blindfolds?",
  "Solution_8": "Have you ever tried to play chess on multiple boards, like with 3 to 5 boards at a time and each turn, a person can move 1 piece, but it can be on any board though. whoever wins the most, out of the number of boards you have, wins.",
  "Solution_9": "My chess rating is 1711. I havent played since 6th grade. When I was in 5th grade, I won the KY State chess championship. I moved to NC and I got 10th in the K-8 in 6th. Havent played an official tournament since. :(",
  "Solution_10": "[quote=\"goldendomer\"]When I was in 5th grade, I won the KY State chess championship.[/quote]\r\n\r\nWOW NICE!!",
  "Solution_11": "[quote=\"7h3.D3m0n.117\"][quote=\"goldendomer\"]When I was in 5th grade, I won the KY State chess championship.[/quote]\n\nWOW NICE!![/quote]\r\n\r\nThanks. I was tied for first with the another 5th grader but I was declared the champion because my opponents were stronger.",
  "Solution_12": "mine is 1456 :ninja:"
}
{
  "Tag": [],
  "Problem": "Aazlae yek Mosallas be toolhaye sahihe $ k,m,n$ hastand, Farz konid $ k > m > n$ va:\r\n\r\n$ \\left\\{\\frac{3^{k}}{10^{4}}\\right\\}\\equal{}\\left\\{\\frac{3^{m}}{10^{4}}\\right\\}\\equal{}\\left\\{\\frac{3^{n}}{10^{4}}\\right\\}$\r\n\r\nmatloob ast kamtarin meghdare mohite in mosallas. :)",
  "Solution_1": "Soale khoobie faghat ye khorde kharkari dare :) , avval az hame tavvajoh konin ke az sharte masale natije mishe :\r\n$ 3^{m}\\equiv 3^{n}\\equiv 3^{k}(mod 10000)$ . hala farz konin $ Ord 3_{10000}\\equal{} a$ ke ba yekam mohasebat ( hamoon kharkarishe :D) dar miad $ a \\equal{} 500$ . amma az chizi ke ghabl dar ovorde boodim natije mishe $ m\\equiv n\\equiv k (mod 500)$\r\npas :\r\n$ n \\equal{} 500a_{1}\\plus{}i$ , $ m \\equal{} 500a_{2}\\plus{}i$ , $ k \\equal{} 500a_{3}\\plus{}i$ va $ a_{1}< a_{2}< a_{3}$ .\r\nhala tebghe ghazie hemar darim :\r\n$ 500a_{1}\\plus{}i\\plus{}500a_{2}\\plus{}i > 500a_{3}\\plus{}i$ ke natije mide $ 500(a_{1}\\plus{}a_{2})\\plus{}i > 500a_{3}$ hala dohalat dar nazar begirin :\r\n1)$ i \\equal{} 0$ dar in halat az namosavie bala natije mishe $ a_{1}\\plus{}a_{2}\\geq a_{3}\\plus{}1$ . hala mohite mosalas mishe\r\n$ 500(a_{1}\\plus{}a_{2}\\plus{}a_{3})\\geq 500(2a_{3}\\plus{}1)$ amma az oonjayike $ a_{3}> a_{2}> a_{1}\\geq 0$ pas agar $ a_{3}\\equal{} 2$ oon moghe $ a_{2}\\equal{} 1$ va $ a_{1}\\equal{} 0$ ke ba namosavie $ a_{1}\\plus{}a_{2}\\geq a_{3}\\plus{}1$ tanaghoz dare , dar gheyre in soorat\r\n$ a_{3}\\geq 3$ ke natije mide mohite mosalas hadeaghal $ 3500$ hast .\r\n2)$ i\\geq 1$ ke tebghe hamoon namosavi avvalie natije mishe $ a_{1}\\plus{}a_{2}\\geq a_{3}$ \r\nhala mohite mosalas mishe $ 500(a_{1}\\plus{}a_{2}\\plus{}a_{3})\\plus{}3i\\geq 1000a_{3}\\plus{}3$\r\nhala mesle bala mitoonin sabet konin ke $ a_{3}\\geq 3$ ke natije mide mohite mosalas hadeaghal\r\n$ 3003$ hastesh . \r\nva mesal baraye $ 3003$ ine ke $ n \\equal{} 501$ , $ m \\equal{} 1002$ va $ k \\equal{} 1501$"
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "ratio",
    "percent"
  ],
  "Problem": "In the rectangle shown, the ratio of width to length is 1:4. What percent of the rectangle is shaded?\n\n[asy]pair A,B,C,D;\nA = (0,0); B = (4,0); C=(0,1); D = (1,1);\ndraw(A--B--B+C--C--A--D--B);\nfill(A--B--D--cycle);[/asy]",
  "Solution_1": "The information of ratios is unnecessary. Draw a line through and you'll see parts match up, and the answer is $ 50%\n$.",
  "Solution_2": "That is what I would do.  This is why it works though.\r\nThe height of the triangle is the short side of the rectangle, while the base is the long side of the rectangle.  If we call the long side $ b$ and the short one $ h$, we can find that the area of the rectangle is $ bh$, while the triangle is $ \\frac {1}{2} bh$.  So we know that the triangle is $ \\frac {1}{2}$ the area of the rectangle, so we have $ \\boxed {50}$, like pinkmuskrat said."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "If $a,b,c$ are positive integer satisfying \r\n\r\n$2ab+2ac+2bc=abc$ \r\n\r\nFind the ordered triples $(a,b,c)$\r\n\r\n :)",
  "Solution_1": "rewrite the given equation as \r\n\r\n$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{2}$\r\n\r\nLet WLOG $a\\leq b\\leq c$ then $\\frac{1}{a}\\geq\\frac{1}{b}\\geq\\frac{1}{c}$.    So $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\leq\\frac{3}{a}$ \r\n\r\nthat is $a\\geq6$ \r\n\r\nOn the other hand $\\frac{1}{2}-\\frac{1}{a}=\\frac{1}{b}+\\frac{1}{c}>0$\r\n\r\nthat is $a>2$ or $a\\geq3$. Finally $3\\leq a\\leq6$ so $a=3 \\ \\text {or}\\ 4 \\ \\text {or}\\ 5 \\ \\text {or}\\ 6$\r\n\r\nnow substitute the values of $a$. the solution is not hard from here \r\n\r\nWhat do you think?? :)",
  "Solution_2": "[quote=\"socrates\"]rewrite the given equation as \n\n$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{2}$\n\nLet WLOG $a\\leq b\\leq c$ then $\\frac{1}{a}\\geq\\frac{1}{b}\\geq\\frac{1}{c}$.    So $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\leq\\frac{3}{a}$ \n\nthat is $a\\geq6$ \n\nOn the other hand $\\frac{1}{2}-\\frac{1}{a}=\\frac{1}{b}+\\frac{1}{c}>0$\n\nthat is $a>2$ or $a\\geq3$. Finally $3\\leq a\\leq6$ so $a=3 \\ \\text {or}\\ 4 \\ \\text {or}\\ 5 \\ \\text {or}\\ 6$\n\nnow substitute the values of $a$. the solution is not hard from here \n\nWhat do you think?? :)[/quote]\r\n\r\nWell , socrates , the inequality you obtained are $a\\geq 3$ and $a\\geq 6$ which combine and gives you $a\\geq 6$ instead  ;)",
  "Solution_3": "[quote=\"shyong\"]\nWell , socrates , the inequality you obtained are $a\\geq 3$ and $a\\geq 6$ which combine and gives you $a\\geq 6$ instead  ;)[/quote]\r\n\r\nNo, you definitely get $a \\leq 6$.",
  "Solution_4": "[quote=\"Arne\"][quote=\"shyong\"]\nWell , socrates , the inequality you obtained are $a\\geq 3$ and $a\\geq 6$ which combine and gives you $a\\geq 6$ instead  ;)[/quote]\n\nNo, you definitely get $a \\leq 6$.[/quote]\r\n\r\nOh ya , you are right Arne  :blush:  I just follows what socrates wrote but didnt notice that he has a typo there  :blush: I should have had a thorough checking next time  :lol:"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "angle bisector",
    "geometry proposed"
  ],
  "Problem": "$P,Q$ are two fixed points on a circle centered at $O$, and $M$ is an interior point of the circle that differs from $O$. $M,P,Q,O$ are concyclic. Prove that the bisector of $\\angle PMQ$ is perpendicular to line $OM$.",
  "Solution_1": "$M \\in \\gamma_1$ where $\\gamma_1$ is the circumference circumscripted to OPQ which is; using chord-angles, we note that $\\angle PMQ$ and $\\angle QMO$ are fixed: so also $\\angle QMO+0.5\\angle PMQ$ is fixed: now prove the case of $M\\equiv O$ m=o, and see that it's $\\frac{\\pi}{2}$",
  "Solution_2": "MP, MQ, MO and the bisector line of PMQ are an armonic fascicul. :) The problem becomes a known theorem :)",
  "Solution_3": "Bravo, Flavian ! It is and my proof.",
  "Solution_4": "Let $MH$, with $H$ on the circumscribed circle of $POMQ$, be the angle bisector of $\\measuredangle PMQ$. And $PQ \\cap MH = K$. Connect $O$ with $H$. It is $OH \\bot PQ$. So $\\measuredangle OPM + \\measuredangle MKQ = 90^{\\circ}$.\r\n$\\measuredangle MKQ = 180^{\\circ} - (\\measuredangle KQM + \\measuredangle KMQ) = 180^{\\circ} - (180^{\\circ} - \\measuredangle POM + \\measuredangle KMQ) = \\measuredangle POM - \\measuredangle KMQ = \\\\ = 180^{\\circ} - \\measuredangle OPM - \\measuredangle PMO - \\measuredangle KMQ$. \r\nThen $90^{\\circ} - \\measuredangle OPM = 180^{\\circ} - \\measuredangle OPM - \\measuredangle PMO - \\measuredangle KMQ$. And finally $90^{\\circ} = \\measuredangle PMO + \\measuredangle KMQ = \\measuredangle PMO + \\measuredangle PMK$.",
  "Solution_5": "let $\\omega$ be the circumcircle of $PMOQ$,given $OP=OQ$,Let $OL$ is the bisector of $O$($L$ on circumcircle),then $OL$ is the diameter,also bisector of $\\angle PMQ$ meet at $L$,hence done.",
  "Solution_6": "Dear Mathlinkers,\nI saw this terminology:\nO, L is the first, second M-perpoint of triangle MPQ....\nSincerely\nJean-Louis",
  "Solution_7": "That's really easy. Just assume $\\angle\\{POQ}=2\\theta\\$ and proceed.\n\n\n\nOil is the food of the hair; Rice is the food of the Bengalis; Maths is the food of my soul.",
  "Solution_8": "Let $ \\omega $ denote the circumcircle of $ POMQ $, and let $ S $ be the intersection of the angle bisector of $ PMQ $ and $ \\omega $. Let $ \\angle SMP = \\angle SMQ = \\alpha $. Note that $ \\angle POQ = \\angle PMQ = 2\\alpha $. Let $ T $ be the point diametrically opposite $ Q $. Then $ \\angle POT = 180 - 2\\alpha $. Since $ POT $ is isosceles, $ \\angle PTQ = \\alpha $, so $ \\angle PQT = 90 - \\angle PTQ = 90 - \\alpha $. Thus $ \\angle PMO = \\angle PQO = \\angle PQT = 90 - \\alpha $, so $ \\angle PMO + \\angle SMP = (90 - \\alpha) + \\alpha = 90 $ and we are done.",
  "Solution_9": "[quote=k2c901_1]$P,Q$ are two fixed points on a circle centered at $O$, and $M$ is an interior point of the circle that differs from $O$. $M,P,Q,O$ are concyclic. Prove that the bisector of $\\angle PMQ$ is perpendicular to line $OM$.[/quote]\n\nI would like to present the following solution which basically trivializes the entire problem. In my opinion it is very fast and elegant :D .\n\n[b][u][i]Solution:[/i][/u][/b] Denote by $C_1$ the circle with centre $O$ and by $C_2$ the other circle .Drop the perpendicular from point $O$ to segment $\\bar{PQ}$ and extend it to the point diametrically opposite $O$, naming them $A$ and $X$ respectively. Clearly, since $OA \\bot AI$, it suffices to prove that $O,A,I,M$ are concyclic. Now for the best part, consider an inversion with respect to $C_1$ ($O$ is the centre of inversion), This it suffices to prove that $A^*,I^*,M^*$ are collinear but this clearly follows from the fact that both $OX^*$ and $OM^*$ are diameters and thus $\\measuredangle{OI^*X^*}= \\measuredangle{M^*I^*O}= 90^o $. Thus done  :D .\n\nSincerely,\nAayam",
  "Solution_10": "Dear Mathlinkers,\n\n[url=http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%207.pdf] here [/url]\n\nProblem 7, p. 20\n\nSincerely\nJean-Louis"
}
{
  "Tag": [
    "calculus",
    "Pythagorean Theorem",
    "geometry"
  ],
  "Problem": "I am helping someone but have a hard time with this calculus problem:\r\n\r\nA person is walking due north from point P at a rate of 8m/s. After 10 min a boy walks from point M (1000 ft away from P) towards point P  at a rate of 10 m/s. After 15 mintues how far will be boy and the person? If some can make me undersatnd using calculus I would appreciate it.",
  "Solution_1": "You can't really solve it unless you know where m is (you don't specify a direction)\r\n\r\nthe key to related rates problems is to find the relationship between two different quantities in terms of a third. in problems like this one, i would expect this relationship to be the pythagorean theorem"
}
{
  "Tag": [
    "trigonometry",
    "trig identities",
    "Law of Cosines",
    "geometry solved",
    "geometry"
  ],
  "Problem": "consider a quadrilateral such that all the lengths of the sides and diagonals are integers. prove that one of the six integers is divisible by 3.\r\n\r\n[i]Hungarian TST 2003[/i]",
  "Solution_1": "Don't expect me to give a solution or anything :D It's from the 2'nd Hungarian TST for the 43'rd IMO. This point (a). \r\n\r\nPoint (b) (more general):\r\n\r\nWe have n>=5 points in the plane s.t. all the distances between 2 of these pts are natural. Prove that at least 1/6 of all the distances are multiples of 3.",
  "Solution_2": "And, if I'm not wrong, it was first given in a bulgarian competitions. I also don't have a solution except the official one.",
  "Solution_3": "and what is that official solution?",
  "Solution_4": "This is one of the problems and solutions impossible to forgot. I will reproduce the official solution, so please do not ask me why it is so:\r\n  Let ABCD the quadrilateral. Lt us suppose that none of AB, BC, CD, DA, AC, BD is a multiple of 3. We can assume that BAD=BAC+CAD (as angles). Let BAC=x and CAD=y. Let alfa=2AB*AC*cosx, beta=2AD*Ac*cosy, gamma=2Ab*AD*cos(x+y). Using law of cosines in ABC, ACD and ABD, it follows that BC^2=AB^2+C^2-alfa, CD^2=AD^2+ AC^2-beta, BD^2=AB^2+AD^2-gamma. Since the square of an integer is 1 modulo3, it follows that alfa, beta and gamma are integers congruent to 1 modulo 3. Since 2AC^2*gamma=4AC^2*AB*AD*cos(x+y)=alfa*beta-4AC^2*AB*AD*sinx*siny, it follows that 4AC^2*AB*AD*sinx*siny is an in teger congruent to 2. So, sinx*siny is a rational number that written in lowest term has the numerator not a multiple of 3. \r\n    Now let p=2AB*AC and q=2AD*AC. Thus, cosx=alfa/p and cosy=beta/q. Since sinxsiny=sqrt((p^2-alfa^2)(q^2-beta^2))/pq ia rationla, the numerator must be an integer. This numerator is a multiple of 3 (since 3|p^2-alfa^2), but the denominator is not a multiple of 3. Thus, when written in lowest terms, the numerator of sinx*siny is a multiple of 3, false."
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope",
    "geometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "1.\r\nWrite the equation of a line that passes through the indicated points: O(2,3) and P(-1,1)\r\n\r\n2.\r\nFind the equation of a line that passes through (-2,-3) and parallel to the line whose equation is 2x - 5y = 6.\r\n\r\n3.\r\nGiven the points P(1,-1), Q(4,2), and R(6,4),\r\na. Find the slopes of line segments PQ and PR.\r\nb. From your answer to part a, what can you conclude regarding the points P, Q, and R?\r\nc. Find the equation of the line that passes through P and R.\r\nd. Does the point S(10,8) lie on the line PR? Why or why not?\r\n\r\nThe underlined letters are supposed to be lined on top.",
  "Solution_1": "i dont think this problem belongs in the olympiad section. it is trivial and routine co-ordinate geometry that you would learn at school.",
  "Solution_2": ":rotfl:  :rotfl:"
}
{
  "Tag": [
    "function",
    "conics",
    "parabola",
    "quadratics",
    "calculus",
    "algebra",
    "system of equations"
  ],
  "Problem": "Let $ f(x) \\equal{} ax^2 \\plus{} bx \\plus{} c\\ \\left(a\\neq 0,\\ x > \\minus{} \\frac {b}{2a}\\right)$. Suppose that $ f^{ \\minus{} 1}(0) \\equal{} \\frac {4}{3},\\ f^{ \\minus{} 1}(2) \\equal{} 2,\\ f^{ \\minus{} 1}(10) \\equal{} 3$.\r\n\r\n(1) Find the values of coefficients $ a,\\ b,\\ c$.\r\n\r\n(2) Use the values of coefficients $ a,\\ b,$ found in (1) and let the coefficient $ c$ vary, find the values of $ c$ such that $ f(x)$ touches to $ f^{ \\minus{} 1}(x)$.",
  "Solution_1": "[hide=\"1\"]From the 3 equations $ f^{-1}(0)=\\frac 43$, $ f^{-1}(2)=2$, and $ f^{-1}(10)=3$, we get: $ f(\\frac 43)=0$, $ f(2)=2$, and $ f(3)=10$, from which we get:$  \\{\\begin{array}{c} (\\frac 43)^2a+(\\frac 43)b+c = 0\\\\\n    4a+2b+c = 2\\\\\n    9a+3b+c = 10 \\end{array}  \\;$\nWe can solve this to get $ (a,b,c)=\\boxed{(3,-7,4)}$\n[/hide]\n[hide=\"2\"]\nSince $ f(x)$ and $ f^{-1}(x)$ can only intersect on the line $ y=x$, they're common point of tangency must also be on that line. Therefore, we want a $ c$ such that $ y=x$ is tangent to $ f(x)$. This occurs when $ f(x)=3x^2-7x+c=x$, and $ f'(x)=6x-7=1$. So, $ x=\\frac 43$ and $ c=\\boxed{\\frac{16}3}$.\n[/hide]",
  "Solution_2": "[hide]\ni'm getting something different for #2\n\nlike schnoggin said, the intersection must be on the line y=x\nso now solve y=x and f(x) like any other system of equations\nsubstitute y for x in the original equation\nit has only one solution now, but the equation remains quadratic\ntherefore it is a double root\nnow use the discriminant = 0\n$ b^2\\minus{}4\\cdot a\\cdot c\\equal{}49\\minus{}4\\cdot 3\\cdot c\\equal{}0$\n$ c\\equal{}\\frac{49}{12}$\n\nmy answer looks too ugly\n\n[/hide]",
  "Solution_3": "once u start taking calculus, its just really hard to solve some questions without it. anyway, you're method works, there's just a slight flaw. There should only be one solution to the equation $ f(x)\\equal{}x\\rightarrow 3x^2\\minus{}8x\\plus{}c\\equal{}0$. So the discriminant is $ 64\\minus{}4(3)(c)\\equal{}0\\rightarrow c\\equal{}\\boxed{\\frac{16}3}$.",
  "Solution_4": "That's right. I would add one thing: You can use $ D/4$."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Is there a set of real nos u,v,w,x,y,z such that\\[ \\begin{array}{l}\r\n u^2  \\plus{} v^2  \\plus{} w^2  \\plus{} 3(x^2  \\plus{} y^2  \\plus{} z^2 ) \\equal{} 6, \\\\ \r\n ux \\plus{} vy \\plus{} wz \\equal{} 2 \\\\ \r\n \\end{array}\r\n\\]",
  "Solution_1": "[quote=\"anorithdialga\"]Is there a set of real nos u,v,w,x,y,z such that\n\\[ \\begin{array}{l} u^2 \\plus{} v^2 \\plus{} w^2 \\plus{} 3(x^2 \\plus{} y^2 \\plus{} z^2 ) \\equal{} 6, \\\\\nux \\plus{} vy \\plus{} wz \\equal{} 2 \\\\\n\\end{array}\n\\]\n[/quote]\r\n\r\n[hide=\"solution\"]\nSuppose there exist such reals,\n\nLet $ a\\equal{}u^2\\plus{}v^2\\plus{}w^2,b\\equal{}x^2\\plus{}y^2\\plus{}z^2$,and we see $ a\\plus{}3b\\equal{}6$.\n\nBy Cauchy-schwartz inequality we have\n\n$ 4\\equal{}(ux\\plus{}vy\\plus{}wz)^2\\le (u^2\\plus{}v^2\\plus{}w^2)(x^2\\plus{}y^2\\plus{}z^2)\\equal{}ab\\equal{}(6\\minus{}3b)b\\equal{}3\\minus{}3(b\\minus{}1)^2 \\le 3$ \n\n:Contradiction!! So it's done.\n\n[/hide]"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Good evening.\r\n\r\n$A \\in \\mathcal{M}_n(\\mathbb{C})$ Demonstrate the equivalence between\r\ni)$A\\bar{A}$ has positive eigenvalues\r\nii)There exists $U$ unitary and $T$ triangular superior such that ${A=UT^{t}U}$",
  "Solution_1": "[quote=\"Maxbl\"]\n(I am sorry but I didnt find a way to write tU where t denotes \"transposition\"  (I dont know the english term)).[/quote]\r\n\r\nHi! :)\r\n\r\nI think this is what you're looking for : ^{t}U $^{t}U$.",
  "Solution_2": "It is known result which you can read for example in Horn.Matrix analysis in the chapter Hermitian and symmetric matrices(I'm not sure whether it is called exactly so in the original English version-i don't have this one).\r\nIf you won't find it and nobody solve it  i can post a solution from the book.",
  "Solution_3": "I found the book but had only a glimpse of the proof, it appears to be recursive on n.\r\nI'll look at it better and post it later.",
  "Solution_4": "For Maxbl \r\nWich page of this book ?\r\n\r\n---------------------------\r\nEugene could you post the solution from the book ?",
  "Solution_5": "Just as eugene said , in the chapter Hermitian and symmetric matrix, quite far in the chapter part 4 or something.\r\nIt is in a bookshop near my school but I had not enough time when I looked to fully understand it.\r\nI'll look back in a few days.",
  "Solution_6": "Here we are:\r\n\r\nii)=>i) is quite evident it is only calculus.\r\n Now suppose $A \\in \\mathcal{M}_n(\\mathbb{C})$ verifies $A\\bar{A}$ has only positive eigenvalues.\r\n\r\nLet $\\lambda$ be an eigenvalue and $v$ an nonzero eigenvector for $A\\bar{A}$.\r\nThen either $A\\bar{v}$ and $v$ are linked either not.\r\nIf they are and $A\\bar{v}=\\mu v$ then $A\\bar{A}v=A\\bar{\\mu v}=|\\mu|^2v$\r\n\r\nIf not then let's choose $\\mu$ such than $|\\mu|^2=\\lambda$ and consider $y=A\\bar{v}+\\mu v$:\r\n$A\\bar{y}=\\lambda v+\\bar{\\mu}A\\bar{v}=\\bar{\\mu}y$ with $y$ non zero. Hence we can always find $v$ with $||v||=1$ such than $A\\bar{v}=\\mu v$ with $|\\mu|^2=\\lambda$\r\n By choosing a correct $e^{i\\theta}$ and multiplying $v$ with it we can even afford that $A\\bar{v}=\\mu v$ with $\\mu \\in \\mathbb{R}$.\r\nNow consider an orthonormal basis $(v,v_2,...,v_n)$ of matrix $V_i$ and look to\r\n$V_i^{*}A\\bar{V_i}$. Its first colum has only its first term non zero and you only have to recur to finish."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c>0$ with $ abc\\equal{}1$ and the numbers $ A\\equal{}\\frac{1}{1\\plus{}a\\plus{}b}\\plus{}\\frac{1}{1\\plus{}b\\plus{}c}\\plus{}\\frac{1}{1\\plus{}c\\plus{}a}$ and \r\n$ B\\equal{}\\frac{1}{1\\plus{}a\\plus{}bc}\\plus{}\\frac{1}{1\\plus{}b\\plus{}ca}\\plus{}\\frac{1}{1\\plus{}c\\plus{}ab}$ .\r\n\r\nFind which of the numbers $ A,B$ is bigger  :wink:",
  "Solution_1": "Nice.\r\n\r\n[i]solution[/i]\r\n\r\nWe will prove that $ A \\le 1 \\equal{} B$.\r\n\r\nFor $ A$ setting $ a \\equal{} x^3,b \\equal{} y^3,c \\equal{} z^3$ we take $ xyz \\equal{} 1$ and \r\n$ A \\equal{} \\frac {1}{x^3 \\plus{} y^3 \\plus{} 1} \\plus{} \\frac {1}{y^3 \\plus{} z^3 \\plus{} 1} \\plus{} \\frac {1}{z^3 \\plus{} x^3 \\plus{} 1} \\le \\frac {1}{xy(x \\plus{} y) \\plus{} xyz} \\plus{} \\frac {1}{yz(y \\plus{} z) \\plus{} xyz} \\plus{} \\frac {1}{zx(z \\plus{} x) \\plus{} xyz} \\equal{} \\frac {x}{x \\plus{} y \\plus{} z} \\plus{} \\frac {y}{x \\plus{} y \\plus{} z} \\plus{} \\frac {z}{x \\plus{} y \\plus{} z} \\equal{} 1$.\r\n\r\nSo $ A\\le1$.\r\n\r\nIn $ B$ setting $ a \\equal{} \\frac {x}{y},b \\equal{} \\frac {y}{z},c \\equal{} \\frac {z}{x}$ we take easily enough that $ B \\equal{} 1$.\r\n\r\nDone.",
  "Solution_2": "It's Moscow olympiad 97",
  "Solution_3": "[quote=\"Dimitris X\"]Nice.\n\n[i]solution[/i]\n\nWe will prove that $ A \\le 1 \\equal{} B$.\n\nFor $ A$ setting $ a \\equal{} x^3,b \\equal{} y^3,c \\equal{} z^3$ we take $ xyz \\equal{} 1$ and \n$ A \\equal{} \\frac {1}{x^3 \\plus{} y^3 \\plus{} 1} \\plus{} \\frac {1}{y^3 \\plus{} z^3 \\plus{} 1} \\plus{} \\frac {1}{z^3 \\plus{} x^3 \\plus{} 1} \\le \\frac {1}{xy(x \\plus{} y) \\plus{} xyz} \\plus{} \\frac {1}{yz(y \\plus{} z) \\plus{} xyz} \\plus{} \\frac {1}{zx(z \\plus{} x) \\plus{} xyz} \\equal{} \\frac {x}{x \\plus{} y \\plus{} z} \\plus{} \\frac {y}{x \\plus{} y \\plus{} z} \\plus{} \\frac {z}{x \\plus{} y \\plus{} z} \\equal{} 1$.\n\nSo $ A\\le1$.\n\nIn $ B$ setting $ a \\equal{} \\frac {x}{y},b \\equal{} \\frac {y}{z},c \\equal{} \\frac {z}{x}$ we take easily enough that $ B \\equal{} 1$.\n\nDone.[/quote]\r\nWrong solution !!\r\n\r\nClearly $ B \\neq 1$ ( and $ B \\leq 1$ is also obvious from AM-GM's inequality )\r\n\r\nBTW,I've already proved that $ \\sum \\frac{1}{a\\plus{}2} \\geq \\sum \\frac{1}{a\\plus{}bc\\plus{}1}$ by AM-GM,I'll try to find later a nice ( AM-GM's ) proof for the stronger ( $ A \\geq B$ )  :)",
  "Solution_4": "Oooops sorry for my wrong solution.\r\n\r\nI was under the impression that $ B\\equal{}\\sum\\frac{1}{ab\\plus{}a\\plus{}1}$. :blush:",
  "Solution_5": "[url=http://www.artofproblemsolving.com/community/c6h301938p1634675]Bulgaria 1997:[/url]\nLet $ a,b,c$ be positive numbers having product $ 1.$ Prove that\n$ \\frac{1}{a+b+1}+\\frac{1}{b+c+1}+\\frac{1}{c+a+1} \\le \\frac{1}{a+2}+\\frac{1}{b+2}+\\frac{1}{c+2}.$"
}
{
  "Tag": [
    "trigonometry",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "[color=darkred]In figure, two cicles $(O)$ and $(O')$ are external tangent. $OH\\perp MN,\\,HT=HP$ and $HT$ is a tangent of circle $(O)$\n\nProve that: $\\angle MPN=\\angle HOT$[/color]",
  "Solution_1": "It's a very nice problem, also $\\overline{MP}\\parallel\\overline{O'O}$",
  "Solution_2": "[color=red]Please post the complete solution! [/color]\r\n\r\nAnd I don't think $MP\\parallel OO'$  :wink:",
  "Solution_3": "$[HP] \\cap C(O)=S$. Let $K$ be on $[HN]$ such that $HK=OT$. Since $\\triangle HTO \\equiv \\triangle PHK$ it follows that $\\angle HOT = \\angle PKH$. It's sufficient to prove that $\\triangle KMP \\sim \\triangle PMN$. Denote $r_{O}_'$ $=r$, $r_{O}=R$, $MH=x$ and $HS=y$. $(r+x)^{2}+(y+R)^{2}=(r+R)^{2}$ $(1)$.\r\n$\\triangle KMP \\sim \\triangle PMN$ $\\Leftrightarrow$ $MP^{2}=MK \\cdot MN$ $\\Leftrightarrow$ $x^{2}+y \\cdot (y+2R)=(R-x) \\cdot 2r$ $\\Leftrightarrow$ $(1)$.",
  "Solution_4": "[color=darkblue][b][u]Thank you,[/color] [color=darkred]April[/color][color=darkblue], for your nice proposed problem ! Thank you,[/color] [color=darkred]SonicX[/color][color=darkblue], for your nice proof ![/u][/b]\n\n[b][u]Remark.[/u][/b] Denote the second intersection $R$ between the line $PM$ and the circle $(O')$. Then the quadrilateral $PNRH$ is cyclically and $R\\in TH$ ![/color]",
  "Solution_5": "[quote=\"April\"][color=darkred]In figure, two cicles $C(O,R)$ and $C(O',r)$ are external tangent. $OH\\perp MN,\\,HT=HP$ and $HT$ is a tangent of circle $(O)$ Prove that $\\widehat{MPN}=\\widehat{HOT}$.[/color][/quote] [color=darkblue][b][u]Proof.[/u][/b] Denote $HP=HT=y$, $HM=x$. Observe that $O'O^{2}=O'H^{2}+OH^{2}=O'H^{2}+HT^{2}+OT^{2}$ $\\implies$ $(R+r)^{2}=(r+x)^{2}+y^{2}+R^{2}$ $\\implies$ $\\boxed{2Rr=x^{2}+y^{2}+2rx}\\ \\ (*)$. Denote $m(\\widehat{MPN})=u$, $m(\\widehat{HOT})=v$, $m(\\widehat{MPH})=w$. Therefore, $\\tan (u+w)=\\frac{HN}{HP}=\\frac{2r+x}{y}$, $\\tan w=\\frac{MH}{HP}=\\frac{x}{y}$ and $\\tan v=\\frac{HT}{OT}=\\frac{y}{R}$. Must prove $u=v$, i.e. $\\tan [(u+w)-w]=\\tan v$ $\\Longleftrightarrow$ $\\frac{\\tan (u+w)-\\tan w}{1+\\tan (u+w)\\tan w}=\\tan v$ $\\Longleftrightarrow$ $\\frac{\\frac{2r+x}{y}-\\frac{x}{y}}{1+\\frac{2r+x}{y}\\cdot\\frac{x}{y}}=\\frac{y}{R}$ $\\Longleftrightarrow$ $(*)$.[/color]",
  "Solution_6": "[color=indigo]Thanks for all of your nice solutions and thank you for your nice [b]Remark,[/b] [b]Virgil Nicula![/b]\nAnd now, I'll present my solution (like [b]Virgil Nicula's[/b])\n\n[u][b]Proof.[/b][/u] We have:\n\\[\\tan\\widehat{MPN}=\\tan\\left(\\widehat{HPN}-\\widehat{HPM}\\right)=\\frac{\\tan\\widehat{HPN}-\\tan\\widehat{HPM}}{1+\\tan\\widehat{HPN}\\tan\\widehat{HPM}}\\\\ =\\frac{\\frac{HN}{HP}-\\frac{HM}{HP}}{1+\\frac{HN}{HP}\\cdot\\frac{HM}{HP}}=\\frac{MN\\cdot HP}{HP^{2}+HM\\cdot HN}=\\frac{2R'\\cdot HP}{HO^{2}-R^{2}+HO'^{2}-R'^{2}}\\\\ =\\frac{2R'\\cdot HT}{OO'^{2}-R^{2}-R'^{2}}=\\frac{2R'\\cdot HT}{2RR'}=\\frac{HT}{OT}=\\tan\\widehat{HOT}\\]\nHence: $\\widehat{MPN}=\\widehat{HOT}$[/color]",
  "Solution_7": "I'm not sure if Nicula's statements are correct...?\r\n[img]8726[/img]"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "A 1001-digit number begins with 6. The number formed by any two adjacent digits is divisible by 17 or 23. Write down the last six digits.",
  "Solution_1": "692346.\r\n\r\nMultiples of 17: 17,34,51,68,85\r\nMultiples of 23: 23,46,69,92\r\n\r\nAs 7 doesnt occur in tens place, so numbers with 7 can't occur => numbers with 1 => numbers with 5 => numbers with 8 cant occur\r\nThe final numbers are: 69,92,23,34,46\r\n\r\nThe number 69234 repeats itself"
}
{
  "Tag": [
    "conics",
    "trigonometry",
    "geometry",
    "rhombus",
    "HMMT",
    "email",
    "AMC"
  ],
  "Problem": "Post your results here. I understand the Hopkins (Varsity) is giving Choate a run for its money this year...impressive.",
  "Solution_1": "66363\r\n60336\r\n36613\r\n\r\n\r\nApparently, the juniors this year are really crappy, we have the top junior I think and most of the sophomores are better than the juniors, and we have someone who's tied for the top senior (all perfect scores) I think the Hopkins two freshmen are the only ones in top at all.\r\n\r\nI think the first two meets, Hopkins lost one by 2 points, and one by 3 points.\r\n\r\nNow from what I hear, Choate brings like, people who've already been through college from other countries, and just want a high school diploma so they're going from one year which IMO is a bit unfair, considering they Choate scored a 104/105 on the most recent meet which was the highest score ever in GNHML.",
  "Solution_2": "Actually, I don't think that's true (the part about Choate). As far as I can see, all their best members have been there all 4 years, or  whatever year they are. I'm referring to Rosen, Cao, Watanaprakornkul, Choi, Rajagopalan, etc.",
  "Solution_3": "[quote=\"Phelpedo\"]Actually, I don't think that's true (the part about Choate). As far as I can see, all their best members have been there all 4 years, or  whatever year they are. I'm referring to Rosen, Cao, Watanaprakornkul, Choi, Rajagopalan, etc.[/quote]\r\n\r\nMaybe it's just a joke I guess, I know for a fact that they do offer a program like that though.",
  "Solution_4": "You're probably thinking about a post-graduate program (my school has one too), which does fit your description, but most all PGs are for sports, not math...except maybe for exeter.",
  "Solution_5": "Well, our coach says this is probably Hopkins' 2nd best team ever.",
  "Solution_6": "what is the GNHML?\r\n-jorian",
  "Solution_7": "[quote=\"jhredsox\"]what is the GNHML?\n-jorian[/quote]\r\n\r\nGreater New Haven Mathematics League",
  "Solution_8": "ahhhh!, i live in glastonbury\r\n\r\ndo u know of any state wides? or any hartford ones?\r\n-jorian",
  "Solution_9": "[quote=\"jhredsox\"]ahhhh!, i live in glastonbury\n\ndo u know of any state wides? or any hartford ones?\n-jorian[/quote]\r\n\r\nUmm.. there's one in Hartford and there's the state-wide CSAML.",
  "Solution_10": "do most high schools participate\r\n\r\nif so, i can't wait!\r\n-jorian",
  "Solution_11": "[quote=\"jhredsox\"]do most high schools participate\n\nif so, i can't wait!\n-jorian[/quote]\r\n\r\nI don't live in Hartford so I wouldn't know. Around here I think there's about 15 that participate.",
  "Solution_12": "66363\r\n60336\r\n36614\r\n66644\r\n\r\nHopkins tied with Hamden last week which is pretty sad. Do individuals count your top 4 or top 5 scores? As if it's top 5 I'm screwed (due to not being put on rounds I wanted 1st 2 meets and careless errors) but if it's top 4, I still have a chance as the coach is listening to me now after seeing the 1st two meets and the fact that I've worked on my careless errors.",
  "Solution_13": "For individuals, it's top 4 scores.\r\n\r\nChoate has won the last 2 contests by 101-90 and 92-78.",
  "Solution_14": "66643\r\n61463\r\n66643\r\n66666\r\n\r\nI'm trying to give Choate's sophomores and freshmen a run for their money this year.",
  "Solution_15": "beanwholeaworth: Why not the juniors?",
  "Solution_16": "Only 3 of the numbers are counted...so what's your official score for each round?",
  "Solution_17": "Actually, the thing about Choate is true, so far, two members of their team have all perfect scores. The next person has a score of 59. The two members are Sarita Bunsupha and Phumpong Watanaprakornkul. The former is a 2006 IMO Bronze Medalist and the latter a 2006 IMO Gold Medalist both from Thailand.",
  "Solution_18": "Hm, well this makes me feel better that we are beating Choate at the neml .",
  "Solution_19": "66643\r\n61463\r\n66643\r\n66666 \r\n36663\r\n\r\nI'm in third!",
  "Solution_20": "Yifan, how did you do?\r\n\r\nThe two PGs from choate are in their Thai Scholars program. Two students come every year.",
  "Solution_21": "33636\r\n60336\r\n36614\r\n66644 \r\n632\r\n\r\nRound 1 #3 was just stupid.",
  "Solution_22": "YAY FOR PHUMPONG WATANAKARUNAPDFGADPFOGVKADL;FVKRE\r\n\r\n\r\n\r\n33636\r\n60336\r\n36614\r\n66644\r\n632\r\n666",
  "Solution_23": "Strangely, even Choate's Thai scholars didn't do very well on the last one. Must have been hard..,",
  "Solution_24": "No, our coach was making fun of them. He was like IMO Medalists can't even figure out conic sections. Cause they both failed the Conic Sections round last meet and probably failed the Trig round.",
  "Solution_25": "Oh and they don't know what a rhombus is either.kn",
  "Solution_26": "Uhh, what's the point of the last post? Yeah, they sometimes struggle in rounds with a lot of english terms, maybe b/c they're not american? Just putting it out there.\r\n\r\nI don't think they did too badly on the trig round. I mean, our A team did get 18/18 for that round, but who knows.\r\n\r\nBtw, phelpedo, how do you get such accurate info on GNHML? I'm just curious, because it's pretty impressive, and I don't even know some of the stuff you find out.",
  "Solution_27": "I keep an eye out  :wink: \r\n\r\nhttp://studentweb.choate.edu/mathteam/menu.htm",
  "Solution_28": "yifan why was everybody laughing when you won",
  "Solution_29": "Final scores:\r\n66643\r\n61463\r\n66643\r\n66666\r\n36663\r\n6-560\r\n\r\nIf I just hadn't screwed up Round 3 #1 I would have tied for third!\r\n\r\nOh well, first for underclassmen is enough. Hopkins did pretty well  in terms of awards. Teamwise, the things that killed us were Choate's Thai Scholars and the lack of anyone good who could do the later rounds (Genevieve didn't do so well for the most part).",
  "Solution_30": "[b]33[/b]6[b]3[/b]6\r\n[b]603[/b]36\r\n[b]366[/b]14\r\n[b]666[/b]44\r\n[b]632[/b]\r\n[b]666[/b]",
  "Solution_31": "New year!\r\n\r\n5-[b]666[/b]: 18\r\n\r\n...unless the people decide to take 2 points for me not putting a dollar sign on 3/2 (the entire problem was talking in dollars, but apparently I still needed to put it in the answer after \"13\"... sigh).\r\n\r\nEdit: So apparently they announced you had to put units... but I wasn't there because I was challenging 1/3... just so stupid.\r\n\r\nToo bad Gardner's gone.",
  "Solution_32": "5-[b]664[/b]: 16\r\n3[b]66[/b]-[b]4[/b]: 16\r\n\r\nBecause of course 2 isn't an integer between 0 and 5.\r\nProbably not gonna 72 this thing this year. Oh well. I still have two more years.\r\n\r\nTied Choate this meet. Looks like we might give you guys a run for your money this year, Yifan. Look out :D\r\n\r\nI didn't particularly like these problems. 3/3 I essentially got by guesswork - I saw 30 and thought 5-12-13.\r\n\r\nEdit: This forum is dead. Yuck.",
  "Solution_33": "As an interesting anecdote, look up the definition of \"inchoate.\"",
  "Solution_34": "Third meet!\r\n\r\n5-[b]664[/b]: 16\r\n3[b]66[/b]-[b]4[/b]: 16 \r\n66[b]363[/b]: 12\r\n\r\nYuck. Oh well, we beat Choate (72-71, now down 2 on the year). I'm happy. It'll drop anyway. Maybe I'll be able to drop one of those 16s...\r\n\r\nNext meet December 13 at Hillhouse. BTW, I started a [url=http://hs.facebook.com/group.php?gid=2376655891]facebook group[/url] if anyone wants to join... not that anyone ever reads this.",
  "Solution_35": "Lessee.. If we can get Jon, Yifan, and Phelpedo posting again... and you...\r\n\r\nThat's 5 I guess.\r\n\r\nAnyway, back on topic, Yifan needs to post his scores.",
  "Solution_36": "5-[b]664[/b]: 16\r\n3[b]66[/b]-[b]4[/b]: 16\r\n66[b]363[/b]: 12\r\n--[b]615[/b]: 12\r\n\r\nChoate: 75\r\nHopkins: 70\r\n\r\nGG to me.",
  "Solution_37": "So are any of you guys coming up for HMMT?",
  "Solution_38": "[quote=\"Phelpedo\"]So are any of you guys coming up for HMMT?[/quote]\r\nWish Hopkins was... sadly, it's hard to get the administration moving on much of anything.\r\n\r\nEdit: Meet 5!\r\n\r\n5-[b]664[/b]: 16\r\n3[b]66[/b]-[b]4[/b]: 16\r\n66[b]363[/b]: 12\r\n--[b]615[/b]: 12\r\n5-[b]666[/b]: 18\r\n\r\nYay!\r\n\r\nChoate is 9 points ahead going into the final meet... not going to happen this year. Sadly.\r\n\r\nChoate: 393\r\nHopkins: 384",
  "Solution_39": "SO I GOT AN EMAIL THAT SAID I GOT A PM AND I REMEMBERED THIS FORUM\r\nuh i dunno my exact scores\r\nBUT MY CULMULATIVE SCORE IS THE SAME AS WAWB'S\r\nbest 4 are something stupid like 1 lower or some **|* like that\r\nI HAVE 2 13S AND 1 18 AS WELL AS A 15!!!!!!!!!!!!!!!!!!! A 13 WAS DROPPED!!!!!!!!!\r\n\r\nso itd look something like this 16+16+12+12+18=74-13-13-18-15=15 so my scores are 13 13 15 15 18.\r\n\r\nCHOATE CHEATS.\r\n\r\nwawb: will you do itest comp sci with me?\r\n\r\nPHELPEDO: HAVE YOU SEEN HOTEL RWANDA?",
  "Solution_40": "HEY GUYS NEW YEAR\r\n\r\n6[b]566[/b]6\r\n\r\nHopkins > Choate this year I think.",
  "Solution_41": "Are you sure??",
  "Solution_42": "did Rajopolan make USAMO?? or any of em?",
  "Solution_43": "I recently basically got kicked off Choate A for no reason.... well, maybe it was b/c I thought 72/2=9. But still. At least there will be rotation.\r\nThen for this meet I phailed by doing round five with no memory of absolute value inequalities.\r\nIf I'd done ANY other round I would have had 18.\r\ninstead, 12. ..... :blush: \r\nAren't Choate and Hopkins tied for this meet???\r\nDid Hopkins get a perfect team round too?\r\nBy the way, I think Stephanie Choi, Andrew? Kim, and the Thai scholar made USAMO in 2008. New Thai scholars this year are awesome ppl!",
  "Solution_44": "[quote=\"academicsrule\"]I recently basically got kicked off Choate A for no reason.... well, maybe it was b/c I thought 72/2=9. But still. At least there will be rotation.\nThen for this meet I phailed by doing round five with no memory of absolute value inequalities.\nIf I'd done ANY other round I would have had 18.\ninstead, 12. ..... :blush: \nAren't Choate and Hopkins tied for this meet???\nDid Hopkins get a perfect team round too?\nBy the way, I think Stephanie Choi, Andrew? Kim, and the Thai scholar made USAMO in 2008. New Thai scholars this year are awesome ppl![/quote]\r\n\r\nStephanie Choi = Most annoying person ever\r\n\r\nHow did she do today?"
}
{
  "Tag": [
    "probability",
    "geometry",
    "3D geometry",
    "sphere"
  ],
  "Problem": "A point in space $ (x,y,z)$ is randomly selected so that $ \\minus{}1\\le x \\le 1$,$ \\minus{}1\\le y \\le 1$,$ \\minus{}1\\le z \\le 1$.  What is the probability that $ x^2\\plus{}y^2\\plus{}z^2\\le 1$?",
  "Solution_1": "Because $ x^2\\plus{}y^2\\plus{}z^2 \\le 1$, the point must lie in the unit sphere centered at $ (0,0,0)$.  The volume of this is $ \\frac{4}{3}\\pi$.  The total volume of the cube is $ 2^3\\equal{}8$, so the probability is $ \\frac{\\frac{4}{3}\\pi}{8}\\equal{}\\boxed{\\frac{\\pi}{6}}$.",
  "Solution_2": "Why is it in the unit sphere??",
  "Solution_3": "Distance formula.",
  "Solution_4": "It is similar to how a unit circle centered at the origin is $ x^2\\plus{}y^2\\equal{}1$.",
  "Solution_5": "ahhhh...",
  "Solution_6": "Two comments:\n\nIt's nice the way they tie in some simple geometry concepts to a three dimensional graph.\n\nBut why, why, why, was this a level 25 problem?  Actually, the fact that the problem requires geometry and this is C&P might account for it, but still...",
  "Solution_7": "[quote=cw357]Two comments:\n\nIt's nice the way they tie in some simple geometry concepts to a three dimensional graph.\n\nBut why, why, why, was this a level 25 problem?  Actually, the fact that the problem requires geometry and this is C&P might account for it, but still...[/quote]\n\nik its ez",
  "Solution_8": "because it's a little hard for some people to notice that the region we want (x,y,z) in is a sphere",
  "Solution_9": "Yeah, geometric probability can be a hard one sometimes. \n\nI once got the 3 people going to meet in the restaurant from 2 to 4 for a meeting and i [size=50]almost[/size] killed myself.",
  "Solution_10": "Yes visualizing 3d for that problem is hard [s]until you decide to use integrals to overkill the problem[/s]",
  "Solution_11": "The fact that this is level 25 shows that many people taking CP don't know the area of a sphere formula.",
  "Solution_12": "How come I never see any prealgebra questions in these alcumus problems? All of these are like, geometry or highschool stuff.",
  "Solution_13": "How do integrals help except if you forgot formula for volume of sphere?",
  "Solution_14": "[hide=solution]The region that will satisfy $ x^2+y^2+z^2 \\le 1,$ is a unit sphere centered at $(0, 0, 0),$ so the volume is $\\frac{4}{3}\\pi.$ The possible region is a cube with side length $2,$ so it has volume $8,$ and the desired answer is $ \\frac{\\frac{4}{3}\\pi}{8}=\\boxed{\\frac{\\pi}{6}}$. [/hide]",
  "Solution_15": "This is a really nice problem :D",
  "Solution_16": "why is this #25??? bro alcumus puts AIME #15s and AMC 10 #1's in the same category",
  "Solution_17": "I guess it has to do with the skill level of the people doing a certain subject. AIME #15's would be more likely to be in IA or Precalc, which are the intermediate topics (and very hard ngl).",
  "Solution_18": "[quote=judgefan99]This is a really nice problem :D[/quote]\n\nReally had to think for this one"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "function",
    "limit",
    "logarithms",
    "derivative"
  ],
  "Problem": "(1) Find the minimum value of $ \\int_{ \\minus{} \\pi}^{\\pi} (x \\minus{} a\\sin x)^2dx$ \r\n\r\n(2) Evalute $ \\int_{ \\minus{} 1}^{1} |e^{x} \\minus{} 1|dx$\r\n\r\n(3) Evaluate $ \\int_{ \\minus{} \\pi}^{\\frac {\\pi}{2}} |\\sin x|dx$\r\n\r\n(4) Find the continuous function $ f(x)$ satisfying $ f(x) \\equal{} x \\plus{} \\int_0^{\\pi} f(t)\\sin t\\ dt$\r\n\r\n(5) Find $ \\frac {d}{dx}\\int_0^{x} (x \\minus{} t)e^{t}dt$\r\n\r\n(6) $ \\lim_{x\\rightarrow 0} \\int_0^{x} \\frac {(1 \\plus{} \\sin t)^2}{x}dt$\r\n\r\n(7) Evaluate $ \\int_0^1 xe^{x}dx$\r\n\r\n(8) Evaluate $ \\int_1^{e} x\\ln x\\ dx$\r\n\r\n(9) Evaluate $ \\int_0^{\\frac {\\pi}{2}} |\\cos 2x|dx$\r\n\r\n(10) Evaluate $ \\int_0^{\\pi} |\\sin x \\plus{} \\cos x|dx$",
  "Solution_1": "[hide=\"7)\"]$ \\int_{0}^{1}xe^{x}dx  \\\\ \\text{Using Integration by parts,we get} \\\\ (x\\minus{}1)e^x |_{0}^{1} \\\\ \\equal{}1$[/hide]\n[hide=\"8)\"]$ \\int_{1}^{e}x\\ln xd\\ x \\\\ \\text{Using integration by parts,we get} \\\\ \\frac{x^2}{4}(2\\ln x\\minus{}1)|_{1}^{e} \\\\ \\equal{}\\frac{e^2\\plus{}1}{4}$[/hide]",
  "Solution_2": "5) $ \\int_{0}^{x}xe^{t}\\minus{}te^{t}\\,dt\\equal{}e^{t}(x\\minus{}t\\plus{}1)\\bigg|_{0}^{x}\\equal{}e^{x}\\minus{}x\\minus{}1$.  Taking the derivative yields $ e^{x}\\minus{}1$ (assuming I didn't make a careless arithmetic mistake while doing the integral).",
  "Solution_3": "[hide=\"9)\"]$ \\int_{0}^{\\frac{\\pi}{2}}|\\cos 2x|dx$\n\n$ \\cos 2x$ is positive in the interval $ [0,\\frac{\\pi}{4}]$ and negative in $ (\\frac{\\pi}{4},\\frac{\\pi}{2}]$\n\nHence the integral can be written as $ \\int_{0}^{\\frac{\\pi}{4}}\\cos 2xdx- \\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}}\\cos 2x dx$\n\n$ \\Longrightarrow \\frac{1}{2} (\\sin 2x|_0^{\\frac{\\pi}{4}} - \\sin 2x|_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} ) \\\\ =\\frac{1}{2} (1+1) \\\\ =1$\n\nTherefore $ \\int_{0}^{\\frac{\\pi}{2}}|\\cos 2x|dx =1$[/hide]\n\n[hide=\"My doubt\"]Does a definite integral denote the area under the given curve in the given interval ?(just got confused)[/hide]",
  "Solution_4": "I can't find so far any mistake in your solutions. :) \r\n\r\n@picture74  A definite integral doesn't always denote the area under the given curve.\r\n\r\nFor (5), we can use Fundamental Thorem of Calculus.\r\n\r\nFor (9), we can remark the periodicity for the integral of Triangular function with absolute value, in this case set $ 2x \\equal{} t$.",
  "Solution_5": "[hide=\"10)\"]$ \\int_{0}^{\\pi}|\\sin x\\plus{}\\cos x|dx \\\\ \\equal{}\\int_{0}^{\\pi} |\\sin \\left(x\\plus{}\\frac{\\pi}{4}\\right)|dx$\n\n$ \\sin \\left(x\\plus{}\\frac{\\pi}{4}\\right)$ is positive in $ \\left[0,\\frac{3\\pi}{4}\\right]$ and negative in $ \\left(\\frac{3\\pi}{4},\\pi\\right]$\n\nHence,the given integral can be written as $ \\int_{0}^{\\frac{3\\pi}{4}}\\sin \\left(x\\plus{}\\frac{\\pi}{4}\\right)dx \\minus{} \\int_{\\frac{3\\pi}{4}}^{\\pi}\\sin \\left(x\\plus{}\\frac{\\pi}{4}\\right)dx$\n\nSolving the above integral yields $ 1\\plus{}\\sqrt{2}$\n\nHence $ \\int_{0}^{\\pi}|\\sin x\\plus{}\\cos x|dx\\equal{}1\\plus{}\\sqrt{2}$[/hide]\r\n\r\n[quote=\"Kunny\"]@picture74 A definite integral doesn't always denote the area under the given curve.[/quote]\r\nOther than area,what else does it represent?",
  "Solution_6": "[quote=\"picture74\"]\n[quote=\"Kunny\"]@picture74 A definite integral doesn't always denote the area under the given curve.[/quote]\nOther than area,what else does it represent?[/quote]\r\n\r\nSometimes denotes the volume, the length of arc, e.t.c.\r\n\r\nYour answer to (10) is incorrect.",
  "Solution_7": "[quote=\"kunny\"]Sometimes denotes the volume, the length of arc, e.t.c.[/quote]\nThank you  :) \n[quote=\"kunny\"]Your answer to (10) is incorrect.[/quote]\r\nI couldn't figure out the mistake in my solution,Could you locate it?",
  "Solution_8": "$ a\\sin x \\plus{} b\\cos x \\equal{} \\sqrt {a^2 \\plus{} b^2}\\sin (x \\plus{} \\alpha)$ for a fixed angle $ \\alpha$.",
  "Solution_9": "@picture74\r\n[quote]$ \\int_{0}^{\\pi}|\\sin x\\plus{}\\cos x|dx \\\\ \\equal{}\\int_{0}^{\\pi} |\\sin \\left(x\\plus{}\\frac{\\pi}{4}\\right)|dx$[/quote]\r\nThat's the trouble. Actually, \r\n$ |\\sin x \\plus{}\\cos x|\\equal{}\\sqrt{2}|\\sin\\left(x\\plus{}\\dfrac{\\pi}{4}\\right)|$\r\nFurther, the final step is incorrect. The integral in the final step will add up to 2.\r\nAs such the required answer is $ 2\\sqrt{2}$",
  "Solution_10": "[quote=\"kaymant\"]@picture74\n[quote]$ \\int_{0}^{\\pi}|\\sin x \\plus{} \\cos x|dx \\\\\n\\equal{} \\int_{0}^{\\pi} |\\sin \\left(x \\plus{} \\frac {\\pi}{4}\\right)|dx$[/quote]\nThat's the trouble. Actually, \n$ |\\sin x \\plus{} \\cos x| \\equal{} \\sqrt {2}|\\sin\\left(x \\plus{} \\dfrac{\\pi}{4}\\right)|$\nFurther, the final step is incorrect. The integral in the final step will add up to 2.\nAs such the required answer is $ 2\\sqrt {2}$[/quote]\r\n\r\nThat's right.\r\n\r\n$ \\int_0^{\\pi} |\\sin x \\plus{} \\cos x|dx \\equal{} \\sqrt {2}\\int_0^{\\pi} \\sqrt {2}\\left|\\sin \\left(x \\plus{} \\frac {\\pi}{4}\\right)\\right|dx$\r\n\r\n$ \\equal{} \\sqrt {2}\\int_{\\frac {\\pi}{4}}^{\\pi \\plus{} \\frac {\\pi}{4}} |\\sin x|dx$\r\n\r\n$ \\equal{} \\sqrt {2}\\int_0^{\\pi} |\\sin x|dx \\equal{} 2\\sqrt {2}.$",
  "Solution_11": "Thank you kunny and kaymant,for explaining my mistake!  :)",
  "Solution_12": "[hide=\"#2\"]\n\\[ \\varphi \\int_{ \\minus{} 1}^{1}|e^x \\minus{} 1|\\text{ d}x\\]\n$ f(x) \\equal{} e^x \\minus{} 1$ has a zero at $ x \\equal{} 0$ & $ f^{\\prime}(0) > 0$. Therefore we compute\n\\[ \\varphi \\equal{} \\int^1_0 (e^x \\minus{} 1)\\text{ d}x \\minus{} \\int^0_{ \\minus{} 1} (e^x \\minus{} 1)\\text{ d}x\\]\nThis has a value of\n\\[ \\boxed{\\varphi \\equal{} e \\plus{} \\frac {1}{e} \\minus{} 2}\\]\n[/hide]\n[hide=\"#4\"]\n\\[ f(x) \\equal{} x \\plus{} \\int_{0}^{\\pi}f(t)\\sin{t}\\text{ d}t\\]\nAssuming $ f(x) \\equal{} x \\plus{} \\delta$ because $ f'(x) \\equal{} 1$, I'll evaluate the definite integral\n\\[ \\int^{\\pi}_0 f(t)\\sin{t}\\text{ d}t \\equal{} \\int^{\\pi}_0 (t\\sin{t} \\plus{} \\delta \\sin{t})\\text{ d}t \\equal{} \\minus{} t\\cos{t} \\plus{} \\sin{t} \\minus{} \\delta \\cos{t} \\plus{} C \\equal{} \\pi \\plus{} 2\\delta\\]\nNow,\n\\[ x \\plus{} \\delta \\equal{} f(x) \\equal{} x \\plus{} \\pi \\plus{} 2\\delta \\Longrightarrow \\delta \\equal{} \\minus{} \\pi\\]\n\n\\[ \\Leftrightarrow \\boxed{f(x) \\equal{} x \\minus{} \\pi}\\]\nThis is my favorite problem here. I don't know how to solve it without that assumption, but I'd love to see it.\n[/hide]\n[hide=\"#6\"]\n\\[ \\varphi \\equal{} \\lim_{x\\rightarrow 0}\\int_{0}^{x}\\frac {(1 \\plus{} \\sin t)^{2}}{x}dt\\]\nExpand, substitute & integrate to find the indefinite representation\n\\[ \\varphi \\equal{} \\frac {6t \\minus{} 8\\cos{t} \\minus{} \\sin{2t}}{4x} \\plus{} C\\]\nbefore evaluation. Plug in $ t \\equal{} x,0$, but hold the limit on $ x$ to obtain\n\\[ \\varphi \\equal{} \\frac {8 \\plus{} 6x \\minus{} 8\\cos{x} \\minus{} \\sin{2x}}{4x}\\]\nApply L'Hopital:\n\\[ \\lim_{x\\to 0}\\frac {3 \\plus{} 4\\sin{x} \\minus{} \\cos{x}}{2} \\equal{} \\boxed{1}\\]\n[/hide]",
  "Solution_13": "Those answers are correct. :lol: \r\n\r\nAre there someone who solve the problem #6 without using l'Hopital?",
  "Solution_14": "6) Without L'Hospital:\r\nSince $ x$ is constant as far as the integration is concerned, we have\r\n$ \\int_0^x (1\\plus{}\\sin t)^2\\ \\mathrm dt\\equal{}\\dfrac{1}{2}(4\\plus{}3x\\minus{}4\\cos x\\minus{}\\cos x\\sin x)$\r\nHence, the required limit is\r\n$ \\lim_{x\\to0}\\dfrac{4\\plus{}3x\\minus{}4\\cos x\\minus{}\\cos x\\sin x}{2x}$\r\n$ \\equal{}\\lim_{x\\to 0}\\left(\\dfrac{2(1\\minus{}\\cos x)}{x}\\plus{}\\dfrac{3}{2}\\minus{}\\dfrac{\\cos x\\sin x}{2x}\\right)$\r\nNow, note that \r\n$ \\lim_{x\\to0}\\dfrac{1\\minus{}\\cos x}{x}\\equal{}\\lim_{x\\to0}\\dfrac{2\\sin^2(x/2)}{x}\\equal{}2\\lim_{x\\to0} \\dfrac{x}{4}\\cdot \\left(\\dfrac{\\sin(x/2)}{(x/2)}\\right)^2$\r\n$ \\equal{}\\dfrac{1}{2}\\lim_{x\\to0}x\\cdot \\lim_{x\\to0}\\left(\\dfrac{\\sin(x/2)}{(x/2)}\\right)^2\\equal{}\\dfrac{1}{2}\\cdot0\\cdot1\\equal{}0$\r\nAnd $ \\lim_{x\\to0}\\dfrac{\\cos x\\sin x}{2x}\\equal{}\\dfrac{1}{2}\\lim_{x\\to0}\\cos x \\cdot \\lim_{x\\to0}\\dfrac{\\sin x}{x}\\equal{}\\dfrac{1}{2}\\cdot1\\cdot1\\equal{}\\dfrac{1}{2}$\r\nAs such\r\n$ \\lim_{x\\to 0}\\left(\\dfrac{2(1\\minus{}\\cos x)}{x}\\plus{}\\dfrac{3}{2}\\minus{}\\dfrac{\\cos x\\sin x}{2x}\\right)$\r\n$ \\equal{}2\\lim_{x\\to0}\\dfrac{1\\minus{}\\cos x}{x}\\plus{}\\lim_{x\\to0}\\dfrac{3}{2}\\minus{}\\lim_{x\\to0}\\dfrac{\\cos x\\sin x}{2x}$\r\n$ \\equal{}0\\plus{}\\dfrac{3}{2}\\minus{}\\dfrac{1}{2}\\equal{}\\boxed{1}$",
  "Solution_15": "Jus missed my chance , anyways thank you  sir .",
  "Solution_16": "hello for 3) we get $ \\int_{\\minus{}\\pi}^{\\frac{\\pi}{2}}|\\sin(x)|\\,dx\\equal{}\\int_{\\minus{}\\pi}^{0}\\minus{}\\sin(x)\\,dx\\plus{}\\int_{0}^{\\frac{\\pi}{2}}\\sin(x)\\,dx\\equal{}3$.\r\nSonnhard.",
  "Solution_17": "Anyway use  of L'Hospital really simplifies it.\r\n$ \\lim_{x\\to0}\\dfrac{\\int_0^x(1\\plus{}\\sin t)^2\\ \\mathrm dt}{x}$\r\n$ \\lim_{x\\to0}\\left(\\int_0^x(1\\plus{}\\sin t)^2\\  \\mathrm dt\\right)^\\prime\\equal{}\\lim_{x\\to0}(1\\plus{}\\sin x)^2\\equal{}1$",
  "Solution_18": "Let $ \\int (1+\\sin x)^2dx=F(x)+C$,\r\n\r\n$ \\lim_{x\\rightarrow 0} \\frac{\\int_0^{x}(1+\\sin t)^2dt}{x}$\r\n\r\n$ =\\lim_{x\\rightarrow 0} \\frac{[F(t)]_0^{x}}{x}$\r\n\r\n$ =\\lim_{x\\rightarrow 0} \\frac{F(x)-F(0)}{x}$\r\n\r\n$ =F'(0)$ $ \\because \\text{The definition of differential coefficient}$\r\n\r\n$ =1$. $ \\because F'(x)=(1+\\sin x)^2$"
}
{
  "Tag": [
    "function",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "if $ (X, \\Sigma , \\mu)$ is a finite measure space and $ f_n$ is an arbitrary sequence of real val $ \\Sigma$-measurable functions on X.  \r\n\r\nshow for every $ \\epsilon > 0$ \r\nthere exists $ E \\in \\Sigma$ with $ \\mu (E) < \\epsilon$ \r\nand a sequence $ \\{a_n\\}$ of positive reals with $ \\lim_{n \\to \\infty} a_n f_n (x) \\equal{} 0$ for every $ x \\in E^c$",
  "Solution_1": "taking a random stab at this (since i have to take a test on monday)\r\n\r\nmaybe for every positive number M, and positive int n we can define\r\n\r\n$ Q_{n,M} \\equal{} \\{E: |f_n|>M\\}$ so we see $ Q^c_{n,M} \\to X$ as $ M \\to \\infty$\r\n\r\nso $ \\mu(Q_{n,M}) \\to 0$ as $ M \\to \\infty$\r\n\r\nso fix $ \\epsilon > 0$ then for each n, there exists $ M_n$ with $ \\mu(Q_{n,M_n})  < \\frac{\\epsilon}{2^n}$\r\n\r\nletting $ E\\equal{}\\cup_{n \\in N} Q_{n,M_n}$ we have $ \\mu (E) < \\epsilon$\r\n\r\nand hopefully... we can define $ a_n \\equal{} \\frac{1}{nM_n}$ so we get\r\n\r\n$ |\\lim_{n \\to \\infty} a_nf_n(x)| \\le \\lim_{n \\to \\infty} \\frac{|f_n(x)|}{nM_n} \\le \\lim_{n \\to \\infty} n^{\\minus{}1} \\equal{}0$\r\n\r\nlegit?"
}
{
  "Tag": [],
  "Problem": "If $x^2=x+3$ then $x^5$ is equal to?",
  "Solution_1": "[quote=\"Melissa\"]If $x^2=x+3$ then $x^5$ is equal to?[/quote]\r\nTake x= sqrt(x+3) then take it to the fifth?",
  "Solution_2": "[hide=\"Hint\"]$x^5=x^2\\times{x^2}\\times{x}$\n[hide=\"Further hint if you are stuck\"]$x^5=(x+3)(x+3)x$[/hide][/hide]",
  "Solution_3": "Melissa's second hint gave it away. \r\n\r\n[hide]answer: \n\n$x^3+6x^2+9x$. [/hide]",
  "Solution_4": "[quote=\"236factorial\"]Melissa's second hint gave it away. \n\n[hide]answer: \n\n$x^3+6x^2+9x$. [/hide][/quote]\nWell, you can go further.\n[hide]\n$x^3+6x^2+9x$\n$=x^2(x+6)+9x$\n$=(x+3)(x+6)+9x$\n$=x^2+18x+18$\n$=(x+3)+18x+18$\n$=19x+21$ etc...[/hide]",
  "Solution_5": "[quote=\"frt\"][quote=\"236factorial\"]Melissa's second hint gave it away. \n\n[hide]answer: \n\n$x^3+6x^2+9x$. [/hide][/quote]\nWell, you can go further.\n[hide]\n$x^3+6x^2+9x$\n$=x^2(x+6)+9x$\n$=(x+3)(x+6)+9x$\n$=x^2+18x+18$\n$=(x+3)+18x+18$\n$=19x+21$ etc...[/hide][/quote]\r\nI think that there's no need for an et cetera  ;) $\\gcd (19,21)=1$.",
  "Solution_6": "$19x+21$ is the answer :)",
  "Solution_7": "[quote=\"Melissa\"]If $x^2=x+3$ then $x^5$ is equal to?[/quote]\r\n[hide]\n$x^5=x^2*x^2*x$\n$x^2=x+3$\n$x^5=(x+3)(x+3)(x)$\n$x^5=x^2+6x+9$\n[/hide]",
  "Solution_8": "IntrepidMath, please read the previous posts. We've already got the answer. Also, you forgot to multiply the expression by $x$. (Then that gives the same solution as 236factorial's, which can be simplified further.)"
}
{
  "Tag": [
    "function",
    "topology",
    "advanced fields",
    "advanced fields theorems"
  ],
  "Problem": "Just to remind myself... :\r\n\r\nLet $ X,Y$ be topological spaces, $ A,B\\subseteq X$. Is it true that if\r\n$ f: A\\rightarrow Y$, $ g: B\\rightarrow Y$ are continuous functions such that $ f\\equal{}g$ on $ A\\cap B$ then $ h$ defined as $ h\\equal{}f$ on $ A$ and $ h\\equal{}g$ on $ B$ is continuous on $ A\\cup B$?",
  "Solution_1": "no. but it is sufficient that $ A$ and $ B$ are open.",
  "Solution_2": "Ok, and when we have $ X: \\equal{} X\\times [0,1]$ and $ A \\equal{} X\\times [0,b]$, $ B \\equal{} X\\times [b,1]$ then it is true ?",
  "Solution_3": "It's also sufficient that $ A$ and $ B$ are both closed.\r\n\r\nThe proof, because it's a bit tricky:\r\nIt suffices to show that a set $ U$ which has $ U_{A}\\equal{}U\\cap A$ and $ U_{B}\\equal{}U\\cap B$ both relatively open is itself open. Let $ A'$ be the complement of $ A$ and $ B'$ be the complement of $ B$ in $ X$; these are both open.\r\n\r\nBy hypothesis, $ U_{A}\\equal{}V_{A}\\cap A$ and $ U_{B}\\equal{}V_{B}\\cap B$ for some open sets $ V_{A},V_{B}$ in $ X$. We now write $ U$ as the union of three open sets:\r\n$ U\\equal{}(V_{A}\\cap V_{B})\\cup (V_{A}\\cap B')\\cup (V_{B}\\cap A')$\r\nThe first set is contained in $ U$, and contains $ U\\cap A\\cap B$. The second is exactly the part of $ U$ in $ A$ but not $ B$, and the third is the part of $ U$ in $ B$ but not $ A$.",
  "Solution_4": "I think it's correct (assuming $ A\\cup B\\equal{}X$).",
  "Solution_5": "Yes, I did assume that. We obviously can't get any continuity information outside the union of $ A$ and $ B$, so I restricted to that union.",
  "Solution_6": "[quote=\"-oo-\"]no. but it is sufficient that $ A$ and $ B$ are open.[/quote]\r\n\r\nI agree.\r\n\r\nTo let a function to be continuous, is that it maps every open set to another open set and every open set in Y is mapped from an open set in X.\r\n\r\nWhen $ A$ and $ B$ are open,\r\nfor an open set $ S\\subseteq A\\cup B$,\r\nwe can find two open set $ S_{A}$ and $ S_{B}$ that are open in $ A$ and $ B$ ($ S_{A}= S\\cap A$ and $ S_{B}= S\\cap B$), where $ S_{A}\\cup S_{B}= S$\r\nand then $ f[S] = f[S_{A}\\cup S_{B}] = f[S_{A}]\\cup f[S_{B}]$.\r\nand $ f[S_{A}]\\cup f[S_{B}]$ is known to be open.\r\n\r\nthe opposite is similar...",
  "Solution_7": "[quote=\"darken\"]To let a function to be continuous, is that it maps every open set to another open set ...[/quote]\r\n\r\nI don't think so  :)",
  "Solution_8": "It is ONE of the DEFINITION of continuous mapping...\r\n\r\nOtherwise, how do you define \"continuous\" in topology (especially discreet topology.)",
  "Solution_9": "[quote=\"darken\"]It is ONE of the DEFINITION of continuous mapping...\n\nOtherwise, how do you define \"continuous\" in topology (especially discreet topology.)[/quote]\r\n\r\n$ f: X\\to\\mathbb R$ with $ f(x) \\equal{} 1$ is continuous but $ f(X) \\equal{}\\{1\\}$ is not open in $ \\mathbb R$. Many counterexample can be found easily  :wink:",
  "Solution_10": "A function is continuous if the [i]preimage[/i] of every open set is open, i.e. if $ f^{\\minus{}1}$ takes open sets to open sets."
}
{
  "Tag": [
    "inequalities",
    "induction",
    "inequalities unsolved"
  ],
  "Problem": "\uff08m,n\u2208N, x\u2208[0,1]\uff09\r\nprove that: (1-x^n)^m+(1-(1-x)^m)^n\u22651",
  "Solution_1": "EDIT: Sorry about that [hide=\"Another proof\"]If $ x\\equal{}1\\vee x\\equal{}0$ the inequality is true. Assume $ 0\\neq x\\neq 1$ .\n\nDenote : $ f(x) \\equal{} 1 \\minus{} (1 \\minus{} x)^m$. The inequality is then equivalent to $ f(x)^{n}\\geq f(x^n)$. I will show using induction:\n \n$ (1 \\minus{} (1 \\minus{} x)^m)^2 \\geq 1 \\minus{} (1 \\minus{} x^2)^m \\Rightarrow \\frac {(1 \\minus{} x^2)^m \\plus{} ((1 \\minus{} x)^2)^m}{2} \\geq (1 \\minus{} x)^m \\Rightarrow \\frac {(1 \\plus{} x)^m \\plus{} (1 \\minus{} x)^m}{2} \\geq 1$ , true by Power mean (for every $ m \\geq 1$).\nAssume that it is true for some integer $ n\\equal{}k>1$ , for $ n\\equal{}k\\plus{}1$ :\n\n$ f(x)^{k \\plus{} 1} \\equal{} f(x)f(x)^k \\geq f(x)f(x^{k})$\n\nit'll be enough to prove that $ f(x)f(x^k)\\geq f(x^{k \\plus{} 1})$ , equivalent to:\n\n$ (1 \\minus{} x^{k \\plus{} 1})^m \\plus{} (1 \\minus{} x \\minus{} x^k \\plus{} x^{k \\plus{} 1})^m \\geq (1 \\minus{} x^k)^m \\plus{} (1 \\minus{} x)^m$\n\n, true by Karamata .\n\nSo it is true for every $ m \\geq 1$ , and natural $ n$.[/hide] May be is correct now .\r\n\r\nAlso probalistic interpretation [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=245389&search_id=282459874]here[/url]",
  "Solution_2": "[quote=\"alex2008\"][hide=\"Proof\"]Denote $ f(x) \\equal{} (1 \\minus{} x^n)^m \\plus{} (1 \\minus{} (1 \\minus{} x)^m)^n$, $ \\forall x \\in [0,1]$.\n\nSince $ f(0) \\equal{} 2$ and $ f(1) \\equal{} 1$ to prove $ f(x) \\ge 1$ in $ [0,1]$, we just need to show that $ f$ decreases in the interval.\n\n$ f'(x) \\equal{} \\minus{} n \\cdot m \\cdot x^{n \\minus{} 1} \\cdot (1 \\minus{} x^n)^{m \\minus{} 1} \\plus{} n \\cdot m \\cdot (1 \\minus{} x)^{m \\minus{} 1} \\cdot (1 \\minus{} (1 \\minus{} x)^m)^{n \\minus{} 1} < 0$\n$ \\iff \\left(\\frac {1 \\minus{} (1 \\minus{} x)^m}x \\right)^{n \\minus{} 1} < \\left(\\frac {1 \\minus{} x^n}{1 \\minus{} x} \\right)^{m \\minus{} 1}$,\n\nequivalent :\n$ (1 \\minus{} x \\plus{} .. \\plus{} m( \\minus{} x)^{n \\minus{} 1})^{m \\minus{} 1} < (1 \\plus{} x \\plus{} .. \\plus{} x^{m \\minus{} 1})^{n \\minus{} 1}$.[/hide]\nf(1)=f(0)=1  !!!!!!!!!\nAlso probalistic interpretation [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=245389&search_id=282459874]here[/url][/quote]",
  "Solution_3": "[quote=\"alex2008\"]EDIT: Sorry about that [hide=\"Another proof\"]If $ x \\equal{} 1\\vee x \\equal{} 0$ the inequality is true. Assume $ 0\\neq x\\neq 1$ .\n\nDenote : $ f(x) \\equal{} 1 \\minus{} (1 \\minus{} x)^m$. The inequality is then equivalent to $ f(x)^{n}\\geq f(x^n)$. I will show using induction:\n \n$ (1 \\minus{} (1 \\minus{} x)^m)^2 \\geq 1 \\minus{} (1 \\minus{} x^2)^m \\Rightarrow \\frac {(1 \\minus{} x^2)^m \\plus{} ((1 \\minus{} x)^2)^m}{2} \\geq (1 \\minus{} x)^m \\Rightarrow \\frac {(1 \\plus{} x)^m \\plus{} (1 \\minus{} x)^m}{2} \\geq 1$ , true by Power mean (for every $ m \\geq 1$).\nAssume that it is true for some integer $ n \\equal{} k > 1$ , for $ n \\equal{} k \\plus{} 1$ :\n\n$ f(x)^{k \\plus{} 1} \\equal{} f(x)f(x)^k \\geq f(x)f(x^{k})$\n\nit'll be enough to prove that $ f(x)f(x^k)\\geq f(x^{k \\plus{} 1})$ , equivalent to:\n\n$ (1 \\minus{} x^{k \\plus{} 1})^m \\plus{} (1 \\minus{} x \\minus{} x^k \\plus{} x^{k \\plus{} 1})^m \\geq (1 \\minus{} x^k)^m \\plus{} (1 \\minus{} x)^m$\n\n, true by Karamata .\n\nSo it is true for every $ m \\geq 1$ , and natural $ n$.[/hide] May be is correct now .\n\nAlso probalistic interpretation [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=245389&search_id=282459874]here[/url][/quote]\r\n\r\nbut f(1)=f(0)=1   ......",
  "Solution_4": "http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=75884&hilit=Czech+inequality"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "function",
    "induction",
    "functional equation",
    "algebra unsolved"
  ],
  "Problem": "Find all the functios f: Z ---> Z such that\r\n\r\nf(x+y+ f(y)) = f(x) + 2y",
  "Solution_1": "f(x)=x was saw.\u0130 study this question.",
  "Solution_2": "Ok, but i believe is difficult to prove.\r\nPlease, iff solve this problem. posted!!",
  "Solution_3": "It has 2 solutions. $f(x)=x$ and $f(x)=-2x$. The problem is from St. Petersburg MO 2000. As soon as I get free time, I'll post my complete solution.",
  "Solution_4": "Please, posted  your solution!!!! :oops: \r\nIm very interesting!!!",
  "Solution_5": "first, let $x=y=0$ gives $f(f(0))=f(0), f(0)=0$. Now let $x=0$ to give $f(y+f(y))=2y$ so $f(1+f(1))=2$. $f(1)=a$ $f(1+a)=2$ $f(2+a)=a+2$ (let $x=y=1$ in functional equation)\r\nNow I'm not sure about this but can't you assume that it is a polynomial, then show that it must be linear? Once you do that, we know it has exactly one zero (i. e. one $x$ such that $f(x)=0$) and we know this x is zero from above, so $f(x)=cx$, and we found a point $x=2+a$ above where $f(2+a)=2+a$ so $c(2+a)=2+a$ means $c=1$ of $a=-2$. If $c=1$ we have $f(x)=x$ if $a=-2$ we have $f(1)=-2(1)$ so $f(x)=-2x$\r\nthats my answer, probably faulty somewhere :wink:",
  "Solution_6": "[quote=\"me@home\"]first, let $x=y=0$ gives $f(f(0))=f(0), f(0)=0$. Now let $x=0$ to give $f(y+f(y))=2y$ so $f(1+f(1))=2$. $f(1)=a$ $f(1+a)=2$ $f(2+a)=a+2$ (let $x=y=1$ in functional equation)\nNow I'm not sure about this but can't you assume that it is a polynomial, then show that it must be linear? Once you do that, we know it has exactly one zero (i. e. one $x$ such that $f(x)=0$) and we know this x is zero from above, so $f(x)=cx$, and we found a point $x=2+a$ above where $f(2+a)=2+a$ so $c(2+a)=2+a$ means $c=1$ of $a=-2$. If $c=1$ we have $f(x)=x$ if $a=-2$ we have $f(1)=-2(1)$ so $f(x)=-2x$\nthats my answer, probably faulty somewhere :wink:[/quote]\r\n\r\nyou are wrong.I have solved this problem many times.I know two solutions.And also this function is injective.Now I post what came to my mind in 5 minutes(sorry I don't have enough time I want to look over other problems.)\r\nI will show that $f(0)=0$. Putting $x=0;y=0$ and denoting $f(0)=a$ we get $f(f(0))=f(0)=a$. so $f(a)=a$.Now put $y=0;x=a$ \r\nWe get $f(2a)=a$,now put $x=0;y=a$ we get $f(2a)=3a$.so we get $3a=a$ so $a=0$.Now continue yourself.....",
  "Solution_7": "so is my method wrong (some logical error in proof) or is my answer wrong? i'm not sure which...",
  "Solution_8": "No I mean your claims about the form of the function(polynomial) are wrong and (also if it is right you must prove)\r\nin this case it is wrong.",
  "Solution_9": "Please finish the solution Spider Boy!!",
  "Solution_10": "This is my solution.\r\nby induction,\r\n$f(x+k(y+f(y)))=f(x)+2kx$ for all integer k.\r\nk:$z+f(z)$ ->$f(x+(y+f(y))(z+f(z)))=f(x)+2(z+f(z))y=f(x)+2(y+f(y))z$\r\ntherefore, $\\frac{f(y)}{y}=\\frac{f(z)}{z}=c$ (y, z \u22600) , \r\nand\r\n$f(x+f(0))=f(x)$, If f(0) isn't 0, then f is a periodic funcion, so range of f is finite set. but $f(x+y+f(y))=f(x)+2y$ y-> 1, 2, 3, 4, ... contradiction!\r\n\u2234 $f(0)=0$ -> $f(0)=c\\cdot 0$\r\n\u2234$f(x)=cx$ for all integer x.\r\n$c(c+1)=2$ -> $c=1,-2$.\r\nSo $f(x)=x,-2x$.\r\n\r\nSorry, My english is bad.",
  "Solution_11": "At last I post the solution :blush: \r\nSet $x=-f(y)$ we get $f(y)=f(-f(y))+2y$\r\nNow set $y=-f(y)$ we get $f(x-f(y)+f(-f(y)))=f(x-f(y)+f(y)-2y)=f(x)-2f(y)$.\r\nso $f(x-2y)=f(x)-2f(y)$.\r\nset $y=1$ we get $f(x-2)=f(x)-2f(1)$. So $f(x)=f(x-2)+2f(1)$.\r\nNow $f(2n)=f(2n-2)+2f(1)=f(2n-4)+4f(1)=. . . . . . =f(2n-2n)+2nf(1)=2nf(1)$ because $f(0)=0$.look to my earlier post.\r\nNow we are setting $x=2n+1$ we get $f(2n+1)=f(2n-1)+2f(1)=f(2n-3)+4f(1)=. . . . . . . =f(2n-2n+1)+2nf(1)=(2n+1)f(1)$. So for all naturals $n$ we get $f(n)=nf(1)$.\r\nNow we go back to the equation $f(x-2y)=f(x)-2f(y)$.\r\nHere let us take $y$ as negative integer and $x$ as natural number: We get\r\n$xf(1)-2yf(1)=xf(1)-2f(y)$ so for all integers we get $f(z)=zf(1)$.Checking we get the answers \r\n$f(x)=x$ and $f(x)=-2x$."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "How many square inches are in the area of a square inscribed in a circle of radius 6 inches?",
  "Solution_1": "Diagonal measures 12, so 2s^2=144\r\n\r\n s^2=72, which is our answer.",
  "Solution_2": "Can any one explain in detail",
  "Solution_3": "We know that if the square is inscribed in a circle then the diameter of the circle is equal to the diagonal of the square. Since the radius of the circle is 6, the diameter is 12, which means the diagonal of the square is also 12. There is a trick to solving this problem or anyone similar. If you know the diagonal of a square and you need to find the area, just square the diagonal and divide by 2. So 12(12) = 144. 144 / 2 equals 72, which is the area of the square."
}
{
  "Tag": [
    "geometry",
    "projective geometry",
    "geometry unsolved"
  ],
  "Problem": "A convex quadrilateral ABCD with AC \u2260 BD is inscribed in a circle with center O. Let E be the intersection of diagonals AC and BD.  If P is a point inside ABCD such that $ \\angle PCB\\plus{}\\angle PAB\\equal{}\\angle PDC\\plus{}\\angle PBC\\equal{}90^{o}$. Prove that O, P and E are collinear.",
  "Solution_1": "this is an old problem,maybe it has been posted before.\r\nanyway,u can solve it using PASCAL's Theorem,or another way is to use the properties of orthogonal circles."
}
{
  "Tag": [
    "vector",
    "trigonometry"
  ],
  "Problem": "A particle has velocity vector(V) and there is a force vector(F) acting upon it.\r\nFiind acceleration vector(a).",
  "Solution_1": "$F=\\gamma^{3}ma$",
  "Solution_2": "Wrong answer.and you must fiind Vector(a) as a expression of other 2 vectors.i repeat,a relation between vectors",
  "Solution_3": "Is the force constant?",
  "Solution_4": "fiind the instant acceleration",
  "Solution_5": "Just a tiny modification of [b]Namiqfizik[/b]'s answer:\r\n\\[\\vec a = \\frac{\\vec F}{m}\\left(1-\\frac{v^{2}}{c^{2}}\\cos^{2}\\phi\\right)^{3/2}, \\]\r\nwhere $\\phi = \\measuredangle(\\vec v, \\vec F)$.\r\n\r\n[b]edit[/b]: lapsus calami :(",
  "Solution_6": "you must have vector v too.the final answer must contain vector F and V ,m (moovement velocity)and c",
  "Solution_7": "$\\vec v$ is in the formula as well, since $\\phi$ depends on its direction.",
  "Solution_8": "What you should reach is vec(a)=vec(F)/m-[vec(F)*vec(v)]*vec(v)/(m*sqr(c))",
  "Solution_9": "Ok, can you please post the solution?",
  "Solution_10": "I'll write it with M.Ecuations.",
  "Solution_11": "I don't get your second line. There's a mistake there, it should be $\\vec v \\cdot \\frac{\\text d\\vec p}{\\text dt}$, but I don't get the following step. Could you explain how you got it?",
  "Solution_12": "yes there was a mistake.sorry.here is more explicitely.",
  "Solution_13": "Um, could you attach that ppt file again? There seems to be a bug of some sort, I can't download it.",
  "Solution_14": "Ok, I get it now, thanks a lot!"
}
{
  "Tag": [],
  "Problem": "What does this mean\r\n\r\na2,+ a2,+ ...................+,a2\r\n 1    2                               n\r\n     \r\nThanks.",
  "Solution_1": "You're adding $ a_2$ to itself $ n$ times?",
  "Solution_2": "Do you mean:\r\n\r\n$ a_1^2 \\plus{} a_2^2 \\plus{} \\dots a_n^2$?\r\n\r\nLol yeah, I remember being confused about that before. Its just\r\n$ a_1^2 \\plus{} a_2^2 \\plus{} \\dots a_n^2 \\equal{} (a_1)^2 \\plus{} (a_2)^2 \\plus{} \\dots \\plus{} (a_n)^2$\r\n\r\nWe're just two lazy to use parenthesis. The $ _i^j$ isnt a new notation or anything. $ a_i$ is its own term, and its being raised to the second power.",
  "Solution_3": "tHanks thunder thats what u meant exceptit didnt come ou properly.",
  "Solution_4": "Q:A piece of equipment cost a certain factory Rs600,000.If it depriciates in value 15% the first,13.5% the next year,12% the third year,and so on.What will be its value at the end of 10 years ,all percentages applying to the original cost?"
}
{
  "Tag": [],
  "Problem": "This is a challenge problem so if you are above this level, leave this to Getting Started.\r\n\r\nFind the ordered pair of positive integers $(a,b)$ for which $a^2-b^2 = 37$.\r\n\r\nFurther challenge: Find ALL ordered pair of integers $(a,b)$ for which $a^2-b^2 = 37$.",
  "Solution_1": "[hide]Treating $a$ as $x$ and $b$ as $y$, the solution can be written as 4 ordered pairs$: (19, -18) , (19, 18) , (-19, 18) , (-19, -18)$ \n\n$a^2 - b^2 = 37$ can be rewritten as $(a+b)(a-b) = 37$\nSince $37$ is prime, it only has $2$ factors. Thus a and b both being integers must produce $2$ integers which are $1$ and $37$ (the two factors of $37$) as their sum or difference.\n\n$a + b = 37$\n$a - b = 1$\n\n$a = 19, b = 18$\n\n$a + b = 1$\n$a -b = 37$\n\n$a = 19, b = -18$\n\nSwitch the signs of the two pairs to make the sum and differences negatives so the equation becomes $2$ negative factors of $37$, which still yields $37$. \n[/hide]",
  "Solution_2": "I don't understand the difference between the two. Should the first one be \"Find [b]one[/b] such pair\"?\r\n\r\n[hide=\"Answer\"]Second part: $a^2-b^2=(a+b)(a-b)=37$, so one is $\\pm 37$ and the other is $\\pm 1$. If $a+b=37$, then $a-b=1$, so we have $(a,b)=(19,18)$. If $a+b=1$, then $a-b=37$, so we have $(a,b)=(19,-18)$. We can simply switch the signs of these for the other two answers, so we have $(-19,-18)$ and $(-19,18)$.[/hide]",
  "Solution_3": "The first one is positive integers and the second one is integers.",
  "Solution_4": "Silverfalcon, do you live in Nassau county and participate in this competition, I do, tell me if you do so I can look for you on the top 50.",
  "Solution_5": "[quote=\"mna851\"]Silverfalcon, do you live in Nassau county and participate in this competition, I do, tell me if you do so I can look for you on the top 50.[/quote]\r\n\r\nNah.  :D \r\n\r\nI don't even live there.. But I just know the ways to solve them.  :lol:",
  "Solution_6": "[hide]We factor to get that (a+b)(a-b) = 37. Because a and b are both integers, a+b and a-b must also be integers. \n\nSince 37 is prime, one of the factors have to be $\\pm37$ and the other factor has to be $\\pm1$. If the factors are 37 and 1: \n\nIf a+b=37, then a-b=1, and a=19, b=18. \nIf a-b=37, a+b=1, a=19, b=-18. \n\nNow let the factors be -37 and -1: \n\nIf a+b=-37, then a-b=-1, and a=-19, b=-18. \nIf a+b=-1, a-b=-37, then a=-19, b=18. \n\nThe positive pair is (19,18). \nAll pairs are (19,18), (19,-18), (-19,-18), (-19,18). [/hide]",
  "Solution_7": "[quote=\"Silverfalcon\"]This is a challenge problem so if you are above this level, leave this to Getting Started.\n\nFind the ordered pair of positive integers $(a,b)$ for which $a^2-b^2 = 37$.\n\nFurther challenge: Find ALL ordered pair of integers $(a,b)$ for which $a^2-b^2 = 37$.[/quote]\r\n[hide]\nFactor the eqn. into $(a+b)(a-b)=37$.\n37=1*37.\n2 cases, $a+b=37$ and $a-b=1$ OR $a-b=37$ and $a+b=37$\nCASE 1:\n$a+b=37$\n$a-b=1$\n$2a=38$\n$a=19$\n$19-b=1$\n$b=18$\n$(19,18)$\nCASE 2:\n$a-b=37$\n$a+b=1$\n$2a=38$\n$a=19$\n$19-b=37$\n$-b=18$\n$b=-18$\n$(19,-18)$\n----END HARD WORK----\nThese are the only 2 integer solutions.\nThe answer to number 1 is (19,18), and the answer to number 2 is (19,-18) and (19,18).\nthank you and have a nice day.\n[/hide]",
  "Solution_8": "[quote=\"IntrepidMath\"][quote=\"Silverfalcon\"]This is a challenge problem so if you are above this level, leave this to Getting Started.\n\nFind the ordered pair of positive integers $(a,b)$ for which $a^2-b^2 = 37$.\n\nFurther challenge: Find ALL ordered pair of integers $(a,b)$ for which $a^2-b^2 = 37$.[/quote]\n[hide]\nFactor the eqn. into $(a+b)(a-b)=37$.\n37=1*37.\n2 cases, $a+b=37$ and $a-b=1$ OR $a-b=37$ and $a+b=37$\nCASE 1:\n$a+b=37$\n$a-b=1$\n$2a=38$\n$a=19$\n$19-b=1$\n$b=18$\n$(19,18)$\nCASE 2:\n$a-b=37$\n$a+b=1$\n$2a=38$\n$a=19$\n$19-b=37$\n$-b=18$\n$b=-18$\n$(19,-18)$\n----END HARD WORK----\nThese are the only 2 integer solutions.\nThe answer to number 1 is (19,18), and the answer to number 2 is (19,-18) and (19,18).\nthank you and have a nice day.\n[/hide][/quote]\r\nYou forgot some other possible cases...\r\nDid you read what JesusFreak197 and 236factorial wrote?"
}
{
  "Tag": [
    "inequalities",
    "inequalities open"
  ],
  "Problem": "a,b,c,d>0\r\n  prove that\r\n        (a+b)^3(b+c)^3(c+d)^3(a+d)^3>=16(abcd)^2(a+b+c+d)^4",
  "Solution_1": "$ a,b,c,d>0$\r\nProve that \r\n$ (a+b)^{3}(b+c)^{3}(c+d)^{3}(a+d)^{3}\\geq 16(abcd)^{2}(a+b+c+d)^{4}$",
  "Solution_2": "If it is true it can surely be proved by newton/maclaurin inequalities... No time for this now",
  "Solution_3": "Let us put x=abc,y=bcd,z=cda,w=dab  \r\nand homogenize to a+b+c+d=1\r\n(a+b)(b+c)(c+d)(d+a)=(ac-bd)^2+(a+b+c+d)(x+y+z+w)\r\nand xyz+yzw+zwx+wxy=(abcd)^2(a+b+c+d)\r\n\r\ntherefore enough to prove \r\n(x+y+z+w)^3>= 16(xyz+yzw+zwx+wxy)  \r\nbut this is trivial by AM-GM. :wink:",
  "Solution_4": "so then this problem doesnt belong to this section Alisher,this section is for open questions:[b]An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not.[/b]\r\nyou should have posted this problem into [b]Inequalities Unsolved Problems[/b]`s section: [url]http://www.mathlinks.ro/index.php?f=51[/url] which is for [b]Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it.[/b]..."
}
{
  "Tag": [
    "calculus",
    "integration",
    "inequalities",
    "trigonometry",
    "Cauchy Inequality",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $f:\\mathbb{R}\\to\\mathbb{R}$ be an integrable function.Prove the following inequality:\r\n\r\n         $\\left(\\int_{a}^{b} f(x)sinx dx\\right)^2$ $+$ $\\left(\\int_{a}^{b} f(x)cosx dx\\right)^2\\leq\\left(\\int_{a}^{b} |f(x)|dx \\right)^2$",
  "Solution_1": "I am sorry to announce you, but you should change the source and say \"classical\". Actually no, you should rather write \"wrong inequality\".",
  "Solution_2": "Is it wrong now? well if you say it is classical can you give me some refferences? :what?:",
  "Solution_3": "Making the face  :what?:  really does not impress me. And please don't ask me to give refference to Cauchy inequality (I choosed this one since it solves this problem) for example, since all of us know it very well. Of course, one can insist and call it new and even more spectacular \"cool\", but it remains classical.",
  "Solution_4": "[quote=\"harazi\"]And please don't ask me to give refference to Cauchy inequality (I choosed this one since it solves this problem) for example, since all of us know it very well.[/quote]\r\nHehe, it was one of my homework problems. Apply Cauchy-Schwarz intequality,\r\n$\r\n\\left(\\int_{a}^{b} f(x)\\sin x dx\\right)^2 + \\left(\\int_{a}^{b} f(x)\\cos x dx\\right)^2\r\n\\leq\r\n\\left(\r\n\\int_a^b \\sqrt{|f(x)|}\\cdot \\sqrt{|f(x)|}|\\sin x|dx\r\n\\right)^2+\r\n\\left(\r\n\\int_a^b \\sqrt{|f(x)|}\\cdot \\sqrt{|f(x)|}|\\cos x|dx\r\n\\right)^2\r\n\\leq \\int_a^b |f(x)|dx \\int_a^b |f(x)|\\sin^2 x dx + \\int_a^b |f(x)|dx \\int_a^b |f(x)|\\cos^2 x dx\r\n= \\left(\\int_a^b |f(x)|dx\\right)^2\r\n$\r\n\r\np.s. Is it Schwarz or Schwartz? I know they are both mathematicians, which one is the correct name here?",
  "Solution_5": "Schwarz."
}
{
  "Tag": [],
  "Problem": "Hi, :blush:\r\nI registered today\r\nI can't write English well, sorry\r\n...... $x=x$ oh It's cool,,\r\nAnybody teach me english with E-mail or MSN messenger??\r\nI'm looking for pen pal.",
  "Solution_1": "cool, you already know a bit of latex. Where are you really from? :)",
  "Solution_2": "He's from Earth, The Galaxy\r\n\r\nCan't you tell?",
  "Solution_3": "Do you know Kimchi?\r\nIm from Republic of Korea.\r\nIm a middle school student, $9^{0}.5$grade.",
  "Solution_4": "Who's KimChi?",
  "Solution_5": "kimchi is food.",
  "Solution_6": "I've got it! Are you from North Korea?\r\n\r\n[edit] I'm lovin' kimchi! My mom makes it. Even though we are in USA.",
  "Solution_7": "oh no :lol:  I'm from South Korea\r\nI don't like Kim Jeong Ill.",
  "Solution_8": "[quote=\"nutz_for2.718281828\"]I've got it! Are you from North Korea?[/quote]\r\nI don't think they have computers in North Korea. :|  :fool:",
  "Solution_9": "[quote=\"lotrgreengrapes7926\"][quote=\"nutz_for2.718281828\"]I've got it! Are you from North Korea?[/quote]\nI don't think they have computers in North Korea. :|  :fool:[/quote]\r\n\r\nIs North Korea communist?",
  "Solution_10": "[quote=\"SplashD\"][quote=\"lotrgreengrapes7926\"][quote=\"nutz_for2.718281828\"]I've got it! Are you from North Korea?[/quote]\nI don't think they have computers in North Korea. :|  :fool:[/quote]\n\nIs North Korea communist?[/quote]\r\nI don't know about communist, but I know it's dictated by Kim Jeong Ill, who's caused a lot of problems last week. :noo:  :help:",
  "Solution_11": "[quote=\"Inspired By Nature\"]cool, you already know a bit of latex. Where are you [b]really[/b] from? :)[/quote]\n[quote=\"SplashD\"]He's from Earth, The Galaxy\nCan't you tell?[/quote]\r\nI think you should read your signature splashD. :ninja: \r\nclick the smilie in case you're still confused.--->\r\n[hide=\" :D \"] joking [/hide]",
  "Solution_12": "Yes NK is communist.\r\n\r\nIt sucks to be so close to them.\r\n\r\nMy parents occasionally eat kimchi but I don't wanna try. I'm cantonese btw.",
  "Solution_13": "I think I've seen North Korea before.  :o I was 8 or something and I went on a boat ride in that gulf thing between China and North Korea.",
  "Solution_14": "There was a front cover on the economist making fun of Kimmy.\r\n\r\nHOW DO YOU NOT LIKE KIM CHI????"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "I've been hearing a lot about all the math people, or at least some having extremely low GPAs. I heard that all they do well in is math, and since they spend all their time on math, somehow their GPAs drop. I don't believe it (since good at math and good at school should be proportional), but many people say so. So is this bs?",
  "Solution_1": "This isn't BS. When I started focusing on math, everything else besides maybe science just wasn't important to me anymore. If I didn't care so much about my grades (for my college application) I would've completely blown off all my other subjects to focus on my math.",
  "Solution_2": "[quote=\"moogra\"]all the math people[/quote]\r\nDo you mean current students or historical mathematical figures?  There are certainly some anecdotes about the latter, but as to the former I have a suspicion that most competent math students come from backgrounds that predispose them to care about their grades  :P",
  "Solution_3": "i didn't care about my grades until AFTER i started doing math",
  "Solution_4": "I recently discovered math and now I'm finding it hard to cope with all the Honors classes while being active here.  It's hard to do homework at the same time as AoPS.  AoPS is not like MySpace/Facebook/Youtube/...; you have to actually concentrate on what you're doing on AoPS.",
  "Solution_5": "[quote=\"archimedes1\"]I recently discovered math and now I'm finding it hard to cope with all the Honors classes while being active here.  It's hard to do homework at the same time as AoPS.  AoPS is not like MySpace/Facebook/Youtube/...; you have to actually concentrate on what you're doing on AoPS.[/quote]\r\n\r\nWow... that is like the same thing I was thinking. \r\n\r\nActually, this year, I've done far less math than I did last year as an 8th grader. Instead, I am beginning to focus a lot more on school subjects, but maybe that is due to the fact that I'm not completely adjusted to high school yet? It's impossible for me to do as much of each (school subjects + extra math) as you would like to do on a weekly basis.",
  "Solution_6": "[quote=\"moogra\"]I've been hearing a lot about all the math people, or at least some having extremely low GPAs. I heard that all they do well in is math, and since they spend all their time on math, somehow their GPAs drop. I don't believe it (since good at math and good at school should be proportional), but many people say so. So is this bs?[/quote]\r\n\r\nThis is largely a myth.  There are certainly counterexamples, but many of your USAMO winners are also valedictorians.  Sure, there are some examples of math-only types who do math to the exclusion of all else and don't have any interest in grades, but most math superstars are at or near the top of their school in nearly everything (particularly science and computer science, as you might expect).\r\n\r\nThat said, to become really good at something you love, you do have to make some trade-offs.  The one I recommend is noting that maybe it's not so important to take 13 AP exams, including 9 in subjects you don't care at all about.  And if you hate playing <whatever instrument or sport here> at age 13 or 14, then maybe it's time to reasses whether or not it's worth putting 10 hours a week into it.  You have to make choices at some point. . .   If important work is suffering because of your dedication to something you enjoy, then you have to reasses: is that important work so important after all (maybe, maybe not), or is the thing you enjoy really worth spending that much time on (maybe, maybe not).",
  "Solution_7": "A lot of the young people who are prominently good at math (i.e. do very well on contests) are so because they can think very quickly. This is a skill that's applicable in almost every field of study, although it is often most useful in math and science (since it is usually best expressed by the ability to do calculations and come to logical conclusions quickly). But when someone is taking a large number of difficult classes in high school, it helps a lot to be able to plow through the homework quickly (often during other classes) so that there's time at home for doing more math. \r\n\r\nThat's certainly how my friends and I in high school maintained a high GPA, took many diverse AP tests, and still managed to do decently in a number of different math and science competitions. \r\n\r\nAnother possible trait is the ability to devote all free time to working. I certainly can't do this, but some people are able to come home and just spend every waking minute doing homework or studying math and science (or whatever else interests them). Even if you don't work very quickly, if you spend all of your time working, you will surely get a lot done. Personally, I don't see how someone could do this and still be happy, but I'm not going to make judgments about what's the \"'right\" way to go about studying and doing homework.",
  "Solution_8": "[quote]Personally, I don't see how someone could do this and still be happy[/quote]\r\n\r\n... so true."
}
{
  "Tag": [],
  "Problem": "\u6c42\u901a\u89e3\u516c\u5f0f\u7684\u8868\u8fbe\u5f0f\r\n\r\n\u7b97\u6cd5\u6bd4\u62fc\u4e2d\r\n\u00ad\r\nimi00\u7f51\u53cb\u7684\u76f4\u63a5for\u5faa\u73af\u6cd5\u2014\u2014\r\n\u00ad\r\npublic   class   test   { \r\npublic   static   void   main(String[]   args)   { \r\nint   count   =   0   ; \r\nfor(int   i   =   1;   i   <100;   i++   ) \r\nfor(int   j   =   i   +   1;   j   <   100;   j++   ) \r\nfor(   int   k   =   j   +   1;   k   <   100;   k++   ) \r\nfor   (int   m   =   k   +   1;   m   <   100;   m++   ) \r\nfor(int   n   =   m   +   1   ;   n   <100;   n++   ) \r\nif(i   +   j   +   k   +   m   +   n   <   100){ \r\ncount++; \r\n// System.out.println(   i   +   \"   \"   +   j   +   \"   \"   \r\n// +   k   +   \"   \"   +   m   +   \"   \"   +   n   +   \"   sum: \"   +   (i   +   j   +   k   +   m   +   n)   ); \r\n} \r\nSystem.out.println(count); \r\n} \r\n}\r\n\u6700\u76f4\u63a5\u6700\u539f\u59cb\u7684\u7b97\u6cd5\uff1a\u7b54\u6848\u4e3a455175\uff0c\u5faa\u73af\u6b21\u6570\u4e3a71523114\r\n\u00ad\r\nwqs1106\u7f51\u53cb\u7684\u6539\u826f\u65b9\u6848\r\n\u2014\u2014\u5982\u679c\u6ca1\u6709\u6570\u5b66\u516c\u5f0f\u53ef\u4ee5\u76f4\u63a5\u8ba1\u7b97,\u6211\u60f3\u6211\u8fd9\u4e2d\u65b9\u6cd5\u5e94\u8be5\u7b97\u662f\u5faa\u73af\u6b21\u6570\u6bd4\u8f83\u5c11\u7684\u4e86,\u6709\u591a\u5c11\u79cd\u7ec4\u5408\u5c31\u5faa\u73af\u591a\u5c11\u6b21,\u4e00\u6b21\u4e0d\u591a\u4e00\u6b21\u4e0d\u5c11.\r\n\u00ad\r\npublic       class       Test3       {   \r\npublic       static       void       main(String[]       args)       {   \r\nint       count       =       0       ;   \r\nfor(int   i=1;i <86;i++)   \r\n{ \r\nfor(int       j=i+1;j <(88-i);       j++       )   \r\n{ \r\nfor(       int   k=   j+1;k <(91-i-j);       k++       ) \r\n{ \r\nfor       (int       m       =k+   1;m <(95-i-j-k);       m++       )   \r\n{ \r\nfor(int       n       =   m+     1;n <(100-i-j-k-m);       n++       )   \r\n{ \r\nif(i       +       j       +       k       +       m       +       n       <       100) \r\n{   \r\ncount++;   \r\n}   \r\n} \r\n} \r\n} \r\n} \r\n} \r\nSystem.out.println(count);   \r\n}   \r\n} \r\n\u7b54\u6848\u4e3a455175\uff0c\u5faa\u73af\u6b21\u6570\u4e3a455175\u3002\r\n\u00ad\r\njseasidej\u7f51\u53cb\u7684\u8bd5\u9519\u6cd5\r\n\u00ad\r\npublic   class   less100   \r\n{ \r\npublic   static   void   main(String[]   args)   \r\n{ \r\nint   a1,a2,a3,a4,a5; \r\nint   sum=0; \r\nint   time=0; \r\nfor(a1=1;a1 <=17;a1++) \r\n{ \r\nfor(a2=a1+1;a2 <=23;a2++) \r\n{ \r\nif((a2*4+6+a1)> =100)   break; \r\nelse \r\nfor(a3=a2+1;a3 <=31;a3++) \r\n{ \r\nif((a3*3+3+a1+a2)> =100)   break; \r\nelse \r\nfor(a4=a3+1;a4 <=46;a4++) \r\n{ \r\nif((a4*2+1+a1+a2+a3)> =100)   break; \r\nelse \r\n{ \r\ntime++; \r\n                                                          sum=sum+(99-a1-a2-a3-a4)-a4; \r\n} \r\n} \r\n} \r\n} \r\n} \r\nSystem.out.println( \"\u7ed3\u679c\u662f \"+sum+ \"       \u5faa\u73af\u6b21\u6570\uff1a \"+time); \r\n} \r\n} \r\n\u8fd0\u884c\u7ed3\u679c\uff1a       \u7ed3\u679c\u662f455175       \u5faa\u73af\u6b21\u6570\uff1a   24260 \r\n\u00ad\r\njacaranda\u7f51\u53cb\u7684\u9012\u5f52\u6cd5\u2014\u2014\u6bd4\u8f83\u5168\u9762\u7684\u5206\u6790\u554a\r\n\u00ad\r\nC#   code \r\nusing   System; \r\nusing   System.Collections.Generic; \r\nusing   System.Text; \r\nnamespace   Test1 \r\n{ \r\n        class   Program \r\n        { \r\n                static   void   Main(string[]   args) \r\n                { \r\n                        int   total=0; \r\n                        int   num   =   0; \r\n                        for   (int   i   =   1;   i   <   18;   i++)   //\u7b2c\u4e00\u4e2a\u6570\u6700\u5927\u53ea\u80fd\u53d6\u523017   \r\n                                for   (int   j   =   i   +   1;   j   <   24;   j++)   //\u7b2c\u4e8c\u4e2a\u6570\u6700\u5927\u53ea\u80fd\u53d6\u523023   \r\n                                        for   (int   k   =   j   +   1;   k   <   32;   k++)   //\u7b2c\u4e09\u4e2a\u6570\u6700\u5927\u53ea\u80fd\u53d6\u523031 \r\n                                                for   (int   l   =   k   +   1;   l   <   47;   l++)   //\u7b2c\u56db\u4e2a\u6570\u6700\u5927\u53ea\u80fd\u53d6\u523046   \r\n                                                { \r\n                                                        //num++;   \u653e\u5230if   \u91cc\u9762 \r\n                                                        int   m   =   99   -   i   -   j   -   k   -   l; \r\n                                                        if   (m   >   l) \r\n                                                        { \r\n                                                                num++;//\u8fd9\u91cc \r\n                                                                total   =   total   +   (m   -   l); \r\n                                                        } \r\n                                                } \r\n                        System.Console.WriteLine(total); \r\n                        System.Console.WriteLine(num); \r\n                        System.Console.ReadLine(); \r\n                } \r\n        } \r\n} \r\n\u8f93\u51fa\uff1a       \r\n455175               //\u7b54\u6848       \r\n24260                   //\u5faa\u73af\u6b21\u6570",
  "Solution_1": "[1 X X X X]\r\n\r\n[1 2 A A A]=1*1+2*3+4*4+5*6+...+43*43,\r\n[1 3 A A A]=2*2+3*4+5*5+6*7+...+41*41,\r\n[1 4 A A A]=1*2+3*3+4*5+6*6+...+39*39,\r\n[1 5 A A A]=1*1+2*3+4*4+5*6+...+37*37,\r\n[1 6 A A A]=2*2+3*4+5*5+6*7+...+35*35,\r\n[1 7 A A A]=1*2+3*3+4*5+6*6+...+33*33,\r\n.....\r\n.....\r\n.....\r\n[1 21 A A A]=2*2+3*4+5*5=41,\r\n[1 22 A A A]=1*2+3*3=11,\r\n[1 23 A A A]=1*1=1."
}
{
  "Tag": [],
  "Problem": "While traveling home from school, Alyssa fell asleep halfway through the journey. When she awoke, she still needed to travel one-fourth of the distance that she had traveled while sleeping. For what part of the journey was she awake?",
  "Solution_1": "[hide]she was awake for the first $\\frac 12$, so if she was awake in the last half for $x$ time, then the total amount she was asleep was $4x$ so $5x=\\frac 12$ and $x=\\frac{1}{10}$. She was asleep for $\\frac{2}{5}$, so she was awake for ${\\boxed{\\frac 35}}$[/hide]",
  "Solution_2": "You answered the wrong thing. ;)",
  "Solution_3": "[hide]She was awake for half of the journey.  Set the part when she was asleep $x$.  If she was asleep for $x$ of the journey, she was awake $\\frac{1}{2} + \\frac{1}{4}x$ of the journey.  Also, $x+\\frac{1}{4}x=\\frac{1}{2}$ because that was the second half of the journey. Solving gives $x=\\frac{2}{5}$. Therefore, she was awake $\\frac{1}{2}+\\frac{1}{10}$ or [size=200]$\\frac{3}{5}$[/size].[/hide]",
  "Solution_4": "[hide]1/4x + x = 1/2, 5/4 x = 1/2, so she was asleep for 2/5 of the journey x. Therefore she was awake for 3/5.[/hide]",
  "Solution_5": "[quote=\"JesusFreak197\"]You answered the wrong thing. [/quote] :fool:  :ewpu: \r\nEdited"
}
{
  "Tag": [
    "trigonometry",
    "inequalities",
    "limit",
    "geometric sequence"
  ],
  "Problem": "find all real x, such that\r\ncos x, cos 2x, cos 4x, cos 8x, ...\r\nare all negative.",
  "Solution_1": "[deleted]",
  "Solution_2": "Did you keep in mind that cos x < 0 for 90 + k.360 < x < 270 + k.360?\r\n[k is an integer]\r\n\r\nBy the way, cos 2x = 2 cos<sup>2</sup> x - 1.",
  "Solution_3": "I really hope that you have a solution for a beginner or that you joke. As far as I remember, this is a russian problem and from what I've seen it's not at all easy. I would like to see a solution for a beginer!",
  "Solution_4": "the problem is asked by one of my classmates.. and my teacher didn't solve it. But i don't think it is difficult.",
  "Solution_5": "If the story I heard about it is true, then it is really difficult. I heard that when it was proposed on the russian olympiad, it was marked as a very difficult problem.",
  "Solution_6": "Because of the geometric progression of the number in front of x, wouldn't it just keep cycling around and around?  I don't think you could find a value (which would need to be between -pi/2 and pi/2, where cosine is negative) such that multiplying it by those numbers would keep the cosine negative",
  "Solution_7": "It isn't that simple, and the set in question is not empty. An example of such a number: $\\frac\\pi3.$",
  "Solution_8": "[hide]I think its just a matter of solving these two inequalities (if |cosx| is more than 1/2 then cos2x > 0):\n\n[tex] \\displaystyle 0 > \\cos{x} \\ge -\\frac{1}{2} \\\\\n|2\\cos^2{x}-1| \\le \\frac{1}{2} \\\\ [/tex]\n\nSolving the second inequality, we have [tex] \\displaystyle \\frac{\\pi}{6} \\le x \\le \\frac{\\pi}{3} [/tex]. However, including the first inequality we know that [tex] \\displaystyle \\frac{\\pi}{2} < x \\le \\frac{2\\pi}{3} [/tex] or [tex] \\displaystyle \\frac{4\\pi}{3} \\le x < \\frac{3\\pi}{2} [/tex]. So then there are no solutions, right? [/hide]\r\n\r\nbtw, [tex] \\displaystyle \\cos{\\frac{\\pi}{3}} = \\frac{1}{2} > 0 [/tex] so that doesn't work.\r\n\r\nEDIT: nvm, i thought about it, my method is not right...",
  "Solution_9": "[hide=\"Hint\"]Look at the base 2 representation of $\\frac{x}{2\\pi}$.[/hide]\r\n\r\nI think jmerry meant $\\frac{2\\pi}{3}$.",
  "Solution_10": "[quote=\"Ravi B\"][hide=\"Hint\"]Look at the base 2 representation of $\\frac{x}{2\\pi}$.[/hide]\n[/quote]\r\nCute! (I read the hint!)",
  "Solution_11": "I think I have a solution.but I'm not sure about it.can somebody check it for me :) ?\r\nwe can know $ \\cos x,\\cos 2x, \\cdots ,cos 2^{n}x,\\cdots $ are all negetive is equal to $ \\cos x,\\cos \\frac{x}{2}, \\cdots ,\\cos \\frac{x}{2^n},\\cdots $ are all negetive.\r\nso we can get $\\cos x<0$\r\n$-\\frac{\\sqrt{2}}{2}<cos \\frac{x}{2}<0$\r\n$\\cdots $\r\nlet ${a_0=0,a_{n+1}=-\\sqrt {\\frac {a_{n}+1}{2}}}$\r\nwe can get $a_n<cos \\frac{x}{2^n}<a_{n-1}$\r\nbecause $\\lim_{n\\rightarrow \\infty } a_n= -\\frac{1}{2}$\r\nthen we get $cos \\frac{x}{2^n}= -\\frac{1}{2}$\r\nso $\\cos x= -\\frac{1}{2}$\r\nam I right?",
  "Solution_12": "This is where I posted my solution: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343568#p343568[/url]\r\nI think the proof is deeply related to the base 2 representation of $\\alpha/2\\pi$."
}
{
  "Tag": [
    "LaTeX",
    "modular arithmetic",
    "Euler",
    "function"
  ],
  "Problem": "What are the last two digits of $2^{66}-4$?",
  "Solution_1": "do you mean $2^{66}-4$? :huh:",
  "Solution_2": "[hide]I think he meant $2^{66}-4$.\n\nYou should try to realize people mistakes and then solve the question.\n\nWe see that there is a pattern every 20.  I won't type them all out, so this means we are just looking at $2^{6}-4=62$.\n\n$62$[/hide]",
  "Solution_3": "[quote=\"SplashD\"]do you mean $2^{66}-4$? :huh:[/quote] yes I did.\r\nand mathgenius is wrong.",
  "Solution_4": "[hide]Yes, I am wrong.\n\nI menat to say 64-4=60, not 62.\n\nSo yes, $60$[/hide]",
  "Solution_5": "make a table of this and you can see tat there is a pattern among the answers\r\n\r\n(no, i will not do latex)\r\n\r\ns= square\r\nv= value\r\n\r\ns v\r\n0 1\r\n1 2\r\n2 4\r\n3 8\r\n4 16\r\n5 32\r\n6 64\r\n...\r\n\r\nafter all tat, u can see tat the pattern tat 2^66-4=...60",
  "Solution_6": "There is a way to do this without the pattern ;)",
  "Solution_7": "[quote=\"bpms\"]There is a way to do this without the pattern ;)[/quote]\r\n\r\nyeah, all-out multiplication",
  "Solution_8": "fermat's last theorem?",
  "Solution_9": "I think you mean Fermat's Little Theorem.\r\n\r\nLast would be useless in this problem.",
  "Solution_10": "[quote=\"bpms\"]What are the last two digits of $2^{66}-4$?[/quote]\r\n\r\n[hide] $2^{66}\\equiv x\\pmod{100}$\n\n$2^{66}\\equiv0\\pmod{4}$\n$2^{66}\\equiv x\\pmod{25}$\n\nFrom Euler's Totient Theorem:\n$\\phi(25)=25(1-\\frac{1}{5})=20$.\nSo, $2^{66\\mod20}\\equiv2^{6}=64$\n\n$x=64$\n$64-4=\\boxed{60}$[/hide]",
  "Solution_11": "I made it so that this is easily factored...\r\nWhat is Euler's totient theorem...",
  "Solution_12": "It is just a fancy name for Euler-Fermat, I believe.\r\n\r\nIt just uses the $\\phi$ function, in it.",
  "Solution_13": "omg i dont kno any of these stuff\r\n\r\nwat r these theorems u speak of",
  "Solution_14": "[quote=\"mathgeniuse^ln(x)\"]I think you mean Fermat's Little Theorem.\n\nLast would be useless in this problem.[/quote]\r\n\r\noh right, lol",
  "Solution_15": "I would like to see how.",
  "Solution_16": "My method without any fancy formulae\r\n[hide]$2^{1}=02$\n$2^{2}=04$\n$2^{4}=16$\n$2^{8}=56$\n$2^{16}=36$\n$32=96$\n$64=16$\n$(2^{64})(2^{2})=64$\n$64-4=60$[/hide]\r\nTook me under the allotted 80 sec.",
  "Solution_17": "[quote=\"mathgeniuse^ln(x)\"]I think you mean Fermat's Little Theorem.\n\nLast would be useless in this problem.[/quote]\r\n\r\nwhat are those two fromulas?",
  "Solution_18": "[quote=\"slamchess\"][quote=\"mathgeniuse^ln(x)\"]I think you mean Fermat's Little Theorem.\n\nLast would be useless in this problem.[/quote]\n\nwhat are those two fromulas?[/quote]\r\nFermat's last $a^{x}+b^{x}$isn't equal to$c^{x}$, for any natural numbers a, b, c, and x if x>2.\r\nFermat's little $a^{p}=a(mod\\ p)$ if a is prime.",
  "Solution_19": "[quote=\"bpms\"][quote=\"slamchess\"][quote=\"mathgeniuse^ln(x)\"]I think you mean Fermat's Little Theorem.\n\nLast would be useless in this problem.[/quote]\n\nwhat are those two fromulas?[/quote]\nFermat's last $a^{x}+b^{x}$isn't equal to$c^{x}$, for any natural numbers a, b, c, and x if x>2.\nFermat's little $a^{p}=a(mod\\ p)$ if a is prime.[/quote]\r\n\r\nfermat's little sounds important\r\n\r\nthnx bpms",
  "Solution_20": "[quote]fermat's little sounds important [/quote]\r\n\r\nbut worth barely anything in mc",
  "Solution_21": "[quote=\"akalra1\"][/quote][quote=\"bpms\"][quote=\"slamchess\"][quote=\"mathgeniuse^ln(x)\"]I think you mean Fermat's Little Theorem.\n\nLast would be useless in this problem.[/quote]\n\nwhat are those two fromulas?[/quote]\nFermat's last $a^{x}+b^{x}$isn't equal to$c^{x}$, for any natural numbers a, b, c, and x if x>2.\nFermat's little $a^{p}=a(mod\\ p)$ if a is prime.[/quote][quote=\"akalra1\"]\n\nfermat's little sounds important\n\nthnx bpms[/quote]\r\nPlease, please, please do not use chatspeak. I find it very hard to read some your posts because of the chatspeak. Thanks.  :wink:",
  "Solution_22": "1:2\r\n2:4\r\n3:8\r\n4:16\r\n5:32\r\n6:64\r\n7:28\r\n8:56\r\n9:12\r\n10:24\r\n11:48\r\n12:96\r\n13:92\r\n14:84\r\n\r\nok yo uget the point (it does go to 20)\r\n64-4=60",
  "Solution_23": "[quote=\"moogra\"]1:2\n2:4\n3:8\n4:16\n5:32\n6:64\n7:28\n8:56\n9:12\n10:24\n11:48\n12:96\n13:92\n14:84\n\nok yo uget the point (it does go to 20)\n64-4=60[/quote]\r\nMy solution has alot less computations and doesn't require knowledge of the pattern.",
  "Solution_24": "$2^{66}-4$\r\nFirst take the 4th root of $(2^{6}6)$ because $2^{x}$ is a 4peat.\r\n\r\nThen you get $2^{2}$.\r\n\r\nFinally the units digit will be $2^{2}-4=0$\r\n\r\nSo 0 is the units digit.\r\n\r\nNext we have to figure out the tens digit.\r\n\r\nHmm, gimme some time im not sure if there is a pattern to find or not.\r\n\r\nI found out that the tens digit of the number has a pattern of every 20.\r\n\r\nSo then it wouuld be $2^{6}$ which is 64 so its tens digit is 6 and units digit is 0. So 60.\r\n\r\nPS i didnt read it right i thought it was the units digit not the last 2 digits"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Can you show\r\n\r\n$ \\int_{\\minus{}\\infty}^\\infty \\frac{\\sin(\\pi x^2)}{\\sinh^2(\\pi x)}\\, dx\\equal{}\\frac{2\\minus{}\\sqrt{2}}{2}$ ?",
  "Solution_1": "Or can you show\r\n\r\n$ \\int_0^\\infty \\frac {2x\\cos \\pi x^2}{e^{2\\pi x} \\minus{} 1}\\, dx \\equal{} \\frac {2 \\minus{} \\sqrt {2}}{8}$\r\n\r\n$ \\int_0^\\infty \\frac {2x\\sin \\pi x^2}{e^{2\\pi x} \\minus{} 1}\\, dx \\equal{} \\frac {\\sqrt {2}}{8} \\minus{} \\frac {1}{2\\pi}$\r\n\r\n$ \\int_0^\\infty \\frac {2x\\cos 2\\pi x^2}{e^{2\\pi x} \\minus{} 1}\\, dx \\equal{} \\frac {1}{16}$\r\n\r\n\r\n$ \\lim_{N\\in\\Bbb{N}} \\int_0^N 4x\\, \\cos \\pi x^2\\, \\sin(2\\pi ux)\\, dx\\equal{}\\sqrt{2}\\, u\\, \\left(\\sin\\pi u^2\\minus{}\\cos \\pi u^2\\right)$\r\n\r\n$ \\lim_{N\\in\\Bbb{N}}\\int_0^{2N} \\frac {2x\\sin \\pi x^2}{1 \\minus{} e^{ \\minus{} 2\\pi x}}\\, dx \\equal{} \\frac {\\sqrt {2}}{8} \\minus{} \\frac {1}{2\\pi}$\r\n\r\n$ \\lim_{N\\in\\Bbb{N}}\\int_0^{2N \\plus{} 1} \\frac {2x\\sin \\pi x^2}{1 \\minus{} e^{ \\minus{} 2\\pi x}}\\, dx \\equal{} \\frac {\\sqrt {2}}{8} \\minus{} \\frac {3}{2\\pi}$\r\n\r\n$ \\lim_{N\\in\\Bbb{N}}\\int_0^{N \\plus{} \\frac12} \\frac {2x\\sin \\pi x^2}{1 \\minus{} e^{ \\minus{} 2\\pi x}}\\, dx \\equal{} \\frac {\\sqrt {2}}{8} \\minus{} \\frac {1 \\minus{} \\sqrt {2}}{2\\pi}$\r\n\r\n\r\n$ \\lim_{N\\in\\Bbb{N}}\\int_0^{N} \\frac {2x\\cos \\pi x^2}{1 \\minus{} e^{ \\minus{} 2\\pi x}}\\, dx \\equal{} \\frac {2 \\minus{} \\sqrt {2}}{8}$ ?\r\n\r\nAnd what is $ \\lim_{N\\in\\Bbb{N}}\\int_0^{N \\plus{} \\frac12} \\frac {2x\\cos \\pi x^2}{1 \\minus{} e^{ \\minus{} 2\\pi x}}\\, dx$ ?",
  "Solution_2": "[quote=Doe John]Can you show\n\n$ \\int_{\\minus{}\\infty}^\\infty \\frac{\\sin(\\pi x^2)}{\\sinh^2(\\pi x)}\\, dx\\equal{}\\frac{2\\minus{}\\sqrt{2}}{2}$ ?[/quote]\n\nhttps://math.stackexchange.com/questions/61605/tough-integral-involving-sinx2-and-sinh2-x",
  "Solution_3": "Let's evaluate $ \\int_0^\\infty \\frac {2x\\sin \\pi x^2}{e^{2\\pi x} - 1}\\, dx $\n$$I=\\int_0^\\infty \\frac {2x\\sin \\pi x^2}{e^{2\\pi x} - 1}\\, dx =\\int_0^\\infty \\frac {4x\\sin (\\pi x^2/2)\\cos (\\pi x^2/2)}{e^{2\\pi x} - 1}\\, dx =\\frac{2}{\\pi}\\int_0^\\infty \\frac {d\\Bigl(\\sin^2(\\pi x^2/2)\\Bigr)}{e^{2\\pi x} - 1}\\, =4\\int_0^\\infty \\frac {\\sin^2(\\pi x^2/2)e^{2\\pi x}}{(e^{2\\pi x} - 1)^2}\\, dx$$\n$$=\\int_0^\\infty \\frac {\\sin^2(\\pi x^2/2)}{\\sinh^2\\pi x}\\, dx=\\frac{1}{4}\\int_{-\\infty}^\\infty \\frac {1-\\cos\\pi x^2}{\\sinh^2\\pi x}\\, dx=\\frac{1}{4}\\Re \\,J\\,, \\,\\,\\text {where} \\,J=\\int_{-\\infty}^\\infty \\frac {1-e^{i\\pi x^2}}{\\sinh^2\\pi x}\\, dx$$\n$x=0$ is a removable singular point, therefore we can consider both terms in the principal value sense\n$$J_1=\\int_{-\\infty}^{-r} \\frac {1}{\\sinh^2\\pi x}\\, dx+\\int_{r}^{\\infty} \\frac {1}{\\sinh^2\\pi x}\\, dx=-\\frac{1}{\\pi}\\Bigl(\\coth |_{-\\infty}^{-r}+\\coth |_{r}^{\\infty}\\Bigr)=-\\frac{2}{\\pi}\\,+\\,\\frac{2}{\\pi^2 r}$$\nFollowing the approach [url=https://math.stackexchange.com/questions/2717244/is-there-a-rapider-or-more-elegant-way-to-evaluate-int-0-infty-frac-cos?noredirect=1](here)[/url] we introduce function $f(x)=\\frac{e^{\\pi i x^2}e^{\\pi x}}{\\sinh(2\\pi x)\\sinh(\\pi x)}$. It is easy to check that $f(x)-f(x+i)=\\frac{e^{\\pi i x^2}}{\\sinh^2(\\pi x)}$\nNext, we integrate along rectangle in the comples plane: from $-R$ to $R$, from $R$ to $R+i$, from $R+i$ to $-R+i$ and from $-R+i$ to $-R$ ($R\\to\\infty$). We also add two small semi-circles of radius  $r\\to0$ (going clockwise) above $x=0$ and below $x=i$ - to close the contour.\n$$J_2=\\int_{-\\infty}^\\infty \\frac {e^{i\\pi x^2}}{\\sinh^2\\pi x}\\, dx=\\int_{-\\infty}^\\infty \\bigl(f(x)-f(x+i)\\bigr) dx=\\oint\\frac{e^{\\pi i x^2}e^{\\pi x}}{\\sinh(2\\pi x)\\sinh(\\pi x)}dx -\\int_{C_1}-\\int_{C_2}-(1)-(2)$$\nwhere $\\int_{C_1}$ and $\\int_{C_2}$ are integrals along small semi-circles, and (1) and (2) - integrals along right and left path (from $R$ to $R+i$ and from $-R+i$ to $-R$; (1) and (2) $\\to0$ as $R\\to\\infty$).\nWe have a single simple pole inside the contour at $x=\\frac{i}{2}$,  therefore\n$$\\oint=2\\pi i Res_{x=i/2}\\frac{e^{\\pi i x^2}e^{\\pi x}}{\\sinh(2\\pi x)\\sinh(\\pi x)}=2\\pi i\\frac{e^{-\\pi i/4} e^{\\pi i/2}}{-2\\pi i}=-\\frac{1+i}{\\sqrt2}$$\n$\\int_{C_1}=-\\frac{1}{\\pi^2r}-\\frac{i}{2}\\,$ and $\\int_{C_2}=-\\frac{1}{\\pi^2r}-\\frac{i}{2}$\n$$J_2=-\\frac{1+i}{\\sqrt2}+\\frac{2}{\\pi^2r}+i$$\nTaking all together\n$$I=\\frac{1}{4}\\Re\\,J=\\frac{1}{4}\\Re\\,(J_1-J_2)=-\\frac{1}{2\\pi}+\\frac{1}{4\\sqrt2}$$\n\nAs a bonus,\n$$\\int_{-\\infty}^\\infty \\frac {\\sin{\\pi x^2}}{\\sinh^2\\pi x}\\, dx=\\Im \\int_{-\\infty}^\\infty \\frac {e^{i\\pi x^2}}{\\sinh^2\\pi x}\\, dx= -\\Im J=\\Im J_2=1-\\frac{1}{\\sqrt2}$$"
}
{
  "Tag": [
    "logarithms",
    "calculus",
    "calculus computations"
  ],
  "Problem": "$ x_{1} = 2$ and $ x_{n + 1} = ln{\\frac {x_{n}^{2}}{1 + ln n}}$\r\nFind $ lim x_{n}$",
  "Solution_1": "If exist lim and equal a, then $ e^a\\equal{}\\frac{a^2}{1\\plus{}ln n}\\plus{}o(1)\\to a\\equal{}\\minus{}\\infty$.",
  "Solution_2": "[quote=\"Rust\"]If exist lim and equal a, then $ e^a \\equal{} \\frac {a^2}{1 \\plus{} ln n} \\plus{} o(1)\\to a \\equal{} \\minus{} \\infty$.[/quote]\r\n\r\nWhy  :wink:",
  "Solution_3": "[quote=\"Rust\"]If exist lim and equal a, then $ e^a \\equal{} \\frac {a^2}{1 \\plus{} ln n} \\plus{} o(1)\\to a \\equal{} \\minus{} \\infty$.[/quote]\r\n\r\nWould you care to explain if you let $ n \\rightarrow \\infty$, how come $ \\frac{1}{1\\plus{}\\ln n}$ sticks? Thank you! :)"
}
{
  "Tag": [],
  "Problem": "If $a+b=3$ and $a^2+b^2=8$ find $a^4+b^4$.",
  "Solution_1": "[hide]$a^4+b^4=(a^2+b^2)^2-2a^2b^2$.  So, we have to find $a^2b^2$.  $(a+b)^2-(a^2+b^2)=2ab$.  Plugging numbers into that gives 9-8=2ab.  ab=$\\frac12$.  so, $a^4+b^4=64-2(ab)^2=63\\frac12$.[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "limit",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "Calculate the integral:\r\n\r\n\\[ \\int\\limits_{0}^{\\infty}\\int\\limits_{0}^{\\infty}\\frac{\\sin^{3}(x\\plus{}y)}{(x\\plus{}y)^{3}}dxdy.\\]",
  "Solution_1": "My first thought is to apply the change of variables $ x\\equal{}uv,y\\equal{}u(1\\minus{}v),$ so $ 0<u<\\infty,$ $ 0<v<1,$ $ dx\\,dy\\equal{}u\\,du\\,dv,$ and $ x\\plus{}y\\equal{}u.$ This leads to\r\n\r\n$ \\int_0^{\\infty}\\frac{\\sin^3 u}{u^2}\\,du.$\r\n\r\nWell, it's a first step. I don't see the next step yet. At least it's clear that the integral converges absolutely.",
  "Solution_2": "A change of variables $ (u, v) = (x + y, x - y)$ would give the same result $ \\int_{0}^{\\infty} \\frac {\\sin^3 u}{u^2} \\, du$. Then we can do like this:\r\n\r\n\\begin{eqnarray*} \\int_{0}^{\\infty} \\frac {\\sin^3 u}{u^2} \\, du & = & \\int_{0}^{\\infty} \\sin^3 u \\left( \\int_{0}^{\\infty} t \\, e^{ - ut} \\, dt \\right) \\, du \\\\\r\n& = & \\int_{0}^{\\infty} t \\left( \\int_{0}^{\\infty} \\sin^3 u \\, e^{ - tu} \\, du \\right) \\, dt \\\\\r\n& = & \\int_{0}^{\\infty} t \\left( \\int_{0}^{\\infty} \\frac {3\\sin u - \\sin 3u}{4} \\, e^{ - tu} \\, du \\right) \\, dt \\\\\r\n& = & \\int_{0}^{\\infty} \\frac {t}{4} \\left( \\frac {3}{1 + t^2} - \\frac {3}{9 + t^2} \\right) \\, dt \\\\\r\n& = & \\lim_{R\\to\\infty} \\int_{0}^{R} \\frac {3}{8} \\left( \\frac {2t}{1 + t^2} - \\frac {2t}{9 + t^2} \\right) \\, dt \\\\\r\n& = & \\lim_{R\\to\\infty} \\frac {3}{8} \\left[ \\log\\left( \\frac {1 + t^2}{9 + t^2} \\right) \\right]_{0}^{R} \\\\\r\n& = & \\frac {3}{4} \\log 3 \\end{eqnarray*}"
}
{
  "Tag": [
    "Support",
    "videos",
    "inequalities"
  ],
  "Problem": "Hi guys. I will try to keep this as simple as possible, so that it is more or less comprehensible, but I am afraid this is not going to work very well. I will appreciate a lot if you could read this till the end and advise me what to do.\r\n\r\nFor past year-year and half or so I have been having some advances in 'logistics' (if that's the correct word). I gradually improved my analytical and logical skills by questioning the reasons behind various happenings and things in the world around me. I managed to discover some interesting things, for example:\r\n\r\n-That people do pretty much everything to achieve some sense of comfort, on conscious or sub-conscious level.\r\n\r\n-Believing that there is no God (basically being an atheist) is not 'smarter' than believing in one since both sides lack evidence to support their opinions (hence making the debate pointless).\r\n\r\n-That people who oppose racism, sexism, and religionism based on an ideology of treating people fairly regardless of their 'predefined' characteristics (or 'predefined' in some cases, such as religion) should be opposing the idea of nationalism (putting one nationality/ethnicity above/below other(s), stereotyping based on their nationality/ethnicity etc.), or they are illogical - it's quite easy to find a person with 'Western moral values' who is anti-racist/sexist/etc. but is yet patriot of his own country and 'own people', as opposed to being patriotic of the whole planet Earth. \r\n\r\n-Moral values vary from society to society not because one is 'wrong' and the other one is 'correct', but because different groups of people have different aims. If one aims to build a society with mutual respect and peaceful coexistence between its members, so that this society can spend time in more meaningful way (exploring unknown, i.e. space) that it is now (wars?) then it perfectly makes sense to advocate Unity in Diversity, respect to people regardless of their skin color and similar attributes, and so on.\r\n\r\nThese are just several things I picked, and they may sound not as convincing as they originally did (I will mention later why exactly, and that's actually the reason I created this topic). But in a nutshell, I progressed a lot over the last year or two. If 5 years ago I would call a \"you give money to poor people only to make yourself feel better by helping people on a sub-conscious level\" absurd, I would give it a full consideration and analyze it now. If 5 years ago I would argue just to defend my point regardless of it being logical or not, now I would use any argument as a source of learning, and I would change my opinion if presented some logical evidence against it. I came to understand that people are not equal, they are different, but that's what makes us so unique, and what we should embrace, and that people should get fair treatment regardless of their pre-defined attributes/relation to some group that may cause stereotypes, and should rather be judged (if ever) on their own views and opinions, as opposed to the things mentioned earlier. I have learned to prevent my own nationality/previous opinions/race/gender/etc. from affecting my thinking, as I tried to look at things from a perspective; now I can understand virtually any opinion/action (by that I mean understanding the reasons that led to it), without necessarily supporting it. Some people called it an emotionless 'computer' approach, but I enjoyed it because this 'neutrality of opinion' led me to explore and understand A LOT. I have explored the 'logistics' further, and was very disappointed to find out that logic is not absolute either - it relies on basic axioms used to build a 'tree'. For example, for most people 'human beings are born with a right of being treated fairly' sounds logical, but there is no proof that it actually is logical, since it's pretty much a fundamental statement/axiom we use to build more complex 'trees' (there probably is a way to prove it using even more fundamental statement(s), but I am not going into there because I only wanted to show a basic example). So, based on  the aforementioned statement, one can say 'gay people are born with a right of being treated fairly, since they are human beings as well.'\r\n\r\nI have just re-read what I wrote, and it has a lot of 'I's. Oh well.\r\n\r\nLong story short, after waking up about three days ago, I realized that my 'skills' were gone. While I enjoyed logical debates before this incident, now I started to encounter a 'mental block' each time I tried to have a similar conversation. \r\n\r\n[quote]Before:\n\nPerson x: Abortion is wrong because it is against human rights.\nMe: The pregnant person is also a human being, and since the baby is in her body, she has rights what to do with her body too\nPerson x: Killing fetus is immoral because you by doing so you prevent birth of a potential human being\nMe: Then, would you consider using condoms immoral, since they do the very exact thing?\n\nNow:\n\nPerson x: Abortion is wrong because it is against human rights.\nMe: Uhh... Oh... *DEAD END*.[/quote]\r\n\r\nSince there are a lot of math-oriented people, I will try to explain it this way - before 'waking up' on that day, I could foresee several possible solutions/answers to the debate question - my mind was automatically analyzing everything faster that I spoke, so I could reply instantly, at the same time predicting several replies I could get back from the person I am talking to, and prepare answers for them too, etc. Kind of like in a chess game - if you succeed in 'brute forcing' your opponent and foreseeing his combinations farther than he does yours, you win. Or at a math olympiade, you know the theory needed to solve the problem, but since you need to work fast, you need to 'load' the problem in your head and figure out where to approach the problem from, sometimes trying a certain solution intuitively, noticing something specific, or so. Up until this time, this worked well - I could 'load' pretty much anything, and then analyze and discuss it immediately. Now, it's the exact opposite - when I try to debate, I open my mouth expecting the ideas to flow as usual, but it's not happening. \r\n\r\nAs mentioned above, I became anti-patriotic/anti-nationalistic after analyzing and deciding that nation/ethnicity are not more important than race/gender in any way, and should not be used to discriminate people. I started to strongly dislike pretty much any nationalistic symbols after this. I couldn't stand patriotic brainwashy ads on TV or flags and national anthems sung in solemnity (why saying I am proud to be white is OhMyGod, but saying I am proud to be from [insert country's name] is great?) After waking up on [b]that morning[/b], I didn't care. I turned on the tv, found a nationalistic ad, and was surprised to notice that I didn't feel any disgust. Same reaction followed when encountering several pretentious people (I prefer, or should I say used to prefer, simple/down-to-earth people, and dislike pretentiousness and pompousness). Now, I didn't care at all - some kind of apathy. I asked several friends to debate with me on various topics - I am still able to argue 'logically' about the topics we have covered before, mainly because I remember what I used to say to respond to their claims, but if we debate about something new - I am TOTALLY lost. It took a while for some of my friends to realize I am not kidding or pretending. I have tried some iq tests and math problems -- my performance on those didn't change. I have tried to read several pages, and it looks like while my reading speed didn't change, my comprehension is lagging - I can't immediately grasp the hidden points as I used to. \r\n\r\nRecalling Leonardo Da Vinci's quote 'There are three types of people, those who see, those who see when shown, and those who do not see', I can say that I moved from the first category to the second one, for no apparent reason. And that's what terrifies me.  I can still understand the previous logical debates I had, debates of other people, but I can no longer 'see' or analyze as fast as I used to. If I take significantly more time to think, I might get it, but might encounter a mental block. \r\n\r\nRereading the whole thing I wrote, I now recall Einstein's 'If you cannot explain it simply then you do not understand it well enough.' What I wrote is not simple for sure, and several days ago I would have been able to fit this in 3 times less symbols.\r\n\r\nDo any of you guys have any idea what I could do?  :maybe: Try brain food such as fish oil?  :lol: I am feeling both weird and funny at the same time - I have always thought this happens only in movies. I feel like a basketball player who woke up and realized that he can't dribble for his life, or like a math major who has realized he can't solve anything beyond arithmetic. And as I said before - I didn't do anything different that could have possibly lead to a change in me, it just suddenly came on me, and now I can't even find logical solutions to everyday arguments such as 'omg my boyfriend doesn't spend enough time with me' which would be extremely easy for me ~3-4 days ago. Should I treat this as if I was 1-2 years ago, and try to begin everything from level 0? Thanks for reading this until the end. :)",
  "Solution_1": ":lol:  I didn't have the patience to read your entire post, so I can't label it (it is unfair too) is as being less \"productive\". I think you should be prepared to face some \"snark\" and brickbats from our distinguished moderator [b] JBL [/b].  :D",
  "Solution_2": "[quote=\"Gen8\"] I think you should be prepared to face some \"snark\" and brickbats from our distinguished moderator [b] JBL [/b].  :D[/quote]\r\nI think you should stop being bitter and passive-aggressive and think about why you get negative responses to your posts.  (Were you trying to say \"bric-a-brac\"?)\r\n\r\nAs for the OP's question, your brain may just be tired out.  If rest doesn't work, you might want to see a doctor; this might be a neurological symptom of some disorder.",
  "Solution_3": "[quote] I think you should stop being bitter and passive-aggressive and think about why you get negative responses to your posts. (Were you trying to say \"bric-a-brac\"?) [/quote]\r\n\r\n[b] @ t0rajir0u [/b] : I never expected such a knee jerk reaction from you! Well, I was not bitter or passive-aggressive as you may think. I didn't understand what you meant by -- \"get negative response to your posts\". (I acknowledge that some of my queries in the Round Table were highly hypothetical or sometimes silly in nature, so, I might not have got fruitful/elaborate replies)\r\n\r\nI was saying \"brickbat\". You may want to refer to the link below: \r\n\r\n[url]http://www.thefreedictionary.com/brickbat[/url]\r\n\r\nI was saying that \"posts\" that exhibited the [b] OP [/b] \"post pattern\" (lengthy, etc)were sometimes mocked.",
  "Solution_4": "[quote=\"Gen8\"]I was saying that \"posts\" that exhibited the [b] OP [/b] \"post pattern\" (lengthy, etc)were sometimes mocked.[/quote]\r\nI'm not sure what posts you're referring to that didn't have other problems.",
  "Solution_5": "Maybe a little of topic?\r\n\r\nI am confused too. I can't think of a possible reason. You probably could have a week ago  :lol: .",
  "Solution_6": "After thinking about it, i have to ask how did it start? Did it start overnight or gradually?",
  "Solution_7": "Ignore Gen8.\r\n\r\nThis skill is incredible. To tell you the truth I envy you. :)\r\nDid you do anything different on that fateful night? Change of diet, living conditions, etc.?",
  "Solution_8": "[b] @ 3100 : [/b] Hey! My comment was not meant to offend you (or anything like that), I was just cautioning you. If I hurted you, I apologize.  :( \r\n\r\n[b] @ 1=2: [/b] I haven't done anything \"offensive\", so it would be better if you desisted from giving unwarranted suggestions (The first line of your post) :ninja:",
  "Solution_9": "Let's put an end to this meta-discussion, shall we?  Those of you brave enough to read through the entire treatise (I am not among your number) and come out believing there's something worth saying in response, feel free to carry on with actual discussion.",
  "Solution_10": "It's ok, I am not mad or anything. I am afraid there is nothing to be jealous of -- the 'skill' is gone :(. I woke up in the morning, did my daily routines, then when talking to my friend sometime after noon I realized that I couldn't find a simple solution for 'omg my bf doesn't want to spend enough time with me' or something along the lines. I then started to hectically recall some logical puzzles to see if something is wrong with my mind, and I encountered 'mental blocks' when trying to solve them at some point. By 'mental block' I mean some feeling of laziness coupled with unease of thinking about any of the problems. I can't think of doing something differently, I tried to recall all the 'different' things I have done, and the only thing that comes to my mind is physical workouts which I began to do. Those were rapid, but not really tiring, and I stopped doing them as soon as I got these 'mental blocks' to check if there was any relation - no improvements so far. I am not sure it's because my brain is tired out either - I wasn't really loading it over in last year, and all the analysis I have done up until this point was more like a hobby - it was very easy to do, and I didn't even need to concentrate much. I probably would have to concentrate and tire myself if I would go in-depth trying to find the very fundamentals of logic and how they are related to human psychology, but I wasn't going there just yet (and I can't imagine it being that tiring either). The fact that it happened rapidly as opposed to gradually is the one that frightens me the most - if it was something gradual, like me getting 'dumber' for past 2 months, then I might have been able to think of a reason. But this way it just left me totally puzzled. The irony is that if it was a physical disability of some sort then I could have asked a doctor to help me, I can't imagine psychologist listening to me seriously and advising me what to do... I am actually more [b]normal[/b] person now than I was several days ago. The only places I can ask for ANY type of advice is probably some high iq society board or this forum, and hope for a miracle. :maybe: Losing my mind has been one of the things I am actually afraid of, and I guess it happened now. I will try consuming some 'brain food' such as fish oil, nuts, etc. and try to treat this situation as if I was 2 years back from now, and had to build all of my 'skills' from zero.",
  "Solution_11": "I can tell you that if you waste money seeing a shrink, now would be the moment to stop doing so.\r\n\r\nI highly doubt your problem is a 'simple' physical deal...I would not venture to say that it is purely neurological, or a direct reprecussion of eating patterns.\r\n\r\nI think you are lost. But not in the way you think you are. Lemme explain..\r\n\r\nHave you ever looked back at a time in your childhood, i.e., watched home videos, read something you wrote back then with all ernesty, but now you realize that it was just a simple thought from a simpler time? When you refocus things, put things into a new perspective, your point of view is bound to change.  Oftentimes, this takes a rather long duration of time, a new accumulation of life experiences, etc. This is one possibility....I would guess that you've come to look back at your opinions, no matter how recently formed, and you've started to question.\r\n\r\nI'm guessing that you're questioning your logic, your insight, your clear-mindedness that was 'second-nature' until that faithful morning when, *sound effect*, everything went KABLOOEY!\r\n\r\nQuestioning is good if it doesn't completely rip to shreds what little sanity human beings have to start with.\r\n\r\n---\r\n\r\nBut at this time of the post, I think you should take some deep breaths, and take a break for a while, and not freak out about this. That was a run-on sentence, but don't let that distract you from the relatively sound advice.  You can't hear it, so, assume it's sound advice.  Try not to think about this as a crisis of sorts, just a 'phase' that you're going through.  Try to relax and do other things besides well reasoned arguments.\r\n\r\n---\r\n\r\nSo now, perhaps you [i]should[/i] start back at 'level 0'. \r\n\r\nI guess, in essence, what I'm trying to say is that you requestioning everything you believe in doesn't necessarily indicate something is wrong with you.  It's part of this inevitable little process called change, and it's perfectly normal.  I think you should think about your current confidence level, and re-evaluate priorities and experiences that led to your 'breakthrough' conclusions in the first place.  I think you would find that helpful, and although it would not entirely stop your mind from playing tricks on you, it's where I would start.\r\n\r\nAlright. Ending my ramblings. Lemme know how that goes for you. :)",
  "Solution_12": "Thanks, The Scintillator. You might be correct, although I would expect a change in my views to occur much less rapidly. I have tried to pull my stuff together and analyze all the possible ways to get back on track (I don't worry about the cause anymore, hoping it doesn't happen again), and it looks like (surprisingly) SAT prep back in Nov-Dec-Jan has helped me quite some bit. I am talking about reading passages as a 'robot' and neglecting the common emotional reactions you get in favor of more analytical, 'computer', critical reading. Seems to be helping me a little by little -- who would ever think standardized testing can be of such use. :)",
  "Solution_13": "Here's a lossless and much more readable paraphrase of the original post. I would advise the OP to replace his/her original post with this so that people might actually care to read the thread. Note that I used hide tags for a quasi-relevant part of the summary.\r\n\r\n[quote]For past year-year and half or, I\u2019ve gradually improved my analytical and logical skills by questioning/pondering about the world around me. [hide=\"Click here to see some of my conclusions so far\"]- Almost everything people do is in pursuit of physical, emotional/moral*, or spiritual comfort, whether or not they are aware of it. That is, people are selfish.\n- Atheists should shut up since they are equally lacking in evidence.\n- Nationalism is a form of discrimination, just like racial/gender/religious/orientation discrimination\n- Morality is only relevant within a particular societal/cultural context. We can\u2019t say anyone\u2019s morals are right or wrong; that would be discrimination.[/hide]\n\nThree days ago, I started getting a mental block every time I get into a debate. When I try to debate, I open my mouth expecting the ideas to flow as usual, but it's not happening. I can hold familiar debates, since I remember the points I\u2019ve used in the past, but I can\u2019t think of new points on the fly.\n\nAs mentioned above, I became anti-patriotic/anti-nationalistic after analyzing and deciding that nation/ethnicity are not more important than race/gender in any way, and should not be used to discriminate people. I became very anti-nationalist. Now, I feel apathy \u2013 I don\u2019t care anymore.\n\nI\u2019ve tried IQ tests and math problems, and my performance hasn\u2019t changed. But when I read, it takes longer for me to grasp the underlying meaning of the author.\n \nWhat do I do?[/quote]",
  "Solution_14": "[quote]\nAs mentioned above, I became anti-patriotic/anti-nationalistic after analyzing and deciding that nation/ethnicity are not more important than race/gender in any way, and should not be used to discriminate people. I became very anti-nationalist. Now, I feel apathy \u2013 I don\u2019t care anymore.\n[/quote]\r\n\r\nI'm antipatriotic and antinationalist too....\r\nI am Greek but i'm feeling that my homeland is the earth and not Greece....I love Greeks the same as Albanians,Indians,Italians etc...\r\nI hate when people believe that they are above all the others because of their nationality and i strongly believe that ''when your sleeping the night patriot you wake up fascist''.\r\n\r\nBUT i cant understand what do you  mean that you feel apathy.You feel apathy against...?\r\n\r\nOn the other hand when i read about these ideas and make them my own i became more sensitive about :\r\nsocial inequalities, refugees etc and i'm 'fighting' to pretend their rights....\r\nSo just make to yourself clear what you are believe and why you believe this and you will feel better :wink: \r\n\r\nI cant expreess myself as i would like because of my mediocre English but i believe that you can understand what i want say...\r\nDimitris",
  "Solution_15": "About how time consuming was this hobby of yours?\r\nIt could be overload: you have become obsessed with this certain hobby and tried to learn as much as possible so that you burnt yourself out. \r\n\r\nWhat types of discussions are you having with your friends now? Are they more complex than before? Perhaps you don't know where to go with your \"guideline\" for thinking that you outlined in your testimony. Maybe this is an indication of necessary adaptation: experience/read more of life so that you can improve your former assumptions, because maybe you have outgrown your conclusions as scintillator pointed out. \r\n\r\nGood luck!",
  "Solution_16": "[quote=\"3100\"]\n-Believing that there is no God (basically being an atheist) is not 'smarter' than believing in one since both sides lack evidence to support their opinions (hence making the debate pointless).\n[/quote]\r\nthe average atheist, at least among the people i know, does not specifically believe that thier is not a god, the just don't specifically believe in any god.\r\n\r\nalso the existence of no gods is more likely then the existence of gods by occam's razor since it is a simpler system",
  "Solution_17": "Play tennis or something? No? Well, let's pretend you do.\r\n\r\nMatch point, ad-out, your serve, what's the worst thing that could happen? You realize what situation you're in. - oh crap if i miss the first serve which i've been doing at a frequency of about 60% this game (goddam first serve's not working) then i'll have to rely on my second serve which he's been constantly pounding down the alley but what if he takes it cross-court and catches me off guard but crap i haven't even served yet i should do that but etc.\r\n\r\nI don't have some secret window into your life, every situation is different, and so I may be off. In the end, only you can properly diagnose the problem because only you are completely immersed in its context. Which is the problem. I think you are being too self-conscious about your lack of logic. Instead of tuning your mind into solving the problem, you tune your mind into why you aren't solving the problem. You realize the situation and then your mind pops up a level and focuses at a paradigm it shouldn't be focusing on. You may not realize you're doing this - certainly the test problem you give yourself after reading this post will be all too self-aware about your self-awareness - but it's happening on some level which is fundamentally inhibiting the process. I can only see it be more affected when that is the unconscious level, which is usually reserved for grabbing different associations between ideas but in your case is too focused on how it should be grabbing ideas. Started off simply enough - you tried to solve a problem, noticed your ability, and from then on you always noticed your ability (and now lack thereof).\r\n\r\nHow to fix this, if this is indeed the case? Other than putting yourself into a Skinner box, I don't know. I'm afraid that a diagnosis probably only worsens the symptoms. You can try convincing yourself that you don't have an ability. This is why so many people are so humble concerning themselves (e.g. omg i failed OR ugly solution, above is better, how did you think of that?) By consciously deflecting their thoughts away from how strong their abilities are - or used to be, in your case - they don't ever fall into the trap of self-awareness. The mindset is: how good you are is trivial, what matters is the solution. So convince yourself you never had anything, and it might just come back.",
  "Solution_18": "@ Latios315: The definition of \"atheism\" is \"the assertion that dieties do not exist\". In other words, it's saying there is no god.\r\n\r\nNot being sure of the existence of a god is agnostism (did I spell that right?).\r\n\r\nUh... I forgot the word that means \"believing in more than one god\"..... (sweatdrops)\r\n\r\nSimply not really asserting/believing anything is being \"non-religious\". It is not the same thing as \"atheist\". \r\n\r\nAt least, that's what I think. Other than that, I hav absolutely nothing to contribute to the discussion. Carry on.  :lol:",
  "Solution_19": "I think you'll figure yourself out with time and with experience living life. If that's not fast enough for you, then go out and try something new. Maybe you'll figure yourself out better. I feel like you're too focused on tapping this one, hyper-analytical part of yourself, and have a very narrow idea of how your analytical mind can contribute to your life.\r\n\r\nI remember my mathematical abilities sort of stagnated after 10th grade, when I stopped having time to invest into reading math books and such. A huge boost in my competition math abilities came, to my surprise, from doing mock trial. Going back and connecting the dots, it all makes sense. Another example -- I have yet to have a technical job. I remember, in particular, working at a courthouse the summer after my senior year, where I helped people file claims/motions/cases, which involved some customer service type stuff and giving out some legal advice. Because this involved a lot of dealing with lots of people and communicating analytical thoughts to less analytical people, I developed these skills, which I now find invaluable when dealing with professors, looking for research, etc.\r\n\r\nWhile neither of these examples have directly to do with your debating ability and only tangentially to your analytical mind, they have the relevant message -- branch out, and then go back and connect the dots. The dots [i]will[/i] connect, and you'll be surprised at what you'll be able to do.\r\n\r\nInterestingly, I heard Steve Jobs say something similar at his speech at Stanford's commencement. While I won't guarantee that dropping out of college and taking calligraphy classes will lead to you becoming like Steve Jobs, I will guarantee that if you explore a bit, open your mind, and let your gut guide you, the dots will connect.",
  "Solution_20": "The serious reply here is that you're obsessing way too much over this. Judging by the length of your posts and points raised, it would seem you're quite concerned. If you actively pick up on every fault you make, every time you're not quick enough (by some imaginary standards), every time you can't remember something then you will find many things indeed. It sounds to me a lot like fairly ordinary anxiety/obsession. A similar example would be a person who feels depersonalised and is constantly thinking about this, judging every experience by this.  \r\n\r\nOf course I am probably reading too much into your posts, given you've only felt this way for a few days, but you can rest assured that you don't need to punish yourself by pondering this kind of stuff. Nothing has happened. Your abilities and skills are intact, a \"neurological symptom\" is ridiculously unlikely. That your \"ability of comprehension\" could have changed in 3 days is just such total nonsense if you break it down; I am 100 % you are just worrying (\"imagining things\").",
  "Solution_21": "So 3100, care to give us a status report?\r\n\r\nMany others, (too lazy to list everyone) have been suggesting lots of wonderful ideas, such as going out and trying new things, taking a break, and just in general not being too concerned with your problem.\r\n\r\nI think stuck said it best:\r\n[quote=\"stuck\"]In the end, only you can properly diagnose the problem because only you are completely immersed in its context. Which is the problem. I think you are being too self-conscious about your lack of logic.[/quote]\r\n\r\nSo the best way, in my opinion? Take some more time off and let us know how you feel. :)",
  "Solution_22": "[quote=\"dreambold22\"]@ Latios315: The definition of \"atheism\" is \"the assertion that dieties do not exist\". In other words, it's saying there is no god.\n\nNot being sure of the existence of a god is agnostism (did I spell that right?).\n\nUh... I forgot the word that means \"believing in more than one god\"..... (sweatdrops)\n\nSimply not really asserting/believing anything is being \"non-religious\". It is not the same thing as \"atheist\". \n\nAt least, that's what I think. Other than that, I hav absolutely nothing to contribute to the discussion. Carry on.  :lol:[/quote]\r\nI know that, which is why when someone asks my religion i say i am agnostic. however many people use the term \"atheist\" to describe anyone that does not believe in god\r\n\r\n@OP that's a very strange situation, If i were you i would go to the doctor just to make sure you don't have a brain tumor or other disease as the suddenness of this is very wierd",
  "Solution_23": "That would seem more like a power than a burden, right? I mean, its a pretty cool thing that your so insightfull now! :lol: \r\n\r\nI agree with The_Scintillator. We need a status report.",
  "Solution_24": "TZF, thanks for shortening it, and the message after that one. It is quite helpful.\r\n\r\n[quote]I'm antipatriotic and antinationalist too....\nI am Greek but i'm feeling that my homeland is the earth and not Greece....I love Greeks the same as Albanians,Indians,Italians etc...\nI hate when people believe that they are above all the others because of their nationality and i strongly believe that ''when your sleeping the night patriot you wake up fascist''.\n\nBUT i cant understand what do you mean that you feel apathy.You feel apathy against...?\n\nOn the other hand when i read about these ideas and make them my own i became more sensitive about :\nsocial inequalities, refugees etc and i'm 'fighting' to pretend their rights....\nSo just make to yourself clear what you are believe and why you believe this and you will feel better Wink\n\nI cant expreess myself as i would like because of my mediocre English but i believe that you can understand what i want say...\nDimitris[/quote]\r\n\r\nNice to see a fellow anti-nationalist, Dimirtis. I was trying to say that if a day before I was all sure about my position (since I could logically back it), a day after I did not feel 'allergic' to patriotic ads and other brainwashing techniques which would make me 'vomit' several days before this accident. I became apathetic, and that was frightening. :) Keep it up, and your English is very good, don't worry about it.\r\n\r\nLightsAbove, it was pretty time consuming -- I was applying to everything I did in my life, but I think it's more that I can't get the conclusions with the same ease as I used to - I can't 'see' as fast right now.\r\n\r\nlatios315, if I do not mistake, occam's razor principle states that one should make as less assumptions as possible, so assuming that there is(are)/is(are) no God(s) is equally 'wrong' according to this principle. Not to assume anything, you would have to state that you 'don't know anything on this issue, since there is no solid evidence to support either of the opinions.'\r\n\r\nstuck, thanks, your way to solve this problem might actually work. I will try it out (actually, I have already been trying it out, albeit unconsciously.)\r\n\r\nKalle, I wish I was imagining things :) I know you probably feel that I am just going crazy -- if I had read something like that myself I would be like 'what the hell is he talking about?' The situation is so absurd and unbelievable, sometimes I wonder if I am dreaming or not. But, drawing an analogy, I went from solving multi-variable calc problems to not being able to factor the (a+b)^2. I used to 'think logically' with such an inspiration, that I couldn't find something hard enough to keep me engaged for a long while, and I can't think of something that could prevent me from solving something trivial I had stumbled upon.\r\n\r\n\r\n\r\n\r\nI have thought of several things, and well, it might have been one specific national dish that got me in this 'mental block' position -- I am recalling feeling sleepy and kind of zoned out. The other possible reason might have been that some building right across my room's window was being renewed, and I had all this plaster powder and like flying in the direction of my room, and I assume it's not all that good for one's health. Can't think of any other reasons  :blush: I have stopped eating that dish and changed my room, and it seems that the 'mental block' is gone -- I haven't regained my abilities, but now I feel like I can do it if I try, that might happen (previously, it was like mental block -- you can't do it.) This was an interesting occurrence though, like a chance from god (metaphorically speaking) to understand people who are not trying to be 100% logical. Earlier, when I told people that crying when someone dies is pointless, because it doesn't change anything and it is [b]normal[/b] to die, I was surprised that majority of them didn't agree with me (despite not coming up with counter-arguments.) Now I kinda understand them, yes it is logical, but looks like people are not all about logic, and nature/animal instincts still play big part in our responses to the things around us. I am kind of back to the level I was at a year or two ago; I have reluctantly found out that now I care what others people say, what I wear, where I leave, what my social status is, and so on. Two months ago, I wouldn't mind leaving in a tub as Diogenes did (for example), but now I am kind of worried that this will not be viewed 'correctly' by the society in the scope of [b]current[/b] moral values and established views. I hope I will manage to prescind from the everyday routines once again and focus on more 'global' issues, like before. I want to thank everyone who tried to help me -- looks like I have achieved some progress by following several advices suggested on here.",
  "Solution_25": "[quote=\"3100\"]\nlatios315, if I do not mistake, occam's razor principle states that one should make as less assumptions as possible, so assuming that there is(are)/is(are) no God(s) is equally 'wrong' according to this principle. Not to assume anything, you would have to state that you 'don't know anything on this issue, since there is no solid evidence to support either of the opinions.'\n[/quote]\r\noccam's razor says that the system making the least assumptions is most likely, since you make less assumptions in believing that there is no god, then believing in god, the existence of no god is more likely then the existence of a god, that is all i was saying. I agree with you that it is best to make no assumptions and be agnostic",
  "Solution_26": "This conversation is so deep, I think that 3100 should just carry on and start from scratch.\r\n\r\nYou know what they say, have fun in life, nobody ever makes it out alive.",
  "Solution_27": "dreambold, the word for believing in more than one god is:\r\n\r\npolytheism.\r\n\r\nThe noun form is polytheist.\r\n\r\nBut isn't this whole conversation derived from the words:\r\n\r\nwhy\r\n\r\nhow\r\n\r\nwho\r\n\r\nwhat\r\n\r\nwhere\r\n\r\nwhen",
  "Solution_28": "After reading through all that, I can only agree with the poster who told you to take a break. Maybe you're trying too hard to see something/think of something when before your power lay in your ability to just think and go with the flow of thought.\r\n\r\nI'd also like to address \"atheists should shut up because they have just as little evidence\". Although technically the definition of atheism is belief that no god exists, many choose it as the only logical choice after pointing out that it makes no sense for \"my god\" to be better than \"your god\". Atheism tends to be more \"anti-religious-ism\" than the actual technical definition itself (referring to the common person, at least in my experience).\r\n\r\nSorry for the off topicness -- feel free to not respond.",
  "Solution_29": "@tcjy: Thank you for telling me the word for \"believeing in more than one god\". And I know my post was extremely off-topic... That's why I added the last two sentences. (\"Other than that, I hav absolutely nothing to contribute to the discussion. Carry on.\") I'm not really sure what you're trying to tell me, but I'm inferring you want me to either shut up or get on topic. I think I'll shut up now. (unless someone brings this particular subject about religions up again.)\r\n\r\nOkay. I am now officially shutting up. Carry on.",
  "Solution_30": "Off topic: I envy dreambold, she is lost in a sushi.\r\n\r\nOn topic: I am an anti-nationalist too [hide=\"(conversation)\"]Someone else: July the 4th. This country was born.\nMe: Whatever.[/hide], and read my response to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286761]this[/url] topic for my religion. I think you should just relax about the loss of ability, let it go for a few days, and maybe it will come back.",
  "Solution_31": "Oh--and one more thing.\r\nI was reading this one book, and it was kind of interesting....it might explain why people aren't agreeing with you:\r\n\r\nPure logic won't always win the competition--you need to be persuasive. Which sounds weird and manipulative, but most people don't respond well to cold logic.  Even people on this site. Just a thought.\r\n\r\nI am quite pleased to hear you are doing better. Be happy! :lol:",
  "Solution_32": "[quote=\"The_Scintillator\"]\nPure logic won't always win the competition--you need to be persuasive. [/quote]\r\nActually, even more than that. If you want to convince somebody, you should play on his feelings, desires, and other subconscious things much more than on his reasoning. That's why the TV commercials (most of which are perfect examples of ultimate idiocy as far as the pure logic is concerned) are so effective. Besides, beyond some elementary cases, correct application of pure logic requires complete information about the subject, which is almost always absent in the real life arguments. \r\n\r\nAs to \"nationalism\" you dislike so much, there are many faces of it too. I do not mean here the existence of some mild forms for which the word \"patriotism\" is more often used. Rather I mean the state of Israel. In under 20 years the Jews built a thriving country in what is technically a mountainous desert with a narrow coastal strip and without any significant natural resources. That was quite a feat by all standards and most people who were in the leading positions were extreme nationalists. They just were clever enough not to equate equate the preservation of cultural identity with the world domination. I do not see anything logically wrong with the former, it is just the latter that is questionable (though even here there is a huge difference between the Victorian British empire and the Nazi Germany). Most ironically, the abhorrence of all kinds of nationalism often comes from the standard logical mistake, which is \"If A is B and C is B, then A is C\" with B being \"nationalism\", C being \"German nazism\" and A everything else.",
  "Solution_33": "you should be passive and nice and friendly",
  "Solution_34": "I think the Scintillator is right, and you should get some rest :wink:",
  "Solution_35": "with a very modesty i may think that : \r\n1. the deductive proof scheme we use is natural(mechanical) like breathing or say walking or running\r\n2. you are using this proof scheme consciously. So it gives up functionning mecanically since you took it in your hand(your mental hand).\r\n3. If you concentrate on you breath or on the moment you fall asleep, or on your walking you will make a disorder on them.(many poeple had breathing and sexual desorders by practicing tantra yoga and concentrationg on the prana)\r\n3.But i dont think what happened to you is   irreversible. It seems to me that if you stop doing so(forcing the deductive proof scheme) and let him fall from your hand, it will take its normal function.\r\n4. I think also that the deductive proof scheme is the last think to trust, because the fundamental think to have is big datta and information ground because fast deduction can be damaging. So it's better for us if we concentrate on checking informations \"pertinece\" according to the different situations and take time to say that way be we are wrong instead of fast deductions that kills the audience virtually only.\r\nIn your case you have erased virtually all your backing by experiencing fast deductin that made you see  that you  can have arguments against anything, even if they are contradictory, so there's no reason for backing for you so when you sollicitated it, it doesnt emerge,(but be sur that it is in your memory).",
  "Solution_36": "Sorry to revive, but when reading this, I did come up with an interesting analogy:\r\n\r\nConsider your situation to when you start noticing your breathing (which you should after reading this). If you think about it, it doesn't come naturally anymore and you are suddenly cognizant of it for the next few minutes right?. In time, hopefully, your mind will drift so that you breath more naturally and don't have to worry about breathing, but I think this is the case with your thinking.\r\nPerhaps that day was when you became more cognizant of your ability to think rapidly? Hopefully the same principle applies: that you will regain your natural thought structure back. But in this case, there is something that I hope you learned:\r\n\r\n[quote=\"3100\"]\nNow I kinda understand them, yes it is logical, but looks like people are not all about logic, and nature/animal instincts still play big part in our responses to the things around us.\n[/quote]\r\n\r\nthat logic cannot run everything. Yes, logic is the purest form of reason, but that is not all our thoughts and actions are limited to (hopefully, as long as you don't believe in determinism). That additional factor of soul and emotions must also play a factor in any conclusion.\r\n\r\nAlso, since its been so long, what's happened since then 3100?"
}
{
  "Tag": [],
  "Problem": "if (x\u00b2+x+2)(x+3)= ax^3+bx^2+cx+d, for all values of x, what is a+b+c+d?\r\n\r\nI just need an answer for this one\r\n\r\nif x and y are positive integers, x^2-y^2= 35 and x^2+2xy+y^2= 49, what is x-y?\r\n\r\nexplain this one please",
  "Solution_1": "[hide] For number one, the answer is 16. You multiply, and corresponde to the second equation, and then add them up.[/hide]",
  "Solution_2": "[quote=\"Forte\"][hide] For number one, the answer is 16. You multiply, and corresponde to the second equation, and then add them up.[/hide][/quote]\r\n\r\nand can you do the 2nd one too? \r\nalso\r\n\r\ny=x^2-6x+9 what is the least value of y?\r\n\r\nthanks in advance",
  "Solution_3": "[hide=\"The First One\"]Do a manuever which is almost like Foil-ing.\n\n$ x^2 \\times x + 3x^2 + x \\times x + 3x + 2x + 2 \\times 3 = x^3 + 4x^2 + 5x + 6$  Thus, the answer is $ 1 + 4 + 5 + 6$, or $ \\boxed{16}$.[/hide]\n\n[hide=\"The Second One\"]Factor both equations:\n\n$ (x - y)(x + y) = 35$\n\n$ (x + y)(x + y) = 49$\n\nTake the square root of the second equation.\n\n$ x + y = 7$\n\nPlug this into the first.\n\n$ 7(x - y) = 35$\n\n$ x - y = \\boxed{5}$[/hide]",
  "Solution_4": "[quote=\"ernie\"][hide=\"The First One\"]Do a manuever which is almost like Foil-ing.\n\n$ x^2 \\times x + 3x^2 + x \\times x + 3x + 2x + 2 \\times 3 = x^3 + 4x^2 + 5x + 6$  Thus, the answer is $ 1 + 4 + 5 + 6$, or $ \\boxed{16}$.[/hide]\n\n[hide=\"The Second One\"]Factor both equations:\n\n$ (x - y)(x + y) = 35$\n\n$ (x + y)(x + y) = 49$\n\nTake the square root of the second equation.\n\n$ x + y = 7$\n\nPlug this into the first.\n\n$ 7(x - y) = 35$\n\n$ x - y = \\boxed{5}$[/hide][/quote]\r\n\r\nLook at the way you factored y= x\u00b2-6x+9.\r\n\r\nCorrect answer- The smallest value occurs when x-3= 0. Therefore the answer is 0.",
  "Solution_5": "im confused from ernie's post  :(",
  "Solution_6": "[quote=\"PiLuvah\"]im confused from ernie's post  :([/quote]\r\n\r\nDon't be, he just messed up.  :P",
  "Solution_7": "If you're confused by the third question, I edited it out."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Prove that:\r\n$ \\sqrt{(a^{2}b\\plus{}b^{2}c\\plus{}c^{2}a)(ab^{2}\\plus{}bc^{2}\\plus{}ca^{2})}\\geq abc\\plus{}\\sqrt[3]{(a^{3}\\plus{}abc)(b^{3}\\plus{}abc)(c^{3}\\plus{}abc)}$,for all $ a,b,c>0$.",
  "Solution_1": "Very nice Erken . :) .\r\n\r\nMy solution :\r\n\r\nThe inequality is equivalent to :\r\n\r\n$ \\sqrt{(m\\plus{}n\\plus{}p)\\left(\\frac{1}{m}\\plus{}\\frac{1}{n}\\plus{}\\frac{1}{p}\\right)}\\ge 1\\plus{}\\sqrt [3]{(\\frac{m}{n}\\plus{}1)(\\frac{n}{p}\\plus{}1)(\\frac{p}{m}\\plus{}1)}(1)$.\r\n\r\n($ m \\equal{}\\frac{a}{b},n \\equal{}\\frac{b}{c},p \\equal{}\\frac{c}{a}$ then $ mnp \\equal{} 1$ )\r\n\r\nLet $ t \\equal{}\\frac{m}{n}\\plus{}\\frac{n}{m}\\plus{}\\frac{m}{p}\\plus{}\\frac{p}{m}\\plus{}\\frac{n}{p}\\plus{}\\frac{p}{n}$\r\n\r\nSo $ t\\ge 6$.\r\n\r\n$ (1)\\Leftrightarrow\\sqrt{t\\plus{}3}\\ge 1\\plus{}\\sqrt [3]{t\\plus{}2}(1)$\r\n\r\nLet $ t \\equal{} x^{2}\\minus{}3,x\\ge 3$.\r\n\r\n$ (2)\\Leftrightarrow x\\minus{}1\\ge\\sqrt [3]{x^{2}\\minus{}1}\\Leftrightarrow x(x\\minus{}1)(x\\minus{}3)\\ge 0$\r\n\r\nDone . :)",
  "Solution_2": "I think that you mean $ t \\equal{}\\frac{m}{n}\\plus{}\\frac{n}{p}\\plus{}\\frac{p}{m}\\plus{}\\frac{m}{p}\\plus{}\\frac{p}{n}\\plus{}\\frac{n}{m}$.\r\nBut anyway your solution is very nice.My solution is longer and isn't so nice,i solved it using ABC theorem.",
  "Solution_3": "[quote=\"Erken\"]I think that you mean $ t \\equal{}\\frac{m}{n}\\plus{}\\frac{n}{p}\\plus{}\\frac{p}{m}\\plus{}\\frac{m}{p}\\plus{}\\frac{p}{n}\\plus{}\\frac{n}{m}$.\nBut anyway your solution is very nice.My solution is longer and isn't so nice,i solved it using ABC theorem.[/quote]\r\n\r\nSorry, i have edited it . :D . \r\n\r\nBy ABC -theorem , we must consider only case $ a \\equal{} b$, but i think , it is long solution . Please post it , Erken  :wink:",
  "Solution_4": "[quote=\"quangpbc\"][quote=\"Erken\"]I think that you mean $ t \\equal{}\\frac{m}{n}\\plus{}\\frac{n}{p}\\plus{}\\frac{p}{m}\\plus{}\\frac{m}{p}\\plus{}\\frac{p}{n}\\plus{}\\frac{n}{m}$.\nBut anyway your solution is very nice.My solution is longer and isn't so nice,i solved it using ABC theorem.[/quote]\n\nSorry, i have edited it . :D . \n\nBy ABC -theorem , we must consider only case $ a \\equal{} b$, but i think , it is long solution . Please post it , Erken  :wink:[/quote]\r\nDo you see my post about ABC theorem?I used 4th Consequences from there.But if $ a\\equal{}b$ then the inequality is obvious,isn't it?\r\nAnyway i'll post my solution later. :) \r\nIs there any other solutions to this inequality?",
  "Solution_5": "by cauchy schwarz and Am-Gm \r\n$ \\sqrt{(a^{2}b\\plus{}b^{2}c\\plus{}c^{2}a)(b^{2}a\\plus{}c^{2}b\\plus{}a^{2}c)}\\equal{}\\frac{1}{2}\\sqrt{[b(a^{2}\\plus{}bc)\\plus{}c(b^{2}\\plus{}ca)\\plus{}a(c^{2}\\plus{}ab)] [c(a^{2}\\plus{}bc)\\plus{}a(b^{2}\\plus{}ca)\\plus{}b(c^{2}\\plus{}ab)]}\\geq\\frac{1}{2}(\\sqrt{bc}(a^{2}\\plus{}bc)\\plus{}\\sqrt{ca}(b^{2}\\plus{}ca)\\plus{}\\sqrt{ca}(b^{2}\\plus{}ca))\\geq\\frac{3}{2}\\sqrt [3]{(\\sqrt{bc}(a^{2}\\plus{}bc)\\sqrt{ca}(b^{2}\\plus{}ca)\\sqrt{ab}(c^{2}\\plus{}ab))}\\equal{}\\frac{1}{2}\\sqrt [3]{(a^{3}\\plus{}abc)(b^{3}\\plus{}abc)(c^{3}\\plus{}abc)}\\plus{}\\sqrt [3]{(a^{3}\\plus{}abc)(b^{3}\\plus{}abc)(c^{3}\\plus{}abc)}\\geq\\frac{1}{2}\\sqrt [3]{2\\sqrt{a^{4}bc}2\\sqrt{b^{4}ca}2\\sqrt{c^{4}ab}}\\plus{}\\sqrt [3]{(a^{3}\\plus{}abc)(b^{3}\\plus{}abc)(c^{3}\\plus{}abc)}\\equal{} abc\\plus{}\\sqrt [3]{(a^{3}\\plus{}abc)(b^{3}\\plus{}abc)(c^{3}\\plus{}abc)}$",
  "Solution_6": "WOW !Very very nice short and beautiful solution!Thank you very much! :lol:",
  "Solution_7": "[quote=\"Erken\"]Prove that:\n$ \\sqrt{(a^{2}b\\plus{}b^{2}c\\plus{}c^{2}a)(ab^{2}\\plus{}bc^{2}\\plus{}ca^{2})}\\geq abc\\plus{}\\sqrt [3]{(a^{3}\\plus{}abc)(b^{3}\\plus{}abc)(c^{3}\\plus{}abc)}$,for all $ a,b,c > 0$.[/quote]\r\nSee also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=90015",
  "Solution_8": "Just want to provide an alternate proof for the inequality $ a,b,c > 0$, $ abc = 1$, then \r\n$ \\sqrt{(a+b+c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)}\\ge 1+\\sqrt[3]{(a+b)(b+c)(c+a)}$\r\n\r\nNote that\r\n$ 1+\\sqrt[3]{(a+b)(b+c)(c+a)}$\r\n$ =\\sqrt{\\left(1+\\sqrt[3]{(a+b)(b+c)(c+a)}\\right)\\left(1+\\sqrt[3]{\\left(\\frac{1}{a}+\\frac{1}{b}\\right)\\left(\\frac{1}{b}+\\frac{1}{c}\\right)\\left(\\frac{1}{c}+\\frac{1}{a}\\right)}\\right)}$\r\n$ \\leq\\sqrt{1+\\frac{2(a+b+c)}{3}}\\sqrt{1+\\frac{2\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)}{3}}$\r\n$ \\leq\\sqrt{(a+b+c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)}$"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "For $ n \\in \\mathbb{N}$, let $ f(n)\\equal{}1^n\\plus{}2^{n\\minus{}1}\\plus{}3^{n\\minus{}2}\\plus{}...\\plus{}n^1$. Determine the minimum value of: $ \\frac{f(n\\plus{}1)}{f(n)}.$",
  "Solution_1": "we [color=#f00][b]Claim:-[/b][/color]$\\frac{f(n+1)}{f(n)}>3$ $\\forall$ $n>4$\\\\\n\n[color=#00f][b]Proof:-[/b][/color] we need to show that $1^{n+1}+2^{n}+\\cdots n+1>3(1+2^{n-1}+\\cdots+n)$\\\\\n\nnow what we observe is the series of these inequalities:\\\\\n\n$1^{n+1}+2^{n}+\\cdots+(n-1)^3+n^2+n+1>1^{n+1}+2^{n}+3^{n-1}+(4^{n-2}+5^{n-3}+\\cdots+(n-1)^3+n^2)$\\\\\n\nnow $(4^{n-2}+5^{n-3}+\\cdots+(n-1)^3+n^2)>3(4^{n-3}+5^{n-4}+\\cdots+(n-1)^2+n)$\\\\\n\nso our goal is to prove that $1^{n+1}-3\\cdot 1^{n}+2^{n}-3\\cdot 2^{n-1}+3^{n-1}-3^{n-1}+4^{n-2}-3\\cdot 4^{n-3}+5^{n-3}-3\\cdot 5^{n-4}+\\cdots+n^2-3\\cdot n >0$ which is true for $\\forall$ $n>4$ , hence we get\\\\\n\n$f(n+1)>3f(n)$ $\\forall$ $n>4$ , claim proved $\\blacksquare$\\\\\n\nso we check for $n=1,2,3,4$ where the minimum is getting attained, clearly by inspection we observe for $n=2,  \\left(\\frac{f(n+1)}{f(n)} \\right)_{\\text{min}}=\\boxed{\\frac{8}{3}}$ $\\blacksquare$",
  "Solution_2": "Excellent idea for n>4! Congratulations!"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "Ok, I'm trying to solve this problem and would appreciate some help.\r\n\r\nHere it is:\r\n[b]A Circular Massacre[/b]\r\na)  100,000 sailors are arranged in a circle on board a ship.  Every other one is shoved overboard and the last one standing gets to live.  Where should you stand to survive (ex. 4th position).\r\nb)Answer the same question if every third person is thrown overboard.\r\nc)Do the problem with every 4th person being eliminated.\r\n\r\nOk, so ya, I have to find a pattern in each one.  I have already completed a and b, but am not totally sure on the answers.  If you people know a formula/solution/link/strategy/whatever It would be greatly appreciated.\r\n[img]http://instagiber.net/smiliesdotcom/games/diablo/KNTWALK.GIF[/img]",
  "Solution_1": "a. Maybe... [hide] We can see that each time the people gone are 2^n + 1 (mod 2^(n+1)), such as 2 (mod 2), 3 (mod 4), 5 (mod 8), etc. So we want the maximum n such that 2^n - 1 < 100,000 which is 17. So the last person living is 2^17 + 1 = 65537 ? Sounds a little too simple... [/hide]",
  "Solution_2": "Ok guys I really need help on this.  So plz don't just blow me off.  Anyway I think I got an answer for a b and c.\r\n\r\na)68929\r\nb)89644\r\nc)89689\r\n\r\nIs that what you guys think is right?  I really need someone to post a formula/answer/link to a site that has this.\r\n\r\nCan you at least tell me what a math problem that requires elimination is called?  If I know that I can plug it into a search enine and see if I can find a formula.  Plz some1 help.",
  "Solution_3": "I think this is what you're looking for: \r\nhttp://mathworld.wolfram.com/JosephusProblem.html\r\n\r\nMarcos",
  "Solution_4": "[quote=\"marcos\"]I think this is what you're looking for: \nhttp://mathworld.wolfram.com/JosephusProblem.html\n\nMarcos[/quote]\n\nwow thank you soooo much.  Anyway could someone help me know with solving the equation?  K here is what I got:\nL(n,m)=1+2n-2^(1+[lgn])\n\nlg is log base 2, n is survivor is every mth man is killed.\n\nWhat I really don't get is this [quote]Given a group of n men arranged in a circle under the edict that every mth man will be executed going around the circle until only one remains, find the position L(n,m) in which you should stand in order to be the last survivor[/quote]\r\n\r\nposition L(n,m)????   I understand logs etc, but could you help me with this part, or maybe even solve it for me?  I need it with:\r\n\r\nevery other person, every third, and every 4th.  They are all for 100,000 people.  Tnx again"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Let $ ABC$ be an equilateral triangle, and $ P$ an arbitrary point within the triangle. Perpendiculars PD, PE, PF are drawn to the three sides of the triangle. Show that, no matter where $ P$ is chosen,\r\n\r\n$ \\frac {PD \\plus{} PE \\plus{} PF}{AB \\plus{} BC \\plus{} CA} \\equal{} \\frac {1}{2\\sqrt {3}}$.\r\n\r\n[hide=\"triv solution\"]Recall our formula for the area of an equilateral triangle in terms of one of its side lengths. Draw a diagram and see that PD, PE, and PF are the altitudes of three triangles whose area sums to the area of ABC. Equate![/hide]",
  "Solution_1": "[hide]Let the side length of $ \\triangle ABC$ be $ s$.  We connect $ PA$, $ PB$, and $ PC$, forming three triangles in $ \\triangle ABC$ with these three semgents as altitudes.  Finding $ [ABC]$ in two ways, we get: $ \\frac {s^2\\sqrt {3}} {4}\\equal{}\\frac {s(PD\\plus{}PE\\plus{}PF)} {2}.$  \nSolving for $ PD\\plus{}PE\\plus{}PF$, we get that $ PD\\plus{}PE\\plus{}PF\\equal{}\\frac {s\\sqrt {3}} {2}$.  Thus, $ \\frac {PD\\plus{}PE\\plus{}PF} {AB\\plus{}BC\\plus{}AC}\\equal{}\\frac {1} {2\\sqrt {3}}$.[/hide]\r\nIsn't this kinda, as you say, trivial/as I say, uber-easy? :P",
  "Solution_2": "[quote=\"Math Geek\"][hide]Let the side length of $ \\triangle ABC$ be $ s$.  We connect $ PA$, $ PB$, and $ PC$, forming three triangles in $ \\triangle ABC$ with these three semgents as altitudes.  Finding $ [ABC]$ in two ways, we get: $ \\frac {s^2\\sqrt {3}} {4} \\equal{} \\frac {s(PD \\plus{} PE \\plus{} PF)} {2}.$  \nSolving for $ PD \\plus{} PE \\plus{} PF$, we get that $ PD \\plus{} PE \\plus{} PF \\equal{} \\frac {s\\sqrt {3}} {2}$.  Thus, $ \\frac {PD \\plus{} PE \\plus{} PF} {AB \\plus{} BC \\plus{} AC} \\equal{} \\frac {1} {2\\sqrt {3}}$.[/hide]\nIsn't this kinda, as you say, trivial/as I say, uber-easy? :P[/quote]\r\n\r\nindeed, however it was on the 1969 Canadian National Olympiad, so go-figure."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "What is the minimum number of equilateral triangles, of side length 1 unit, needed to cover an equilateral triangle of side length 10 units?",
  "Solution_1": "We're spreading an area (square unit) with a given measurement of side length (linear unit). Thus the answer is $ 10^2\\equal{}\\fbox{100}$."
}
{
  "Tag": [
    "ratio",
    "vector"
  ],
  "Problem": "Two bugs begin their journey at a vertex of an equilateral triangle with side length $3$, and travel a distance of $1$ towards a second vertex. Bug $A$ continues on its way, turning $60^\\circ$ towards the third vertex and traveling $\\frac{2}{3}$ of the distance it just traveled. It continues turning in the same direction, with the same ratio, until it approaches one location and gets really dizzy :P . Bug $B$, on the other hand, travels $\\frac{2}{3}$ of its previous distance [u]directly towards[/u] the third vertex, then the first, then the second, and so on until it too approaches one location and gets quite dizzy.\r\n\r\nNow, what is the distance from their originating vertex to each of the bugs? Which is farther away?",
  "Solution_1": "merr...whats the point of two of them...once you have one, the other is annoying\r\n\r\nOMG I THINK WE NEED TO BUST OUT SOME VECTORS",
  "Solution_2": "This problem becomes trivial once you use third roots of unity."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $a,b,c$  be positive real number such that: $abc=1$. Prove that:\r\n$\\sqrt[4]{\\frac{a}{15b+1}}+\\sqrt[4]{\\frac{b}{15c+1}}+\\sqrt[4]{\\frac{c}{15a+1}}\\geq \\frac{3}{2}$",
  "Solution_1": "[quote=chien than]Let $a,b,c$  be positive real number such that: $abc=1$. Prove that:\n$\\sqrt[4]{\\frac{a}{15b+1}}+\\sqrt[4]{\\frac{b}{15c+1}}+\\sqrt[4]{\\frac{c}{15a+1}}\\geq \\frac{3}{2}$[/quote]\n[url]https://artofproblemsolving.com/community/c6h554702p3222734[/url]\n[url]https://artofproblemsolving.com/community/c6h451457p2538878[/url]\n"
}
{
  "Tag": [
    "Columbia",
    "probability",
    "geometry",
    "perimeter",
    "floor function"
  ],
  "Problem": "The problems are here [url]http://www.columbiacollegesc.edu/academics/math/sample_level_II_questions.pdf[/url] but there are no answers as far as I could find.  It'd be amazing if someone could check to see if the ones I answered are right and help me with the couple that I haven't gotten.\r\n\r\n1. [hide][b]C.[/b] Dimensions are 6*4*8[/hide]\n\n2. Is this one even a question?  If it is, it has eluded me. @_@\n\n3. [hide][b]D. [/b] 4 odd and 1 even, or 2 odd and 3 even, which leads to 5*4 + 10*4.[/hide]\n\n4. [hide][b]H.[/b] There are 210 total paths from A to B, 20 from A to P, and 4 from P to B.  So 20*4/210.[/hide]\n\n5. [hide][b]A.[/b] If the first card is an ace, the probability of getting a 2nd ace is (4/52)(3/51).  If the first card wasn't an ace, it's (48/52)(4/51).  Adding those together yields 1/13.[/hide]\n\n6. Problematic Problem (aka, I don't know.  Help, please?)\n\n7. [hide][b]B.[/b]  r^2 = 1.1(xr)^2 --> x=.95 [/hide]\n\n8. I knew how to do this once upon a time...but now I think I'm over-compensating for repeated multiples. @_@\n\n9. [hide][b]B.[/b] a-b=5, ab=24.  a^2 - 2ab + b^2 = 25 --> a^2 + b^2= 25+48=73. \na^2 + 2ab + b^2 = x^2, x=a+b \na^2 + b^2 = x^2-48 = 73.  \nx^2=121, x=11 [/hide]\n\n10. [hide][b]H.[/b] [/hide]\n\n11. [hide][b]B.[/b] xyz=125, wxyz=625*4, w=(wxyz)/(xyz)=20 [/hide]\n\n12. [hide][b]F.[/b] Side of the square = s. Perimeter of square=4s, as does circumference of the circle. \nRadius of circle: 4s/(2pi).\nArea of square: s^2\nArea of circle: pi(2s/pi)^2= (4s^2)/pi\nSquare:Circle :: pi:4 [/hide]\n\n13. [hide][b]F.[/b] 2/x=a.  \na^2 - 2a + 1 = 0 --> (a-1)^2=0 --> a=1 [/hide]\n\n14. [hide][b]B.[/b] (1000/25)6=240 classes.  240/5=48 teachers. [/hide]",
  "Solution_1": "[quote=\"avcamethyst\"]\n6. Problematic Problem (aka, I don't know.  Help, please?)[/quote]\nOne brute force method is: you can determine $ y$ and $ z$ in terms of $ x$ from the first two equations, and if you insert $ y$ and $ z$ into the third, you'll get a value for $ x$. From that you can get the values of $ y$ and $ z$.\n\n[quote]8. I knew how to do this once upon a time...but now I think I'm over-compensating for repeated multiples. @_@[/quote]\r\nUse the inclusion-exclusion principle.\r\n[hide=\"Solution.\"]\nI will denote the first 1000 positive integers with [1000].\n\n$ A_2 \\equal{} \\{ n \\in [1000] \\colon 2|n\\}$, $ A_5 \\equal{} \\{ n \\in [1000] \\colon 5|n\\}$, $ A_7 \\equal{} \\{ n \\in [1000] \\colon 7|n\\}$. Therefore, $ |A_2| \\equal{} \\lfloor 1000/2 \\rfloor$, $ |A_5| \\equal{} \\lfloor 1000/5 \\rfloor$, $ |A_7| \\equal{} \\lfloor 1000/7 \\rfloor$, $ |A_2 \\cap A_5| \\equal{} \\lfloor \\frac {1000}{2\\cdot 5} \\rfloor$, $ |A_2 \\cap A_7| \\equal{} \\lfloor \\frac {1000}{2\\cdot 7} \\rfloor$, $ |A_5 \\cap A_7| \\equal{} \\lfloor \\frac {1000}{5\\cdot 7} \\rfloor$, $ |A_2 \\cap A_5 \\cap A_7| \\equal{} \\lfloor \\frac {1000}{2\\cdot 5\\cdot 7} \\rfloor$, where $ \\lfloor\\cdot\\rfloor$ is the floor function.\n\n$ |A_2\\cup A_5 \\cup A_7| \\equal{} |A_2| \\plus{} |A_5| \\plus{} |A_7| \\minus{} |A_2 \\cap A_5| \\minus{} |A_2 \\cap A_7| \\minus{} |A_5 \\cap A_7| \\plus{} |A_2 \\cap A_5 \\cap A_7|\\equal{}$ $ 657$.[/hide]",
  "Solution_2": "[hide=\"One approach for Number 6:\"]\nTake the reciprocal of each equation to get\n$ \\frac {x \\plus{} y}{xy} \\equal{} \\frac {1}{x} \\plus{} \\frac {1}{y} \\equal{} 7$\n\n$ \\frac {x \\plus{} z}{xz} \\equal{} \\frac {1}{x} \\plus{} \\frac {1}{z} \\equal{} 15$\n\n$ \\frac {y \\plus{} z}{yz} \\equal{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} 16$\n\nSubtracting the first equation from the second gives\n$ \\frac {1}{z} \\minus{} \\frac {1}{y} \\equal{} 8$\n\nAdding this to the third equation we get\n$ \\frac {2}{z} \\equal{} 24$. So $ \\frac {1}{z} \\equal{} 12$, and back substituting leads to $ \\frac {1}{y} \\equal{} 4$ and $ \\frac {1}{x} \\equal{} 3$.\n\nSo $ x \\equal{} \\frac {1}{3}$, $ y \\equal{} \\frac {1}{4}$, and $ z \\equal{} \\frac {1}{12}$; and $ x \\plus{} y \\plus{} z \\equal{} \\frac {2}{3}$.[/hide]"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "Let $A \\in M_2(C) $ with two distincts eigenvalues\r\n\r\nB=\r\n(A  A  A  A)\r\n(A  0  0   A)\r\n(A  0  0   A)\r\n(A  A  A   A) \r\n\r\nProve that B is diagonalisable\r\n\r\n\r\n$ou \\beta ino \\lambda $",
  "Solution_1": "Let $T=\\left(\\begin{array}{cccc} I_2& 0&0&0 \\\\ 0&0&0&I_2 \\\\ 0&0&I_2&0\\\\ 0&I_2&0&0 \\end{array}\\right)$.\r\n\r\nThen $T^{-1}=T$ and $B_1=T^{-1}BT=\\left(\\begin{array}{cccc} A&A&A&A \\\\ A&A&A&A \\\\ A&A&0&0\\\\ A&A&0&0 \\end{array}\\right)$\r\n\r\nLet $U^{-1}AU=D$ such that D is diagonal. This is possible because the eigenvalues of A are distinct, so A is diagonalizable.\r\n\r\nLet $R=\\left(\\begin{array}{cccc} U&0&0&0 \\\\ 0&U&0&0 \\\\ 0&0&U&0\\\\ 0&0&0&U \\end{array}\\right)$. Then $R^{-1}B_1R=\\left(\\begin{array}{cccc} D&D&D&D \\\\ D&D&D&D \\\\ D&D&0&0\\\\ D&D&0&0 \\end{array}\\right)$\r\n\r\nThe last is similar to $\\left(\\begin{array}{cccc}  xD&0&0&0 \\\\ 0&xD&0&0 \\\\ 0&0&yD&0\\\\ 0&0&0&yD \\end{array}\\right)$ because $\\left(\\begin{array}{cc} 1&1\\\\1&0\\end{array}\\right)$ is similar to $\\left(\\begin{array}{cc} x&0\\\\0&y\\end{array}\\right)$, where $(X-x)(X-y)=X^2-X-1$."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Let $a,b,c,d,e,f$ be six real numbers with sum 10, such that \\[ (a-1)^2+(b-1)^2+(c-1)^2+(d-1)^2+(e-1)^2+(f-1)^2 = 6. \\] Find the maximum possible value of $f$. \r\n\r\n[i]Cyprus[/i]",
  "Solution_1": "Let the a,b,c,d, and e expressions be all treated as (x-1)^2\r\nbecause they're not being asked to be maximized.\r\n\r\n5(x-1)^2 + (f-1)^2 = 6\r\n\r\nAnd because we have the variables adding up to 10, then we have:\r\n\r\n5x + f = 10\r\n\r\nThen f = 10 - 5x\r\n\r\nSubstituting,\r\n\r\n5(x-1)^2 +(9 - 5x)^2 = 6\r\n\r\n>\r\n>\r\n>  Steps left out here\r\n>\r\n>\r\n\r\n30x^2 - 100x + 86 = 6\r\n\r\n30x^2 - 100x + 80 = 0\r\n\r\n3x^2 - 10x + 8 = 0\r\n\r\n(3x - 4)(x - 2) = 0\r\n\r\nx = 4/3,  x = 2\r\n\r\nf = 10/3, f = 0\r\n\r\nThe maximum f = 10/3.",
  "Solution_2": "You can do it the same way as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47256]1978 USAMO, #1[/url], but the numbers do not seem to come out as nice.",
  "Solution_3": "Sorry to revive, but here is a solution, similar to what nsato was probably describing.\n\n[hide=\"Solution\"]Let $A=a-1$, $B=b-1$, $C=c-1$, $D=d-1$, $E=e-1$, $F=f-1$. Then $A^2+B^2+C^2+D^2+E^2+F^2=6$ and $A+B+C+D+E+F=4$.\n\nThen $A+B+C+D+E=4-F$ AND $A^2+B^2+C^2+D^2+E^2=6-F^2$. By Cauchy-Schwarz, we have $5(A^2+B^2+C^2+D^2+E^2) \\ge (A+B+C+D+E)^2 \\Rightarrow 5(6-F^2) \\ge (4-F)^2$. Expanding gives $30-5F^2 \\ge F^2-8F+16$ so $6F^2+8F-14 \\le 0 \\Rightarrow 3F^2+4F-7=(F-1)(3F+7) \\le 0$. Therefore $-\\frac{7}{3} \\le F \\le 1$ so $F$ is at most $1$. Therefore, $f$ is at most $1+1=\\boxed{2}$.[/hide]",
  "Solution_4": "[quote=\"AwesomeToad\"]Let $A=a-1$, $B=b-1$, $C=c-1$, $D=d-1$, $E=e-1$, $F=f-1$. Then $A^2+B^2+C^2+D^2+E^2+F^2=6$ and $A+B+C+D+E+F=4$.\n\nThen ... so $F$ is at most $1$.[/quote]\nHow about $(A,B,C,D,E,F) = (0,0,0,1,1,2)$ ?",
  "Solution_5": "[quote=\"AwesomeToad\"]... Expanding gives $30-5F^2 \\ge F^2-8F+16$ so $6F^2+8F-14 \\le 0 \\Rightarrow 3F^2+4F-7=(F-1)(3F+7) \\le 0$. Therefore $-\\frac{7}{3} \\le F \\le 1$ so $F$ is at most $1$. Therefore, $f$ is at most $1+1=\\boxed{2}$.[/quote]\nThe error stems from the fact that from $30-5F^2 \\ge F^2-8F+16$ follows $6F^2-8F-14 \\le 0$, i.e. $3F^2-4F-7=(F+1)(3F-7) \\le 0$. Therefore $-1 \\le F \\le \\frac{7}{3}$.\n\nEquality occurs when $A=B=C=D=E = \\frac {1} {3}$ and $F = \\frac {7} {3}$, thus for $a=b=c=d=e=\\frac {4} {3}$ and $f = \\frac {1} {3}$."
}
{
  "Tag": [
    "geometry",
    "trigonometry"
  ],
  "Problem": "Suppose that $ ABC$ is an isosceles triangle with $ AB \\equal{} AC$. Let $ P$ be the point on side $ AC$ so that $ AP \\equal{} 2CP$. Given that $ BP \\equal{} 1$, determine the maximum possible area of $ ABC$.",
  "Solution_1": "[hide=\"Solution by Mass Points\"]\nDraw a symmetric trisector from $ C$ to side $ AB$ to form point $ Q$, and let $ O \\equal{} \\overline{BP} \\cap \\overline{CQ}$. Now, draw in ray $ \\vec{AO}$ and let $ R \\equal{} \\overline{BC} \\cap \\vec{AO}$.\n\nNow, apply a mass of $ 1$ to point $ A$. Then, $ B$ and $ C$ have masses of $ 2$ because of trisection. Thus, the masses of $ P,Q,R$ are $ 3,3,4,$ respectively. \n\nFrom our masses we see that $ BO \\equal{} OC \\equal{} \\frac {3}{5} BP \\equal{} \\frac {3}{5}$ and that $ RO \\equal{} \\frac {1}{5} AR \\implies [ABC] \\equal{} 5[AOC]$.\n\nBy the sine formula for area, $ [ABC] \\equal{} 5 \\cdot \\frac {1}{2} \\cdot AO \\cdot OC \\cdot \\sin \\angle AOB \\equal{} \\frac {5}{2} \\left( \\frac {3}{5} \\right) ^2 \\sin \\angle AOB \\equal{} \\frac {9}{10} \\sin \\angle AOB$\n\nThis is clearly maximized when $ \\sin \\angle AOB \\equal{} 1 \\implies [ABC] \\equal{} \\frac {9}{10}$[/hide]\r\n\r\nI thought this was both a nice contest problem and teaching problem. However, I think it should've been placed earlier in the round  :)"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "algebra",
    "domain",
    "logarithms",
    "trigonometry"
  ],
  "Problem": "Hello... Good morning! \r\n\r\n\r\nI'm trying to solve these 4 integrals using the calculus of residues.\r\nPlease help.   :huh: \r\n\r\n\r\n(1) $\\int_{0}^{\\infty}\\frac{\\ln(1+x)}{1+x^{2}}dx$\r\nPS: I know how to solve \r\n$\\int_{0}^{\\infty}\\frac{\\ln(x)}{1+x^{2}}dx$\r\nwhich equals 0, where ln(z) is a multiple-valued function \r\nin the complex domain with branch point z=0. \r\nBut I didn't know what contour to use for (1) since\r\nthe branch point of ln(1+z) is at z=-1. If I indent it\r\nat z=-1 and use a similar shaped contour to that of\r\nln(z), I get the contribution from -1 to 0 in addition to\r\nthe contribution from 0 to $\\infty$, which \r\nthrows me off.\r\n\r\n\r\n(2) $\\int_{0}^{\\infty}\\frac{\\ln(1+x+x^{2})}{1+x^{2}}dx$\r\nPS: I know how to solve \r\n$\\int_{0}^{\\infty}\\frac{\\ln(1+x^{2})}{1+x^{2}}dx$\r\nwhich equals $\\pi \\ln 2$, but the presence of \"x\"\r\nwithin $\\ln(1+x+x^{2})$ in (2) is giving me a hard time.\r\n\r\n\r\n(3) $\\int_{0}^{\\infty}\\frac{\\ln^{3}(1+x^{2})}{1+x^{2}}dx$\r\n\r\n\r\n(4) $\\int_{0}^{\\infty}\\frac{\\ln(1+x^{3})}{1+x^{3}}dx$\r\n\r\n\r\nThank you.\r\n---Nasim",
  "Solution_1": "$ I(\\alpha) \\equal{} \\int_0^\\infty \\frac {\\ln(1 \\plus{} 2 \\sin\\alpha\\,\\, x \\plus{} x^2)}{1 \\plus{} x^2}\\, dx$ with $ 0<\\alpha<\\pi$ is after \r\n\r\nsubstitution $ x\\mapsto \\tan x \\qquad \\int_0^{\\frac {\\pi}{2}} \\ln\\left(\\frac {1}{\\cos^2 x} \\plus{} 2\\sin\\alpha\\, \\tan x\\right) dx$\r\n\r\n$ \\equal{} \\underbrace{\\int_0^{\\frac {\\pi}{2}} \\ln\\left(1 \\plus{} 2\\sin\\alpha\\, \\sin x\\, \\cos x\\right) dx}_{ \\equal{} : A} \\plus{} \\underbrace{\\int_0^{\\frac {\\pi}{2}} \\minus{} \\ln\\left(\\cos^2 x\\right) dx}_{\\pi\\ln 2}$\r\n\r\n$ A \\equal{} \\int_0^{\\frac {\\pi}{2}} \\ln\\left(1 \\plus{} \\sin \\alpha\\, \\sin 2x\\right) dx \\equal{} \\frac12 \\int_0^\\pi \\ln\\left(1 \\plus{} \\sin\\alpha \\, \\sin x\\right) dx$\r\n\r\n$ \\equal{} \\frac12 \\sum_{k \\equal{} 1}^\\infty \\frac {( \\minus{} 1)^{k \\minus{} 1}}{k}\\, (\\sin \\alpha)^k \\int_0^\\pi (\\sin x)^k \\, dx$\r\n\r\n$ \\equal{} \\frac12 \\sum_{k \\equal{} 1}^\\infty \\frac { \\minus{} 1}{2k}\\, (\\sin \\alpha)^{2k} \\int_0^\\pi (\\sin x)^{2k} dx \\plus{} \\frac12 \\sum_{k \\equal{} 0}^\\infty \\frac {1}{2k \\plus{} 1}\\, (\\sin \\alpha)^{2k \\plus{} 1} \\int_0^\\pi (\\sin x)^{2k \\plus{} 1} dx$\r\n\r\nwhereat $ \\int_0^\\pi (\\sin x)^{2k}\\, dx \\equal{} \\frac {\\pi}{2^{2k}} {2k\\choose k}$ and $ \\int_0^\\pi (\\sin x)^{2k \\plus{} 1}\\, dx \\equal{} \\frac {2^{2k \\plus{} 1}}{2k \\plus{} 1} \\frac {1}{{2k\\choose k}}$ .\r\n\r\nSo $ A \\equal{} \\frac {\\pi}{2} \\underbrace{\\sum_{k \\equal{} 1}^\\infty \\frac { \\minus{} 1}{2k} \\frac {(\\sin \\alpha)^{2k}}{2^{2k}} {2k\\choose k}}_{ \\equal{} \\ln \\left(\\cos^2 \\frac {\\alpha}{2}\\right)} \\plus{} \\underbrace{\\frac12 \\sum_{k \\equal{} 0}^\\infty \\frac {(2\\sin \\alpha)^{2k \\plus{} 1}}{(2k \\plus{} 1)^2} \\frac {1}{{2k\\choose k}}}_{ \\equal{} : B}$\r\n\r\nTo calculate $ B$ first differentiate $ (\\arcsin z)^2 \\equal{} \\frac12 \\sum_{k \\equal{} 1}^\\infty \\frac {(2z)^{2k}}{k^2\\, {2k\\choose k}}$ to obtain\r\n\r\n$ \\frac {\\arcsin z}{\\sqrt {1 \\minus{} z^2}} \\equal{} \\sum_{k \\equal{} 1}^\\infty \\frac {(2z)^{2k \\minus{} 1}}{k\\, {2k\\choose k}} \\equal{} \\sum_{k \\equal{} 0}^\\infty \\frac {(2z)^{2k \\plus{} 1}}{(k \\plus{} 1) {2k \\plus{} 2 \\choose k \\plus{} 1}} \\equal{} \\sum_{k \\equal{} 0}^\\infty \\frac {(2z)^{2k}\\, z}{(2k \\plus{} 1) {2k\\choose k}}$\r\n\r\nHence $ B \\equal{} \\int_0^{\\sin \\alpha} \\frac {\\arcsin z}{z\\, \\sqrt {1 \\minus{} z^2}}\\, dz \\equal{} \\int_0^\\alpha \\frac {z}{\\sin z}\\, dz \\equal{} \\left[z\\, \\ln\\left(\\tan \\frac {z}{2}\\right)\\right]_0^\\alpha \\minus{} \\int_0^\\alpha \\ln\\left(\\tan \\frac {z}{2}\\right) dz$\r\n\r\nFor the last integral use the fourier expansion $ \\minus{} \\ln\\left(\\tan \\frac {z}{2}\\right) \\equal{} 2\\sum_{k \\equal{} 0}^\\infty \\frac {\\cos (2k \\plus{} 1) z}{2k \\plus{} 1}$ .\r\n\r\n$ \\minus{} \\int_0^\\alpha \\ln\\left(\\tan \\frac {z}{2}\\right) dz \\equal{} 2\\sum_{k \\equal{} 0}^\\infty \\frac {\\sin (2k \\plus{} 1) \\alpha}{(2k \\plus{} 1)^2}$\r\n\r\nWith that $ I(\\alpha)$ can be written as $ \\frac {\\pi}{2}\\ln\\left[\\left(2\\cos \\frac {\\alpha}{2}\\right)^2\\right] \\plus{} \\alpha \\ln\\left(\\tan \\frac {\\alpha}{2}\\right) \\plus{} 2\\, \\sum_{k \\equal{} 0}^\\infty \\frac {\\sin (2k \\plus{} 1) \\alpha}{(2k \\plus{} 1)^2}$\r\n\r\nIn particular for $ \\alpha \\equal{} \\frac {\\pi}{6}$\r\n\r\n$ 2\\sum_{k \\equal{} 0}^{3N} \\frac {\\sin\\left((2k \\plus{} 1)\\frac {\\pi}{6}\\right)}{(2k \\plus{} 1)^2} \\equal{} \\sum_{k \\equal{} 0}^{3N} \\frac {( \\minus{} 1)^k}{(2k \\plus{} 1)^2} \\plus{} 3\\sum_{k \\equal{} 0}^{N \\minus{} 1} \\frac {( \\minus{} 1)^k}{(6k \\plus{} 3)^2}\\to K \\plus{} \\frac {K}{3}$ .\r\n\r\nTherefore $ I\\left(\\frac {\\pi}{6}\\right) \\equal{} \\int_0^\\infty \\frac {\\ln(1 \\plus{} x \\plus{} x^2)}{1 \\plus{} x^2} \\, dx$\r\n\r\n$ \\equal{} \\frac {\\pi}{2} \\ln\\Big(2 \\plus{} \\sqrt {3}\\Big) \\plus{} \\frac {\\pi}{6} \\ln\\Big(2 \\minus{} \\sqrt {3}\\Big) \\plus{} \\frac {4K}{3} \\equal{} \\frac13 \\left(\\pi \\ln\\Big(2 \\plus{} \\sqrt {3}\\Big) \\plus{} 4K\\right)$ .",
  "Solution_2": "[quote=\"misan\"]Hello... Good morning! \n\n\nI'm trying to solve these 4 integrals using the calculus of residues.\nPlease help.   :huh: \n\n[/quote]\r\n\r\n@Dr. Doe: I didn't really read your entire solution, but it didn't appear that you used residues. \r\n\r\nI'm curious as well how to define a contour to do these with residues. Does anyone know how to do it this way?"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "How many ordered pairs of integers (a,b) satisfy the inequalities:\r\n\r\n$ a^2\\plus{}b^2<16$\r\n$ a^2\\plus{}b^2<8a$\r\n$ a^2\\plus{}b^2<8b$ \r\n\r\n?\r\n\r\n[hide=\"Graphical solution\"]\nIf you graph the three inequalities they are three circles with different center, and the ordered pairs that satisfy are their intersection. This gives 6\n[/hide]\r\nProblems with an algebraic solution?\r\nIf you subtract the second inequality from the first, you get $ 0<16\\minus{}8a \\rightarrow a<2$. However (2,2) is clearly a solution. Where's the error?",
  "Solution_1": "From $ a<b$ and $ c<d$ it doesn't necessarily follow that $ a\\minus{}c<b\\minus{}d$.  For example, take $ (a,b,c,d)\\equal{}(0,1,\\minus{}666666666,1)$.",
  "Solution_2": "[hide]$ a$ and $ b$ are positive from the 2nd and 3rd inequalities.\n\n$ b < 4$ and $ a < 4$\nA little case work\n\nand we have $ (1,1),(1,2),(2,1),(2,2),(2,3),(3,2)$[/hide]"
}
{
  "Tag": [
    "function",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all functions $f:\\mathbb N\\to\\mathbb Z$ for which $f(1)=1$, $f(2)=0$, and \n\\[f(m)^2-f(n)^2=f(m+n)f(m-n),\\quad \\forall m,n \\in \\mathbb N, m>n.\\]",
  "Solution_1": "[quote=\"TTsphn\"]Find all function such that : \n$ f: N\\to Z$ \n$ f(m)^2 \\minus{} f(n)^2 \\equal{} f(m \\plus{} n)f(m \\minus{} n),\\forall m > n$\n$ f(1) \\equal{} 1 ,f(2) \\equal{} 0$[/quote]\r\n\r\nLet $ P(m,n)$ the property : $ f(m)^2 \\minus{} f(n)^2 \\equal{} f(m \\plus{} n)f(m \\minus{} n)$\r\n\r\n$ P(n\\plus{}2,n)$ gives $ f(n\\plus{}2)^2\\equal{}f(n)^2$ which implies $ f(2p)\\equal{}0$ and $ f(2p\\plus{}1)^2\\equal{}1$\r\n$ P(2p\\plus{}1,2p)$ gives then $ 1\\equal{}f(4p\\plus{}1)$\r\n$ P(2p\\plus{}2,2p\\plus{}1)$ gives $ \\minus{}1\\equal{}f(4p\\plus{}3)$\r\n\r\nAnd so we have a full definition of $ f(n)$ :\r\n$ f(2p)\\equal{}0$\r\n$ f(4p\\plus{}1)\\equal{}1$\r\n$ f(4p\\plus{}3)\\equal{}\\minus{}1$\r\n\r\nAnd it is easy to verify that these necessary conditions are also sufficient.",
  "Solution_2": "More than find all function satisfy conditon : \r\n$ f: N\\to Z$ such that : \r\n$ f(m)^2\\minus{}f(n)^2\\equal{}f(m\\plus{}n)f(m\\minus{}n)$"
}
{
  "Tag": [
    "limit",
    "trigonometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Compute the following limit: \r\n\r\n$ \\lim_{x\\rightarrow 0}{\\dfrac{(x\\minus{}\\sin{x})^2}{\\left(e^{x^{2}}\\minus{}1\\right)x^4}}$",
  "Solution_1": "Since there is $ x^4$ in the denominator, we substitute \r\n\r\n$ sinx \\equal{} x\\minus{}\\frac{x^4}{4!}$ \r\n\r\nTaking $ x^2\\equal{}k$, and applying L'Hospital, we arrive to the limit $ \\frac{1}{12}$.",
  "Solution_2": "Near zero, we have:\r\n\r\n$ \\sin x \\sim x \\minus{} \\frac {x^3}{6}$\r\n\r\n$ e^{x^2} \\sim 1 \\plus{} x^2$\r\n\r\nThis gives us:\r\n\r\n$ \\lim_{x\\rightarrow 0}{\\frac {(x \\minus{} \\sin{x})^{2}}{\\left(e^{x^{2}} \\minus{} 1\\right)x^{4}}} \\equal{} \\lim_{x\\rightarrow 0}{\\frac {(x \\minus{} x \\plus{} \\frac {x^3}{6})^{2}}{\\left(1 \\plus{} x^2 \\minus{} 1\\right)x^{4}}} \\equal{} \\frac {1}{36}$\r\n\r\nEDIT:\r\n\r\n[quote=\"soumik\"]Since there is $ x^4$ in the denominator, we substitute \n\n$ sinx \\equal{} x \\minus{} \\frac {x^4}{4!}$ [/quote]\r\n\r\nThis makes no sense... You can approximate $ \\sin x$ only in ways it is possible, not in ways it would be convenient in a given situation. What was your reasoning that let you derive this?",
  "Solution_3": "My one comment is that \"L'H\u00f4pital rule application\" is precisely the title that I [i]wouldn't[/i] give to this problem. I spend a significant amount of effort telling my students [i]not[/i] to use L'H\u00f4pital most of the time, and to instead use either general knowledge of size - what grows or shrinks faster than what else - and power series. This very problem would be a very good problem for me to use in this campaign - a problem for which L'H\u00f4pital's rule will be nightmarishly messy but for which power series methods will be straightforward and easily followed.\r\n\r\nWhich is to say that my actual solution would be functionally identical to the first part of milin's post, with insignificant differences in notation and intermediate steps.",
  "Solution_4": "hello, the right answer is $ \\frac{1}{36}$.\r\nSonnhard."
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "derivative",
    "logarithms",
    "function",
    "inequalities solved"
  ],
  "Problem": "It looks very easy.\r\nBut it is very hard.\r\nIt takes me two weeks.\r\nI just give a special form:\r\n(n=2)\r\nx^x+y^y >= x^y+y^x(x,y>0)\r\n(anlysis)\r\n\r\n[i]Edited by Myth[/i]",
  "Solution_1": "I have proved $n=2$\r\nfirst we can assume $x>=y$\r\nlet $f(t)=t^x-t^y$      now we should prove $f(x)>=f(y)$\r\nso $f'(t)=ln t(t^x-t^y)$\r\nit is obvious that $f'(t)>=0$\r\nso $f(x)>=f(y)$",
  "Solution_2": "do you mean that\r\n$x_{1}^x_{1}+x_{2}^x_{2}+...+x_{n}^x_{n} >= x_{1}^x_{2}+x_{2}^x_{3}+...+x_{n}^x_{1}$",
  "Solution_3": "I proved it too, with differentiation:\r\n\r\nWLOG $y\\geq x$\r\n\r\nLet $F(y)=x^x+y^y-x^y-y^x$\r\n\r\nThen $\\frac {dF}{dy}=y^y(\\ln y +1)-\\frac {y^xx}{y}-x^y\\ln x$\r\n\r\nNow divide into cases where $y\\geq 1$ and $y<1$.\r\n\r\nIn all cases we have $F'(y)\\geq 0$ with equality iff $y=x$.  This leads to $F(y)\\geq 0$, which is the desired result.",
  "Solution_4": "Your derivate $F'$ is wrong.  ;)",
  "Solution_5": "Myth, thank you for pointing out my all time mathematical low - getting a derivative wrong when calculating in Maple.\r\n\r\nAnyways, it's clear now.",
  "Solution_6": "He made a mistake!!!!!\r\nI think:\r\nwhen\r\nx >= y ,x>= 1\r\n1>=x >= y >= (e^-1)\r\nit is quiet easy (function:x^a-x^b)\r\nwhen\r\n1 >= x >= y ;\r\ny <= (e^-1)\r\nWe have to try another way.",
  "Solution_7": "[quote=\"blahblahblah\"]Myth, thank you for pointing out my all time mathematical low - getting a derivative wrong when calculating in Maple.\n\nAnyways, it's clear now.[/quote]\r\n\r\n :lol: Hehe, I wouldn't claim that as my all time mathematical low. Do you beat this: 2+2=5? Here we got Bush arithmetic!!! :rotfl: :rotfl:  :D",
  "Solution_8": ":D  :D  :D \r\n^_^",
  "Solution_9": "[quote=\"zxingtan\"]I have proved $n=2$\nfirst we can assume $x>=y$\nlet $f(t)=t^x-t^y$      now we should prove $f(x)>=f(y)$\nso $f'(t)=ln t(t^x-t^y)$\nit is obvious that $f'(t)>=0$\nso $f(x)>=f(y)$[/quote]\r\n\r\n :blush:",
  "Solution_10": "I'm quite sorry",
  "Solution_11": "Anyone disagree with my 'solution'?",
  "Solution_12": "[quote=\"TonyCui\"]It looks very easy.\nBut it is very hard.\nIt takes me two weeks.\nI just give a special form:\n(n=2)\nx^x+y^y >= x^y+y^x(x,y>0)\n(anlysis)\n\n[i]Edited by Myth[/i][/quote]\r\n\r\n\r\n\r\n\r\ncan you show it?",
  "Solution_13": "I think you can Finsh it!\r\nGook Luck  :blush:",
  "Solution_14": "I prove that:\r\n(x/y)^(x-y)>=1,x>0,y>0\r\n so:x^x*y^y>=x^y*y^x.\r\nlemma:a>=b>=c>=d>0,ad>=bc ,than:a+d>=b+c \r\nso we have:x^x+y^y>=x^y+y^x",
  "Solution_15": "[quote=\"huhuhu\"]I prove that:\n(x/y)^(x-y)>=1,x>0,y>0\n so:x^x*y^y>=x^y*y^x.\nlemma:a>=b>=c>=d>0,ad>=bc ,than:a+d>=b+c \nso we have:x^x+y^y>=x^y+y^x[/quote]\r\nI think you are creavity!",
  "Solution_16": "I have something wrong in my solution,but I have a new one.\r\nlet x>=y,it is easy when x >=1,so we have 1>x>=y and x^x*y^y>=x^y*y^x\r\nlemma:if:(x^x)^(2^n)+(y^y)^(2^n)>=(x^y)^(2^n)+(y^x)^(2^n),(x^x)*(y^y)>=(x^y)*(y^x),then:(x^x)^(2^(n-1))+(y^y)^(2^(n-1))>=(x^y)^(2^(n-1))+(y^x)^(2^(n-1)).\r\nthe lemma is easy to prove.\r\nwhen n is very very big,we have (x^x)^(2^n)+(y^y)^(2^n)=(x^y)^(2^n)+(y^x)^(2^n)-->0.\r\nunder the lemma,we have:x^x+y^y>=x^y+y^x\r\nIs that OK?",
  "Solution_17": "My second solution is similer with yours.\r\nI used to want it be a secret.HaHa!\r\n\r\nI think your solution doesn't have any mistake.I will not chack it.\r\nBy the way.I will tell you.Only this problem cost me four days.\r\nI only find two ways to prove it.\r\nI think you are creavity.\r\nYou are very strong! :D",
  "Solution_18": "Clear it have been contributed.\r\nIt is not the last edition.(The last eaidion I have forgot it)\r\nSay sorry to Friends who is not from China.\r\nI will translate it soon.\r\nIt USE CCJ instead of CJK.",
  "Solution_19": "In this edition I only use one way to prove\r\nBut In the finally edition I use two way to prove it\r\nI'm on my father's computer, so I can't find it.\r\nSorry!!!!!!!!"
}
{
  "Tag": [
    "ratio",
    "geometric series",
    "geometric sequence"
  ],
  "Problem": "What is the arithmetic mean of all the values of $2^x$ where $x$ is a whole number and $0 \\le x \\le 4$?  Express your answer as a decimal.",
  "Solution_1": "[hide]$2^0=1, 2^1=2, 2^2=4, 2^3=8, 2^4 = 16, 1+2+4+8+16 = 31.\\ \\  31/5 =6.2$[/hide]",
  "Solution_2": "[hide]\r\n2^0=1\r\n2^1=2\r\n2^2=4\r\n2^3=8\r\n2^4=16\r\nthan find the mean\r\n1+2+4+8+16=3\r\n31/5=6.2\r\n :D",
  "Solution_3": "There is actually a pretty neat trick for this... You'll like it when you see it.\r\n\r\nWhat is the arithmatic mean of all the vlaues of $2^x$ where $x$ is a whole number and $0\\le x\\le 11$?\r\n\r\nSeems a bit more difficult than the first problem...doesn't it?\r\n\r\n[hide]\n\nLets do this the long way and begin by adding up the numbers.\n\n$2^0 \\Rightarrow 1$ Sum: 1\n$2^1 \\Rightarrow 2$ Sum: 3\n$2^2 \\Rightarrow 4$ Sum: 7\n$2^3 \\Rightarrow 8$ Sum: 15\n$2^4 \\Rightarrow 16$ Sum: 31\n...\n\n\nIs it just me, or is there a pattern of sum sort? The sum is always one less then the next number in the series!\nLet's proof this shall we?\n\nHow can we begin?\nWhy don't we use the formula for the sum of a geometric series!\n$\\frac{a-ar^n}{1-r}$\nWhere $a$ is the first term, $r$ is the rate and $n$ is the number of the final term.\n\nPluging in the definate values in any case gives us a first number of 1 and a rate of two yeilding:\n$\\frac{1-1 \\cdot 2^n}{1-2} \\Rightarrow \\frac{1-2^n}{-1} \\Rightarrow 2^n-1$\n\nPlease notice that zero is supposedly the first number...so we count from zero up...making one the second number two the third and so on.\n\nUsing that little trick we can find the sum of the first eleven powers of two as $2^12-1 \\Rightarrow 4096-1 \\Rightarrow 4095.\n\nThen we divide by the number of terms which is from 0 to 11 inclusive giving us twelve...so the answer in this case is  341.25\n\nThat little trick comes in handy for speed and is definately better than having to add up all the values :lol:.\n\nSo remember the sum is the next power of 2 minus 1.\n[/hide]",
  "Solution_4": "[quote=\"MCrawford\"]What is the arithmetic mean of all the values of $2^x$ where $x$ is a whole number and $0 \\le x \\le 4$?  Express your answer as a decimal.[/quote]\r\n[hide=\"solution\"]\n(1+2+4+8+16)/5\n31/5\n[b]6.2[/b]\n[/hide]",
  "Solution_5": "Farazy, the entire formula is $\\frac {p^{n+1} - 1}{p-1}$ for the sum $p^0 + p^1 + p^2 + ... + p^n$. Just some extra info, unless of course you allready knew it  :lol:.\r\nThe proof is basically the same thing except $p$ is substituted in for $2$.\r\n[hide]Sum of finite geometric sequence: $a \\frac {1-r^n}{1-r}$. If we want to find the sum of the geometric sequenece $p^0 + p^1 + p^2 + ... + p^n$, then, $p$ is the common ratio, $1$ is the first term, and the number of terms is $n+1$. Plugging in these values we get $\\frac {1-p^{n+1}}{1-p} = \\frac {p^{n+1}-1}{p-1}$.[/hide]",
  "Solution_6": "[hide=\"da answer\"] 6.2[/hide]"
}
{
  "Tag": [
    "abstract algebra",
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "let $ G$ be both Noetherian and artinian abelian group ,prove that $ G$ is finite.",
  "Solution_1": "modules which are artinian and noetherian have a (finite) composition series. thus it's enough to show that abelian simple groups are finite (and in fact, cyclic of prime order). but this is very clear.",
  "Solution_2": "What about non-abelian groups?\r\nDoes there exist an infinite simple group, in which every proper subgroup is contained in some finite maximal subgroup?"
}
{
  "Tag": [],
  "Problem": "Q is the midpoint of PR. $ \\overline{PR}$ is 12 cm longer than $ \\overline{PQ}$. What is the number of centimeters in QR?\n\n[asy]draw((-1.2,0)--(1.2,0),Arrows); unitsize(1inch);\npair P,Q,R; P = (-1,0); R = (1,0);\ndot(P);  dot(Q); dot(R);\nlabel(\"P\",P,S); label(\"Q\",Q,S); label(\"R\",R,S);[/asy]",
  "Solution_1": "Since Q is the midpoint of PR, we find that PQ is 12 cm. This is the same as QR, so our answer is $ 12$."
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "perimeter"
  ],
  "Problem": "The number of inches in both the length and the width of a rectangle are prime numbers. The area of the rectangle is 391 square inches. What is the number of inches in the perimeter of the rectangle?",
  "Solution_1": "[hide]391 =17*23, so the perimeter is 2(17+23) =80[/hide]",
  "Solution_2": "[hide]I just know this one\n\n$17*23$\n\n[/hide]-jorian",
  "Solution_3": "Is there any quick way to do this other than guess and check? :?:",
  "Solution_4": "[quote=\"ckck\"]Is there any quick way to do this other than guess and check? :?:[/quote]\r\nOkay, a couple comments.\r\n1) There's no systematic formula or equation that gives a prime number so this is somewhat convoluted. Writing an equation isn't that bad.\r\n2) [hide]17 is the 7th prime and 2, 3, 5, 7, 11, and 17 can each be checked under 10 seconds with certain tricks[/hide]",
  "Solution_5": "Because 391 is 400 - 9, we can factor 391 using difference of squares.",
  "Solution_6": "[quote=\"Ravi B\"]Because 391 is 400 - 9, we can factor 391 using difference of squares.[/quote]\r\nThat is very clever... It gives 20-3, or 17, one of the factors. Never thought of that! Very Cool!  :lol:",
  "Solution_7": "Glad you liked it.  Here's a related problem from the national countdown round about 3 years ago.  The two students took a while to solve it, but now you can solve it quickly.\r\n\r\nWhat is the largest prime factor of 221?",
  "Solution_8": "[quote=\"Ravi B\"]Glad you liked it.  Here's a related problem from the national countdown round about 3 years ago.  The two students took a while to solve it, but now you can solve it quickly.\n\nWhat is the largest prime factor of 221?[/quote]\r\n[hide]17*13\nPS: When I saw this the first time in practice I checked all the primes up to 11 then I buzzed in and said 221 because I thought it was prime :D[/hide]"
}
{
  "Tag": [
    "vector",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "given a subspace spanned by three vectors, how do i find a a vector in r3 that is orthogonal to it?\r\n\r\nspecifically, the subspace is spanned by { (1, -1, 3), (5, -4, -4), (7, -6, 2) }.\r\n\r\nany help regarding the methodology of solving this problem would be greatly appreciated.\r\n\r\nthanks.\r\n\r\ns.",
  "Solution_1": "If $S$ is a subspace generated by $a_1,\\ldots a_m$, then solve $\\langle x,a_i\\rangle=0\\ \\forall i=\\overline{1,m}$. This would be the most obvious methdology.\r\nIn this case, let $S$ be the subspace generated by $(1,-1,3),(5,-4,-4),(7,-6,2)$. $(7,-6,2)=(5,-4,-4)+2(1,-1,3)$, so $S$ is the same as the subspace generated by $(1,-1,3),(5,-4,-4)$. Now solving $\\left\\{\\begin{array}{cc}a-b+3c=0\\\\5a-4b-4c=0\\end{array}\\right.$ yields $a=16c,b=19c$. So for examaple $(16,19,1)$ is orthogonal to $S$.",
  "Solution_2": "We are to find the normal vector of the plane which is determined by $A(1,-1,3),B(5,-4,-4),C(7,-6,2)$.\r\nSince $\\vec{AB}=(4,-3,-7),\\vec{AC}=(6,-5,-1)$, we get $\\vec{AB}X\\vec{AC}=(-32,-38,-2)\\parallel (16,19,1)$"
}
{
  "Tag": [
    "function",
    "quadratics",
    "algebra",
    "polynomial",
    "geometry",
    "geometric transformation",
    "functional equation"
  ],
  "Problem": "For $ x,y \\in {R}$. We have\r\n$ f(x)\\plus{}f(y)\\equal{}f(x\\plus{}y)\\minus{}xy\\minus{}1$\r\n$ f(1)\\equal{}1$, find $ n \\neq 1$ such that $ f(n)\\equal{}n$",
  "Solution_1": "Plugging in $ x\\equal{}1$ we get $ y\\plus{}2\\plus{}f(y)\\equal{}f(y\\plus{}1)$\r\n\r\nHence, $ f(0)\\equal{}\\minus{}1$, $ f(\\minus{}1)\\equal{}\\minus{}2$, and $ f(\\minus{}2)\\equal{}\\minus{}2$. Hence, we have $ n\\equal{}\\minus{}2$",
  "Solution_2": "[hide=\"A more complete solution\"]\nLet $ y\\to 1$ to get $ f(x)\\plus{}f(1)\\equal{}f(x\\plus{}1)\\minus{}x\\minus{}1\\implies f(x\\plus{}1)\\equal{}f(x)\\plus{}x\\plus{}2$, or $ f(x)\\equal{}f(x\\minus{}1)\\plus{}x\\plus{}1$.\n$ f(1)\\equal{}f(0)\\plus{}2\\implies f(0)\\plus{}2\\equal{}1\\implies f(0)\\equal{}\\minus{}1$.\n$ f(2)\\equal{}f(1)\\plus{}2\\plus{}1\\equal{}4$\n$ f(3)\\equal{}f(2)\\plus{}3\\plus{}1\\equal{}8$\n$ f(4)\\equal{}f(3)\\plus{}4\\plus{}1\\equal{}13$\n$ f(n)\\equal{}f(n\\minus{}1)\\plus{}n\\plus{}1\\equal{}f(n\\minus{}2)\\plus{}2n\\plus{}1\\equal{}f(n\\minus{}3)\\plus{}3n\\equal{}\\cdots$\n\nAt this point, I noticed that the coefficient of $ n$ is the value being subtracted from $ n$ in the recursive definition, and the trailing constant follows a pattern like $ 1\\plus{}1\\minus{}1\\plus{}1\\minus{}2\\plus{}1\\minus{}3\\plus{}\\cdots$, or $ (1\\minus{}0)\\plus{}(1\\minus{}1)\\plus{}(1\\minus{}2)\\plus{}(1\\minus{}3)\\plus{}\\cdots$.\n\n$ f(n)\\equal{}f(n\\minus{}n)\\plus{}n^2\\plus{}\\sum_{k\\equal{}0}^{n\\minus{}1}(1\\minus{}k)$\n\nNow, $ \\sum_{k\\equal{}0}^{n\\minus{}1}(1\\minus{}k)\\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1}1\\minus{}\\sum_{k\\equal{}0}^{n\\minus{}1}k\\equal{}n\\minus{}\\frac{n(n\\minus{}1)}{2}\\equal{}\\frac{n(3\\minus{}n)}{2}$.\n\nSo we have $ f(n)\\equal{}f(0)\\plus{}n^2\\plus{}\\frac{n(3\\minus{}n)}{2}\\equal{}\\frac{n^2\\plus{}3n\\minus{}2}{2}$\n\nIt is easy to check that\n$ f(x)\\plus{}f(y)\\equal{}\\frac{x^2\\plus{}3x\\minus{}2\\plus{}y^2\\plus{}3y\\minus{}2}{2}\\equal{}\\frac{x^2\\plus{}y^2\\plus{}3x\\plus{}3y\\minus{}4}{2}$,\n$ f(x\\plus{}y)\\minus{}xy\\minus{}1\\equal{}\\frac{x^2\\plus{}2xy\\plus{}y^2\\plus{}3x\\plus{}3y\\minus{}2\\minus{}2xy\\minus{}2}{2}\\equal{}\\frac{x^2\\plus{}y^2\\plus{}3x\\plus{}3y\\minus{}4}{2}$\nand so our condition is met.\n\nSo we need to solve the equation $ n^2\\plus{}3n\\minus{}2\\equal{}2n\\implies n^2\\plus{}n\\minus{}2\\equal{}0\\implies n\\equal{}\\frac{\\minus{}1\\pm3}{2}\\equal{}1, \\boxed{\\minus{}2}$\n[/hide]",
  "Solution_3": "You would have saved time if you had written the functional equation as $ f(x) \\minus{} f(x\\minus{}1) \\equal{} \\nabla_1 [f](x) \\equal{} x\\plus{}1$, and noted that a function over the integers with linear finite differences must be a quadratic. You can quickly find the particular solution by comparing coefficients.\r\n\r\nFor example, in this case, the functional equation is solved by any function of the form $ x^2/2 \\plus{} C x \\plus{} 1$\r\n\r\nIncluding the initial condition, we find that $ C \\equal{} \\minus{}\\frac{1}{2}$",
  "Solution_4": "As [b]TZF[/b] mentioned, $ f(x) \\minus{} f(x \\minus{} 1) \\equal{} x \\plus{} 1$ implies that f is a quadratic, a reasonably quick way of finding the function is as follows,\r\n\r\n$ \\sum_{x\\equal{}2}^{y} \\left(f(x) \\minus{} f(x \\minus{} 1)\\right) \\equal{} \\sum_{x\\equal{}2}^{y} \\left(x \\plus{} 1\\right)$\r\n\r\n$ f(y) \\minus{} 1 \\equal{} \\frac{y(y\\plus{}1)}{2} \\minus{} 1 \\plus{} (y \\minus{} 1)$\r\n\r\nWhich holds for all y in $ \\mathbb{R}$ by the identity theorem.",
  "Solution_5": "That's not at all the case, and I'm not sure what theorem you're using.  We aren't given that $ f$ is a polynomial, so the only thing we know about $ f$ is its values at a translation of the integers.",
  "Solution_6": "You're right. I should've said if we consider x an integer, then the function f on the integers must be a quadratic polynomial, in which case we know that the equation I wrote above holds for y in the integers."
}
{
  "Tag": [
    "search",
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "For $ x,y,z$ positive reals\r\n\r\n$ \\frac {2(yz \\plus{} zx \\plus{} xy)(x^2 \\plus{} y^2 \\plus{} z^2)}{x \\plus{} y \\plus{} z} \\geq yz\\sqrt {2y^2 \\plus{} 2z^2} \\plus{} zx\\sqrt {2z^2 \\plus{} 2x^2} \\plus{} xy\\sqrt {2x^2 \\plus{} 2y^2}.$",
  "Solution_1": "Look at \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=1614159#1614159\r\n\r\n\r\nfor wonderful Ji Chen's proof. :wink:",
  "Solution_2": "A related inequality is\r\n\r\n$ yz\\sqrt {2y^2 \\plus{} 2z^2} \\plus{} zx\\sqrt {2z^2 \\plus{} 2x^2} \\plus{} xy\\sqrt {2x^2 \\plus{} 2y^2}$\r\n\r\n$ \\leq2(\\sqrt{2}\\minus{}1)(x^3 \\plus{} y^3 \\plus{} z^3) \\plus{}(3\\minus{}2\\sqrt{2})[x^2(y \\plus{} z) \\plus{} y^2(z \\plus{} x) \\plus{} z^2(x \\plus{} y)] \\plus{} 6(\\sqrt{2}\\minus{}1)xyz$\r\n\r\nholds for nonnegative real numbers $ x,y,z.$"
}
{
  "Tag": [
    "ARML",
    "MATHCOUNTS",
    "AMC",
    "AIME"
  ],
  "Problem": "On Thursday (Feb 8), we will be hosting a Math Jam to discuss the harder problems from the AMC 10A and 12A.  As we have done in the past, we will present full solutions to problems 21-25 on both contests, and then (time permitted) do other solutions upon request.\r\n\r\nThe Math Jam will be tomorrow (Thursday) at 7:00 PM Eastern / 4:00 PM Pacific.  [b]Note the early start time![/b]\r\n\r\nTo join the Math Jam, you must be logged in to your Art of Problem Solving account.  Then, up to 15 minutes before the Jam starts, simply click on the \"Classroom\" link on the left side of the screen.  Participation is free!\r\n\r\n[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for the complete Math Jam schedule.[/url]",
  "Solution_1": "I won't be able to join the Math Jam tomorrow due to ARML tryouts; when will a transcript of the Jam be available?",
  "Solution_2": "All of our Math Jams are transcripted.  Highlight on the \"Math Jams\" tab above, then click the \"Transcripts\" option.",
  "Solution_3": "[quote=\"DPatrick\"]All of our Math Jams are transcripted.  Highlight on the \"Math Jams\" tab above, then click the \"Transcripts\" option.[/quote]\r\n\r\nI know; I'm just wondering [i]when[/i] the transcript will be online. Is it instantaneous, or does it take a while to be processed?\r\n\r\nThanks.  :)",
  "Solution_4": "Not instantaneous, but probably later that evening.  Friday morning at the latest.",
  "Solution_5": "Great. Thank you!  :)",
  "Solution_6": "I'm excited for this. This will be the first MathJam i am attending! :D",
  "Solution_7": "Are there going to be any game mathcounts math jams?\r\n\r\nAlso, just wondering, but the floor is constant for both AIMEs, right?",
  "Solution_8": "[quote=\"bpms\"]Are there going to be any game mathcounts math jams?[/quote]We have not planned any, but that may change.  \n\n[quote]Also, just wondering, but the floor is constant for both AIMEs, right?[/quote]Not necessarily.  Often it turns out that one version of the AIME is much harder than the other.  In that case, adjustments can be made.",
  "Solution_9": "The transcript for the Math Jam is now available.  [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=161]Click here.[/url]",
  "Solution_10": "does anyone have the link to the list of colleges administering the amc 12?",
  "Solution_11": "[url]http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml[/url]",
  "Solution_12": "Will transcripts ever be deleted?",
  "Solution_13": "[quote=\"buzzer11\"]Will transcripts ever be deleted?[/quote]I wouldn't say \"ever\" (ever is a long time), but we haven't deleted any Math Jam transcripts so far, and don't expect to.",
  "Solution_14": "If you ever decide to delete some, are you going to inform us?",
  "Solution_15": "[quote=\"DPatrick\"][url]http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml[/url][/quote]\r\n\r\nthanks for the link, but I can't find any schools that are close to DC."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "parallelogram",
    "geometry solved"
  ],
  "Problem": "$ABC$ is a given triangle. $ABDE$ and $ACFG$ are the squares drawn outside of the triangle. The points $M$ and $N$ are the midpoints of $DG$ and $EF$ respectively. Find $MN: BC$",
  "Solution_1": "Nice little problem. A slight generalization:\r\n\r\n[color=blue][b]Problem.[/b] On the sides CA and AB of a triangle ABC, parallelograms CAGF and ABDE are erected. Let M and N be the midpoints of the segments DG and EF. Show that $\\overrightarrow{MN}=\\frac{\\overrightarrow{BC}}{2}$.[/color]\r\n\r\n[i]Solution.[/i] Since CAGF and ABDE are parallelograms, $\\overrightarrow{DE}=\\overrightarrow{BA}$ and $\\overrightarrow{GF}=\\overrightarrow{AC}$. Since M is the midpoint of the segment DG, we have $\\overrightarrow{AM}=\\frac{\\overrightarrow{AD}+\\overrightarrow{AG}}{2}$; similarly, $\\overrightarrow{AN}=\\frac{\\overrightarrow{AE}+\\overrightarrow{AF}}{2}$. Thus,\r\n\r\n$\\overrightarrow{MN}=\\overrightarrow{AN}-\\overrightarrow{AM}=\\frac{\\overrightarrow{AE}+\\overrightarrow{AF}}{2}-\\frac{\\overrightarrow{AD}+\\overrightarrow{AG}}{2}=\\frac{\\left(\\overrightarrow{AE}-\\overrightarrow{AD}\\right)+\\left(\\overrightarrow{AF}-\\overrightarrow{AG}\\right)}{2}$\r\n$=\\frac{\\overrightarrow{DE}+\\overrightarrow{GF}}{2}=\\frac{\\overrightarrow{BA}+\\overrightarrow{AC}}{2}=\\frac{\\overrightarrow{BC}}{2}$,\r\n\r\nwhat solves the problem.\r\n\r\n  Darij"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $a_n=y^n+z^n$ and $ b_n=x^n+z^n$ and $c_n=x^n+y^n$ with $x,y,z$ positive real numbers. Prove that\r\n\r\n$\\frac{a_5+a_2}{(a_1)(a_2)+1}+\\frac{b_5+b_2}{(b_1)(b_2)+1}+\\frac{c_5+c_2}{(c_1)(c2)+1}< a_2+b_2+c_2$",
  "Solution_1": "There's a mistake I think - instead of $x_n , y_n , z_n$ there should be $a_n , b_n , c_n$"
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let $ ABC$ be an equilateral triangle, and $ P,Q,R$ three points in its interior satisfying\r\n\\[ \\measuredangle PCA \\equal{} \\measuredangle CAR \\equal{} 15^{\\circ},\\ \\measuredangle RBC \\equal{} \\measuredangle BCQ \\equal{} 20^{\\circ},\\  \\measuredangle QAB \\equal{} \\measuredangle ABP \\equal{} 25^{\\circ}.\\] Compute the angles of triangle $ PQR$.",
  "Solution_1": "This is a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=107969 . I have started the new topic just for the resources section.\r\n\r\n  Darij from the Pentacracy"
}
{
  "Tag": [
    "ratio",
    "limit",
    "geometry",
    "rectangle",
    "quadratics",
    "algebra",
    "quadratic formula"
  ],
  "Problem": "what exactly is the golden ratio and what is the purpose of it \r\nand how is it derived ?",
  "Solution_1": "Let [tex]F_k[/tex] denote the kth Fibionacci (probably not spelled right) number.  The golden ratio phi is defined as\r\n\r\n[tex]\r\n\\begin{eqnarray*}\r\n\\displaystyle\\phi & = & \\lim_{k \\to \\infty} \\frac{F_k}{F_{k-1}}\\\\\r\n \\displaystyle      & = & \\lim_{k \\to \\infty} \\frac{F_{k-2} + F_{k-1}}{F_{k-1}}\\\\\r\n  \\displaystyle     & = & 1 + \\lim_{k \\to \\infty} \\frac{F_{k-2}}{F_{k-1}}\\\\\r\n   \\displaystyle    & = & 1 + \\frac{1}{\\phi}\r\n\\end{eqnarray*}\r\n[/tex]\r\n\r\nSolving for [tex]\\phi[/tex] gives [tex]\\phi = \\frac{1+\\sqrt{5}}{2}[/tex]  Note that we want the positive solution since the limit is a ration of two positive numbers.  As for the use, it is used in a formula for the kth Fibionacci number, and the Greeks thought that the best looking rectangles had a side ratio 1:[tex]\\phi[/tex]",
  "Solution_2": "All I know is the golden ratio is exactly one more then its recipricoal and it is:\r\n(1+ :rt5: )/2\r\nYou  can get it by:\r\nx=(1/x)+1\r\nx :^2:-x-1=0\r\nand then you do the quadratic formula and get:\r\nx=(1 :pm:  :rt5: ) /2",
  "Solution_3": "The Golden Ratio is traditionally derived by answering the question:\r\n\r\nWhat is the side of a rectangle whose other side is 1, such that if a square of side 1 is removed from the rectangle the remaining rectangle is similar to the original?"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "vector",
    "linear algebra unsolved"
  ],
  "Problem": "a)Let $A$ be an $n*n$ symmetric real matrix with$a_{ii}=1$ for all $i$ and $|a_{ij}|\\leq n^{\\frac{-1}{2}}$ for all $j\\neq i$.prove that $A$ has rank at least $\\frac{n}{2}$.\r\nb)Let $A$ be an $n*n$ real matrix of rank $d$,let $k$ be a positive integer,and let $B$ be an $n*n$ matrix with  $b_{ij}=a_{ij}^k$.prove that the rank of $B$ is at most ${k+d\\choose k}$.\r\nenjoy :)",
  "Solution_1": "Obs: deleted after saying a great non-sense!",
  "Solution_2": "After the great non-sense I said before, here is what I've done for a). Consider any complex matrix. I want to show that if all columns have length equal to 1 relatively to the norm imposed by the hermitian product $x\\cdot y=C(y_1)x_1+...+C(y_n)x_n$ where $C(y_i)$ is the conjugate of $y_i$, then the rank of A is at least $\\sum_{i}{|a_{ii}|^2}$.  It will obviously follow that for any matrix A we have $%Error. \"rankA\" is a bad command.\n\\geq\\sum_{i}{\\frac{|a_{ii}|^2}{{|a_{i1}^2|}+...+|a_{in}^2|}}$ and from here we will have $rankA\\geq\\frac{n}{\\frac{n-1}{n}+1}$, from where $rankA\\geq\\frac{n^2}{2n-1}>n/2$. \r\n    Now, the proof of the first claim. Let $c_i$ the column vectors of $A$ and suppose that they all have lenght $ l(c_i)=1$. Let $ (e_1,...,e_n)$ the canonical base in $C^n$ and let $r=rankA$. By Gram-Schmidt theorem we can find $r$ ortonormals vectors $v_1,...,v_r$ with the same vectorial space generated as the $c_i$. Then we have $c_i=\\sum_{t=1,r}{(c_i\\cdot v_t)v_t}$ and thus we have $|a_{ii}^2|=|c_i\\cdot e_i|^2\\leq\\sum_{j=1,r}{|c_i\\cdot v_j|^2}\\cdot\\sum_{j=1,r}{|v_j\\cdot e_i|^2}$ (Cauchy-Schwarz inequality). Now, sum this over i and observe that the last sum is exactly r.",
  "Solution_3": "The second problem is really wierd, since again I have a better estimate, but it is surely wrong, as it turned out to be with the first part of the problem. \r\n     Suppose thus that $rankA=d$ and regard its columns. We may assume the first d among them are a basis in the vectorial subspace generated by the columns. Then there exist $ x_{11},...,x_{1d},...,x_{n-d,1},...,x_{n-d,d}$ such that $c_{d+1}=x_{11}c_1+...+x_{1d}c_d,..., c_n=x_{n-d,1}c_1+...+x_{n-d,d}c_d$. Let $A=(a_{ij})$. If we manage to prove that the last $n-d$ columns of the matrix $a_{ij}^k$ generate a subspace of dimension at most ${k+d\\choose k}-d$, then we are done, since then all columns will generate a subspace of dimension at most ${k+d\\choose k}$. \r\n     But using the multinomial formula we have that the columns $C_{i+d}$ of the matrix $a_{ij}^k$ verify $ C_i=^{t}( (x_{i1}a_{11}+...+x_{id}a_{1d})^k,...,(x_{i1}a_{n1}+...+x_{id}a_{nd})^k)$=$\\sum_{j_1+...+j_d=k, j_s\\geq 0}{\\frac{k!}{j_1!...j_d!}\\cdot x_{i1}^{j_1}...x_{id}^{j_d}C_{j_1,...,j_d}}$ where $C_{j_1,...,j_d}=^{t}(a_{11}^{j_1}...a_{1d}^{j_d},..., a_{n1}^{j_1}...a_{nd}^{j_d})$. Thus the vectorial space generated by $C_i$ is in the vectorial space generated by $C_{j_1,...,j_d}$, which has dimension at most the number of solutions of $j_1+...+j_d=k$, which is smaller than ${k+d\\choose k}-d$ for k at least 2.\r\n   Could anyone please show me the mistake?"
}
{
  "Tag": [
    "probability",
    "floor function"
  ],
  "Problem": "Fifty ping-pong-balls are numbered 1, 2, 3, ..., 49, 50. A ball is\nchosen at random. What is the probability that the number on\nthe selected ball is a multiple of 5 or 7 or both? Express your\nanswer as a common fraction.",
  "Solution_1": "There are $ \\left\\lfloor \\frac{50}{5}\\right\\rfloor \\equal{}10$ multiples of $ 5$ from $ 1$ to $ 50$ and $ \\left\\lfloor \\frac{50}{7} \\right\\rfloor \\equal{}7$ multiples of $ 7$, so there are $ 10\\plus{}7\\equal{}17$ numbers that are multiples of $ 5$ or $ 7$... not really.\r\n\r\nWe've counted the multiples of $ 5$ [b]and[/b] $ 7$ twice, but we only want to count them once.  A multiple of $ 5$ and $ 7$ is a multiple of $ 35$, and there is $ \\left\\lfloor \\frac{50}{35}\\right\\rfloor \\equal{}1$ multiple of $ 35$, so there are $ 17\\minus{}1\\equal{}16$ numbers that are multiples of $ 5$ or $ 7$, or both.\r\n\r\nFinally, the probability is $ \\frac{16}{50}\\equal{}\\boxed{\\frac{8}{25}}$"
}
{
  "Tag": [
    "college",
    "geometry"
  ],
  "Problem": "I'm going to a summer class in UC Berkeley, and I've been assigned some homework pages. I'm taking a geometry class so i can skip to Algebra II in High-Shool. I'm 14 years old.\r\n\r\nI'm having trouble on this problem. It's about angles, and I can't quite get it. I can only guess the angles. [u]I KNOW HOW TO DRAW IT OUT[/u], i just don't know how you can find the angles with given measures. Please help if you can.\r\n\r\nThank You\r\n\r\n36. line OC bisects angle AOB, line OD bisects angle AOC, line OE bisects angle AOD, line OF bisects angle AOE, and line OG bisects angle FOC.\r\n\r\na. If (measure of) angle BOF = 120, then (measure of) angle DOE= ____ ?\r\n\r\nb. If (measure of) angle COG = 35, then (measure of) angle EOG = ____ ?\r\n\r\nHere is an image of the problem:\r\n[url]Http://img382.imageshack.us/my.php?image=06152007101832pmrn3.jpg[/url]",
  "Solution_1": "Due to the difficulty, this problem belongs in the \"Getting Started\" forums. I suspect that most problems for an introductory geometry class would go in either the \"Getting Started\" or the \"Intermediate Topics\" forum."
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hello there,\r\n\r\nI am interested in a effective solution for a \"permutation\" i'm not entirely sure what type of permutation it is, but i do believe it is s repetative one.\r\n\r\nThere is 3 different objects:\r\n\r\n1)      x\r\n2)      o\r\n3)      v\r\n\r\nAnd there are six different digits eg: ? ? ? ? ? ?\r\n                                                    1 2 3 4 5 6\r\n\r\n\r\nExample:\r\n\r\n1)     x x x x x x\r\n2)     x x x x x o\r\n3)     x x x x o o\r\n4)     x x x o o o\r\n5)     x x o o o o\r\n6)     x o o o o o\r\n7)     o o o o o o\r\n8)     o o o o o v\r\n9)     o o o o v v\r\n10)   o o o v v v\r\n11)   o o v v v v\r\n12)   o v v v v v\r\n13)   v v v v v v\r\n14)   v v v v v x\r\n15)   v v v v x x\r\n16)   v v v x x x\r\n17)   v v x x x x\r\n18)   v x x x x x\r\n\r\n20)   x o x o x o\r\n21)   o x o x o x\r\n22)   v o v o v o\r\n23)   o v o v o v\r\n\r\n24)   x v o x v o\r\n25)   o v x o v x\r\n26)   v x o v x o\r\n27)   x o v x o v\r\n\r\n28)   o x v v x x\r\n29)   o x v v v x\r\n30)   o x v v v v\r\n\r\nI figured that if the permutation is repeatable then the result should be 216 different permutations but does anybody have a quick formula to tell me how many there are just to be sure for a second opinion? and secondly possibly a print out of each single permutation result.\r\n\r\nThanks.",
  "Solution_1": "you have three options for each digits.\r\n\r\n$3^{6}=729$",
  "Solution_2": "[quote=\"Geehoon\"]you have three options for each digits.\n\n$3^{6}=729$[/quote]\r\n\r\n\r\n\r\n\r\n\r\nThankyou, you have been a great help."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "given 1000 points in a place contain 800 red points,200 blue points.Prove that:There exists a line partitionned a place into 2 half such that each haft contains 400 red points,100 blue points",
  "Solution_1": "can 3 pts be collinear?",
  "Solution_2": "Sorry for my poor presentation.\r\n\r\nIf no 3 pts are collinear, let L(x) be the line with inclination x with the horizontal line which partition red pts into harves and the minimum distance from the line to red pts in one side of the line is equal to the minimum distance from the line to red pts in another side. Let f(x) be the number of blue pts on the left side of L(x). Note that f(x) and f(x+pi) refer to different side, so f(x) = 200-f(x+pi). Since f(x) is \"continuous\" (| lim x + ->y f(x) - lim x - ->y f(x) | <= 1), therefore there exist y such that f(y) = 100."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "AMC 12",
    "AMC 10",
    "MATHCOUNTS"
  ],
  "Problem": "who thinks that AIME is a lot easier than the AMC 12???\r\ncuz i HATE the time limit on the AMC's...i'd rather have 15 HARDER problems on 3 hours than 10 easy problems, 10 okay problems, and 5 hard problems in 1 hour 15 minutes...with an average of 3 minutes per problem. Most of the \"hard\" problems require some sort of pre-learned trick. On the AIME, we have time to actually think things out and find our own ways to do problems.",
  "Solution_1": "AIME is easier because the answers are from 1-999.",
  "Solution_2": "I disagree that the AIME is easiest. I think the USAMO is [u]much[/u] easier. (come on, we get 1.5 hours per question!)  \r\n\r\n :P Seriously though, the questions on the AMC 12 level are much easier than those on the AIME level. AMC 12 problems (especially the first 20 or so) are much more straight forward, while AIME questions tend to invoke much more thought that you will probably end up needing 12+ minutes per question. Indeed, if you work on the first 20 something questions at a decent rate, then you should have plenty of time to think out the last several questions.",
  "Solution_3": "[quote=\"hunter34\"]AIME is easier because the answers are from 1-999.[/quote]\r\n\r\numm thats 1000 choices (its 000-999)...\r\n\r\nand technically all answers can be applied, they would just make you transform it.\r\n\r\nAnyways, AMC its from A-E, 5 choices..\r\n\r\nIll take my chance with the 5 choices...even if you get counted off for guessing because the reward is much higher than with 1000 choices since most of the answers can be narrowed down.\r\n\r\nAnd for the AMC, who cares about time?  When you say 10 easy, 10 okay, 5 hard, compared to the AIME thats 20 easy, 4 okay, 1 hard (and the first 10 questions can be done in like 15 minutes so its really 15 questions for 60 minutes) and all AIME problems would go from hard-impossible...\r\n\r\nPlus with 3 hours, most of the questions, no matter how long you think, youll never get it.\r\n\r\nEDIT: wait azjps just said (basically) the last part of what I said...but anyways thats two people...sorry i didnt real azjps popst after reading that thing about USAMO.",
  "Solution_4": "well maybe it's just cuz i don't know any of the \"concepts\" you guys are talking about, and I'm okay with reasoning everything out on the AIME =P",
  "Solution_5": "WHAT? AMC12 harder? Wow...I don't quite see how that's possible\r\nFor me, I can get quite decent scores on AMC12s (130s I'm hoping 140+ on the real thing coming up) and I don't find the time limit a problem at all. AIME is different, I find that there is NOT ENOUGH TIME :( Well I sorta suck so that's no wonder but...",
  "Solution_6": "There is some overlap in difficulty. The very easiest AIME problems (usually the first few on the exam) are often easier than the hardest back-page AMC 12 problems. But that's just an overlap. It's not the whole test in either case.\r\n\r\nThe first page of the AMC 12 (or AMC 10) is going to contain problems that should be accessible to just about anyone, including people with no problem solving background. It is in the interest of the test writers not to totally intimidate those who come into the test relatively casually. It's really better if we can have people walking away thinking \"At least I got [i]something[/i] right.\"\r\n\r\nThat motivation is not there on the AIME. Those taking the AIME have already proven themselves. The AIME seeks to separate those student on the basis of some somewhat more difficult challenges. And more difficult challenges do require more time. That's built in.",
  "Solution_7": "hmmm\r\nmaybe i mean it a different way\r\ni enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.",
  "Solution_8": "[quote=\"Narcissa\"]hmmm\nmaybe i mean it a different way\ni enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]\r\nYes I agree on that, not necessarily the formula oriented part but the more fun part. That doesn't have anything to do with difficulty though...",
  "Solution_9": "Here is whats going on guys.\r\n\r\nWith each test, there are different expectations for success. Like that is literally all there is to it.\r\n\r\nAMC - 130 is considered a pretty good score. which is like.. 21ish questions right, or roughly 80% of the questions.\r\n\r\nAIME - 8/9 is considered a pretty decent score. It is usually more than enough to pass. Thats roughly thats like between 50 and 60%, which is alot better than AMC.\r\n\r\nOmigosh.\r\n\r\nIt definitely helps that the first few AIME questions from the last few years have been super approachable by even those with limited mathematical experience. I personally enjoy doing AIME problems alot more than AMC because AMC problems generally wind up being alot more nitpicky/tricky than AIME problems. \r\n\r\nI never really like the comparison of the last 5 of AMC to first 5 of AIME or whatever. The first 5 of the AIME are generally pretty nice and straight forward, whereas the last 5 of an AMC 12 are generally as unapproachable as possible. It has become kind of a standard to do like the first 5-6ish AIME problems in the first hour, 3-4ish in the second hour, and 1-2ish in the last hour, along with a nice nap somewhere in there. \r\n\r\nI dunno. AMC is just a qualifying test, as is the AIME. Just do well enough to pass well and move on to the next test.",
  "Solution_10": "[quote=\"Pakman2012\"]Here is whats going on guys.\n\nWith each test, there are different expectations for success. Like that is literally all there is to it.\n\nAMC - 130 is considered a pretty good score. which is like.. 21ish questions right, or roughly 80% of the questions.\n\nAIME - 8/9 is considered a pretty decent score. It is usually more than enough to pass. Thats roughly thats like between 50 and 60%, which is alot better than AMC.\n\nOmigosh.\n\nIt definitely helps that the first few AIME questions from the last few years have been super approachable by even those with limited mathematical experience. I personally enjoy doing AIME problems alot more than AMC because AMC problems generally wind up being alot more nitpicky/tricky than AIME problems. \n\nI never really like the comparison of the last 5 of AMC to first 5 of AIME or whatever. The first 5 of the AIME are generally pretty nice and straight forward, whereas the last 5 of an AMC 12 are generally as unapproachable as possible. It has become kind of a standard to do like the first 5-6ish AIME problems in the first hour, 3-4ish in the second hour, and 1-2ish in the last hour, along with a nice nap somewhere in there. \n\nI dunno. AMC is just a qualifying test, as is the AIME. Just do well enough to pass well and move on to the next test.[/quote]\r\nHmmmm...interesting, I don't find the last 5 of the AMC12 nearly as hard as some medium AIME questions (thanks in part to multiple choice, I can save time and work by checking the range of the answers they give :D I usually get 3+ right), but the first 7 or so AIME questions are generally pretty straightforward. The second half is a nightmare for me though...\r\nTo me AMC12 is not that different from AMC10 which is not that much different from mathcounts and mathcounts is super basic stuff...",
  "Solution_11": "Well, I did some past AIMEs and AMCs, and their difficulty really can vary from year to year. (I got 12 on 2003 AIME, 8 on 2004 AIME.) :ewpu:",
  "Solution_12": "[quote=\"Narcissa\"]hmmm\nmaybe i mean it a different way\ni enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]\r\n\r\nWell that's different.\r\n\r\nDo you like the USAMO even better?",
  "Solution_13": "[quote=\"Narcissa\"]hmmm\nmaybe i mean it a different way\ni enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]\r\n\r\nwhat do you mean formula-oriented.\r\n\r\nNo AMC or AIME problem should be strictly formula oriented..that should be for school math.\r\n\r\nThey are equally \"formula-oriented\" or lack there of.\r\n\r\nWhen you say you enjoy the a lot more, you are probably only talking about problems 1-4ish because after that, you wont enjoy them at all.",
  "Solution_14": "[quote=\"dnkywin\"]Well, I did some past AIMEs and AMCs, and their difficulty really can vary from year to year. (I got 12 on 2003 AIME, 8 on 2004 AIME.) :ewpu:[/quote]\r\nThat's true, although my scores hover around 6 :(",
  "Solution_15": "[quote=\"dnkywin\"]Well, I did some past AIMEs and AMCs, and their difficulty really can vary from year to year. (I got 12 on 2003 AIME, 8 on 2004 AIME.) :ewpu:[/quote]\r\n\r\nI feel like these competitions all tend to get little more difficult each year. The AMC actually seems to stay relatively stable, though. (Is it true that the alternate, AMC12-B, is harder than the AMC12-A?)",
  "Solution_16": "for me, thinking under pressure = hard. how bout that :P\r\n\r\n[quote]Do you like the USAMO even better?[/quote]no, cuz i can't prove for life.\n\n[quote]When you say you enjoy the a lot more, you are probably only talking about problems 1-4ish because after that, you wont enjoy them at all.[/quote]i dont know, like the first 10?",
  "Solution_17": "[quote=\"azjps\"]\nAMC 12 problems (especially the first 20 or so) are much more straight forward, while AIME questions tend to invoke much more thought that you will probably end up needing 12+ minutes per question. Indeed, if you work on the first 20 something questions at a decent rate, then you should have plenty of time to think out the last several questions.[/quote]\r\n\r\nOkay, I take this back .. on my most recent practice test I spent 10 minutes working on #4, and in the end I only got #2 and 4 wrong. That's just weird  :rotfl:",
  "Solution_18": "[quote=\"firecricket91\"][quote=\"Narcissa\"]hmmm\nmaybe i mean it a different way\ni enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]\n\nwhat do you mean formula-oriented.\n\nNo AMC or AIME problem should be strictly formula oriented..that should be for school math.\n\nThey are equally \"formula-oriented\" or lack there of.\n\nWhen you say you enjoy the a lot more, you are probably only talking about problems 1-4ish because after that, you wont enjoy them at all.[/quote]\r\n\r\nTrue, no contest can be done by school knowledge alone. But the first 15 or so problems of the AMC 12 can be made shorter by knowing shortcuts, while AIME has less apparent problems that seem impossible if you don't know any more than theorems. AIME might be easier due to the time constraint, though I actually find AMC 10 to be more time-pressuring than the 12, so neither the 12 or the AIME really give time problems (except for finding a careless mistake at the last minute). And on the average, AIME is slightly harder. They give you just enough easy problems so that you need to work hard in order to reach your goal."
}
{
  "Tag": [
    "geometry",
    "geometry proposed"
  ],
  "Problem": "[color=darkred]Let $ABC$ be a triangle. The its incircle $C(I)$ touches the lines $AB$, $AC$ in the points $E$, $F$ respectively. The line $EF$ meets the circle $w$ with the diameter $[BC]$ in the points $X$, $Y$ so that the line $AI$ separates the points $X$, $F$. The same circle $w$ cuts the lines $AB$, $AC$ in the points $M$, $N$ respectively. Denote the intersections $K\\in MX\\cap NY$ and $L\\in NX\\cap MY$. Prove that $K\\in AI$ and $L\\in HI$, where the point $H$ is the orthocenter of $\\triangle ABC$.[/color]",
  "Solution_1": "Here's my main ideas. ( I'm busy now so I can't post the full solution, sorry :( )\r\n- Prove that $BY$ and $CX$ are pass through $I$ - only calculating with Menelaus's theorem.\r\nUse this lemma: ( I think it's well-known )\r\nTak 6 points $A,B,C,X,Y,Z$ in an given circle. Then the intersections of $AY$ and $BX$, $BZ$ and $CY$, $CX$ and $AZ$ are collinear.\r\n+ With $M,Y,C,N,X,B$: We have $K,I,A$ are collinear\r\n+ With $M,X,B,N,Y,C$: We have $L,I,H$ are collinear",
  "Solution_2": "Here's my solution. I'm sorry to post a new post...but I can't edit my above post now  :blush: \r\n[b]Lemma:[/b]\r\nTake on a given circle six points $A_{1},A_{2},A_{3}$ and $B_{1},B_{2},B_{3}$. Then the intersections of $A_{1}B_{2}$ and $A_{2}B_{1}$ - called as $C_{3}$ , $A_{2}B_{3}$ and $A_{3}B_{2}$ - called as $C_{1}$ , $A_{1}B_{3}$ and $A_{3}B_{1}$ - called as $C_{2}$ are collinear.\r\nI think it's well-known, so I only post my main ideas to prove it:\r\n- Prove that: If $A,B,C,D,M$ lie on the circle $(O)$, then $(M.ABCD)$ is constant when $M$ vary\r\n- Let $A_{2}B_{1}$ intersects $A_{1}B_{3}$ at  $M$, $A_{2}B_{3}$ intersects $A_{3}B_{1}$ at $N$\r\nBecause $(A_{1}.B_{1}B_{2}B_{3}A_{2})=(A_{3}.B_{1}B_{2}B_{3}A_{2})$ so $(B_{1}C_{3}MA_{2})=(NC_{1}B_{3}A_{2})$\r\nBut we've known that $C_{2}B_{1},C_{2}C_{1},C_{2}M,C_{2}B$ intersects $A_{2}B_{3}$ at $N,C'_{1}, B_{3}, A_{2}$, respectively, then $(B_{1}C_{3}MA_{2})=(NC'_{1}B_{3}A_{2})$. It means that $C_{1}\\equiv C'_{1}$, and $C_{1},C_{2},C_{3}$ are collinear.\r\n[hide=\"Figure\"][img]http://i108.photobucket.com/albums/n9/PDatK40SP/735656425.jpg[/img][/hide]\r\nTo Mods: I'm quite busy now so I post my figure I've created before from Photobucket, and don't like ImageShack, it will not be cleared  :wink: \r\n[b]Back to the problem:[/b]\r\n$BI$ intersects $EF$ at $Y'$. We have $\\hat{Y'FC}= 180^{o}-\\hat{EFA}= 90^{o}+\\frac{1}{2}\\hat{BAC}= \\hat{BIC}$ so $I,Y',F,C$ are concyclic. Then, $\\hat{BY'C}= \\hat{IY'C}= \\hat{IFC}= 90^{o}$, it means that $Y'$ lies on $w$, or $Y \\equiv Y'$. So $BY$ passes through $I$, and similarly, $CX$ passes through $I$\r\nUse the lemma with $M,Y,C$ and $N,X,B$ we have $K,I,A$ are collinear \r\nUse the lemma with $M,X,B$ and $N,Y,C$ we have $L,I,H$ are collinear\r\nP/s to Mods: Could you delete my above post for me please ? Thank you  :)",
  "Solution_3": "[quote=\"PDatK40SP\"]\n$BI$ intersects $EF$ at $Y'$. We have $\\hat{Y'FC}= 180^{o}-\\hat{EFA}= 90^{o}+\\frac{1}{2}\\hat{BAC}= \\hat{BIC}$ so $I,Y',F,C$ are concyclic. Then, $\\hat{BY'C}= \\hat{IY'C}= \\hat{IFC}= 90^{o}$, it means that $Y'$ lies on $w$, or $Y \\equiv Y'$. So $BY$ passes through $I$, and similarly, $CX$ passes through $I$\n[/quote]\r\n\r\ni think you might mixed up some notations, but the proof is ok.\r\n\r\nanother proof for the collinearity of $B,I,Y$ etc you can see here the lemma i used to solve megus's problem : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132338[/url]"
}
{
  "Tag": [
    "induction",
    "algebra",
    "polynomial",
    "algebra solved"
  ],
  "Problem": "Let F(n) be the n-th Fibonnacci number, where F(1) = F(2) = 1, F(n+2) = F(n+1) + F(n). Let p(x) be a polynomial of degree 990 such that p(k) = F(k) for k = 992, 993, ... , 1982. Show that p(1983) = F(1983) - 1.",
  "Solution_1": "Unfortunately, on http://www.kalva.demon.co.uk/short/sh83.html\r\nu cannot find the solution\r\n:)\r\n:) \r\n:D",
  "Solution_2": "here is a proof using newton interpolation:\r\n\r\nwe have P(992)=F(992), P(993)=F(993),..,P(1982)=F(1982)\r\nimagine another polynome Q(x) , Deg(Q)=991, such that\r\nQ(992)=F(992), Q(993)=F(993),..,Q(1982)=F(1982),Q(1983)=F(1983)\r\n\r\nsuppose\r\nP(x)=c_0+c_1*(x-992)+c_2*(x-992)(x-993)+...+c_990*(x-992)(x-993)..(x-1981)\r\n\r\nwe have:\r\nP(992)=c_0=F(992)\r\n   -> c_0=F(992)\r\nP(993)=F(992)+c_1*1=F(993)  \r\n   -> c_1=F(991)\r\nP(994)=F(992)+F(991)*2+c_2*2*1=F(994)\r\n   -> c_2=(1/2)*F(990)\r\nP(995)=F(992)+F(991)*3+(1/2)*F(990)*3*2+c_3*3*2*1=F(995)\r\n   -> c_3=(1/3!)*F(989)\r\netc..\r\nyou can easily prove by induction that c_k=(1/K!)*F(992-k)\r\n\r\n moreover it's obvious that\r\nQ(x)=P(x)+(1/991!)*F(1)*(x-992)(x-993)..(x-1981)(x-1982)\r\n\r\nP(1983)=Q(1983)-(1/991!)*F(1)*(1983-992)(1983-993)..(1983-1981)(1983-1982)\r\nso\r\nP(1983)=F(1983)-1\r\n\r\nbye.",
  "Solution_3": "I never heard of \"Newton interpolation\" ..."
}
{
  "Tag": [],
  "Problem": "Hi, my brother is going to be writing the MOEMS competition (Division E) this year. I wanted to know what the average mark in MOEMS is.",
  "Solution_1": "Why don't you put a poll.\r\n\r\nBTW, I think the average mark is 3. The people in my class normally get like 1",
  "Solution_2": "No. The average mark is *not* 3. From last years' statistics, something like 40% of the test-takers on each round got none right. The average was about 2.",
  "Solution_3": "[quote=\"solafidefarms\"]No. The average mark is *not* 3. From last years' statistics, something like 40% of the test-takers on each round got none right. The average was about 2.[/quote]\r\n\r\nIt depends on the year. But are you serious that 40% get 0 :P",
  "Solution_4": "i know that when i did it i always got all correct (25/25)\r\n-jorian \r\n\r\nthe average in my class was a 18, 18/5= 3.6 (I was also in an advanced math class though)",
  "Solution_5": "so. u got the george lenchner award :P",
  "Solution_6": "[quote]No. The average mark is *not* 3. From last years' statistics, something like 40% of the test-takers on each round got none right. The average was about 2.[/quote]\r\n\r\nWhere did you get the statistics from? I tried to look for it over the web, but couldn't find anything useful.",
  "Solution_7": "[quote=\"satyam\"][quote]No. The average mark is *not* 3. From last years' statistics, something like 40% of the test-takers on each round got none right. The average was about 2.[/quote]\n\nWhere did you get the statistics from? I tried to look for it over the web, but couldn't find anything useful.[/quote]\r\n\r\nThe MOEMS people send out statistics for the last round with each new set of tests to the teachers. My teacher would read the statistics to us to encourage us. That's why I'm not quite sure on the numbers."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "let $ a,b,c>0$. Prove that\r\n$ a\\sqrt{\\frac{b}{c}}\\plus{}b\\sqrt{\\frac{c}{a}}\\plus{}c\\sqrt{\\frac{a}{b}}\\geq \\sqrt{3(ab\\plus{}bc\\plus{}ca)}$",
  "Solution_1": "[quote=\"VIMF\"]let $ a,b,c > 0$. Prove that\n$ a\\sqrt {\\frac {b}{c}} \\plus{} b\\sqrt {\\frac {c}{a}} \\plus{} c\\sqrt {\\frac {a}{b}}\\geq \\sqrt {3(ab \\plus{} bc \\plus{} ca)}$[/quote]\r\nIt is equivalent to the old and easy problem:\r\n\r\nIf $ x,$ $ y,$ $ z$ are positive real numbers, then\r\n$ \\frac{x^2}{y}\\plus{}\\frac{y^2}{z}\\plus{}\\frac{z^2}{x} \\ge \\sqrt{3(x^2\\plus{}y^2\\plus{}z^2)}.$"
}
{
  "Tag": [
    "trigonometry",
    "LaTeX",
    "logarithms",
    "limit"
  ],
  "Problem": "Set $ \\sin x+\\sin y=\\cos x+\\cos y=\\frac{1}{2}$ , So $ \\sin 2x=?$",
  "Solution_1": "$ Sin(y)=1/2-Sin(x)$\r\n$ Cos(y)=1/2-Cos(x)$\r\nby squaring both equations and adding them we get:\r\n$ Sin^{2}(y)+Cos^{2}(y)=1/4+Sin^{2}(x)-Sin(x)+1/4+Cos^{2}(x)-Cos(x)$\r\napplying that $ Cos^{2}(a)+Sin^{2}(a)=1$\r\n$ 1=1/2+1-Sin(x)-Cos(x)$\r\n$ 1/2=Sin(x)+Cos(x)$\r\nSquaring again and from $ Sin(2x)=2Sin(x)Cos(x)$ we get:\r\n$ 1/2=Sin^{2}(x)+Sin(2x)+Cos^{2}(x)$\r\nthen $ Sin(2x)=-1/2$",
  "Solution_2": "LaTeX hint: \\sin and \\cos give $ \\sin$ and $ \\cos$, which looks much nicer than what you were typing.\r\n\r\n(Other similar codes: \\max, \\min, \\log, \\ln, \\tan, \\cot, \\lim, etc.)",
  "Solution_3": "possible typo in the last 2 lines ... should be 1/4 (with answer something like -3/4)"
}
{
  "Tag": [],
  "Problem": "So when did/does your break start?\r\n\r\nExtra credit to anyone who realizes why the break couldn't've started on Monday :)\r\n\r\nThe third option is to specify whether or not you have a half day or not, also if you're school doesn't have Friday off [size=59]that's too bad :twisted: [/size] I mean please say so :)\r\n\r\nAlso wondering if there's a school who doesn't celebrate Thanksgiving Day in the US (I mean I'm guessing some religious schools don't but who knows?)",
  "Solution_1": "at 2:58 Pm Pst today! :D",
  "Solution_2": "Hehe.  It could have started on Monday.  They have [b]Sunday[/b] school at church, you know.   :wink: </jk>\r\n\r\nAnyway, we have a half day on wednesday, and then this big potluck thing, and then we're home free.",
  "Solution_3": "we had school all day today, and have Thursday and Friday off...my teachers, however, felt the need to ruin break by giving too much homework...grrrr...",
  "Solution_4": "Wednesday for me! Well... it actually started Tues. because we went to Classics Day at OU  and didn't go to school.",
  "Solution_5": "Who's the lucky duck who got to have a whole week off? :)",
  "Solution_6": "10:50 wed.",
  "Solution_7": "Over here middle schoolers get their break by Wed...break for high school starts Thurs.",
  "Solution_8": "darn thanksgiving is over.:(\r\nschool will start on monday....\r\npumpkin pie all gone.....:cry:",
  "Solution_9": "Not to mention the fact that there's no major break until Christmas...*sigh* Did you know that most people suicide on New Year's Day (not Chinese :)) out of federal holidays? OK that was random...*sigh*",
  "Solution_10": "I do not celebrate thanksgiving",
  "Solution_11": "@above, please refrain from necroposting. This thread had not been posted in for nineteen years.",
  "Solution_12": "Other. I live in Canada, where Thanksgiving is in October.",
  "Solution_13": "November 19th",
  "Solution_14": "November 23rd, so Wednesday.",
  "Solution_15": "my break is November 23rd to 25th\n\nappreciate how my school gives 3 days of vacation.",
  "Solution_16": "[quote=MathWinner121]I do not celebrate thanksgiving[/quote]\n\nbro this post is older than twice your age stop bumping these threads",
  "Solution_17": "My thanksgiving break's in October lol\n\nI'm in Canada",
  "Solution_18": "[quote=mzarony0]My thanksgiving break's in October lol\n\nI'm in Canada[/quote]\n\nHello, fellow Canadian!!! I really love how Thanksgiving is in October, because the weather is not gruesomely snowy.",
  "Solution_19": "My thanksgiving break is on Wednesday."
}
{
  "Tag": [
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Calculate\r\n\\[\\sum_{n=0}^{\\infty}{\\frac{n}{n^{4}+n^{2}+1}}\\]",
  "Solution_1": "\\[\\frac{2n}{n^{4}+n^{2}+1}=\\frac{1}{(n-1)^{2}+(n-1)+1}-\\frac{1}{n^{2}+n+1},\\]\r\nso the sequence telescopes and the sum to infinity is $\\frac{1}{2}$.",
  "Solution_2": "\\[\\sum_{n=0}^{\\infty}{\\frac{n}{n^{4}+n^{2}+1}=\\frac{1}{2}\\sum_{n}(\\frac{1}{n^{2}-n+1}-\\frac{1}{n^{2}+n+1})=\\frac{1}{2},}\\]\r\nbecause $(n+1)^{2}-(n+1)+1=n^{2}+n+1.$"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "trigonometry",
    "incenter",
    "analytic geometry",
    "IMO Shortlist"
  ],
  "Problem": "Let $ Q$ be the centre of the inscribed circle of a triangle $ ABC.$ Prove that for any point $ P,$\r\n\\[ a(PA)^2 \\plus{} b(PB)^2 \\plus{} c(PC)^2 \\equal{} a(QA)^2 \\plus{} b(QB)^2 \\plus{} c(QC)^2 \\plus{} (a \\plus{} b \\plus{} c)(QP)^2,\r\n\\]\r\nwhere $ a \\equal{} BC, b \\equal{} CA$ and $ c \\equal{} AB.$",
  "Solution_1": ".Let $w=C(I,r)$ be the incircle of the triangle $ABC$.\r\nThen there are the following wellknown relations:\r\n\r\n$\\blacksquare \\ 1^{\\circ}.\\ a\\cdot IA^2+b\\cdot IB^2+c\\cdot IC^2=abc.$\r\n$\\blacksquare \\ 2^{\\circ}.\\ \\left(\\forall\\right) P,\\ a\\cdot PA^2+b\\cdot PB^2+c\\cdot PC^2=(a+b+c)\\cdot PI^2+abc.$\r\n\r\n[u]Examples.[/u] \r\n$Ex\\ 1.\\ \\ P: =I\\Longrightarrow 1^{\\circ}.$\r\n$Ex\\ 2.\\ \\ P: =O\\Longrightarrow 2sR^2=2s\\cdot OI^2+4Rsr\\Longrightarrow OI^2=R^2-2Rr.$\r\n\r\nLet $ABC$ be a fixed triangle, a fixed point $F(u,v,w)$ and the mobile point $M: =M(x,y,z)$\r\n(with barycentrical coordinates). I note the power $p(X)$ of the point $X$ w.r.t. the circumcircle\r\nof the triangle $ABC$. Then for $\\left(\\forall\\right) M$ there are the following relations:\r\n\r\n$1^{\\circ}.\\ p(M)=-(yza^2+zxb^2+xyc^2).$\r\n$2^{\\circ}.\\ x\\cdot MA^2+y\\cdot MB^2+z\\cdot MC^2=-p(M).$\r\n$3^{\\circ}.\\ u\\cdot MA^2+v\\cdot MB^2+w\\cdot MC^2=MF^2-p(F).$\r\n$4^{\\circ}.\\ FM^2=-[(y-v)(z-w)a^2+(z-w)(x-u)b^2+(x-u)(y-v)c^2]\\ge0.$\r\n\r\nThe [u]remarcable points[/u] for a triangle $ABC: \\ A(1,0,0),\\ B(0,1,0),\\ C(0,0,1);$\r\n\r\n$G(\\frac 13,\\frac 13,\\frac 13),\\ I\\left(\\frac{a}{2s},\\frac{b}{2s},\\frac{c}{2s}\\right),\\ N\\left(\\frac{s-a}{s},\\frac{s-b}{s},\\frac{s-c}{s}\\right), I_a\\left(\\frac{-a}{2(s-a)},\\frac{b}{2(s-a)},\\frac{c}{2(s-a)}\\right);$\r\n\r\n$H(\\cot B\\cot C,\\coth\\cot A,\\cot A\\ cot B),\\ O\\left(\\frac{R^2}{2S} \\sin 2A,\\frac{R^2}{2S} \\sin 2B,\\frac{R^2}{2S}\\sin 2C\\right);$\r\n\r\n$\\Gamma\\left(\\frac{(s-b)(s-c)}{r(4R+r)},\\frac{(s-c)(s-a)}{r(4R+r},\\frac{(s-a)(s-b)}{r(4R+r)}\\right)$ (the [u]Gergogne's point[/u]);\r\n\r\n$L\\left(\\frac{a^2}{a^2+b^2+c^2},\\frac{b^2}{a^2+b^2+c^2},\\frac{c^2}{a^2+b^2+c^2}\\right)$ (the [u]Lemoine's point[/u] or the [u]simedian center[/u]).\r\n\r\n[u]Remark.[/u] You observe that: $x+y+z=1,\\ \\frac{[MBC]}{|x|}=\\frac{[AMC]}{|y|}=\\frac{[ABM]}{|z|}$,\r\nwhere $[XYZ]$ is the area of the triangle $ABC.$",
  "Solution_2": "Here is a solution with vectors.  The claim is clearly true for $ P\\equal{}Q$.  No we shall show that given a point $ P$ satisfying the property, another point $ R$ will satisfy the property as well.  That is, we must prove\r\n\r\n$ a(PA)^{2}\\plus{}b(PB)^{2}\\plus{}c(PC)^{2}\\equal{} a(QA)^{2}\\plus{}b(QB)^{2}\\plus{}c(QC)^{2}\\plus{}(a\\plus{}b\\plus{}c)(QP)^{2}$\r\n\r\n$ \\Rightarrow a(RA)^{2}\\plus{}b(RB)^{2}\\plus{}c(RC)^{2}\\equal{} a(QA)^{2}\\plus{}b(QB)^{2}\\plus{}c(QC)^{2}\\plus{}(a\\plus{}b\\plus{}c)(QR)^{2}$\r\n\r\nSo we must prove the respective differences are the same:\r\n\r\n$ \\sum a(PA^2\\minus{}RA^2)\\equal{}(a\\plus{}b\\plus{}c)(QP^2\\minus{}QR^2)$\r\n\r\nWe use vectors:\r\n\r\n$ \\sum a(PA^2\\minus{}RA^2)\\equal{}(a\\plus{}b\\plus{}c)(QP^2\\minus{}QR^2)$\r\n\r\n$ \\Leftrightarrow \\sum a(PA\\minus{}RA)(PA\\plus{}RA)\\equal{}(a\\plus{}b\\plus{}c)(QP\\minus{}QR)(QP\\plus{}QR)$\r\n\r\n$ \\Leftrightarrow \\sum a(P\\minus{}R)(P\\plus{}R\\plus{}2A)\\equal{}(a\\plus{}b\\plus{}c)(P\\minus{}R)(P\\plus{}R\\plus{}2Q)$\r\n\r\n$ \\Leftrightarrow \\sum a(P\\plus{}R\\plus{}2A)\\equal{}(a\\plus{}b\\plus{}c)(P\\plus{}R\\plus{}2Q)$\r\n\r\n$ \\Leftrightarrow \\sum a(2A)\\equal{}(a\\plus{}b\\plus{}c)(2Q)$\r\n\r\nNow we simply have the vector definition of Q, the incenter of the triangle.",
  "Solution_3": "Place masses $a, b, c$ on vertices $A, B, C$. It is clear that the centre of gravity is $Q$. By parallel axis theorem, the solution follows.",
  "Solution_4": "[b]My solution uses $I$ for incenter instead of $Q$:[/b]\n\nLet $I$ be the origin to all the vectors I will work with. First off, note that $\\vec{I}=0$ if we take the origin as $I.$\n\nWe know that $\\vec{PA}=\\vec{IP}+\\vec{IA}.$\n\nThe same holds true for the rest as well. So, the left hand side of our equation becomes $$a(\\vec{IP}+\\vec{IA})^2+b(\\vec{IP}+\\vec{IB})^2+c(\\vec{IC}+\\vec{IP})^2.$$\nWe expand that to get $$(a \\cdot PA^2 + b \\cdot PB^2 + c \\cdot PC^2 = a \\cdot IA^2 + b \\cdot IB^2 + c \\cdot IC^2 + (a + b + c) \\cdot IP^2)+2\\vec{IP}(a\\vec{IA}+b\\vec{IB}+c\\vec{IC}).$$\nAll that remains is showing that $2\\vec{PI}(a\\vec{IA}+b\\vec{IB}+c\\vec{IC})=0.$\n\nFirst off, since we took $I$ as the origin, we know that $a\\vec{IA}+b\\vec{IB}+c\\vec{IC}$ is the same as $a\\vec{A}+b\\vec{B}+c\\vec{C}.$\n\nNow, this reminds us of the barycentric coordinates of incenter of $\\triangle ABC.$ Namely, we know that $$\\dfrac{a}{a+b+c}\\vec{A}+\\dfrac{b}{a+b+c}\\vec{B}+\\dfrac{c}{a+b+c}\\vec{C}=\\vec{I}=0.$$Multiplying through by $a+b+c$ gives $a\\vec{A}+b\\vec{B}+c\\vec{C}=0$ as desired.\n\nSo, we have shown that $a \\cdot PA^2 + b \\cdot PB^2 + c \\cdot PC^2 = a \\cdot IA^2 + b \\cdot IB^2 + c \\cdot IC^2 + (a + b + c) \\cdot IP^2.$",
  "Solution_5": "[hide=Solution]\nLet $P$ be the origin, and let $I=Q$. Then, we need to show $aA\\cdot A+bB\\cdot B+cC\\cdot C=aIA^2+bIB^2+cIC^2+(a+b+c)I\\cdot I$. We have\n\\begin{align*}\nIA^2&=\\frac1{(a+b+c)^2}(bB+cC-(b+c)A)\\cdot(bB+cC-(b+c)A)\\\\\n&=\\frac1{(a+b+c)^2}((b+c)cb^2+(b+c)bc^2-bca^2)\\\\\n&=\\frac1{(a+b+c)^2}{bc(b+c)^2-bca^2}\\\\\n&=\\frac{bc(b+c+a)(b+c-a)}{(a+b+c)^2}\\\\\n&=\\frac{bc(b+c-a)}{a+b+c}.\\end{align*}\nTherefore, $aIA^2+bIB^2+cIC^2=abc$. Now, we have\n\\begin{align*}\n(a+b+c)I\\cdot I&=\\frac1{a+b+c}(aA+bB+cC)\\cdot(aA+bB+cC)\\\\\n&=\\frac1{a+b+c}(a^2A\\cdot A+b^2B\\cdot B+c^2C\\cdot C+2abA\\cdot B+2bcB\\cdot C+2caC\\cdot A)\\\\\n&=\\frac1{a+b+c}(A\\cdot A(a^2+ab+ac)+B\\cdot B(ba+b^2+bc)+C\\cdot C(ca+cb+c^2)-abc(a+b+c))\\\\\n&=aA\\cdot A+bB\\cdot B+cC\\cdot C-abc.\\end{align*}\nTherefore, we get $aA\\cdot A+bB\\cdot B+cC\\cdot C=aIA^2+bIB^2+cIC^2+(a+b+c)I\\cdot I$.",
  "Solution_6": "[hide=Solution]Let $I=Q.$\nNote that $$a\\cdot IA^2+b\\cdot IB^2+c\\cdot IC^2=a\\cdot(\\overrightarrow{A}-\\overrightarrow{I})^2+b\\cdot(\\overrightarrow{B}-\\overrightarrow{I})^2+b\\cdot(\\overrightarrow{C}-\\overrightarrow{I})^2=$$$$a\\cdot(\\overrightarrow{A}\\cdot\\overrightarrow{A}-2\\overrightarrow{A}\\cdot\\overrightarrow{I}+\\overrightarrow{I}\\cdot\\overrightarrow{I})+b\\cdot(\\overrightarrow{B}\\cdot\\overrightarrow{B}-2\\overrightarrow{B}\\cdot\\overrightarrow{I}+\\overrightarrow{I}\\cdot\\overrightarrow{I})+c\\cdot(\\overrightarrow{C}\\cdot\\overrightarrow{C}-2\\overrightarrow{C}\\cdot\\overrightarrow{I}+\\overrightarrow{I}\\cdot\\overrightarrow{I})=$$$$a\\cdot|\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{C}|^2-2(a\\cdot\\overrightarrow{A}\\cdot\\overrightarrow{I}+b\\cdot\\overrightarrow{B}\\cdot\\overrightarrow{I}+c\\cdot\\overrightarrow{C}\\cdot\\overrightarrow{I})+(a+b+c)\\cdot\\overrightarrow{I}\\cdot\\overrightarrow{I}=$$$$a\\cdot|\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{C}|^2-2\\cdot\\overrightarrow{I}\\cdot(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+(a+b+c)\\cdot\\overrightarrow{I}\\cdot\\left(\\frac{a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C}}{a+b+c}\\right)=$$$$a\\cdot|\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{C}|^2-\\overrightarrow{I}\\cdot(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})$$ Also, $$(a+b+c)IP^2=(a+b+c)|\\overrightarrow{IP}|^2=(a+b+c)|\\overrightarrow{P}-\\overrightarrow{I}|^2=$$$$(a+b+c)(\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{I}+\\overrightarrow{I}\\cdot\\overrightarrow{I})=$$$$(a+b+c)(\\overrightarrow{P}\\cdot\\overrightarrow{P})-2\\cdot\\overrightarrow{P}(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+(a+b+c)\\cdot\\overrightarrow{I}\\cdot\\overrightarrow{I}=$$$$(a+b+c)\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+(a+b+c)\\cdot\\overrightarrow{I}\\cdot\\left(\\frac{a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C}}{a+b+c}\\right)=$$$$(a+b+c)\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+\\overrightarrow{I}\\cdot(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C}).$$ In addition, $$a\\cdot PA^2+b\\cdot PB^2+c\\cdot PC^2=a\\cdot|\\overrightarrow{PA}|^2+b\\cdot|\\overrightarrow{PB}|^2+c\\cdot|\\overrightarrow{PC}|^2=$$$$a\\cdot|\\overrightarrow{P}-\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{P}-\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{P}-\\overrightarrow{C}|^2=$$$$a\\cdot(\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{A}+\\overrightarrow{A}\\cdot\\overrightarrow{A})+b\\cdot(\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{B}+\\overrightarrow{B}\\cdot\\overrightarrow{B})+c\\cdot(\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{C}+\\overrightarrow{C}\\cdot\\overrightarrow{C})=$$$$(a+b+c)\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+(a\\cdot|\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{C}|^2).$$ We conclude that $$a\\cdot IA^2+b\\cdot IB^2+c\\cdot IC^2=a\\cdot(\\overrightarrow{A}-\\overrightarrow{I})^2+b\\cdot(\\overrightarrow{B}-\\overrightarrow{I})^2+b\\cdot(\\overrightarrow{C}-\\overrightarrow{I})^2+(a+b+c)IP^2=$$$$a\\cdot|\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{C}|^2-\\overrightarrow{I}\\cdot(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+(a+b+c)\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+\\overrightarrow{I}\\cdot(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})=$$$$(a+b+c)\\cdot\\overrightarrow{P}\\cdot\\overrightarrow{P}-2\\cdot\\overrightarrow{P}(a\\cdot\\overrightarrow{A}+b\\cdot\\overrightarrow{B}+c\\cdot\\overrightarrow{C})+(a\\cdot|\\overrightarrow{A}|^2+b\\cdot|\\overrightarrow{B}|^2+c\\cdot|\\overrightarrow{C}|^2)=$$$$a\\cdot PA^2+b\\cdot PB^2+c\\cdot PC^2=a\\cdot|\\overrightarrow{PA}|^2+b\\cdot|\\overrightarrow{PB}|^2+c\\cdot|\\overrightarrow{PC}|^2,$$ as desired.\n[/hide]"
}
{
  "Tag": [
    "geometry",
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $a,b,c$ be the side lengths of a triangle of area $S$. Prove that \r\n\\[2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})\\ge 4\\sqrt 3S \\]",
  "Solution_1": "See here:\r\nstronger than Klamkin:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31598[/url]",
  "Solution_2": "hey, i got all those results :mad:"
}
{
  "Tag": [],
  "Problem": "How many distinct integers are in the solution set of $ \\left|\\frac{1}{3}m \\plus{} 2\\right| \\leq 1$?",
  "Solution_1": "\\begin{eqnarray*}\r\n\\left|\\frac{1}{3}m+2\\right|\\leq 1\r\n& \\Rightarrow & \\left|\\frac{1}{3}m+2\\right|^2\\leq 1^2 \\\\\r\n& \\iff & \\frac{1}{9}m^2+\\frac{4}{3}m+3\\le0 \\\\\r\n& \\iff & \\frac{1}{9}(m+3)(m+9)\\le0 \\\\\r\n& \\iff & -9\\le m \\le-3\r\n\\end{eqnarray*}\r\n\r\nThus, $ \\boxed{7}$.",
  "Solution_2": "$ \\left|\\frac {1}{3}m \\plus{} 2\\right|\\leq 1$\r\n\r\n$ \\frac {1}{3}m \\plus{} 2 \\leq 1$ or $ \\frac {1}{3}m \\plus{} 2 \\geq \\minus{} 1$\r\n\r\n\r\n$ \\frac {1}{3}m \\plus{} 2 \\leq 1$\r\n$ m \\leq \\minus{} 3$\r\n\r\n$ \\frac {1}{3}m \\plus{} 2 \\geq \\minus{} 1$\r\n$ m \\geq \\minus{} 9$\r\n\r\n$ \\minus{} 9 \\leq m \\leq \\minus{} 3$ so there are $ \\boxed{7}$ solutions.\r\n\r\nEDIT: Noo beaten again.",
  "Solution_3": "Darn much easier way:\nThe interval is 2 so 2/(1/3)+1(because it's $\\le$)=7"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "S is a set of 100 stones. f(S) is the set of integers n such that we can find n stones in the collection weighing half the total weight of the set. What is the maximum possible number of integers in f(S)?",
  "Solution_1": ":?: So where do you define the \"weight\"?"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "ARML",
    "USA(J)MO",
    "USAMO",
    "articles"
  ],
  "Problem": "Do either one of these help you qualify for the ARML???\r\n\r\nHow do you qualify for the USAMO???",
  "Solution_1": "For the second question:  see the Frequently Asked Questions, USAMO #5\r\nhttp://www.unl.edu/amc/f-miscellaneous/faq.shtml\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions",
  "Solution_2": "The AMC has essentially no relationship with ARML. Your state/school/region coaches may use AMC scores to help pick teams, there is no official qualification mechanism.",
  "Solution_3": "For the first question: ARML team selection is governed locally. It all depends on what information the coaches in your state collect and act on. That information might include AMC/AIME scores, and it might not. Ask about it locally.\r\n\r\n[Or essentially what worthawholebean said.]",
  "Solution_4": "In addition to above posts, I recommend you read this Wiki article that thoroughly covers the selection process.\r\n\r\nhttp://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Please solve the following problem taken from Greece 95\r\n\r\nEquation    ax^2 + (c-b)x  + (e-d) = 0  has two real roots  each of them > 1\r\nProve that equation  ax^4 + bx^3 + cx^2 + dx + e + 0 has at least one real root.\r\n\r\n Thanks for Your help. :)",
  "Solution_1": "Does anybody have some idea?"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $n$ be a positive integer. The set $\\{1,2,\\ldots,3n\\}$ is partitioned in three subsets $A,B,C$, each one containing $n$ elements.\r\nProve that there is $x \\in A$, $y\\in B$ and $z \\in C$ such that one of these integers is the sum of the two others.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=521113]All problems of the France TST 2006[/url]",
  "Solution_1": "recycling,reclycling,... see this http://www.mathlinks.ro/Forum/viewtopic.php?t=3781"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "advanced fields",
    "advanced fields solved"
  ],
  "Problem": "Let S the subset of $ \\mathbb{C}^4$ given by the points $ \\{(n,2^n,3^n,6^n)\\}$ for n in $ \\mathbb{N}$. Find generators for the ideal $ I(S)$ of polynomials in $ \\mathbb{C}[X,Y,Z,W]$ that are zero in $ S$, and find the dimension of the associated affine variety $ V(I(S))$ (the Zariski closure of $ S$).",
  "Solution_1": "$ W\\minus{}YZ$ belongs to $ I(S)$.\r\nLet $ P(X,Y,Z,W)\\in I(S)$. By using that $ W\\minus{}YZ\\in I(S)$ we can assume that $ P\\in\\mathbb C[X,Y,Z]$ (just divide by W-YZ and take the residue).\r\n\r\nWe have $ P(n,2^n,3^n)\\equal{}0$ for all $ n$. Look at $ P$ as a polynomial in $ Y,Z$ with coefficients in $ \\mathbb C[X]$. Let $ d$ be the degree of $ P$. If we assume for a contradiction that $ P\\neq 0$, then $ d$ must be a well defined integer. Let $ k$ be the greatest integer such that $ Z^kY^{d\\minus{}k}$ appears in $ P$ with nonzero coefficient. \r\nIt should now be clear that if we consider $ Q\\equal{}P/(Z^kY^{d\\minus{}k})$ then $ lim_{n\\to\\infty}Q(n,2^n,3^n)$ is nonzero which contradicts $ P\\in I(S)$.\r\n\r\nTherefore $ I(S)\\equal{}(W\\minus{}YZ)$. The dimension of the closure is 3."
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Prove that in $ \\triangle\\ \\mathrm{ABC}$ we have $ \\boxed {\\ \\mathrm N\\Gamma\\ \\parallel\\ \\mathrm{BC}\\ \\Longleftrightarrow\\ a(b \\plus{} c) \\equal{} b^2 \\plus{} c^2\\ \\Longleftrightarrow\\ \\mathrm F\\in (\\mathrm{AM})\\ }$ \r\n\r\nwhere  $ \\mathrm M$ - [u]midpoint[/u] of $ [\\mathrm{BC}]$ and $ \\mathrm N$ - [b]Nagel[/b]'s , $ \\Gamma$ - [b]Gergonne[/b]'s , $ \\mathrm F$ - [b]Feuerbach[/b]'s [u]points[/u] .",
  "Solution_1": "Seems it so difficult ?! I don't think. This problem was proposed for \r\n the RMG - contest (shortlist !) by [b]Evgeniy Kulanin, Rusia[/b]. I have \r\na short metrical proof of it. Maybe someone has a synthetical proof ...",
  "Solution_2": "I think it is nice,\r\n\r\nLet $ AN\\cap BC\\equal{}A_1,A\\Gamma\\cap BC\\equal{}A_2$ thus \\[ N\\Gamma\\parallel BC\\Leftrightarrow \\frac{AN}{AA_1}\\equal{}\\frac{A\\Gamma}{AA_2}\\Leftrightarrow \\frac{p\\minus{}b\\plus{}p\\minus{}c}{p\\minus{}a\\plus{}p\\minus{}b\\plus{}p\\minus{}c}\\equal{}\\frac{1/(p\\minus{}b)\\minus{}1/(p\\minus{}c)}{1/(p\\minus{}a)\\plus{}1/(p\\minus{}b)\\plus{}1/(p\\minus{}c)}\\Leftrightarrow\\frac{a}{p}\\equal{}\\frac{a(p\\minus{}a)}{\\sum (p\\minus{}b)(p\\minus{}c)}\\Leftrightarrow a(b\\plus{}c)\\equal{}b^2\\plus{}c^2\\]\r\n\r\nWe have $ F(a\\minus{}a\\cos(B\\minus{}C),..,..)$ therefore $ F\\in AM\\Leftrightarrow b\\minus{}b\\cos(C\\minus{}A)\\equal{}c\\minus{}c\\cos(A\\minus{}B)\\Leftrightarrow a(b\\plus{}c)\\equal{}b^2\\plus{}c^2$\r\n\r\nSo combine them we have nice relation $ \\boxed {\\ \\mathrm N\\Gamma\\ \\parallel\\ \\mathrm{BC}\\ \\Longleftrightarrow\\ \\mathrm F\\in (\\mathrm{AM})\\ }$",
  "Solution_3": "[b]Thanks, [u]Encyclopedia[/u] ! [/b] My proof is [url=http://mateforum.ro/viewtopic.php?t=3471][color=red][b]here[/b][/color][/url].",
  "Solution_4": "Thank you for nice relation in triangle :), it is quite hard for me to understand solution in Romanian, but now I understood :oops: !",
  "Solution_5": "You are right, [b]Encyclopedia[/b] ! In at most two or three words ... But no all know barycentrical coordinates. For them is my proof. Thank you."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "cyclic quadrilateral",
    "geometry unsolved"
  ],
  "Problem": "The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $O$.  The circumcircles of triangles $AOB$ and $COD$ intersect again at $K$.  Point $L$ is such that the triangles $BLC$ and $AKD$ are similar and equally oriented.  Prove that if the quadrilateral $BLCK$ is convex, then it is tangent [has an incircle].",
  "Solution_1": "Why no one answered :( Is this to hard? Or because no sketch from me?",
  "Solution_2": "Need better translation as my graphic does not agree with the problem.",
  "Solution_3": "See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=105470[/url].",
  "Solution_4": "Let $P \\equiv OK \\cap BC.$  $\\angle BKP= \\angle BAO= \\angle CDO= \\angle CKP$ $\\Longrightarrow$ $KP$ bisects $\\angle BKC.$ Let the angle bisectors of $\\angle KBL$ and $\\angle KCL$ cut $LC,LB$ at $U,V.$ Then\n\n$\\angle KCL = \\angle KCB + \\angle ADK = \\angle ACB - \\angle KCO + \\angle ADB + \\angle ODK$ \n\n$\\Longrightarrow \\angle KCL= 2\\angle BCA=\\angle 2\\angle KCV$\n\nSimilarly, $\\angle KBU=\\angle CBD.$ Hence $CA,CV$ and $BD,BU$ are isogonals WRT $\\angle KCB$ and $\\angle KBC,$ respectively. If $I \\equiv BU \\cap CV,$ then $O,I$ are isogonal conjugates WRT $\\triangle KBC,$ but since $KO$ bisects $\\angle BKC,$ it follows that $I \\in KP$ $\\Longrightarrow$ Internal angle bisectors of the convex $BLCK$ concur at $I,$ i.e. $BLCK$ has an incircle centered at $I.$",
  "Solution_5": "To prove that $BLCK$ has an incircle, we have to prove that $KB-KC=BL-LC$ by Pitot's Theorem.\nBy Ptolemy in $AKOB$ and $DKOC$, we will have \n$KB-KC=\\frac{OB.AK+KO.AB}{OA}-\\frac{KD.OC+KO.DC}{DO} $\n\nFirst, we have $OB/OA=OC/DO$ by Power of point $O$\nand $AB/OA=DC/DO$ : will simplify the expression:\n$KB-KC=\\frac{OB}{OA}(AK-KD)$\nNow, $\\triangle BLC \\sim \\triangle AKD$ gives us $BC/AD=BL/AK=LC/KD=\\frac{BL-LC}{AK-KD}$\nSo \n$KB-KC=\\frac{OB}{OA}\\cdot \\frac{AD(BL-LC)}{BC}=BL-LC$ ",
  "Solution_6": "Straightforward length computation.\n\nWe need to show that $BK - KC = BL - LC$, so since $BL - LC = \\frac{BC}{AD} (AK - KD)$ from the similar triangle condition, we want to show that\n$$AD \\cdot KC + BC \\cdot KA = AD \\cdot KB + BC \\cdot KD.$$\nSince $\\triangle AKC \\sim \\triangle BKD$, we have that $KC = \\frac{AC}{BD} \\cdot KD, KA = \\frac{AC}{BD} \\cdot KB,$ so that plugging this into the left side of the above means that we wish to show\n$$AD \\cdot \\frac{AC}{BD} \\cdot KD + BC \\cdot \\frac{AC}{BD} \\cdot KB = AD \\cdot KB + BC \\cdot KD,$$\nwhich is equivalent to\n$$(AC \\cdot AD - BC \\cdot BD) \\cdot KD = (DA \\cdot DB - CA \\cdot CB) \\cdot KB.$$\nSince $\\triangle KBA \\sim \\triangle KDC$, we have that $\\frac{KB}{KD} = \\frac{AB}{CD},$ so we only need to show that\n$$(AC \\cdot AD - BC \\cdot BD) \\cdot CD = (DA \\cdot DB - CA \\cdot CB) \\cdot AB.$$\nHowever, if we observe that $[\\triangle ABC] = \\frac{AB \\cdot BC \\cdot CA}{4R}$, where $[]$ denotes area and $R$ is the circumradius of $ABCD$, and three other cyclically symmetric expressions, plugging this into the above expression easily finishes.\n$\\square$ \n",
  "Solution_7": "[b]S[/b]olved with [b]mueller.25[/b], [b]starchan[/b], [b]Siddharth03[/b], [b]AdhityaMV[/b]\n\nLet $I'$ be the isogonal conjugate of $O$ in $\\triangle ADK$. Let $I$ be the corresponding point in $\\triangle BCL$ to $I'$ in $\\triangle ADK$. Since $\\angle AKO = \\angle ABO = \\angle DCO = \\angle DKO$, $K$ lies on the angle bisector, meaning so does $I'$, its isogonal conjugate. As a consequence, $LI$ bisects $\\angle BLC$.\n\nNext, note that $\\angle LBK = \\angle LBC + \\angle KBC = \\angle KAD + \\angle KBC = \\angle KAO + \\angle CAD + \\angle KBC = 2 \\angle DBC$.\n\nAnd $\\angle IBL = \\angle I'AK = \\angle OAD = \\angle DBC$, so $BI$ bisects $\\angle KBL$. Similarly, $CI$ bisects $\\angle KCL$, implying that $I$ is in fact the incenter of quadrilateral $BLCK$ (as it is the intersection of three angle bisectors), so we are done. $\\blacksquare$"
}
{
  "Tag": [
    "Putnam",
    "rotation",
    "analytic geometry",
    "geometry",
    "geometric transformation",
    "college contests",
    "Putnam complex"
  ],
  "Problem": "Let $n$ be a positive integer, $n \\ge 2$, and put $\\theta=\\frac{2\\pi}{n}$. Define points $P_k=(k,0)$ in the [i]xy[/i]-plane, for $k=1,2,\\dots,n$. Let $R_k$ be the map that rotates the plane counterclockwise by the angle $\\theta$ about the point $P_k$. Let $R$ denote the map obtained by applying in order, $R_1$, then $R_2$, ..., then $R_n$. For an arbitrary point $(x,y)$, find and simplify the coordinates of $R(x,y)$.",
  "Solution_1": "[u][b]The author of this posting is : Kent Merryfield[/b][/u]\r\n____________________________________________________________________\r\n\r\nWork in the complex numbers.\r\n\r\nLet $\\omega = e^{i\\theta} = e^{2\\pi i/n}$. Multiplication by $\\omega$ rotates counterclockwise by $\\theta$. The mapping $R_k$ rotates around $k$. To do that, let $z^{\\prime} = R_kz$. Then $z^{\\prime}-k = \\omega (z-k)$, which becomes $z^{\\prime} = R_kz = \\omega z - \\omega k + k$.\r\n\r\nThen $R_2R_1z = \\omega \\left(\\omega z - \\omega + 1\\right) - 2\\omega + 2 = \\omega^2 z - \\omega^2 - \\omega + 2$.\r\n\r\nWe claim that\r\n\r\n$R_kR_{k-1}\\dots R_2R_1z = \\omega^k z - \\omega^k - \\omega^{k-1} - \\cdots - \\omega + k$.\r\n\r\nThis claim can be verified by a straightforward induction.\r\n\r\nHence,\r\n\r\n$R_nR_{n-1}\\dots R_2R_1z = \\omega^n z - \\omega^n - \\omega^{n-1} - \\cdots - \\omega + n$.\r\n\r\nBut $\\omega^n + \\omega^{n-1} + \\cdots + \\omega$ is a geometric sum that adds to zero. Also, $\\omega^n = 1$. Hence,\r\n\r\n\\[R_nR_{n-1}\\dots R_2R_1z = z+n\\].\r\n\r\nThe overall operation is translation by $n$. The image of $\\left(x,y\\right)$ is $\\left(x+n,y\\right)$.",
  "Solution_2": "[u][b]The author of this posting is : Myth[/b][/u]\r\n____________________________________________________________________\r\n\r\nIt is really simple problem.\r\n\r\nWhat are part A and part B?",
  "Solution_3": "[u][b]The author of this posting is : blahblahblah[/b][/u]\r\n____________________________________________________________________\r\n\r\nA are the morning problems, B are the afternoon problems.",
  "Solution_4": "[u][b]The author of this posting is : bugzpooder[/b][/u]\r\n____________________________________________________________________\r\n\r\n[quote=\"Kent Merryfield\"]The overall operation is translation by $n$. The image of $\\left(x,y\\right)$ is $\\left(x+n,y\\right)$.[/quote]\r\n\r\nare you sure this is right? i got (x-n, y)",
  "Solution_5": "[u][b]The author of this posting is : ComplexZeta[/b][/u]\r\n____________________________________________________________________\r\n\r\nCheck it for $n=2$.",
  "Solution_6": "[u][b]The author of this posting is : bugzpooder[/b][/u]\r\n____________________________________________________________________\r\n\r\nunless i understood the problem wrong, (2,0) gets mapped to (0,0), then a rotation about itself gives (0,0).\r\n(x,y) gets mapped to from rotation about (1,0), then we rotate about (0,0) to get (2-x,-y) (x-2,y)",
  "Solution_7": "[u][b]The author of this posting is : zscool[/b][/u]\r\n____________________________________________________________________\r\n\r\nI think you still rotate it around the old point,\r\n\r\nso (2,0) would go to (0,0) and then to (4,0)",
  "Solution_8": "[u][b]The author of this posting is : ComplexZeta[/b][/u]\r\n____________________________________________________________________\r\n\r\nI wrote a rotation of $\\theta$ about $z_0$ as $z\\mapsto \\left(z-z_0\\right)e^{i\\theta} + z_0$. Then it's easier to do computations.",
  "Solution_9": "[u][b]The author of this posting is : bugzpooder[/b][/u]\r\n____________________________________________________________________\r\n\r\ndarn... guess 0 for me on that one!"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "$ a,b,c\\geq 0$  \r\n\r\n$ \\frac{1\\plus{}a}{1\\plus{}b}\\plus{}\\frac{1\\plus{}b}{1\\plus{}c}\\plus{}\\frac{1\\plus{}c}{1\\plus{}a}\\leq 3\\plus{}a^{2}\\plus{}b^{2}\\plus{}c^{2}$",
  "Solution_1": "[quote=\"zaya_yc\"]$ a,b,c\\geq 0$  \n\n$ \\frac {1 \\plus{} a}{1 \\plus{} b} \\plus{} \\frac {1 \\plus{} b}{1 \\plus{} c} \\plus{} \\frac {1 \\plus{} c}{1 \\plus{} a}\\leq 3 \\plus{} a^{2} \\plus{} b^{2} \\plus{} c^{2}$[/quote]\r\nLet $ c \\equal{} \\max\\{a,b,c\\}.$ Then\r\n$ \\frac {1 \\plus{} a}{1 \\plus{} b} \\plus{} \\frac {1 \\plus{} b}{1 \\plus{} c} \\plus{} \\frac {1 \\plus{} c}{1 \\plus{} a}\\leq 3 \\plus{} a^{2} \\plus{} b^{2} \\plus{} c^{2}\\Leftrightarrow$\r\n$ a^2 \\plus{} b^2 \\plus{} c^2\\geq\\frac {(a \\minus{} b)^2}{(1 \\plus{} a)(1 \\plus{} b)} \\plus{} \\frac {(c \\minus{} a)(c \\minus{} b)}{(1 \\plus{} a)(1 \\plus{} c)},$ which true because\r\n $ a^2 \\plus{} b^2\\geq(a \\minus{} b)^2\\geq\\frac {(a \\minus{} b)^2}{(1 \\plus{} a)(1 \\plus{} b)}$ and $ c^2\\geq(c \\minus{} a)(c \\minus{} b)\\geq\\frac {(c \\minus{} a)(c \\minus{} b)}{(1 \\plus{} a)(1 \\plus{} c)}.$"
}
{
  "Tag": [
    "geometry",
    "rectangle"
  ],
  "Problem": "Prove that we cannot arrange the numbers from 1 to 81 on a $9 \\times 9$ chessboard so that on every line the sum of the first 4 numbers is equal to the sum of the last 5 numbers.",
  "Solution_1": "Consider all 9 rows. The rectangle containing the first four numbers of each row has the same sum as the rectangle containing the last five numbers of each row. Since these cover the entire square, the sum of all numbers in the square must be even. But $ 1\\plus{}2\\plus{}\\cdots\\plus{}81 \\equal{} 3321$, which is odd. Contradiction.\r\n\r\nAnother question: What if the numbers were $ 0, 1,\\ldots, 80$ instead?",
  "Solution_2": "[hide=\"0 to 80\"]\nI came up with millions of ways for this to work.\n\nLet all the rows come in the form $ a\\plus{}1,a\\plus{}3,a\\plus{}6,a\\plus{}8,a,a\\plus{}2,a\\plus{}4,a\\plus{}5,a\\plus{}7$, where $ a$ is a multiple of 9. Of course, this is only one of the very many examples that work.\n[/hide]",
  "Solution_3": ":maybe: \r\nThe way I understood that, $ a \\equal{} 9$ gives $ 10, 12, 15, 17, 9, 11, 13, 14, 16$ as a row, and that doesn't work. Also, I think you have to consider the columns as well (the original post specified all lines).\r\n\r\nCan you clarify?",
  "Solution_4": "Wow. I must have been extremely tired when I wrote that, because it's completely not true unless $ a\\equal{}0$. If the original post means all lines, does it mean columns from top to bottom? The diagonal from bottom left to top right?"
}
{
  "Tag": [],
  "Problem": "Can anyone provide me a proof of Descartes' rule of Signs?",
  "Solution_1": "[url]http://www.cut-the-knot.org/fta/ROS2.shtml[/url]\r\n\r\nEnjoy.",
  "Solution_2": "also [url]http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF[/url]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "Calculate $ \\int \\frac {x^2}{{(xsinx \\plus{} cosx)}^{2}} dx$\r\n\r\nP.S. I'm sorry it is already posted.\r\n\r\nInstead if it is not posted - you may try to find:\r\n\r\n$ \\int_{0}^{\\frac {\\pi}{4}} arctg \\sqrt {\\frac {cos2x}{2cos^2x}} dx$",
  "Solution_1": "[quote=\"borislav_mirchev\"]Calculate $ \\int \\frac {x^2}{{(xsinx \\plus{} cosx)}^{2}} dx$[/quote]\r\n\r\n[url=http://www.mathlinks.ro/viewtopic.php?p=1153094#1153094]See 226[/url]",
  "Solution_2": "kunny, it is a problem from a russian contest. The second problem is from the same source.\r\nYou may try it."
}
{
  "Tag": [
    "geometry",
    "symmetry",
    "area of a triangle",
    "Heron\\u0027s formula",
    "angle bisector",
    "geometry unsolved"
  ],
  "Problem": "Two isosceles triangles, which sides are x, x, a and x, x, b respectively, have the same area; a\u2260b. Find x.\r\n\r\nI have a bit of trouble with this... can anyone help plz? (I know it should be fairly simple)\r\n\r\nthx",
  "Solution_1": "by heron's formula you have that \r\n$\\frac{1}{4}\\sqrt{(2x+a)(2x-a)a^2}=\\frac{1}{4}\\sqrt{(2x+b)(2x-b)b^2} \\\\\\Rightarrow 4x^2a^2-a^4=4x^2b^2-b^4 \\\\\\Rightarrow (a^2-b^2)(4x^2-(a^2+b^2))=0$, \r\nand since $a^2-b^2\\neq 0$ we must have that \r\n$x=\\frac{1}{2}\\sqrt{a^2+b^2}$",
  "Solution_2": "Here is my solution:\r\n\r\n   You can avoid  use of Heron's formula.\r\n\r\n   Just cut the triangle along the symmetry line and \r\n\r\n   turn 180 around the midpoint of the base one of the halves \r\n\r\n  so that half of the old base becomes \r\n\r\n  the altitude in the new triangle  \r\n\r\n   and  the old altitude becomes half of the new base.\r\n\r\n   This transformation does not change the area.\r\n\r\n   Now you see what the value of the side X must be.\r\n\r\n\r\n\r\n   Thank you.\r\n\r\n   M.T.",
  "Solution_3": "Let's call $ ADB$ and $ ADC$ the two triangles, having $ BD\\equal{}AD\\equal{}CD$. In order the altitudes from A to the bases $ BD\\equal{}CD$ we need $ AD$ as angle bisector of $ \\angle ADC$, in which case we get $ AB \\equal{} AC$, and this is not the case, or $ B, D$ and $ C$ are collinear and, in this case $ AB\\perp AC$, hence $ 4x^2\\equal{}BC^2\\equal{}AB^2\\plus{}AC^2\\equal{}a^2\\plus{}b^2$, done.\r\n\r\nBest regards,\r\nsunken rock"
}
{
  "Tag": [],
  "Problem": "salam \r\nye soale sede\r\n$2n+1$nafar adam hastand ke har $n$ nafari hade ahgal yek dooste moshtarak darand\r\nsabet konid yek nafar hast ke ba hame doost ast.",
  "Solution_1": "rahnama i:say konid khoshe n+1 rasi besazid tozih khastid begid.",
  "Solution_2": "soale russia grade 9 salaye 2001 ta 2004\nin rahe khushe giri yeki az sari tarin rahashe harchand ye rahe algorithmic khodam daram barash,avalin bar."
}
{
  "Tag": [
    "email",
    "function",
    "set theory",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Hi every body . I have some question about counting operation and set theory.\r\n\r\n1\\ If $X_1,X_2,...$ be countable sets then $\\bigcup_{n=1}^{\\infty}X_{n}$ is countable set.\r\n\r\n2\\[0;1] has the same potency with (0;1)\r\n\r\nI want to know about set theory but I haven't any book or article a bout this problem .\r\nCould you lend me some ebook about this  problem?\r\nThanks .\r\nmy email address k09tranhoan@gmail.com",
  "Solution_1": "I'm so sorry . My English is so poor.\r\n\r\n(1)for example:\r\nX1:x11  x12  x13  x14  ......\r\nx2:x21  x22  x23  x24  ......\r\nx3:x31  x32  x33  x44  .......\r\n..... .... .....   .....  ....  ..... ..\r\n\r\nyou can acquire a new sequence\r\nX:x11, x12, x21, x13, x22, x31, x14,x23,x32,x41....\r\nthen you can  delete some repeated value.\r\nX will be a countable set\r\n\r\n(2)\r\nyou can take a countable set  A from (0,1)\r\nB=AU{0,1} is also a countable set\r\nwe can make a function f\r\nA:{xi},B:{yi},f(xi)=yi\r\nif x in (0,1),but x is not in A,then f(x)=x\r\nso [0;1] has the same potency with (0;1) \r\n\r\nYou can read some books about real variable function."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Calculete integral : $ \\int_{0}^{2\\pi}\\left(\\frac {\\sin 5\\phi}{\\sin\\left(\\frac {\\phi}{2}\\right)}\\right)^{6}d\\phi$\r\n\r\n[hide=\"Answer\"]$ 110504\\pi$[/hide]",
  "Solution_1": "$ I\\equal{}\\int_\\gamma \\frac{(z^5\\plus{}1)^6(z^4\\plus{}z^3\\plus{}z^2\\plus{}z\\plus{}1)^6}{iz^{28}}dz\\equal{}2\\pi i Res(0)\\equal{}110504\\pi$"
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "On Monday, 30 girls and 0 boys in the Crescent Valley Band missed practice, resulting in a $ 2 : 5$ ratio of girls to boys. On Tuesday, 30 girls and 48 boys missed practice, resulting in a $ 2 : 1$ ratio of girls to boys. How many students are in the band?",
  "Solution_1": "48 more boys missed the band practice, and the ratio reduced by 4.\r\n\r\nso ratio 1 is 12 students.\r\n\r\nratio 5, which is 'no boy missed practice', is 12*5=60.\r\n\r\nand ratio 2, which is '30 girls missed practice', is 12*2=24.\r\n\r\nso the answer is 60+24+30=114. (boys = 60, girls = 54)\r\n\r\nanswer : 114"
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "ratio"
  ],
  "Problem": "Given a parallelogram $ABCD$, join $A$ to the midpoints $E$ and $F$ of the opposite sides $BC$ and $CD$. $AE$ and $AF$ intersect the diagonal $BD$ in $M$ and $N$. Prove that $M$ and $N$ divide $BD$ into three equal parts.",
  "Solution_1": "We can show that $\\Delta BEM$ is similar to $\\Delta DAM$, because all angles are equal.\r\n$\\frac{BE}{AD}=\\frac{BM}{DM}=\\frac{1}{2}$, so:\r\n\\[ DN+NM = 2 BM (1) \\]\r\n\r\nIn a similar way, we prove that $\\Delta DNF$ is similar to $\\Delta BNA$, because all angles are equal.\r\n$\\frac{DF}{AB}=\\frac{DN}{BN}=\\frac{1}{2}$, so:\r\n\\[ MN+BM = 2 DN (2) \\]\r\n\r\nsubtracting $(1)$ from $(2)$ gives:\r\n\\[ DN-BM=2BM-2DN \\iff 3 (BM-DN) = 0 \\iff BM=DN \\]\r\n\r\nPlugging this into $(1)$, finally tells us that \\[ DN=MN=BM \\]",
  "Solution_2": "Cool.\r\n\r\nThere is another solution - think of centroids!",
  "Solution_3": "Oh, yeah, I see. $BD$ and $AE$ are two medians of $\\Delta ABC$, and intersect with the ratio $\\frac{1}{2}$. Then the same in $\\Delta ADC$, sum up and get the answer.",
  "Solution_4": "[i]Now it is too late , but tommorow I ' ll try to prove or disprove the converse:[/i]\r\n\r\n << If  we take the midpoints $E$  and $F$ of sides $CB, CD$ of a quadratilateral $ABCD$  and the diagonal $BD$  is divided into three equal parts from the two segments $AE$  and $AF$ , then $ABCD$ is a parallelogram. >>\r\n\r\n[u]Babis[/u]",
  "Solution_5": "[i](Morning) Yes ! It is true![/i]\r\n [b]A hint.[/b]\r\n\r\n $AMCN$must be a parallelogramm , so  $AC$ and  $NM$(and also $BD$ ! ) bisect each other .\r\n\r\n [u]Babis[/u]"
}
{
  "Tag": [
    "probability",
    "MATHCOUNTS"
  ],
  "Problem": "This 0ne stumped me. If there are 52 well shuffled deck of cards in a standard deck and I deal 13 cards to each of the four players (including me) then what is the probability that I as one of the players can get\r\n1) All four aces (the other cards don't matter)\r\n2) All the 13 cards are from the same suite\r\n3) All the 13 cards go from Ace, 2, 3 ... Jack, Queen, King sequentially irrespective of the suite from which they are",
  "Solution_1": "[color=darkred]...[/color]",
  "Solution_2": "Umm... this may be for the test forum or the MATHCOUNTS or Getting Started.\r\n\r\nMove please!!!!",
  "Solution_3": "[quote=\"kyyuanmathcount\"]Umm... this may be for the test forum or the MATHCOUNTS or Getting Started.\n\nMove please!!!![/quote] This is Getting Started.  :|",
  "Solution_4": "2. $\\frac{4}{\\binom{52}{13}}$"
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "analytic geometry",
    "congruent triangles"
  ],
  "Problem": "How many square units are in the area of the convex quadrilateral with vertices (0,0), (3,0), (2,2), and (0,3)\r\n\r\nI know the answer but not the solution thanks\r\n\r\nAnswer\r\n[hide]6 square units[/hide]",
  "Solution_1": "[hide]\nThere are a couple of ways of going about this, you could use the Shoelace Theorem and then you'd have\n0  0\n3  0\n2  2\n0  3\n0  0\n(0 + 6 + 6 + 0) - (0 + 0 + 0 + 0) = 12\nDivide by 2, to get $6$, your answer.\n\nYou could divide the figure into a square, and two triangles. One triangle has the coordinates (2,2) (0,2) and (0,3). Height is 1, base is 2, area of that is 1. The next triangle has the coordinates of (2,2) (3,0) (2,0), height is 2, base is 1, area of that is 1. The square has coordinates (0,0) (0,2) (2,0) and (2,2). The sidelength is 2, square that to get 4. Lastly, add 4, 1, and 1 up to get 6.[/hide]",
  "Solution_2": "Shoe string method:\r\n\r\nhttp://artpad.art.com/?ixhawi10rspo\r\n\r\nNote: I know it's not flawless, but hopefully you'll get it.\r\n\r\nBy the way, I recommend you view it at \"fast\" speed.",
  "Solution_3": "Thank you all so much :D  :)",
  "Solution_4": "You can also divide it into two congruent triangles by drawing the line segment from (0,0) to (2,2). Each one has a base of 3 and a height of 2, so each one has an area of 3*2/2 = 3. Double it and you get [b]6[/b].",
  "Solution_5": "shoe string method???\r\n\r\nnever heard of it....can you give me a link on it? oh and preferably not with flash...\r\n\r\ni can't install the latest version, so it never works :oops:",
  "Solution_6": "Here, in PDF:\r\n\r\nhttp://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf"
}
{
  "Tag": [
    "search",
    "algorithm",
    "logarithms",
    "floor function"
  ],
  "Problem": "Suppose there is a stack of paper with words that needs to be sorted alphabetically, and there are three people in the room. What is the most efficient way to sort the stack?\r\n\r\nMy thought:\r\n\r\n1) Divide the big stack into three or more smaller stacks\r\n2) Each person grabs a smaller stack\r\n3) For each paper, put it into a pile that starts with that alphabet\r\n4) When a person is done with their stack get another stack or take a portion from another person's stack\r\n5) When all papers are in piles (there are at most 26 of them), each person gets a pile and sort it alphabetically\r\n6) When all 26 piles are sorted, simply stack them up to get the final stack\r\n\r\nIs this the most efficient method?\r\n\r\n\r\nNow there is an additional constraint:\r\n\r\nInstead of a room, the three people are sitting in a subway with quite many people moving. They can't lay the paper around to make piles. They can only hold on to their piles. Holding a stack right, everyone could only keep one pile (So there can only be three piles at most). In addition, to avoid papercut, paper can only transfered to another person by placing it on the top of their stack. Pulling a paper from another person is not allowed (even if that page is on top) because that is the easiest way to get papercut. Similar to the original situation, the three people might not work at the same speed.\r\n\r\nWhat is the most efficient way to do the same job?",
  "Solution_1": "This isn't exactly what I've done, but I do have experience with sorting papers- typically, putting student homework papers in alphabetical order by name.\r\nAs a one-person job with some room to work, I typically go with a merge sort. Sort piles of up to about five or six, and then merge those.\r\n\r\nWhat are the key properties of a stack of papers as a data structure? Finding a location is about as costly as a search, which is a binary search or maybe a little better. Inserting something or removing something once you know where to do it is cheap. The storage space cost that matters is the number of stacks, not the size of those stacks.\r\n\r\nThe difficulty in the second case is finding any algorithm that sorts at all. Here's what I can do:\r\nInsertion/bubble sort for three piles:\r\n1. Person A starts with the whole stack, and passes the top paper to C.\r\n2. Look at the top sheet of the pile. If it belongs later than what C is holding,\r\n2a. C passes their paper to B face down.\r\n2b. A passes their top paper to C.\r\nOtherwise, 2c. A passes their top paper to B face down.\r\n3. Repeat step 2 until A is holding nothing.\r\n4. C passes their paper to B face down.\r\n5. B flips their stack and passes it back to A.\r\n6. Repeat steps 1-5 until everything is sorted. This takes up to $ n\\minus{}1$ passes if there are $ n$ papers.",
  "Solution_2": "\"Finding a location is about as costly as a search\" [above]\r\n\"The storage space cost that matters is the number of stacks.\" [above]\r\n\r\nWith these two in mind, I thought of an alternate method of sorting in the first, kind of like a recursive sorting algorithm.\r\n\r\n1) Divide the pile into three arbitrary stacks and give them to the three people\r\n2) For each person: for each paper: put the paper in one of three stacks: \"A-I\",\"J-R\",or \"S-Z\".\r\n3) For each stack:\r\n*Divide the stack into three arbitrary piles and distribute\r\n*Sort the stack into \"first third, second third, and third third\" eg: \"A-C\", \"D-F\", \"H-J\"\r\n*Divide and sort into each letter\r\n*Sort each letter-stack, perhaps with similar methods if the stacks are large\r\n4) Put the sorted stacks on top of each other.\r\n\r\nThis method reduces the amount of time spent searching for a location and perhaps the storage space as well, as there are always 3 active stacks instead of 26.",
  "Solution_3": "Hi, I don't understand the algorithm jmerry said. Could you do a simulation?\r\n\r\nSuppose a stack is like this: Top [ 9 6 7 4 1 3 2 8 5 0 ] Bottom\r\n\r\nInitially:\r\nA: Top [ 9 6 7 4 1 3 2 8 5 0 ] Bottom\r\nB: Top [ ] Bottom\r\nC: Top [ ] Bottom\r\n\r\nA looks at the first page. It is 9 (Suppose they don't know that 9 is the max number).\r\nA gives that to C\r\n\r\nA: Top [ 6 7 4 1 3 2 8 5 0 ] Bottom\r\nB: Top [ ] Bottom\r\nC: Top [ 9 ] Bottom\r\n\r\nNow A sees the page with a 6 on it. It passes it to B faced down:\r\n\r\nA: Top [ 7 4 1 3 2 8 5 0 ] Bottom\r\nB: Top [ 6 ] Bottom\r\nC: Top [ 9 ] Bottom\r\n\r\nNext, A sees page 7, it is still less than the page C has, so A gives B faced down:\r\n\r\nA: Top [ 4 1 3 2 8 5 0 ] Bottom\r\nB: Top [ 7 6 ] Bottom\r\nC: Top [ 9 ] Bottom\r\n\r\nA sees page 4. Now what happens?",
  "Solution_4": "On Jmerry's algorithm:\r\n\r\nWe have\r\nA: Top [ 4 1 3 2 8 5 0 ] Bottom \r\nB: Top [ 7 6 ] Bottom \r\nC: Top [ 9 ] Bottom\r\n\r\nWe will simply continue doing what we have done so far, so A gives \"4\" to B face down (and all other ones as well) - (3)\r\n\r\nA: Top [ ] Bottom\r\nB: Top [ 0 5 8 2 3 1 4 7 6 ] Bottom\r\nC: Top [ 9 ] Bottom\r\n\r\nC then passes to B face down (4):\r\n\r\nA: Top [ ] Bottom\r\nB: Top [ 9 0 5 8 2 3 1 4 7 6 ] Bottom\r\nC: Top [ ] Bottom\r\n\r\nB flips the stack and passes to A (5):\r\n\r\nA: Top [ 6 7 4 1 3 2 8 5 0 9 ] Bottom\r\nB, C are empty.\r\nRepeat steps 1-5:\r\n\r\nA: Top [ 6 4 1 2 3 7 5 0 8 9 ] Bottom\r\nB, C are empty\r\nNote how the \"8\" has moved to its correct location now.\r\n\r\nAnd again and again, and so on.",
  "Solution_5": "It takes way more than n-1 passes that jmerry was saying. Now it looks like n log n or something.",
  "Solution_6": "I think that by \"passes,\" he meant the number of times 1-5 would be repeated.  While this algorithm does take a fair amount of steps, I think it's fairly efficient with respect to the restrictions.",
  "Solution_7": "But do you think that it is a waste of processing from B and C? They are not doing anything. They are just holding the stacks. Wouldn't it be better if they are also doing some sorting?  \r\n\r\nThis isn't about minimizing the total number of comparison, it is about sorting is quickly. Right now it looks like B and C are idling most of the time. \r\n\r\nWhat about this algorithm:\r\n\r\nThe stack of paper have names (like the original prompt). Instead of labeling the people A,B,C, they are now called person 1, 2, 3\r\n\r\nSTEP 1) Initially, the stack is separated into three, and each person gets one.\r\n\r\nSTEP 2) In the first round, each person look through their own stack and extracts a stack that contains A-I. By extract I meant when you see a paper in the range of A-I, put it at the back of the stack you are holding. You go through each paper in your own stack. When you see a name you have already seen, you know you are done. \r\n\r\nSTEP 3) When all three are done, they combine the A-I stacks for Person 1, Person 1 starts to sort this stack on his own\r\n\r\nSTEP 4) While Person 1 is sorting, Person 2 and 3 do similarly to get a stack that is from J-R and another S-Z.\r\n\r\nSTEP 5) When they get those stacks, they sort them independently. \r\n\r\nIn this algorithm they only pass stacks for three rounds. The first round happens when the A-I stack is formed. The second round happens when the J-R and S-Z stacks are formed. The last pass is just combining the three sorted stacks.",
  "Solution_8": "The algorithm I gave was an $ n^2$ algorithm. Yes, it's not efficient with processing time- but you can't really process a stack beyond truly simple bits with one person that can only hold one stack. Even that extraction operation- do you have three hands? You need one to hold the top half of the stack as you look, one to move the paper, and one to hold the bottom. Only two hands? Oops; you just lost your place.\r\nOne person, alone, with no room to lay stacks down, [i]can't[/i] sort a stack.\r\n\r\nNow, can you do better than my algorithm? Sure. If you start by splitting into $ k$ pieces of approximately equal size and then restack, you reduce the work by almost a factor of $ k$. Also, person C can do more with the \"buffer\"; if they keep some of the stack in unaltered order (face down), that can split in half and do a bubble pass on the other half at the same time. Do that right (with different cuts for each run), and it's an $ n\\log n$ algorithm.\r\nHere's one version of it:\r\nAs before, except add a case in step 2:\r\nIf A's top sheet is in a certain set (depending on pass), A passes that to C, who swaps it to the bottom of their stack.\r\nWhat is that set? Here's one version, assuming approximately uniform distribution among letters:\r\nFirst pass: A-M. Second pass: A-F and N-S. Third pass: A-C and G-I and N-P and T-V. Fourth pass: A, D, G, J-K, N, Q, T, W-X. Fifth pass: B, E, H, J, L, O, R, U, W, Y. In later passes, don't bother with anything; you probably couldn't remember anyway.\r\nThen, in step 5, pass back from both B and C with a merge. A will take them face down, and then flip the stack up for the next pass.\r\n\r\nThat gives you the first-letter sort in five passes, while incrementally improving the rest. The number of passes required after that is comparable to the number of papers in the most numerous letter.",
  "Solution_9": "I see what you mean. Practically you could sort the stack in your hands by using your index finger as a bookmark. This matches the original assumption that a person could sort a stack in their hands, if they want to.\r\n\r\nCould we talk more in terms of real people and real paper? I wasn't trying to dress up a computer problem. I was really talking about people sorting paper. I was wondering what algorithm could human use to sort it effectively.",
  "Solution_10": "I would probably get the three people to split the papers into three (hopefully) equal sized piles and then sort, then mergesort those piles one at a time.\r\n\r\n\r\nIf the stack got huge, then just doing it with your fingers would be ghastly.\r\n\r\nIf you have enough people, maybe you can just do it for each letter of the alphabet (and maybe combine some) sort each letter, then merge the piles.",
  "Solution_11": "In general, the most efficient use of time (to try to maximize the number of people active at any time as well as minimize the number of steps used by these people) for the first (unbound) scenario would likely be the mergesort algorithm with $ n$ divisions, where $ n$ is the number of people who are participating.\r\n\r\nWhen the space-papercut restriction is added, however, the most efficient use of time appears to be something based on a mergesort with $ \\lfloor \\frac{n}{2} \\rfloor$ divisions, where $ n$ is again the number of participants.",
  "Solution_12": "[quote=\"Twinbee\"]Could we talk more in terms of real people and real paper?[/quote]\r\nAs I said back in my first post, under real-world conditions with effectively unlimited space, I've used mergesort, with a base of piles of about five or six each, sorted ad hoc. The problem with space-limited algorithms is that the more you try to do with a pile without putting it down, the more likely you are to drop part of the pile on the floor, where it will partially scatter and you'll lose some of your sorting. It's hard to quantify, but I'd be conservative about such things.",
  "Solution_13": "I didn't realize that when you merge the stacks, \r\nyou don't necessarily have to look at all three stacks at once. \r\nI always felt that took me a long time."
}
{
  "Tag": [
    "Support",
    "graph theory",
    "number theory open",
    "number theory"
  ],
  "Problem": "Which is the minimum $ n$ that have the following properties,\r\n \r\n$ a_1,a_2,\\cdots,a_n$ is a permutation of $ 1,2,\\cdots,n$ \r\n\r\n$ a_1 \\plus{} a_2$, $ a_2 \\plus{} a_3$, $ \\cdots$, $ a_{n \\minus{} 1} \\plus{} a_n$ and $ a_n \\plus{} a_1$ are all squares?",
  "Solution_1": "Here's a quick argument showing $ n \\geq 31$. I don't know if $ n\\equal{}31$ actually supports such a sequence, and I can't see a clever way to attack this - just brute force Hamilton circuit search.\r\n\r\nPresumably $ n\\equal{}1$ is too degenerate to matter, so $ n \\geq 2$. The smallest integer which pairs up with 2 to yield a square is 7, so $ n \\geq 7$. But we must have another mate for 7 other than 2, which forces $ n \\geq 9$. Now 9 forces 16 and 16 forces 20. But if $ n \\geq 20$ we must take care of 18, which pairs up with 7 to yield 25 but has no other potential partner until we get to 31.\r\n\r\nLuckily, with $ n\\equal{}31$ we no longer have any isolated numbers: the graph on $ \\{1,\\ldots,31\\}$ where edges connect two numbers if they add up to a square has minimum degree 2, which is a trivial necessary condition for a Hamiltonian cycle. Checking if this graph actually has a Hamiltonian cycle is tedious, but if it doesn't one can hope that $ n\\equal{}32$ or $ n\\equal{}33$ will do better.",
  "Solution_2": "OK, so the answer is 32. As I mentioned before 31 is a lower bound. I was able to find the following near miss for $ n\\equal{}31$:\r\n\r\n5 31 18 7 29 20 16 9 27 22 3 13 12 4 21 28 8 17 19 30 6 10 26 23 2 14 11 25 24 1 15\r\n\r\nThat's a near miss because the first and last elements don't add up to a square, as required. I did this by hand because I was too lazy to program at this stage :blush: but after failing to find a full cycle I gave up a and turned to my trusty computer for help. It confirmed that there is no solution for $ n\\equal{}31$, so unless my code is buggy there really isn't one. \r\n\r\nHowever, for $ n\\equal{}32$ there is a unique full cycle:\r\n\r\n1 8 28 21 4 32 17 19 30 6 3 13 12 24 25 11 5 31 18 7 29 20 16 9 27 22 14 2 23 26 10 15"
}
{
  "Tag": [
    "inequalities",
    "symmetry",
    "inequalities proposed"
  ],
  "Problem": "For any positive real numbers $a,b$ and $c$,\r\n\\[\\frac{1}{\\sqrt{a^{2}+bc}}+\\frac{1}{\\sqrt{b^{2}+ca}}+\\frac{1}{\\sqrt{c^{2}+ab}}\\le\\frac{3}{\\sqrt{2}}\\!\\cdot\\!\\frac{a+b+c}{ab+bc+ca}\\]",
  "Solution_1": "Hi, apply Jensen:\r\n$\\sum \\frac{1}{\\sqrt{a^{2}+bc}}\\leq 3\\sqrt{\\frac{3}{a^{2}+b^{2}+c^{2}+ab+bc+ca}}\\leq \\frac{3(a+b+c)}{\\sqrt{2}(ab+bc+ca)}$  :)",
  "Solution_2": "[quote=\"bibibibi\"]Hi, apply Jensen:\n$\\sum \\frac{1}{\\sqrt{a^{2}+bc}}\\leq 3\\sqrt{\\frac{3}{a^{2}+b^{2}+c^{2}+ab+bc+ca}}$[/quote]\r\n\r\nI think you've got the sign reversed here",
  "Solution_3": "Very nice inequality, Duc.",
  "Solution_4": "By AM-QM, we have\r\n\\[\\sum_{cyc}\\frac{1}{\\sqrt{a^{2}+bc}}\\leq 3\\sqrt{\\frac{ \\sum_{cyc}\\frac{1}{a^{2}+bc}}{3}}\\]\r\nso it's enough to prove that\r\n\\[\\frac{1}{a^{2}+bc}+\\frac{1}{b^{2}+ca}+\\frac{1}{c^{2}+ab}\\leq \\frac{3(a+b+c)^{2}}{2(ab+bc+ca)^{2}}\\]\r\nwhich can be solved easyly but tediously by expanding, which will leave\r\n\\[\\sum_{sym}a^{4}b^{3}c+4\\sum_{cyc}a^{4}b^{2}c^{2}+3\\sum_{cyc}a^{3}b^{3}c^{2}\\leq 3\\sum_{cyc}a^{6}bc+\\sum_{sym}a^{5}b^{3}+4\\sum_{cyc}a^{4}b^{4}. \\]\r\n\r\nAny more beautiful solutions? I'd like to see one.",
  "Solution_5": "There is one more solution, maybe. It's by SOS. The following inequality also holds:\r\n\r\n\\[{1 \\over \\sqrt{a^{2}+b^{2}}}+{1 \\over \\sqrt{b^{2}+c^{2}}}+{1 \\over \\sqrt{c^{2}+a^{2}}}\\leq{3 \\over \\sqrt{2}}\\cdot{a+b+c \\over ab+bc+ca}. \\]",
  "Solution_6": "[quote=\"Lovasz\"]There is one more solution, maybe. It's by SOS.[/quote]\nWhat about posting this SOS solution.\n\n[quote=\"Lovasz\"]\nThe following inequality also holds:\n\\[{1 \\over \\sqrt{a^{2}+b^{2}}}+{1 \\over \\sqrt{b^{2}+c^{2}}}+{1 \\over \\sqrt{c^{2}+a^{2}}}\\leq{3 \\over \\sqrt{2}}\\cdot{a+b+c \\over ab+bc+ca}. \\]\n[/quote]\r\nThis is a trivial consequence of the easy inequality\r\n\\[\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\le\\frac{3(a+b+c)}{2(ab+bc+ca)}\\]\r\nwhich was already mentioned at http://www.mathlinks.ro/Forum/viewtopic.php?t=111171",
  "Solution_7": "[quote=\"ductrung\"][quote=\"Lovasz\"]There is one more solution, maybe. It's by SOS.[/quote]\nWhat about posting this SOS solution.[/quote]\r\nI just rewrite it ina possible form:\r\n\\[\\sum_{sym}{a^{2}+bc \\over b^{2}+ca}+3 \\leq{3(a+b+c)^{2}\\over ab+bc+ca}. \\]",
  "Solution_8": "[quote=\"ductrung\"]For any positive real numbers $a,b$ and $c$,\n\\[\\frac{1}{\\sqrt{a^{2}+bc}}+\\frac{1}{\\sqrt{b^{2}+ca}}+\\frac{1}{\\sqrt{c^{2}+ab}}\\le\\frac{3}{\\sqrt{2}}\\!\\cdot\\!\\frac{a+b+c}{ab+bc+ca}\\]\n[/quote]\r\nTo prove this, we only need to prove that\r\n\\[\\underset{cyc}{\\sum}\\frac{1}{\\sqrt{a^{2}+bc}}\\le \\underset{cyc}{\\sum}\\frac{\\sqrt{2}}{a+b}(1)\\\\ \\underset{cyc}{\\sum}\\frac{\\sqrt{2}}{a+b}\\le \\frac{3}{\\sqrt{2}}.\\frac{a+b+c}{ab+bc+ca}(2) \\]\r\nTo prove (1), we only need to use Chebyshev. To prove (2), we only need AM - GM. :)",
  "Solution_9": "can you post detail?, toanhocmuonmau",
  "Solution_10": "[quote=\"toanhocmuonmau\"][quote=\"ductrung\"]For any positive real numbers $ a,b$ and $ c$,\n\\[ \\frac{1}{\\sqrt{a^{2}+bc}}+\\frac{1}{\\sqrt{b^{2}+ca}}+\\frac{1}{\\sqrt{c^{2}+ab}}\\le\\frac{3}{\\sqrt{2}}\\!\\cdot\\!\\frac{a+b+c}{ab+bc+ca}\\]\n[/quote]\nTo prove this, we only need to prove that\n\\[ \\underset{cyc}{\\sum}\\frac{1}{\\sqrt{a^{2}+bc}}\\le \\underset{cyc}{\\sum}\\frac{\\sqrt{2}}{a+b}(1)\\\\ \\underset{cyc}{\\sum}\\frac{\\sqrt{2}}{a+b}\\le \\frac{3}{\\sqrt{2}}.\\frac{a+b+c}{ab+bc+ca}(2) \\]\nTo prove (1), we only need to use Chebyshev. To prove (2), we only need AM - GM. :)[/quote]\r\n\r\nDear Ductrung, is your solution similar to this one or more elementary?\r\n\r\nThis solution is OK but very hard to think  :( \r\n\r\nThank you very much.",
  "Solution_11": "[quote=\"manlio\"]Dear Ductrung, is your solution similar to this one or more elementary?[/quote]\r\nHi Manlio,\r\nI posted this version just for fun. The actual inequality I proved at that time is the one mentioned by Yimin Ge above, which now is just an easy corollary of [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=131247[/url].\r\nMy solution back then is very simple: split it into two inequalities\r\n\\[ \\frac{a(b+c)}{a^{2}+bc}+\\frac{b(c+a)}{b^{2}+ca}+\\frac{c(a+b)}{c^{2}+ab}\\le\\frac{(a+b+c)^{2}}{ab+bc+ca}\\]\r\nand\r\n\\[ \\frac{bc}{a^{2}+bc}+\\frac{ca}{b^{2}+ca}+\\frac{ab}{c^{2}+ab}\\le\\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\\]\r\nIt is VERY easy to prove the above two inequalities. I am busy at the moment but will post my solutions soon (if nobody does that until then :))",
  "Solution_12": "[quote=\"ductrung\"]My solution back then is very simple: split it into two inequalities\n\\[ \\frac{a(b+c)}{a^{2}+bc}+\\frac{b(c+a)}{b^{2}+ca}+\\frac{c(a+b)}{c^{2}+ab}\\le\\frac{(a+b+c)^{2}}{ab+bc+ca}\\]\nand\n\\[ \\frac{bc}{a^{2}+bc}+\\frac{ca}{b^{2}+ca}+\\frac{ab}{c^{2}+ab}\\le\\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\\]\nIt is VERY easy to prove the above two inequalities. I am busy at the moment but will post my solutions soon (if nobody does that until then :))[/quote]\r\n\r\nVery nice these two inequalities but I cannot solve them in concise manner :oops: using only Cauchy, AM-GM and Shur.\r\n\r\nP.S. I like a lot all inequalities which can be solved in concise manner using only these three elementary inequalities",
  "Solution_13": "[hide=\"Hint for Manlio\"]\n\\[ \\frac{a(b+c)}{a^{2}+bc}+\\frac{b(c+a)}{b^{2}+ca}+\\frac{c(a+b)}{c^{2}+ab}\\leq 3\\leq \\frac{(a+b+c)^{2}}{ab+bc+ca}\\]\nand\n\\[ \\frac{bc}{a^{2}+bc}+\\frac{ca}{b^{2}+ca}+\\frac{ab}{c^{2}+ab}\\leq\\frac32\\leq \\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\\]\n[/hide]\n\n[hide=\";)\"]Vornicu-Schur for number 1...[/hide]\r\n\r\n  Darij",
  "Solution_14": "[quote=\"manlio\"]Thank you very much, Darij. :wink: \n\nI will try.[/quote]\r\n\r\nDear Darij. I think both your inequalities are wrong?!",
  "Solution_15": "OMG, I should learn calculating...  :blush: \r\n\r\nIn fact, the first one is at least correct for a, b, c sides of a triangle.\r\n\r\n  Darij",
  "Solution_16": "\\[ \\frac{bc}{a^{2}+bc}+\\frac{ca}{b^{2}+ca}+\\frac{ab}{c^{2}+ab}\\le\\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\\]\r\n\r\nSol:\r\n\r\nWrite it as \r\n$ \\sum \\frac{a^{2}}{a^{2}+bc}+\\frac{(\\sum a)^{2}}{2\\sum ab}\\geq 3$\r\n\r\nNow use Cauchy \r\n\r\n$ \\sum \\frac{a^{2}}{a^{2}+bc}\\geq \\frac{(\\sum a)^{2}}{\\sum ab+\\sum a^{2}}$\r\n\r\nto reduce our ineq at\r\n\r\n$ \\frac{(\\sum a)^{2}}{\\sum ab+\\sum a^{2}}+\\frac{(\\sum a)^{2}}{2\\sum ab}\\geq 3$\r\n\r\nClearing denominators we are left with\r\n\r\n$ 3(\\sum ab)(\\sum a)^{2}+(\\sum a^{2})(\\sum a)^{2}\\geq 6(\\sum ab)(\\sum a^{2})+6(\\sum ab)^{2}$\r\n\r\nNow use $ (\\sum a)^{2}\\geq 3\\sum ab$ and all is OK (I hope  :maybe: )\r\n\r\nDear Ductrung is your proof the same?\r\n\r\nThank you very much.\r\n\r\nFor the second inequality I tried to use Cauchy in this form\r\n\r\n$ \\frac{a(b+c)}{a^{2}+bc}= \\frac{a(b+c)^{2}}{(b+c)(a^{2}+bc)}\\leq \\frac{ab^{2}}{b(a^{2}+c^{2})}+\\frac{ac^{2}}{c(a^{2}+b^{2})}$\r\n\r\nbut unfortunately this trick doesn't work so I am  :( \r\n\r\nThank you very much for your patience with me.",
  "Solution_17": "[quote=\"manlio\"]\n\\[ \\frac{bc}{a^{2}+bc}+\\frac{ca}{b^{2}+ca}+\\frac{ab}{c^{2}+ab}\\le\\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\\]\nSol:\n\nWrite it as \n$ \\sum \\frac{a^{2}}{a^{2}+bc}+\\frac{(\\sum a)^{2}}{2\\sum ab}\\geq 3$\n\nNow use Cauchy \n\n$ \\sum \\frac{a^{2}}{a^{2}+bc}\\geq \\frac{(\\sum a)^{2}}{\\sum ab+\\sum a^{2}}$\n\nto reduce our ineq at\n\n$ \\frac{(\\sum a)^{2}}{\\sum ab+\\sum a^{2}}+\\frac{(\\sum a)^{2}}{2\\sum ab}\\geq 3$\n\nClearing denominators we are left with\n\n$ 3(\\sum ab)(\\sum a)^{2}+(\\sum a^{2})(\\sum a)^{2}\\geq 6(\\sum ab)(\\sum a^{2})+6(\\sum ab)^{2}$\n\nNow use $ (\\sum a)^{2}\\geq 3\\sum ab$ and all is OK (I hope  :maybe: )\n\nDear Ductrung is your proof the same?\n\nThank you very much.\n\nFor the second inequality I tried to use Cauchy in this form\n\n$ \\frac{a(b+c)}{a^{2}+bc}= \\frac{a(b+c)^{2}}{(b+c)(a^{2}+bc)}\\leq \\frac{ab^{2}}{b(a^{2}+c^{2})}+\\frac{ac^{2}}{c(a^{2}+b^{2})}$\n\nbut unfortunately this trick doesn't work so I am  :( \n\nThank you very much for your patience with me.[/quote]\r\n\r\nFor your patience, I will try to prove nice problem of Duc ... although I'm really busy these days. Actually, I just prove the second result as Duc says. As Darij say, by generalized Schur theorem, we have certainly\r\n\\[ \\sum \\frac{a(b+c)}{a^{2}+bc}-3=\\sum\\frac{(a-b)(a-c)}{a^{2}+bc}\\le 0, \\]\r\nif $ a,b,c$ are three side-lengths of a triangle. Otherwise, we assume that $ a\\ge b+c\\ge b\\ge c$. The inequality becomes\r\n\\[ \\frac{(a+b+c)^{2}}{ab+bc+ca}-3+\\sum \\frac{(a-b)(a-c)}{c^{2}+ab}\\ge 0. \\]\r\nNow, since $ a\\ge b\\ge c$, we refer that\r\n\\[ (b-c)^{2}(ab+bc+ca) \\le (b^{2}-bc)(ab+bc+ca) \\le (b^{2}+ca)(c^{2}+ab). \\]\r\nThus\r\n\\[ \\sum \\frac{(a-b)(a-c)}{c^{2}+ab}\\ge \\frac{(c-a)(c-b)}{c^{2}+ab}+\\frac{(b-a)(b-c)}{c^{2}+ab}\\]\r\n\r\n\\[ \\ge \\frac{(b-a)(c-b)}{c^{2}+ab}+\\frac{(b-a)(b-c)}{c^{2}+ab}\\]\r\n\r\n\\[ =-\\frac{(a-b)(b-c)^{2}(a-b-c)}{(c^{2}+ab)(b^{2}+ac)}\\ge-\\frac{(b-c)^{2}(a-b)^{2}}{(c^{2}+ab)(b^{2}+ac)}\\ge-\\frac{(a-b)^{2}}{ab+bc+ca}. \\]\r\nThe desired result follows by\r\n\\[ \\frac{(a+b+c)^{2}}{ab+bc+ca}-3=\\frac{(a-b)^{2}+(c-a)(c-b)}{ab+bc+ca}\\ge \\frac{(a-b)^{2}}{ab+bc+ca}. \\]",
  "Solution_18": "Please see my solution attached",
  "Solution_19": "Thank you very much, hungkhtn and ductrung for your very nice solutions. I am really very happy to study them. :)",
  "Solution_20": "[quote=\"ductrung\"]For any positive real numbers $ x,y$ and $ z$,\n\\[ \\frac {1}{\\sqrt {x^{2} \\plus{} yz}} \\plus{} \\frac {1}{\\sqrt {y^{2} \\plus{} zx}} \\plus{} \\frac {1}{\\sqrt {z^{2} \\plus{} xy}}\\leq\\frac {3(x \\plus{} y \\plus{} z)}{\\sqrt {2}(yz \\plus{} zx \\plus{} xy)}.\\][/quote]\n[quote=\"Yimin Ge\"]By AM-QM, we have\n\\[ \\sum\\frac {1}{\\sqrt {x^{2} \\plus{} yz}}\\leq \\sqrt {\\sum\\frac {3}{x^{2} \\plus{} yz}}.\\][/quote]\r\n$ \\frac {9(x \\plus{} y \\plus{} z)^2}{2(yz \\plus{} zx \\plus{} xy)^2} \\minus{} \\sum\\frac {3}{x^{2} \\plus{} yz}$\r\n\r\n$ \\equiv \\frac {3F(x,y,z)}{2(x^{2} \\plus{} yz)(y^{2} \\plus{} zx)(z^{2} \\plus{} xy)(yz \\plus{} zx \\plus{} xy)^2}$.\r\n\r\nWithout losing generality, we may assuming $ x \\equal{} \\min\\{x,y,z\\}$.\r\n\r\n$ F(x,y,z) \\equal{} F(x,x \\plus{} s,x \\plus{} t)$\r\n\r\n$ \\equal{} 48(s^2 \\minus{} st \\plus{} t^2)x^6 \\plus{} (s \\plus{} t)(82s^2 \\minus{} 61st \\plus{} 82t^2)x^5$\r\n\r\n$ \\plus{} (57s^4 \\plus{} 91s^3t \\plus{} 71s^2t^2 \\plus{} 91st^3 \\plus{} 57t^4)x^4$\r\n\r\n$ \\plus{} 2(s \\plus{} t)(10s^4 \\plus{} 22s^3t \\plus{} 23s^2t^2 \\plus{} 22st^3 \\plus{} 10t^4)x^3$\r\n\r\n$ \\plus{} (3s^6 \\plus{} 21s^5t \\plus{} 32s^4t^2 \\plus{} 73s^3t^3 \\plus{} 32s^2t^4 \\plus{} 21st^5 \\plus{} 3t^6)x^2$\r\n\r\n$ \\plus{} st(s \\plus{} t)(3s^4 \\plus{} 20s^2t^2 \\plus{} 3t^4)x \\plus{} s^3t^3(s^2 \\plus{} 4st \\plus{} t^2)\\geq 0$.",
  "Solution_21": "[quote=\"toanhocmuonmau\"]\n\\[ \\underset{cyc}{\\sum}\\frac {1}{\\sqrt {a^{2} + bc}}\\le \\underset{cyc}{\\sum}\\frac {\\sqrt {2}}{a + b}\n\\]\n[/quote]\r\nFor along time, we haven't got solution for this problem yet. Now, I will post my solution\r\n[b]Problem[/b] (Vo Quoc Ba Can) Let $ a,b,c$ be nonnegative real numbers. Prove that\r\n\\[ \\underset{cyc}{\\sum}\\frac {1}{\\sqrt {a^{2} + bc}}\\le \\underset{cyc}{\\sum}\\frac {\\sqrt {2}}{a + b}.\r\n\\]\r\n[i]Solution.[/i]\r\nBy the Cauchy schwarz Inequality, we have\r\n\\[{ (\\sum \\frac {1}{\\sqrt {a^2 + bc}})^2 \\le (\\sum \\frac {1}{(a + b)(a + c)})(\\sum \\frac {(a + b)(a + c)}{a^2 + bc})} = \\frac {2\\sum a}{(a + b)(b + c)(c + a)}(\\sum \\frac {a(b + c)}{a^2 + bc} + 3)\r\n\\]\r\nIt suffices to show that\r\n\\[ \\frac {2\\sum a}{(a + b)(b + c)(c + a)}(\\sum \\frac {a(b + c)}{a^2 + bc} + 3) \\le 2( \\sum \\frac {1}{a + b})^2\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum \\frac {a(b + c)}{a^2 + bc} + 3 \\le \\frac {( \\sum a^2 + 3\\sum ab)^2}{(a + b)(b + c)(c + a)\\sum a}\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum \\frac {a(b + c)}{a^2 + bc} - 3 \\le \\frac {( \\sum a^4 - \\sum a^2b^2}{(a + b)(b + c)(c + a)\\sum a}\r\n\\]\r\n\r\n\\[ \\Leftrightarrow \\sum (a - b)(a - c)(\\frac {1}{a^2 + bc} + \\frac {1}{(b + c)(a + b + c)}) \\ge 0\r\n\\]\r\nDue to symmetry, we may assume $ a \\ge b \\ge c$, since $ a - c \\ge \\frac {a}{b}(b - c)$. It suffices to show that\r\n\\[ a(\\frac {1}{a^2 + bc} + \\frac {1}{(b + c)(a + b + c)}) \\ge b(\\frac {1}{b^2 + ca} + \\frac {1}{(a + c)(a + b + c)})\r\n\\]\r\n\r\n\\[ \\Leftrightarrow c(a^2 - b^2)[(a - b)^2 + ab + bc + ca] \\ge 0\r\n\\]\r\nwhich is trivial.\r\nEquality holds if and only if $ a = b = c.$ :)",
  "Solution_22": "WONDERFUL,  can_hang2007!!!\r\n\r\n :)"
}
{
  "Tag": [
    "geometry",
    "geometric transformation"
  ],
  "Problem": "Find the minimal value of integer $ n$ that guarantees: \nAmong $ n$ sets, there exits at least three sets such that any of them does not include any other; or there exits at least three sets such that any two of them includes the other. \n\n$\\textbf{(A)}\\ 4  \\qquad\\textbf{(B)}\\ 5  \\qquad\\textbf{(C)}\\ 6  \\qquad\\textbf{(D)}\\ 7  \\qquad\\textbf{(E)}\\ 8$",
  "Solution_1": "Let $ A \\subset B$ , $ C \\subset D$ ,  and $ A \\cap C \\equal{} \\{\\}$. Now on there are 2 subsets and 2 non-subsets. The set $ E$ can be a non-subset. At that situation $ A,C,E$ are the three sets such that any of them does not include any other. Other possibility is $ E \\subset A$ (of course it is same for $ C$). So $ A$ and $ C$ will include $ E$. The answer is $ B$ which is $ 5$. \r\n\r\nPS: The translation of this problem was very difficult because it is also diffucult to understand the Turkish text.",
  "Solution_2": "This is a special case of [url=http://mathworld.wolfram.com/DilworthsLemma.html]Dilworth's Lemma[/url]: every [[partially ordered set]] with $ mn \\plus{} 1$ elements contains either a chain of length $ m \\plus{} 1$ or an antichain of length $ n \\plus{} 1$.  Here $ m \\equal{} n \\equal{} 2$, where the partial order is \"containment as a subset.\"  (A chain is a sequence of elements that are pairwise comparable, here a sequence of three nested sets; an antichain is a sequence of elements that are pairwise incomparable, here a sequence of three elements such that none contains any other.)"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $ a,b$ be potisive numbers(a<b) and p is a prime number.\r\n      prove that:$ \\large C_{pb}^{pa}$ $ \\equiv$ $ \\large C_{b}^{a}$($ mod$p)\r\n                       $ \\large C_{pb\\plus{}r}^{pa}$ $ \\equiv$ $ \\large C_{pb}^{pa}$ ($ mod$p)\r\n                         where $ r$ be natural number $ r$ $ \\in$ (0,b).",
  "Solution_1": "Lucas theory\r\n$ m\\equal{}c_{0}\\plus{}...\\plus{}c_{t}p^{t}$\r\n$ n\\equal{}b_{0}\\plus{}...\\plus{}b_{t}.p^{t}$\r\n$ C_{m}^{n}\\equal{}\\prod_{i\\equal{}1}^{t}C_{c_{t}}^{b_{t}}$\r\nUse this.",
  "Solution_2": "Can you give the proof of $ Lucas theory$? :D and how applying it to solve this problem? :o",
  "Solution_3": "Write $ p,pa,pb,a,b,r$ in base p and apply this theory."
}
{
  "Tag": [
    "inequalities",
    "quadratics",
    "inequalities proposed"
  ],
  "Problem": "\\[ \\boxed {\\ \\{a,b,c\\}\\subset (0,\\infty )\\Longrightarrow\\left(a+\\frac ca\\right)^2+\\left(b+\\frac cb\\right)^2\\ge \\frac{\\left[(a+b)^2+4c\\right]^2}{2(a+b)^2}\\ }\\ .  \\]\r\n\r\n[b]Remark.[/b] Thanks, Argady ! I corrected.",
  "Solution_1": "Easy inequality, Virgil.",
  "Solution_2": "\\[ \\boxed {\\ \\{a,b,c\\}\\subset (0,\\infty )\\Longrightarrow\\left(a+\\frac ca\\right)^2+\\left(b+\\frac cb\\right)^2\\ge \\frac{\\left[(a+b)^2+4c\\right]^2}{2(a+b)^2}\\ }\\ .  \\]\r\n\r\nQuadratic mean-AM: $(a+\\frac{c}{a})^2+(b+\\frac{c}{b})^2 \\geq \\frac{(a+b+\\frac{c}{a}+\\frac{c}{b})^2}{2}$\r\n\r\nAlso: $\\frac{1}{a}+\\frac{1}{b} \\geq \\frac{4}{a+b}$\r\n\r\nThen we have: $(a+\\frac{c}{a})^2+(b+\\frac{c}{b})^2 \\geq \\frac{[(a+b) + \\frac{4c}{a+b}]^2}{2}$ ; $QED$"
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "perimeter",
    "trigonometry",
    "ratio"
  ],
  "Problem": "The perimeter of a parallelogram $ABCD$ is 40, and its altitudes are 4 and 7.  Compute $\\sin A$.",
  "Solution_1": "[hide]\n11/20\n\nAD times one altitude = area = AB times the other altitude\nAD:AB=4:7 (inverse of ratio between altitudes)\nAD+AB+BC+CD = 2(AD+AB) = 40\n\nlet AD=4x, 22x=40, x = 20/11, AD = 80/11\n\nSinA = 4 over 80/11 = 11/20\n\neven if the diagram is constructed differently, sinA = sin (180-A), so it remains the same. \n\n[/hide]",
  "Solution_2": "Nice use of area."
}
{
  "Tag": [],
  "Problem": "If $ 2^{2x\\minus{}1}\\minus{}3^{x\\plus{}\\frac{1}{2}}\\equal{}3^{x\\minus{}\\frac{1}{2}}\\minus{}4^x$ then $ x\\plus{}x^{\\minus{}1}\\equal{}$\r\n\r\n$ \\textbf{(A)}\\ \\frac{11}{6} \\qquad \\textbf{(B)}\\ \\frac{13}{6} \\qquad \\textbf{(C)}\\ \\frac{17}{6} \\qquad \\textbf{(D)}\\ \\frac{19}{6} \\qquad \\textbf{(E)}\\ \\frac{23}{6}$",
  "Solution_1": "[quote=\"hasan4444\"]If $ 2^{2x \\minus{} 1} \\minus{} 3^{x \\plus{} \\frac {1}{2}} \\equal{} 3^{x \\minus{} \\frac {1}{2}} \\minus{} 4^x$ then $ x \\plus{} x^{ \\minus{} 1} \\equal{}$\n\n$ \\textbf{(A)}\\ \\frac {11}{6} \\qquad \\textbf{(B)}\\ \\frac {13}{6} \\qquad \\textbf{(C)}\\ \\frac {17}{6} \\qquad \\textbf{(D)}\\ \\frac {19}{6} \\qquad \\textbf{(E)}\\ \\frac {23}{6}$[/quote]\r\n\r\nLet $ y\\equal{}x\\minus{}\\frac{1}{2}$, then the equation \r\n$ \\Leftrightarrow 4^y\\minus{}3(3^y)\\equal{}3^y\\minus{}2(4^y)$\r\n$ \\Leftrightarrow 3 \\times 4^y\\equal{}4 \\times 3^y$\r\n$ \\Leftrightarrow (\\frac{4}{3})^{y\\minus{}1}\\equal{}1$\r\n$ \\Leftrightarrow y\\minus{}1\\equal{}x\\minus{}\\frac{3}{2}\\equal{}0$\r\n$ \\Rightarrow x\\equal{}\\frac{3}{2}$\r\n$ \\Rightarrow x\\plus{}x^{\\minus{}1}\\equal{}\\frac{13}{6}$\r\n\r\nSo the answer is $ \\textbf{(B)}$",
  "Solution_2": "$ \\frac{4^{2}}{2}\\plus{}4^{x}\\equal{}\\sqrt{3}\\cdot3^{x}\\plus{}\\frac{3^{x}}{\\sqrt{3}}$\r\n$ \\frac{9}{4}4^{2x}\\equal{}\\frac{16}{3}3^{2x}$\r\n$ 9\\cdot4^{2x\\minus{}1}\\equal{}16\\cdot3^{2x\\minus{}1}$\r\nThus $ 2x\\minus{}1\\equal{}2$ and $ x\\equal{}3/2$. Then, $ 3/2\\plus{}2/3\\equal{}13/6$."
}
{
  "Tag": [
    "inequalities",
    "integration",
    "limit",
    "triangle inequality",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ f$ be of bounded variation on $ [a,b]$.\r\n\r\nShow $ \\int_a^b | f' | \\leq T(f)$ where $ T(f)$ denotes the total variation of $ f$ on $ [a,b]$",
  "Solution_1": "Let $ \\mathcal{P}$ denote the set of partitions of $ [a,b]$.  Given a partition $ P\\in \\mathcal{P}$ of $ [a,b]$ into $ n$ subintervals with $ a\\equal{}t_0<t_1<\\dots <t_n\\equal{}b$, let $ v(f,P) \\equal{} \\sum_{k \\equal{} 1}^n |f(t_k) \\minus{} f(t_{k \\minus{} 1})|$.  By definition, $ T(f) \\equal{} \\sup_{P\\in \\mathcal{P}} v(f,P)$.  Furthermore, $ \\int_a^b|f'| \\equal{} \\lim_{k\\to \\infty}v\\left(f,P_k\\right)$ for some sequence of partitions $ \\{P_k\\}$.  But $ \\{P_k\\}\\subset \\mathcal{P}$, so $ \\int_a^b|f'|\\leq T(f)$ by definition of supremum.",
  "Solution_2": "[quote=\"JoeBlow\"] Furthermore, $ \\int_a^b|f'| \\equal{} \\lim_{k\\to \\infty}v\\left(f,P_k\\right)$ for some sequence of partitions $ \\{P_k\\}$.  [/quote]\r\nCan you justify this?",
  "Solution_3": "Use mean value theorem! \r\n$ v(f,P_n) \\equal{} \\sum_{k \\equal{} 1}^n |f(t_k) \\minus{} f(t_{k \\minus{} 1})| \\equal{} \\sum_{k \\equal{} 1}^n |f'(c_k)|{}|t_k \\minus{} t_{k \\minus{} 1}|\\rightarrow \\int_a^b |f'|$, as $ n\\rightarrow \\infty$, where $ c_k\\in (t_{k \\minus{} 1},t_k)$.",
  "Solution_4": "Thank you Mr.Doe. Can you prove equality when $ f$ is absolutely continuous?",
  "Solution_5": "[quote=\"msaid\"]Let $ f$ be of bounded variation on $ [a,b]$.\n\nShow $ \\int_a^b | f' | \\leq T(f)$ where $ T(f)$ denotes the total variation of $ f$ on $ [a,b]$[/quote]\r\n\r\nWe know that $ f$ has bounded total variation on $ [a,b]$ iff $ f\\equal{}g\\minus{}h$  for $ g$ increasing and $ h$ decreasing.  Now, use the definition of total variation, use the triangle inequality somewhere, you can prove the problem"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "$ a$ , $ b$ and $ c$ $ \\in$ $ {\\mathbb R}^ \\plus{}$  , prove that :\r\n\r\n$ \\frac {1}{2}(\\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {b \\plus{} a}{c}) \\geq \\frac {a \\plus{} b \\plus{} c}{\\sqrt [3]{abc}}$",
  "Solution_1": "Multiply by abc, then muirhead. Q.E.D.",
  "Solution_2": "[quote=\"y-a-s-s-i-n-e\"]$ a$ , $ b$ and $ c$ $ \\in$ $ {\\mathbb R}^ +$  , prove that :\n\n$ \\frac {1}{2}(\\frac {b + c}{a} + \\frac {c + a}{b} + \\frac {b + a}{c}) \\geq \\frac {a + b + c}{\\sqrt [3]{abc}}$[/quote]\r\n\r\n$ \\leftrightarrow \\frac {b + c}{a} + \\frac {c + a}{b} + \\frac {b + a}{c}\\ge \\frac {2(a + b + c)}{\\sqrt [3]{abc}}$\r\n\r\nwe have $ \\sum \\frac {a}{b} \\ge \\frac {a + b + c}{\\sqrt [3]{abc}}(*)$\r\n\r\nand $ \\sum \\frac {a}{c} \\ge \\frac {a + b + c}{\\sqrt [3]{abc}}$\r\n\r\n$  \\frac {b + c}{a} + \\frac {c + a}{b} + \\frac {b + a}{c}\\ge \\frac {a + b + c}{\\sqrt [3]{abc}} + \\frac {a + b + c}{\\sqrt [3]{abc}}\\ge \\frac {2(a + b + c)}{\\sqrt [3]{abc}}$\r\n\r\nwe proven$ (*)$\r\nwe have \r\n$ \\frac {2a}{b} + \\frac {b}{c}\\ge 3\\sqrt [3]{\\frac {a^2}{bc}} = \\frac {3a}{\\sqrt [3]{abc}}$\r\n\r\nanalogue with another inequality \r\n\r\ndone!~ :)",
  "Solution_3": "thank's for the nice prof ..",
  "Solution_4": "[quote=\"y-a-s-s-i-n-e\"]$ a$ , $ b$ and $ c$ $ \\in$ $ {\\mathbb R}^ \\plus{}$  , prove that :\n\n$ \\frac {1}{2}(\\frac {b \\plus{} c}{a} \\plus{} \\frac {c \\plus{} a}{b} \\plus{} \\frac {b \\plus{} a}{c}) \\geq \\frac {a \\plus{} b \\plus{} c}{\\sqrt [3]{abc}}$[/quote]\r\nby Chebyshev : $ LHS \\ge \\frac{1}{3}(\\frac{1}{a}\\plus{}\\frac{1}{b}\\plus{}\\frac{1}{c}) (a\\plus{}b\\plus{}c)$ :)"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "hello,\r\nI would like to have no number associated with Introduction an conclusion.\r\nI want the same result as\r\n\\section*{Introduction}\r\n\r\nUnfortunately, by doing this, \"Introduction\" doesn't appear in the \\tableofcontents.\r\n\r\nIs there a way to force it?\r\nHow do you cope with this problem?\r\n\r\nThanks a lot.",
  "Solution_1": "Before \\chapter*{Introduction}, add \r\n\\addcontentsline{toc}{chapter}{Introduction}\r\nSimilarly,\r\n\\addcontentsline{toc}{chapter}{Conclusion}\r\n\r\nSSN.",
  "Solution_2": "Thanks a lot!\r\n\r\nDid you change your message? I saw something about titlesec !\r\n\r\nAnyway, this solution works fine. I may try to use the titlesec package to improve my documents title and headers."
}
{
  "Tag": [
    "modular arithmetic",
    "quadratics",
    "function",
    "number theory",
    "totient function"
  ],
  "Problem": "Using index arithmetic, solve the following congruences:\r\n\r\n1.) $ 7^x \\equiv 6 (mod 17)$\r\n2.) $ x^8 \\equiv 5 (mod 14)$",
  "Solution_1": "[hide=\"Solution to Problem 2\"]\\[ x^8\\equiv 5 \\pmod{14}\\]\nAlso, we have $ \\phi {\\left( 14\\right)}\\equal{}6$:\n\\[ x^2 \\equiv 5 \\equiv \\pmod{14}\\]There are no solutions to this congruence...Looking at the congruence in mod $ 14$, we have that:\n\\[ 5 \\equiv x^2 \\pmod{14}\\]However, the quadratic residues of $ 14$ are $ 0, 1, 4, 9, 2, 11, 8, 7$. Therefore, there are no solutions.[/hide]",
  "Solution_2": "[hide=\"Solution to Problem 1\"]From Euler's Totient Function, we know that $ 7^{16} \\equiv 1 \\pmod{17}$.\nTherefore, all we need to do is check all the residues of $ 7^x$ in mod $ 17$.\n\nEDIT: Since only $ 7^{13} \\equiv 6 \\pmod{17}$, that is the sole solution.[/hide]",
  "Solution_3": "actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\\equiv 13(\\text{mod}16)$",
  "Solution_4": "[quote=\"SCHNOGGINVOGGINSCHWITZERW\"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\\equiv 13(\\text{mod}16)$[/quote]Sorry, I calculated wrong. I will edit my answer.",
  "Solution_5": "[quote=\"SCHNOGGINVOGGINSCHWITZERW\"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\\equiv 13(\\text{mod}16)$[/quote]\r\n\r\nHow do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?",
  "Solution_6": "[quote=\"Zellex\"][quote=\"SCHNOGGINVOGGINSCHWITZERW\"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\\equiv 13(\\text{mod}16)$[/quote]\n\nHow do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?[/quote]\r\n\r\nsay $ 7^m \\equiv 7^n \\mod 17$\r\nthen $ 7^{m\\minus{}n} \\equiv 0 \\mod 17$\r\nclearly not true",
  "Solution_7": "[quote=\"CircleSquared\"][quote=\"Zellex\"][quote=\"SCHNOGGINVOGGINSCHWITZERW\"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\\equiv 13(\\text{mod}16)$[/quote]\n\nHow do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?[/quote]\n\nsay $ 7^m \\equiv 7^n \\mod 17$\nthen $ 7^{m \\minus{} n} \\equiv 0 \\mod 17$\nclearly not true[/quote]\r\n\r\nIsn't that $ 7^{m\\minus{}n}\\equiv1\\mod17$?",
  "Solution_8": "[quote=\"worthawholebean\"][/quote][quote=\"CircleSquared\"][quote=\"Zellex\"][quote=\"SCHNOGGINVOGGINSCHWITZERW\"]actually, as x goes from 1 to 16, $ 7^x$ will assume the value of each residue mod 17, except 0. In this case, its easy enough to list out the sequence of residues of $ 7^x$ in mod 17, and get 7,15,3,4,11,9,12,16,10,2,14,13,6,8,5,1. The solution is $ x\\equiv 13(\\text{mod}16)$[/quote]\n\nHow do we know that for each $ x$, $ 7^x$ assume the value of each residue mod 17?[/quote]\n\nsay $ 7^m \\equiv 7^n \\mod 17$\nthen $ 7^{m - n} \\equiv 0 \\mod 17$\nclearly not true[/quote][quote=\"worthawholebean\"]\n\nIsn't that $ 7^{m - n}\\equiv1\\mod17$?[/quote]\r\n\r\nmeh, i knew something seemed wrong with my idea\r\nbut the explanation is something simple like that, i'm certain",
  "Solution_9": "It's not that simple.  \r\n\r\n$ 7$ is an example of a [url=http://en.wikipedia.org/wiki/Primitive_root_modulo_n]primitive root[/url] $ \\bmod 17$, and not all residues $ \\bmod 17$ have this property.  The powers of $ 4$, for example, only have the residues $ 1, 4, 16, 13 \\bmod 17$."
}
{
  "Tag": [
    "counting",
    "distinguishability",
    "function",
    "LaTeX"
  ],
  "Problem": "Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand.  The order of rings on each finger is significant, but it is not required that each finger have a ring.  Find the leftmost three nonzero digits of $n.$",
  "Solution_1": "[hide]First choose the rings. There are $\\binom{8}{5}=\\binom{8}{3}=\\frac{8*7*6}{3*2*1}=56$ ways to do this.\n\nThen choose the ordering of the rings and on what fingers. If 1-5 denote the rings and f denotes a change of fingers, then this is the number of permutations of 12345fff, which is given by $\\frac{8!}{3!}=8*7*6*5*4=6720$.\n\nThe number of arrangements is then $6720*56=376320\\Rightarrow \\boxed{376}$.[/hide]",
  "Solution_2": "[hide]$\\binom{8}{5}$ ways to choose the rings, $5!$ ways to order the rings, and $\\binom{8}{3}$ solutions to $a+b+c+d=5$, so the answer is the the first three digits of the product of all these.[/hide]",
  "Solution_3": "[hide]\nThen there are $\\binom{8}{5}=56$ ways to choose the rings, without considering order.\nNow, in some arrangment of the rings on the fingers, let the rings be numbered 1 through 5, where 1 is the bottom ring on the first finger, and 5 the top ring on the last finger.  There are $5!$ ways to order the five rings, and the number of ways to distribute five indistinguishable rings between four fingers is the number of nonnegative integer solutions to $a+b+c+d=5$.  There are \\[ \\binom{4}{1}+4*3+4*3+\\binom{4}{1}\\binom{3}{2}+\\binom{4}{2}\\binom{2}{1}+\\binom{4}{1}=4+12+12+12+12+4=56 \\] such solutions.  So $n=56*56*120=376320$, so our answer is $\\boxed{376}$.\n[/hide]",
  "Solution_4": "The number of solutions to the equation \\[ x_1+x_2+x_3+\\cdots+x_k = n \\] can easily be found by $\\binom{n+k-1}{k-1}$, which you can see by looking at the number of ways to place $k-1$ separators in the list $\\underbrace {\\bullet \\bullet \\cdots \\bullet}_{n \\text { bullets}}$.\r\n\r\n(Consider the case $a+b+c=10$. You must place 2 separators in ********** for each solution, and there are $\\binom{12}{2}$ ways to do this.)",
  "Solution_5": "Oh yeah...forgot about that method...\r\nHmm..but come to think about it, you can place two separators in one space...\r\nYeah...",
  "Solution_6": "[hide]every sequence of 5 rings is $8*7*6*5*4$\n\nwe can distribute these 5 items by placing barriers in between them:  _1_1_1_1_1_ \n\nthere are $\\binom{6}{4}+\\binom{6}{3}+\\binom{6}{2}+\\binom{6}{1}$ since we can have rings on 4,3,2,1 fingers\n\nso we get $(15+20+15+6)8*7*6*5*4=376320 \\implies \\boxed{376}$[/hide]",
  "Solution_7": "Then again...\r\nThe number of solutions of nonnegative integers for $a+b+c+d=5$ is the coefficient of $x^5$ in $(1+x+x^2+x^3+...)^4=(1+2x+3x^2+4x^3+...)^2$ which means the $x^5$ coefficient is $2(1*6+2*5+3*4)=56$.\r\nBut the number of ways to put 3 dividers between 5 objects is $6^3$, if more than one divider can go in the same place, and each possibility corresponds to exactly one triple...",
  "Solution_8": "How could more than one go in the same place? That doesn't make sense.",
  "Solution_9": "[quote=\"chess64\"]How could more than one go in the same place? That doesn't make sense.[/quote]\r\n\r\nIf one of the numbers is $0$...",
  "Solution_10": "It would just be ||, with nothing in the middle. ;)",
  "Solution_11": "[quote=\"chess64\"]It would just be ||, with nothing in the middle. ;)[/quote]\r\n\r\nYeah...\r\n\r\nLike we have 5 objects and 3 dividers\r\n\r\n* * * * *\r\n\r\nWe could do something like\r\n\r\n* * ||| * * *\r\n\r\nwhich gives $(2, 0, 0, 3)$\r\n\r\nBut yet we know $6^3$ is not the right answer...is there something being repeated?\r\n\r\nEDIT: Okay, yeah i found it.  Each possibility is repeated a number of times, because the positions of two dividers can be switched, and it's still the same combination.  Thus, anything with the three dividers in different places is repeated $6$ times; any with exactly two in the same place is repeated $3$ times, and any where they're all in the same place is not repeated.  Thus, our final answer is $\\frac{1}{6}(6*5*4)+\\frac{1}{3}(3*6*1*5)+6$.  The first term is the number of ways to have all of the dividers in different places, the second the number of ways to have  exactly two in the same place, and the third the number of ways to have all three in the same place.  The answer comes out to $56$.",
  "Solution_12": "Why is the case * * ||| * * * any different from *|*|*|**, for example? The calculation $\\binom{8}{3}=56$ will still give us the correct answer. ;)",
  "Solution_13": "[quote=\"chess64\"]Why is the case * * ||| * * * any different from *|*|*|**, for example? The calculation $\\binom{8}{3}=56$ will still give us the correct answer. ;)[/quote]\r\n\r\n* | * | * | ** is (1, 1, 1, 2) while * * ||| * * * is (2, 0, 0, 3).  The first number is the number of stars to the left of the leftmost divider, the second the number of starts between dividers 1 and 2, and so on.\r\n\r\nAnd yeah, now I see why $\\binom{8}{3}$ works...8 total items, considering dividers and stars, and 3 ways to choose which ones are dividers.  I was thinking $\\binom{6}{3}$ at first, and that's where the having two dividers in one place came up.",
  "Solution_14": "[quote=\"K81o7\"][quote=\"chess64\"]Why is the case * * ||| * * * any different from *|*|*|**, for example? The calculation $\\binom{8}{3}=56$ will still give us the correct answer. ;)[/quote]\n\n* | * | * | ** is (1, 1, 1, 2) while * * ||| * * * is (2, 0, 0, 3).  The first number is the number of stars to the left of the leftmost divider, the second the number of starts between dividers 1 and 2, and so on.[/quote]\r\n\r\nNo, no, I know what it means, (after all, i'm the one who explained it to you :D) but I was asking why you would need different cases for them. :)",
  "Solution_15": "[quote=\"K81o7\"]Then again...\nThe number of solutions of nonnegative integers for $a+b+c+d=5$ is the coefficient of $x^5$ in $(1+x+x^2+x^3+...)^4=(1+2x+3x^2+4x^3+...)^2$ which means the $x^5$ coefficient is $2(1*6+2*5+3*4)=56$.\nBut the number of ways to put 3 dividers between 5 objects is $6^3$, if more than one divider can go in the same place, and each possibility corresponds to exactly one triple...[/quote]\r\n\r\nOr instead of the generating function, you can just use $n+r-1Cr-1$ which will give you the same answer\r\n\r\nHow do you do the choose bracket thing for combinations in the latex?\r\n\r\nEDIT: $\\binom{n+r-1}{r-1}$",
  "Solution_16": "\\binom{n}{k} $\\Rightarrow \\binom{n}{k}$",
  "Solution_17": "[quote=\"chess64\"][quote=\"K81o7\"][quote=\"chess64\"]Why is the case * * ||| * * * any different from *|*|*|**, for example? The calculation $\\binom{8}{3}=56$ will still give us the correct answer. ;)[/quote]\n\n* | * | * | ** is (1, 1, 1, 2) while * * ||| * * * is (2, 0, 0, 3).  The first number is the number of stars to the left of the leftmost divider, the second the number of starts between dividers 1 and 2, and so on.[/quote]\n\nNo, no, I know what it means, (after all, i'm the one who explained it to you :D) but I was asking why you would need different cases for them. :)[/quote]\r\n\r\nOh.\r\n\r\nYeah, I've used this method before...it just never occurred to me to use it here...",
  "Solution_18": "[hide]\nWe have $\\binom{8}{5}$ ways to choose the rings. $\\binom{8}{3}$ ways to split # of rings among fingers. $5!$ ways for the colors to be there.\n$\\binom{8}{5}\\binom{8}{3}5!=376320$.\nAnswer:$\\boxed{376}$.[/hide]",
  "Solution_19": "[hide]There are $\\binom{8}{5}$ ways to choose the rings, $\\binom{8}{3}$ ways to distribute the rings among the fingers, and $5!$ distinct color arrangements. \n\nMultiplying gives the answer: $\\binom{8}{5}\\binom{8}{3}5!=376320$, and the three leftmost digits are $\\boxed{376}$.[/hide]",
  "Solution_20": "Sorry to revive.\n\n[quote=\"AoPSWiki-Balls-and-Urns\"]It is used to solve problems of the form: how many ways can one distribute k indistinguishable objects into n bins?.[/quote]\n\nThe problem says that the rings are distinguishable, so why can we use balls and urns here?",
  "Solution_21": "Bump.....?",
  "Solution_22": "[quote=\"4everwise\"][hide]There are $\\binom{8}{5}$ ways to choose the rings, $\\binom{8}{3}$ ways to distribute the rings among the fingers, and $5!$ distinct color arrangements. \n\nMultiplying gives the answer: $\\binom{8}{5}\\binom{8}{3}5!=376320$, and the three leftmost digits are $\\boxed{376}$.[/hide][/quote]\nWhat do you mean by color arrangements?  What are we arranging differently, and how do we know we can do this when the rings are all on different fingers?",
  "Solution_23": "Sorry for the revive; but it seems that FlyingWombat's question has not been answered. You can use balls and urns here because, once you choose the number of ways to distribute the rings, there are $5!$ ways to arrange the rings because there are $5$ fixed positions possible for each ring to occupy. Hopefully this clarifies.",
  "Solution_24": "First we find the number of ways to pick the rings; this is just $\\binom{8}{5}=56$. Now we find the amount of ways to put the rings on $4$ fingers for a total of $5$ rings. By Stars & Bars, this is $\\binom{5+4-1}{4-1}=56$. The amount of ways to arrange the five rings is $5!=120$. Multiplying gives $376320 \\implies \\boxed{376}$.",
  "Solution_25": "Hey... so why wouldn't [hide=this method]There are $\\binom{8}{5}$ ways to choose the five rings we are going to use, and since each ring can go on one of $4$ fingers, we multiply this by $4^5$ to get $57344$, or $573$ as the final answer.[/hide] work in this case? Is it because of the fact that the fingers themselves are not distinguishable, since I think I assumed that they were.",
  "Solution_26": "[quote=vsar0406]Hey... so why wouldn't [hide=this method]There are $\\binom{8}{5}$ ways to choose the five rings we are going to use, and since each ring can go on one of $4$ fingers, we multiply this by $4^5$ to get $57344$, or $573$ as the final answer.[/hide] work in this case? Is it because of the fact that the fingers themselves are not distinguishable, since I think I assumed that they were.[/quote]\n\nThey are distinguishable.\nIf the order of the rings on one finger didn\u2019t matter, then your answer would be correct.\n",
  "Solution_27": "Oh, I see... Even the distinguishable-distinguishable method doesn't account for the order of the items in each container, so we have to do a modified stars and bars approach to find the answer. Cool problem :)",
  "Solution_28": "You first chose which $5$ of the $8$ rings you want, which we can do in $\\binom{8}{5}$ ways. Then we order them, which we can do in $5!$ ways. Lastly, we distribute the rings to the fingers, and this is essentially stars and bars so there are $\\binom{8}{3}$ ways. Hence, the answer is the first three digits of $\\binom{8}{5} \\cdot 5! \\cdot \\binom{8}{3} = 376320,$ which is $\\boxed{376}.$"
}
{
  "Tag": [],
  "Problem": "For how many values of $ 3^{n}\\plus{}81$ the square of an integer",
  "Solution_1": "[quote=\"firecricket91\"]For how many values of $ 3^{n} \\plus{} 81$ the square of an integer[/quote]\r\n\r\nI think its one.  ($ 3^4\\plus{}81\\equal{}18^2$)",
  "Solution_2": "$ 3^4\\plus{}81\\equal{}162\\neq 18^2$...",
  "Solution_3": "None. You can prime factorize the equation to $ 3^n\\plus{}3^4$ This cannot be factorized any further.",
  "Solution_4": "[quote=\"hunter34\"]None. You can prime factorize the equation to $ 3^n \\plus{} 3^4$ This cannot be factorized any further.[/quote]\r\n\r\n3^5+81=324 which is 18^2",
  "Solution_5": "... well\r\n\r\n$ 3^n\\plus{}3^4\\equal{}3^4(3^{n\\minus{}4}\\plus{}1)$\r\n\r\nFor this to be a perfect square $ 3^{n\\minus{}4}\\plus{}1$ must be a perfect square.\r\n\r\n$ 3^{n\\minus{}4}\\equal{}k^2\\minus{}1$\r\n$ 3^{n\\minus{}4}\\equal{}(k\\minus{}1)(k\\plus{}1)$\r\nThis implies that both $ (k\\minus{}1)$ and $ (k\\plus{}1)$ have only factors of three.  The only possible value is $ k\\equal{}2$\r\n\r\nThus, the future's answer is the only one possible."
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "Given the polynomial identity\r\n\r\n\\[x^6+1 = (x^2+1)(x^2+ax+1)(x^2+bx+1)\\],\r\n\r\nwhat is the value of $ab$?\r\n\r\nAs usual, if you're beyond this level, leave this to other beginning problem solvers.",
  "Solution_1": "[hide=\"solution\"]\nI would start by just expanding it and finding values of $a$ and $b$ so that the coefficients of $x^5, x^4, x^3, x^2,\\text{ and }x$ are $0$.\n\nWhen we expand it we get:\n\n$x^6 + (a+b)x^5 + (ab+3)x^4 + (2a+2b)x^3 + (ab+3)x^2 + (a+b)x + 1$\n\nso we have the equations:\n\n$a + b = 0$\nand\n$ab = -3$\n\nSince we want to find $ab$, we are done. The answer is $\\boxed{-3}$. However, should we want to find $a$ and $b$, we would solve the two equations to get that one of them is $\\sqrt 3$ and the other is $-\\sqrt 3$\n\n[/hide]"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "I would like to know  how  can i  write in  columns, like in  excalibur magazine. But  not  in all pages  of same document,  \r\njust  few of them  to  be at columns. Thank you very much!",
  "Solution_1": "you should find what you are looking for here:\r\n\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59948&sid=52c0b3832b9d85594c6a3feed5cc2bfe]www.artofproblemsolving.com/Forum/viewtopic.php?t=59948&sid=52c0b3832b9d85594c6a3feed5cc2bfe[/url]"
}
{
  "Tag": [
    "calculus",
    "search",
    "function",
    "calculus computations"
  ],
  "Problem": "any interesting problems on continuity and differentiability please!",
  "Solution_1": "There are nearly 20000 threads in the calculus section. Look at some of them. Use the search function to narrow it down a little.",
  "Solution_2": "Prove that every continuous open mapping from $ R$ to $ R$ is monotonic."
}
{
  "Tag": [
    "group theory",
    "number theory",
    "relatively prime",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ G$ be a finite cyclic group of order $ n$. Let $ a$ be a generator. Let $ r$ be a nonzero integer and relatively prime to $ n$.\r\n\r\n(i) Show that $ a^r$ is also a generator of $ G$.\r\n(ii) Show that every generator of $ G$ can be written in this form.\r\n(iii) Let $ p$ be a prime number, and $ G$ a cyclic group of order $ p$. How many generators does $ G$ have?",
  "Solution_1": "i) Use Bezout's theorem, i.e. that for coprime $ a,b$ there are integers $ x,y$ such that $ ax\\plus{}by\\equal{}1$.\r\n\r\nii) Every element of $ g$ is of type $ a^r$ for some integer $ r$. You have to show that if $ a^r$ is a generator, than $ n$ and $ r$ are coprime. For this, how many different powers of $ a^r$ are there if $ n,r$ have a common divisor $ d>1$.\r\n\r\niii) You have shown that generators come from integers $ r$ coprime to $ n$. Which $ r$ give the same generators\u00bf",
  "Solution_2": "alternative for (ii): write $ (a^r)^k \\equal{} a$ and use $ n \\equal{} ord(a)$."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "MATHCOUNTS"
  ],
  "Problem": "Find the ratio of the area of an equilateral triangle inscribed in a circle to the area of a square circumscribed about the same circle.",
  "Solution_1": "I don't think the area of a equilateral triangle inscribed in a circle belongs in the Basics. I'll move it to the Mathcounts forum.",
  "Solution_2": "[hide=\"Answer\"]Drawing three radii from the center of the circle to the vertices of the triangle, and the perpendicular bisectors to each side of the triangle, we have six 30-60-90 triangles, each with hypotenuse $r$. Therefore, the area of the inscribed equilateral triangle is $\\frac{3r^2\\sqrt{3}}{4}$. The area of the circumscribed square is $4r^2$, so the ratio of the two is $\\frac{\\frac{3r^2\\sqrt{3}}{4}}{4r^2}=\\boxed{\\frac{3\\sqrt{3}}{16}}$.[/hide]",
  "Solution_3": "thats right"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "[i]Let a triangle $ABC$ , the median $BD$ , the bisector of $\\angle BDA$ $DE$ , and the bisector of  $\\angle BDC$ $DF$. Proove that $EF //AC$[/i]\r\n\r\n[img]http://img50.imageshack.us/img50/9613/fasfaszq0.png[/img]",
  "Solution_1": "[hide=\"hint\"]angle bisector theorem?[/hide]",
  "Solution_2": "[hide]Hint:  angle EDF is a right angle... [/hide]"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "It is possible to have a negative pH?????",
  "Solution_1": "yes! consider for example a solution consisting of 1 dm^3  with 2 moles of HCl. This gives $pH=-lg[H]=-lg 2 = -0.301...$",
  "Solution_2": "it is possible, just not generally available... also, u could have pH >14 (guess that would be too little $H^+$",
  "Solution_3": "Well, 6M HCl is a common concentration; it has pH=-0.778",
  "Solution_4": "[quote=\"Mariag\"]it is possible, just not generally available... also, u could have pH >14 (guess that would be too little $H^+$[/quote]\r\nin what conditions can you have pH>14?",
  "Solution_5": "[quote] in what conditions can you have pH>14?[/quote]\r\nwhen $[H^+] < 10^{-14}$",
  "Solution_6": "Well, I worked with 6M KOH at camp one year (not USNCO camp; not yet at least).  That has pH=14.778",
  "Solution_7": "Once I used 19M NaOH in class.  That breaks 15!\r\n\r\n[quote=\"Mariag\"]it is possible, just not generally available... also, u could have pH >14 (guess that would be too little $H^+$[/quote]\r\n\r\n8+ M HCl is quite easy to get at hardware stores.  It's cheap, too.",
  "Solution_8": "At USNCO camp, they prepared a 125M solution of KOH.  That breaks 16, though not 16!  (this is a math forum, remember...  :P  )",
  "Solution_9": "[quote=\"inteluser\"]At USNCO camp, they prepared a 125M solution of KOH.  That breaks 16, though not 16!  (this is a math forum, remember...  :P  )[/quote]\r\n\r\nOops, sorry.\r\n\r\nI want to go to USNCO camp!  But I don't think I did so well on the test :(.",
  "Solution_10": "i guess i don't have enough experience with crazy concentrated chemicals (i don't recall ever working with any thing with molarity more than 3)  :( \r\ni also hope i was able to do better on the USNCO test, camp sounds like so much fun!",
  "Solution_11": "Things are a little bit more complicated.\r\n\r\nMeasured pH is not -log of H+ concentration, but of H+ activity. The more concentrated solution the higher its ionic strength and the lower activity coefficient - this is described by so called Debye-Huckel theory. And it is not the problem of extremally concentrated solutions - in fact for concentrations above 0.1M even DH theory is not able to properly predict activities. 0.1M HCl should have pH = 1.00, but the result of measurements will be 1.12 - the activity coefficient is 0.75.\r\n\r\nThis all boils down to the fact that having high concentration of acid (or base) doesn't mean you will have a negative pH or pH higher than 14 - due to low activity coefficients. IIRC they are possible and can be measured, but for much more concentrated solutions that one may expect.\r\n\r\nBest,\r\nBorek",
  "Solution_12": "What is the difference between pH and activity? Thanks. :blush:",
  "Solution_13": "http://www.chembuddy.com?left=pH-calculation&right=ionic-strength-activity-coefficients",
  "Solution_14": "u have to meet KAWKABI ,if u do u will be persuaded :rotfl:",
  "Solution_15": "The bottom line is that the activity measures amazingly enough the activity or apparent concentration.  This is what is measured by a pH meter, or pH paper.  The concentration is important if one is reacting a solution to a given endpoint, such as a pH of 7.  Activity is important is one is concerned about strength of the acid/base (will it eat my container?).\r\n\r\nMost every \"law\" or rule that applies math to the real world doesn't work at extreme conditions very well.  These are those conditions when the math would say one has a negative pH.  Even if one got there, it isn't really important.  The only reason to use the log function was to avoid using exponents in talking about acid/base situations.\r\n\r\nIt's just as critical when the temperature goes negative.  (The then retrograde, now Celsius scale was set using the then known highest and lowest temperatures on earth [knowledge was a bit narrower than it now is].)  Negative simply means that the original folks missed the mark.  Kelvin and Rankin (not spelled correctly) start at absolute zero and have been slid fractions of degrees down, as the initial folks didn't know as much as they thought.)\r\n\r\nEven the calendar is wrong, Christ was born a bit off.  It really doesn't matter for anything, but they missed.  The meter was based on a French attempt to measure the distance from the equator to the north pole (or the other way around), which was way off.  As long as we agree and can talk, it really doesn't matter what the units are, or what they are based upon.",
  "Solution_16": "[quote=\"inteluser\"]At USNCO camp, they prepared a 125M solution of KOH.  That breaks 16, though not 16!  (this is a math forum, remember...  :P  )[/quote]\r\n\r\nThey didn't. Solubility of KOH is lower. Even pure melted KOH (if treated as its own solution) will have much lower concentration.",
  "Solution_17": "[quote=\"annornor\"]It is possible to have a negative pH?????[/quote]\r\n\r\npH= -log[H+] then ph=f(x) where x=[H+] and f(x)=-logx\r\nDf=R  :)"
}
{
  "Tag": [
    "geometry",
    "Support",
    "inequalities",
    "articles"
  ],
  "Problem": "I remember my fifth grade days when I was in the Cottage, and Jamal would hang around the windows trying to see homeless people. On our field trip to Philadelphia, they saw a homeless person begging for money when he had a huge wallet full of $ 100$ dollar bills, and a leather jacket. (Our group saw plenty of homeless people needing money as well!) Is it right to laugh at them? Is it right to laugh at them privately, in your head or somewhere far away from them? Is it right to walk right past them without giving them any money? It doesn't seem right.\r\n\r\nI was homeless for about 5 hours when the extermination people came to our house and we couldn't stay there. We went to Wegmans because it was around lunchtime and we were hungry, when there was plenty of food in our kitchen that we could eat. It seemed kind of sad and I decided that I shouldn't at least laugh with my friends at a funny looking vagrant.\r\n\r\nWhat do you think?",
  "Solution_1": "My aunt is the director of Home Aid in Atlanta and that program helps homeless people a lot. I do not think you should laugh at them because 90% of people are born into homelessness and maybe don't know the skills to help themselves. The average age of all homeless people in the U.S. is 9 years old, which I find very sad. I find homelessness to be one of the biggest issues in the U.S. today.",
  "Solution_2": "Wait, 90% of all US born kids are homeless? Perhaps this is a world statistic, and not just US.",
  "Solution_3": "No, 90% of all current homeless people were born homeless.",
  "Solution_4": "9 years old? Wow. \r\n\r\nI'd rather really be struggling on the streets, because my friend said she overheard a classmate say, \"Mom, can we actually have dinner today?\". I'd rather have food in my stomach than a huge apartment where you have some comfort for sleep, that's all. \r\n\r\nThe problem with homelessness is because of rising home prices, which make people more welcoming to apartments, and the apartment prices rising too. They should make a huge mansion (the kind football players have :no:) and make so many huge apartment- sized rooms for some homeless people.",
  "Solution_5": "As part of the organization my aunt directs, they help establish homeless shelters in different communities. I think they are headed in the right direction.",
  "Solution_6": "That is definitely a World Statistic, not a U.S. one. In America, homeless kids go to orphanages or foster care and hence are no longer homeless.",
  "Solution_7": "If it was a world statistic, Africa takes a large chunk of the percentage  :read:",
  "Solution_8": "They aren't homeless so much as tribal at that point. You need a certain amount of development in a country before you can call its people without houses homeless\r\n\r\nI'm thinking India.",
  "Solution_9": "[quote=\"dynamo729\"]... 90% of people are born into homelessness ... The average age of all homeless people in the U.S. is 9 years old[/quote]  Neither of these claims sounds very plausible to me -- can you cite any source for either of them?\n\n[quote=\"demonshadow\"]The problem with homelessness is because of rising home prices, which make people more welcoming to apartments, and the apartment prices rising too.[/quote]  I don't think this is really true: the lack of affordable housing almost everywhere is a serious problem, but it's not a new problem tied to rising home prices.  In some places rising rental costs could be partially to blame, but purchasing a home has always required either a steady income flow or a large amount of savings or both, and homelessness is contraindicative of both.  I would guess that financial instability leading to temporary inability to pay rent is probably a frequent cause of short-term homelessness.  (Long-term homelessness is a different beast: it's often linked to mental illness, for example.)\n\n[quote=\"Anaxerzia\"]In America, homeless kids go to orphanages or foster care[/quote] I doubt that this is true in any sort of generality.\r\n\r\n(Edit: just fixed an errant quote tag.)",
  "Solution_10": "My source is my aunt (not biological, but her husband and my dad have known each other from birth and we do a lot of stuff together and just call them aunt and uncle. My uncle was my dads best man and is my brothers godfather), who is the director of the Home Aid, a program in Atlanta (it may be other places, but she is the director in Atlanta) thats helps build shelters and stuff.  She has been in this program for years; may I ask why you find them not very plausible?",
  "Solution_11": "I find the first not very plausible because significant portions of the homeless population fall into the following two categories:\r\n\r\n1) people who are temporarily homeless, for example because of job loss -- this includes a lot of families with one or multiple children, for example.  Hardly any of these people are born into homelessness, and as individuals they don't stay homeless for very long periods of time (a few months or a year).\r\n\r\n2) people who suffer from serious mental illness (including many veterans, for example) make up a lot of the \"visible\" homeless, and I have no reason to think any significant fraction of them are born into homelessness because in this case there's a different identifiable cause.\r\n\r\n90% is a very large number -- it just seems unlikely to me that the two groups I've named are such a small portion of the homeless population.\r\n\r\nFor the second one, I don't so much doubt it as not understand what it means.  But, I find the following information on the [url=http://www.homeaid.org/index.cfm?tdc=dsp&page=homeless_facts]Facts & Statistics[/url] page of the [url=http://www.homeaid.org/]HomeAid[/url] website:\r\n[quote=\"HomeAid website\"]Officials estimate that, on average, single men comprise 51 percent of the homeless population, families with children 30 percent, single women 17 percent, and unaccompanied youth 2 percent.[/quote]\r\n\r\n\r\nBy the way, HomeAid sounds like a great organization for helping the temporarily homeless.  So, kudos to your aunt for working for such a good cause.",
  "Solution_12": "Hm..... either my aunt got it wrong or I misunderstood (the latter most likely)!  Thanks for the clarification!",
  "Solution_13": "[quote=\"D3m0n Shad0w\"]I'd rather really be struggling on the streets, because my friend said she overheard a classmate say, \"Mom, can we actually have dinner today?\". I'd rather have food in my stomach than a huge apartment where you have some comfort for sleep, that's all.[/quote]\r\n\r\n\r\nWait, what? Are you saying that you'd rather be poor than rich?",
  "Solution_14": "er I think he's saying that he'd rather have a full stomach than a comfortable shelter. idk..\r\n\r\nIf they were asking for some, I'd give them money. It'll at least get them a meal or two, though it would take a whole lot of effort to completely change someone's life. Of course, there are lots of programs that are trying to accomplish just that (i.e. [url=http://www.compassion.com/default.htm]Compassion[/url])",
  "Solution_15": "[quote=\"sailingdog\"][quote=\"JBL\"][quote=\"sailingdog\"]or else the gap between rich and poor would be continually widening.[/quote]  I come across a statement like this and I think, \"What rock are you living under?\"  Income inequality in this country [i]is[/i] growing, and has been for years.[/quote]  \nXD\nI know that. It would accelerate, though.\nBesides, I am still 13 - not really old enough to experience this first hand...[/quote]\r\nsailingdog, you're right. that's why we gotta learn how to deal with these kind of things NOW [practicing, kinda], as what...say 11-year-olds (like me, except im 12 in a few days) before moving on to the teen years. that way we have the knowledge about these situations when we face them in real life.",
  "Solution_16": "Did BOTH of you just completely miss that a growing income inequality is [i]equivalent[/i] to a widening gap between the rich and the poor?\r\n\r\nAnd more than 50% tax on the rich? Do you realize how high that is already?\r\n\r\nSocialism to that degree doesn't work at all. When those who have highly skilled professions earn only a little bit more than those who anybody can do their job, then there is far less incentive to get highly skilled professions. Look at a country like France and its unemployment compared with the U.S.. Even now, many can argue that in America's ghettos people are gaming welfare and having children as teenagers specifically so that they can live solely of off tax dollars. My mom (an OBGYN working in this area) sees this everyday, and this is a TINY community we are talking about. \r\n\r\nGranted, I think most welfare can do good, but just needs more regulation so that things like this do not happen.\r\n\r\nAlso, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.\r\n\r\nI think the best way to get the homeless to work is public infrastructure. America's infrastructure is crumbling, and cheap labor is needed. The homeless should be hungry for such jobs.",
  "Solution_17": "[quote=\"Anaxerzia\"]\nAnd more than 50% tax on the rich? Do you realize how high that is already?\n\nSocialism to that degree doesn't work at all. When those who have highly skilled professions earn only a little bit more than those who anybody can do their job, then there is far less incentive to get highly skilled professions. \n[/quote]\n\nMost people will prefer to get highly skilled professions even without money as an incentive because \"low-level\" jobs are usually pretty boring. \n\nEven if I make less money as a mathematician, I will still prefer to be a mathematician than a worker at MacDonalds.\n\n[quote=\"Anaxerzia\"]\n\nAlso, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.\n[/quote]\r\n\r\n This is because America is very \"socialist\" comparing to many other countries in the world.",
  "Solution_18": "Not all competitive fields are academic. Also, the poor will have less incentive to peform in school if they shall have a decent life subsidies by the government no matter what.",
  "Solution_19": "[quote=\"Anaxerzia\"]And more than 50% tax on the rich? Do you realize how high that is already?\n\nSocialism to that degree doesn't work at all.[/quote]\nFor most of the post-WW2 era, the US top marginal tax rate has been higher than 50% (and, in fact, higher than 70%).  This happened to coincide with some of the most impressive economic growth anywhere ever, growth that did not happen at the expense of increasing income inequality.\n\n(An x% marginal rate means that for people making more than a certain amount, x% of the [i]additional[/i] amount was taxed.  So, for example, our current top marginal rate is 35% and applies only to those making about 350,000 USD annually.  In order to actually owe one third of your income in income taxes, you'd have to make more than 1.2 million dollars.  These numbers are for a single individual -- other marital statuses would give somewhat different numbers.)\n\n[quote=\"Anaxerzia\"]Look at a country like France and its unemployment compared with the U.S.[/quote]  What about it?  The French have made a societal decision to work shorter hours and be a more generous society than we have.  The result is that they're equally productive on a per-hour-worked basis, have a generally powerful economy, and have vastly lower inequality (as measured by the GINI index) than the US does.  (I also remember vaguely that Europe may compute its unemployment rate differently than we do, but I wasn't able to find a citation for that.)\n\n[quote=\"Anaxerzia\"]Even now, many can argue that in America's ghettos people are gaming welfare and having children as teenagers specifically so that they can live solely of off tax dollars.[/quote]  Yes, one could argue that, but then one would be either an idiot or someone intentionally trying to stir up racial resentment.\n\n[quote=\"Anaxerzia\"]Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.[/quote]  Um, no:\n[quote=\"[url=http://en.wikipedia.org/wiki/Economic_mobility#Economic_mobility_worldwide]Wikipedia on economic mobility[/url]\"]Using the ratio of an individual\u2019s current income to that of their parent\u2019s, the United States has much less relative mobility than other industrialized nations. The income of our parents is a great deal more predictive of our own incomes in the United States than other countries. France, Germany, Sweden, Canada, Finland, Norway, and Denmark all have more relative mobility than the US, while only the United Kingdom is shown to have less mobility. According to this study done by Miles Corak, The United States ratio of relative mobility is 1, whereas the other countries mentioned with more mobility have a range of 1.25 (France) to over 3 (Denmark).[/quote]",
  "Solution_20": "[quote=\"Anaxerzia\"]\nAlso, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.[/quote]Woah, hold on. I'm confused: are you saying that it's a bad thing?",
  "Solution_21": "[quote=\"JBL\"]\n\n[quote=\"Anaxerzia\"]Also, people argue that it is not fair. This is not true. America has one of the least biased systems in the world when it comes to this because at birth, ANY poor kid has the opportunity 1) educate himself with public school 2) work hard and go to a decent state college, most likely for free 3) get a job. The vast majority of kids in America have this opportunity, and to suggest that it is not fair for them is not necessarily true.[/quote]  Um, no:\n[quote=\"[url=http://en.wikipedia.org/wiki/Economic_mobility#Economic_mobility_worldwide]Wikipedia on economic mobility[/url]\"]Using the ratio of an individual\u2019s current income to that of their parent\u2019s, the United States has much less relative mobility than other industrialized nations. The income of our parents is a great deal more predictive of our own incomes in the United States than other countries. France, Germany, Sweden, Canada, Finland, Norway, and Denmark all have more relative mobility than the US, while only the United Kingdom is shown to have less mobility. According to this study done by Miles Corak, The United States ratio of relative mobility is 1, whereas the other countries mentioned with more mobility have a range of 1.25 (France) to over 3 (Denmark).[/quote][/quote]\r\n\r\nSo what? The way it is matters zilch. Anaxerzia is talking about the opportunity kids have, you are discounting his argument with facts that show what kids end up doing. Just because a kid doesn't take his life as far as a kid in a similar situation in another country, doesn't mean he had less of an opportunity to be successful growing up.",
  "Solution_22": "i see. thanks!!\r\n\r\n**this is a little off topic, but do posts in the round table count towards your overall post score?",
  "Solution_23": "Shouldn't be too hard to figure out...\r\n\r\nAlso, I found an article some of you might like:\r\n\r\n[url=http://www.nytimes.com/2008/05/30/opinion/30jacoby.html?_r=1&ex=1212811200&en=cc94734646d73502&ei=5070&emc=eta1&oref=slogin]Shortened by moderator.[/url]",
  "Solution_24": "sorry i just figured it out.",
  "Solution_25": "[quote=\"sailingdog\"]So what? The way it is matters zilch. Anaxerzia is talking about the opportunity kids have, you are discounting his argument with facts that show what kids end up doing. Just because a kid doesn't take his life as far as a kid in a similar situation in another country, doesn't mean he had less of an opportunity to be successful growing up.[/quote]\r\n\r\nSo, in other words, America is a great place because we have so much opportunity, but Americans are so hugely incompetent that we don't take advantage of this?  This is what I would describe as \"highly contrived.\"  A more natural explanation would be that people everywhere take roughly equal advantage of their opportunities, and there are simply fewer opportunities available in this country than elsewhere.  This is also consistent with the substantially larger levels of economic inequality here than in the rest of the developed world.",
  "Solution_26": "Ohh... Who is John Galt?\r\n\r\nAre you suggesting that we give away on need instead of ability? That we should sacrifice our own livelihoods to someone who has given up his long ago? That we should pour our minds into alms meant for the people that have lost the capacity to use theirs? \"From each according to his ability, to each according to his need?\" To spend our cash on someone whom we have no vested interest in only breeds resentment. It is much better to finance something that can bring others up, such as a university, than to aimlessly funnel funds into someone's need. If I give to a charity, I want to convince myself that I am giving because my money may help carry someone forward, to my own stature, not that they will become dependent on my handouts, like a parasite. Call it a Horatio Alger wet dream, but if everyone worked for their own self-interest, rather than in the interest of others, then everyone would fall where their ambition placed them.\r\n\r\n<3 rhetorical questions aimed at no one.",
  "Solution_27": "[quote=\"stuck\"]if everyone worked for their own self-interest, rather than in the interest of others, then everyone would fall where their ambition placed them.[/quote]  Of course.  After all, only the unworthy get hit by automobiles.  Only the lazy get cancer.  And it's well-documented that the only victims of discrimination and bad luck are those who deserve it.  \r\n\r\nMeanwhile, those of us who live in the real world seek outcomes that will reduce human suffering, rather than self-serving explanations for why we shouldn't be bothered by it.\r\n\r\nIncidentally, increasing income inequality in this country has little to do with availability of higher education and much more to do with structural changes (e.g. to the tax code) that significantly favor the wealthy.  See, for example, http://www.epi.org/content.cfm/ib232",
  "Solution_28": "there are people who are just lazy and don't work\r\ni guess those are the ppl who you can despise",
  "Solution_29": "[quote=\"beta\"]I agree with calc rulz. Does anyone really NEED >$ \\$$500K per year?[/quote]\r\n\r\nYes."
}
{
  "Tag": [
    "trigonometry",
    "complex numbers",
    "geometric series"
  ],
  "Problem": "Compute the smallest positive angle $ x$ such that $ sin x+sin 3x+...+sin 99x = csc x$,\r\nexpressing your answer in radians.",
  "Solution_1": "[quote=\"mdk\"]Compute the smallest positive angle $ x$ such that $ sin x+sin 3x+...+sin 99x = csc x$,\nexpressing your answer in radians.[/quote]\r\n\r\n[hide=\"Hint\"]Complex numbers, sum of geometric series[/hide]",
  "Solution_2": "[hide=\"hint\"]\nU can show even without compex numbers that $ \\sum_{k=1}^{n}\\cos k\\phi=\\frac{\\sin (n+\\frac{1}{2})\\phi}{2\\sin\\frac{\\phi}{2}}-\\frac{1}{2}$[/hide]",
  "Solution_3": "[hide=\"Solution\"]\nMultiplying the equality with $ 2\\sin x$ we have\n$ 2\\sin^{2}x+2\\sin x\\sin 3x+\\ldots+2\\sin x\\sin 99x=2$\n$ 1-\\cos 2x+\\cos 2x-\\cos 4x+\\ldots+\\cos 98x-\\cos 100x=2$\n$ \\cos 100x=-1\\Rightarrow x=\\frac{(2k+1)\\pi}{100},k\\in\\mathbf{Z}$\nFor $ k=0\\Rightarrow x=\\frac{\\pi}{100}$\n[/hide]"
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "When three fair 6-sided dice are tossed, what is the probability of tossing at least one 6? Express your answer as a common fraction.",
  "Solution_1": "[quote=\"GameBot\"]When three fair 6-sided dice are tossed, what is the probability of tossing at least one 6? Express your answer as a common fraction.[/quote]\r\n\r\nThe probability that we get ZERO 6's is $ \\frac{5}{6}^3 \\equal{} \\frac{125}{216}$. Thus our answer is $ 1 \\minus{} \\frac{125}{216} \\equal{} \\frac{91}{216}$.",
  "Solution_2": "Or if you were me and decided to add up all the probabilities:\n\\[3C1\\left(\\frac{1}{6}\\right)\\left(\\frac{5}{6}\\right)^{2}+3C2\\left(\\frac{1}{6}\\right)^{2}\\left(\\frac{5}{6}\\right)+3C3\\left(\\frac{1}{6}\\right)^{3}\\]\n\\[\\boxed{\\frac{91}{216}}\\]"
}
{
  "Tag": [
    "inequalities",
    "absolute value",
    "inequalities proposed"
  ],
  "Problem": "The problem stated:\r\n\r\nLet $ a,b,c$ be real numbers with $ a^2\\plus{}b^2\\plus{}c^2\\equal{}2.$ Prove that\r\n$ |a^3\\plus{}b^3\\plus{}c^3\\minus{}abc| \\le 2\\sqrt{2}.$\r\n\r\nI wait for the nicest solution.",
  "Solution_1": "Re-homogenizing we see it follows from\r\n\\[ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 1\\implies a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} abc\\le 1\\]\r\nNow observe that\r\n\\[ (a^2 \\plus{} b^2 \\plus{} c^2)^3 \\minus{} (a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} abc)^2 \\equal{}\\]\r\n\r\n\\[ \\equal{} \\sum_{cyc}a^4(b \\plus{} c)^2 \\plus{} 5a^2b^2c^2 \\plus{} \\sum_{cyc}a^2b^2(a \\minus{} b)^2 \\plus{} \\sum_{sym}a^4b^2\\geq 0\\]\r\n:P",
  "Solution_2": "[url=http://www.mathlinks.ro/viewtopic.php?t=56197&search_id=1410260003]here[/url]",
  "Solution_3": "Dear friends, thanks for reply the topic. Now, I will post my solutions, I hope you like them\r\n\r\nFirst way: We apply the Cauchy Schwarz Inequality and the AM-GM Inequality, respectively, so that\r\n$ \\begin{aligned} (a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} abc)^2 & \\equal{} [a(a^2 \\minus{} bc) \\plus{} b^3 \\plus{} c^3]^2 \\le (a^2 \\plus{} b^2 \\plus{} c^2)[(a^2 \\minus{} bc)^2 \\plus{} b^4 \\plus{} c^4] \\\\\r\n& \\equal{} 2(a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} b^2c^2 \\minus{} 2a^2bc)\\le 2[a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} b^2c^2 \\plus{} a^2(b^2 \\plus{} c^2)] \\le 8,\\end{aligned}$\r\nwhere the last inequality holds because\r\n$ a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} b^2c^2 \\plus{} a^2(b^2 \\plus{} c^2) \\le a^4 \\plus{} b^4 \\plus{} c^4 \\plus{} 2(a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2) \\equal{} (a^2 \\plus{} b^2 \\plus{} c^2)^2 \\equal{} 4.$\r\nNow, from this inequality, we can easily deduce that\r\n$ |a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} abc| \\le 2\\sqrt {2},$\r\nas desired. This ends our proof.\r\n\r\nSecond way: Using the property of absolute value and the Cauchy Schwarz Inequality, we have\r\n$ \\begin{aligned} |a^3 \\plus{} b^3 \\plus{} c^3 \\minus{} abc| & \\le |a^3| \\plus{} |b^3| \\plus{} |c^3| \\plus{} |abc| \\\\\r\n&\\le \\sqrt {(a^2 \\plus{} b^2 \\plus{} c^2)(a^4 \\plus{} b^4 \\plus{} c^4)} \\plus{} |abc| \\\\\r\n& \\equal{} \\sqrt {2(a^4 \\plus{} b^4 \\plus{} c^4)} \\plus{} |abc|. \\end{aligned}$\r\nThus, it suffices to prove that\r\n$ \\sqrt {2(a^4 \\plus{} b^4 \\plus{} c^4)} \\plus{} |abc| \\le 2\\sqrt {2}.$\r\nBecause $ 2\\sqrt {2} \\equal{} (a^2 \\plus{} b^2 \\plus{} c^2)\\sqrt {2},$ this comes down to\r\n$ \\sqrt {2} \\left( a^2 \\plus{} b^2 \\plus{} c^2 \\minus{} \\sqrt {a^4 \\plus{} b^4 \\plus{} c^4}\\right) \\ge |abc|,$\r\nwhich is equivalent to\r\n$ \\frac {2\\sqrt {2} (a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2)}{a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} \\sqrt {2(a^4 \\plus{} b^4 \\plus{} c^4)}} \\ge |abc|.$\r\nNote that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} \\sqrt {a^4 \\plus{} b^4 \\plus{} c^4} \\le 2(a^2 \\plus{} b^2 \\plus{} c^2) \\equal{} 4,$ one can see that the last inequality was deduced from\r\n$ a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2 \\ge \\sqrt {2} |abc|.$\r\nNow, using the well-known inequality $ (x \\plus{} y \\plus{} z)^2 \\ge 3(xy \\plus{} yz \\plus{} zx)$ for the triple $ (x,y,z) \\equal{} (a^2b^2,b^2c^2,c^2a^2),$ we get\r\n$ (a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2) \\ge 3a^2b^2c^2(a^2 \\plus{} b^2 \\plus{} c^2) \\equal{} 6a^2b^2c^2 \\ge 2a^2b^2c^2,$\r\nfrom which we deduce that\r\n$ a^2b^2 \\plus{} b^2c^2 \\plus{} c^2a^2 \\ge \\sqrt {2}|abc|,$\r\nas desired. This ends our proof.",
  "Solution_4": "The second proof is natural on the aspect of idea. Anyway, they are both wonderful and manageable.  :)",
  "Solution_5": "Can I think that your solutions are supernatural  :D , I have also an approach by cauchy shwarz :\r\n\r\ndenote : $ [x]= abs(x)$\r\n\r\n$ [( a^3+b^3+c^3-abc)] = [ \\sum{a(a^2-\\frac{bc}{3}}] \\leq  \\sqrt{ (a^2+b^2+c^2)( \\sum{(a^2-\\frac{bc}{3})^2})}$\r\n\r\nand $ \\sum{(a^2-\\frac{bc}{3})^2 = \\sum{a^4} - \\frac{2abc\\sum{a}}{3} + \\frac{\\sum{(ab)^2}}{9}  \\leq   \\sum{a^4)+\\sum{(ab)^2} \\leq  (\\sum{a^2})^2}}$ :)"
}
{
  "Tag": [
    "logarithms",
    "algebra",
    "system of equations",
    "algebra unsolved"
  ],
  "Problem": "sovle this system of equations in $ R$:\r\n\r\n$ \\begin {cases} \\sqrt {8x^3 \\plus{} 1} \\minus{} \\sqrt {y^3 \\plus{} 3y^2 \\plus{} 3y \\plus{} 2} \\equal{} y \\minus{} 2x \\plus{} 1 \\\\\r\ny^3 \\plus{} 4x \\plus{} 1 \\plus{} \\ln{(y^3 \\plus{} 2x)} \\equal{} 0 \\end {cases}$",
  "Solution_1": "show that :$ y\\plus{}1\\equal{}2x$ and $ y\\equal{}\\minus{}1$ is unique root of equation$ y^3\\plus{}2y\\plus{}3\\plus{}ln( y^2\\plus{}y\\plus{}1)\\equal{}0$",
  "Solution_2": "[quote=\"Vincent Gilbert\"]show that :$ y \\plus{} 1 \\equal{} 2x$ and $ y \\equal{} \\minus{} 1$ is unique root of equation$ y^3 \\plus{} 2y \\plus{} 3 \\plus{} ln( y^2 \\plus{} y \\plus{} 1) \\equal{} 0$[/quote]\r\n\r\n you have mistake $ y^{3}\\plus{}2y\\plus{}3\\plus{}ln(y^{3}\\plus{}y\\plus{}1)\\equal{}0$\r\n\r\n$ y\\equal{}\\minus{}1$---> $ ln(y^{3}\\plus{}y\\plus{}1)\\equal{}ln(\\minus{}1)$--> not solution",
  "Solution_3": "^^ .Maybe he had a typo because this problem is exactly alike a problem in a contest at my province"
}
{
  "Tag": [
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "If $f_{n}$ is a convergent sequence in $D'(R)$ (distributions), $F_{n}' = f_{n}$ and if the integral of $\\phi_{o}$ is nonzero such that $(F_{n},\\phi_{o})$ is bounded. Then show that $F_{n}$ has a convergent subsequence.\r\n\r\n------------------\r\n\r\nIt's pretty easy to show that $(F'_{n},\\phi) \\rightarrow (f,\\phi)$ but i'm not sure about the rest.",
  "Solution_1": "As a first step, you need to show this: if $F'=f$ and $G'=f$ in $\\mathcal{D}',$ then there exists a constant $C$ such that $F=G+C.$\r\n\r\nThe definition of $C$ as a distribution is, of course, $(C,\\varphi)=C\\int_{\\mathbb{R}}\\varphi$"
}
{
  "Tag": [
    "floor function",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Show that $ \\sum_{k\\equal{}1}^n \\lfloor \\dfrac{n}{k} \\rfloor \\minus{}\\lfloor \\sqrt{n}\\rfloor$ is even for all positive integers $ n$.",
  "Solution_1": "[hide]\nThe quantity $ \\sum_{k\\equal{}1}^n\\left\\lfloor\\frac{n}{k}\\right\\rfloor$ counts the number of pairs $ (a,b)$ where $ a\\le b\\le n$ and $ b$ is divisible by $ a$, since for a given $ a$, it has $ \\left\\lfloor\\frac{n}{k}\\right\\rfloor$ multiple lss than or equal to $ n$. Now this is equivalent then to $ \\sum_{k\\equal{}1}^n d(k)$, where $ k$ is the total number of divisors that $ k$ has. This number is even except when $ k$ is a perfect square, so this quantity has the same parity as the number of perfect squares less than or equal to $ n$, or $ \\lfloor\\sqrt{n}\\rfloor$.\n[/hide]",
  "Solution_2": "Thanks a lot - I just completely algebra bashed this thing with casework n odd, n even. Your way is far more instructive."
}
{
  "Tag": [
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let A be a ring. \r\n\r\nIf every finitely generated right ideal is genreated by an idempotent then Jac(A) =0. \r\n\r\nHere Jac(A) means jacobson radical.\r\n\r\nProof: Let a be in Jac(A) and pick an idempotent element e such that Ae = Aa, thus a = ae and e=xa for x in A. Hence a =axa so a(1-xa)=0. Since a is in Jac(A) also xa is in Jac(A) \r\nand 1-xa is a unit , hence a=0.\r\n\r\n\r\nMy question is that why is a = ae  please help me with this. \r\n\r\nBecause Ae = Aa, so I will say that ae = a'a, please help.",
  "Solution_1": "$ a\\in Aa\\equal{}Ae\\Rightarrow a\\equal{}be\\Rightarrow ae\\equal{}be^2\\equal{}be\\equal{}a$.",
  "Solution_2": "Thanks, but I was thinking that since Aa =Ae  so I am thinking that a'a =be, what does  elements in Aa look like. \r\n\r\nI think that Aa = {a_0a,a_1a,a_2a,......}\r\nso every element in Aa has a form xa for x in A. Byt now we are intresenting only in a so you pick 1a  =a in Aa where 1 is anutral element in A. \r\n\r\nIs it  correct what I am saying thanks.",
  "Solution_3": "we use also the definition of $ Ae$ ..."
}
{
  "Tag": [
    "MATHCOUNTS",
    "search",
    "geometry",
    "HCSSiM"
  ],
  "Problem": "I know a few strong states, Indiana, California, Texas, for instance. The toughest chapter(s) for 2008 is a California chapter, where one needed to score 42/46 to either make chapter or make state (don't remember which one).\r\n\r\nAnd also that Oklahoma is classified as a \"weak state\" since, for 2008, no one got above 30/46 at the State level. My brother said that Idaho was a \"weak state\" but surely you know better than even my brother.",
  "Solution_1": "Cool, what's the question?\r\n\r\nIn general, yeah. It's kind of unfair that they take 4 from every state to nats, regardless of intelligence or population. I happen to be in a weaker state. How/can you figure out which states are \"good\" and \"weak\" states? The people going to nats last year (including me) had scores of 33,33,32, and 32",
  "Solution_2": "This may be of interest, the team champions from 1984-2008 have come from:\r\nVA, FL, Ca, NY, NY, NC, OH, AL, CA, KS, PA, IN, PA, MA, WI, MA, CA, VA, CA, CA, IL, TX, VA, TX, TX.  \r\nAnd Individual winners from:\r\nTX, OK, FL, TN, IL, PA, AK, MA, OH, KA, IL, KY, PA, WI, MA, WI, IL, NJ, IL, WA, IL, LA, NJ, TX, WA.  \r\n\r\n(This is sourced from Nationals Yearbooks)\r\n\r\nThings like this vary from year to year, in general, it's very hard for small states (such as mine) to field 4 incredibly strong members at Mathcounts Nats, but every once and a while we place fairly high.  We've also had several very strong individuals, 3 black group members, 2 IMOers, one 2nd place Mathcounts.  So, it's not impossible.",
  "Solution_3": "I could tell a \"strong state\" from a \"weak state\" by their state/national cutoff.",
  "Solution_4": "But surely it is easier to make nationals if you live in say Guam (I think they have a team) then Texas or California. I have heard that California has the best students but they somehow fail as a team. This may or may not be true.",
  "Solution_5": "If you live in a \"weak\" state and make the nationals then will the competition be a little harder because you are from a \"weak\" state?",
  "Solution_6": "[quote=\"cyao15\"]If you live in a \"weak\" state and make the nationals then will the competition be a little harder because you are from a \"weak\" state?[/quote]\r\n\r\nNot necessarily.  You yourself could be beast.  But in general, yes.",
  "Solution_7": "You are right. Oklahoma is a horrible state, I'm from Oklahoma. But this year there will be some people (including me) that will probably get a higher score than a 30 at state. \r\nLast year my friend got first with a 29ish...lol :rotfl:",
  "Solution_8": "This is a bit off-topic but does anyone know what national cutoff scores (I mean to get in the top 3) for Virginia have been like in recent years?\r\nAnd BTW, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=561897297&t=84737]This[/url] has some interesting information regarding relative difficulty.",
  "Solution_9": "No score discussing; even if data on one chapter is available on a public website, don't provide the link to that website. As far as the Mathcounts forum is concerned, reality is that there are no public discussions of chapter/school scores until the end of the announcement. Reality may be different in different places ;).",
  "Solution_10": "Aren't they talking about previous years though?",
  "Solution_11": "Only the one who overtly gave the link to the chapter's data for the current year was in direct violation. Everyone else really were discussing goals for this year or past years' scores.",
  "Solution_12": "Hmmm, sorry... I thought our chapter coordinator would know better, though.\r\n\r\nI would have to say that states with larger populations GENERALLY do better.\r\n(more population pool)\r\nAlso those states with higher immigration rates are probably affected in some way (either better or for worse)\r\nWe could corroborate data of illegal immigration, high school dropouts, cost of college tuition, median income, etc. and still cannot predict who will win.\r\n\r\nOh, and MOP/IMO/USAMO =/= Mathcounts...",
  "Solution_13": "[quote=\"james4l\"]Oh, and MOP/IMO/USAMO =/= Mathcounts...[/quote]\r\n\r\nI'm aware of that, I was just making a point that just because your state is small, or in the case of my state, absolutely minuscule, it doesn't mean that there are no smart people or that the team at nats won't do well.",
  "Solution_14": "The only reason as to why a larger state would have a better chance to succeed in MATHCOUNTS/USAMO is only because a larger population pool brings about more intense competition. Still, it cannot explain everything, since VT already had one MATHCOUNTS silver medalist (excuse the expression).",
  "Solution_15": "Right, I think large states doubly benefit themselves.  The have a largely population to draw from, so statistically they are already at an advantage, and, in addition, more competition pushes people to work harder.",
  "Solution_16": "Is West Virginia, Virginia, and Kentucky considered weak, ok, or strong states?",
  "Solution_17": "Virginia is a strong state, but I don't know for the other two.",
  "Solution_18": "Kentucky ranks 26th in state pop, WV is 38th (including Puerto Rico, which goes to Nationals).  \r\nhttp://en.wikipedia.org/wiki/US_States_by_population\r\n\r\nIf I could locate my nats sheets for the past 3 yrs, I could tell you if any of those states made the top 22ish (MC only releases up to that), but I can't find them.  Perhaps someone else who has them?",
  "Solution_19": "Isn't Missouri another example of a weak state...? :maybe:",
  "Solution_20": "NORTH CAROLINA!!!!!!!!! \r\n\r\nachoo. excuse me. :D \r\n\r\ni know that the population principle applies here. for example, in some chapters here, there are entire TEAMS who haven't even heard of mathcounts. but, in other more concentrated places (i.e. Ligon, Lincoln, hopefully Hanes), competition is really intense.",
  "Solution_21": "recently texas has pwned mathcounts :roll:",
  "Solution_22": "texas is a really weak state.  :D",
  "Solution_23": "Honestly, it's one of the strongest states, along with California. The weak states are usually the smaller ones or the least densely populated ones like Wyoming, New Hampshire, DC, Department of Defense, Guam, Puerto Rico, Virgin Islands, Hawaii, Alaska, etc. But some small states are still strong b/c of large mathematical institutions, such as Massachusetts, New Jersey, etc. And some medium states are...uh...medium difficulty (illinois, nebraska, wisconsin, arizona, etc.)",
  "Solution_24": "[quote=\"funtwo\"]texas is a really weak state.  :D[/quote]\r\n\r\nkind of funny that u have a longhorn avatar... :P",
  "Solution_25": "I'm not sure if State matters more than Chapter. For example where I live, the chapter I went to was the peninsula chapter, where 30 made state, but down south in Santa Clara chapter, I think every year half the people going to state or so got perfect. At nationals you really need to be good to win, but some people who could make state but are in a difficult chapter could get disqualified for state, even if they probably would've made it from any other chapter in the state.",
  "Solution_26": "[quote=\"gauss1181\"]But some small states are still strong b/c of large mathematical institutions, such as Massachusetts, New Jersey, etc. [/quote]\r\n\r\nMass and NJ are 15th and 11th in population...\r\n(or maybe he's referring to area?)\r\nI wonder how state qualification happens for larger states.  My state has only 4 chapters, so, each chapter gets at least two, then the next top w/e teams in the whole state qualify.  Then, top indivs whose teams didn't qualify form a team (of 4) from each chapter.",
  "Solution_27": "[quote=\"Quickster94\"]My state has only 4 chapters, so, each chapter gets at least two, then the next top w/e teams in the whole state qualify.  Then, top indivs whose teams didn't qualify form a team (of 4) from each chapter.[/quote]\r\n\r\nIt seemed just as easy to get to state (or nationals) as it is for Oklahoma.",
  "Solution_28": "I live in MA, and its not too difficult. However, the chapter i live in (lexington clarke, diamond, mccall) is really tough. For chap all 8 clarke students made states w/ individual scores, and they had 6 ppl in countdown. their lowest score on chap was a 40 i think",
  "Solution_29": "where can i find practice tests, like 1998 and 1999 state tests?",
  "Solution_30": "[quote=\"mathlovers1731\"]where can i find practice tests, like 1998 and 1999 state tests?[/quote]\r\n\r\nAsk in the resources page."
}
{
  "Tag": [
    "AMC 10",
    "AMC"
  ],
  "Problem": "What exactly are you allowed to take?  I looked at one of the threads for an AMC 10 problem, where someone had arrived at the answer by using a protractor.\r\n\r\nI know you are not allowed to use a calculator, but what [i]are[/i] you allowed to use?\r\n\r\nEDIT: Fixed spelling.",
  "Solution_1": "I think basically anything else?\r\n\r\nA compass and a ruler would probably be helpful for geometry.",
  "Solution_2": "Front cover of the AMC contest booklets, item 5.\r\nSee the Teachers' Manual, pages 20, and 21\r\nhttp://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC1012/2010AMC1012A-TM.pdf\r\n\r\nSteve Dunbar\r\nMAA Direcotr, American Mathematics Competitions",
  "Solution_3": "It says you can use scratch paper, graph paper, rulers, protractors, and erasers."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "An empty cilindrical can floats on water with 6 cm of its height above the level of the water. The base of the can has an area of 50 cm2, and from the center of it a block B of lead (density = 11.2 g/mL, weight = 560 g)) is suspended by a small rope. We have something like this:\r\n\r\n[code]                        _\n          |          |   |\n          |          |   |-  h = 6cm\n          |          |  _|\n----------|          |------------\n  Water   |          |\n          |          |\n          |__________|\n                | \n                |\n               _|_\n              |   |\n              | B |\n              |___|\n[/code]\n\nNext, the lead block B is taken out from the base of the can and is placed inside the can:\n\n[code]                        _\n          |          |   |\n          |          |   |-  h\u2019 = ?\n          |          |  _|\n----------|    ___   |------------\n  Water   |   |   |  |\n          |   | B |  |\n          |___|___|__|\n                \n[/code]\r\n\r\nCompute the new height h' of the can that now stays above the level of the water.",
  "Solution_1": "Are you sure that's all the information?\r\n\r\nIt seems that we need to know the height of the cylinder...\r\nI got that h'=5+6/x, where x is the height of the cylinder, but I'm not sure if that's right.",
  "Solution_2": "[quote=\"Kate25\"]Are you sure that's all the information?[/quote]\r\n\r\nYes I am sure: we don't have to know the height of the cylinder. It seems that you are not using all the information that is given. Also, your formula \"h'=5+6/x\" is wrong.",
  "Solution_3": "Problem still not solved.",
  "Solution_4": "did you do the free body diagram? in both situations?",
  "Solution_5": "if I have time, I'll write the solution tomorrow",
  "Solution_6": "Yes, in problems like this its common practice to draw the applied forces on each body.",
  "Solution_7": "I made it last night. Send me and e-mail and I will send you the solution.\r\njonysatie@gmail.com",
  "Solution_8": "I don't need the solution, because I know perfectly how to solve this problem. I just posted it because I find it interesting and perhaps other people studying introductory Physics could also like it.",
  "Solution_9": "ok ... and yes, it's very interesting, but I have seen better ones.",
  "Solution_10": "I wish we had one of these problems when we were studying fluids. It's nice, but simple when you think about it.\r\n\r\n[hide=\"Wrong Solution\"]\nFirst, I set up the equation to get the height of the cylinder under water.\n$50 \\cdot 10^{-3}\\cdot 9.8+9.8 \\cdot 50 \\cdot 10^{-2}\\cdot h = 5.488$\nThis is just equating the force of buoyancy with  the weight of the lead, assuming the can's weight is  negligible.\nTherefore, the height under the water is $1.02 m$ That's a lot, so it's possible I messed up somewhere.\nThis is the more difficult problem of the two, but it gives the total can height as 1.08 m.\n\nFor the answer, I set up the equation\n$5.488 = 50 \\cdot 10^{-2}\\cdot 9.8 \\cdot h'$\n$h' = 1.12 m$ This is larger than the previous answer, so that checks. However, it is also larger than the total can height, so the can sinks.\n\nI think that my order of magnitude may be incorrect somewhere. Can someone please verify my answer?\n[/hide]",
  "Solution_11": "Stuck, your answer is wrong.",
  "Solution_12": "In the solution below I did not take the mass of the can into account,\r\nwhich is bad since nothing is said about its weight, but it should not be too hard to\r\nadd it to the solution, perhaps I'll do it later.\r\n[hide=\"my try =)\"]\n[hide=\"notations\"]\nLet the total height of can be $x$ and its bottom area $A$.\nAlso $\\rho_{W}$ is water density and $\\rho_{B}$ density of lead.\nAccordingly $V_{B}$ is volume of $B$ $\\ldots$\n[/hide]\n\nArchimedes law gives the buoyancy to be (force)\n$\\underbrace{V_{B}\\rho_{water}g}_{lead thing}+\\underbrace{(x-h)Ag\\rho_{water}}_{can}=m_{B}g$\nsolving for $x$ yields\n\\[x=\\frac{m_{B}}{\\rho_{W}}\\frac{1}{A}\\left(1-\\frac{\\rho_{W}}{\\rho_{B}}\\right)+h \\]\nUnits are right at least  :lol: \n\nNow the other scenario gives an equation in which $h'$ is unknown but know we know $x$\n$(x-h')A\\rho_{W}g=m_{B}g$\nwhich gives\n$h'=x-\\frac{m_{B}}{A\\rho_{W}}=\\frac{m_{B}}{\\rho_{W}}\\frac{1}{A}\\left(1-\\frac{\\rho_{W}}{\\rho_{B}}\\right)+h-\\frac{m_{B}}{A\\rho_{W}}=h-\\frac{m_{B}}{\\rho_{W}}\\frac{1}{A}\\left(\\frac{\\rho_{W}}{\\rho_{B}}\\right)$\n[/hide]\n\nSo It becomes $\\approx 5 \\rm{cm}$\n\n[hide=\"with mass of can\"]\nJust add $m_{can}g$ to $RHS$ and it gives $x=\\frac{m_{B}+m_{A}-V_{B}\\rho_{W}}{A\\rho_{W}}$\nnow simplify and proceed as above, BUT $m_{A}$ is not given.\n\n[/hide]",
  "Solution_13": "Your answer is correct."
}
{
  "Tag": [
    "logarithms",
    "calculus",
    "integration",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let be given a real number $\\alpha$.Prove that:\r\n$\\alpha>1\\Longleftrightarrow \\exists k>0$ such that $\\sum_{i=2}^n\\frac{1}{i (\\ln i)^\\alpha}<k \\forall n$\r\nWhen $\\ln x=\\log_e x$",
  "Solution_1": "We'll use the contrapositive both ways, and also use implicitly the fact that a sum of positive numbers either converges to a finite number or diverges to infinity.\r\n\r\nIf $a\\leq 1$, the series diverges by comparison to the series $\\sum \\frac{1}{i\\ln i}$ (think integral test or condensation test).\r\n\r\nConversely, if the sum on the left hand side diverges, then $a\\leq 1$, since\r\n\\[ \\sum^{\\infty}_{i=1}2^i\\cdot\\frac{1}{2^i(\\log 2^i)^{\\alpha}}=\\frac{1}{(\\log 2)^{\\alpha}}\\sum^{\\infty}_{i=1}\\frac{1}{i^\\alpha}  \\]\r\nand we apply the condensation test again."
}
{
  "Tag": [
    "floor function",
    "Divisibility Theory"
  ],
  "Problem": "Determine the least possible value of the natural number $n$ such that $n!$ ends in exactly $1987$ zeros.",
  "Solution_1": "As there are always more factors of 2 than 5 in $n!$ ($n \\geq 2$), we need to find $n!$ with exactly $1987$ factors of 5\r\n\r\nNow, amongst integers $\\leq 8000$, there are-\r\n\r\n$1600$ multiples of $5$\r\n$320$ multiples of $5^2=25$\r\n$64$ multiples of $5^3=125$\r\n$12$ multiples of $5^4=625$\r\n$2$ multiples of $5^5=3215$.\r\n\r\nThis totals 1998, which is 11 too many.\r\n\r\nSo we have $n$ slightly less than 8000- if we reduce it to $7960$, then we have one fewer multiple of 125, two fewer multiples of 25 and 8 fewer multiples of 5, giving us $1987$ zeros on the end of $7960!$.\r\n\r\nWe can't reduce this any further, because $7959!$ only has $1986$ zeros on the end, so the answer is \r\n\r\n$\\boxed{7960}$",
  "Solution_2": "we must find n such that\n$\\sum_{i=1}^{k} [\\frac {n}{5^{i}}] =1987$ with $5^{k}\\leq n$\nuse that $[x] \\leq x$ we have that $1987\\leq \\frac {n}{4}.(1- \\frac{1}{5^k}) \\leq \\frac {n-1}{4}$ \nwe have that $n\\geq 7949$ then easy find n=7960.",
  "Solution_3": "power of 2(say a) in n! is always less than or equal to power of 5(say b) in n!\n\nand 2*5=10 so the number n! will end with exactly b many zeros.\n\nso we need to find n least such that $\\ s(n)=[\\frac{n}{5}]+[\\frac{n}{5^{2}}]+.....=1987$ if at all such n exist.\n\nbut then see if $\\ n=5^{6},s(n)=\\frac{5^{6}-1}{4}=3906 $\nand $\\ n=5^{5},s(n)=\\frac{5^{5}-1}{4}=781 $\nso if such n exists it is $\\ 5^{5} < n < 5^{6} $\nso let we take base 5 expansion for n to get $\\ n=a_{5}5^{5}+a_{4}5^{4}+...a_{0} $ where $\\  0\\leq a_{i} \\leq 4 $\nthen see\n$\\ s(n)=781a_{5}+156a_{4}+31a_{3}+6a_{2}+a_{1} =1987 $ if possible\nto make n least we try to keep a_{i} 's with the larger i's as small as possible.\n\nsee a_{5} is at least 1. if we set a_{5}=1 then s(n) is at max 1557 so a_{5} have to be atleast 2.\n\nfor a_{5}=2 we see $\\ 156a_{4}+31a_{3}+6a_{2}+a_{1} =425 $\nsee 156*3=468>425 so a_{4} have to be less than 3 but if a_{4}=1 then see\ns(n) can be atmost 308 so a_{4}=2\nso $ \\ 31a_{3}+6a_{2}+a_{1} =113 $\nthen by similar arguement $ \\ a_{3}=3$\nso $\\ 6a_{2}+a_{1} =20 $\nsee by same arguement $\\ a_{2}=3 $\nthen $\\ a_{1}=2 $\nas we want the least no so we set $\\ a_{0}=0$\nso the desired no is  $\\ n=2.5^{5}+2.5^{4}+3.5^{3}+3.5^{2}+2.5+0=7960 $\n$ \\ n=7960$",
  "Solution_4": "[hide=\"A106\"]Note that the number of zeroes at the end of $n!$ is an increasing function. The number of zeroes at the end of $n!$ is $\\lfloor frac{n}{5} \\floor +\\lfloor \\frac{n}{5^{2}} \\rfloor+..... \\approx \\frac{\\frac{n}{5}}{\\frac{4}{5}}$. Therefore, $n$ is about $7948$. Note that $7948$ contains $1589$ $5$s, $317$ additional from $5^2$, $63, 12, 2$ for the rest. This totals $1983$ zeroes. Increasing to $7950$ adds an additional $2$ ($5^2||7950)$, and it is clear that we need 2 more, the first from 7955 and the final from 7960. Therefore, the smallest such number is $7960$.[/hide]"
}
{
  "Tag": [
    "number theory",
    "greatest common divisor",
    "algorithm",
    "algebra",
    "polynomial",
    "Euclidean algorithm"
  ],
  "Problem": "The numbers in the sequence 101, 104, 109, 116, $\\dots$ are of the form $a_n = 100 + n^2$, where $n = 1$, 2, 3, $\\dots$.  For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n + 1}$.  Find the maximum value of $d_n$ as $n$ ranges through the positive integers.",
  "Solution_1": "[hide=\"Solution - incomplete\"]\nAfter experimenting for a bit, we can confidently make the conjecture that for sequences $x_n$, $\\{k+1^2,k+2^2,k+3^2,\\ldots\\}$, the maximum value of the greatest common divisor of consecutive terms is $\\text{gcd}(x_{2k},x_{2k+1})=4k+1$. This suggests that when $k=100$, our answer will be $401$. But how can we prove this conjecture? :?\n[/hide]",
  "Solution_2": "Let $a_{2k} = 100+4k^2$, and $a_{2k+1}=4k^2+4k+1$. This notation is fine because any two consecutive terms can be expressed as the nth even and nth odd term. By the Euclidean Algorithm, we have \r\n\r\n\\begin{eqnarray*}(4n^2+100, 4n^2+4n+101) &=& (4n^2+100, 4n+1)\\\\ &=&(100-n, 4n+1)\\\\ &=&(3n+101, 4n+1)\\\\ &=&(3n+101, n-100)\\\\ &=&(401, n-100)\\end{eqnarray*}\r\n\r\nSo our maximum GCD possible is $401$, the answer. The GCD occurs between the $1002$ and $1003$ terms.",
  "Solution_3": "That is, of course, the right answer.\r\n\r\nHere's how I did it\r\n\r\n[hide]Solution:\nLet $d$ be the divisor of two consecutive terms.\n$d| n^2+100$ and $d| n^2+2n+101$.\nTherefore, taking the difference, $d| 2n+1$\nSquaring, $d| 4n^2+4n+1$\n\nSince $d| n^2+100$, $d| 4n^2+400$\n\nSince $d| 4n^2+400$ and $d| 4n^2+4n+1$, we subtract to achieve $d| 399-4n$\n\nSince $d| 2n+1$, multiply by 2 to get $d| 4n+2$\n\nSince $d| 4n+2$ and $d| 399-4n$, we add to achieve $d| 401$\n\nThis means d is a factor of $401$. Since $401$ is prime, d must always be either $1$ or $401$.\n\nIn order for $401$ to work, we must find an $n$ such that $401| 2n+1$\n\nOur first instinct is to try $401=2n+1$, which yeilds $n=200$. This works, so the answer is $\\boxed{d=401}$\n[/hide]",
  "Solution_4": "that is in essence the same solution as the one above it...",
  "Solution_5": "wow, the solution in Zeits is way longer than that, i suppose he was trying to prove a point though...",
  "Solution_6": ":? What's the Eucledian Algorithm?   :fool:",
  "Solution_7": "yes, I am also wondering what it is",
  "Solution_8": "What about googling?  :roll: \r\nhttp://www.google.com/search?q=euclidean+algorithm",
  "Solution_9": "did you guys try out that euclidean algorithm java app? it is funny watching it take 20 steps to figure out that the gcd of two numbers is two.",
  "Solution_10": "http://en.wikipedia.org/wiki/Euclidean_algorithm",
  "Solution_11": "v. interesting!",
  "Solution_12": "yea this is in acops",
  "Solution_13": "[quote=\"Elemennop\"]Let $a_{2k} = 100+4k^2$, and $a_{2k+1}=4k^2+4k+1$. This notation is fine because any two consecutive terms can be expressed as the nth even and nth odd term. By the Euclidean Algorithm, we have \n\n\\begin{eqnarray*}(4n^2+100, 4n^2+4n+101) &=& (4n^2+100, 4n+1)\\\\ &=&(100-n, 4n+1)\\\\ &=&(3n+101, 4n+1)\\\\ &=&(3n+101, n-100)\\\\ &=&(401, n-100)\\end{eqnarray*}\n\nSo our maximum GCD possible is $401$, the answer. The GCD occurs between the $1002$ and $1003$ terms.[/quote]\r\n\r\nA few questions:\r\n\r\nWhy did you choose to use $2k$ and $2k+1$ rather than simply $k$ and $k+1$?  Also, doesn't the notation assume that the first of the two consecutive integers is even?  It seems to me that the notation overlooks half of the possibilities.",
  "Solution_14": "For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\r\n\r\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)",
  "Solution_15": "[quote=\"matt276eagles\"][quote=\"Elemennop\"]Let $a_{2k} = 100+4k^2$, and $a_{2k+1}=4k^2+4k+1$. This notation is fine because any two consecutive terms can be expressed as the nth even and nth odd term. By the Euclidean Algorithm, we have \n\n\\begin{eqnarray*}(4n^2+100, 4n^2+4n+101) &=& (4n^2+100, 4n+1)\\\\ &=&(100-n, 4n+1)\\\\ &=&(3n+101, 4n+1)\\\\ &=&(3n+101, n-100)\\\\ &=&(401, n-100)\\end{eqnarray*}\n\nSo our maximum GCD possible is $401$, the answer. The GCD occurs between the $1002$ and $1003$ terms.[/quote]\n\nA few questions:\n\nWhy did you choose to use $2k$ and $2k+1$ rather than simply $k$ and $k+1$?  Also, doesn't the notation assume that the first of the two consecutive integers is even?  It seems to me that the notation overlooks half of the possibilities.[/quote]\r\n\r\nI chose $2k$ and $2k+1$ because I already tried $k$ and $k+1$ in my head and it got nowhere (ie. you get stuck with fractions/unremovable variable). You can try it yourself. \r\n\r\nAlso, it doesn't overlook half the possibilities because the algorithm is commutative, ie. it doesn't matter which one is first. We could just as well have done the odd term first and we would have arrived at the same answer.",
  "Solution_16": "But if you actually did the polynomial division between the $n$ and $n+1$ terms, you would get the same answer.",
  "Solution_17": "I don't understand how repeated subtraction could get rid of the squared terms.",
  "Solution_18": "[quote=\"mathmanman\"]What about googling?  :roll: \nhttp://www.google.com/search?q=euclidean+algorithm[/quote]\r\n\r\nYeah! Google loves you",
  "Solution_19": "Ah, I see, he taking the remainder of one divided by the other is the same as subtracting repeatedly.",
  "Solution_20": "[quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\r\n\r\nand we learned it in kindergarden in china.",
  "Solution_21": "[quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\r\nand you just spammed",
  "Solution_22": "[quote=\"junggi\"][quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\nand you just spammed[/quote]\r\n\r\nand u did too",
  "Solution_23": "[quote=\"Pineappleperson\"][/quote][quote=\"junggi\"][quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\nand you just spammed[/quote][quote=\"Pineappleperson\"]\n\nand u did too[/quote]\r\nthank you",
  "Solution_24": "[quote=\"junggi\"][/quote][quote=\"Pineappleperson\"][/quote][quote=\"junggi\"][quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\nand you just spammed[/quote][quote=\"Pineappleperson\"]\n\nand u did too[/quote][quote=\"junggi\"]\nthank you[/quote]\r\n\r\nhuh\r\n\r\n??????",
  "Solution_25": "[quote=\"Elemennop\"][/quote][quote=\"junggi\"][/quote][quote=\"Pineappleperson\"][/quote][quote=\"junggi\"][quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\nand you just spammed[/quote][quote=\"Pineappleperson\"]\n\nand u did too[/quote][quote=\"junggi\"]\nthank you[/quote][quote=\"Elemennop\"]\n\nhuh\n\n??????[/quote]\r\nand you just spammed",
  "Solution_26": "[quote=\"chess64\"][/quote][quote=\"Elemennop\"][/quote][quote=\"junggi\"][/quote][quote=\"Pineappleperson\"][/quote][quote=\"junggi\"][quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\nand you just spammed[/quote][quote=\"Pineappleperson\"]\n\nand u did too[/quote][quote=\"junggi\"]\nthank you[/quote][quote=\"Elemennop\"]\n\nhuh\n\n??????[/quote][quote=\"chess64\"]\nand you just spammed[/quote]\r\n\r\nI dont get it",
  "Solution_27": "[quote=\"Elemennop\"][/quote][quote=\"chess64\"][/quote][quote=\"Elemennop\"][/quote][quote=\"junggi\"][/quote][quote=\"Pineappleperson\"][/quote][quote=\"junggi\"][quote=\"Pineappleperson\"][quote=\"pkerichang\"]For those of you who's wondering, Euclidean Algorithm is that you find the GCD of two numbers be repeatingly subtracting two numbers from one another until you have a 0 - then the other number is the GCD.\n\nThis is true because the divisor of two numbers must also divide the difference, so repeatingly taking the difference will give you the GCD.  (we learned that in Taiwan in elementary school...)[/quote]\n\nand we learned it in kindergarden in china.[/quote]\nand you just spammed[/quote][quote=\"Pineappleperson\"]\n\nand u did too[/quote][quote=\"junggi\"]\nthank you[/quote][quote=\"Elemennop\"]\n\nhuh\n\n??????[/quote][quote=\"chess64\"]\nand you just spammed[/quote][quote=\"Elemennop\"]\n\nI dont get it[/quote]\r\n\r\nyou're supposed to say 'and u did too'\r\n\r\nso i'll just assume you said that\r\n\r\nthank you",
  "Solution_28": "[quote=\"Elemennop\"][quote=\"matt276eagles\"][quote=\"Elemennop\"]Let $a_{2k} = 100+4k^2$, and $a_{2k+1}=4k^2+4k+1$. This notation is fine because any two consecutive terms can be expressed as the nth even and nth odd term. By the Euclidean Algorithm, we have \n\n\\begin{eqnarray*}(4n^2+100, 4n^2+4n+101) &=& (4n^2+100, 4n+1)\\\\ &=&(100-n, 4n+1)\\\\ &=&(3n+101, 4n+1)\\\\ &=&(3n+101, n-100)\\\\ &=&(401, n-100)\\end{eqnarray*}\n\nSo our maximum GCD possible is $401$, the answer. The GCD occurs between the $1002$ and $1003$ terms.[/quote]\n\nA few questions:\n\nWhy did you choose to use $2k$ and $2k+1$ rather than simply $k$ and $k+1$?  Also, doesn't the notation assume that the first of the two consecutive integers is even?  It seems to me that the notation overlooks half of the possibilities.[/quote]\n\nI chose $2k$ and $2k+1$ because I already tried $k$ and $k+1$ in my head and it got nowhere (ie. you get stuck with fractions/unremovable variable). You can try it yourself. \n\nAlso, it doesn't overlook half the possibilities because the algorithm is commutative, ie. it doesn't matter which one is first. We could just as well have done the odd term first and we would have arrived at the same answer.[/quote]\n\nCorrect me if I'm wrong, but I believe it does overlook half the possibilities.  Your expression assumes that the odd term is one more than the even term.  So (2,3), (4,5), (6,7) fit your expression but pairs like (3,4) and (5,6) do not.  Luckily this doesn't really affect the answer because:\n\n\\begin{eqnarray*}(4n^{2}+100, 4n^{2}-4n+101) &=& (4n^{2}+100, -4n+1)\\\\ &=&(100+n, -4n+1)\\\\ &=&(100 + n, 401)\\end{eqnarray*}\n\n and n cannot be negative.",
  "Solution_29": "I have a very similar solution and it's probably worst but oh well this thread seems pretty spammed anyways :P\n\n[hide=\"solution\"] \n\nWe apply EA To $(100+n^2, 100+n^2+2n+1) = (100+n^2, 2n+1)$ Note that $2$ is coprime to $2n+1$ meaning that $(100+n^2,2n+1) = (200+2n^2,2n+1)$\n\nApplying EA again $(200+2n^2 - 2n^2 - n ,2n+1) = (200-n,2n+1)$ Again, we see that $2$ is coprime to $2n+1$ and again we have $(200-n,2n+1) = (400-2n,2n+1)$\n\nApplying EA again we get that $(400-2n,2n+1) = (400-2n+2n+1, 2n+1) = (401, 2n+1)$ which is obviously maxed at $401$. To test if this can work we set $n=200$. \n\n\nWe see that $100+200^2 = 40100$ which is obviously divisible by $401$. We use EA again to check for $100+201^2$ as it is a multiple of 401 if and only if $100+201^2 - 100-200^2$ is a multiple of $401$. But we see that this is the same as $201^2 - 200^2 = (201+200)(201-200) = 401$. Thus $100+201^2$ is also a multiple of $401$. \n\nThus the maximum is $\\boxed{401}$\n\n[/hide]",
  "Solution_30": "INFORMAL SOLUTION\nLet $a_n=1+n^2$ instead. Then, $a_n=2,\\mathbf{5,10},17,26,37,50,65,82,101,122...$, and we speculate that $\\gcd(a_n,a_{n+1})$ is $5$.\n\nNow, instead let $a_n=2+n^2$.Then, $a_n=3,6,11,\\mathbf{18,27},38,51,66,83,102,123...$, and we speculate that $\\gcd(a_n,a_{n+1})$ is $9$.\n\nNow, instead let $a_n=3+n^2$.Then, $a_n=4,7,12,19,28,\\mathbf{39,52},67,84,103,124...$, and we speculate that $\\gcd(a_n,a_{n+1})$ is $13$.\n\now, instead let $a_n=4+n^2$.Then, $a_n=5,8,13,20,29,40,53,\\mathbf{68,85},104,125...$, and we speculate that $\\gcd(a_n,a_{n+1})$ is $17$.\n\nWe speculate that that if $a_n = 100 + n^2$, then the maximum $\\gcd(a_n,a_{n+1})$ is $\\boxed{401}$.\n\nFORMAL SOLUTION\nWe need to find the maximum value of $\\gcd(100 + n^2,100+n^2+2n+1)$. By the Euclidean Algorithm, we can re-write it as $\\gcd(2n+1, 100+n^2)$ or $\\gcd(2n+1, 100+n^2-100(2n+1))$, which is simplified to $\\gcd(2n+1, n^2-200n)$. Noting that $\\gcd(n,2n+1)=1$, we can re-write our answer as $\\gcd(2n+1, n-200)$, $\\gcd(2n+1-2(n-200), n-200)$ or $\\gcd(401,n-200)$. Hence, the maximum of $\\gcd(100 + n^2,100+n^2+2n+1)$ is $\\boxed{401}$.",
  "Solution_31": "we wish to find $\\gcd(100+n^2 , 100+n^2+2n+1=\\gcd(100+n^2 , 1+2n)=\\gcd(400+4n^2 , 2n+1)$ since $2n+1$ will be always odd so we have $\\gcd(400+4n^2, 2n+1)=\\gcd(401, 2n+1)$ so maximum gcd occurs as $\\boxed{401}$ for $n=200$."
}
{
  "Tag": [],
  "Problem": "which is [b]not[/b] a stimulatory plant  hormone?\r\n\r\nzeatin\r\nIAA\r\nGA3\r\nethylene\r\ngibberellins\r\n\r\nim thinking is IAA, but not really sure about it...\r\nafter reading the other hormone plants i got that conclusion does someone knows more about hormone plants?\r\nthank's",
  "Solution_1": "I'd guess GA3 by POE\r\nbecause\r\nzeatin is the plant hormone for Cytokinine, IAA is the plant hormone for Auxin, ethylene is the gas produced by plants for fruit ripening and gibberlins wake seeds from dormancy.",
  "Solution_2": "Indolacetic acid (IAA) was the first auxin discovered in plants.  I've never heard of zeatin (not in campbell's), but I'll take agolsme's word that it's a cytokinin."
}
{
  "Tag": [
    "trigonometry",
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "If $n\\in N*$ prove that\r\n\r\n$0< \\tan (\\pi\\sqrt {9n^2+n})< \\dfrac{1}{\\sqrt 3}$",
  "Solution_1": "Nice indeed, and easy too :D\r\n\r\nHere's my solution:\r\n\r\n$\\dfrac{1}{\\sqrt{3}}=tan(\\dfrac{\\pi}{6})$, and because the period of the tangent function is $\\pi$ and the function is increasing over the period, we need to prove that \\[0<\\{\\sqrt{9n^2+n}\\}<\\frac{1}{6},\\]\r\n\r\nwhere $\\{x\\}=x-[x]$.\r\n\r\nThe smallest integer less than $\\sqrt{9n^2+n}$ is clearly $\\sqrt{9n^2}=3n$. Then all we have to prove is\r\n\r\n\\[\\sqrt{9n^2+n}-3n<\\frac{1}{6}    \\Leftrightarrow\\]\r\n\\[\\Leftrightarrow    9n^2+n<9n^2+n+\\frac{1}{6}\\]which is true.",
  "Solution_2": "[quote=\"portoseb\"]\n and because the period of the tangent function is $\\pi$ and the function is increasing over the period, we need to prove that \\[ 0<\\{\\sqrt{9n^2+n}\\}<\\dfrac{1}{6}, \\]\n[/quote]\r\n\r\nThere is a theorem which says this?",
  "Solution_3": "Well, we need to prove that\r\n\r\n$\\tan(\\pi\\cdot 0)< \\tan (\\pi\\cdot\\sqrt {9n^2+n})< \\tan (\\pi \\cdot\\dfrac{1}{6})$.\r\n\r\nAnd since $\\tan \\pi a+b=\\tan b$, if $a \\in Z$ and $0\\leq b<\\pi$, we have\r\n\r\n$\\tan(\\pi\\cdot 0)< \\tan (\\pi\\cdot\\sqrt {9n^2+n})< \\tan (\\pi \\cdot\\dfrac{1}{6}) \\Leftrightarrow$\r\n\r\n$\\Leftrightarrow \\tan(\\pi\\cdot 0)< \\tan (\\pi\\cdot\\{\\sqrt {9n^2+n}\\})< \\tan (\\pi \\cdot\\dfrac{1}{6}) \\Leftrightarrow$\r\n\r\n$\\Leftrightarrow 0<\\{\\sqrt{9n^2+n}\\}<\\dfrac{1}{6}$, because $\\tan x$ is increasing over $[0, \\dfrac{\\pi}{6}]$ and it is also a bijection over $[0, \\pi)$. We are allowed to analyze only the interval $[0, \\pi)$ because we applied the above formula: $\\tan \\pi a+b=\\tan b$."
}
{
  "Tag": [],
  "Problem": "Solve in integers $x^2(y-1)+y^2(x-1)=1$.\r\nIn the book, An Introduction to Diophanitine Equations, the author makes the substitutions $x=u+1$ and $y=v+1$, thus the new equation is $(u+1)^2v+(v+1)^2u=1$.\r\nSince the author wants to solve this via decomposition, he says that the equation is equivilent to $(u+v+4)(uv+1)=5$.\r\nCan someone explain how you would know to add 4 to both sides, and somehow factor it?\r\nSince me calculator (TI-89) was able to factor the equation after I added 4, I know there must be some logical way.\r\nThanks.",
  "Solution_1": "Are you asking about the factoring technique of completing the square?",
  "Solution_2": "I'm basically asking, how would one know to proceede in that direction from $(u+1)^2v+(v+1)^2u=1$.",
  "Solution_3": "[quote=\"mna851\"]Are you asking about the factoring technique of completing the square?[/quote]\r\n\r\nThis isn't completing the square....",
  "Solution_4": "$\\displaystyle (u+1)^2v+(v+1)^2u=1$\r\n$\\displaystyle uv(u+v) + 4uv + (u+v)=1$\r\n$\\displaystyle (u+v)(uv+1) + 4uv=1$\r\n\r\nby this point we know that we have to add something, m,  to factor it with 4uv, either by u+v or uv+1. since 4uv is a multiplication, u+v doesnt seem to work. SO, uv+1 is the factor we are going to factorize 4uv+m with. \r\n\r\n$4uv+m = n(uv+1)$\r\n$n=4, m = 4$\r\n\r\ntherefore, we add four to both sides...\r\n\r\n$\\displaystyle (u+v+4)(uv+1) = 5$\r\n\r\ni hope that kinda explained the concept behind adding numbers to factor.... just like you, i do not know what to add until i factor things out and organize in an according manner.",
  "Solution_5": "Thanks I think I get it now."
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "logarithms",
    "function",
    "calculus computations"
  ],
  "Problem": "$\\int \\frac{x^\\frac{1}{2}}{(x+1)^\\frac{1}{2}}dx$",
  "Solution_1": "i used brute force to solve it... with substitution $x = \\tan^{2}t$.\r\n\r\n$\\int \\frac{x^{\\frac{1}{2}}}{(x+1)^{\\frac{1}{2}}}\\, dx$\r\n$= \\sqrt{x(x+1)}-\\ln ( \\sqrt{x}+\\sqrt{x+1})+C$",
  "Solution_2": "Me too, but it's not so nasty\r\n\r\nWith trig. substitution you got $2\\int{\\sec u\\tan^{2}udu}= 2\\int{\\sec^{3}udu}-2\\int{\\sec udu}$\r\n\r\nThat's not a nasty integral but with this you got it.\r\n\r\n[hide=\"...\"]\nI prefer $u$  :rotfl: [/hide]",
  "Solution_3": "I use next substitution $x=sh^{2}(y)$ then $\\sqrt{1+x}$=$ch(y)$\r\nand $dsh^{2}(y)=2sh(y)(ch(y)dy$ and our integral is equivalent \r\n$2\\int(sh^{2}(y))dy$ which is simpl to find",
  "Solution_4": "[quote=\"Extremal\"]I use next substitution $x=sh^{2}(y)$ then $\\sqrt{1+x}$=$ch(y)$\nand $dsh^{2}(y)=2sh(y)(ch(y)dy$ and our integral is equivalent \n$2\\int(sh^{2}(y))dy$ which is simpl to find[/quote]\r\n\r\nWhat's $sh(y)$ and $ch(y)$?",
  "Solution_5": "Those are hyperbolic functions- the standard abbreviations in the USA are $\\sinh$ and $\\cosh$.",
  "Solution_6": "Ah, I just hadn't seen that notation before.  That's amazing that there is a different notation for them... shouldn't stuff like that be standardized?",
  "Solution_7": "Making standards work worldwide across many languages is hard. Following this forum will show you European variations on many things that you thought were standard.",
  "Solution_8": "sometimes we use $sinhx$ but it is long and therefore we use $shx$ :lol:"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $ x $ and $y$ be a positive naturel numbers such that: $ 2x^2-1=y{}^1{}^5$\r\nProve that $5|x$",
  "Solution_1": "[quote=\"kam\"]Let $ x $ and $y$ be a positive naturel numbers such that: $ 2x^2-1=y{}^1{}^5$\nProve that $5|x$[/quote]\r\nAre you sure?\r\nWhat about $(x,y)=(1,1)$ ?  ;)",
  "Solution_2": "Yes there's a mistake somewhere. I tried to solve it but I only get $x\\equiv 1\\left( %Error. \"func\" is a bad command.\n{mod}5\\right) $ or $x\\equiv 4\\left( %Error. \"func\" is a bad command.\n{mod}5\\right) $  :? \r\nCan you check it again please ?  ;)",
  "Solution_3": "well this is the from this years russian olympiad  :D",
  "Solution_4": "Yes i forgot the condition $x>1$.",
  "Solution_5": "then it was already solved by Harazi!  :D",
  "Solution_6": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=5&t=35316]solution[/url]\r\n\r\nit was posted before as ARO 2005."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let $a , b , c > 0$ , with $a , b ,c \\in \\mathbb{R}$ , prove that : \r\n\r\n\\[\\sum \\frac{a^{6}+b^{6}}{a+b}\\ge 3a^{2}b^{2}c^{2}\\]",
  "Solution_1": "It wrong for $a=b=c>1$"
}
{
  "Tag": [
    "vector",
    "function",
    "linear algebra",
    "matrix",
    "analytic geometry",
    "calculus",
    "calculus computations"
  ],
  "Problem": "I want to calculate $ \\nabla\\times[\\vec{F}(r)]$ and $ \\nabla^2[\\vec{F}(r)]$ where F if a function that depends of r, and $ r \\equal{} \\sqrt {x^2 \\plus{} y^2 \\plus{} z^2}$ (the magnitude, what I want to say is that it depends of x, y and z)\r\n\r\nI know that...\r\n$ 1)\\nabla \\times \\vec A \\equal{} \\left|\\begin{matrix} \\mathbf{\\hat{i}} & \\mathbf{\\hat{j}} & \\mathbf{\\hat{k}} \\\\\r\n \\\\\r\n{\\frac {\\partial}{\\partial x}} & {\\frac {\\partial}{\\partial y}} & {\\frac {\\partial}{\\partial z}} \\\\\r\n \\\\\r\nA_x & A_y & A_z \\end{matrix}\\right|$\r\n$ 2)\\nabla^2\\vec A \\equal{} \\nabla(\\nabla\\cdot\\vec A) \\minus{} \\nabla\\times(\\nabla\\times\\vec A)$\r\n\r\nMy attempt to a solution\r\n$ \\left( \\frac {\\partial}{\\partial y}\\vec F_{z}(r) \\minus{} \\frac {\\partial}{\\partial z}\\vec F_{y}(r)\\right)\\hat{i} \\plus{} \\left( \\frac {\\partial}{\\partial x}\\vec F_{z}(r) \\minus{} \\frac {\\partial}{\\partial z}\\vec F_{x}(r)\\right)\\hat{j} \\plus{} \\left( \\frac {\\partial}{\\partial x}\\vec F_{y}(r) \\minus{} \\frac {\\partial}{\\partial y}\\vec F_{x}(r)\\right)\\hat{k}$\r\n\r\nI need help calculating the partials I tried this:\r\n$ \\frac {1}{r}\\left(y \\minus{} z \\right) \\plus{} \\frac {1}{r}\\left(x \\minus{} z \\right) \\plus{} \\frac {1}{r}\\left(x \\minus{} y \\right) \\Rightarrow \\frac {2x \\minus{} 2z}{r}$ What happened to the i, j, k? I don't know =P! I suppose I'm wrong for vanishing them...\r\n\r\nWhen I'm trying to calculate the vectorial laplacian I get stuck when I have to calculate $ \\nabla\\times\\vec {F}(r)$ so...\r\n\r\nedit:\r\n$ \\vec{F}(r)$ is an arbitrary vector function in cartesian form that depends of x, y and z. Maybe I was wrong to say that $ r \\equal{}  \\sqrt{x^2\\plus{}y^2\\plus{}z^2}$ maybe $ \\left( \\frac{\\partial}{\\partial y}\\vec F_{z}(r)\\minus{}\\frac{\\partial}{\\partial z}\\vec F_{y}(r)\\right)\\hat{i} \\plus{} \\left( \\frac{\\partial}{\\partial x}\\vec F_{z}(r)\\minus{}\\frac{\\partial}{\\partial z}\\vec F_{x}(r)\\right)\\hat{j}\r\n\\plus{} \\left( \\frac{\\partial}{\\partial x}\\vec F_{y}(r)\\minus{}\\frac{\\partial}{\\partial y}\\vec F_{x}(r)\\right)\\hat{k}$ could be the general asnwer to calculate the curl of any function that depends of r, for example: $ \\left( \\frac{r^2\\sqrt{csc^2(r)}}{ ln (3r)}\\right)\\hat i \\plus{} \\left( 3\\sqrt[5]{r^2}\\cdot e^{3r^2}\\right)\\hat j \\plus{} \\left( \\frac{6r^3}{cot^3(r)}\\right) \\hat k$ (or any other)",
  "Solution_1": "Wouldn't it be easier, using spherical coordinates  ?",
  "Solution_2": "I'm not very familiar with spherical coordinates, it's easier? i found the gradient, the divergence and the scalar laplacian for that type of functions, and I need to find the curl before I can find the vector laplacian..."
}
{
  "Tag": [
    "LaTeX",
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Question 1:\r\nProove that if \u03bb is an eigenvalue of [A], then 1/\u03bb is an eigenvalue of [A]T\r\n\r\n\r\nQuestion 2\r\nProove that a square matrices [A] and [A]T have the same Eigenvalues.\r\n\r\n\r\nQuestion 3:\r\nShow that |det(A)| is the product of the absolute values of the eigenvalues of \r\n[A]\r\n\r\nQuestion 4:\r\nLet A and B be nonsingular nxn matrices. show that AB^-1 and B^-1A have the same eigenvalues.\r\n\r\nI am not asking for complete solutions to the problems, it will suffice if you can point me to some theorem or give a clue that will help toward finding the solution.\r\nI am new to the forum and I don't know how to format the mathematical notations, if there a reference file I can consult?",
  "Solution_1": "Use the definition to find the answers to your questions.\r\n\r\nAnd $ \\text{\\LaTeX}$ is the typesetting language for math on this forum. You can click the \"Latex Help\" button on top of this forum, if you are new to it.\r\n\r\nAnd shouldn't $ A^T$ be $ A^{-1}$ in Question 1?",
  "Solution_2": "Hints/stronger statements:\r\n#1: If $ Av\\equal{}\\lambda v$, $ A^{\\minus{}1}v\\equal{}\\frac1{\\lambda}v$\r\nThis works over an arbitrary field.\r\n\r\n#2: If $ (A\\minus{}\\lambda I)v\\equal{}0$, $ v^T(A^T\\minus{}\\lambda I)\\equal{}0$. Use the various rank/nullity theorems to show that the left and right eigenspaces of $ A^T$ have the same dimension. This implies that $ A$ and $ A^T$ have the same eigenvalues, with the same geometric multiplicity.\r\nThis also works over an arbitrary field.\r\n\r\n#3: Drop the absolute values, and count with algebraic multiplicity. This is obviously true if $ A$ is upper triangular. Over an algebraically closed field, every matrix is similar to an upper triangular matrix; for $ \\mathbb{C}$, this is Schur's lemma.\r\n\r\n#4: $ AB^{\\minus{}1}$ and $ B^{\\minus{}1}A$ are similar. This works over an arbitrary field.\r\n\r\nWhy does #3 require an algebraically closed field, where the others don't? #3 needs the existence of a full list of eigenvalues, while the others are true even if there are no eigenvalues at all."
}
{
  "Tag": [],
  "Problem": "Did anyone take the 3 hour 7 question proof test of the MAML level two yesterday?\r\n\r\nI couldn't figure out like a third of the questions, but I'm hoping i get partial credit for a lot of the stuff i wrote down... :| \r\n\r\nBtw, did anyone get how to do the first two parts of number 6, or the last part of number 7?",
  "Solution_1": "I'm guessing that this is the Massachusetts-specific competition, so perhaps this should be moved to the Massachusetts forum. But yeah, I did the first part of number 6 through contradiction, and I did the second part proving that every number had to be of the form x/999, y/990, or z/900"
}
{
  "Tag": [
    "calculus",
    "derivative",
    "function",
    "limit",
    "calculus computations"
  ],
  "Problem": "Show that the function $ f: \\mathbf{R}^2\\rightarrow \\mathbf{R}$ defined by \\[ f(x,y) \\equal{} \\left\\{ \\begin{array}{cc}\\frac{x^3y}{x^4\\plus{}y^2}   & (x,y)\\neq (0,0)\\\\ \r\n0& (x,y)\\equal{}(0,0) \\end{array}\\right.\\]\r\n\r\nhas $ \\nabla_{0,0}f(u)\\equal{}0$ for all $ u$ but is not differentiable at $ (0,0)$.\r\n\r\nI'm completely stuck on this... for some reason I can't wrap my head around it.\r\n\r\nThanks.",
  "Solution_1": "$ f$ is differentiable at a point $ (x,y)$ iff $ \\lim_{(h,k)\\to(0,0)}\\frac1{\\sqrt{h^2\\plus{}k^2}}(f(x\\plus{}h,y\\plus{}k)\\minus{}f(x,y)\\minus{}f_x(x,y)h\\minus{}f_y(x,y)k)\\equal{}0.$\r\n\r\nVisually, that's the statement that the surface that is the graph of $ f$ has a tangent plane at that point.\r\n\r\nIn this particular case, that would mean that \r\n\r\n$ \\lim_{(x,y)\\to(0,0)}\\frac{f(x,y)}{\\sqrt{x^2\\plus{}y^2}}\\equal{}0.$\r\n\r\nNow, look at points on the curve $ y\\equal{}x^2.$ For such points,\r\n\r\n$ f(x,y)\\equal{}\\frac{x^3x^2}{x^4\\plus{}x^4}\\equal{}\\frac{x}2.$\r\n\r\nFor small $ x,$, we have $ \\sqrt{x^2\\plus{}y^2}\\equal{}\\sqrt{x^2\\plus{}x^4}\\approx \\sqrt{x^2}\\equal{}|x|.$\r\n\r\nThat means that the limit we were looking at is $ \\frac12$ along the first quadrant portion of this curve and $ \\minus{}\\frac12$ along the second quadrant portion of the curve; either way, that's not zero.\r\n\r\nIf you approach the origin along any ray, we do have $ \\frac{f(x,y)}{\\sqrt{x^2\\plus{}y^2}}\\to 0.$ That's how we got all the directional derivatives to be zero. But the different curve of approach gives a different result. There's no tangent plane.",
  "Solution_2": "Suddenly I remember doing this when I was in Multi, but that was long enough ago that I forgot about the \"curve\" approach (this problem actually is for analysis hw)\r\n\r\nThanks a lot. \r\n\r\nDoes this geometric view of \"differentiable\" expand to $ R^n$ at all, or is it too hard to visualize in the higher dimensions?"
}
{
  "Tag": [
    "MATHCOUNTS",
    "ARML",
    "AMC",
    "AMC 10",
    "email"
  ],
  "Problem": "Not nearly complete... If you have any more details, please post them here. I only have personal experience with the Montgomery County Middle School Math League and the Univ. of MD competitions. This is a list of competitions exclusive to maryland, so Mathcounts and AMC's and MAO and Mandelbrot, etc. are implied.\r\n\r\nName: Montgomery County Middle School Math League\r\nSince: ?\r\nType: Individual & team event for middle schools. Used to select ARML.\r\nWhen: On a wednesday, 4 meets throughout the year\r\nWhere: Random middle schools\r\nEligibility: 1 team of 6 middle schoolers, plus 2 secondary teams. Chosen by coach.\r\nStructure: 6 Individual short answers in 24 minutes. 6 team questions in 20 minutes. 2 relay questions. (questions where one person's result feeds into the next person)\r\nInformation and contact: ?\r\n\r\nName: Montgomery County Interscholastic Mathematics League.\r\nType: Used to select ARML\r\nElegibililty: High schoolers chosen by coach\r\n(No Details)\r\n\r\nName: Howard County Math League\r\nSince: 1978\r\nType: Individual & team event for high schools. Used to select ARML.\r\nWhen: On a wednesday, 8 meets throughout the year\r\nWhere: Random high schools\r\nEligibility: 2 teams of high schoolers. Chosen by coach\r\nStructure: 6 Individual short answers in 30 minutes. 5 team questions in 15 minutes.\r\nInformation and contact: http://www.howard.k12.md.us/math/hcml/hcml.html\r\n\r\nName: Univ. of MD math competition: Part I\r\nSince: 1979\r\nType: Individual event for high schoolers.\r\nWhen: October 10 this year\r\nWhere: Local high schools\r\nEligibility: Anyone not graduated from high school\r\nStructure: 25 AMC10 style questions, 75 minutes\r\nInformation and contact: http://www.math.umd.edu/highschool/mathcomp/\r\n\r\nName: Univ. of MD math competition: Part II\r\nSince: 1979\r\nType: Individual event for high schoolers.\r\nWhen: December 3 this year\r\nWhere: Local high schools OR Univ. MD campus\r\nEligibility: Top scorers (usually around 230) in Part I\r\nStructure: 5 proof style problems, 2 hours\r\nInformation and contact: http://www.math.umd.edu/highschool/mathcomp/",
  "Solution_1": "There is also a Johns Hopkins Mathematics Competition that is run by some very smart people.\r\n\r\nConsider looking into that!  If you have any questions about JHMT, email \r\n\r\nmathclub@lists.tripleintegral.org"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Even though I'm registered, my name doesn't appear at the bottom of the forum.\r\nI think I am classified \"hidden\" because I'm not in the registered users nor am I a guest.  How do I become \"unhidden\"?",
  "Solution_1": "Go in your profile (click profile in the links at the top of the forum when you are logged in).  The second line of your preferences sets your hidden status - you can change it there.",
  "Solution_2": "read this faqa:[url]http://www.artofproblemsolving.com/wforum/faq.php#3[/url]",
  "Solution_3": "Thank you very much for your help!"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "If $x,y$ are reals prove that \r\n$4(x^2+xy+y^2)^3-27x^2y^2(x+y)^2\\geq 0$",
  "Solution_1": "$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\sqrt{\\frac{x}{y}}=a $ and  $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\sqrt{\\frac{y}{x}}=b $\r\nWe have ab=1.\r\nwe must pove $ 4.(a^2+b^2+1)^3 \\le 27(a+b)^2 $\r\n$ \\Leftrightarrow 4.( a^6 + b^6 +1 + 3(a^2+1)(b^2+1)(a^2+b^2)) \\geq 27.(a^2+b^2 +2) $\r\nwe have:\r\n$ 3(a^2+1)(b^2+1)(a^2+b^2) \\geq 12.(a^2 +b^2) $\r\nSo \r\n$  4.( a^6 + b^6 +1 + 3(a^2+1)(b^2+1)(a^2+b^2)) \\geq 4.a^6 + 4.b^6 + 4 + 48(a^2+b^2) \\geq 27.(a^2+b^2 +2) $",
  "Solution_2": "[quote=\"silouan\"]If $x,y$ are reals prove that \n$4(x^2+xy+y^2)^3-27x^2y^2(x+y)^2\\geq 0$[/quote]\r\n\r\nWell, put $x+y=a$ and $xy=b$\r\nThen the inequality is equivalent to \r\n$4(a^2-b)^3-27a^2b^2\\geq0$ \r\n\r\nNote that\r\n\r\n             $\\begin{alligned} &  4(a^2-b)^3-27a^b^2=4a^6-4b^3-12a^2b(a^2-b)-                  27a^2b^2=4a^6-4b^3-12a^4b-15a^2b^2=(4a^6-16a^4b)+(4a^4b-16a^2b^2)+(a^2b^2-4b^3)=4a^4(a^2-4b)+4a^2b(a^2-4b)+b^2(a^2-4b)=(a^2-4b)(4a^4+4a^2b+b^2)\\geq0 \\end{alligned}$\r\n\r\n :)",
  "Solution_3": "My solution is \r\n$4(x^2+xy+y^2)^3-27x^2y^2(x+y)^2=(x-y)^2(2x^2+2y^2+5xy)^2\\geq 0$\r\nwhich is true",
  "Solution_4": "Together your solution is not true because $x,y$ can be negatives , so your squareroots may not exists ;)"
}
{
  "Tag": [],
  "Problem": "The time on a digital clock is 5:55. How many minutes will pass before the clock next shows a time with all digits identical?",
  "Solution_1": "[hide]11:11 is the next time.  So there are 5 minutes until 6:00, then 5 hours until 11:00, and finally 11 minutes until 11:11.  So the answer is 5+300+11=[b]316[/b][/hide]",
  "Solution_2": "[hide]Since the next time the digits will align is 11:11 (there's no such 6:66, etc.) the time passed will be 316 minutes. From 5:55 to 10:55 is 5 hours or 300 minutes. Add 16 minutes to get to 11:11 to get a total of 316 minutes elapsed from 5:55 to 11:11. (Whew, my first post!)[/hide]",
  "Solution_3": "Correct."
}
{
  "Tag": [
    "linear algebra",
    "linear algebra unsolved"
  ],
  "Problem": "I jusr caome up with a difference equation : \r\n\r\n$x_{k+1}=(3k+5)x_{k}+5$ with $x_{0}=\\frac{1}{2}$\r\n\r\nCan anyone tell me how to solve.\r\n\r\nIn general, how could we solove $x_{k+1}=f(k)x_{k}+g(k)$?",
  "Solution_1": "[quote=\"bchui\"]In general, how could we solove $x_{k+1}=f(k)x_{k}+g(k)$?[/quote]\nConsider the special case $f(k)\\equiv 1$: you want a formula for the partial sums of a given series. Too much to ask for.\n\n[quote]It is not known how to solve a general recurrence equation to produce an explicit form for the terms of the recursive sequence[/quote] from [url=http://mathworld.wolfram.com/RecursiveSequence.html]Wikipedia[/url]\r\n\r\nOf course, some ad hoc approach may work for your sequence. For example, let $b_{k}=\\prod_{j=0}^{k-1}(3j+5)$: then $\\frac{x_{k+1}}{b_{k+1}}=\\frac{x_{k}}{b_{k}}+\\frac{5}{b_{k+1}}$.  Therefore, $x_{k}/b_{k}$ can be expressed in terms of the partial sums of $\\sum b_{k}^{-1}$; I don't know if they have a closed form.",
  "Solution_2": "Here is the solution:\r\n\r\nThe solution of  ${ x_{k+1}=A_{k}x_{k}+b_{k}~~,~~x_{0}=z}$  is given by\r\n\r\n${ x_{k}=\\left\\lgroup \\prod^{k-1}_{i=0}A_{i}\\right\\rgroup z+\\sum_{i=0}^{k-1}\\prod_{j=i}^{k-1}A_{j}b_{i}}$    :)\r\n\r\nThe problem remains on how to compute  ${\\prod_{i=0}^{m}(i+\\lambda) }$ and ${\\sum_{i=0}^{m}\\prod_{j=i}^{m}(j+\\lambda) }$"
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "All digits of a number $ N$ are equal to $ 1$. Prove that if $ 7$ divides $ N$, then so\r\ndoes $ 13$.",
  "Solution_1": "Write $ N\\equal{}\\sum_{k\\equal{}0}^n10^k$ for some $ n\\in\\mathbb{N}\\cup\\{0\\}$. Now, $ 0\\equiv N\\bmod{7}$ implies $ 3^{n\\plus{}1}\\equiv 1\\bmod{7}\\Longrightarrow 6|n\\plus{}1$. Moreover, we have $ N\\equiv\\frac{1\\minus{}(\\minus{}3)^{n\\plus{}1}}{4}\\equiv 0\\bmod{13}$, the last step, because $ (\\minus{}3)^6\\equiv 1\\bmod{13}$.",
  "Solution_2": "my solution was completley different (i would be suspicious)\r\n\r\nUsing the divisibility rule for 7 , we will construct a multiple of 7\r\n\r\nstarting with one 1 and continuing all the way up to six ones we find that the first multiple is 111111\r\nwe can generalize for all the multiples of 7 that; the number of ones will be a multiple of 6.\r\n\r\nnow, we find that 13 also evenly divides 111111. since it evenly divides 111111, it must also divide all the other numbers in which the number of digits is a multiple of 6(all the digits are 1s)\r\n\r\nthe ONLY way for 7 to divide N is if the number of digits is a multiple of 6, we found that if the number of digits is a multiple of 6, 13 divides N so, if 7divides N, so does 13.\r\n\r\n\r\n\r\n\r\n\r\ncould you PLZ send feedback on this solution\r\n\r\n3q!",
  "Solution_3": "Well what you did wasn't exactly rigorous since you only computed the values up to 111111. To generalize you would need to construct the sequence 1, 11, 111,... and take it mod 7, then mod 13, or something similar.",
  "Solution_4": "Uhh, repunit with $ n$ ones is equal to $ \\frac{10^{n}\\minus{}1}{9}$. Consider only the top since the divided by 9 won't matter... $ 10^n\\minus{}1\\equiv 0 \\pmod 7 \\Rightarrow 6|n$. Then show $ 10^{6k}\\minus{}1\\equiv 0 \\pmod {13}$, which is pretty easy since $ 10^6 \\equiv 1 \\pmod {13}$..."
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Is there any way to draw border around the page using latex?\r\n\r\nwhat should i do?",
  "Solution_1": "You posted this message twice, look at your first message."
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a, b,$ and $c,$ and that the roots of $x^3+rx^2+sx+t=0$ are $a+b, b+c,$ and $c+a.$  Find $t.$",
  "Solution_1": "[hide]$t=-(a+b)(b+c)(c+a)$\n\n$=-(-3-c)(-3-a)(-3-b)$\n\n$=(3+a)(3+b)(3+c)$\n\n$=27+9(a+b+c)+3(ab+bc+ca)+abc$\n\n$=27+9(-3)+3(4)+11$\n\n$=\\boxed{23}$\n[/hide]",
  "Solution_2": "[hide]$t$ being the constant term, we know that $-t$ is the product of the roots. We also know that $abc=11$ and $a+b+c=-3$, so $t=-(a+b)(b+c)(c+a)=(3+a)(3+b)(3+c)$ after substituting in from $a+b+c$, so after that we foil everything out, we have $27+9c+9b+9a+3bc+3ac+3ab+abc\\Longrightarrow 27+9(a+b+c)+3(ac+bc+ab)+abc$, and substituting with everything we know, we have $27+9(-3)+3(4)+11=\\boxed{23}$. Note that $ab+bc+ac$ is always equal to the third term in a polynomial.[/hide]",
  "Solution_3": "[hide=\"Answer\"]$-t=(a+b)(b+c)(c+a)=ab(a+b)+ac(a+c)+bc(b+c)+2abc$\nFrom the first equation, $abc=11$, $ab+ac+bc=4$, and $a+b+c=-3$, so then $a+b=-(c+3)$, $a+c=-(b+3)$, and $b+c=-(a+3)$. Substituting, we have $ab(c+3)+ac(b+3)+bc(a+3)-2abc=3(ab+ac+bc)+abc=3(4)+11=\\boxed{23}$.[/hide]",
  "Solution_4": "Let $ y\\equal{}b\\plus{}c\\equal{}a\\plus{}b\\plus{}c\\minus{}a\\equal{}\\minus{}3\\minus{}x$, so substitute $ x\\equal{}\\minus{}3\\minus{}y$. Then $ \\minus{}p(\\minus{}3\\minus{}y)\\equal{}y^3\\plus{}6y\\plus{}13y\\plus{}23$...so the answer is $ \\boxed{023}$.",
  "Solution_5": "hmmm.\r\ncouldn't you recognize \r\n$ \\minus{}(\\minus{}3\\minus{}a)(\\minus{}3\\minus{}b)(\\minus{}3\\minus{}c)$ as evaluating the initial polynomial at $ x \\equal{} \\minus{}3$ and then negating it. no foil that way.",
  "Solution_6": "$ \\minus{}t\\equal{}2abc\\plus{}\\sum a^2b \\equal{} 2abc \\plus{} \\sum(a\\plus{}b)ab$\r\n\r\n$ \\equal{}2abc\\plus{}\\sum(\\minus{}3\\minus{}c)ab\\equal{}2abc\\plus{}\\sum(\\minus{}3ab\\minus{}abc)$\r\n\r\n$ \\equal{}2*11\\plus{}(\\minus{}3*4\\minus{}3*11)\\equal{}\\minus{}23$",
  "Solution_7": "[hide=My sol]\n\nBy vietas, $t$ is the negative product of $(a+b)(b+c)(c+a)$. We will neglect the negative for now for simplicity. Expanding $(a+b)(b+c)(c+a)$ we get $2abc+\\sum a^2b$. We can cleverly introduce an extra $abc$ term to make it factorable; which gives $(a+b+c)(\\sum ab)-abc$. Since we know all of these three factors/terms from vietas on the given cubic, we plug and get $-3(4)-11$. Notice that $t$ is the negative of this by vietas. Thus, the answer is $-(-23)=\\boxed{23}$.\n[/hide]\n\n@mathwizarddude AIME answers are non negative.\n\nOops sorry for bump but I felt that the answer should be corrected.",
  "Solution_8": "We know that $(a+b)(b+c)(c+a) = ((a+b+c) - a)((a+b+c)-b)((a+b+c) - c) = f(a+b+c) = f(-3) = -23$. Note that since $-t = -(a+b)(b+c)(c+a)$, $t = \\boxed{23}$.",
  "Solution_9": "[hide=My Solution]\n\nBy Vieta's Formula, we have that:\n$$\\implies a + b + c = -3$$\n$$\\implies ab + bc + ca = 4$$\n$$\\implies abc = 11$$\n\nBy Vieta's Formula:\n$\\implies t = -{(a + b)(b + c)(c + a)} = -{(-3 - c)(-3 - a)(-3 - b)} = (a + 3)(b + 3)(c + 3) = abc + 3(ab + bc + ca) + 9(a + b + c) + 27 = \\boxed{23}$\n$\\blacksquare$",
  "Solution_10": "Hmm, [url]https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_5[/url]",
  "Solution_11": "[hide= Solution]\nQuick inspection yields that $(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c) - abc = -12 - 11 = -23.$ This yields $t=23$.\n[/hide]",
  "Solution_12": "[quote=joml88]Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a, b,$ and $c,$ and that the roots of $x^3+rx^2+sx+t=0$ are $a+b, b+c,$ and $c+a.$  Find $t.$[/quote]\n\nwhy was this on my AMSP test :huh:",
  "Solution_13": "[hide=Solution]$t=-(a+b)(b+c)(c+a)=abc-(a+b+c)(ab+bc+ca)=11-(-3)(4)=11+12=\\boxed{23}$.[/hide]",
  "Solution_14": "Note that $t = -(a+b)(b+c)(c+a)=-(-3-a)(-3-b)(-3-c) = (3+a)(3+b)(3+c) = abc+3(ab+ac+bc)+9(a+b+c)+27$, or $11+12-27+27 = \\boxed{23}$.",
  "Solution_15": "Note that $f(x)=x^3+3x^2+4x-11=(x-a)(x-b)(x-c)$. We have that $t=-(a+b)(b+c)(c+a)=-(-3-a)(-3-b)(-3-c)=-f(-3)=\\boxed{23}$."
}
{
  "Tag": [
    "geometry",
    "area of a triangle"
  ],
  "Problem": "I couldn't find the way to solve this one too fast...\r\n\r\n\\[A, B, C, D \\in \\bigcirc(O, R),\\] \\[AB=a, BC=b, CD=c, AD=d, \\]\r\n\\[AB+BC+CD+AD=2p,\\]\r\n\\[AC=2R.\\]\r\nProve that:\r\n\\[(p-a)(p-b)(p-c)(p-d)=\\frac{(ad+bc)^2}{4}\\]",
  "Solution_1": "[quote=\"Cauchy_Schwartz\"]I couldn't find the way to solve this one too fast...\n\n\\[A, B, C, D \\in \\bigcirc(O, R),\\] \\[AB=a, BC=b, CD=c, AD=d, \\]\n\\[AB+BC+CD+AD=2p,\\]\n\\[AC=2R.\\]\nProve that:\n\\[(p-a)(p-b)(p-c)(p-d)=\\frac{(ad+bc)^2}{4}\\][/quote]\r\n\r\nActually, the last formula should be\r\n\r\n$\\left( p-a\\right) \\left( p-b\\right) \\left( p-c\\right) \\left( p-d\\right) =\\frac{\\left( ab+cd\\right) ^{2}}{4}$.\r\n\r\nIt's not very clear to me whether you have solved this or not, so I post my solution:\r\n\r\nSince AC = 2R, the segment AC is a diameter of the circle (O, R), and since the points B and D lie on this circle, we have < ABC = 90 and < ADC = 90. Hence, the triangles ABC and ADC are right-angled, and their areas are\r\n\r\n$\\left[ ABC\\right] =\\frac{AB\\cdot BC}{2}=\\frac{ab}{2}$\r\n\r\nand\r\n\r\n$\\left[ ADC\\right] =\\frac{CD\\cdot DA}{2}=\\frac{cd}{2}$,\r\n\r\nrespectively, where [XYZ] denotes the area of a triangle XYZ. Now, the area of the quadrilateral ABCD is\r\n\r\n$\\left[ ABCD\\right] =\\left[ ABC\\right] +\\left[ ADC\\right] =\\frac{ab}{2}+\\frac{cd}{2}=\\frac{ab+cd}{2}$;\r\n\r\nhence,\r\n\r\n$\\left[ ABCD\\right] ^{2}=\\left( \\frac{ab+cd}{2}\\right) ^{2}=\\frac{\\left( ab+cd\\right) ^{2}}{4}$.\r\n\r\nOn the other hand, after the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15973]Brahmagupta formula[/url],\r\n\r\n$\\left[ ABCD\\right] ^{2}=\\left( p-a\\right) \\left( p-b\\right) \\left( p-c\\right) \\left( p-d\\right) $.\r\n\r\nThus,\r\n\r\n$\\left( p-a\\right) \\left( p-b\\right) \\left( p-c\\right) \\left( p-d\\right) =\\frac{\\left( ab+cd\\right) ^{2}}{4}$\r\n\r\nas intended to prove.\r\n\r\nNice problem, thanks!\r\n\r\n  Darij",
  "Solution_2": "I enjoyed the problem too, but when I had to solve it for a countest, I didn't have any clue... :D"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "calculus",
    "derivative",
    "trigonometry",
    "function",
    "search"
  ],
  "Problem": "[color=darkred]Let $ z\\in\\mathbb C$ for which exists $ n\\in\\mathbb N$ so that $ n\\ge 2$ and $ 2^{n \\minus{} 1}\\cdot\\left(z^n \\plus{} 1\\right) \\equal{} (z \\plus{} 1)^n$ . Prove that $ |z| \\equal{} 1$ .[/color]",
  "Solution_1": "For $ n\\geq 2$, the polynomial $ 2^{n\\minus{}1}(x^n\\plus{}1)\\minus{}(x\\plus{}1)^n$ has nonzero $ x^n$ coefficient, so it has $ n$ complex roots counting multiplicities. Now we can see (by taking derivatives, for example) that 1 is a double root of the polynomial, so it remains to find $ n\\minus{}2$ other complex roots with absolute value 1. (If $ n\\equal{}2$, we're already done.)\r\n\r\nSuch roots will have the form $ e^{\\theta i}$ for $ \\frac{\\theta}{2\\pi}\\not\\in \\mathbb{Z}$. Now we have\r\n\\[ e^{\\theta i}\\plus{}1\\equal{}e^\\frac{\\theta i}{2}(2\\cos \\frac{\\theta}{2})\\]\r\nso after dividing out $ 2^n e^\\frac{n\\theta i}{2}$, it's equivalent to\r\n\\[ \\cos \\frac{n\\theta}{2}\\equal{}\\cos^n \\frac{\\theta}{2}\\]\r\nWe need to find $ n\\minus{}2$ solutions in $ (0,2\\pi)$. To show this, let $ f(\\theta)\\equal{}\\cos \\frac{n\\theta}{2}\\minus{}\\cos^n \\frac{\\theta}{2}$. Then we have $ f\\left( \\frac{2\\pi k}{n}\\right)\\equal{}(\\minus{}1)^k\\minus{}\\cos^n \\frac{\\pi k}{n}$ for every integer $ k$. For the case $ 1\\leq k\\leq n\\minus{}1$, $ f$ is strictly negative for $ k$ odd and strictly positive for $ k$ even. That's $ n\\minus{}2$ sign changes, so the continuous $ f$ necessarily has $ n\\minus{}2$ zeroes in $ \\left[ \\frac{2\\pi}{n},\\frac{2\\pi (n\\minus{}1)}{n}\\right]$.\r\n\r\nSo if $ z$ were a complex number satisfying $ 2^{n\\minus{}1}(z^n\\plus{}1)\\equal{}(z\\plus{}1)^n$ for some $ n\\geq 2$, then $ z$ is the root of a polynomial with all of its complex roots having absolute value 1. This forces $ |z|\\equal{}1$, as desired.",
  "Solution_2": "[b][u]Please[/u], without continue functions or derivatives[/b], if it is possible ! \n\nI have an own such proof. See and the exercises from [url=http://www.artofproblemsolving.com/viewtopic.php?t=186913][color=red][b]here[/b][/color][/url] and from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1438694044&t=42951][color=red][b]here[/b][/color][/url].",
  "Solution_3": "I\"ll present here my proof [b][u]without Darboux's property or derivatives[/u].[/b]\n\n[color=darkred]Let $ z\\in\\mathbb C$  for which exists $ n\\in\\mathbb N$ so that $ n\\ge 2$ and $ 2^{n \\minus{} 1}\\left(z^n \\plus{} 1\\right) \\equal{} (z \\plus{} 1)^n$ . Prove that $ |z| \\equal{} 1$ .[/color]\n\n[b][u]Proof.[/u][/b] Observe that $ z\\ne 0\\ (z\\equal{}0\\implies n\\equal{}1)$ . Suppose w.l.o.g. $ z\\ne 1$ . Denote $ |z| \\equal{} r$ . Thus, $ \\overline z \\equal{} \\frac {r^2}{z}$ and \n\n$ 2^{n \\minus{} 1}\\left(z^n \\plus{} 1\\right) \\equal{} (z \\plus{} 1)^n$ $ \\Longleftrightarrow$ $ 2^{n \\minus{} 1}\\left(\\overline z^n \\plus{} 1\\right) \\equal{} (\\overline z \\plus{} 1)^n$ $ \\Longleftrightarrow$ $ 2^{n \\minus{} 1}\\left(r^{2n} \\plus{} z^n\\right) \\equal{} \\left(r^2 \\plus{} z\\right)^n$ .\n\nSuppose $ \\mathrm {ad\\ absurdum}$ that $ r\\ne 1$ . Therefore, $ \\left\\{\\begin{array}{c} 2^{n \\minus{} 1}\\left(z^n \\plus{} 1\\right) \\equal{} (z \\plus{} 1)^n \\\\\n \\\\\n2^{n \\minus{} 1}\\left(r^{2n} \\plus{} z^n\\right) \\equal{} \\left(r^2 \\plus{} z\\right)^n\\end{array}\\ \\right\\|\\ \\boxed \\minus{} \\ \\uparrow$ $ \\implies$ \n\n$ 2^{n \\minus{} 1}\\left(r^{2n} \\minus{} 1\\right) \\equal{} \\left(r^2 \\plus{} z\\right)^n \\minus{} (z \\plus{} 1)^n$ $ \\implies$ $ \\frac {2^{n \\minus{} 1}\\left(r^{2n} \\minus{} 1\\right)}{r^2 \\minus{} 1} \\equal{} \\left|\\sum_{k \\equal{} 1}^n\\left(r^2 \\plus{} z\\right)^{n \\minus{} k}(z \\plus{} 1)^{k \\minus{} 1}\\right|\\le$ \n\n$ \\sum_{k \\equal{} 1}^n\\left(r^2 \\plus{} r\\right)^{n \\minus{} k}(r \\plus{} 1)^{k \\minus{} 1} \\equal{}$ $ (r \\plus{} 1)^{n \\minus{} 1}\\cdot\\sum_{k \\equal{} 1}^nr^{n \\minus{} k}$ $ \\implies$  $ \\frac {2^{n \\minus{} 1}\\left(r^{2n} \\minus{} 1\\right)}{r^2 \\minus{} 1}\\le (r \\plus{} 1)^{n \\minus{} 1}\\cdot \\frac {r^n \\minus{} 1}{r \\minus{} 1}$ $ \\implies$ \n\n$ \\boxed {\\ 2^{n \\minus{} 1}\\left(r^n \\plus{} 1\\right)\\le (r \\plus{} 1)^{n}\\ \\ (*)\\ }$ because $ \\frac {r^n \\minus{} 1}{r \\minus{} 1} > 0$ . From the relation $ (*)$ and the well-known inequality  \n\n$ 2^{n \\minus{} 1}\\left(r^n \\plus{} 1\\right)\\ge (r \\plus{} 1)^{n}$ obtain $ 2^{n \\minus{} 1}\\left(r^n \\plus{} 1\\right) \\equal{} (r \\plus{} 1)^{n}$ , i.e. $ r \\equal{} 1$ , what is $ \\mathrm {absurd}$ . In concluzion, \n\n$ 2^{n \\minus{} 1}\\left(z^n \\plus{} 1\\right) \\equal{} (z \\plus{} 1)^n$ $ \\implies$ $ |z| \\equal{} 1$ .\n\n[b][u]Remark.[/u][/b] Here is an equivalent enunciation :\n\n[color=darkred]Let $ \\{u,v\\}\\subset \\mathbb C$ for which exists $ n\\in\\mathbb N$ so that $ n\\ge 2$ and $ 2^{n \\minus{} 1}\\left(u^n \\plus{} v^n\\right) \\equal{} (u\\plus{} v)^n$ . Prove that $ |u| \\equal{} |v|$ .[/color]"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "geometric transformation",
    "reflection",
    "geometry unsolved"
  ],
  "Problem": "Let $ P$ be an interior point of triangle $ ABC$, the cevian triangle of $ P$ is $ LMN$, determine $ P$ such that the perimeter of triangle $ LMN$ attains minimum.",
  "Solution_1": "It is well- known that this occurs for the orthocenter. \r\nSketch: We will actually show that the triangle with vertices as the feet of the altitudes gives the minimal perimeter of any triangle with vertices on the sides. Fix L on AB, let P and Q be the reflections of it over AC and BC, and X and Y the intersections of PQ with AC and AB. PX=LX and QY=LY then, so we see the perimeter of the inscribed triangle equals PQ, which is clearly minimized when M=X and N=Y. So we have to find the best choice for L. Since CP=CQ=CL, and angle PCQ is fixed, we find this occurs when CL is minimum, or L is the foot of the perpendicular from C to AB. We just have to show now that in this case, M and N also turn out to be the feet of the perpendiculars.\r\n\r\nThis was first given by Hungarian mathematician L Fejer in 1900, it is given in the book Geometric Problems on Maxima and Minima by Titu Andreescu, Oleg Mushkarov, and Luchezar Stoyanov",
  "Solution_2": "Exercise: Find all mistakes in this solution.\r\n\r\n  darij",
  "Solution_3": "Is it because the triangle has to be acute  :P ? Or are there more mistakes?",
  "Solution_4": "[quote=\"dgreenb801\"]Is it because the triangle has to be acute  :P ?[/quote]\r\n\r\nYes, this is the point where one starts wondering why the solution yields a wrong answer.\r\nSee http://www.cip.ifi.lmu.de/~grinberg/errata.html for details (scroll to \"Page 169, \u00a7264, Proof of the last Theorem\" in Johnson).\r\n\r\n  darij"
}
{
  "Tag": [
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "I am trying to find a noncommutative ring without a  unity...please help... thanks",
  "Solution_1": "noncommutative rings can easily be found by taking the ring of n dimensional square matrices over a ring $R$ , we denote this by $M_{n}(R)$ \r\n\r\nnow for what you ask i suggest you take the ring of all 2 by 2 matrices, with entries even integers",
  "Solution_2": "Thank you!!"
}
{
  "Tag": [],
  "Problem": "The number 64 has the property that it is divisible by its units\ndigit. How many whole numbers between 10 and 50 have this property?",
  "Solution_1": "let x be the tens digit, and y the unit digit.\r\n\r\nThen, 10x+y is divisible by y.\r\n$ \\frac{10x\\plus{}y}{y} \\equal{} \\frac{10x}{y} \\plus{} y$ has to be an integer.\r\nSo $ \\frac{10x}{y}$ is an integer.\r\n\r\nWe systematically check every integer between 10 and 50 for the property that 10x/y is an integer.\r\nThere aren't that many... the ones that work are:\r\n11, 12, 15, 21, 22, 24, 25, 31, 32, 33, 35, 36, 41, 42, 44, 45, 48\r\n\r\nTherefore, 17 integers satisfy this property.\r\n\r\nEDIT: misread problem... corrected\r\nThis is a somewhat slow method... does anyone have there a better way to solve this problem?",
  "Solution_2": "Why does 31 work though...it is not divisible by 3...",
  "Solution_3": "$ 1$ is the units digit for $ 31$, and the question stipulates divisibility by the units digit, which is obviously true for $ 31$.",
  "Solution_4": "If you think about it, you can tell that some numbers are always divisible by their units digit. Every integer is divisible by 1, so all integers ending in 1 will work. Every integer ending in 2 will work because all integers ending in 2 are even which means that they are divisible by 2. Every integer ending in 5 will work because all integers ending in 5 are divisible by 5.\r\n\r\nSimilarly, none of the integers ending in 0 will work because integers are not divisible by 0.",
  "Solution_5": "[hide=Alcumas Answer]Twelve numbers ending with 1, 2, or 5 have this property. They are 11, 12, 15, 21, 22, 25, 31, 32, 35, 41, 42, and 45. In addition, we have 33, 24, 44, 36, and 48, for a total of $\\boxed{17}$. (Note that 20, 30, and 40 are not divisible by 0, since division by 0 is not defined.)[/hide]",
  "Solution_6": "[quote=BarbarianKing][hide=Alcumas Answer]Twelve numbers ending with 1, 2, or 5 have this property. They are 11, 12, 15, 21, 22, 25, 31, 32, 35, 41, 42, and 45. In addition, we have 33, 24, 44, 36, and 48, for a total of $\\boxed{17}$. (Note that 20, 30, and 40 are[b] not divisible by 0, since division by 0 is not defined.)[/hide][/quote]\n\nYou spelled alucmas wrong. It is alc[b][b][b]u[/b][/b][/b]mus.",
  "Solution_7": "I think @BarbarianKing does it on purpose...it's been pointed out before -.-",
  "Solution_8": "[quote=rubik]let x be the tens digit, and y the unit digit.\n\nThen, 10x+y is divisible by y.\n$ \\frac{10x\\plus{}y}{y} \\equal{} \\frac{10x}{y} \\plus{} y$ has to be an integer.\nSo $ \\frac{10x}{y}$ is an integer.\n\nWe systematically check every integer between 10 and 50 for the property that 10x/y is an integer.\nThere aren't that many... the ones that work are:\n11, 12, 15, 21, 22, 24, 25, 31, 32, 33, 35, 36, 41, 42, 44, 45, 48\n\nTherefore, 17 integers satisfy this property.\n\nEDIT: misread problem... corrected\nThis is a somewhat slow method... does anyone have there a better way to solve this problem?[/quote]\n\nI did it this way too. But seems only slightly better than brute force. Amy other way to do it?",
  "Solution_9": "the answer is 17",
  "Solution_10": "Is there a faster way to do this problem maybe with an equation because this problem was on the amc 8 and you only have 40 minutes to answer 25 questions on the amc 8\n                    Thank you so much.",
  "Solution_11": "This way can be pretty fast, with practice. \n[hide]We automatically see that numbers that end in 1,2, or 5 qualify, so that is 12 numbers so far. Then you can just check each of the other numbers by using the divisibility rules. You get 5 more possibilities, and 12+5=17. [/hide]",
  "Solution_12": "why isn't 24 in this? 24 is divisible by 4"
}
{
  "Tag": [
    "limit",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "if $ a_{1}\\equal{}\\frac{1004}{3}$ and $ \\frac{a_{1}}{1}\\plus{}\\frac{a_{2}}{2}\\plus{}...\\plus{}\\frac{a_{n}}{n}\\equal{}\\frac{n\\plus{}1}{2}\\bullet a_{n}$ .  Find $ lim_{n\\minus{}>\\propto}(2008\\plus{}n)a_{n}\\equal{}?$",
  "Solution_1": "We have $ \\frac {n \\plus{} 1}{2}a_n \\plus{} \\frac {a_{n \\plus{} 1}}{n \\plus{} 1} \\equal{} \\frac {n \\plus{} 2}{2}a_{n \\plus{} 1}$ then $ a_{n \\plus{} 1} \\equal{} \\frac {(n \\plus{} 1)^2}{n(n \\plus{} 3)}a_n$.\r\nFrom this we have $ a_n \\equal{} \\frac {3n}{(n \\plus{} 1)(n \\plus{} 2)}a_1$\r\nThen $ \\lim_{n\\rightarrow\\infty}(2008 \\plus{} n)a_n \\equal{} 1004$"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Given a triangle ABC,knowing that there exists 3 points X,Y,Z lie on BC,CA ,AB respectively such that AX=BY=CZ and angle BAX=CBY=ACZ.Prove that ABC is equilateral.",
  "Solution_1": "[quote=\"SUPERMAN2\"]Given a triangle ABC,knowing that there exists 3 points X,Y,Z lie on BC,CA ,AB respectively such that AX=BY=CZ and angle BAX=CBY=ACZ.Prove that ABC is equilateral.[/quote]\r\nHelp me with this problem,please."
}
{
  "Tag": [
    "function",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Find the least natural number $ n$ ($ n\\ge3$) with the following property: for any coloring in 2 colors of $ n$ distinct collinear points $ A_1,A_2,\\ldots,A_n$ such that $ A_1A_2\\equal{}A_2A_3\\equal{}\\cdots\\equal{}A_{n\\minus{}1}A_n$ there exist three points $ A_i,A_j,A_{2j\\minus{}1}$ ($ 1\\le i\\le2j\\minus{}i\\le n$) which are colored in the same color.",
  "Solution_1": "Let $ f: A_i\\rightarrow \\{1,2\\}$ be the function that associates every $ i$ th point $ A_i$ to one of the colors 1 or 2. Consider the triple $ (A_1,A_1,A_n)$ (which is actually a double $ (A_1,A_n)$), if $ f(A_1)\\equal{}1$ wlog then $ f(A_2)\\equal{}2$. Similarly, $ f(A_3)\\equal{}1$ by considering the triple$ (A_2,A_2,A_3)$.\r\n\r\nConsider the sequence $ \\{a_\\infty \\}\\equal{}\\{3,5,9,\\dots \\}$ which obeys the recurrence relation $ a_{n\\plus{}1}\\equal{}2a_n \\minus{}1$. We shall color the points so that  \r\nif $ f(A_i)\\equal{}1$ then $ f(A_{i\\plus{}1})\\equal{}2$\r\nor if $ f(A_i)\\equal{}2$ then $ f(A_{i\\plus{}1})\\equal{}1$.\r\n\r\nAs we move from left of $ \\mathbf{N}$ from 3 onwards we observe that either $ k$ th point belongs to a sequence else a new sequence starts from there (for example like $ A_4$) with the same recurrence relation (for example like $ 4,7,13,\\dots$). Since every pair of form $ (j,2j\\minus{}1)$ have different colours, we cant find a triple with all of them having same colour.\r\n\r\nWe are left to prove that no two sequences have a common element. Let $ \\{a_\\infty\\}$ and $ \\{b_\\infty\\}$ be two different sequences. Let $ a_k\\equal{}b_r$ which implies $ a_1\\equal{}b_{r\\minus{}k\\plus{}1}$ which means we are trying to start a sequence from a certain point when its already a part of sequence $ b$. Since this is not possible, we conclude that no two sequences have a common element.\r\n\r\nHence there does not exist a least natural number $ n$.",
  "Solution_2": "The coloring BWWBBWWB doesnt have any triple of points satisfying the condition.\r\n\r\nIf there are 9 points, there are 5 points in one color, let's say black. We can't color three consecutive points on the same color, and if there are no adjacent black points the arrangement is BWBWBWBWB which clearly satisfies the condition. So there are two adjacent points in black. Then, there are three cases  (WE DONT KNOW THE COLOR X):\r\n\r\n- BBW --> BBWB --> BBWBXWW --> BBWBBWWXX --> B[b]B[/b]WB[b]B[/b]WW[b]B[/b]X  Contradiction.\r\n\r\n- BBW --> BBWW --> BBWWB --> BBWWBXXWW--> BBW[b]WBXBW[/b]W If X = B or W there is BBB or WBWBW. Contradiction.\r\n\r\n- WBBW --> WBBWXXB --> WBBWBXB --> WB[b]B[/b]W[b]B[/b]W[b]B[/b] Contradiction.\r\n\r\nThen n=9"
}
{
  "Tag": [
    "geometry",
    "perimeter"
  ],
  "Problem": "What is the smallest possible perimeter, in units, of a triangle whose side-length measures are consecutive integer values?",
  "Solution_1": "start trying some-\r\n1,2,3, but that is degenerate\r\n2,3,4 works\r\n\r\n$ 2\\plus{}3\\plus{}4\\equal{}9$"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation",
    "reflection",
    "AwesomeMath",
    "summer program",
    "pigeonhole principle"
  ],
  "Problem": "Find the six configurations that are unique under rotation and reflection of four distinct points on the plane such that among the $ \\binom{4}{2} \\equal{} 6$ line segments that may be drawn with the points as endpoints, there are only 2 distinct lengths. Prove that these are the only six configurations. (For example, four points on the vertices of a square would be one configuration, having only the lengths $ k$ and $ k\\sqrt {2}$.)\r\n\r\nEDIT: I know 6 configurations, but not how to prove that they are the only ones.",
  "Solution_1": "No responses?! :(  Is the problem worded poorly? Is the problem trivial? Is it too difficult?\r\n\r\n(Problem from Zuming Feng, AwesomeMath 2008, don't know the original source though)",
  "Solution_2": "This seems like an ugly casework problem...\r\n\r\nTake one vertex, it has 5 edges emanating from it. Since we have two lengths, we must have at least 3 edges of the same length by the pidgeonhole principle. From here, it should be pretty easy to finish: the endpoints of these edges make an isosceles triangle, etc."
}
{
  "Tag": [
    "factorial",
    "algebra",
    "binomial theorem"
  ],
  "Problem": "proove that:n!/r!(n-r)!+n!/(r+1)!(n-r+1)=(n+1)!/r!(n-r+1)! :wink:",
  "Solution_1": "[quote=\"21 7 15\"]Prove that: $ \\frac{n!}{r!(n\\minus{}r)!}\\plus{}\\frac{n!}{(r\\plus{}1)!(n\\minus{}r\\plus{}1)!}\\equal{}\\frac{(n\\plus{}1)!}{r!(n\\minus{}r\\plus{}1)!}$[/quote]\r\n\r\n[hide=\"solution\"]$ \\frac{n!}{r!(n\\minus{}r)!}\\equal{}\\binom{n}{r}$\n\n$ \\frac{n!}{(r\\plus{}1)!(n\\minus{}r\\plus{}1)!}\\equal{}\\binom{n}{r\\plus{}1}$\n\n$ \\frac{(n\\plus{}1)!}{r!(n\\minus{}r\\plus{}1)!}\\equal{}\\binom{n\\plus{}1}{r}$\n\n[[Pascal's Identity]][/hide]",
  "Solution_2": "In the original problem, shouldn't $ (r \\plus{} 1)!$ be $ (r \\minus{} 1)!$ instead?",
  "Solution_3": "hey plz explain how does (n/r)+(n/(r+1)=(n+1)/r",
  "Solution_4": "yes sorry(r-1)!instead :|",
  "Solution_5": "[hide=\"A combinatorial approach\"]We are to count the number of teams of $ r$ members that can be formed from a group of $ n \\plus{} 1$ people. Choose any one of the people and call him $ X$. The number of teams that already has $ X$ as a member is $ \\binom{n}{r \\minus{} 1}$, and the number of teams that does not have $ X$ as a member is $ \\binom{n}{r}$. Together they sum up to the total number of teams $ \\binom{n \\plus{} 1}{r}$.[/hide]",
  "Solution_6": "Or you can just do it algebraically..",
  "Solution_7": "The original identity should be written $ {n \\choose r} \\plus{} {n \\choose r\\minus{}1} \\equal{} {n\\plus{}1 \\choose r}$.  The combinatorial proof is instructive, but the algebraic proof stays valid when $ n$ isn't a positive integer. $ n$ can actually be any complex number, where\r\n\r\n$ {n \\choose r} \\equal{} \\frac{n(n\\minus{}1)...(n\\minus{}(r\\minus{}1))}{r!}$\r\n\r\nand $ r$ is a non-negative integer.  This allows us to state the [i]generalized Binomial Theorem[/i], which is the expansion of $ (x \\plus{} 1)^n$ for $ n$ not a non-negative integer (this expansion is infinite)."
}
{
  "Tag": [],
  "Problem": "Solve these two equations to obtain f(x).\r\n\r\n$ f(x) \\plus{} f\\left(\\frac{1}{x}  \\right ) \\equal{} \\frac{1}{a\\plus{}b}\\left [ x \\plus{} \\frac{1}{x}\\minus{}10 \\right ]$\r\n\r\n$ f(x)\\minus{} f\\left ( \\frac{1}{x} \\right )\\equal{} \\frac{1}{a\\minus{}b}\\left [ \\frac{1}{x}\\minus{}x \\right ]$\r\n\r\nAnswer : $ \\frac{1}{a^2\\minus{}b^2}\\left \\{ \\frac{a}{x} \\minus{} (b\\times x) \\right \\}\\minus{} \\frac{5}{a\\plus{}b}$\r\n\r\nThank you.",
  "Solution_1": "[quote=\"ninni\"]Solve these two equations to obtain f(x).\n\n$ f(x) \\plus{} f\\left(\\frac {1}{x} \\right ) \\equal{} \\frac {1}{a \\plus{} b}\\left [ x \\plus{} \\frac {1}{x} \\minus{} 10 \\right ]$\n\n$ f(x) \\minus{} f\\left ( \\frac {1}{x} \\right ) \\equal{} \\frac {1}{a \\minus{} b}\\left [ \\frac {1}{x} \\minus{} x \\right ]$\n\nAnswer : $ \\frac {1}{a^2 \\minus{} b^2}\\left \\{\\frac {a}{x} \\minus{} (b\\times x) \\right \\} \\minus{} \\frac {5}{a \\plus{} b}$\n\nThank you.[/quote]\r\n\r\n[url=http://www.servimg.com/image_preview.php?i=264&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk13.jpg[/img][/url]\r\n :coolspeak:",
  "Solution_2": "[quote=\"almonkez\"][quote=\"ninni\"]Solve these two equations to obtain f(x).\n\n$ f(x) \\plus{} f\\left(\\frac {1}{x} \\right ) \\equal{} \\frac {1}{a \\plus{} b}\\left [ x \\plus{} \\frac {1}{x} \\minus{} 10 \\right ]$\n\n$ f(x) \\minus{} f\\left ( \\frac {1}{x} \\right ) \\equal{} \\frac {1}{a \\minus{} b}\\left [ \\frac {1}{x} \\minus{} x \\right ]$\n\nAnswer : $ \\frac {1}{a^2 \\minus{} b^2}\\left \\{\\frac {a}{x} \\minus{} (b\\times x) \\right \\} \\minus{} \\frac {5}{a \\plus{} b}$\n\nThank you.[/quote]\n\n[url=http://www.servimg.com/image_preview.php?i=264&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk13.jpg[/img][/url]\n :coolspeak:[/quote]\r\n\r\n\r\nThank you. I got that. I was solving it some different way, by that I could not reach to the solution. thanks. :-)"
}
{
  "Tag": [],
  "Problem": "i got this calculator cable for TI-83 plus and i dont get how to use it\r\n\r\ni downloaded the TI connect...now what",
  "Solution_1": "*sigh*...\r\n\r\nwhat a n00b...\r\n\r\n\r\n\r\nanyway, you then plug the calc in if you need to download something or another.",
  "Solution_2": "one end goes in the calculator, the other end goes in the USB port. then magic happens. or actually, use TI-Connect to send the downloads to the calculator.",
  "Solution_3": "well obviouslyi know how to plug my calculator in :D  :mad: \r\n\r\ni figured it out but i didnt do it the way you did...",
  "Solution_4": "now on my TI-89 titanium, how do i activate programs?",
  "Solution_5": "Has anybody have had trouble syncing their TI-89 Titanium to their computer? I plug in the USB cable, and the computer recognizes it but the software doesn't, reinstalled the softare already, any hints?",
  "Solution_6": "[quote=\"elston\"]Has anybody have had trouble syncing their TI-89 Titanium to their computer? I plug in the USB cable, and the computer recognizes it but the software doesn't, reinstalled the softare already, any hints?[/quote]\r\n\r\nAre you using TI Connect?  Sometimes the computer takes a couple seconds to recognize that the calculator is plugged in.  If it doesn't start working I would download the most recent version of TI Connect (I think its freely available on the net) and install that.\r\n\r\nI use TI Connect with the silver link cable on a TI-89 and I've never experienced any problems."
}
{
  "Tag": [
    "calculus",
    "LaTeX",
    "integration",
    "search",
    "\\/closed"
  ],
  "Problem": "I want to give a soluiton for a calculus problem that will be a reference item\r\n\r\nHow are readers supposed to find the reference if they can't see it in the subject line??  For instance I just tried to post the latex code for$\\int cos\\theta d\\theta$ in the subject line but it will not display",
  "Solution_1": "[quote]How are readers supposed to find the reference if they can't see it in the subject line?? [/quote]\r\nThere is a search button. Many people, including myself, consider it a bad form to use math symbols in the title of a math paper.",
  "Solution_2": "For reasons explained above (bad idea in a math paper to do that) and also for display reason that feature will not be implemented."
}
{
  "Tag": [],
  "Problem": "Who here is going to PAMO this year?\r\nI just heard that I am going :wow:",
  "Solution_1": "congrats  :D \r\n\r\nour team hasn't been selected  yet , but i think i'am going too :lol:",
  "Solution_2": "Sorry for the question, but where can I find problems of this olympiad???? Thanks...........",
  "Solution_3": "You can find them on this site's resources page :) \r\n[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=28[/url]",
  "Solution_4": "[quote=\"Melissa\"]You can find them on this site's resources page :) \n[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=28[/url][/quote]\r\n\r\nThanks!!! Nice problems",
  "Solution_5": "I'm going!!!!!!! Like you didn't know that :D\r\ncan't wait!!"
}
{
  "Tag": [
    "email",
    "limit",
    "trigonometry",
    "LaTeX",
    "calculus",
    "derivative",
    "calculus computations"
  ],
  "Problem": "I will paypal someone 20 dollars if they can work out these 13 problems and show the work step by step..... the questions can be found here...   http://www.southwestdental.net/cal.pdf .  I need these answeres by mid afternoon thursday 2-16-06 because my midterm is friday and I need to study them but very few people at my school are willing to help.  Please email me if you would like to do this or think you could possibly help.\r\n\r\nskgallion@gmail.com\r\nSteve",
  "Solution_1": "[quote=\"famtrecrew\"]I will paypal someone 20 dollars [/quote].  \nWrong place...\n[quote=\"famtrecrew\"]\nI need these answeres by mid afternoon thursday 2-16-06\n[/quote]\nWrong time...\n[quote=\"famtrecrew\"]\nmy midterm is friday and I need to study them but very few people at my school are willing to help.  \n[/quote]\r\nThat's a real pity. Maybe you should consider getting a private tutor. Here you can get help any time for free but you need to post problems one per time and let people solve them at their leisure. I will be not surprised if nobody gets interested in your offer but that's not because people are unwilling to help, just because you ask for too much and too soon at once...\r\n\r\nAs to the problems themselves, here are the  solutions for #1 and #13. Maybe somebody else will help you with other ones :)\r\n\r\n#1. The limit of the sum is the sum of the limits (provided that both limits exist). Now $\\lim_{x\\to\\infty}\\frac{x^2+\\sin x}{x^2-2x-3\\cos x}=\\lim_{x\\to\\infty}\\frac{1+\\frac{\\sin x}{x^2}}{1-\\frac 2x-3\\frac{\\cos x}{x^2}}$. Note that every expression in the numerator and in the denominator except $1$ tends to $0$, so the limit is $\\frac 1 1=1$. The second limit $\\lim_{x\\to\\infty} \\frac{\\sin 1/x}{1/x}=\\lim_{y\\to 0}\\frac{\\sin y}{y}=1$ The latter one of the few standard limits you should know by heart. So, the entire expression tends to $1+1=2$.\r\n\r\n#13. Since $f(x)$ is continuous, it is enough to find two points where $f$ takes values of different sign. But the direct computation shows that $f(0)=-3<0$ while $f(2)=7>0$.",
  "Solution_2": "Speaking as a TA:\r\nI wouldn't do this for one of my own students; why would I do it for you?\r\n\r\nThe right way to ask for help on these:\r\n[b]Work them first yourself[/b]. Then, when you get stuck on a few of them, ask for specific help on those problems. You'll find people are much more likely to answer a few specific questions than to do the whole list of practice problems for you.",
  "Solution_3": "[quote=\"jmerry\"]\nThe right way to ask for help on these:\n[b]Work them first yourself[/b]. Then, when you get stuck on a few of them,...[/quote]\r\nAn excellent advice! I would sign under it except I'm afraid that we are dealing with the worst case scenario here, i.e., famtrecrew studied so well during the term that he got stuck on [b]all[/b] of them. Whether something can (or even should) be done to help a person in such a case, is a topic for the Round Table rather than for Computations and Tutorials. I'm inclined to give almost everybody a second chance, so, sometimes I would do it for my students. (Actually, I usually run a review session before each exam and they are allowed to ask questions about how to solve the practice problems. Of course, the assumption is that they have tried their best to solve them themselves, but I never try hard to verify this assumption). Anyway, as I said, the place and the time are wrong but if some people have free time and mercy, why not to give famtrecrew one time what he wants? If it helps him to study, that'll be great; if he is beyond any help by now, it wouldn't make the situation any worse. Not that I myself have time right away to write lengthy explanations for 11 problems (though, if he chooses just 2 more of them, I'll try to help).",
  "Solution_4": "[quote=\"jmerry\"]Speaking as a TA:\nI wouldn't do this for one of my own students; why would I do it for you?\n\nThe right way to ask for help on these:\n[b]Work them first yourself[/b]. Then, when you get stuck on a few of them, ask for specific help on those problems. You'll find people are much more likely to answer a few specific questions than to do the whole list of practice problems for you.[/quote]\r\n\r\nI did all of them myself...but with no answers to compare it to how do i know if they are right??? I would gladly post all the problems that I worked out but i dont know how to show the work in this forum the way you guys do....",
  "Solution_5": "You don't have to use $\\LaTeX$ to post here- we can read plain text just fine, with appropriate parentheses, ^ for exponents, etc. Mathematical meaning is more important than looks.",
  "Solution_6": "ok so i have worked through the majority of the practice test and the problem that is holding me up the most is 4 and 9.  4 states: Find the second derivative d^2y/dx^2 , to the parametric curve defined by x = (2t+1)^(1/2) , y= 2t^2+2t when t=4.  When i work this problem out I am using the quotent rule and coming up with something like this=> (4t+2)/[(1/2)(2)^(-1/2) but past that (which i think is probably wrong anyway) i dont know what to do........  \r\n\r\nOn problem 9 it states.  A drop of mist is a perfect sphere.  The drop picks up moisture at a rate proportional to its surface area.  Show that under these circumstances the drop's radius increases at a constant rate.  I am completly confused on how to even approach this problem.  If someone could post a hint on where to start I would appreciate it.. it does not have to be a full answer just a starting point would help.",
  "Solution_7": "[quote=\"famtrecrew\"]\nI did all of them myself...but with no answers to compare it to how do i know if they are right??? [/quote]\r\nIf all you need is answers, that is easy (but why didn't you say so in the first place? :?)\r\nHere goes:\r\n\r\n2) $f'(x)=\\frac{x-3}{(x-1)x}+\\frac{x+2}{(x-1)x}-\\frac{(x+2)(x-3)}{(x-1)^2 x}-\\frac{(x+2)(x-3)}{(x-1)x^2}$. Linearization at $2$: $y=\\frac 92(x-2)-2$ (of course, your answers may be written in a different form, but they should be the same up to algebraic manipulations)\r\n\r\n3) a) No\r\n    b) The limit is $0$\r\n\r\n4) $54$\r\n\r\n5) $2x+\\cos((x^2+1)^2)\\cdot 2(x^2+1)\\cdot 2x$ (the expression in the problem is written in an ambiguous way. I understood it as $\\sin((x^2+1)^2)$. If  you interpreted it as $(\\sin(x^2+1))^2$, the answer becomes $2x+2\\sin(x^2+1)\\cos(x^2+1)\\cdot 2x$.\r\n\r\n6)  $\\frac{16}3$ feet per second.\r\n\r\n7) $-36$\r\n\r\n8) a) $x=1, y=1$ or $x=-1, y=1$.\r\n    b)  There must be some misprint here: there are no real $x$ such that $x^2+1+1=1$\r\n\r\n10) The tangent line is $y-2=-\\frac 14(x-1)$; the normal is $y-2=4(x-1)$\r\n\r\n11) About $1\\%$\r\n\r\n12) $2$\r\n\r\nI hope I haven't made a stupid mistake in my computations. If some of your answers disagrees with mine, we can go over the corresponding problem in detail. This leaves you with 2 unanswered questions: the third part of 3 and 9. Before answering them, I'd like to see how you did them.",
  "Solution_8": "4: $\\frac{dy}{dx}=\\frac{dy/dt}{dx/dt}=\\frac{4t+2}{2\\cdot\\frac12(2t+1)^{-\\frac12}}= 2(2t+1)^{\\frac32}$\r\nNow $\\frac{d^2y}{dx^2}=\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)=\\frac{d}{dt} \\left(\\frac{dy}{dx}\\right)\\cdot\\frac1{\\frac{dx}{dt}}=4\\cdot\\frac32\\cdot(2t+1)^{\\frac12}\\cdot \\frac1{(2t+1)^{-\\frac12}}=6(2t+1)$\r\n\r\n9: Write $V=\\frac{4\\pi}3r^3$. Differentiate with respect to $t$. Write $A=4\\pi r^2$. Notice anything?"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find all pairs of integers $ (m,n)$ such that:\r\n\r\n$ |(m^2\\plus{}2000m\\plus{}999999)\\minus{}(3n^3\\plus{}9n^2\\plus{}27n)|\\equal{}1.$",
  "Solution_1": "Let $ x\\equal{}m\\plus{}1000$, then $ f(x,n)\\equal{}x^2\\minus{}1\\minus{}3n(n^2\\plus{}3n\\plus{}9), |f(x,n)|\\equal{}1$. If $ 3\\not x$, then $ 3|f(x,n)$ - contradition with $ f(x,n)\\equal{}\\pm 1$.\r\nIf $ 3|x$, then $ 3|n, 9|x$ , and $ x\\equal{}9t,n\\equal{}3y$ give $ t^2\\equal{}y(y^2\\plus{}y\\plus{}1)$.\r\nIf $ t\\equal{}0$ we get solution $ m\\equal{}\\minus{}1000,n\\equal{}0$,else $ y\\equal{}k^2, k^4\\plus{}k^2\\plus{}1\\equal{}l^2$ have not solution, because\r\n$ k^2<l<k^2\\plus{}1$."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "this is not a very new problem, but I can't seem to solve.  can someone help?\r\n\r\nThere are $100$ towns in a country and some of them are connected by airlines.  It is known that one can reach every town from any other (perhaps with several intermediate steps).  Prove that you can fly around the country and visit all the towns making no more than\r\na) $198$ flights;\r\nb) $196$ flights.\r\n\r\nThanks.",
  "Solution_1": "We have a connected graph of $n$ vertices and want to touch them all with at most $2n-4$ steps. Obviously, if it is possible for a graph $G$, it's possible for any graph obtained from $G$ by adding edges, so we can remove edges from $G$ till we get a tree (if we don't have a tree, find a cycle and take away an edge; it remains connected).\r\n\r\nSo it's enough to prove it for trees. But now it's obvious: since a tree has $n-1$ edges, the sum of the degrees is $2n-2$, so there's a vertex with degree at most $1$, call it $X$ and let its neighbor be $Y$. Remove $X$ with its edge and take the path that covers the graph with $n-1$ vertices in $2n-6$ steps; to this path, add edges $YX, XY$ where $Y$ appears in the sequence of steps, and we're done.",
  "Solution_2": "why does it follow that from having a degree sum of $2n-2$ that there is a vertex with degree at most $1$? (I know that is true for trees in general but you seemed to derive it from the fact that the sum of degrees was $2n-2$). how do we know that the $n-1$ vertices has a path of $2n-6$?  sorry for being a noob :).\r\nthanks.",
  "Solution_3": "It's ok. Well, if the sum of the degrees is $2n-2$, since there are $n$ vertices and each has degree at least $1$ (there are no isolated vertices), there has to be a vertex of degree $1$, as otherwise the sum would be at least $2 \\times n>2n-2$. The existence of a path of length $\\leq 2n-6$ for the graph with $n-1$ vertices is just inductive hypothesis (note that $2n-6=2(n-1)-4$).",
  "Solution_4": "cool thanks. btw, do you know what $n$ it is first true for?",
  "Solution_5": "For $n=2$, we need a flight, but $2n-4=0$. For $n=3$, we can do with $2$ flights, and $2n-4=6-4=2$, so $3$ is the smallest $n$."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "analytic geometry",
    "symmetry",
    "algebra",
    "polynomial",
    "combinatorics open"
  ],
  "Problem": "I thought I posted this before, but I couldn't find it, so I apologize if this a double-post.\r\n\r\nConsider the unit cube placed at the origin so that each vertex of the cube can be uniquely specified by coordiantes $ (x,y,z)$ where each $ x,y,z$ are either $ 0$ or $ 1$.\r\n\r\nThe problem: how many ways are there to traverse from the vertex $ (0,0,0)$ to $ (1,1,1)$ without passing through the same vertex twice.\r\n\r\nIn this case of the 3-dimensional cube, the solution is easy. By symmetry, we may assume our first step is:\r\n\r\n1). $ (0,0,0)$ to $ (1,0,0)$\r\n\r\nand then multiply our result by 3. Similarly, we may assume our second step is:\r\n\r\n2). $ (1,0,0)$ to $ (1,1,0)$\r\n\r\nand then multiply our result by 2. Given this, it is easy to count by simply looking at a picture that there are 3 ways to travel from $ (1,1,0)$ to $ (1,1,1)$ without going through either $ (0,0,0)$, $ (1,0,0)$, or $ (1,1,0)$. Thus, the answer is $ 3\\cdot 2\\cdot 3\\equal{}18$.\r\n\r\n\r\n\r\n\r\nNow, the real problem: how many ways are there to do this for the 4-dimensional hypercube? That is, how many ways are there to travel from $ (0,0,0,0)$ to $ (1,1,1,1)$ without passing through a vertex twice?",
  "Solution_1": "I don't think anybody really knows how to solve self-avoiding walk problems in polynomial time.  I tentatively suggest that the problem for $ d$ dimensions is exponential in $ d$.  See, for example, [url=http://cat.inist.fr/?aModele=afficheN&cpsidt=1020105]this paper[/url]."
}
{
  "Tag": [
    "calculus",
    "derivative",
    "invariant",
    "geometry",
    "geometric transformation",
    "rotation",
    "algebra"
  ],
  "Problem": "Can anything general be said about the solutions of a functional equation of the form:\r\n$ f^{n}(x)+a_{n-1}f^{n-1}(x)+\\dots+a_{2}f^{2}(x)+a_{1}f(x)+a_{0}x = 0$ (f - real-valued, $ f^{k}(x) = f(f(f(\\dots(x))\\dots)))$ without additional assumptions?",
  "Solution_1": "If the \"characteristic\" equation $ \\lambda^{n}+a_{n-1}\\lambda^{n-1}+\\dots+\\alpha_{0}=0$ has a real root $ \\lambda_{0}$, then $ f(x)=\\lambda_{0}x$ solves the functional equation. This condition is also necessary for the existence of a solution that fixes $ 0$ and has a derivative there. \r\n\r\nLet's consider an example. If $ f(f(x))+x=0$, then the graph of $ f$ must be invariant under clockwise rotation by $ 90$ degrees -- indeed, along with each point $ (x,y)$ the graph contains $ (y,-x)$. Rotating it twice, we find that the graph is centrally symmetric, which means that $ f$ is an odd function. But then $ f\\circ f$ is even*, a contradiction.\r\n\r\nDoes anyone have an example where the \"characteristic\" equation has no real roots, yet the functional equation has a solution?\r\n\r\n(*) Okay, that is wrong. I still think that $ f(f(x))+x=0$ has no solutions."
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Prove that if function $ f: [0,1]\\rightarrow \\mathbb{R}$ satisfy conditions \r\n\r\n$ i) f(0) \\equal{} f(1) \\equal{} 0$ \r\n\r\n$ ii) f\\left(\\frac {a \\plus{} b}{2} \\right) \\leq f(a) \\plus{} f(b) ,\\forall a,b \\in [0,1]$\r\n\r\nthen equation $ f(x) \\equal{} 0$ have infinitely many roots in $ [0,1]$ .Is there any such function different from $ 0$ function",
  "Solution_1": "Of course not. Take for example $ f(x)\\equal{}\\minus{} \\sqrt {x(1 \\minus{} x)}$",
  "Solution_2": "I edited my post ,FelixD ,and sorry for my translating  mistake.",
  "Solution_3": "[quote=\"enndb0x\"]Prove that if function $ f: [0,1]\\rightarrow \\mathbb{R}$ satisfy conditions \n\n$ i) f(0) \\equal{} f(1) \\equal{} 0$ \n\n$ ii) f\\left(\\frac {a \\plus{} b}{2} \\right) \\leq f(a) \\plus{} f(b) ,\\forall a,b \\in [0,1]$\n\nthen equation $ f(x) \\equal{} 0$ have infinitely many roots in $ [0,1]$ .Is there any such function different from $ 0$ function[/quote]\r\n\r\n$ \\forall_{a\\in [0,1]}b \\equal{} a\\Longrightarrow 0\\leq f(a)$ , $ \\ \\forall_{a\\in [0,1]}b \\equal{} 0 \\Longrightarrow 0\\leq f(\\frac a2) \\leq f(a)$ , $ \\ \\forall_{a\\in [0,1]}b \\equal{} 1\\Longrightarrow 0\\leq f(\\frac {a\\plus{}1}2) \\leq f(a)$ \r\nNow it's easy to see that forall $ n$ (natural number) we have  $ f(\\frac 1{2^n}) \\equal{} 0$\r\nExemple  of a such function : $ f(x) \\equal{} 0$ if $ (x\\in [0,1]\\cap Q)$ (rational), $ \\ f(x) \\equal{} 1$ if $ x\\in ([0,1] \\minus{} Q)$ (irational)\r\n :cool:"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Suppose $\\triangle{ABC}$ is a triangle with 3 acute angles A, B, and C. Then the point $(\\cos{B}-\\sin{A}), (\\sin{B}-\\cos{A})$ lies in which quadrants?",
  "Solution_1": "[hide=\"Answer\"]I kind of know how to prove it rigorously, but can't figure out how to express it. Testing gives $\\boxed{\\text{II}}$ as the answer.[/hide]"
}
{
  "Tag": [
    "function",
    "inequalities",
    "complex analysis",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "if $||xy||\\leq C||yx||$ for every pair of elements $x,y\\in B$,where $B$ is banach algebra ,with some constant $C$ ,then $B$ is commutative algebra.",
  "Solution_1": "Hint: this has practically nothing to do with superior algebra, it's analysis.  ;)",
  "Solution_2": "solution?  :lol:",
  "Solution_3": "isn't this standard ? e.g.,\r\n\r\nLet $ x,y$ be arbitrary elements of the banach algebra and pick any linear functional $ \\phi$ of $ B$. Now define an entire function via $ f(z)\\equal{}\\phi\\left(e^{zx}ye^{\\minus{}zx}\\right)$ and observe the following inequality:\r\n\r\n$ \\left|f(z)\\right|\\leq \\parallel \\phi\\parallel \\,\\left|\\left|e^{zx}ye^{\\minus{}zx}\\right|\\right|\\leq C\\,\\parallel \\phi\\parallel \\,\\parallel ye^{\\minus{}zx}e^{zx}\\parallel \\equal{}C\\,\\parallel \\phi\\parallel \\,\\parallel y\\parallel $\r\n\r\nAs a bounded entire function $ f$ must be a constant, i.e. $ 0\\equal{}f'(z)\\equal{}\\phi\\left(e^{zx}(xy\\minus{}yx)e^{\\minus{}zx}\\right)$ and setting $ z\\equal{}0$ we see that the commutator $ xy\\minus{}yx$ must vanish on each linear functional, which is only possible for $ xy\\equal{}yx$."
}
{
  "Tag": [],
  "Problem": "Can anyone prove in plain english why the angle inscribed in a semicircle is a right angle?",
  "Solution_1": "Join the angle with the centre of the semi-circle. Two isoceles triangles appear. Calulcate the angles.",
  "Solution_2": "You can also complete the circle and use the fact that it cuts off a 180 degree arc, and since the point is on the circle, the angle is 180/2 = 90 degrees.",
  "Solution_3": "Since the inscribed angle is one half the arc it intercepts, 1/2*180 = 90."
}
{
  "Tag": [
    "algebra",
    "function",
    "domain",
    "inequalities"
  ],
  "Problem": "prove that :\r\n$ \\sqrt{\\frac{a}{b\\plus{}c}}\\plus{}\\sqrt{\\frac{b}{c\\plus{}d}}\\plus{}\\sqrt{\\frac{c}{d\\plus{}a}}\\plus{}\\sqrt{\\frac{d}{a\\plus{}b}}>2$",
  "Solution_1": "The strict condition is not true -- consider $ (a,b,c,d) \\equal{} (1,0,1,0)$",
  "Solution_2": "Presumably the domain is meant to be positive reals (so that the left-hand side is always meaningful).",
  "Solution_3": "[quote=\"The Zuton Force\"]The strict condition is not true -- consider $ (a,b,c,d) \\equal{} (1,0,1,0)$[/quote]\r\nsorry\r\n$ a,b,c,d \\in R^\\plus{}$",
  "Solution_4": "[quote=\"mathson\"]prove that :\n$ \\sqrt {\\frac {a}{b \\plus{} c}} \\plus{} \\sqrt {\\frac {b}{c \\plus{} d}} \\plus{} \\sqrt {\\frac {c}{d \\plus{} a}} \\plus{} \\sqrt {\\frac {d}{a \\plus{} b}} > 2$[/quote]\r\nwe have $ a \\plus{} b \\plus{} c \\plus{} d \\geq 2\\sqrt {a(b \\plus{} c \\plus{} d)}$ leads to $ \\frac {1}{\\sqrt{a(b \\plus{} c \\plus{} d)}} \\geq \\frac {2}{a \\plus{} b \\plus{} c \\plus{} d}$ which means $ \\sqrt {\\frac {a}{b \\plus{} c \\plus{} d}} \\geq \\frac {2a}{a \\plus{} b \\plus{} c \\plus{} d}$\r\nsetting up 3 other similar one we have the following inequality\r\n$ \\sqrt {\\frac {a}{b \\plus{} c \\plus{} d}} \\plus{} \\sqrt {\\frac {b}{c \\plus{} d \\plus{} a}} \\plus{} \\sqrt {\\frac {c}{d \\plus{} a \\plus{} b}} \\plus{} \\sqrt {\\frac {d}{a \\plus{} b \\plus{} c}} > 2$\r\nhowever , notice that $ \\sqrt {\\frac {a}{b \\plus{} c }} > \\sqrt {\\frac {a}{b \\plus{} c \\plus{} d}}$ ...\r\n\r\nfinally we have desired result :)",
  "Solution_5": "[quote=\"HTA\"] leads to $ \\frac {1}{a(b \\plus{} c \\plus{} d)} \\geq \\frac {2}{a \\plus{} b \\plus{} c \\plus{} d}$ [/quote]\r\n\r\nI think you missed something :wink:",
  "Solution_6": "[quote=\"polskimisiek\"][quote=\"HTA\"] leads to $ \\frac {1}{a(b \\plus{} c \\plus{} d)} \\geq \\frac {2}{a \\plus{} b \\plus{} c \\plus{} d}$ [/quote]\n\nI think you missed something :wink:[/quote]\r\nthanks polskimisiek , it was a typo , i edited it ."
}
{
  "Tag": [],
  "Problem": "1. a balloon has a volume of 5.00 liters of 22 degree celsius and 1.00 atmosphere of pressure.  What will be the balloons's volume in liters if the pressure is increased to 3.00 atmospheres at 22 degree celsius?\r\n\r\na. 1.67    b. 0.600  c. 1.30  d. 0.800  i chose a.\r\n\r\n#2.  a 3.2 liter sample of gas is at 40 degree celsius and 1.0 atmosphere of pressure. If the temperature decreases to 20 degree celsius and the pressure decreases to 0.60 atmospheres, what is the new volume in liters.\r\n\r\na. 5.7   b. 5.0    c. 1.8   d. 0.20   i chose b.\r\n\r\n\r\n#3  A gas sample has a volume of 500ml at 20 degree celsius.  What celsius temperature would the gas have a volume half as large.\r\n\r\na 10  b. 147    c.  -147      d.   -127  i chose d, but wasn't sure how to work problem",
  "Solution_1": "All your answers are correct. In problem #3, use Charles' Law:\r\n\r\n$\\frac{V_{1}}{T_{1}}= \\frac{V_{2}}{T_{2}}$.",
  "Solution_2": "thanks alot",
  "Solution_3": "One can generalize these situtations with the ideal gas law:\r\n\r\n$PV = nRT$\r\n\r\nIt follows from this that:\r\n\r\n$\\frac{P_{1}V_{1}}{n_{1}T_{1}}= \\frac{P_{2}V_{2}}{n_{2}T_{2}}$"
}
{
  "Tag": [
    "articles"
  ],
  "Problem": "[url=http://www.foreignpolicy.com/story/cms.php?story_id=3376]http://www.foreignpolicy.com/story/cms.php?story_id=3376[/url]\r\n\r\nMy take: this stands to be a dominant historical force over the next few decades.  (Perhaps longer, but I'm optimistic that advanced technology can eventually step in and radically change the rules of the game.)",
  "Solution_1": "But even if technology steps in and all that, there might be a huge debate on the ethics and all that stuff of using technology to alter life by itself..... {just look at the debate on human cloning...l..}......So I kind of feel two decades is too short....that stuff might quite continue for at least about 50-60 years from now.",
  "Solution_2": "[quote=\"Dog of Justice\"][url=http://www.foreignpolicy.com/story/cms.php?story_id=3376]http://www.foreignpolicy.com/story/cms.php?story_id=3376[/url]\n\nMy take: this stands to be a dominant historical force over the next few decades.  (Perhaps longer, but I'm optimistic that advanced technology can eventually step in and radically change the rules of the game.)[/quote]\n\nFrom the article\n\n[quote]\nThe greatly expanded childless segment of contemporary society, whose members are drawn disproportionately from the feminist and countercultural movements of the 1960s and 70s, will leave no genetic legacy. Nor will their emotional or psychological influence on the next generation compare with that of their parents.[/quote]\n\nIronically these are the same people who have been calling their opponents `dinosaurs'.\n\n[quote]In addition to the greater fertility of conservative segments of society, the rollback of the welfare state forced by population aging and decline will give these elements an additional survival advantage, and therefore spur even higher fertility.[/quote]\r\n\r\nWelfare leads to more conservatives. Put that in your pipe and smoke it, hippies! (that was a joke)\r\n\r\nGeneral comments:\r\n\r\nThere seems to be no good argument in the article for why `patriarchy', and not some wider cause, is connected to high birthrates. He seems to be assuming implicitly that only women who are controlled by men reproduce in any significant proportion, which does not seem obvious\r\n\r\nanyway, to help fight the problem of educated people not reproducing enough, I intend to become a sperm donor. Hopefully my progeny will be as strong and intelligent and agressive as I am, but more importantly, if they are as fruitful as me, well, there just might be hope.....\r\n\r\n\r\n\r\nas for the return of patriarchy-- you make it sound like that's a [i]bad[/i] thing.  :lol: \r\n\r\nAnyway, a more serious and related topic is demographic conquest-- certain groups of people are being encouraged by their leaders to reproduce in greater numbers, and this not only may threaten our western `liberal`values, but our conservative ones as well.",
  "Solution_3": "If one assumes that a downward sloping demand curve applies to children as well as almost everything else ever analyzed, one can ask why the US tax law provides economic incentive to have children, unless one has income beyond a certain level.\r\n\r\nThe only logical assumptions that can be behind this are:\r\n\r\nA) Politicians have so little backbone and class warfare is so strong that they refuse to do the right thing at political cost today,\r\n\r\nB) Those drafting the tax code don't understand basic economics and how the tax code influences behavior,\r\n\r\nC) Politicians believe that the economic system in the US is random and there are no inheritable characteristics that are preferentially accumulated in higher income families, or \r\n\r\nD) Politicians want to have the leadership of this country taken over by immigrants.\r\n\r\nPlease let me know if you can find another logic alternative that I've overlooked.\r\n\r\nThis is dismal.  The government has few tools to address this issue and they are using one to make it worse rather than reduce the significance of the problem with respect to the higher income \"subclass\" of their country."
}
{
  "Tag": [
    "modular arithmetic",
    "number theory"
  ],
  "Problem": "Let [i]p[/i] be an odd prime and a be any integer which is not congruent to 0 modulo [i]p[/i]. Prove that the congruence x^2 = -a^2 mod p has solutions if and only if p=1 mod 4\r\n\r\nI don't have any idea how to tackle this problem...",
  "Solution_1": "$x^{2}\\equiv-a^{2}\\pmod{p}\\Longleftrightarrow (\\frac{-1}{p})=1\\Longleftrightarrow p\\equiv 1\\pmod{4}.$",
  "Solution_2": "Thank you TUAN!!!\r\n\r\ncan i have some more detailed explanation..?",
  "Solution_3": "$x^{2}\\equiv-a^{2}\\pmod{p}\\Longleftrightarrow (xa^{-1})^{2}\\equiv-1\\pmod{p}.$ ($a^{-1}$ exist because $\\gcd (a,p)=1$).",
  "Solution_4": "Thank you a thounsand times !!!",
  "Solution_5": "[quote=\"N.T.TUAN\"]$x^{2}\\equiv-a^{2}\\pmod{p}\\Longleftrightarrow (\\frac{-1}{p})=1\\Longleftrightarrow p\\equiv 1\\pmod{4}.$[/quote]\r\n\r\nI do not understand what you mean by $(\\frac{-1}{p})=1$ because that implies p=-1... or am I missing notation? Please explain.",
  "Solution_6": "It is the Legendre Symbol, not the usual division.  :wink:",
  "Solution_7": "[quote=\"bthings_2000\"][quote=\"N.T.TUAN\"]$x^{2}\\equiv-a^{2}\\pmod{p}\\Longleftrightarrow (\\frac{-1}{p})=1\\Longleftrightarrow p\\equiv 1\\pmod{4}.$[/quote]\n\nI do not understand what you mean by $(\\frac{-1}{p})=1$ because that implies p=-1... or am I missing notation? Please explain.[/quote]\r\n\r\nThat fraction-like thingy is actually a Legendre symbol:\r\n\\[\\left( \\frac{a}{p}\\right) =\\left\\{ \\begin{array}{rl}0,&p|a\\\\ 1,&p\\not| a\\text{ and }x^{2}\\equiv a\\pmod{p}\\text{ has solution}\\\\-1,&\\text{otherwise}\\\\ \\end{array}\\right.\\]\r\n(I've actually seen the Legendre symbol and a fraction appear in the same line in a proof...  :rotfl: )\r\n\r\nThe key in the problem is that $x^{2}\\equiv-1\\pmod{p}$ has solution for odd prime $p$ if and only if $p\\equiv 1\\pmod{4}$. For the \"only if\" part, raise both sides of $x^{2}\\equiv-1$ to the $\\frac{p-1}{2}$ power. For the \"if\" part, you can construct a square root using primitive roots (if $g$ a primitive root modulo $p$, then $g^\\frac{p-1}{4}$ is a square root of -1) or Wilson's Theorem (when $p\\equiv 1\\pmod{4}$, we have $\\left( \\frac{p-1}{2}\\right) !$ (that's a fraction here) as a square root of -1)."
}
{
  "Tag": [
    "geometry",
    "incenter",
    "geometry solved"
  ],
  "Problem": "Hello!\r\n\r\n  Let ABCD be a convex quadrilateral. Let P be the intersection point of AB and CD and let I_1 and I_2 be the incenters in the triangles ABC and BCD respectively.Let E be the intersection point of I_1I_2 and AB and F the intersection point of I_1I_2 and CD.\r\nProve that PE=PF if and only if ABCD is cyclic.\r\n\r\nCan you help me with this problem, please?",
  "Solution_1": "This is from CWMO 2004.\r\nTry to prove $I_1I_2CB$ is cyclic.",
  "Solution_2": "Or look at http://www.mathlinks.ro/Forum/viewtopic.php?t=58690 (at least for the direction $PE=PF$ $\\Rightarrow$ the quadrilateral ABCD is cyclic; the other direction is analogous).\r\n\r\n  Darij"
}
{
  "Tag": [
    "limit",
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "Evaluate the limit:\r\n\r\n$ \\lim_{n \\to \\infty} \\left( \\sum_{k \\equal{} 0}^{n}{\\frac {1}{\\binom{n}{k}} } \\right) \\equal{} ?$\r\n\r\nWhere $ \\binom{n}{k}\\equal{} \\frac{n!}{k!(n\\minus{}k)!}$.",
  "Solution_1": "http://mathworld.wolfram.com/BinomialSums.html\r\n\r\n(31) and on.",
  "Solution_2": "Thanks :).\r\n\r\nSomebody knows a proof without using integrals?",
  "Solution_3": "\\begin{eqnarray*} \\left(\\sum_{l = 0}^{n} \\frac {1}{\\binom{n}{l}} \\right) & = & \\left( \\sum_{l = 0}^{n}\\frac {k!(n - l)!}{n!}\\right) = \\\\\r\n& = & \\frac {0!n!}{n!} + \\frac {1!(n - 1)!}{n!} + \\cdots + \\frac {(n - 1)!1!}{n!} + \\frac {n!0!}{n!} = 1 + \\frac {1!(n - 1)!}{n!} + \\cdots + \\frac {(n - 1)!1!}{n!} + 1 = \\end{eqnarray*}\r\nSuppose $ n = 2k + 1$ (as $ n \\to \\infty ; k \\to \\infty$):\r\n\\begin{eqnarray*} & = & 1 + \\frac {1!(2k)!}{(2k+1)!} + \\cdots + \\frac {(2k )!1!}{(2k+1)!} + 1 = \\\\\r\n& = & 1 + 2 \\left( \\frac {1!(2k)!}{(2k+1)!} + \\cdots + \\frac {(k - 1)!(k + 1)!}{(2k+1)!} \\right) + 1 = \\\\\r\n& = & 1 + 2 \\left( \\frac {1}{(2k+1)!} \\left(1!(2k)! + 2!(2k - 1)! + \\cdots + (k - 1)!(k + 1)! \\right) \\right) + 1 = \\\\\r\n& = & 1 + 2 \\left( \\frac {1}{(2k+1)!}(2k)! \\left( 1! + \\frac {2!}{2k} + \\cdots + \\frac {1\\cdot2\\cdot3\\cdots i}{(2k - i + 1)\\cdots(2k)} + \\cdots + \\frac {1\\cdot2\\cdot3\\cdots k - 1}{(k + 2)(2k - i + 1)\\cdots(2k)} \\right) \\right) + 1 < \\\\\r\n& < & 1 + 2 \\left( \\frac {1}{(2k+1)!}(2k)! \\left(1 + (\\frac {1}{2}) + \\cdots + (\\frac {1}{2})^{i-1} + \\cdots + (\\frac {1}{2})^{k-1} \\right)\\right) + 1 = \\\\\r\n& = & 1 + 2 \\left( \\frac {1}{(2k+1)!}(2k)! \\left(\\frac {1}{1 - \\frac {1}{2}} \\right)\\right) + 1 = \\\\\r\n& = & 1 + 2 \\left( \\frac {1}{(2k+1)!}(2k)!2 \\right) + 1 = \\\\\r\n& = & 1 + \\left( \\frac {4(2k)!}{(2k+1)!} \\right) + 1 \\end{eqnarray*}\r\nHence\r\n\\begin{eqnarray*} \r\n2< lim_{n \\to \\infty}\\sum_{l = 0}^{n} \\left( \\frac {1}{\\binom{n}{l}} \\right) & < & lim_{k \\to \\infty} (1 + \\left( \\frac {4(2k)!}{(2k+1)!} \\right) + 1) = \\\\\r\n& = & lim_{k \\to \\infty} (1 + \\left( \\frac {4}{(2k+1)} \\right) + 1) = \\\\\r\n& = & 2. \\end{eqnarray*}\r\nFor $ n = 2k$ we have the same calculation as above, just that there is one more part  $ (\\frac {(k)!(k)!}{2k!})$, which is smaler than $ \\left( \\frac {(2k - 1)!}{(2k)!}\\right)$, hence we get $ lim_{k \\to \\infty} (1 + \\left( \\frac {5(2k - 1)!}{(2k)!} \\right) + 1) = 2$.\r\n\r\nSo $ \\sum_{l = 0}^{n} \\left( \\frac {1}{\\binom{n}{l}} \\right) = 2$.",
  "Solution_4": "[hide=\" Another approach... (click here)\"]Let $ a_n \\equal{} \\sum_{k \\equal{} 0}^n {\\frac {1}{\\binom{n}{0} }}$.\n\nIt is clear that $ a_n \\geq 2$.\n\nWe have that $ a_n \\equal{} 1 \\plus{} \\frac {1}{\\binom{n}{1}} \\plus{} \\frac {1}{\\binom{n}{2}} \\plus{} \\dots \\plus{} \\frac {1}{\\binom{n}{n \\minus{} 2}} \\plus{} \\frac {1}{\\binom{n}{n \\minus{} 1}} \\plus{} 1$.\n\n But then $ a_n < 2 \\plus{} \\frac {2}{n} \\plus{} \\frac {n \\minus{} 3}{\\binom{n}{2}} \\equal{} 2 \\plus{} \\frac {2}{n} \\plus{} \\frac {(n \\minus{} 3) \\cdot 2! \\cdot (n \\minus{} 2)!}{n!} \\equal{} 2 \\plus{} \\frac {2}{n} \\plus{} \\frac {2(n \\minus{} 3)}{n(n \\minus{} 1)}$.\n\nSo,\n\\[ 2 < \\lim_{n \\to \\infty}{a_n} < \\lim_{n \\to \\infty} \\left(2 \\plus{} \\frac {2}{n} \\plus{} \\frac {2(n \\minus{} 3)}{n(n \\minus{} 1)} \\right) \\equal{} 2.\n\\]\nThen $ \\lim_{n \\to \\infty}{a_n} \\equal{} 2$[/hide]"
}
{
  "Tag": [
    "LaTeX",
    "absolute value"
  ],
  "Problem": "Let W =Sqrt(3) +i , be a fixed number in the complex plane.\r\n\r\nZ is a variable complex number such that  I (Z-W) I <IWI  (where I...I denotes the complex absolute value or modulus)\r\n\r\nFind a value of A such that  A< argZ < A+(Pi)",
  "Solution_1": "you can use |x| for modulus/absolute value, on querty it should be shift+\\. Or in latex you can do $|x|$ or $\\left| x \\right|$ as well."
}
{
  "Tag": [
    "induction",
    "function",
    "algebra",
    "polynomial",
    "inequalities",
    "trigonometry"
  ],
  "Problem": "The functions $ T_n (x)$, $ n \\equal{} 0, 1, 2, \\ldots$, satisfy the recurrence relation\r\n\\[ T_{n\\plus{}1} (x) \\minus{} 2x T_n(x) \\plus{} T_{n\\minus{}1} (x) \\equal{} 0.\r\n\\]\r\nShow by induction that \r\n\\[ (T_n(x))^2 \\minus{} T_{n\\minus{}1}(x) T_{n\\plus{}1}(x) \\equal{} f(x),\r\n\\] where $ f(x) \\equal{} (T_1(x))^2 \\minus{} T_0(x) T_2(x)$.",
  "Solution_1": "Induction?  The useful idea here is to think of $ x$ as a constant and apply standard techniques from the theory of linear recurrences.  The really useful idea here is a certain identity I'll motivate by pointing you [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=175257]here[/url] or alternately you can see why this problem isn't really any harder than proving the corresponding identity for Fibonacci numbers by reading some of the relevant comments [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].\r\n\r\nIf you're wondering, these functions have a name and a purpose:  after giving them certain initial conditions, they are called the [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]Chebyshev polynomials[/url] of the first or second kind.  (Fun things happen to the recursion when you set $ x \\equal{} \\cos \\theta$ for $ \\theta$ a rational multiple of $ \\pi$, which is a way to motivate some of the results you'll find.)",
  "Solution_2": "Thank you! That was a big help"
}
{
  "Tag": [
    "pigeonhole principle",
    "greatest common divisor",
    "number theory",
    "relatively prime"
  ],
  "Problem": "In each square of an $ 8$ by $ 8$ board, Michael writes a positive integer not exceeding $ 10$ such that each two numbers that appear in adjacent or diagonally adjacent squares are relatively prime. Prove that some number appears at least $ 11$ times.",
  "Solution_1": "[hide=\"solution\"]Divide the 8x8 grid into 16 2x2 squares. If two numbers that are both even appear in the same 2x2 square, they are not relatively prime but they are adjacent. Similiarly, only one number divisible by 3 can appear in each 2x2 square. Therefore, even numbers can only appear 16 times. Numbers divisible by 3 can also only appear in total 16 times, so the other numbers appear 64-16-16=32 times. The other numbers are 1, 5, and 7. There are 3 of these numbers. Using pigeonhole, one number must appear 11 times[/hide]",
  "Solution_2": "I'm surprised that the bound is so sharp.\r\n\r\nOne thing that would be appropriate is to show that you can actually do such a tiling. It isn't totally obvious that such a numbering exists.",
  "Solution_3": "grey = 8\r\nwhite=2\r\nyellow =3\r\nred=9\r\ngreen=1\r\nblue=5\r\npurple=7",
  "Solution_4": "I just realized that it is possible to put $ 1$ in every square so I guess it is trivial.",
  "Solution_5": "[quote=\"pythag011\"][hide=\"solution\"]The other numbers are 1, 5, and 7. There are 3 of these numbers. Using pigeonhole, one number must appear 11 times[/hide][/quote]\r\n\r\nBut isn't every number NOT relatively prime to 1? I thought it'd be impossible to put a 1 on there.",
  "Solution_6": "Two numbers are relatively prime if their gcf is 1, so 1 is relatively prime with all integers.",
  "Solution_7": "Then why does $ \\phi(p) \\equal{} p\\minus{}1$ for all primes p?",
  "Solution_8": "$ 1, 2, 3, ... p\\minus{}1$ are relatively prime to $ p$.  What is the issue?",
  "Solution_9": "Oops nevermind...1 being relatively prime to everything would let it be on the board I guess."
}
{
  "Tag": [
    "geometry",
    "geometry unsolved"
  ],
  "Problem": "Find a point(s)  M inside triangle ABC so that the  3 \r\n  incircles  of  AMB,BMC,CMA  have a common tangent line.\r\n\r\n\r\n\r\n   Pestich.",
  "Solution_1": "Does such point M really exist?",
  "Solution_2": "It follows from continuity reasons. Take $M$ near to $A$ and then move it toward to $BC$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "linear algebra",
    "matrix",
    "vector",
    "complex numbers",
    "linear algebra theorems"
  ],
  "Problem": "It exists I know. It's a theorem. The book I use is Sheldon-Axler he uses the factorization of a polynomial linear transform to prove this. I was wondering how I can see this more intuitively... can anyone suggest a way to do this?",
  "Solution_1": "You mean complex linear transformation.  Axler's proof is considered, at least by Axler, to be more intuitive than the usual proof, which is based on the fact that the characteristic polynomial has a complex root.  You should probably develop your intuition about linear transformations.",
  "Solution_2": "o.O this is really interesting stuff. Could you elaborate on complex linear transformations? Sorry I just had a course in introductory linear algebra and am now fascinated by the topic.",
  "Solution_3": "A linear transform (from a space into itself) is a type of operator. Every linear transform can be represented in matrix form in straightforward manner. Thus, when we refer to \"linear transform\" we are really referring to a matrix being treated as a transform (i.e., we care about its \"transformative\" properties, like eigenvalues). The adjective \"complex\" means (specifically) that the base field of this matrix (that is, the base field of the vector space we are transforming) is the complex numbers.\r\n\r\nHowever, if I understand correctly which theorem is being spoken about, the reliance on the field of complex numbers is superficial, as any algebraically closed field will do."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "rotation"
  ],
  "Problem": "Does anyone have any information about the contested problem last year like what number it was, why it was contested, and what other answers were accepted?",
  "Solution_1": "If I remember correctly it was what is the smallest two digit number $x$ for which the last two digits of $x^3$ is $x$. The common answer was $24$. However, Texas pointed out that negative numbers weren't considered. So $24$ was NOT accepted. ONLY $-99$ was. Most answer keys, such as the $1$-year competition set, have this changed.",
  "Solution_2": "The one posted above was one although it wasn't truly contested. As soon as everyone saw the answer i'm sure they didn't argue.\r\n\r\nAnyway, the one that WAS contested was this:\r\n\r\nSome random person randomly wanted to find out how many ways the first two turns of tic-tac-toe could be played. She randomly decided that reflections and rotations wouldn't count, but switching X and O would. How many ways could the first two turns be played?\r\n\r\nThe answer key had the wrong answer on it, but they were grading it with the right answer. We sent in a complaint even though we got it wrong both ways -_-.\r\n\r\nPS. If it wasn't uberly obvious, I changed the wording a bit.",
  "Solution_3": "Sorry to bump the thread, but does anyone know what the correct answer was supposed to be?"
}
{
  "Tag": [
    "MATHCOUNTS",
    "email",
    "LaTeX",
    "probability",
    "AMC",
    "AIME",
    "geometry"
  ],
  "Problem": "I have NO idea if this will work, but if anyone from the JLS MathCounts Club followed directions and signed up for AoPS, this topic is for discussing the week 1 class and homework problems, or other club-related material. Also feel free to start your own topic if you think it's appropriate. Um. Also tell me how I can be a better teacher. I have the strangest suspicion that you'll have LOTS of ideas there. Um. Yeah.",
  "Solution_1": "Eh, so you've taken up teaching? :)",
  "Solution_2": "[quote=\"solafidefarms\"]Eh, so you've taken up teaching? :)[/quote]\r\nYeah...I'm supposed to be JLS's assistant coach. I spent today trying to explain permutations and combinations. I'm not sure if I made sense to anyone who didn't know them already, but I tried.",
  "Solution_3": "um...SURE. Anyway. I'm following directions for once, but I'm supposed to be doing other stuff.  :maybe: \r\n\r\nSO. Now what? I still don't really get this site. And the title of this forum's going to expire VERY SOON. Any bright ideas, bookaholic?  :?: \r\n\r\nI think you and I are going to be the only ones for awhile, unless \"blondeazn\" is going to actually register.",
  "Solution_4": "[quote=\"Bluestalk\"]um...SURE. Anyway. I'm following directions for once, but I'm supposed to be doing other stuff.  :maybe: \n\nSO. Now what? I still don't really get this site. And the title of this forum's going to expire VERY SOON. Any bright ideas, bookaholic?  :?: \n\nI think you and I are going to be the only ones for awhile, unless \"blondeazn\" is going to actually register.[/quote]\r\nIf she doesn't, make her. I'll probably remind her by email, too. Make the other MathCounts people you know do it too.\r\n\r\nWhat specifically don't you get about this site?",
  "Solution_5": "...just a question...what happens if we don't do our homework?",
  "Solution_6": "[quote=\"Edz504\"]...just a question...what happens if we don't do our homework?[/quote]\r\nI can't force you to, but in the future I'm going to assume that you know everything I've already talked about, so unless the problems aren't challenging at all, you should probably make sure you can do them so you don't get lost in the future. If the homework is too hard I can help you.",
  "Solution_7": "oh lol bookaholic you made a hw discussion topic? :lol: \r\nomg I wanna be a coach too\r\nedit:I'm coming next week",
  "Solution_8": "no, im just wondering...",
  "Solution_9": "1. who r Edz504 and junggi? i MIGHT think i know who u are, but from past experience....NO.\r\n\r\n2. WHAT IS LATEX, HOW DO U USE IT, and WHAT GOOD IS IT?\r\n(gotta do social studies, so....i can't really read the site...)\r\n\r\n3. where are the practice problems, and how do you use them?",
  "Solution_10": "[quote=\"junggi\"]oh lol bookaholic you made a hw discussion topic? :lol: \nomg I wanna be a coach too\nedit:I'm coming next week[/quote]\nGaaaah. Why did I ever tell you two? Mean bullies. Okay, you can come if you stay very silent and do not bother me with questions like \"If we have four choices for each letter in the word, why aren't there 4+4+4+4 instead of 4*4*4*4 different words?\" That and generally don't laugh at me. Please.\n\n[quote]1. who r Edz504 and junggi? i MIGHT think i know who u are, but from past experience....NO. [/quote]\nJunggi was the single non-Chinese person on the school team last year. I don't know Edz504.\n\n[quote]2. WHAT IS LATEX, HOW DO U USE IT, and WHAT GOOD IS IT? \n(gotta do social studies, so....i can't really read the site...)[/quote] \nLaTeX is a program that lets you type math expressions and symbols that you can't do that well with just the regular keyboard. To use it, you first have to go to the LaTeX page on AoPS and download everything it tells you to. Then go through the tutorials. You don't need to know it that well for it to be useful.\n\n[quote]3. where are the practice problems, and how do you use them?[/quote]\r\nThere are links with all kinds of practice problems in the Resources section. The other forums on the message board also have a lot of problems. Also, have you borrowed one of Ms. Beddoes' binders with old tests yet? If not you should.",
  "Solution_11": "yay. Thanks!!! I'll do my best.....",
  "Solution_12": "[quote=\"bookaholic\"]\n[quote]1. who r Edz504 and junggi? i MIGHT think i know who u are, but from past experience....NO. [/quote]\nJunggi was the single non-Chinese person on the school team last year. I don't know Edz504.\n[/quote]\r\n\r\nhaha...I feel special\r\n\r\nPOP QUIZ\r\nFILL IN THE BLANK\r\nEdz504=____\r\nBluestalk=_____\r\n\r\nIm guessing Bluestalk is susan or linda?\r\nedit:\r\nmy sister's(kimberly) username on aops is juyoung or something. I don't think she ever posted",
  "Solution_13": "I'm going to use this topic to criticize your teaching. :D  You need to articulate yourself better.  You were basicly just going through the calculations as if everyone knew what you were doing.  If I really had that question (why do you multiply the ways?  why not add?), your explanation really wouldn't have cut it.  You could have \"illustrated\" the method by drawing a tree diagram or something of the sort.\r\n\r\nJust a few thoughts on what I saw.\r\n\r\nHave fun torturing people with Mathcounts. :)\r\n\r\n[quote]...just a question...what happens if we don't do our homework?[/quote]\r\n\r\nYou die a painful death.  It's your decision to not do your homework, but you will have to live (or die) with the consequences.\r\n\r\nedit: added crapcrap pun",
  "Solution_14": "[quote=\"ccy\"][quote]...just a question...what happens if we don't do our homework?[/quote]\n\nYou die a painful death.  It's your decision to not do your homework, but you will have to live with the consequences.[/quote]\r\n :rotfl: \r\n\r\nbtw, if you want to do more practice problems, try the mathcounts forum. I havent been there much but it should be good for practice.\r\nBUT! ignore every post by anirudh. He is the best spammer ive seen in my lifetime(except that guy who sent me spam mail with a picture of spam in it)",
  "Solution_15": "Lol  Susan, you practically [i]live[/i] on AoPS now. :P",
  "Solution_16": "SHUT UP. It's not like I WANT TO. \r\n\r\nPLUS since AIM and YIM are blocked, pming is the closest thing I have. JEEZ.",
  "Solution_17": "how is it blocked?",
  "Solution_18": "no we suck no one is posting",
  "Solution_19": "3.1415926535 is 10 digits (decimal) remmber that",
  "Solution_20": "I think that's a new record! Triple post! :rotfl: \r\n\r\n$\\pi\\approx$ 3.14159265358979323846264338327950288419716939937510...",
  "Solution_21": "yeah no one posted for 3 hours so i umpec it",
  "Solution_22": "this is getting TOO SPAMMY OMG.\r\nmoogra, PLEASE stop spamming this forum AND the other forums. PLEASE.\r\n\r\nok to make this a non spam post:\r\ntry this one\r\n\r\nA license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible(AMC 10 A 2006)?",
  "Solution_23": "I did that and totally guessed. Hold on, I'm going to see if I can do it. \r\n\r\nUm.....\r\n\r\nHow are we supposed to do the part where the letters are next to eachother?\r\n\r\n((shows you just how sad I am))",
  "Solution_24": "[quote=\"Bluestalk\"]I did that and totally guessed. Hold on, I'm going to see if I can do it. \n\nUm.....\n\nHow are we supposed to do the part where the letters are next to eachother?\n\n((shows you just how sad I am))[/quote]\r\n[hide=\"hint\"]make the letters into a block. ie consider the 2 adjacent letters as one thingy.\nor you can just write down the cases since theres not much[/hide]\r\nps. GD my wording sucks",
  "Solution_25": "[quote=\"junggi\"][quote=\"Bluestalk\"]I did that and totally guessed. Hold on, I'm going to see if I can do it. \n\nUm.....\n\nHow are we supposed to do the part where the letters are next to eachother?\n\n((shows you just how sad I am))[/quote]\n[hide=\"hint\"]make the letters into a block. ie consider the 2 adjacent letters as one thingy.\nor you can just write down the cases since theres not much[/hide]\nps. GD my wording sucks[/quote]\nNOT MUCH? :wallbash:  :stretcher:\n\n[hide=\"Wild Guess\"]$6600000$[/hide]",
  "Solution_26": "[quote=\"i_like_pie\"][quote=\"junggi\"][quote=\"Bluestalk\"]I did that and totally guessed. Hold on, I'm going to see if I can do it. \n\nUm.....\n\nHow are we supposed to do the part where the letters are next to eachother?\n\n((shows you just how sad I am))[/quote]\n[hide=\"hint\"]make the letters into a block. ie consider the 2 adjacent letters as one thingy.\nor you can just write down the cases since theres not much[/hide]\nps. GD my wording sucks[/quote]\nNOT MUCH? :wallbash:  :stretcher:\n\n[hide=\"Wild Guess\"]$6600000$[/hide][/quote]\r\nuh no. and leave it in factored form its much nicer :) \r\nand uh by \"not much\" i was referring to where the letters can be(eg ####AA, ###AA#,##AA## BLABLABLA.)",
  "Solution_27": "[hide]There are 10000 ways for the digits to be arranged $\\left(10^{4}\\right).$\n\nThe letters can be arranged in 3380 ways $\\left(26^{2}\\cdot5\\right).$\n\nMultiplying that, we get $33800000$ total combinations.[/hide]",
  "Solution_28": "i know that nubmer is greater than 6!",
  "Solution_29": "[hide]10^4*26^5 yeah. ok i read i like pie's post but i get it now. I was lazy and didn't do it :( [/hide]\r\n\r\noh adn 400th post! :)"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Find an expression for the radius x of a circle externally tangent to 3 tangent circles of radius a, b, c.\r\nEquivalent statement: In a triangle with sides a+b, b+c, and a+c, find x such that segments of length a+x, b+x, c+x drawn from the vertices of the triangle are concurrent (with bounds on x of course, such that this is possible).\r\n\r\nI had a wierd idea for solving last year's USAMO number 2, but I couldn't get this...",
  "Solution_1": "http://mathworld.wolfram.com/SoddyCircles.html",
  "Solution_2": "Oh snap thanks, I didn't know that's what Soddy Circles were...\r\nHmmm okay knowing that makes number 2 on last year's usamo really easy then...[/hide]",
  "Solution_3": "Many people used Descartes' theorem (which Soddy's formula is based on) on the USAMO last year. A lot of them did not get full credit because it is a little tricky to reduce the problem to the case where three circles of the covering are tangent."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "rhombus",
    "similar triangles",
    "geometry proposed"
  ],
  "Problem": "A circle is tangent to the sides of a square AB, BC, CD, DA at K, L, M, N respectively. Let U be on (DM) and V on (DN) s.t. BU||KV. Prove that UV is tangent to the circle (the circle inscribed in the square). It's kind of easy, but I posted it anyway  :D",
  "Solution_1": "It is enough to prove NV+UM=VU.\r\nIf DU=x then \r\nUM=1/2 - x;\r\nUC=1-x ==> AV=1/(2-2x) (since BCU and AKV are similar triangles)\r\nNV=1/(2-2x) - 1/2\r\nVD=1-1/(2-2x)\r\n\r\nThus\r\nUV=\\sqrt(VD<sup>2</sup>+DU<sup>2</sup>)=1/(2-2x)-x=NV+UM.",
  "Solution_2": "would be to prove that, if $ O$ is the center of the square, then $ \\angle VOU \\equal{} 45^\\circ$.\r\nFor this issue we note $ AB\\equal{}2a$, $ UM\\equal{}x$, $ VD\\equal{}y$ and $ \\frac{x\\plus{}a}{2a} \\equal{} m$ ( 1 ). From (1) we easily get $ \\frac{x}{a}\\equal{}2m\\minus{}1$ ( 2 ) and $ \\frac{a\\minus{}y}{a}\\equal{}\\frac{1}{m}\\minus{}1$ ( 3 ), but (2) and (3) represent $ tan \\angle MOU$ and $ tan \\angle VON$ respectively and, compiling we get $ tan (\\angle MOU \\plus{}\\angle VON) \\equal{} 1$, i.e. $ \\angle VON \\plus{} \\angle MOU \\equal{} 45^\\circ$, q.e.d.\r\n\r\nBest regards,\r\nsunken rock",
  "Solution_3": "[quote=\"grobber\"] [color=darkred]A circle $ w = C(O)$ is tangent to the sides $ [AB]$ , $ [BC]$ , $ [CD]$ , $ [DA]$ of the square $ ABCD$ at $ K$ , $ L$ , $ M$ , $ N$ \n\nrespectively. Consider  $ U\\in (DM)$ and $ V\\in (DN)$ so that  $ BU\\parallel KV$ . Prove that $ UV$ is tangent to $ w$ .[/color] [/quote]\r\n\r\n[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $ AK = 1$ . Denote $ T\\in UV$ so that $ OT\\perp UV$ and $ DV = x$ , $ DU = y$ . Therefore, \n\n$ BU\\parallel KV\\Longleftrightarrow$ $ \\triangle AKV\\sim\\triangle CUB\\Longleftrightarrow$ $ \\frac {AK}{AV} = \\frac {CU}{CB}\\Longleftrightarrow$ $ \\frac {1}{2 - x} = \\frac {2 - y}{2}\\Longleftrightarrow$ $ \\boxed {2 + xy = 2(x + y)}\\ \\ (*)$ .\n\nSince $ [ONDM] = [ONV] + [OUV] + [UVD] + [OMU]$ obtain that $ UV$ is tangent to the circle $ w$ $ \\Longleftrightarrow$ $ OT = 1$ $ \\Longleftrightarrow$ \n\n$ 2 = (1 - x) + \\sqrt {x^2 + y^2} + xy + (1 - y)$ $ \\Longleftrightarrow$ $ x + y - xy = \\sqrt {x^2 + y^2}$ $ \\Longleftrightarrow$ $ \\boxed {2 + xy = 2(x + y)}$ .\n\n[b]Remark.[/b] Denote $ m(\\angle NOV) = \\alpha$ , $ m(\\angle MOU) = \\beta$ , $ m(\\angle UOV) = \\phi$ $ \\implies$ $ \\alpha + \\beta = 90^{\\circ} - \\phi$ . Therefore, $ \\tan\\alpha = 1 - x$ , \n\n$ \\tan\\beta = 1 - y$ and $ \\tan\\phi = \\frac {1 - \\tan\\alpha\\tan\\beta}{\\tan\\alpha + \\tan\\beta} =$ $ \\frac {1 - (1 - x)(1 - y)}{(1 - x) + (1 - y)}\\stackrel{(*)}{\\ = \\ }$ $ \\frac {(x + y) + 2 - 2(x + y)}{2 - (x + y)} = 1\\implies$ $ \\phi = 45^{\\circ}$ .\n\n[u]Otherwise[/u], the point $ O$ is the $ D$-exincentre of $ \\triangle UDV$ . Thus, $ m(\\widehat {UOV}) = 90^{\\circ} - \\frac 12\\cdot m(\\widehat {UDV}) = 45^{\\circ}$ .\n\nWe can extend this problem to a rhombus ! I\"ll come back soon. [/color]"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "The following identity holds for every \"x\"  real number:\r\n$(a+x)^{2}+(b+x)^{2}=(c+x)^{2}+(d+x)^{2}$\r\n\r\nProve that:\r\na)$a^{2}+b^{2}=c^{2}+d^{2}$\r\nb)$a+b=c+d$\r\nc)$a^{n}+b^{n}=c^{n}+d^{n}$\r\n\r\nA,b,c,d are real numbers and n is natural.\r\n\r\nThis problem was given at a Romanian olympiad in 1993.",
  "Solution_1": "Isn't this very easy for a Romanian test?  :maybe: \r\nUnless I am missing something, here's a sketch of my solution:\r\n\r\n[hide]It's equivalent to \\[a^{2}+b^{2}+2x(a+b) = c^{2}+d^{2}+2x(c+d)\\]\nFor a) just set $x=0$\nFor b) we set $x=\\frac{1}{2}$ and use a)\n\nThen c)\n$(a+b)^{2}-(a^{2}+b^{2}) = (c+d)^{2}-(c^{2}+d^{2})$ so $ab = cd$\n\nNow we can use induction on $n$\nIf $a^{i}+b^{i}= c^{i}+d^{i}$ for $1\\leq i\\leq n$, then \\[(a^{n}+b^{n})(a+b)=(c^{n}+d^{n})(c+d)\\]\n\\[\\iff a^{n+1}+b^{n+1}+ab(a^{n-1}+b^{n-1})=c^{n+1}+d^{n+1}+cd(c^{n-1}+d^{n-1})\\]\nSo finally also $a^{n+1}+b^{n+1}= c^{n+1}+d^{n+1}$\n[/hide]",
  "Solution_2": "It was a 7th grade district round  :P .\r\n\r\n\r\nGrrrrr, i was unsure if we can set x a certain number...   :blush:",
  "Solution_3": "[quote=\"alex__ib\"]i was unsure if we can set x a certain number...   :blush:[/quote]\r\n\r\nCertainly if an identity holds for every real $x$ then it must hold for any specific real $x$   :P"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c$ be nonnegative real numbers, prove that\r\n\\[ \\left( \\frac{2a}{c}\\minus{}1\\right)(a\\minus{}b)^2 \\plus{} \\left( \\frac{2c}{b}\\minus{}1\\right)(c\\minus{}a)^2  \\plus{}\\left( \\frac{2b}{a}\\minus{}1\\right)(b\\minus{}c)^2  \\ge 0.\\]\r\n:)",
  "Solution_1": "[quote=\"can_hang2007\"]Let $ a,b,c$ be nonnegative real numbers, prove that\n\\[ \\left( \\frac {2a}{c} \\minus{} 1\\right)(a \\minus{} b)^2 \\plus{} \\left( \\frac {2c}{b} \\minus{} 1\\right)(c \\minus{} a)^2 \\plus{} \\left( \\frac {2b}{a} \\minus{} 1\\right)(b \\minus{} c)^2 \\ge 0.\n\\]\n:)[/quote]\r\nFor more special, $ 2$ is also the smallest constant so that the inequality\r\n\\[ \\left( \\frac {ka}{c} \\minus{} 1\\right)(a \\minus{} b)^2 \\plus{} \\left( \\frac {kc}{b} \\minus{} 1\\right)(c \\minus{} a)^2 \\plus{} \\left( \\frac {kb}{a} \\minus{} 1\\right)(b \\minus{} c)^2 \\ge 0\r\n\\]\r\nholds. :)"
}
{
  "Tag": [
    "summer program",
    "Mathcamp",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO",
    "MathPath"
  ],
  "Problem": "Have you guys received any letters from AMC or the contributors of AMC? Because you did a good job on the contest?\r\nI received one invitation from Mathcamp last week.\r\nI hope I can receive a letter from AMC, because there is one in the Teacher's Manual.\r\n\r\nWhen we filled in the answer form for AMC and AIME for \"Do you want some univ or orgnizations to collect your info from the test?\", does it only mean those unimportant orgnizations? Or will some well-known univ or scholarship orgs may send us some letters? Or they may have already sent some, but I didn't receive? What have you guys recently received? \r\n\r\nAny PAST(excluding this year) USAMO qualifier or AIME High-scorer, what's your opinion and experience? (This is my first year to make it.) Are we going to receive something that is really valuable? \r\nThanks.",
  "Solution_1": "Yeah I received a Canada/USa Mathcamp pamhplet and the letter said \"Dear AMC top-scorer\"",
  "Solution_2": "I know Mathpath & Mathcamp both send invites to top scorers on various math competitions, but I haven't heard about any letters from AMC's sponsors.",
  "Solution_3": "I got  a \"Dear USAMTS top-scorer,\" letter from Mathcamp last week.",
  "Solution_4": "[quote=\"nat mc\"]Yeah I received a Canada/USa Mathcamp pamhplet and the letter said \"Dear AMC top-scorer\"[/quote]\r\n\r\nMe too.",
  "Solution_5": "I got one for Mathcamp saying \"Dear USAMTS top scorer\"",
  "Solution_6": "[quote=\"chess64\"][quote=\"nat mc\"]Yeah I received a Canada/USa Mathcamp pamhplet and the letter said \"Dear AMC top-scorer\"[/quote]\n\nMe too.[/quote]Yep, same here.",
  "Solution_7": "Doesn't look like I'll quallify for USAMO if they have too many 7's then, because I haven't gotten one.",
  "Solution_8": "Would the AMC people contact the USAMO qualifiers individually by email before they released the USAMO qualifier list?",
  "Solution_9": "That question is completely irrelevant to this topic!  :lol:",
  "Solution_10": "[quote=\"solafidefarms\"]Would the AMC people contact the USAMO qualifiers individually by email before they released the USAMO qualifier list?[/quote]\r\n\r\nI'm pretty sure the answer is no.",
  "Solution_11": "Didn't every one who qualified for the aime get a letter from mathcamp?",
  "Solution_12": "[quote=\"pieterminate\"]Didn't every one who qualified for the aime get a letter from mathcamp?[/quote]\r\n\r\nHmm, I qualified for AIME, but only got a letter saying \"Dear top USAMTS scorer,\" not one saying \"Dear top AMC scorer.\"",
  "Solution_13": "[quote=\"calc rulz\"][quote=\"pieterminate\"]Didn't every one who qualified for the aime get a letter from mathcamp?[/quote]\n\nHmm, I qualified for AIME, but only got a letter saying \"Dear top USAMTS scorer,\" not one saying \"Dear top AMC scorer.\"[/quote]\r\n\r\nMaybe they don't like wasting stamps and only send 1 per person?",
  "Solution_14": "Well they don't send an invite to every AIME qualifier, probably just the top couple hundred for both AMC's.",
  "Solution_15": "I just received a letter from National Honor Roll, but it doesn't say anything about AMC. Did you guys also receive one? What is it? Ad? or Real things?",
  "Solution_16": "National Honor Roll is the biggest scam... They get you to fill out a detailed form, even down to spots for you to list Grandparents and Grandparent's Addresses. Then they send ads to said grandparents to buy the book their child is in. Same goes for parents. And did I mention that they use you in the book, but you have to buy the book at $30 to see it?! \r\n\r\n:D",
  "Solution_17": "[quote=\"solafidefarms\"]National Honor Roll is the biggest scam... They get you to fill out a detailed form, even down to spots for you to list Grandparents and Grandparent's Addresses. Then they send ads to said grandparents to buy the book their child is in. Same goes for parents. And did I mention that they use you in the book, but you have to buy the book at 30 to see it?! \n\n:D[/quote]\r\n\r\nSo it's truly an ad?",
  "Solution_18": "I got one from Mathcamp and the other from Awesome math program :D",
  "Solution_19": "[quote=\"kimby_102\"]I got one from Mathcamp and the other from Awesome math program :D[/quote]\r\n\r\nDid you receive that recently?\r\nWhat is Awesome math?\r\nAnd who can qualify that? Very Top AMC scorers? above 140?",
  "Solution_20": "I also got one from Awesome Math.  I don't think there are any restrictions on who can apply.  My AMC score was 130, by the way.",
  "Solution_21": "[quote=\"Go Beyond\"][quote=\"kimby_102\"]I got one from Mathcamp and the other from Awesome math program :D[/quote]\n\nDid you receive that recently?\nWhat is Awesome math?\nAnd who can qualify that? Very Top AMC scorers? above 140?[/quote]\r\n\r\nAwsome Math is an excellent math camp as well (looking at faculty).\r\nThe letter is just an invitation to apply, and many probably received the letter(since even i got 1).",
  "Solution_22": "[quote=\"kimby_102\"]I got one from Mathcamp and the other from Awesome math program :D[/quote]\r\n\r\nMe too. \r\n\r\nAn August snowball fight sounds cool since in GA, we don't even have December snowball fights. \r\n\r\nBTW: A dodecahedron is not a hair accessory (No offence).\r\n\r\n$\\$$3000 dollars is a little out of my price range, so If I don't qualify for MOSP, I will have to go with my parents to some place in Asia for no apparent reason.",
  "Solution_23": "Got one from Mathcamp and an email from the ARML people... :)",
  "Solution_24": "I just received a letter from [i]Who's Who American High School Students[/i], due to AMC or AIME or USAMO qualifer, I am not sure.\r\n\r\nWhen we filled in the answer form for AMC and AIME for \"Do you want some univ or orgnizations to collect your info from the test?\", does it only mean those unimportant orgnizations? Or will some well-known univ or scholarship orgs may send us some letters? Or they may have already sent some, but I didn't receive? What have you guys recently received?\r\n\r\nAny PAST(excluding this year) USAMO qualifier or AIME High-scorer, what's your opinion and experience? (This is my first year to make it.) Are we going to receive something that is really valuable?\r\nThanks. :)",
  "Solution_25": "I'm thinking that you got Who's Who from PSAT, which is where you'll get college letters, and acedemic stuff like NAtional Honor Roll.\r\n\r\nBut stuff you get from getting good score s on the AMC/AIME is usually math camp type stuff, which is usually worth it.\r\n\r\nBut yea, so far, I've only gotten a letter from MathCamp.",
  "Solution_26": "so what did you get on usamo, Go Beyond?\r\ni wonder if mathpath or mathcamp would invite me",
  "Solution_27": "[quote=\"nirvanatear\"]so what did you get on usamo, Go Beyond?\ni wonder if mathpath or mathcamp would invite me[/quote]\r\n\r\nnothing important yet,\r\nso far, I have got mathcamp invitation, national honor roll, who'who American High school Students, AMC congratulation letter with student number, college magazine.",
  "Solution_28": "I just received one invitation from HCSSiM.",
  "Solution_29": "I got a 126 on AMC 12 A, and 137 on AMC 12 B, and 9 on AIME.... I haven't gotten any letters... boo!\r\n\r\n :rotfl: \r\n\r\nAh well, its not like I was really planning on attending any camps this summer, I've got classes to worry about.",
  "Solution_30": "BTW, you don't need an invitation to go anywhere.",
  "Solution_31": "I just received one from Awesome Math.",
  "Solution_32": "Joining it, I got one from mathcamp, one from Awsome math, and one from HCSSIM.  And my amc scores were terrible, so... :roll:",
  "Solution_33": "just received from Awesome Math.",
  "Solution_34": "I got ones from Mathpath, Mathcamp, and AwesomeMath. I was already aware of all of the programs, but I probably wouldn't have applied to Mathcamp without the pamphlet. (Was going to apply to Mathpath only.) Having a hard copy of the admissions problems around while I was trying to procrastinate on homework motivated me to try the problems :lol:",
  "Solution_35": "[quote=\"pieterminate\"]Joining it, I got one from mathcamp, one from Awsome math, and one from HCSSIM.  And my amc scores were terrible, so... :roll:[/quote]\r\n\r\nTerrible = ?",
  "Solution_36": "hm, my mom said there was some letter from Titu (I guess that would be AwesomeMath?)",
  "Solution_37": "[quote=\"calc rulz\"]\nTerrible = ?[/quote]\r\n\r\nBarely enough.  :(",
  "Solution_38": "[quote=\"MysticTerminator\"]hm, my mom said there was some letter from Titu (I guess that would be AwesomeMath?)[/quote]\r\n\r\nWhoa maybe he's going all like \"be instructor here at AwesomeMath... we caged all turkeys already.\"",
  "Solution_39": "[quote=\"probability1.01\"][quote=\"MysticTerminator\"]hm, my mom said there was some letter from Titu (I guess that would be AwesomeMath?)[/quote]\n\nWhoa maybe he's going all like \"be instructor here at AwesomeMath... we caged all turkeys already.\"[/quote]\r\n\r\noh. yea that would make more sense . . . I was wondering why he'd want me to go. Now maybe I should try to find that letter? hm",
  "Solution_40": "Just received one invitation from Boston U's PROMYS math program.\r\nHowever, I have been accepted by mathcamp, so I will not consider much to go there."
}
{
  "Tag": [],
  "Problem": "For positive integer $ n$, let $ f(n)$ be the number of the L.C.M. of positive integers $ a,\\ b$ with $ a\\leq b$. Find $ f(600)$.",
  "Solution_1": "[quote=\"kunny\"]For positive integer $ n$, let $ f(n)$ be the number of the L.C.M. of positive integers $ a,\\ b$ with $ a\\leq b$. Find $ f(600)$.[/quote]\r\n\r\nI don't think I understand the question. Does $ f(n)$ return the number of distinct values of $ LCM(a,b)$ where $ 0<a\\leq b\\leq n$?"
}
{
  "Tag": [],
  "Problem": "Compute: $ \\left(\\dfrac{4}{5}\\right)^{\\minus{}2}$ Express your answer as a common fraction.",
  "Solution_1": "$ \\left(\\dfrac{4}{5}\\right)^{\\minus{}2}$\r\n\r\n$ \\frac{5^2}{4^2}$\r\n\r\n$ \\frac{25}{16}$"
}
{
  "Tag": [
    "summer program",
    "Mathcamp"
  ],
  "Problem": "Here is a problem from my friend, it seems fun. I'm going to think about it with u. Ok? ;) \r\n\r\nSuppose A is a set of one hundred distinct natural numbers with the property that given a, b, c in A (distinct or not), there is a triangle with no angle greater than a right angle, and with sides of lengths a, b, c. What's the smallest possible value for the largest element of A?\r\n\r\n[i]Mod Edit - Please be careful not to post problems from current competitions. Even if this was unintentional, check your problem sources.[/i]",
  "Solution_1": "-[i]Mod edit - see below[/i]",
  "Solution_2": "- .",
  "Solution_3": "- [i]mod edit - problem censored since it is on Mathcamp quiz[/i]",
  "Solution_4": "I think this thread should be stopped this problem is on the 2005 Mathcamp quiz.\r\n\r\n- [i]Thanks - thread locked. - tetrahedr0n.[/i]"
}
{
  "Tag": [
    "AMC",
    "AMC 8"
  ],
  "Problem": "$ \\textbf{Juan's Old Stamping Grounds}$ \\\\\nJuan organizes the stamps in his collection by country and by the\ndecade in which they were issued.  The prices he paid for them at a\nstamp shop were: Brazil and France, 6 cents each, Peru 4 cents each, and Spain\n5 cents each.  (Brazil and Peru are South American countries and France and Spain are in Europe.)\n[asy]/* AMC8 2002 #8, 9, 10 Problem */\nsize(3inch, 1.5inch);\nfor ( int y = 0; y &lt;= 5; ++y )\n{\n    draw((0,y)--(18,y));\n}\ndraw((0,0)--(0,5));\ndraw((6,0)--(6,5));\ndraw((9,0)--(9,5));\ndraw((12,0)--(12,5));\ndraw((15,0)--(15,5));\ndraw((18,0)--(18,5));\n\ndraw(scale(0.8)*\"50s\", (7.5,4.5));\ndraw(scale(0.8)*\"4\", (7.5,3.5));\ndraw(scale(0.8)*\"8\", (7.5,2.5));\ndraw(scale(0.8)*\"6\", (7.5,1.5));\ndraw(scale(0.8)*\"3\", (7.5,0.5));\n\ndraw(scale(0.8)*\"60s\", (10.5,4.5));\ndraw(scale(0.8)*\"7\", (10.5,3.5));\ndraw(scale(0.8)*\"4\", (10.5,2.5));\ndraw(scale(0.8)*\"4\", (10.5,1.5));\ndraw(scale(0.8)*\"9\", (10.5,0.5));\n\ndraw(scale(0.8)*\"70s\", (13.5,4.5));\ndraw(scale(0.8)*\"12\", (13.5,3.5));\ndraw(scale(0.8)*\"12\", (13.5,2.5));\ndraw(scale(0.8)*\"6\", (13.5,1.5));\ndraw(scale(0.8)*\"13\", (13.5,0.5));\n\ndraw(scale(0.8)*\"80s\", (16.5,4.5));\ndraw(scale(0.8)*\"8\", (16.5,3.5));\ndraw(scale(0.8)*\"15\", (16.5,2.5));\ndraw(scale(0.8)*\"10\", (16.5,1.5));\ndraw(scale(0.8)*\"9\", (16.5,0.5));\n\nlabel(scale(0.8)*\"Country\", (3,4.5));\nlabel(scale(0.8)*\"Brazil\", (3,3.5));\nlabel(scale(0.8)*\"France\", (3,2.5));\nlabel(scale(0.8)*\"Peru\", (3,1.5));\nlabel(scale(0.8)*\"Spain\", (3,0.5));\n\nlabel(scale(0.9)*\"Juan's Stamp Collection\", (9,0), S);\nlabel(scale(0.9)*\"Number of Stamps by Decade\", (9,5), N);[/asy]\nWhat was the average price, in cents, of his 70s stamps?  Round your answer to the\nnearest tenth of a cent.",
  "Solution_1": "The Brazilian stamps cost $ 12 \\times 6 \\equal{} 72$ cents, the French stamps cost $ 12 \\times 6 \\equal{} 72$ cents, the stamps from Peru cost $ 6 \\times 4 \\equal{} 24$ cents, and the Spanish stamps cost $ 13 \\times 5 \\equal{} 65$ cents. \r\n\r\nThe sum of those numbers is 233, divided by 4 is $ 58.2$ cents, to the nearest tenth.",
  "Solution_2": "But the answer is 5.4",
  "Solution_3": "[quote=\"isabella2296\"]The Brazilian stamps cost $ 12 \\times 6 \\equal{} 72$ cents, the French stamps cost $ 12 \\times 6 \\equal{} 72$ cents, the stamps from Peru cost $ 6 \\times 4 \\equal{} 24$ cents, and the Spanish stamps cost $ 13 \\times 5 \\equal{} 65$ cents. \n\nThe sum of those numbers is 233, divided by 4 is $ 58.2$ cents, to the nearest tenth.[/quote]\n\nThe error is in the last line, where izzy divided the cost in cents ($233$) by $4$. Instead, she should have divided it by $43$, which is the total number of stamps, since the question asked for average price per stamp. This will give you the correct answer of $5.4$",
  "Solution_4": "Missed this one because I did not read \"...to the nearest tenth.\" :("
}
{
  "Tag": [
    "modular arithmetic",
    "ratio",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all positive integers $A$ which can be represented in the form: \\[ A = \\left ( m - \\dfrac 1n \\right) \\left( n - \\dfrac 1p \\right) \\left( p - \\dfrac 1m \\right) \\]\r\nwhere $m\\geq n\\geq p \\geq 1$ are integer numbers. \r\n\r\n[i]Ioan Bogdan[/i]",
  "Solution_1": "Solution\r\n[hide]$\\left ( m - \\frac 1n \\right) \\left( n - \\frac 1p \\right) \\left( p - \\frac 1m \\right) = \\frac{(mn-1)(np-1)(pm-1)}{mnp}$\n$\\implies$\n$p \\lvert mn-1 \\implies mn \\equiv 1 \\pmod p$\n$m \\lvert np-1 \\implies np \\equiv 1 \\pmod m$\n$n \\lvert pm-1 \\implies pm \\equiv 1 \\pmod n$\n\nLet \n$p = \\frac{mn-1}{k}$\n$k<m$ to preserve the ordering of $m,n,p$\n\n$n(mn-1) \\equiv k \\pmod m$\n$m(mn-1) \\equiv k \\pmod n$\n\n$n+k \\equiv 0 \\pmod m$\n$m+k \\equiv 0 \\pmod n$\n\n$m+k < 2n \\implies m+k=n$\nTherefore $n < 2m$\nTherefore $n+k < 3k \\implies n+k = 2m \\implies m+k+k=2m \\implies m=2k$\n\n$m\\frac{mn-1}{k} \\equiv 1 \\pmod n \\implies 2(mn-1) \\equiv 1 \\pmod n \\implies -2 \\equiv 1 \\pmod n \\implies n=3$ or $n=1$\n\n$m \\le n \\implies m=2$ or $m=1$\n\nThis leaves us with three ordered pairs $(m,n)$\n$(1,1), (1,3), (2,3)$\n\n$(m,n)=(1,1) \\implies A = 0$\n$(m,n)=(1,3) \\implies p = 1$, contradicting the ordering.\n$(m,n)=(2,3) \\implies p = 5$, and $A=21$\n\nTherefore, the two possible values of $A$ are $0$ and $21$[/hide]",
  "Solution_2": "Alternate solution:\r\n[hide]\nThe product equals $ mnp\\minus{}m\\minus{}n\\minus{}p\\plus{}\\frac{mn\\plus{}np\\plus{}pm\\minus{}1}{mnp}$, or $ mnp\\leq mn\\plus{}np\\plus{}pm\\minus{}1$.  If $ p\\geq3$, it must hold $ 3m\\plus{}3n\\leq 2mn\\leq 3m\\plus{}3n\\minus{}1$, absurd, yielding $ p\\equal{}1$ or $ p\\equal{}2$.\n\nIf $ p\\equal{}2$, then $ 2mn$ must divide $ mn\\plus{}2m\\plus{}2n\\minus{}1$, and if their ratio is at least $ 2$, then $ 2mn\\plus{}2m\\plus{}2n\\leq4mn\\leq mn\\plus{}2m\\plus{}2n\\minus{}1$ absurd, hence $ mn\\equal{}2m\\plus{}2n\\minus{}1$, and $ m,n$ must clearly be odd.  If $ n\\geq5$, then $ 2m\\plus{}2n\\plus{}m\\leq mn\\equal{}2m\\plus{}2n\\minus{}1$, absurd, and $ n\\equal{}3$, yielding $ m\\equal{}5$.  We then find the product equal to $ \\frac{14}{3}\\frac{5}{2}\\frac{9}{5}\\equal{}21$.\n\nIf $ p\\equal{}1$, then $ mn$ must divide $ mn\\plus{}m\\plus{}n\\minus{}1$, or $ mn$ must divide $ m\\plus{}n\\minus{}1$.  If $ n\\geq2$, then $ m\\plus{}n\\leq mn\\leq m\\plus{}n\\minus{}1$, absurd, yielding $ n\\equal{}1$, or one of the factors is $ 0$ and the product is $ 0$.\n\nThe only values are $ 0$ and $ 21$.\n\n[/hide]",
  "Solution_3": "Yes 0 and 21 , nice problem"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "geometry solved"
  ],
  "Problem": "Show that the Cleavance centre is identical with Spieker centre.",
  "Solution_1": "http://mathworld.wolfram.com/Cleaver.html",
  "Solution_2": "dear pleurestique,I know that url but I can not see any proof there...",
  "Solution_3": "Here's a nice argument:\r\n\r\nThe intersection of the cleavers must be the centroid of the border of $ABC$, so in order to prove the desired result, we have to show that the Spieker center is the centroid of the border of $ABC$. If $D,E,F$ are the midpoints of $BC,CA,AB$ respectively, then the centroid of the border of $ABC$ is the centroid of the system formed by the three points $D,E,F$ if we give $D,E,F$ the masses $EF,FD,DE$ respectively. It's easy to see that this means that the point must be the incenter of $DEF$, and this is the Spieker center.",
  "Solution_4": "You can find some properties of cleavers and of the Spieker center of a triangle in the very first pages of Honsberger's \"Episodes in Nineteenth and Twentiethth Century Euclidean Geometry\". These first pages are avaliable online at http://www.amazon.com/exec/obidos/tg/detail/-/0883856395?v=glance (\"sample pages\").\r\n\r\n  Darij"
}
{
  "Tag": [
    "function",
    "calculus",
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ x \\in R$, find minimum of following function:\r\n\r\n$ f(x) \\equal{} \\sqrt {\\frac {1}{2}x^2 \\plus{} 2} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} \\frac {4}{5}x \\plus{} \\frac {8}{5}} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} 4x \\plus{} 10} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} \\frac {16}{5}x \\plus{} \\frac {32}{5}}$\r\n\r\nDon't use calculus!\r\n:)",
  "Solution_1": "[quote=\"nntien\"]Let $ x \\in R$, find minimum of following function:\n\n$ f(x) \\equal{} \\sqrt {\\frac {1}{2}x^2 \\plus{} 2} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} \\frac {4}{5}x \\plus{} \\frac {8}{5}} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} 4x \\plus{} 10} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} \\frac {16}{5}x \\plus{} \\frac {32}{5}}$\n\nDon't use calculus!\n:)[/quote]\r\n$ 2f(x)\\equal{} \\sqrt{(x\\minus{}2)^{2}\\plus{}(x\\plus{}2)^{2}}\\plus{}\\sqrt{(x\\plus{}\\frac{4}{5})^{2}\\plus{}(x\\minus{}\\frac{12}{5})^{2}}\\plus{}\\sqrt{(x\\minus{}6)^{2}\\plus{}(x\\minus{}2)^{2}}\\plus{}\\sqrt{(x\\minus{}\\frac{24}{5})^{2}\\plus{}(x\\minus{}\\frac{8}{5})^{2}}$\r\nBecause $ (\\minus{}2,2),(2,2),(6,2)$ are colinear, and $ (\\minus{}\\frac{4}{5},\\frac{12}{5}),(2,2),(\\frac{24}{5},\\frac{8}{5})$ are colinear, and $ 2f(x)$ is the distance between the point $ (x,x)$ and $ (\\minus{}2,2),(6,2),(\\minus{}\\frac{4}{5},\\frac{12}{5}),(\\frac{24}{5},\\frac{8}{5})$, $ 2f(2)$ is the minimum value of $ 2f(x)$. Therefore, the minimum of $ f(x)$ is $ f(2)$.",
  "Solution_2": "[quote=\"msceok\"][quote=\"nntien\"]Let $ x \\in R$, find minimum of following function:\n\n$ f(x) \\equal{} \\sqrt {\\frac {1}{2}x^2 \\plus{} 2} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} \\frac {4}{5}x \\plus{} \\frac {8}{5}} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} 4x \\plus{} 10} \\plus{} \\sqrt {\\frac {1}{2}x^2 \\minus{} \\frac {16}{5}x \\plus{} \\frac {32}{5}}$\n\nDon't use calculus!\n:)[/quote]\n$ 2f(x) \\equal{} \\sqrt {(x \\minus{} 2)^{2} \\plus{} (x \\plus{} 2)^{2}} \\plus{} \\sqrt {(x \\plus{} \\frac {4}{5})^{2} \\plus{} (x \\minus{} \\frac {12}{5})^{2}} \\plus{} \\sqrt {(x \\minus{} 6)^{2} \\plus{} (x \\minus{} 2)^{2}} \\plus{} \\sqrt {(x \\minus{} \\frac {24}{5})^{2} \\plus{} (x \\minus{} \\frac {8}{5})^{2}}$\nBecause $ ( \\minus{} 2,2),(2,2),(6,2)$ are colinear, and $ ( \\minus{} \\frac {4}{5},\\frac {12}{5}),(2,2),(\\frac {24}{5},\\frac {8}{5})$ are colinear, and $ 2f(x)$ is the distance between the point $ (x,x)$ and $ ( \\minus{} 2,2),(6,2),( \\minus{} \\frac {4}{5},\\frac {12}{5}),(\\frac {24}{5},\\frac {8}{5})$, $ 2f(2)$ is the minimum value of $ 2f(x)$. Therefore, the minimum of $ f(x)$ is $ f(2)$.[/quote]\r\n\r\nThat's right! I use Minkowski's inequality! Thanks!"
}
{
  "Tag": [
    "USAMTS"
  ],
  "Problem": "I sent my entry form in about a month ago via snail mail, but according to my profile on the USAMTS site I haven't sent in an entry form yet.  Should I send in another one?",
  "Solution_1": "[quote=\"matt276eagles\"]I sent my entry form in about a month ago via snail mail, but according to my profile on the USAMTS site I haven't sent in an entry form yet.  Should I send in another one?[/quote]\r\n\r\nNot yet.  We haven't filed the forms we've received yet.  We'll do that in September some time."
}
{
  "Tag": [],
  "Problem": "Compute: $ \\frac{1234}{1235^2\\minus{}(1234)(1236)}$.",
  "Solution_1": "hello, let $ a\\equal{}1234$ then we have $ \\frac{a}{(a\\plus{}1)^2\\minus{}a(a\\plus{}2)}\\equal{}\\frac{a}{a^2\\plus{}2a\\plus{}1\\minus{}a^2\\minus{}2a}\\equal{}a$\r\nSonnhard.",
  "Solution_2": "So that means the answer is 1234."
}
{
  "Tag": [],
  "Problem": "Show that there are infinite numbers of rational numbers in the interval $ (0,1)$",
  "Solution_1": "[hide]\nConsider rational numbers in the form $ \\frac {1}{n} \\longrightarrow n \\in \\mathbb{Z} \\neq 1$\n$ \\Rightarrow 0 < \\frac {1}{n} < 1$\nand there are infinitely many counting numbers $ > 1$\n[/hide]",
  "Solution_2": "Hello\r\nHere is one more proof::\r\nDraw a segment $ AB$ where A correspond to $ (0,0)$ and B correspond to $ (1,0)$.Now draw one more segment $ A'B'$ parallel and above it.Taking $ A'B'$ as diameter draw a circle .If u draw a line through the mid-point of segment $ AB$ it will also pass through the mid-point of segment $ A'B'$ too.\r\nIf u draw a line perpendicular to the segment $ AB$ passing in b/w of midpoint and $ B$.Let this line cut the circumference of semicircle at M.Now Join M to center of semicircle.U can clearly observe that the  line will not cross the segment $ AB$.\r\n  Now take a perpendicular line which tends to pass through $ B$ will touch the circumference at a point $ M'$ which tends to touch the point $ B'$.Now if we join M' and center it will meet x-axis at infinity.\r\nThus every infinite number outside this segment corresponds to a number inside this segment.\r\nThank u.",
  "Solution_3": ":| I dont understand this at all. Could you label your diagram please. Thanks",
  "Solution_4": "hello\r\nHere is my second attempt.\r\nFirst  choose a line segment on x-axis $ AB$ where A is $ (0,0)$ and B is $ (1,0)$.\r\nNow above this line segment draw one more segment $ A'B'$ parallel to AB such that A' comes just above A and B' just above B.\r\nNow taking $ A'B'$ as diameter draw a semi-circle with centre $ O$.\r\nDraw a line perpendicular to $ AB$ passing through its midpoint$ (0.5,0)$.U can easily observe that it will also pass through $ O$ and would bisect the circumference of semi-circle.\r\nNow the main part::\r\n1.Draw II perpendicular line passing $ AB$ through a point in b/w midpoint$ (0.5,0)$ and$ B$$ (1,0)$. Let this line meet the semicircle at point $ M$ .Now if u join $ O$ and $ M$ the line $ OM$ will meet x-axis outside the segment.\r\n2.Draw III perpendicular line passing through a point which is tending to $ B$ and will touch the semicircle at a point $ M'$ which is tending to $ B'$.Now if u join the point $ O$ and $ M'$,the line $ OM'$ is just parallel to $ AB$ and $ A'B'$.Therefore the line $ OM'$ will meet the x-axis at infinity.\r\n\r\n[color=green]Conclusion[/color]\r\nWe now there are infinite numbers outside the segment $ AB$ and each of these infinite numbers correspond to a number inside the segment.Thus we have infinite numbers in this segment.\r\nHope this time my explanation is somehow better.\r\nThank u.\r\nNote::Click on the picture to get a magnified picture."
}
{
  "Tag": [
    "counting",
    "derangement",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Count the number of 2-regular graphs on n vertices.\r\n\r\nLet this number be f(n)",
  "Solution_1": "Labelled vertices, I assume.  If we allow self-loops and multiedges, the answer is $ n!$.  If we allow multiedges but not self-loops, the answer is $ D_n$, the number of derangements of an $ n$-element set (permutations with no fixed points).  If we allow neither multiedges nor self-loops, the answer is the number of permutations of an $ n$-element set with all cycles of length 3 or greater, counted by [url=http://www.research.att.com/~njas/sequences/A038205]A038205[/url] in the OEIS.  There's not a nice closed form, although there is a relatively simple recursion and a very nice exponential generating function."
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Find all functions $ f: \\mathbb{N}\\rightarrow \\mathbb{Z}$ such that\r\n\r\n$ f(x\\plus{}|f(y)|)\\equal{}x\\plus{}f(y)$, for all natural numbers $ x,y$.",
  "Solution_1": "Anybody?  :)",
  "Solution_2": "[hide=\"A rather mean problem if you ask me\"]Suppose there exists a positive integer $ m$ such that $ f(m) \\equal{} 0$. Then $ y \\equal{} m$ gives $ f(x) \\equal{} x$. This is a valid solution and it also contradicts the assumption.\n\nSuppose there exist positive integers $ m,n$ such that $ f(m) \\equal{} \\minus{}n$. Then $ x \\equal{} n, y \\equal{} m$ gives $ f(m\\plus{}n) \\equal{} 0$. Again, by the previous paragraph, contradiction.\n\nTherefore, $ f$ only takes on positive integer values. The equation is now reduced to $ f(x\\plus{}f(y)) \\equal{} x \\plus{} f(y)$ with $ f : \\mathbb{N} \\rightarrow \\mathbb{N}$.\n\nLet $ a$ be the minimum of $ \\{ f(n) | n \\in \\mathbb{N} \\}$. $ m$ exists by the well-ordering principle of $ \\mathbb{N}$. Let $ b$ be a value for which $ f(b) \\equal{} a$. By putting in $ y \\equal{} b$, we know that $ f(x\\plus{}a) \\equal{} x\\plus{}a$. That is, $ f(x) \\equal{} x$ for all $ x > a$\n\nNow for all $ x \\leq a$, we can define $ f(x)$ to be anything greater than or equal to $ a$ without contradiction. Because $ a$ is the minimum value $ f$ takes, $ x \\plus{} f(y) > a$, always. Since we have that $ f$ is fixed for all values greater than $ a$, the original equation will hold.\n\nSo we have a ridiculous amount of answers that I'm not going to list.[/hide]"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "angle bisector"
  ],
  "Problem": "In a triangle $ ABC$, the angle bisector at $ B$ intersects $ AC$ at $ D$ and the circumcircle again at $ E$. The circumcircle of the triangle $ DAE$ meets the segment $ AB$ again at $ F$. Prove that the triangles $ DBC$ and $ DBF$ are congruent.",
  "Solution_1": "We have $ \\angle DBF \\equal{} \\angle DBC$ because $ DB$ is the bisector of $ \\angle FBC$.\r\nWe also have $ \\angle DFB \\equal{} \\angle DEA \\equal{} \\angle BCD$ (because $ DEFA$ and $ BCEA$ are cyclic quadrilaterals).\r\n\r\nHence, triangles $ DBF$ and $ DBC$ are similar. But, since they share the common side $ BD$, they are congruent.",
  "Solution_2": "easy,$\\angle DFA+\\angle DEA=180$,so $\\angle DFB=\\angle C$($\\angle AEB=\\angle C$),$BD$ is the angle bisector,hence done."
}
{
  "Tag": [
    "HMMT",
    "AMC",
    "AIME",
    "MATHCOUNTS"
  ],
  "Problem": "WHERE IS EVERYBODY?!\r\n\r\nWe should start posting things here soon. Math Problems, contest results, you know, stuph like that!\r\n\r\nOr is everybody just [b]antisocial[/b]?",
  "Solution_1": ":huh: What contests have taken place so far? \r\n\r\nHow can we Upstate New Yorkers possibly hope to have such a vibrant, lively forum as NYC?\r\n\r\n-looks- Oh, wait, their forum is dead too  :|",
  "Solution_2": "Contests that have taken place so far: All AMCs, PUMaC, HMMT, MC Chapter. NYSML, AIME, and MC State are coming up soon.\r\n\r\nProposal on how to keep this forum [b]alive[/b]: First of all, we need to check this forum a lot more regularly. Second, we need to post stuff so that this forum doesn't die. Third, we need to point Upstate New York AoPS users to this forum, so that they can use it.\r\n\r\nHere's a list of AoPSers that I know are in Upstate NY:\r\n\r\nme\r\nyou\r\nxgames43\r\npianogirl\r\nSmitSchu\r\nHexahedron\r\nBlueMouse\r\nsailingdog\r\ndilberto\r\nseashell1082\r\n\r\nThese people should know about this forum, and should visit it once every [b]white[/b] moon.",
  "Solution_3": "KEEP THIS FORUM ALIVE!",
  "Solution_4": "OK!\r\nHERE'S A PROBLEM:\r\nGive the first 90 numbers of the Fibonacci Sequence\r\noh and about the signature\r\n12.9+ refers to college level work, meaining that it would be recommended for you to take AP courses in what ever subject that's in or all subjects depending on the exam you recieved the score is",
  "Solution_5": "This is a horrible problem, lol.",
  "Solution_6": "${\\text{Find } 1+1. \\text { Express your answer in terms of } \\pi.}$",
  "Solution_7": "National MATHCOUNTS is coming up soon.",
  "Solution_8": "Does anyone who lives in upstate NY exist anymore?",
  "Solution_9": "the answer is pi-(pi -2).\n\nHow many degrees are in a hexagon?",
  "Solution_10": "[quote=\"anonymous0\"]Does anyone who lives in upstate NY exist anymore?[/quote]\nOf course not.",
  "Solution_11": "I challenge willwang123's signature.",
  "Solution_12": "Nope, all people of Upstate New York mysteriously vanished. Survivors suddenly became antisocial.\n\nProblem: Find the circumference of a circle with diameter $1$. Give your answer rounded to the first 1337 decimal places.",
  "Solution_13": "[hide=\"hmm\"]1.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000$\\pi.$\n[/hide]\n:D",
  "Solution_14": "Hmmm....1=2 doesn't post here anymore either...We should actually post something other than in this INACTIVITY! Bah! thread...",
  "Solution_15": "[quote=\"1=2\"]Here's a list of AoPSers that I know are in Upstate NY:\n\nme\nyou\nxgames43\npianogirl\nSmitSchu\nHexahedron\nBlueMouse\nsailingdog\ndilberto\nseashell1082\n\n[/quote]\nYou didn't know eavqg.whbeg?",
  "Solution_16": "[quote=\"Draco\"]1=2 doesn't post here anymore either[/quote]\nI'm not interested in posting in a forum where less than 5 people are interested in posting.",
  "Solution_17": "1. 1=2\n2. willwang123\n3. anonymous0\n4. draco\n5. hrithikguy",
  "Solution_18": "Hrithikguy lives in Upstate NY? \n\nUpdated list of Upstate New Yorkers\n\nDraco\n1=2\nwillwang123\nAnonymous0\nSammymax\nHanes\njasonion\nTim14\nTrigfan\nTareyza\nJeffshen\n\n...not very many. And half of them don't even post.\n\nAlso, 5 unique people have posted in this forum this year.",
  "Solution_19": "hrthikguy is from Tennessee... anyway,\n\nDraco\n1=2\nwillwang123\nAnonymous0\nSammymax\nHanes\njasonion\nTim14\nTrigfan\nTareyza\nJeffshen\nPattycakechichi\n\nDoes trigfan count as being from upstate New York?",
  "Solution_20": "OK. You three need something productive to do instead of spamming this thread. I created a MathCounts Marathon thread in this forum for you guys to have fun with.",
  "Solution_21": "astuteverbosity, here are your numbers $ 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,............... $ \nFill in the dots [b]by yourself[/b]",
  "Solution_22": "*Bumps forum*",
  "Solution_23": "Hmmm, I don't think the MC marathon worked... Now that competition season is about to start, hopefully we'll have something that's not spam to post about."
}
{
  "Tag": [
    "inequalities",
    "function",
    "triangle inequality",
    "inequalities proposed"
  ],
  "Problem": "Let $a,b,c$ are real numbers.Prove the following inequality.\r\n\r\n\\[\\frac{|a|}{1+|a|}+\\frac{|b|}{1+|b|}\\geq \\frac{|a+b|}{1+|a+b|}\\]",
  "Solution_1": "$1+\\frac{1}{1+|a+b|} \\ge \\frac{1}{1+|a|} + \\frac{1}{1+|b|}$\r\n<=> $\\frac{1}{1+|a+b|} \\ge \\frac{1-|ab|}{1+|a|+|b|+|ab|}$\r\nNow RHS $\\le \\frac{1}{1+|a|+|b|+|ab|} \\le \\frac{1}{1+|a+b|+|ab|} \\le \\frac{1}{1+|a+b|}$\r\nEquality holds when $a$ or $b=0$",
  "Solution_2": "Sorry for that I can't understand.Please explain in detail. :? \r\n\r\nkunny",
  "Solution_3": "$\\frac{|a|}{1+|a|} + \\frac{|b|}{1+|b|} \\ge \\frac{|a+b|}{1+|a+b|}$\r\n$1+ 1- \\frac{|a+b|}{1+|a+b|} \\ge 1 - \\frac{|a|}{1+|a|} +1- \\frac{|b|}{1+|b|}$\r\niff $1+\\frac{1}{1+|a+b|} \\ge \\frac{1}{1+|a|} + \\frac{1}{1+|b|}$\r\niff $\\frac{1}{1+|a+b|} \\ge \\frac{1-|ab|}{1+|a|+|b|+|ab|}$\r\nNow RHS $\\le \\frac{1}{1+|a|+|b|+|ab|}$(because |ab| > 0 \r\n$\\le \\frac{1}{1+|a+b|+|ab|} \\le \\frac{1}{1+|a+b|}$ (because $|a|+|b| \\ge |a+b|$ and $|ab| > 0$)\r\nEquality holds when $a$ or $b=0$",
  "Solution_4": "Thank you for your reply,Soarer.Now I can understand your solution.\r\n\r\nHere is my solution.\r\n\r\nSince $y=\\frac{x}{1+x}=1-\\frac{1}{1+x}$ is an increasing function for $x\\geq 0$, use the triangle inequality,\r\n\r\n$|a+b|\\leq |a|+|b|$, so we have $\\frac{|a+b|}{1+|a+b|}\\leq \\frac{|a|+|b|}{1+|a|+|b|}=\\frac{|a|}{1+|a|+|b|}+\\frac{|b|}{1+|a|+|b|}\\leq \\frac{|a|}{1+|a|}+\\frac{|b|}{1+|b|}$.\r\n\r\nkunny",
  "Solution_5": "Here's another possibly similar solution.\r\n\r\nSince $x/(1+x)$ is increasing for $x\\geq 0$, we may assume that $a,b$ have the same sign, and consequently are both $\\geq 0$.  But then\r\n\r\n$\\frac {x}{1+x}+\\frac {y}{1+y}\\geq \\frac {x+y}{1+x+y}=\\frac {|x+y|}{1+|x+y|}$"
}
{
  "Tag": [
    "conics",
    "hyperbola",
    "rotation",
    "geometry",
    "geometric transformation",
    "reflection",
    "Asymptote"
  ],
  "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14110[/img]\r\n\r\n :(\r\n\r\n[size=150]\\[ \\left\\{ {_{xy \\equal{} a^2 }^{x^2 \\plus{} y^2 \\equal{} d^2 } } \\right.\n\\][/size]",
  "Solution_1": "$ s\\equal{}x\\plus{}y\\equal{}\\sqrt{d^2\\plus{}2a^2}$ that can be costructed by pytagora, so x, y are the solution of the equation $ x^2 \\minus{} sx \\plus{} a^2$ so they are $ \\frac{s \\pm \\sqrt{s^2\\minus{}4a^2}}{2}$ that can be costructed by pytagora.",
  "Solution_2": "i want the [b]s[/b] construction  and  [b]x [/b], [b]y[/b]  construction\r\n\r\n :(",
  "Solution_3": "$ x^2 \\plus{} y^2 \\equal{} d^2$ is a circle centered at the origin $ O$. $ xy \\equal{} a^2$ is a rectangular hyperbola centered at the origin, x-, y-axes are asymptotes. Rotate the hyperbola by $ \\minus{} 45^\\circ.$ The circle does not change, the hyperbola equation becomes $ x^2 \\minus{} y^2 \\equal{} 2a^2.$ Adding the equations, $ 2x^2 \\equal{} d^2 \\plus{} 2d^2.$ Construct intersection(s) of the circle with the hyperbola. ($ E, F$ are foci. You do not have to draw the hyperbola, just the asymptotes.) $ OABC$ is square with the side $ OA \\equal{} a,$ diagonal $ OB \\equal{} a \\sqrt 2.$ $ OL \\perp OB$ and $ OL \\equal{} d.$ $ BL^2 \\equal{} d^2 \\plus{} 2a^2.$ $ K$ is the midpoint of $ BL.$ Perpendicular bisector of $ BL$ cuts the circle $ (K)$ with diameter $ BL$ at $ M, N.$ The distance $ x \\equal{} \\sqrt {\\frac {d^2 \\plus{} 2a^2}{2}} \\equal{} \\frac {BL}{\\sqrt {2}} \\equal{} LM$ has to be transfered along $ OB.$ For example, reflect $ MN$ in the line $ KCA$ into $ UV$ (also a diameter of $ (K)$). Circle with radius $ OU \\equal{} OV$ cuts the ray $ OB$ at $ W$ and perpendicular to $ OB$ at $ W$ cuts the circle $ (O)$ at $ Q$ one intersection of the circcle $ (O)$ with the hyperbola. $ P, R$ are feet of perpendiculars from $ Q$ to the hyperbola asymptotes $ OA, OC.$ Rectangle $ OPQR$ has area $ a^2 \\equal{} OA \\cdot OC \\equal{} OP \\cdot OR$ and diagonal $ d \\equal{} OQ.$",
  "Solution_4": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14740[/img]\r\n\r\n :thumbup:"
}
{
  "Tag": [],
  "Problem": "A building modeled after the Chicago Sears Tower consists of nine square towers arranged in a three by three grid. They have congruent bases, and the heights, in feet, are indicated in the grid to the right. Which of the following is a side view of\nthe building from some direction?\n\n[asy]draw((0,-15)--(0,0)--(15,0)--(15,-15)--cycle);\ndraw((0,-5)--(15,-5));\ndraw((0,-10)--(15,-10));\ndraw((5,0)--(5,-15));\ndraw((10,0)--(10,-15));\nlabel(\"100\",(2.5,0),S);\nlabel(\"200\",(7.5,0),S);\nlabel(\"100\",(12.5,0),S);\nlabel(\"300\",(2.5,-5),S);\nlabel(\"300\",(7.5,-5),S);\nlabel(\"200\",(12.5,-5),S);\nlabel(\"200\",(2.5,-10),S);\nlabel(\"200\",(7.5,-10),S);\nlabel(\"100\",(12.5,-10),S);\nlabel(\"$A$\",(3,1),N);\nlabel(\"$B$\",(9,1),N);\nlabel(\"$C$\",(15,1),N);\nlabel(\"$D$\",(21,1),N);\nlabel(\"$E$\",(26,1),N);\ndraw((2,5)--(5,5)--(5,8)--(4,8)--(4,7)--(2,7)--cycle);\ndraw((3,5)--(3,7));\ndraw((4,5)--(4,7));\ndraw((3,6)--(5,6));\ndraw((7,5)--(10,5)--(10,6)--(9,6)--(9,8)--(7,8)--cycle);\ndraw((10,6)--(8,6));\ndraw((9,5)--(9,8));\ndraw((8,5)--(8,8));\ndraw((7,7)--(8,7));\ndraw((13,5)--(16,5)--(16,6)--(15,6)--(15,8)--(14,8)--(14,7)--(13,7)--cycle);\ndraw((14,5)--(14,7)--(15,7));\ndraw((14,6)--(15,6)--(15,5));\ndraw((20,5)--(23,5)--(23,7)--(22,7)--(22,8)--(20,8)--cycle);\ndraw((20,7)--(22,7)--(22,5));\ndraw((22,6)--(23,6));\ndraw((21,5)--(21,8));\ndraw((25,5)--(28,5)--(28,6)--(27,6)--(27,7)--(26,7)--(26,8)--(25,8)--cycle);\ndraw((25,7)--(26,7)--(26,5));\ndraw((27,5)--(27,6));[/asy]",
  "Solution_1": "It's tough to explain how to do this, but just visualize the building in your head and you see the answer is D."
}
{
  "Tag": [
    "Gauss",
    "function",
    "set theory",
    "advanced fields",
    "advanced fields theorems"
  ],
  "Problem": "Can someone explain me how can we use the concept of infinity with mathematical rigor, because i still cant understand!\r\nThanks!",
  "Solution_1": "I want to hear an explanation from Valentin!",
  "Solution_2": "Infinity is a very subtle thing. I'll give you some ideas and see if this can help out.\r\n\r\nAt an intuitive sense, infinity is like a 'thing' very very VERY large or somewhere far far far... away. Infinity is NOT an exact value. Infinity is rather a mathematical device sort of like denoting something that never ends, like a sequence, like the natural numbers! There are different kinds of infinities but we are not getting into that.\r\n\r\nAt a more formal level (and I believe someone in here can do better at this than I do), infinity is as follows: suppose we have a real number system R. We make an [i]extended real number system[/i] R* as the real number system R with the following 2 'extra' conditions:\r\n\r\n1)For every x \\in R, denote - \\infty to be the greatest lower bound of x.\r\n2)For every x \\in R, denote + \\infty to be the least upper bound of x.\r\n\r\nI defined it under the S-bound properties because those are most commonly used definitions in analysis. If any of the other readers see anything wrong here, let me know. I didn't check this  :)",
  "Solution_3": "Sorry that I'm not Valentin, (is he on now?) but if you insist on getting an explanation from him, it's worth the wait. He explains better than I do  :)",
  "Solution_4": "Well, this is a very controversed subject.It put into difficulty even famous mathematicians.There are two kinds of infinity:actual and potential.For example Gauss and Kronecker never accepted the actual infinity ie there exists actual infinite sets.\r\nCantor dedicated  a big part of his life for rigorously fundamenting the concept of infinity.When studying thi subject he encountered many \"paradoxes\".He found out that   there exists different types of infinity .How many?An infinity!The complicated thing with the infinity is that man's intuition doesnt work very well at this chapter.It is more easy to imagine finite sets than infinite sets,so when working with infinite sets everything wich may seem intuitive clear must be tested by rigorous proof.\r\nFormally, with the basic set theory axioms we can't prove that there exists actual infinite sets, so we must introduce a new axiom:the axiom of infinity! This fundaments the infinite sets.(for more details one can consult som set theory books-.The infinite set with the\" fewest\" elements is the set of natural numbers.Then we can extend the number system-first we construct integer and rationals and then irrationals(this extending to irrationals can be made in many ways-using Cauchy sequences,Dedekind cuts,etc).\r\nAlthough, the problems with infinity don't stop here.We need other axioms such the axiom of choice.If we hadn't the axiom of choice we couldn't prove, for example the equivalence between the classic definition of the limit of a function and the definition with sequences(Heine).\r\nAnother important concepts wich deal with infinity are cardinal numbers wich make possible comparing the infinite sets using bijections.\r\n\r\nSo infinity can be mathematically rigorously founded.\r\nThis is sufficient for the mathemathical use of infinity.If we dont stop here we may get into philosophy.I may give an example:in his last years of life Cantor gave up mathematics and began teaching phylosophy.",
  "Solution_5": "I heard that Cantor went beyond limit and was ... well crazy? ... by constant attacks from Kronecker. Poor fella I can tell you that.",
  "Solution_6": "Cantor couldn't publish his work before his death, because Kronecker had influence at most mathematical magazines of the time and stopped his articles from being published.So he never knew that his work will influence so much the mathematics of XX century.",
  "Solution_7": "That's sort of sad, isn't it? I had heard about the conflict between the two, but I had no idea it was so serious..",
  "Solution_8": "shame on Kronecker.\r\nif he wasn't agree with Cantor, why did he stop his articles ?wasn't there a way\r\nmore kind to \"solve\" this problem ?\r\nthis is not a good way to be famous on mathematics !",
  "Solution_9": "One problem with explaining infinity formally is that the concept of \"infinity\" is used in a lot of completely different contexts mathematically. But each individual instance can actually be defined quite concretely (sort of).\r\n\r\nFor example, in calculus/analysis, when you talk about a limit of a function or sequence as the variable \"goes to infinity\", you're not talking about anything outside of the ordinary real or natural numbers. The standard $n_0$-delta definition of sequence convergence ($\\forall \\epsilon > 0, \\exists n_0, \\forall n>n_0, |x_n - L| < \\epsilon$) is simply a statement about every positive real number. Sorry if you've never seen that sort of thing (it probably looks like complete gibberish if so), but the point is that $\\infty$ is just a shorthand way of saying \"arbitrarily large\". In some way it's all a matter of notation.\r\n\r\nInfinity has a completely different sort of feel in set theory. It's still definable (probably by saying that an infinite set is one that can't be put into one-to-one coorespondence with a set of the form {0,1,2,...,n} for some natural number n (or the empty set)). Of course, there are several different sizes of infinity that sets can be. It gets quite complicated, to say the least. I guess the real question comes down to \"how can set theory be logically definable with such craziness as infinite sets?\". As for that, I have nothing to say other than that axiomatic set theory is quite rigorous and you can still make some kind of living in philosophy.\r\n\r\nI hope some small piece of these ill-concieved ramblings helps you to better comprehend the wonders of infinity.  :) Good luck."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "A base-$ 10$ three-digit number $ n$ is selected at random. Which of the following is closest to the probability that the base-$ 9$ representation and the base-$ 11$ representation of $ n$ are both three-digit numerals?\r\n\r\n$ \\textbf{(A)}\\ 0.3 \\qquad\r\n\\textbf{(B)}\\ 0.4 \\qquad\r\n\\textbf{(C)}\\ 0.5 \\qquad\r\n\\textbf{(D)}\\ 0.6 \\qquad\r\n\\textbf{(E)}\\ 0.7$",
  "Solution_1": "There are a total of 900 three digit base ten numbers. The largest base nine three digit number is 888, which is equivalent to 728. The smallest three digit base eleven number is 100, which is equivalent to 121. The number of integers between 121 and 728 inclusive is 608. So the answer is 608/900 which is equivalent to .675555555... This is closest to E.",
  "Solution_2": "[quote=\"Wickedestjr\"]This is equivalent to .66222222... This is closest to D.[/quote]\r\n\r\n\r\nIt's closer to E)0.7  :wink:",
  "Solution_3": "He's right.\r\n\r\nYour solution was good until the part where you said the smallest three digit base eleven number is 111. It's actually 100. Converting that to base 10, we have 121. So the number are be between 121 and 728, inclusive. So the probability is $ \\frac {608}{900} \\approx \\frac {6}{9} \\equal{} \\frac {2}{3} \\equal{} 0.\\overline{6}$. Thus the closest is $ 0.7 (E)$.",
  "Solution_4": "[hide]\n\nClearly, the largest three digit base 11 number exceeds the largest three digit decimal number, as well as the smallest three digit base 9 number being less than the smallest three digit decimal number, so we only need to count the numbers between $ 100_{11}$ and $ 888_9$.\n\n$ 100_{11}\\equal{}11^2\\equal{}121$\n\n$ 888_9\\equal{}(9^2)\\plus{}8(9)\\plus{}8(1)\\equal{}8(91)\\equal{}728$\n\nHence, there are $ 728\\minus{}121\\plus{}1\\equal{}608$ numbers. $ 608/900\\approx .68\\approx0.7\\Longrightarrow\\mathrm{ (E) \\ }$\n\n[/hide]",
  "Solution_5": "[quote=\"sunehra\"]He's right.\n\nYour solution was good until the part where you said the smallest three digit base eleven number is 111. It's actually 100. Converting that to base 10, we have 121. So the number are be between 121 and 728, inclusive. So the probability is $ \\frac {608}{900} \\approx \\frac {6}{9} \\equal{} \\frac {2}{3} \\equal{} 0.\\overline{6}$. Thus the closest is $ 0.7 (E)$.[/quote]\r\n\r\nOh oops. I can't believe I made two mistakes in one problem.  :blush:"
}
{
  "Tag": [
    "analytic geometry",
    "graphing lines",
    "slope"
  ],
  "Problem": "Do anyone know how to use TI-89 to graph slope field?\r\nsuch as y'=-x-2 and initial condition point(-1,1).",
  "Solution_1": "Look it up in the manual.  It should already be built in as a program so the manual should have it.",
  "Solution_2": "The attachment is an example of graph the solution to the logistic 1st-order differential equation y'=.001y*(100-y).\r\nIn Y=Editor, the manual ask us put in .001*y1*(100-y1).\r\nbut I don't know how to get slope field if y'=10x+2y which has both x and y.\r\nI have try just put 10*x+2*y in the calculator, it tells me undefined variable.",
  "Solution_3": "There said \"undefined variable\". Do anyone know if I can define the variable in calculator?",
  "Solution_4": "Does it have to be on a calculator?  These are rather simple slope fields to do by hand.",
  "Solution_5": "[quote=\"JRav\"]Does it have to be on a calculator?  These are rather simple slope fields to do by hand.[/quote]\r\nYes, it have to be done on the calculator because I want to know how to graph it by both calculator and hand and that is the reason for me to buy TI-89 calculator."
}
{
  "Tag": [],
  "Problem": "\u03a0\u03b1\u03b9\u03b4\u03b9\u03ac \u03c3\u03b1\u03c2 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03ce \u03bb\u03af\u03b3\u03b7 \u03b2\u03bf\u03ae\u03b8\u03b5\u03b9\u03b1 \u03b3\u03b9\u03b1 \u03c4\u03b9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03ba\u03ac\u03c4\u03c9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2.\r\n[url=http://img216.imageshack.us/my.php?image=mathimageec0.png][img]http://img216.imageshack.us/img216/7239/mathimageec0.th.png[/img][/url]",
  "Solution_1": "1. (\u03b1\u03c5\u03bc\u03b2\u03bf\u03bb\u03af\u03b6\u03c9 \u03bc\u03b5 l \u03c4\u03bf \u03bb)\r\n\u0388\u03c3\u03c4\u03c9 A,B,C \u03bf\u03b9 \u03b1\u03bd\u03b1\u03c0\u03b1\u03c1\u03b1\u03c3\u03c4\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c4\u03c9\u03bd \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ce\u03bd \u03c3\u03c4\u03bf \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03cc \u03b5\u03c0\u03af\u03c0\u03b5\u03b4\u03bf \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1 \u03c4\u03cc\u03c4\u03b5:\r\n$ \\vec{AB} \\equal{} k\\vec{BC}(k\\not \\equal{} 0) \\Leftrightarrow ( \\minus{} 1, \\minus{} 1) \\equal{} k(4 \\minus{} 2l,5 \\minus{} l \\minus{} 7)$ \u03ac\u03c1\u03b1\r\n$ k(4 \\minus{} 2l) \\equal{} \\minus{} 1$\r\n\u03ba\u03b1\u03b9\r\n$ \\minus{} k(l \\plus{} 2) \\equal{} \\minus{} 1$\r\n\u03ac\u03c1\u03b1\r\n$ 2l \\minus{} 4 \\equal{} l \\plus{} 2 \\Leftrightarrow l \\equal{} 6$\r\n\r\n2.\u03ad\u03c3\u03c4\u03c9 $ x \\equal{} 2 \\plus{} 3i$ \u03c4\u03cc\u03c4\u03b5 $ z \\equal{} x^{4n} \\plus{} ( \\minus{} i\\bar{x})^{4n} \\equal{} x^{4n} \\minus{} (\\bar{x})^{4n} \\equal{} x^{4n} \\minus{} \\bar{x^{4n}} \\in \\mathbb{R}$\r\n\u03c9\u03c2 \u03ac\u03b8\u03c1\u03bf\u03b9\u03c3\u03bc\u03b1 \u03c3\u03c5\u03b6\u03b7\u03b3\u03ce\u03bd\r\n\r\n4. $ |w| \\equal{} 1\\Leftrightarrow w\\bar{w} \\equal{} 1 \\Leftrightarrow \\frac {(z \\plus{} i)(\\bar{z} \\minus{} i)}{(z \\minus{} i)(\\bar{z} \\plus{} i)} \\equal{} 1 \\Leftrightarrow$\r\n\r\n$ z\\bar{z} \\plus{} i(\\bar{z} \\minus{} z) \\plus{} 1 \\equal{} z\\bar{z} \\minus{} i(\\bar{z} \\minus{} z) \\plus{} 1 \\Leftrightarrow z \\plus{} \\bar{z} \\equal{} \\minus{} (z \\plus{} \\bar{z})\\Leftrightarrow z \\plus{} \\bar{z} \\equal{} 0 \\Leftrightarrow z \\in \\mathbb{R}$\r\n\r\n5. \u03ad\u03c3\u03c4\u03c9 $ w \\equal{} z \\plus{} \\frac {1}{z}$ \u03c4\u03cc\u03c4\u03b5 $ w \\plus{} \\bar{w} \\equal{} z \\plus{} \\frac {1}{z} \\plus{} \\bar{z} \\minus{} \\frac {1}{\\bar{z}} \\equal{}$\r\n\r\n$ z \\plus{} \\bar{z} \\plus{} \\frac {z \\plus{} \\bar{z}}{z\\bar{z}} \\equal{} z \\plus{} \\bar{z} \\plus{} \\frac {z \\plus{} \\bar{z}}{|z|} \\equal{} z \\plus{} \\bar{z} \\plus{} z \\plus{} \\bar{z} \\equal{}$\r\n(\u03b5\u03c0\u03b5\u03b9\u03b4\u03ae $ z \\in \\mathbb{C} \\Leftrightarrow z \\equal{} \\minus{} \\bar{z}$) $ z \\minus{} z \\plus{} z \\minus{} z \\equal{} 0 \\Leftrightarrow w \\in \\mathbb{R}$\r\n\r\n6. $ |1 \\minus{} z| \\equal{} 1 \\Leftrightarrow (1 \\minus{} z)(1 \\minus{} \\bar{z}) \\equal{} 1 \\Leftrightarrow 1 \\minus{} z \\minus{} \\bar{z} \\plus{} z\\bar{z} \\equal{} 1 \\Leftrightarrow$\r\n$ 1 \\minus{} z \\minus{} \\bar{z} \\plus{} 1 \\equal{} 1 \\Leftrightarrow z \\plus{} \\bar{z} \\equal{} 1 \\Leftrightarrow (z \\equal{} x \\plus{} iy) x \\plus{} iy \\plus{} x \\minus{} iy \\equal{} 1 \\Leftrightarrow x \\equal{} \\frac {1}{2}$\r\n\r\n$ \\Leftrightarrow y^2 \\equal{} \\frac{3}{4}$ \u03ac\u03c1\u03b1 $ z\\equal{}\\frac{1\\plus{}\\sqrt{3}}{2}$ \u03ae $ z\\equal{}\\frac{1\\minus{}\\sqrt{3}}{2}$ \r\n\r\n10. $ w \\equal{} z \\plus{} \\frac {1}{z} \\equal{} z \\plus{} \\frac {\\bar{z}}{|z|} \\equal{} z \\plus{} \\frac {\\bar{z}}{2} \\equal{}$ \r\n\r\n$ (z \\equal{} x \\plus{} iy) x \\plus{} iy \\plus{} \\frac {x \\minus{} iy}{2} \\equal{} \\frac {3}{2}x \\plus{} i\\frac {1}{2}y$\r\n\r\n\u03ac\u03c1\u03b1 $ w_x^2 \\plus{} 9w_y^2 \\equal{} 9 \\Leftrightarrow \\frac {w_x^2}{3^2} \\plus{} \\frac {w_y^2}{1^2} \\equal{} 1$\r\n\r\n\u03ac\u03c1\u03b1 \u03c4\u03bf $ w$ \u03ba\u03b5\u03af\u03c4\u03b5\u03c4\u03b1\u03b9 \u03c0\u03ac\u03bd\u03c9 \u03c3\u03b5 \u03ad\u03bb\u03bb\u03b5\u03b9\u03c8\u03b7\r\n\r\n\r\n\r\n\u0395\u03c0\u03ad\u03bb\u03b5\u03be\u03b1 \u03c4\u03c5\u03c7\u03b1\u03af\u03b1 \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03bb\u03cd\u03c3\u03b7. \u0398\u03b1 \u03c3\u03c5\u03bc\u03c0\u03bb\u03b7\u03c1\u03ce\u03c3\u03c9 \u03c4\u03b9\u03c2 \u03c5\u03c0\u03cc\u03bb\u03bf\u03b9\u03c0\u03b5\u03c2 \u03b1\u03cd\u03c1\u03b9\u03bf \u03b1\u03bd \u03c3\u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03b4\u03b5\u03bd \u03ad\u03c7\u03b5\u03b9 \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03b5\u03af \u03ba\u03b1\u03bd\u03ad\u03bd\u03b1\u03c2. \u0393\u03b5\u03bd\u03b9\u03ba\u03ac \u03c0\u03ac\u03bd\u03c4\u03c9\u03c2 \u03b1\u03c5\u03c4\u03bf\u03cd \u03c4\u03bf\u03c5 \u03b5\u03af\u03b4\u03bf\u03c5\u03c2 \u03bf\u03b9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03ba\u03ac\u03c0\u03bf\u03b9\u03b1 \u03b5\u03bd\u03c4\u03c1\u03af\u03c6\u03b7\u03c3\u03b7 \u03bc\u03b1\u03b6\u03af \u03c4\u03bf\u03c5\u03c2 \u03c6\u03b1\u03bd\u03c4\u03ac\u03b6\u03bf\u03c5\u03bd \u03b5\u03cd\u03ba\u03bf\u03bb\u03b5\u03c2. \u03a4\u03bf \u03cc\u03bb\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b9\u03c2 \u03c4\u03bf \u03ba\u03cc\u03bb\u03c0\u03bf.  :lol:",
  "Solution_2": "Andrew \u03ba\u03ac\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03af\u03b4\u03b9\u03b5\u03c2 \u03ae \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03bc\u03b9\u03ba\u03c1\u03ae \u03b1\u03bb\u03bb\u03b1\u03b3\u03ae \u03c3\u03c4\u03b1 \u03bd\u03bf\u03cd\u03bc\u03b5\u03c1\u03b1 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u03b2\u03b9\u03b2\u03bb\u03af\u03bf.\r\n\r\n\u0393\u03b9\u03b1 \u03c0\u03b1\u03c1\u03ac\u03b4\u03b5\u03b9\u03b3\u03bc\u03b1 \u03b7 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 10 \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc ( \u03c3\u03b5\u03bb102 \u0391\u03c3\u03ba\u03b7\u03c3\u03b79)\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 7 \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03ac\u03c2 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03b7 \u03b5\u03c6\u03b1\u03c1\u03bc\u03bf\u03b3\u03ae 2 \u03c3\u03c4\u03b7 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1 93 ( \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc )\r\n\r\n\u0397 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 5 \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03ac\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c4\u03b7 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1 101 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 3 ( \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc )\r\n\r\n\u039d\u03b1 \u03b4\u03b9\u03bf\u03c1\u03b8\u03ce\u03c3\u03c9 \u03bb\u03af\u03b3\u03bf \u03c4\u03bf\u03bd \u039c\u03b1\u03bd\u03ce\u03bb\u03b7.\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 (1) \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03ac\u03c2  \u03b1\u03bd \u03c0\u03bf\u03cd\u03bc\u03b5 \u0391(3,4) , \u0392(4,5), C( 2\u03bb-4,\u03bb+2) \u03c4\u03b9\u03c2 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b5\u03c2 \u03c4\u03c9\u03bd \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ce\u03bd z1, z2, z3 \u03c4\u03cc\u03c4\u03b5\r\n\r\n $ \\vec{AB}$ = (1,1),  \u03ba\u03b1\u03b9  $ \\vec{BC}$ = ( 2\u03bb-4,\u03bb+2) .\r\n\r\n\u0393\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c5\u03b8\u03b5\u03b9\u03b1\u03ba\u03ac \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 det$ (\\vec{AB},\\vec{BC})$=0 =>  \u03bb = 6\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd 6 \u03c0\u03bf\u03c5 \u03c1\u03c9\u03c4\u03ac\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c4\u03b1\u03c5\u03c4\u03cc\u03c7\u03c1\u03bf\u03bd\u03b1\r\n\r\n|z|=1  kai  |z-1|=1  \u03b1\u03c0\u0384\u03cc\u03c0\u03bf\u03c5 \u03b1\u03bd \u03b8\u03ad\u03c3\u03bf\u03c5\u03bc\u03b5  z=x+yi \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03c4\u03bf \u03c3\u03cd\u03c3\u03c4\u03b7\u03bc\u03b1 \r\n\r\n$ x^2 \\plus{} y^2 \\equal{} 1$,  kai  $ (x \\minus{} 1)^2 \\plus{} y^2 \\equal{} 1$ \u03cc\u03c0\u03bf\u03c5 \u03b4\u03af\u03bd\u03b5\u03b9 \u03bb\u03cd\u03c3\u03b5\u03b9\u03c2 \r\n\r\n(\u03c7,y) = ( 1/2 , $ \\sqrt {3}$/2 ) kai (\u03c7,y) = ( 1/2 , $ \\minus{} \\sqrt {3}$/2 )\r\n\r\n\u03ac\u03c1\u03b1 \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03c3\u03c7\u03ad\u03c3\u03b7 |z|=|z-1|=1 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03cd\u03bf\u03c5\u03bd \u03bf\u03b9 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03af\r\n\r\n  z1 = 1/2 +$ \\sqrt {3}$/2  kai  z2 = 1/2 $ \\minus{} \\sqrt {3}$/2\r\n\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd (4) \u03c0\u03bf\u03c5 \u03b8\u03ad\u03bb\u03b5\u03b9\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \r\n\r\n$ \\left| \\frac {z \\plus{} i}{z \\minus{} i}\\right| \\equal{} 1$ <=> |z-i|=|z+i|\r\n\r\n\u039f\u03b9 \u03b5\u03b9\u03ba\u03cc\u03bd\u03b5\u03c2 \u03c4\u03c9\u03bd \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03ce\u03bd  z  \u03c0\u03bf\u03c5 \u03b5\u03c0\u03b1\u03bb\u03b7\u03b8\u03b5\u03cd\u03bf\u03c5\u03bd \u03c4\u03b7\u03bd \u03b5\u03be\u03af\u03c3\u03c9\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae \u03cc\u03bc\u03c9\u03c2 \u03b1\u03bd\u03ae\u03ba\u03bf\u03c5\u03bd  \u03c3\u03c4\u03b7 \u03bc\u03b5\u03c3\u03bf\u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c4\u03bf\u03c5 \u03b5\u03c5\u03b8\u03cd\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf\u03c5 \u03c4\u03bc\u03ae\u03bc\u03b1\u03c4\u03bf\u03c2 \u03bc\u03b5 \u03ac\u03ba\u03c1\u03b1 \u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u0391(0,1) \u039a\u0391\u0399 \u0392(0,-1). \u0397 \u03bc\u03b5\u03c3\u03bf\u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03b1\u03c5\u03c4\u03ae \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf \u03ac\u03be\u03bf\u03bd\u03b1\u03c2 \u03c7\u03c7\u0384 \u03c3\u03c5\u03bd\u03b5\u03c0\u03ce\u03c2 \u03bf z \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03bf R.\r\n\r\nH \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c5\u03c4\u03ae ( \u03bc\u03b5 \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03b5\u03c4\u03b9\u03ba\u03ac \u03bd\u03bf\u03cd\u03bc\u03b5\u03c1\u03b1 ) \u03b2\u03c1\u03af\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c3\u03c4\u03b7 \u03c3\u03b5\u03bb\u03af\u03b4\u03b1 101 , \u0391\u03c3\u03ba\u03b7\u03c3\u03b7 5 \u0391\u0384\u03bf\u03bc\u03ac\u03b4\u03b1\u03c2 \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 (\u03b1).\r\n\r\n\u0393\u03b5\u03bd\u03b9\u03ba\u03ac \u03bd\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03ce \u03cc\u03c4\u03b9 \u03cc\u03bb\u03b5\u03c2 \u03bf\u03b9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03b1\u03c5\u03c4\u03ad\u03c2 \u03b2\u03c1\u03af\u03c3\u03ba\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03af\u03b4\u03b9\u03b5\u03c2 \u03ae \u03bc\u03b5 \u03bc\u03b9\u03ba\u03c1\u03ad\u03c2 \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b1\u03b3\u03ad\u03c2 \u03c3\u03b5 \u03b5\u03be\u03c9\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ac \u03b2\u03bf\u03b7\u03b8\u03ae\u03bc\u03b1\u03c4\u03b1. :)",
  "Solution_3": "\u039d\u03b1 \u03b2\u03bf\u03b7\u03b8\u03ae\u03c3\u03c9 \u03ba\u03b1\u03b9 \u03c3\u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 9.  \u039a\u03b1\u03c4 \u03b1\u03c1\u03c7\u03ae\u03bd \u03bd\u03b1 \u03c3\u03bf\u03c5 \u03c0\u03ce \u03cc\u03c4\u03b9 \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b7\u03bd \u03b5\u03ba\u03c6\u03ce\u03bd\u03b7\u03c3\u03b7.  \u0398\u03b1 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03c4\u03bf \u03bb \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf  ( \u03bb>0). \u0391\u03bd \u03c4\u03bf \u03bb \u03b1\u03bd\u03ae\u03ba\u03b5\u03b9 \u03c3\u03c4\u03bf R* \u03c4\u03cc\u03c4\u03b5 \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03bb = -1 \u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 $ |z_1\\plus{}z_2|^2 \\le 0$\r\n\r\n\u0397 \u03b1\u03c1\u03c7\u03b9\u03ba\u03ae \u03b3\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 :\r\n\r\n$ (z_1\\plus{}z_2)(\\bar{z_1}\\plus{}\\bar{z_2}) \\le |z_1|^2\\plus{}\\lambda |z_1|^2 \\plus{}|z_2|^2\\plus{}\\frac{1}{\\lambda}|z_2|^2$ <=>\r\n\r\n$ |z_1|^2\\plus{}|z_2|^2\\plus{}z_1\\bar{z_2}\\plus{}z_2\\bar{z_1} \\le |z_1|^2\\plus{}\\lambda |z_1|^2 \\plus{}|z_2|^2\\plus{}\\frac{1}{\\lambda}|z_2|^2$  \r\n\r\n\u03b4\u03b9\u03ce\u03c7\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b1 $ |z_1|^2,|z_2|^2$, \u03ba\u03b1\u03b9 \u03c0\u03bf\u03bb\u03bb/\u03b6\u03bf\u03c5\u03bc\u03b5 \u03b5\u03c0\u03af  \u03bb>0 \u03ba\u03b1\u03b9 \u03bc\u03ad\u03bd\u03b5\u03b9 \u03bd\u03b1 \u03b4.\u03bf.\r\n\r\n$ \\lambda^2|z_1|^2 \\plus{} |z_2|^2\\minus{}\\lambda z_1\\bar{z_2}\\minus{}\\lambda z_2\\bar{z_1} \\ge0$  \u03c0\u03bfu \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03cd\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \r\n\r\n$ |\\lambda z_1\\minus{}z_2|^2 \\ge 0$ :)",
  "Solution_4": "\u03a0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9, \u03ad\u03c7\u03c9 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b7\u03bd 6 (\u03bc\u03c0\u03ad\u03c1\u03b4\u03b5\u03c8\u03b1 \u03c4\u03b1 \u03c3\u03cd\u03bd\u03bf\u03bb\u03b1 C \u03ba\u03b1\u03b9 I! \u0391\u03bd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b4\u03c5\u03bd\u03b1\u03c4\u03cc\u03bd!  :o ).  \u0393\u03b9\u03b1 \u03c4\u03b7\u03bd 1, \u03ad\u03c7\u03c9 \u03c0\u03ac\u03c1\u03b5\u03b9 \u03c4\u03b1 \u03b4\u03b9\u03b1\u03bd\u03cd\u03c3\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03b1\u03bd\u03ac\u03c0\u03bf\u03b4\u03b7 \u03c6\u03bf\u03c1\u03ac \u03b1\u03bb\u03bb\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 2\u03bb - ... = \u03bb ... \u03c0\u03bf\u03c5 \u03c4\u03bf \u03ad\u03ba\u03b1\u03bd\u03b1 3\u03bb = ... (\u0395\u03af\u03c7\u03b1 \u03ba\u03b1\u03b9\u03c1\u03cc \u03bd\u03b1 \u03ba\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c4\u03b9 \u03c4\u03ad\u03c4\u03bf\u03b9\u03bf \u03bd\u03b1 \u03b3\u03b5\u03bb\u03ac\u03c3\u03bf\u03c5\u03bc\u03b5 \u03bb\u03af\u03b3\u03bf.) \u03a4\u03ce\u03c1\u03b1 \u03c4\u03b1 \u03ad\u03c7\u03c9 \u03b4\u03b9\u03bf\u03c1\u03b8\u03ce\u03c3\u03b5\u03b9."
}
{
  "Tag": [
    "topology",
    "inequalities",
    "geometric sequence",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $X$ be a metric space. We say that $N = \\{x_{i}| i \\in I\\}$ is an $\\epsilon$-net on $X$ if \r\n*for any distinct $x_{1}, x_{2}\\in N$, $d(x_{1}, x_{2}) > \\epsilon$\r\n*the collection of balls $B_{i}= B(x_{i}, \\epsilon)$ centered at $x_{i}$ with radius $\\epsilon$ cover $X$.\r\n\r\nLet $\\#(Y, \\epsilon)$ be the cardinality of an $\\epsilon$-net on $Y$. Suppose there exist constants $C, r > 0$ (depending only on $X$, and $C$ depending on $r$) such that\r\n$\\#(B_{R}, R\\epsilon) \\le \\frac{C}{\\epsilon^{r}}$\r\nfor any ball $B_{R}$ of radius $R$ in $X$ and all $0 < \\epsilon \\le \\frac{1}{2}$ (we consider $B_{R}$ as a subspace of $X$ and find a net in it). \r\n\r\nThen let $\\alpha = \\inf\\{ r | \\text{the above holds}\\}$. This is called the Assouad dimension of $X$. \r\n\r\nShow that $\\alpha \\ge \\rm{dim}_\\mathcal{H}(X)$, the Hausdorff dimension. Is there a non-horrific example of a space where the inequality is strict?\r\n\r\nedit: In fact, $\\alpha$ is at least as big as the Minkowski dimension, and there are easy examples of sets where the Minkowski dimension is bigger than the Hausdorff dimension. But I think spaces exist where the Hausdorff dimension (and also the Minkowski dimension) are finite, but the inequality above does not hold for any $r$. \r\n\r\nI also read that there exist sets where the Assouad dimension increases under projection! :o  \r\n\r\nNote: it is sometimes called the Bouligand dimension?",
  "Solution_1": "This is quite a fancy notion. Let's play just with sequences of points. Consider first a geometric progression $E=\\{2^{-k}\\}_{k\\ge 0}$. It is not hard to check that $\\alpha(E)=0$. On the other hand, if $E\\subset \\mathbb R$ contains arbitrarily long arithmetic progressions (the steps of the progressions do not matter), then $\\alpha(E)=1$ (if $x_{0}+kd$, $k=1,\\dots,N$ is our progression, just take $R=Nd$, $\\varepsilon=1/N$ and conclude that we must have $N\\le CN^{r}$ for all $N>0$). \r\n\r\nThis immediately gives an example of the situation when $\\alpha$ increases under projection. Just consider any sequence $x_{k}$ satisfying $0<x_{k}<2^{-k}$ and containing arbitrarily long arithmetic progressions and the set $E\\subset\\mathbb R^{2}$ consisting of all points $(x_{k},2^{-k})$. Then $\\alpha(E)=0$ but the projection of $E$ to the $x$-axis has $\\alpha=1$.\r\n\r\nAccording to Wikipedia, Bouligand dimension is the same as Minkowski dimension.",
  "Solution_2": "[quote=\"fedja\"]According to Wikipedia, Bouligand dimension is the same as Minkowski dimension.[/quote]\r\nI saw that, but in a review of a paper I read (I read the review, but not the paper, since I couldn't find the paper) the reviewer said that Bouligand dimension was Assouad dimension. In such an obscure matter I'm more inclined to trust something I read on an AMS page than on Wikipedia, but I still put the question mark just in case. ;)",
  "Solution_3": "Seems like you are right and Wikipedia is wrong. I was too lazy to dig up the original Bouligand paper of 1928 but quite a few authors of modern papers available online agree with you. :) Olson in [url=http://nyjm.albany.edu:8000/PacJ/p/2002/202-2-12.pdf]this paper[/url] discusses the properties of the  \"Assouad-Bouligand\" dimension in detail including the formula for the dimension of the Cartesian product and an example of a projection increasing the dimension.",
  "Solution_4": "Assouad, Assouad-Bouligand, whatever. There is also \"Assouad-Nagata dimension\" which is actually different. I'll try to motivate the definition.\r\n \r\nA metric space $X$ is doubling if there is a constant $N$ such that any ball $B_{X}(x,R)$ can be covered by at most $N$ balls of radius $R/2$. This can be interpreted as being \"finite dimensional\" in a metric sense. For example, if $X$ has a bi-Lipschitz embedding into some $\\mathbb R^{n}$, then $X$ is doubling. [The converse is false, but counterexamples are not obvious.] This is nice enough, but the doubling constant $N$ has no obvious relation to the embedding dimension $n$, and is very hard to compute even for $\\mathbb R^{n}$ itself. \r\n\r\nHowever, if we efficiently cover $B^{n}(x,R)$ with balls of radius $\\epsilon R$, we should expect to have about $\\epsilon^{-n}$ smaller balls. That is, the asymptotic behavior of the number of balls is geometrically meaningful, and this is where the Assouad dimension comes from. Although this dimension may increase under Lipschitz maps, it is bi-Lipschitz invariant.\r\n\r\nIn conclusion: a metric space is doubling iff its Assouad dimension is finite; the value of Assouad dimension is more interesting than the value of a doubling constant.",
  "Solution_5": "Ok, I think I see why the Assouad dimension is so big. Since the net is a collection of points in the space, you're a lot more restricted than in the other dimensions (Hausdorff dimension, Minkowski dimension) where you are covering with sets from a more general collection. So a net you use for the Assouad dimension will give you a covering that you could also use for Hausdorff or Minkowski dimension, so $\\alpha \\ge \\dim_{M}$, but we don't necessarily have equality because it's possible that a covering you have for the Minkowski dimension won't translate easily to a net (as the centers of the boxes might be outside of the set if you are bilipschitz-embedding it in $\\mathbb{R}^{n}$ and looking at it as a subset of $\\mathbb{R}^{n}$, which is basically the way I think about this stuff...). \r\n\r\nSo, sets like the ones fedja was talking about are probably a good way to get examples of strict inequality. But I read that you can find a space with finite Hausdorff dimension and infinite/undefined Assouad dimension! I can't imagine that!  :huh:",
  "Solution_6": "You are slightly misleaded. The problem is not \"designing a net\" for the entire space as opposed to just \"covering\" but designing a net/covering at every scale and [b]for every ball[/b], not only for the whole space. This \"locality\" of the definition is what makes it really different from the Minkowski dimension. As to the example of a set with zero Hausdorff and infinite Assouad dimension, read Olson's article I mentioned in my previous post. I think it'll clear up a lot of things for you. ;)",
  "Solution_7": "ok :( I will read it then",
  "Solution_8": "The solution to $\\sqrt{x}= 5$ is $x=25$."
}
{
  "Tag": [
    "integration",
    "function",
    "algebra",
    "rational function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "transform to solve y(s)\r\n\\[2\\int_{0}^{t}(t-u)^{2}y(u)du+y'(s)=(t-1)^{2}\\quad t\\geq0 \\]\r\n\r\nI get (Y is transf of y) that $Y(s)(\\frac{4}{s^{3}}+s)=2e^{-s}\\frac{1}{s^{3}}$\r\nthis means that we must find function with transform\r\n\\[\\frac{2e^{-2}}{4+s^{4}}\\]\r\n\r\nBut I cant do that. Anyone who can, or have I made a mistake?? Please help.",
  "Solution_1": "Note that the function whose Laplace transform is $\\frac{2e^{-s}}{s^{3}}$ is not $(t-1)^{2}$ for $t>0$ but rather $(t-1)^{2}H(t-1)$ where $H$ is the Heaviside unit step function.\r\n\r\nIf $f(t)=(t-1)^{2}=t^{2}-2t+1$ for $t>0,$ then\r\n\r\n$F(s)=\\frac2{s^{3}}-\\frac2{s^{2}}+\\frac1s.$\r\n\r\nNow, as for finding the inverse transform of a function with $s^{4}+4$ in the denominator, use the factorization\r\n\r\n$s^{4}+4=(s^{2}-2s+2)(s^{2}+2s+2).$\r\n\r\nYou can find in closed form the inverse Laplace transform of any rational function of $s$ under one condition: you must be able to factor the denominator."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "search",
    "USA(J)MO",
    "USAMO",
    "geometry",
    "Support"
  ],
  "Problem": "I don't mean for this to sound like a complaint because I blame any of my mistakes on myself and did okay on the AIME, but I was just wondering what kind of a test environment everyone else has?  (To clarify, I'm not trying to use my situation as an excuse for my score.  I got what I got and don't really care about \"if this hadn't happened...\")\r\n\r\nPersonally, for the first half of the AIME II, I was in a room with a teacher who was incessantly and extremely animatedly talking on the phone for at least 20-30 minutes combined.  I didn't want to cause any trouble and seriously contemplated changing rooms, but due to the strange organization of my school (math wing right next to gym and connected to the main building by a bridge), I didn't want to waste time in search of an empty classroom, of which there aren't many.\r\n\r\nAt about halfway through, I had to change rooms, and the second half was a lot better, though some technology specialist came in for a few minutes to talk with the teacher.\r\n\r\nSo is my situation unique?  Or does everyone else have to suffer like this?  I remember last year I had to switch to a room with a horrible clock (There's nothing like a constant *tick*...*tock*... to drive you insane).  I just want to know whether or not I'm in a position to perhaps request a better room, \"proctor,\" maybe even request a single room rather than a transition in the middle.  Thanks for any feedback.",
  "Solution_1": "My school generally has 10+ people taking it, so it's fine. We're in a basement room and there is a chair-moving party at the beginning of every period above us, but other than that I have no complaints.",
  "Solution_2": "We took the AIME II in the conference room of our library. The only disturbances were the bells that rang between every period, but they were useful in basically warning us of each passing hour.",
  "Solution_3": "i was in an office with one other asian kid (i didnt take it at my school, i took it at a ghetto school)\r\nsince it was an office, random teachers were talking, but i am not the easily distracted type",
  "Solution_4": "Our school had about 10 people taking it. It was pretty quiet but the room was cold or maybe it was just my nerves.",
  "Solution_5": "Conference room. There were a few distractions (one instance of drilling, morning announcements, some muted talking), but overall it was fine.",
  "Solution_6": "Classroom after school had ended. Almost complete silence. \r\nTaking the AMC was a bit harder. We were inside a the school theater and had no desks, which was bad for me, because i use a lot of paper and like to spread it out.",
  "Solution_7": "Including me, there were two kids in my school taking it. We each had like three conference room desks to work with.",
  "Solution_8": "I took the AIME I in the principal's office, oddly enough, he wasn't there that day, so the math dept head decided to use it.  She was the proctor for almost the whole thing.  I had plenty of space, a large desk (only one other person was taking it).  As for distractions, the bells, the staff talking outside (the office is right off of the main office), and the proctor's knitting was about it.  I was pretty happy in general (way better than taking AMC, where I got less space and people were leaving intermittently).  Last year was better for me though, I took it in a conference room (of the superintendent's office, oddly enough), and I had a huge table to myself, though a test with one proctor and one student is somewhat odd.",
  "Solution_9": "I took it in the school library.  No distractions at all (except for bells and muted talking), and a big table all to myself.",
  "Solution_10": "I took it in the school library, no distractions other than the usual police chasing car-jacking people and pointing guns at random people, and the announcements telling everyone to stay calm and not panic since the carjacking people were on the loose in our campus",
  "Solution_11": "Only 2 people from my school qualified (and we make up a 6th of the state) so we almost got stuck taking it on a 4 ft. diameter table, and about a square ft. desk respectively. Luckily, our teacher managed to get us the conference room, so we had more space, and it was quiet the whole time. I love that our school doesn't have bells.",
  "Solution_12": "[quote=\"barbieboy\"]I took it in the school library, no distractions other than the usual police chasing car-jacking people and pointing guns at random people, and the announcements telling everyone to stay calm and not panic since the carjacking people were on the loose in our campus[/quote]\r\n\r\nuhm, whao? I guess that sucks.",
  "Solution_13": "There were about 10 people that took the test at my school. It was in the basement, so it was pretty quiet in general, if I remember correctly (I'm used to distractions, though, so I can't be sure).\r\n\r\n[b]Edit:[/b] I forgot about the morning announcements. Those were really annoying.",
  "Solution_14": "I took it in our library. It was fine for the most part except random bells, morning anouncements, student laughing loudly in the hall ways, people talking loudly in the library, and a false fire alarm which probably DQs our school.",
  "Solution_15": "[quote=\"MCTP\"][quote=\"MCTP\"]At local and state math contests I've been to, the conditions were far better than those I described in my earlier post, and the reason is easy enough to see: there were more people there. [/quote][/quote]\r\n\r\nI completely agree.  The more people there are, the more that teachers and a school will probably care about them.  For some reason, however, many teachers don't even consider the AMC tests as math competitions, which is the main problem.  During the AIME, I asked my noisy \"proctor\" is she could speak a little softly (instead of \"HAHAHAHAHAHAHA...\" on the phone), and the next block she told her class, \"I had this really arrogant student working on some stuff last block who told me to be quiet.  Well, this is my room, and I'll do what I want.\"  The teachers simply don't care about math, ironic because that's the subject that they teach.  I mean, if a teacher won't shut up that's one thing, but telling her class afterwards about an \"arrogant student\" who's just trying to take the AIME in peace?",
  "Solution_16": "I'd be pretty annoyed if that happened here. The teachers/administrators at my school have generally always been very accommodating to these kinds of things. I'm grateful to have teachers that care.",
  "Solution_17": "My school had A LOT of AIME qualifiers (about 160), so we had to use the cafeteria. There weren't any distractions. At all. It was nice.",
  "Solution_18": "I posted earlier, but forgot to tell of someone I knew who went to a school with a dress code and even though they were sequestered in a room by themselves, they were \"forced\" to wear an oxford shirt and tie (and dress pants and shoes) the entire 2 days of the test! They protested to the administration of the school and lost. Now, how uncomfortable would any of you be if you had to test with these clothes??? I know that I'd be miserable if I couldn't wear comfy clothes...",
  "Solution_19": "Our school had about 120 students taking the AIME I...\r\n\r\n(Korean Minjok Leadership Academy)\r\n\r\ndivided into 2 conference rooms, and testing environment was OK.\r\n\r\nWas able to concentrate As much as I wanted to....\r\n\r\n\r\n\r\nWhat matters for me is the USAMO environment.\r\n\r\nJust like last year, as an international participant, I have to abide by the continental US time.\r\n\r\nWhich...makes my schedule[u][b]  1:30~6:00AM, 4/29~4/30.[/b][/u]\r\n\r\nVery nice....very nice to test and improve my concentration skills :D\r\n\r\nWell, I did it last year, and I'm doing it this year too, so....I'm hoping for the best of luck.\r\n\r\nIt's actually not too bad if you get some sleep during the day and get excuses.\r\n\r\n\r\n\r\nAs for the environment you were in, I think there MUST BE SOMETHING DONE, because AIME is not just\r\n\r\na trivial test. You must be guaranteed the best environment that suits you, esp. if you are one of the few\r\n\r\nqualifiers.\r\n\r\nTry contacting the school through official ways - After all, a AIME Qualifier is an honor for the school too!",
  "Solution_20": "[quote=\"physisnu\"]Try contacting the school through official ways - After all, a AIME Qualifier is an honor for the school too![/quote]\r\nUnfortunately, some schools don't see it this way and regard AIME qualifiers (USAMO qualifiers more so, I would imagine) as a nuisance. At least that's what I gather from this topic and from my own experience; if a school thought that it was an honor to have a student taking the AIME, they would treat them better.",
  "Solution_21": "[quote=\"MCTP\"]\nUnfortunately, some schools don't see it this way and regard AIME qualifiers (USAMO qualifiers more so, I would imagine) as a nuisance. [/quote]\r\n\r\nI don't agree with it, but I get how some schools could think so, there's a lot of paperwork to wade through for this stuff (especially USAMO, the manual is 28 pgs worth of instructions and forms...)  For teachers who don't understand how much of an honor it is to be invited to take USAMO, it could appear as a nuisance...",
  "Solution_22": "[quote]the manual is 28 pgs worth of instructions and forms [/quote]\r\n\r\nDon't scare people, it's not THAT bad, we're not the IRS!\r\n\r\nThe USAMO part is \r\n  -- 3 pages (I admit two-column) about half just general info, about half instructions, \r\n  - - then 3 pages of forms for the teacher, \r\n       --- one before the USAMO, and \r\n       --- one each day of the USAMO\r\n  -- then two pages of info for the student and parents\r\n       --- one form for the student and parent to return\r\n  -- then 6 pages of blank templates for the answers.\r\n\r\nWe do make an effort to keep the directions complete, but concise, and we try not to ask for information we don't use.\r\nI hope it's not a nuisance, it's an effort to make sure we can do our job for you quickly and efficiently.\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions",
  "Solution_23": "$ (6561r^5\\minus{}1)/(r\\minus{}1)$",
  "Solution_24": "I think two much larger reasons for the grouping of students are affluence and training. Affluent schools tend to have more resources to cater to gifted students, and so they get support from the administration, and the schools which have historically done well have developed programs to train their students.",
  "Solution_25": "[quote=\"physisnu\"]\n\nJust like last year, as an international participant, I have to abide by the continental US time.\n\nWhich...makes my schedule[u][b]  1:30~6:00AM, 4/29~4/30.[/b][/u]\n\n[/quote]\r\n\r\nwow. :O that really sucks. good luck!",
  "Solution_26": "At TJ, we usually have well over 100 qualifiers, so most years they place us in the cafeteria where it's silent and away from all the classrooms. Another nice thing is that we don't have any bell system at TJ, so that never bothers us :)",
  "Solution_27": "When I was taking the AIME I this year, distractions were minimal (it was taken in the lobby of a guidance office), namely people entering the guidance office, bells, and morning announcements. Room was ample, and paper was unlimited (very useful: I had about 25 pages of work). I definitely needed the good conditions that provided an equivalent of 177 minutes of distraction-free time, because in the last 10-20 minutes I was fervently rechecking my answer to problem number 14 (ironically, my final answer to that problem was not right but that affected nothing).\r\n\r\nI believe the school I attend gave as close to an ideal testing environment as it had available, something for which I am glad.",
  "Solution_28": "[quote=\"herefishyfishy1\"]My school had A LOT of AIME qualifiers (about 160), so we had to use the cafeteria. There weren't any distractions. At all. It was nice.[/quote]\r\nwhat???????!!!!!!!!!!!!!!!\r\nwhere do u go?",
  "Solution_29": "[quote=\"physisnu\"]\n\nJust like last year, as an international participant, I have to abide by the continental US time.\n\nWhich...makes my schedule[u][b]  1:30~6:00AM, 4/29~4/30.[/b][/u]\n\n[/quote]\r\n\r\n1:30 to 6:00 AM? You're not going to be going to school for the week after that... that's brutal.\r\n\r\nAt least ours (which is from 12:30 to 5:00, still not very comfortably fitting into the school day) is during the daytime hours."
}
{
  "Tag": [
    "quadratics",
    "geometry",
    "angle bisector"
  ],
  "Problem": "Prove the Steiner-Lehmus theorem in as many ways as possible. :D",
  "Solution_1": "[quote=\"Inspired By Nature\"]Prove the Steiner-Lehmus theorem in as many ways as possible. :D[/quote]\r\n\r\nCan you restate it? What is it?",
  "Solution_2": "[quote=\"drunner2007\"][quote=\"Inspired By Nature\"]Prove the Steiner-Lehmus theorem in as many ways as possible. :D[/quote]\nCan you restate it? What is it?[/quote]\r\nSure. The Steiner-Lehmus theorem states that any triangle that has two equal angle bisectors (measured from the vertex to the opposite side) is isosceles. :)",
  "Solution_3": "Proof #1:\r\n\r\nYou see that the triangles formed by the bisectors are similar because their share the base of the original triangle. The bisectors are also equal, so by the SS similarity theorem, the triangles are similar. thus, the base angles are equal by the definition of similarity and the triangle is symettric.",
  "Solution_4": "[quote=\"neelnal\"]Proof #1:\n\nYou see that the triangles formed by the bisectors are similar because their share the base of the original triangle. The bisectors are also equal, so by the SS similarity theorem, the triangles are similar. thus, the base angles are equal by the definition of similarity and the triangle is symettric.[/quote]\r\n\r\nit's SAS simililarity, you did not prove the included angles are congruent",
  "Solution_5": "Oops. Sorry forgot that is was SAS.\r\n\r\nWell, I'll work on that later. Another method you can use is indirect proof. Start backwards.",
  "Solution_6": "let a be the base, and b,c be the alleged legs, using the angle bisector length theorem, (after clearing the denominators)\r\n\r\n$ac(a+b)^{2}(a+c-b)(a+c+b)=ab(a+c)^{2}(a+b-c)(a+b+c)$\r\n\r\nlet $p=a+b+c$\r\n\r\n$f(p)=c(p-c)(p-2b)-b(p-b)(p-2c)=0$\r\n\r\n$f(p)$ is a quadratic, $f(a+2b)=0$ and $f(a+2c)=0$ since those imply $a=b$, making the expression 0, there are only two roots, so it factors\r\n$f(p)=k*(c-b)(b-c)=0$\r\n\r\nso $b=c$",
  "Solution_7": "Couldn't k=0?"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "1. Let $ a,b,c$ be sidelengths of a triangle. Prove that\r\n\\[ \\frac{a^2\\minus{}bc}{\\sqrt{4a^2\\plus{}bc}}\\plus{}\\frac{b^2\\minus{}ca}{\\sqrt{4b^2\\plus{}ca}}\\plus{}\\frac{c^2\\minus{}ab}{\\sqrt{4c^2\\plus{}ab}}\\le0\\]\r\nWhen equality occurs?",
  "Solution_1": "Clearly when a=b=c the inequality holds\r\nand by SOS we can prove the inequality,i think",
  "Solution_2": "[quote=\"dduclam\"]1. Let $ a,b,c$ be sidelengths of a triangle. Prove that\n\\[ \\frac {a^2 \\minus{} bc}{\\sqrt {4a^2 \\plus{} bc}} \\plus{} \\frac {b^2 \\minus{} ca}{\\sqrt {4b^2 \\plus{} ca}} \\plus{} \\frac {c^2 \\minus{} ab}{\\sqrt {4c^2 \\plus{} ab}}\\le0\\]\nWhen equality occurs?[/quote]\r\nYes, it's easy SOS.\r\n$ 2\\sum_{cyc}\\frac {a^2 \\minus{} bc}{\\sqrt {4a^2 \\plus{} bc}}\\equal{}\\sum_{cyc}\\frac{(a\\minus{}b)(a\\plus{}c)\\minus{}(c\\minus{}a)(a\\plus{}b)}{\\sqrt {4a^2 \\plus{} bc}}\\equal{}$\r\n$ \\equal{}\\sum_{cyc}(a\\minus{}b)\\left(\\frac{a\\plus{}c}{\\sqrt {4a^2 \\plus{} bc}}\\minus{}\\frac{b\\plus{}c}{\\sqrt {4b^2 \\plus{} ac}}\\right)\\equal{}$\r\n$ \\equal{}\\sum_{cyc}\\frac{c(a\\minus{}b)^2(c^2\\minus{}2(a\\plus{}b)c\\plus{}a^2\\minus{}7ab\\plus{}b^2)}{\\sqrt{(4a^2\\plus{}bc)(4b^2\\plus{}ac)}\\left((a\\plus{}c)\\sqrt{4b^2\\plus{}ac}\\plus{}(b\\plus{}c)\\sqrt{4a^2\\plus{}bc}\\right)}.$\r\nId est, it remains to prove that\r\n$ c^2\\minus{}2(a\\plus{}b)c\\plus{}a^2\\minus{}7ab\\plus{}b^2\\leq0,$ which is obvious.",
  "Solution_3": "Nice, Arqady.  Some same form problems are\r\n\r\n2. Let $ a,b,c$ be sidelengths of a triangle. Prove that \r\n\\[ \\frac{a^{2}\\minus{}bc}{\\sqrt{15a^{2}\\plus{}(b\\plus{}c)^2}}\\plus{}\\frac{b^{2}\\minus{}ca}{\\sqrt{15b^{2}\\plus{}(c\\plus{}a)^2}}\\plus{}\\frac{c^{2}\\minus{}ab}{\\sqrt{15c^{2}\\plus{}(a\\plus{}b)^2}}\\le0\\]\r\n\r\n3. Let $ a,b,c$ be sidelengths of a triangle. Prove that \r\n\\[ \\frac{a^{2}\\minus{}bc}{\\sqrt{7a^{2}\\plus{}b^2\\plus{}c^2}}\\plus{}\\frac{b^{2}\\minus{}ca}{\\sqrt{7b^{2}\\plus{}c^2\\plus{}a^2}}\\plus{}\\frac{c^{2}\\minus{}ab}{\\sqrt{7c^{2}\\plus{}a^2\\plus{}b^2}}\\le0\\]",
  "Solution_4": "[quote=\"dduclam\"]1. Let $ a,b,c$ be sidelengths of a triangle. Prove that\n\\[ \\frac {a^2 \\minus{} bc}{\\sqrt {4a^2 \\plus{} bc}} \\plus{} \\frac {b^2 \\minus{} ca}{\\sqrt {4b^2 \\plus{} ca}} \\plus{} \\frac {c^2 \\minus{} ab}{\\sqrt {4c^2 \\plus{} ab}}\\le0\\]\nWhen equality occurs?[/quote]\r\n\r\nNotice that we have the following Inequality with the same condition\r\n\\[ \\frac {a^2 \\minus{} bc}{4a^2 \\plus{} 5bc} \\plus{} \\frac {b^2 \\minus{} ca}{4b^2 \\plus{} 5ca} \\plus{} \\frac {c^2 \\minus{} ab}{4c^2 \\plus{} 5ab}\\ge0\\]"
}
{
  "Tag": [
    "Olimpiada de matematicas"
  ],
  "Problem": "aca un problema que parece conocido   :D \r\n\r\nhalle todas las ternas $(x,y,z)$ los cuales pertenecen a los reales positivos tales que :\r\n\r\n         $x+y+z = 6$\r\n    y\r\n         $xyz= 20-(xy+yz+zy)$",
  "Solution_1": "[quote=\"mhuarancca\"]  $xyz= 20-(xy+yz+zy)$-->??? no es as\u00ed: \n$xyz= 20-(xy+yz+zx)$[/quote].\r\nconsiderando eso.\r\nahora agamos el siguiente cambio de variable:\r\n$a=x+1, b=y+1, c=z+1. \\Rightarrow abc=(1+x)(1+y)(1+z)=1+x+y+z+xy+yz+xz+xyz=1+6+20=27 \\to abc=27$ \r\naplicando $MA\\geq MG$, a los terminos $a,b,c$.\r\n$(a+b+c)/3\\geq (abc)^{1/3}$\r\n$(x+y+z+3)/3\\geq (27)^{1/3}$->$x+y+z+3\\geq 9$\r\n$x+y+z\\geq 6$, ->$x+y+z=6$, entonces se cumple la igualdad pero en medias se cumple la igualdad cuando se da la igualdad entre sus terminos por lo tanto $x=y=z$, remplasando $3x=6$, y, $x^{3}=20-3x^{2}$\r\nde aqui $x=y=z=2$.",
  "Solution_2": "[quote=\"mhuarancca\"]aca un problema que parece conocido   :D \n\nhalle todas las ternas $(x,y,z)$ los cuales pertenecen a los reales positivos tales que :\n\n         $x+y+z = 6$\n    y\n         $xyz= 20-(xy+yz+xz)$[/quote]\r\n\r\nDesde que $x^{2}+y^{2}+z^{2}\\geq xy+yz+xz$ se sigue que $(x+y+z)^{2}\\geq 3(xy+yz+xz)$ con lo que $36\\geq 3(xy+yz+xz)$ y de ahi que $12\\geq (xy+yz+xz)$. Del dato $xyz= 20-(xy+yz+xz)$ con lo que $xyz\\geq 8$. De la desigualdad $MA-MG$ tenemos que $x+y+z\\geq 3\\sqrt[3]{xyz}$ con lo que $xyz\\leq 8$ luego tenemos que $xyz=8$ con lo que $x=y=z=2$."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Let A be a matrix in Mn(C) s.t. Aii = n-1 and Aij \u0404 {-1, 1} for i \u2260 j.\r\nIf det(A) = 0 then A is symmetric.",
  "Solution_1": "If the determinant is 0 the coresponding system of ecuations attached to the matrix must have a non-trivial solution $(x_1,x_2,...x_n)$. Assume WLOG that $|x_1| \\ge |x_i|\\forall i$. Then, the equation $(n-1)x_1+x_2+x_3+...+x_n=0\\Rightarrow \\sum|x_i|\\le(n-1)|x_1|=|\\sum x_i|\\le\\sum|x_i|$. We can get easily that $x_1=x_2=...=x_n$. Hence, the matrix is symetric."
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Determine all functions $ f: R\\rightarrow R$ such that $ f(x \\minus{} f(y) ) \\equal{} f( f(y) ) \\plus{} x f(y) \\plus{} f(x) \\minus{} 1$ for all $ x, y \\in R$.",
  "Solution_1": "[color=green]Problem  : Find all functions $ f: \\mathbb{R}\\longrightarrow \\mathbb{R}$ such that:\n\\[ f(x \\minus{} f(y)) \\equal{} f(f(y)) \\plus{} xf(y) \\plus{} f(x) \\minus{} 1\n\\]\nfor all $ x,y$ in the real numbers.\n\nSolution: Let $ P(x,y)$ the property $ f(x \\minus{} f(y)) \\equal{} f(f(y)) \\plus{} xf(y) \\plus{} f(x) \\minus{} 1.$\n\nObviously $ \\exists \\;y_0$ such that $ f(y_0)\\neq 0$.\n\nThen, in $ P(x,y_0)$, let $ x\\rightarrow \\plus{} \\infty$ : $ \\text{RHS}$ is not bounded, and so $ \\text{LHS}$ can't be bounded. \n\n$ \\star\\;P(f(x),x)$ implies $ f(0) \\equal{} 2f(f(x)) \\plus{} f(x)^2 \\minus{} 1$ and so\n\\[ f(f(x)) \\equal{} \\frac {f(0) \\plus{} 1}{2} \\minus{} \\frac {f(x)^2}{2}\n\\]\nOr\n\\[ f(x) \\equal{} \\frac {f(0) \\plus{} 1}{2} \\minus{} \\frac {x^2}{2}, \\forall x\\in \\{f(x), x\\in\\mathbb{R}\\}\n\\]\n$ \\star\\;P(2f(x),x)$ implies $ f(f(x)) \\equal{} f(f(x)) \\plus{} 2f(x)^2 \\plus{} f(2f(x)) \\minus{} 1$ and so\n\n   Then\n\\[ f(2f(x)) \\equal{} 1 \\minus{} \\frac {(2f(x))^2}{2}\n\\]\nOr\n\\[ f(x) \\equal{} 1 \\minus{} \\frac{x^2}{2}\\;\\text{for all x such that}\\;\\frac{x}{2}\\in \\{f(x), x\\in\\mathbb{R}\\}\n\\]\nSince $ f(x)$ is not bounded , and since the set of discontinuity points of $ f(x)$ is included in an interval $ (a,b)$, it is possible to find two values $ u$ and $ v$ such that $ f(v) \\equal{} 2f(u)$\n\nBecause $ f(f(v)) \\equal{} \\frac {f(0) \\plus{} 1}{2} \\minus{} \\frac {f(v)^2}{2}$\n\nAnd  $ f(f(v)) \\equal{} 1 \\minus{} \\frac {f(v)^2}{2}$\n\nSo $ \\frac {f(0) \\plus{} 1}{2} \\equal{} 1$ or $ f(0) \\equal{} 1$\n\nNow we prove that : $ f(x) \\equal{} 1 \\minus{} \\frac {x^2}{2}, \\forall x\\in\\mathbb{R}$\n\nLet $ g(x) \\equal{} f(x) \\minus{} (1 \\minus{} \\frac {x^2}{2})$. $ P(x,y)$ becomes : $ g(x \\minus{} f(y)) \\equal{} g(f(y)) \\plus{} g(x)$\n\nLet then $ x \\equal{} f(y)$, we have $ g(f(y)) \\equal{} \\frac {g(0)}{2} \\equal{} 0.$ And so\n\\[ g(x \\minus{} y) \\equal{} g(x) \\forall x\\in\\mathbb{R}, \\forall y\\in \\{f(x), x\\in\\mathbb{R}\\}\n\\]\nSince we know that it exists an intervall $ [ \\minus{} \\infty,x_1)$ or $ (x_2, \\plus{} \\infty)$ included in $ \\{f(x), x\\in\\mathbb{R}\\}$, we can conclude that $ g(x) \\equal{} \\text{contant} \\equal{} 0$ or $ f(x) \\equal{} 1 \\minus{} \\frac {x^2}{2}, \\forall x\\in\\mathbb{R}$\nAnd hence the result.\n\nIt took me 30 minutes to type  :wallbash_red:  :wacko:. [/color]",
  "Solution_2": "Since f(x) is not bounded , and since the set of discontinuity points of f(x) is included in an interval (a,b), it is possible to find two values u and v such that f(v) = 2f(u) \r\n\r\nCan you clarify please?"
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "What are the last two digits of 3 to the 1234 power?",
  "Solution_1": "every 20 numbers, the last two digits repeats. 1234/20 is something r 14. 'that should mean the last two digits is 69. \r\n\r\nof course, i probably made a silly mistake.",
  "Solution_2": "it repeats 03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67,01,and over again. 1234/20 leaves remainder 14 and the 14th pair of numbers is \r\n[hide]69[/hide]",
  "Solution_3": "i agree with Kchande",
  "Solution_4": "[quote=\"math92\"]What are the last two digits of 3 to the 1234 power?[/quote]\r\n\r\n\\[3^{1234} \\equiv 3^{34} \\pmod {100}\\]\r\n\\[3^{34} \\equiv (3^2)^{17} \\pmod {100}\\]\r\n\\[9^{17} \\equiv 9^9 \\cdot 9^8 \\pmod {100}\\]\r\n\\[9^9 \\cdot 9^8 \\equiv 89 \\cdot 21 \\pmod {100}\\]\r\n\\[89 \\cdot 21 \\equiv \\fbox {69} \\pmod {100}\\]\r\n\r\nWhy is this topic called \"adress\"?"
}
{
  "Tag": [],
  "Problem": "from KMO\r\n\r\nIf $x_1$, $x_2$, $x_3$, $...$, $x_n$ are positive reals, and $x_1$ + $x_2$ + $x_3$ + $...$ + $x_n$=1\r\n\r\nProve: $\\dfrac{1}{1 + x_1}$ + $\\dfrac{1}{1 + x_1 + x_2}$ + ... + $\\dfrac{1}{1 + x_1 + x_2 + ... + x_n}$ < $\\sqrt{\\dfrac{2}{3}(\\dfrac{1}{x_1} + \\dfrac{1}{x_2} + ... + \\dfrac{1}{x_n})}$",
  "Solution_1": "sry, i can't solve it  :(  :(  :("
}
{
  "Tag": [],
  "Problem": "I need help with these problems:\r\n\r\n1) Consider an object that has a mass, $ m$, and a weight, $ W$, at the surface of the moon. If we assume\r\nthe moon has a nearly uniform density, which of the following would be closest to the object\u2019s\r\nmass and weight at a distance halfway between Moon\u2019s center and its surface?\r\n\r\nA) $ \\frac12 m$ & $ \\frac12W$     B) $ \\frac14 m$ & $ \\frac14 W$      C) $ 1 m$ & $ 1 W$      D) $ 1 m$ & $ \\frac12 W$      E) $ 1 m$ & $ \\frac14 W$\r\n\r\n2)One object at the surface of the Moon experiences the same gravitational force as a second\r\nobject at the surface of the Earth. Which of the following would be a reasonable conclusion?\r\n\r\n[A] both objects would fall at the same acceleration\r\n[B] the object on the Moon has the greater mass\r\n[C] the object on the Earth has the greater mass\r\n[D] both objects have identical masses\r\n[E] the object on Earth has a greater mass but the Earth has a greater rate of rotation.\r\n\r\n3)Two artificial satellites, $ 1$ and $ 2$, are put into circular orbit at the same altitude above Earth\u2019s\r\nsurface. The mass of satellite $ 2$ is twice the mass of satellite $ 1$. If the period of satellite $ 1$ is $ T$,\r\nwhat is the period of satellite $ 2$?\r\n\r\nA. $ T/4$ B. $ T/2$ C. $ T$ D. $ 2T$ E. $ 4T$\r\n\r\n4)A hypothetical planet orbits a star with mass one-half the mass of our sun. The planet\u2019s\r\norbital radius is the same as the Earth\u2019s. Approximately how many Earth years does it take for\r\nthe planet to complete one orbit?\r\n\r\nWhen solving problems like these, what formulas or concepts are important to remember?\r\nAnd I would also appreciate it if someone points me toward good websites for more practice.\r\n\r\nThanks.",
  "Solution_1": "Here's some help on the first one:\r\n\r\nWhat is mass?  Does mass ever change?  If so, when?  (Assuming stationary-it technically changes at relativistic speeds but let's ignore that).  How about weight?  What is it?  What does Newton's Law of Gravitation state?\r\n\r\nIf you can answer these questions, there's no reason you can't answer your question.",
  "Solution_2": "Well I know that the mass of the object doesn't change in this case. Also, the weight of the object is the force of gravity acting between the object and the moon.\r\nSo, if the object is at the surface, the weight is $ W_1 \\equal{} \\frac {Gm_{obj}m_{moon}}{r^2}$.\r\nIf the object is halfway between the center and the surface, the weight is $ W_2 \\equal{} \\frac {Gm_{obj}m_{moon}}{(0.5r)^2} \\equal{} 4W_1$. So, the weight would be $ 4$ times larger in if it's in the middle.\r\nBut that's not one of the answers. What am I doing wrong here?",
  "Solution_3": "[quote=\"kcn2rivers\"]Well I know that the mass of the object doesn't change in this case. Also, the weight of the object is the force of gravity acting between the object and the moon.\nSo, if the object is at the surface, the weight is $ W_1 \\equal{} \\frac {Gm_{obj}m_{moon}}{r^2}$.\nIf the object is halfway between the center and the surface, the weight is $ W_2 \\equal{} \\frac {Gm_{obj}m_{moon}}{(0.5r)^2} \\equal{} 4W_1$. So, the weight would be $ 4$ times larger in if it's in the middle.\nBut that's not one of the answers. What am I doing wrong here?[/quote]\r\nIf the object is inside the moon,then the outer shell of it doesn't exert force on the object,so $ m_{moon}$ should be $ \\frac {1}{8}$ as it used to be.Thus the force is $ \\frac {1}{2}$ smaller. The answer's $ \\boxed{D}$,isn't it?",
  "Solution_4": "Yes that's right, thank you.\r\n\r\nCan someone tell me if this is right for #3?\r\n\r\n$ R^3 \\equal{} kT_1^2 \\equal{} kT_2^2$, where $ R$ is the mean distance of the two satellites. $ k$ is the same value because the satellites are both orbiting around Earth. So, $ T_1\\equal{}T_2$, $ \\boxed{C}$ ?\r\n\r\nIf I am right, the mass of a satellite doesn't affect its period if the radius is kept constant?",
  "Solution_5": "[quote=\"kcn2rivers\"]\nCan someone tell me if this is right for #3?\n\n$ R^3 \\equal{} kT_1^2 \\equal{} kT_2^2$, where $ R$ is the mean distance of the two satellites. $ k$ is the same value because the satellites are both orbiting around Earth. So, $ T_1 \\equal{} T_2$, $ \\boxed{C}$ ?\n\nIf I am right, the mass of a satellite doesn't affect its period if the radius is kept constant?[/quote]\r\nI think it's right."
}
{
  "Tag": [
    "vector",
    "analytic geometry",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "I've totally lost in this question, Any help is highly appreciated.\r\n\r\nGiven \r\n      \r\n          (U`,U``)+(V`,V``)=(U`+V`+1, U``+V``-1)\r\n         \r\n           k(U`,U``)=(kU`+k-1, kU``-k+1)\r\n\r\nLet vecter U=(1,0),vecter V=(-1,0)\r\n\r\na) Show that {U,V} form a basis in V\r\n\r\nb)Find the coordinate vectors of (2,0) abd (0,0) and (-1,1) relative to this basis.",
  "Solution_1": "We have $kU=(2k-1,-k+1)$ and $lV=(-1,-l+1)$, $kU+lV=(2k-1,-k-l+1)$. Threfore if $kU+lV=(a,b)$ then $k=\\frac{a+1}{2}$ and $l=-b-\\frac{a+1}{2}+1$. We see that each $(a,b)$ can be uniquely expressed.",
  "Solution_2": "[quote=\"Myth\"]We have $kU=(2k-1,-k+1)$ and $lV=(-1,-l+1)$, $kU+lV=(2k-1,-k-l+1)$. Threfore if $kU+lV=(a,b)$ then $k=\\frac{a+1}{2}$ and $l=-b-\\frac{a+1}{2}+1$. We see that each $(a,b)$ can be uniquely expressed.[/quote]\r\n\r\nHow do you deduce such a scalar multiplication. It doesn't seem clear, at least not for me.\r\n\r\nThanks. Best,",
  "Solution_3": "I don't know what you mean. I don't see any scalar multiplication in my post.",
  "Solution_4": "[quote=\"Myth\"]I don't know what you mean. I don't see any scalar multiplication in my post.[/quote]\r\n\r\nWhat does it mean then $kV$, because if $kV=V+V+V+\\cdots+V$ $k$ times, then you don't have a representation for every vector. Note that otherwise, you are multiplying a vector for an apriori real, not necessarely ingeter, scalar!!\r\n\r\nBest regards,",
  "Solution_5": "But who are speaking about integer multipliers? See problem: there is a rule for multiplication by any real number.",
  "Solution_6": "[quote=\"Myth\"]But who are speaking about integer multipliers? See problem: there is a rule for multiplication by any real number.[/quote]\r\n\r\nOK, I missread, but anyway, the problem itself is written wrong, because it does not define a real vector space (a vector space has to have a scalar multiplication, that is, a rule of multiplication for elements on the field) but the one they propose does not hold with the distributive property, because, in general, $kX+kY\\neq k(X+Y)$.\r\n\r\nBest,",
  "Solution_7": "That's because this is an exercise with a silly disguise. Isomorphism from $V$ to $F^2,$ where $F$ is the field the elements are taken from: $(a,b)\\mapsto (a+1,b-1)$"
}
{
  "Tag": [
    "combinatorics open",
    "combinatorics"
  ],
  "Problem": "infinite grid is partitioned into dominoes. Coloring the set of dominoes in 2 colors is good\r\nif any point belongs to at most 2\r\none-coloured dominoes. Is there a good coloring for any partition?",
  "Solution_1": "Are there any ideas?"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "find real solutions of the system\r\n$\\sin x+2\\sin(x+y+z)=0$\r\n$\\sin y+3\\sin(x+y+z)=0$\r\n$\\sin z+4\\sin(x+y+z)=0$",
  "Solution_1": "This should help:\r\n\r\n[hide]sin(x+y+z) = sin(x)cos(y+z) + cos(x)sin(y+z)\n               = sin(x)cos(y)cos(z) - sin(x)sin(y)sin(z) + cos(x)\n                            sin(y)cos(z) + cos(x)cos(y)sin(z)\n\nTherefore, we have the following equations:\nsin(x) + 2cos(x)cos(y)sin(z) + 2cos(x)sin(y)cos(z)\n         + 2sin(x)cos(y)sin(z) - 2sin(x)sin(y)sin(z) = 0\nsin(y) + 3cos(x)cos(y)sin(z) + 3cos(x)sin(y)cos(z)\n         + 3sin(x)cos(y)sin(z) - 3sin(x)sin(y)sin(z) = 0\nsin(z) + 4cos(x)cos(y)sin(z) + 4cos(x)sin(y)cos(z)\n         + 4sin(x)cos(y)sin(z) - 4sin(x)sin(y)sin(z) = 0\n\nThrough a simple rearrangement and combination, \nsin(x)/2 = sin(y)/3 = sin(z)/4\n           = -cos(x)cos(y)sin(z) - cos(x)sin(y)cos(z)\n               - sin(x)cos(y)sin(z) + sin(x)sin(y)sin(z)[/hide]\r\n\r\nEdit: I'm not a newbie anymore!!!  :lol:"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "quadratics",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Given some prime $p$, let $a, b \\in \\mathbb{Z}_p$. If the polynomial $x^3 + ax + b$ is irreducible over $\\mathbb{Z}_p$, prove that $-4a^3 - 27b^2$ is a square in $\\mathbb{Z}_p$.",
  "Solution_1": "Here's the advanced but easy way to do it:\r\nAny irreducible polynomial of degree $n$ over $F_p$ splits completely in the field with $p^n$ elements. If the discriminant of a polynomial is not a square, the splitting field of that polynomial contains the square root of the discriminant, and therefore the splitting field contains a quadratic extension of the base field. An extension of degree $n$ can only contain an extension of degree $m$ if $m|n$. Since $2$ does not divide $3$, the splitting field of an irreducible cubic over $F_p$ cannot contain the square root of a nonsquare element of $F_p$, and the discriminant must already be a square.\r\n\r\nIn fact, a cubic polynomial with a nonsquare discriminant must factor as the product of a linear polynomial and an irreducible quadratic.\r\n\r\nIs there a reasonable elementary solution?",
  "Solution_2": "what's the discriminant of a polynomial? i know what it is for $n=2$ ( $\\Delta=b^2-4ac$ ) but how do we define it for bigger degrees?",
  "Solution_3": "[quote=\"perfect_radio\"]what's the discriminant of a polynomial? i know what it is for $n=2$ ( $\\Delta=b^2-4ac$ ) but how do we define it for bigger degrees?[/quote]\r\nYou need to know the general solution to a cubic equation.\r\n$x^3+ax^2+bx+c=0$\r\nSubstitute $y=x+\\frac{a}{3}$, the equation becpomes\r\n$y^3+py+q=0$ where $p=\\frac{a^2}{3}+b$ and $q=\\frac{2a^3}{27}-\\frac{ab}{3}+c$\r\nThe solution is\r\n$y=\\sqrt[3]{-\\frac{q}{2}+\\sqrt{\\frac{q^2}{4}+\\frac{p^3}{27}}}+\\sqrt[3]{-\\frac{q}{2}-\\sqrt{\\frac{q^2}{4}+\\frac{p^3}{27}}}$.\r\n(This is found by letting $y=u+v$ such that $3uv=-p$)\r\nThe discriminant is $\\frac{q^2}{4}+\\frac{p^3}{27}$.\r\nAnother definition of discriminant is given [url=http://en.wikipedia.org/wiki/Cubic_equation#The_nature_of_the_roots]here[/url].",
  "Solution_4": "Normally, you define the discriminant of a polynomial as $\\left(\\prod_{1 \\leq i < j \\leq n} (a_i-a_j)\\right) ^2$ where the $a_k$ are the $n$ roots of the polynomial. E.g., the discriminant is $0$ iff two roots are identical."
}
{
  "Tag": [
    "Support",
    "puzzles"
  ],
  "Problem": "At a party there were:\r\n\r\n14 adults,\r\n17 children,\r\n12 males,\r\n19 females.\r\n\r\nThen X arrived and the number of man-woman couples possible became equal to the number of boy-girl couples possible. (e.g.: If there are 6 men and 8 women, there are 48 possible man-woman couples)\r\n\r\nIs X a man, woman, boy, or girl?",
  "Solution_1": "X is a [hide]girl OR X is a woman[/hide]because of the following . . .\n[hide]Let 12 - x = # of male children\nThen x = # of male adults\nLet 19 - y =  # of female children\nThen y = # of female adults\n=========================================\n\nCase I -- Suppose X is a boy.\n\nAdd 1 to 12 - x.\n(13 - x)(19 - y) = xy  \nThis can become:\n247 = 13y + 19x\nWhich can become:\n13 - x = (13/19)y\nNo solutions in positive integers x and y.\n\n------------------------------------------------\n\nCase II -- Suppose X is a girl.\n\nAdd 1 to 19 - y.\n(12 - x)(20 - y) = xy  \nThis can become:\n60 = 3y + 5x\nHere, x = 3, y = 15; x = 6, y = 10; x = 9, y = 5\nThis supports X is a girl. \n\n----------------------------------------------------\n\nCase III -- Suppose X is a man.\n\nAdd 1 to x.\n(12 - x)(19 - y) = (x + 1)y\nThis can become:\n228 = 13y + 19x\nWhich can become:\n12 - x = (13/12)y\nNo solutions in positive integers x and y\n\n----------------------------------------------\n\nCase IV -- Suppose X is a woman.\n\nAdd 1 to y.\n(12 - x)(19 - y) = x(y + 1)\nThis can become:\n57 = 3y + 5x\nHere, x = 6, y = 9\nThis supports X is a woman.\n\n\\_/---\\_/---\\_/---\\_/---\\_/---\\_/---\\_/---\\_/---\\_/\n\nTherefore, X can be either a girl or a woman.\n[/hide]",
  "Solution_2": "Arrange your tan, you've neglected one of the constraints of the problem (that there be 14 adults and 17 children); as a result, 3 of your 4 proposed possible answers don't actually satisfy the requirements of the question.",
  "Solution_3": "Amending that, from my case II \r\n(where one supposes the addition of a girl),\r\nit would have to be that x = 9 and y = 5. \r\n\r\nThen there would be 9 men, 5 women, 3 boys,\r\nand 14 girls before the one girl was added."
}
{
  "Tag": [
    "abstract algebra",
    "algebra",
    "polynomial",
    "induction",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let M = {f(x) in Q[x] | f(Z) is a subset of Z} (here Q is the rationals and Z is the integers). Show that M is a free Z-module of countably infinte rank with basis {(x i)|i=0,1,2,...} Here (x i) is the binomial coefficient.",
  "Solution_1": "It is readily that $M$ is $\\mathbb{Z}$-module, so we actually just need to verify that the polynomials  form a base in this $\\mathbb{Z}$-module. In order to do this we can use the following fact:\r\nLet $p$ be a polynomial of degree $\\leq k$ such that $p(\\mathbb{Z}) \\subset \\mathbb{Z}$, then $p(x)= \\sum_{i=0}^k c_0 \\left(\\begin{array}{c}x\\\\k-i \\end{array} \\right)$ (*) with $c_i \\in \\mathbb{Z}$.\r\nProof. It is easy to readily that $\\left(\\begin{array}{c}x\\\\k-i \\end{array} \\right), i=0,1,2...,$ form a basis in $\\mathbb{R}[X]$ and thus there exist such $c_i$ as in (*), we only need to prove that $c_i$ are actually integers.\r\nWe can do this by induction on the degree of polynomial $k$. The base is trivial. Suppose this result is prove for all the polynomials of degree $\\leg k$, and let $p(x)= \\sum_{i=0}^{k+1} c_0 \\left(\\begin{array}{c}x\\\\k-i \\end{array} \\right)$, consider $\\Delta p(x) =p(x+1)-p(x)=\\sum_{i=0}^k c_0 \\left(\\begin{array}{c}x\\\\k-i \\end{array} \\right)$ - it takes integer values at integers piints, so by induction hypothesis $c_i,i=0,1,...,k \\in \\mathbb{Z}$ and $c_{k+1}=p(n)- \\sum_{i=0}^k c_0 \\left(\\begin{array}{c}n\\\\k+1-i \\end{array} \\right)$ is also integer, so we are done."
}
{
  "Tag": [
    "blogs",
    "\\/closed"
  ],
  "Problem": "In my Blog Options, here is the exact CSS in the \"Custom CSS\" box:\r\n\r\n[code]#main { \nbackground: none;\n}\n\n#wraper {\nbackground: none;\n}[/code]\n\n(yes I am quite aware that it should be \"#wrapper\")\n\nI want it to be\n\n[code]#main { \nbackground: transparent;\n}\n\n#wrapper {\nbackground: transparent;\n}[/code]\r\n\r\nHowever, when I apply these changes and click \"Submit\", when I click \"Click Here to set additional options for your blog,\" I go back and see that the changes were [b]not[/b] made and the original CSS is still there. Is this a bug or am I just doing something wrong?",
  "Solution_1": "Bug confirmed. Pretty sure it's a permission problem on the folders.",
  "Solution_2": "Bug fixed."
}
{
  "Tag": [
    "trigonometry",
    "AMC",
    "AIME"
  ],
  "Problem": "What is the smallest positive value of $\\theta$ where $\\sin \\theta$ is the same in degrees and radians?",
  "Solution_1": "Draw it out. You'll see that when the radian graph comes back down, it must intersect with the degrees graph.\r\n\r\n$\\sin k-\\sin (180-k)$\r\n\r\nWe have:\r\n$\\sin \\theta = \\sin \\left( 180-\\frac{\\pi}{180}\\theta \\right) \\implies \\theta = \\frac{180 \\pi}{180+\\pi}$",
  "Solution_2": "This was an AIME problem from a few years back!\r\nI think the question asked to find the two smallest values of theta.",
  "Solution_3": "In fact its already [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=714637#p714637]here[/url] :wink:\r\nI guess it never says there but thats AIME 2002 II #10",
  "Solution_4": "[quote=\"pkothari13\"]This was an AIME problem from a few years back!\nI think the question asked to find the two smallest values of theta.[/quote]\r\nOh wow, I didn't even know if this question was possible so I posted it here.\r\nHow do we know for sure that the two posted at the other link are the smallest?",
  "Solution_5": "Sketch the graph; it goes a long way."
}
{
  "Tag": [
    "quadratics",
    "number theory",
    "prime factorization"
  ],
  "Problem": "[b]WHAT IS THE LEAST POSSIBLE NUMBER OF POSITIVE INTEGERS SUCH THAT THE SUM OF THEIR SQUARES EQUALS1995 :?: [/b]",
  "Solution_1": "The sum of the squares of the FIRST $ n$ integers is $ \\frac{n(n\\plus{}1)(2n\\plus{}1)}{6}$\r\n\r\nSo what we'll do is find an upper bound first.\r\n\r\n$ n(n\\plus{}1)(2n\\plus{}1) \\equal{} 11970$\r\n$ n \\equal{} 17.660604$ \r\n\r\nSo our upper bound is 18 integers.\r\n\r\nCan you modify this to reach a better result?",
  "Solution_2": "Sums of squares of consecutive positive integers have little to do with sums of squares of a collection of integers (and even if they did, the calculation you performed does not imply an upper bound).  The actual upper bound for [i]any[/i] positive integer is $ 4$ by the Four Square Theorem.",
  "Solution_3": "Hmm. Could you explain that theorem a little more?\r\nEDIT: I can do it in three:\r\n\r\n$ 37^2 \\plus{} 25^2 \\plus{} 1^2 \\equal{} 1995$\r\nI don't think it's possible to go lower than that.",
  "Solution_4": "Correct.  It's not possible to write $ 1995$ as a sum of two squares.  \r\n\r\n[hide=\"Generalizations\"] A positive integer $ n$ is expressible as a sum of two squares iff any prime congruent to $ 3 \\bmod 4$ in the prime factorization of $ n$ has even exponent.\n\nA positive integer $ n$ is expressible as a sum of three squares iff it is not of the form $ 4^n (8k \\plus{} 7)$ for non-negative integers $ n, k$. \n\n[i]Every[/i] positive integer is expressible as a sum of four squares. [/hide]",
  "Solution_5": "[quote=\"t0rajir0u\"]\nA positive integer $ n$ is expressible as a sum of three squares iff it is not of the form $ 4^n (8k \\plus{} 7)$ for non-negative integers $ n, k$. \n[/quote]\r\n\r\nWhere'd you get THAT from?\r\n\r\nThat seems a little arcane to me. I followed the other two lines pretty well.",
  "Solution_6": "As I recall, the proof of that result is actually more difficult than the proof of the Four Square Theorem.  It's easy to verify that numbers congruent to $ 7 \\bmod 8$ can't be written as the sum of three squares (since squares are congruent to $ 0, 1, 4 \\bmod 8$).  It's also easy to verify that if a sum of three squares is congruent to $ 0 \\bmod 4$ then each square is divisible by $ 4$.  In other words, the hard part is showing that numbers not of that form [i]are[/i] representable as a sum of three squares.",
  "Solution_7": "For the 3-squares-theorem, you either need a lot about quadratic forms, or the theorem of Hasse-Minkowski, or some work and Dirichlet's theorem on primes in arithmetic progression...\r\nIt's the hardest case, all other number of squares are much more elementary.",
  "Solution_8": "[quote=\"t0rajir0u\"]A positive integer $ n$ is expressible as a sum of three squares iff it is not of the form $ 4^n (8k \\plus{} 7)$ for non-negative integers $ n, k$. [/quote]\r\n\r\nCan you provide an example of this? I'm not sure what $ k$ stands for.",
  "Solution_9": "$ k$ is a non-negative integer.  The set of positive integers that can't be written as a sum of three squares can be ordered first according to the value of $ k$ and then according to the value of $ n$.  In other words, the list of all such integers can be written\r\n\r\n$ n \\equal{} 0:  7, 15, 23, 31, 39, 47, 55, ...$\r\n$ n \\equal{} 1:  4 \\cdot 7, 4 \\cdot 15, 4 \\cdot 23, 4 \\cdot 31, 4 \\cdot 39, 4 \\cdot 47, 4 \\cdot 55, ...$\r\n$ n \\equal{} 2:  16 \\cdot 7, 16 \\cdot 15, 16 \\cdot 23, 16 \\cdot 31, 16 \\cdot 39, 16 \\cdot 47, 16 \\cdot 55, ...$\r\n$ n \\equal{} 3:  64 \\cdot 7, 64 \\cdot 15, 64 \\cdot 23, 64 \\cdot 31, 64 \\cdot 39, 64 \\cdot 47, 64 \\cdot 55, ...$\r\n$ n \\equal{} ...$\r\n\r\nand so forth."
}
{
  "Tag": [],
  "Problem": "A line segment of length 15 + 5sqroot of 3 is folded to create a 30-60-90 right triangle. Two of these are put together to form a equilateral triangle. Four of these equilateral triangles are put together to form a tetrahedron. Two of these are put together to form a hexahedron. What is the volume of this final hexahedron?[/img]",
  "Solution_1": "Dude, this was already posted.\r\nTry figuring out some on your own, Chris.",
  "Solution_2": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116080\r\n\r\nPlagiarism=very bad. Reposting a problem when the original is in the top 10 topics=very bad."
}
{
  "Tag": [],
  "Problem": "5. Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath\r\nhe cuts is 28 inches wide, but he overlaps each cut by 4 inches to make sure that\r\nno grass is missed. He walks at the rate of 5000 feet per hour while pushing\r\nthe mower. Which of the following is closest to the number of hours it will take\r\nMoe to mow his lawn?\r\n\r\n$ (A) 0.75 \\qquad (B) 0.8 \\qquad (C) 1.35 \\qquad (D) 1.5 \\qquad (E) 3$",
  "Solution_1": "file attached\r\n\r\nif you question, emal to me \r\nthstjdus0629@naver.com",
  "Solution_2": "I had no clue what all that swath stuff meant, so I just did stuff. It would regularly take him 2.7 hours, and 1.35 is half of that, so I chose it."
}
{
  "Tag": [
    "conics",
    "irrational number",
    "puzzles"
  ],
  "Problem": "http://www.stripcreator.com/comics/prove_me_wrong/446941",
  "Solution_1": "No offense, but I don't get it at all...\r\n\r\n$ \\frac{1}{10}$",
  "Solution_2": "It's like the classic problem \"I have two coins that have a total value of 15 cents.  One of the coins is not a dime.  What are the two coins?\"  The answer is a nickel and a dime because the problem states that [b]one[/b] coin (the nickel) is not a dime.  It doesn't say anything about the other coin.",
  "Solution_3": "Hmm... surd raised to a surd implies that the surds are the same,... but $ (\\sqrt 2)^{\\sqrt 2} \\neq 2$... Perhaps the surd you are looking for is $ 1.55961046946$ :)",
  "Solution_4": "Yongyi781,\r\n\r\nno, it does not impy that the surds are the same!  The phrase \r\n\"a surd raised to a surd\" literally means an irrational number\r\nraised to an irrational number.  There is nothing in the English\r\nthat stated/states that they are the same surd.  You worked \r\nthe English, which was/is not ambiguous or otherwise, incorrectly.\r\n\r\nOne of the surds is the the square root of two.  The other\r\nsurd is the square root of two raised to the square root of\r\ntwo.",
  "Solution_5": "No, the other surd is 2.\r\n\\[ (\\sqrt 2)^2 \\equal{} 2\r\n\\]\r\nBut 2 is not irrational, contradiction.\r\n\r\nBut, you can go the other way.\r\n\\[ \\left(\\sqrt 2^{\\sqrt 2}\\right)^{\\sqrt 2} \\equal{} 2\r\n\\]\r\nWhich gives the desired result.",
  "Solution_6": "The only thing I find funny is that you are now banned from StripCreator.",
  "Solution_7": "[quote=\"QuantumTiger\"]The only thing I find funny is that you are now banned from StripCreator.[/quote]\r\nThat doesn't make any sense, unless you know that being banned from\r\nStripCreator is actually a badge of honor, because the users and moderators\r\nof it are tyrants who want comics to made a certain way (cookie-cutter style), \r\nand the moderators don't tolerate any dissent.  (If you want to keep going \r\noff topic, away from the merits of the comic strip, then do so, but you'll not\r\nbe making any valid arguments that way.)",
  "Solution_8": "I'm sorry for going off topic, but I just happened to find that funny, just as you happened to find your comic funny. Sorry, but I know nothing about StripCreator. Please forgive me.  :roll:"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Any one have an idea?\r\n:)\r\nHere's the Question:  \r\n\"Let $p$ and $q$ be distinct primes and suppose that $G$ is a finite group with exactly $p+1$ $p$-Sylow subgroups and exactly $q+1$ $q$-Sylow subgroups.  Prove that there exists a $p$-Sylow subgroup $P$ and a $q$-Sylow subgroup $Q$ of $G$ such that the direct product $P\\times Q$ is also a subgroup of $G$.\"\r\n\r\n\r\nNote: we are assuming that $p$ and $q$ are not $2$ and $3$.\r\n\r\nI was given a hint to consider the normalizer of one of the Sylow subgroups (that is $N_G(P)$ or $N_G(Q)$) to help show there is an injective group homomorphism from $P\\times Q$ into $G$.\r\n\r\nOr to consider the action of $P$ on the conjugates of $Q$.  That either approach should help solve the problem.  But I'm not having much luck yet.  \r\n\r\nAny help would be appreciated.",
  "Solution_1": "Prove that there exists a $p$-Sylow subgroup $P$ and a $q$-Sylow subgroup $Q$ of $G$ such that the direct product $P\\times Q$ is also a subgroup of $G$.\r\n\r\nWhat exactly do you mean here, up to isomorphism?",
  "Solution_2": "I don't know if I understand your question.  \r\nIn my question, we know there are exactly $p+1$, $p$-Sylows and $q+1$, $q$- Sylows of the group $G$.  \r\nAnd we have to show that the direct product of some $p$-Sylow and some $q$-Sylow is a subgroup of $G$.\r\nSo I need help because I don't know HOW to show that $P\\times Q$, the direct product of a $p$-Sylow and a $q$-Sylow of $G$, is a subgroup of $G$.\r\n\r\nI don't know if 'up to isomorphism' has anything to do with this problem.\r\n\r\nPlease let me know if any clarification is needed.  \r\n\r\nI would love some input on this problem.\r\n\r\nThanks!"
}
{
  "Tag": [
    "algebra solved",
    "algebra"
  ],
  "Problem": "x1,x2,.....,xn real numbers\r\n\r\nProve that \r\n\r\nx1/(1+x1^2)+x2/(1+x1^2+x2^2)+....+xn/(1+x1^2+...+xn^2)< \\sqrt n",
  "Solution_1": "Try to solv :D \r\nIt's trivial",
  "Solution_2": "yeah i know this problem was one of the problems in the 2001 imo short list - proposed by Bogdan Enescu, Romania. \r\n\r\na hint for the solution: first apply Cauchy and then try to re-arrange the terms to obtain a telescopical sum (i.e. a sum of which some consecutive terms have the sum 0)."
}
{
  "Tag": [
    "geometry unsolved",
    "geometry",
    "BritishMathematicalOlympiad"
  ],
  "Problem": "In triangle $ABC$, $D$ is the midpoint of $AB$ and $E$ is the point of trisection of $BC$ nearer to $C$. Given that $\\angle ADC=\\angle BAE$ find $\\angle BAC$.",
  "Solution_1": "Using Menelaos we obtain that DL=LC, where L is the intersection of AE and DC. That means that m(LAC)=m(LCA). Using that m(DAL)=m(LAD),we find DAC=90. :lol:",
  "Solution_2": "How to use Menelaos?",
  "Solution_3": "That's ok.\r\n$\\frac{CL}{CD}\\frac{DA}{AB}\\frac{BE}{EC}=1\\Rightarrow CL=LD$.",
  "Solution_4": "[quote=\"PaulS\"]In triangle $ABC$, $D$ is the midpoint of $AB$ and $E$ is the point of trisection of $BC$ nearer to $C$. Given that $\\angle ADC=\\angle BAE$ find $\\angle BAC$.[/quote]\r\n\r\n[b] A simpler solution,but the same nice ![/b]\r\n\r\n If $M$ is the midpoint of $BE$ then $BM=ME = EC$ and \r\n   [color=red]a) [/color]from triangle $BAE$ : $DM // AE$\r\n [color=red]  b) [/color]from triangle $CDM$: $LD= LC$, since $EL // MD$ and $E$ is the midpoint of $MC$.\r\n  But , if $L$ is midpoint of $CD$  , then  $LA=LD=LC$ , so triangle $BAC$ must be right at vertex $A$.\r\n \r\n [u]Babis[/u]"
}
{
  "Tag": [
    "Euler",
    "function",
    "number theory",
    "prime numbers"
  ],
  "Problem": "Calculate the Euler function of m for m= p (a prime number), where m=p1p2 where pi and p2 are prime numbers, m=95. Find alln that satisfies that The Euler function for n is even.",
  "Solution_1": "This is a very confused post.\r\n\r\n[hide=\"Sorting out\"] There are many common lemmas about $\\varphi(n)$, two of which are that\n\n$gcd(a, b) = 1 \\implies \\varphi(ab) = \\varphi(a) \\varphi(b)$\n$\\varphi(p) = p-1$\n\nHence\n\n$\\varphi(p_{1}p_{2}) = (p_{1}-1)(p_{2}-1)$\n\nFor $m = 95$ we have\n\n$\\varphi(5 \\cdot 19) = 4 \\cdot 18 = 72$\n\nAs for the last question, $\\varphi(n)$ is even for all $n$ except $n = 2$.  This is because $gcd(a, n) = 1 \\implies gcd(n-a, n) = 1$.  Since $a \\neq \\frac{n}{2}$ except for $n = 2$ it follows $a \\neq n-a$. [/hide]"
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "Find the remainder on dividing $ x^{100} \\minus{} 2x^{51} \\plus{} 1$ by $ x^2 \\minus{} 1$",
  "Solution_1": "$ x^2\\equiv 1\\pmod{x^2\\minus{}1}$, so we have $ 1\\minus{}2x\\plus{}1\\equal{}2\\minus{}2x$.",
  "Solution_2": "I don't think that's the right solution\r\nThis is what the book says but I don't understand.\r\n\r\n[hide=\"Solution\"]\n$ x^{100} \\minus{} 2x^{51} \\plus{}1\\equal{} (x^2 \\minus{} 1)q(x) \\plus{} ax \\plus{} b$. Putting $ x \\equal{} 1$ into this relation gives  $ b \\equal{} 0$. Putting $ x \\equal{} \\minus{} 1$ gives $ a \\equal{} \\minus{} 4$. Thus the remainder is $ \\minus{} 4x$\n[/hide]",
  "Solution_3": "[quote=\"TheHeinrich\"]I don't think that's the right solution\nThis is what the book says but I don't understand.\n\n[hide=\"Solution\"]\n$ x^{100} \\minus{} 2x^{51} \\plus{} 1 \\equal{} (x^2 \\minus{} 1)q(x) \\plus{} ax \\plus{} b$. Putting $ x \\equal{} 1$ into this relation gives  $ b \\equal{} 0$. Putting $ x \\equal{} \\minus{} 1$ gives $ a \\equal{} \\minus{} 4$. Thus the remainder is $ \\minus{} 4x$\n[/hide][/quote]\r\n\r\nCheck the substitution. Putting $ x\\equal{}1$ gives $ a\\plus{}b\\equal{}0$ and putting $ x\\equal{}\\minus{}1$ gives $ \\minus{}a\\plus{}b\\equal{}\\minus{}4$. Solve that simple system, and you get what I said.",
  "Solution_4": "sorry :blush:",
  "Solution_5": "$ f(x)\\equal{}x^{100}\\minus{}2x^{51}\\plus{}1\\equal{}(x^2\\minus{}1)Q(x)\\plus{}l(x)$, $ l(x)\\equal{}ax\\plus{}b$\r\n$ \\Longleftrightarrow f(x)\\minus{}l(x)\\equal{}(x\\plus{}1)(x\\minus{}1)Q(x)$, which means that $ y\\equal{}f(x)$ has intersection points with $ y\\equal{}l(x)$ at $ x\\equal{}\\pm 1$, or\r\n\r\n$ y\\equal{}l(x)$ passes through two points $ (\\minus{}1,\\ 4),\\ (1,\\ 0)$, yielding $ l(x)\\equal{}\\minus{}2x\\plus{}2$."
}
{
  "Tag": [
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "$(tan pi/13).(tan 2pi/13).(tan 3pi/13)....tan (6pi/13)=?$",
  "Solution_1": "Let $f(x)=\\frac{x^n-1}{x-1}=\\prod_{j=1}^{n-1} (x-\\exp{ \\frac{2\\pi ji}{n}})$. Let n is odd.\r\nWe have $f(-1)=1=(-1)^{n-1} 2^{1+(-1)^n} \\prod_{j=1}^{(n-1)/2} (2\\cos{\\frac{j\\pi}{n}})^2\\exp{\\frac{\\pi ji}{n}\\exp{\\frac{-\\pi ji}{n}}=2^{n-1}M_1^2}$,\r\nWere $M_1=\\prod_{j=1}^{(n-1)/2}\\cos{\\frac{\\pi j}{n}}$. Therefore $M_1=2^{-(n-1)/2}$.\r\nWe have $n=f(1)=\\prod_{j=1}^{(n-1)/2} (2\\sin{\\frac{j\\pi}{n}})^2=2^{n-1}M_2^2$,\r\nwere $M_2=\\prod_{j=1}^{(n-1)/2} \\sin{\\frac{j\\pi}{n}}=2^{-(n-1)/2}\\sqrt n.$\r\nTherefore $M_3=\\prod_{j=1}^{(n-1/2} \\tan{\\frac{j\\pi}{n}}=\\frac{M_2}{M_1}=\\sqrt n$."
}
{
  "Tag": [],
  "Problem": "How many integers belong to the arithmetic sequence 13, 20, 27, 34, ..., 2008?",
  "Solution_1": "[quote=\"GameBot\"]How many integers belong to the arithmetic sequence 13, 20, 27, 34, ..., 2008?[/quote]\r\n\r\n$ 2008\\equal{}13\\plus{}(n\\minus{}1)7\\implies n\\equal{} \\frac {2008\\minus{}13}{7}\\plus{}1\\equal{}286$"
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "incenter",
    "inequalities proposed"
  ],
  "Problem": "The geometrical inequality $ \\sum \\frac {a}{b + c - a}\\ge\\sum \\frac {b + c - a}{a}$ is [u]well-known[/u]. \r\n\r\n[b]Prove that[/b] $ \\{\\begin{array}{cc} 1\\blacktriangleright & \\sum\\frac {a}{b + c - a}\\ge\\sum\\frac {b + c - a}{a} + \\frac {R - 2r}{R}\\ . \\\\\r\n \\\\\r\n2\\blacktriangleright & \\sum\\frac {a}{b + c - a}\\ge\\sum\\frac {b + c - a}{a} + \\frac {R - 2r}{R}\\cdot\\frac {8R - r}{4(2R - r)}\\ .\\end{array}$",
  "Solution_1": "\\[ \\sum\\frac{a}{b\\plus{}c\\minus{}a}\\equal{}\\frac{2R\\minus{}r}{r}\\]\r\n\r\n\\[ \\sum\\frac{b\\plus{}c\\minus{}a}{a}\\equal{}\\sum(\\frac{a\\plus{}b\\plus{}c}{a}\\minus{}2)\\equal{}\\frac{(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)}{abc}\\minus{}6\\equal{}\\frac{p^2\\plus{}r^2\\plus{}4Rr}{2Rr}\\minus{}6\\equal{}\\frac{p^2\\plus{}r^2\\minus{}8Rr}{2Rr}\\]\r\n\r\nIt is known that $ p^2\\le 4R^2\\plus{}4Rr\\plus{}3r^2$ so \r\n\r\n\\[ \\sum\\frac{b\\plus{}c\\minus{}a}{a}\\equal{}\\frac{p^2\\plus{}r^2\\minus{}8Rr}{2Rr}\\le\\frac{4R^2\\minus{}4Rr\\plus{}4r^2}{2Rr}\\equal{}\\frac{2R\\minus{}r}{r}\\minus{}\\frac{R\\minus{}2r}{R}\\equal{}\\sum\\frac{a}{b\\plus{}c\\minus{}a}\\minus{}\\frac{R\\minus{}2r}{R}\\]\r\n\r\nWe are done, I haven't solve the second, but the first equivalent to $ p^2\\le 4R^2\\plus{}4Rr\\plus{}3r^2$ so the second stronger than $ p^2\\le 4R^2\\plus{}4Rr\\plus{}3r^2$, I think it isn't true?",
  "Solution_2": "Hi! I am new to geometric inequalities. Can you please show me how to prove \"well-known\" inequality:\r\n\r\n\\[ \\sum \\frac {a}{b \\plus{} c \\minus{} a}\\ge\\sum \\frac {b \\plus{} c \\minus{} a}{a}\r\n\\]\r\n\r\nThank you!\r\n\r\nP.S: in encyclopedia's solution, he uses:\r\n\r\n\\[ \\sum \\frac {a}{b \\plus{} c \\minus{} a}\\equal{}\\frac{2R\\minus{}r}{r}\r\n\\]\r\nHow do you get this kind of identities?",
  "Solution_3": "[quote=\"shaaam\"]Hi! I am new to geometric inequalities. Can you please show me how to prove \"well-known\" inequality:\n\\[ \\sum \\frac {a}{b \\plus{} c \\minus{} a}\\ge\\sum \\frac {b \\plus{} c \\minus{} a}{a}\n\\]\nThank you!\n\nP.S: in encyclopedia's solution, he uses:\n\\[ \\sum \\frac {a}{b \\plus{} c \\minus{} a} \\equal{} \\frac {2R \\minus{} r}{r}\n\\]\nHow do you get this kind of identities?[/quote]\r\nFor the last question:\r\nWe have $ 4RT \\equal{} abc$ and $ T \\equal{} rs$, where $ T$ is the areal of the triangle with side $ a,b,c$. ($ T^2 \\equal{} s(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)$ where $ s \\equal{} \\frac{a\\plus{}b\\plus{}c}{2}$)\r\nNow $ \\frac{2R\\minus{}r}{r} \\equal{} \\frac{abc \\minus{} 2(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}{2(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}$. And thus it comes down to proving that:\r\n$ \\sum \\frac {a}{2(s\\minus{}a)} \\equal{} \\frac{abc \\minus{} 2(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}{2(s\\minus{}a)(s\\minus{}b)(s\\minus{}c)}$, which should be possible.\r\n\r\nFor the inequality:\r\nLet $ a\\equal{}x\\plus{}y$, $ b\\equal{}y\\plus{}z$ and $ c\\equal{}z\\plus{}x$. Then $ x,y,z > 0$.\r\nThen we just have to prove:\r\n$ \\sum_{cyc} \\frac{z\\plus{}x}{2y} \\ge \\sum_{cyc} \\frac{2y}{z\\plus{}x}$. Then just notice that $ z\\plus{}x \\ge 2\\sqrt{zx}$. Thus it suffices to prove that:\r\n$ \\sum_{cyc} \\frac{z\\plus{}x}{2y} \\ge \\sum_{cyc} \\frac{y}{\\sqrt{xz}}$. Which follows by muirhead since $ (1,0,\\minus{}1) \\succ (1,\\minus{}\\frac{1}{2},\\minus{}\\frac{1}{2})$.",
  "Solution_4": "Thanks Mathias!  :) \r\nwe can also finish off your proof differently. we have to prove:\r\n\r\n\\[ \\sum_{cyc} \\frac{z\\plus{}x}{2y} \\ge \\sum_{cyc} \\frac{2y}{z\\plus{}x}\r\n\\]\r\n\r\nFrom Cauchy-Schwarz:\r\n\\[ \\sum_{cyc} \\frac{2y}{z\\plus{}x}\\leq \\frac{1}{2}\\cdot \\sum_{cyc}\\left(\\frac{y}{x}\\plus{}\\frac{y}{z}\\right)\\equal{}\\sum_{cyc} \\frac{z\\plus{}x}{2y}\r\n\\]\r\n\r\n :)",
  "Solution_5": "How you prove that $ p^2\\leq 4R^2\\plus{}4Rr\\plus{}3r^2$",
  "Solution_6": "[quote=\"Flakky\"]How you prove that $ p^2\\leq 4R^2 \\plus{} 4Rr \\plus{} 3r^2$[/quote]\r\nIf $ I$ is incenter and $ H$ is orthocenter, then $ IH^2\\equal{}4R^2\\plus{}4Rr\\plus{}3r^2\\minus{}p^2\\ge 0\\ .$",
  "Solution_7": "[quote=\"Virgil Nicula\"]The geometrical inequality $ \\sum \\frac {a}{b + c - a}\\ge\\sum \\frac {b + c - a}{a}$ is [u]well-known[/u]. \n\n[b]Prove that[/b] $ \\{\\begin{array}{cc} 1\\blacktriangleright & \\sum\\frac {a}{b + c - a}\\ge\\sum\\frac {b + c - a}{a} + \\frac {R - 2r}{R}\\ . \\\\\n \\\\\n2\\blacktriangleright & \\sum\\frac {a}{b + c - a}\\ge\\sum\\frac {b + c - a}{a} + \\frac {R - 2r}{R}\\cdot\\frac {8R - r}{4(2R - r)}\\ .\\end{array}$[/quote]\r\nWe know the distancies $ \\|\\begin{array}{c} \\mathrm {IH}^2 = 4R^2 + 4Rr + 3r^2 - s^2\\ge 0 \\\\\r\n \\\\\r\n\\mathrm H\\Gamma^2 = 4R^2[1 - \\frac {2s^2(2R - r)}{R(4R + r)^2}]\\ge 0\\end{array}\\|$ and we can show easily \r\n\r\nthat $ \\|\\begin{array}{c} \\sum\\frac {a}{b + c - a} = \\frac {2R - r}{r} \\\\\r\n \\\\\r\n\\sum\\frac {b + c - a}{a} = \\frac {s^2 + r^2 - 8Rr}{2Rr}\\end{array}\\|$ . Thus, $ \\boxed {\\ s^2\\stackrel {(*)}{\\ \\le\\ } \\frac {R(4R + r)^2}{2(2R - r)}\\le 4R^2 + 4Rr + 3r^2\\ }$ \r\n\r\nand $ \\sum\\frac {a}{b + c - a} - \\sum\\frac {b + c - a}{a} =$ $ \\frac {2R - r}{r} - \\frac {s^2 + r^2 - 8Rr}{2Rr} = \\frac {4R^2 + 6Rr - r^2 - s^2}{2Rr}\\stackrel {(*)}{\\ \\ge\\ }$\r\n\r\n$ \\frac {4R^2 + 6Rr - r^2 - \\frac {R(4R + r)^2}{2(2R - r)}}{2Rr} =$ $ \\frac {2(2R - r)(4R^2 + 6Rr - r^2) - R(4R + r)^2}{4Rr(2R - r)} =$ $ \\frac {(R - 2r)(8R - r)}{4R(2R - r)}\\ge 0$\r\n\r\n$ \\implies$ $ \\boxed {\\ \\sum\\frac {a}{b + c - a}\\ge\\sum\\frac {b + c - a}{a} + \\frac {(R - 2r)(8R - r)}{4R(2R - r)}\\ge\\sum\\frac {b + c - a}{a} + \\frac {R - 2r}{R}\\ge\\sum\\frac {b + c - a}{a}\\ }\\ .$"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Here are the correct answers compiled by some kids at my school for the Barbra Nunn that Algebra II and Geometry students took today. If you disagree with an answer, post the problem number and your solution.\r\n\r\n[list]\n1. B\n2. A\n3. D\n4. C\n5. C\n6. C\n7. E\n8. C\n9. A\n10. E\n11. A\n12. D\n13. D\n14. D\n15. B\n16. A\n17. C\n18. B\n19. B\n20. A\n21. C\n22. B\n23. D\n24. B\n25. C\n26. C\n27. C\n28. D\n29. C\n30. D\n[/list]",
  "Solution_1": "Didnt take it.... better place to post this though is on the mualpha message board - just a suggestion.",
  "Solution_2": "Farazy, don't you go to Seminole?",
  "Solution_3": "[quote=\"WarpedKlown1335\"]Farazy, don't you go to Seminole?[/quote]\r\n\r\nWassup Yannik, Nice to talk to you again, this is David."
}
{
  "Tag": [
    "geometry solved",
    "geometry"
  ],
  "Problem": "Show that the sum of the inradii of a triangulated cyclic polygon is independant of the triangulation.",
  "Solution_1": "http://agutie.homestead.com/files/Geoproblem_B.htm#Carnot\r\n\r\nSee 6. Sangaku Problem",
  "Solution_2": "I never solved this prb, but I think using the fact that cosA+cosB+cosC=1+r/R works, combined with the fact that the polygon is cyclic. I'm not at all sure, though."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "rotation"
  ],
  "Problem": "The Riverside Bed and Breakfast Inn serves three married couples breakfast at their round table which seats exactly six people. The hostess wants to seat guests so that no husband and wife sit next to each other and the guests alternate male and female. How many different arrangements are possible? A rotation of an arrangement is not considered a different arrangement.",
  "Solution_1": "[quote=\"ccy\"]The Riverside Bed and Breakfast Inn serves three married couples breakfast at their round table which seats exactly six people. The hostess wants to seat guests so that no husband and wife sit next to each other and the guests alternate male and female. How many different arrangements are possible? A rotation of an arrangement is not considered a different arrangement.[/quote]\r\n\r\n[hide]29? there are 6*6-7the[/hide]",
  "Solution_2": "Remember, it's a round table, so you want circular permutations.",
  "Solution_3": "fine then 4?",
  "Solution_4": "[hide]There are two different combinations for the women to sit at. After they are seated, there are two combinations for the men to sit at. Answer:4[/hide]",
  "Solution_5": "Are you sure about the second part?",
  "Solution_6": "[hide]Suppose a husband and his wife sits down, it doesn't matter where, but this stops the rotations of the table.\n\nNext to the husband, women #2 and #3 can either sit on his left or his right, so there are a total of $\\boxed{2}$ ways.[/hide]",
  "Solution_7": "[hide]Supposing the couples are (#1, #2) (#3, #4) (#5, #6)\n#1 sits down first. Now on either side of him can only be #4 and #6.\nAfter all the possibilitys, there are only two ways.\nPossibilitys (clockwise) : \n[b]#1, #4, #5, #2, #3, #6, \n#1, #6, #3, #2, #5, #4,[/b] :lol: \n  [/hide]",
  "Solution_8": "[hide]\nSince we want changing between male and female and couples are together, we can have each couple be its own individual unit.\nTHen there are 3 units and for circular permutations the formula is $(n-1)!$\nso the answer is $2!=2$[/hide]"
}
{
  "Tag": [],
  "Problem": "http://www.artofproblemsolving.com/Forum/memberlist.php\r\nLook at #1",
  "Solution_1": "heh. nice q.",
  "Solution_2": "He has 100% of the posts.",
  "Solution_3": "[quote=\"moogra\"]He has 100% of the posts.[/quote]\r\n\r\nYeah, it used to be three people. Now it's someone else.",
  "Solution_4": "What the...",
  "Solution_5": "Someone is a bit keen  :P .",
  "Solution_6": "yeah.. i saw three people with that many number of posts... :D",
  "Solution_7": "One was a test by me (Chromotron)  :wink: \r\nThis is just a bug, not harmful, just a bit disturbing.",
  "Solution_8": "Wow congratulations. [/sarcasm]\r\n\r\nThis is the $ n$th time this bug has been mentioned. The admins are fixing it."
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "If $ y \\equal{} \\log_{a}{x}$, $ a > 1$, which of the following statements is incorrect?\n\n$\\textbf{(A)}\\ \\text{If }x=1,y=0 \\qquad\\\\\n\\textbf{(B)}\\ \\text{If }x=a,y=1 \\qquad\\\\\n\\textbf{(C)}\\ \\text{If }x=-1,y\\text{ is imaginary (complex)} \\qquad\\\\\n\\textbf{(D)}\\ \\text{If }0<x<z,y\\text{ is always less than 0 and decreases without limit as }x\\text{ approaches zero} \\qquad\\\\\n\\textbf{(E)}\\  \\text{Only some of the above statements are correct}$",
  "Solution_1": "[hide]A is correct since $ a^0\\equal{}1$. B is correct because $ a^1\\equal{}a$. C is correct since if $ x\\equal{}\\minus{}1$, $ y$ must be imaginary. D is also correct...so E[/hide][/hide]",
  "Solution_2": "First of all, what is z? Assuming z is infinity, then A is right because numbers to the zeroth power are 1, and B is right because numbers to the first power are just themselves, and C is right because positive numbers to complex powers can be negative, but D is wrong because what if let's say, a is 2 and x is 4, then y is 2 and isn't negative, but it does decrease without limit as x approaches zero. Still, in this way, D is wrong, so E is right, so answer = D. ",
  "Solution_3": "That is a typo, the original test has\n$\\textbf{(D)}\\ \\text{If }0<x<1,y\\text{ is always less than 0 and decreases without limit as }x\\text{ approaches zero} \\qquad$\n\nThis is certainly true, if you are not convinced, you can look at a graph.",
  "Solution_4": "[quote=ItsAmeYushi]That is a typo, the original test has\n$\\textbf{(D)}\\ \\text{If }0<x<1,y\\text{ is always less than 0 and decreases without limit as }x\\text{ approaches zero} \\qquad$\n\nThis is certainly true, if you are not convinced, you can look at a graph.[/quote]\n\nOK this makes a lot more sense, if on D, 0<x<1, then D is right and E is wrong. Thanks!"
}
{
  "Tag": [
    "search",
    "number theory solved",
    "number theory"
  ],
  "Problem": "Given any 2n integers, prove that there are n of them whose sum is divisible by n.\r\n\r\nThanks.",
  "Solution_1": "The result still holds for a collection of $2n-1$ integers. This appeared many times here.\r\nSearch for Erd\u00f6s-Gizburg-Ziv theorem.\r\n\r\nPierre.",
  "Solution_2": "[quote=\"pbornsztein\"]The result still holds for a collection of $2n-1$ integers. This appeared many times here.\nSearch for Erd\u00f6s-Gizburg-Ziv theorem.\n\nPierre.[/quote]\r\n\r\nThis is also  Ucraina Olympiad"
}
{
  "Tag": [
    "geometry",
    "rotation",
    "Pythagorean Theorem"
  ],
  "Problem": "The figure below shows a right triangle $ \\triangle ABC$.\r\n[asy]unitsize(15);\n  pair A = (0, 4);\n  pair B = (0, 0);\n  pair C = (4, 0);\n  draw(A -- B -- C -- cycle);\n  pair D = (2, 0);\n  real p = 7 - 3sqrt(3);\n  real q = 4sqrt(3) - 6;\n  pair E = p + (4 - p)*I;\n  pair F = q*I;\n  draw(D -- E -- F -- cycle);\n  label(\"$A$\", A, N);\n  label(\"$B$\", B, S);\n  label(\"$C$\", C, S);\n  label(\"$D$\", D, S);\n  label(\"$E$\", E, NE);\n  label(\"$F$\", F, W);[/asy]\r\nThe legs $ \\overline{AB}$ and $ \\overline{BC}$ each have length $ 4$.  An equilateral triangle $ \\triangle DEF$ is inscribed in $ \\triangle ABC$ as shown.  Point $ D$ is the midpoint of $ \\overline{BC}$.  What is the area of $ \\triangle DEF$?",
  "Solution_1": "[hide=\"Bash\"]Let $ A \\equal{} ( \\minus{} 2,2), B \\equal{} ( \\minus{} 2,0), C \\equal{} (2,0), D \\equal{} (0,0)$, and let $ F \\equal{} ( \\minus{} 2,a)$. We rotate $ F$ about the origin $ D$ -60 degrees counterclockwise to find $ E \\equal{} ( \\minus{} 1 \\plus{} \\frac {\\sqrt {3}}{2}a,\\sqrt {3} \\plus{} \\frac {1}{2}a)$. $ E$ lies on the line $ AC$, which is $ x \\plus{} y \\equal{} 2$. Thus, we have $ \\minus{} 1 \\plus{} \\frac {\\sqrt {3}}{2}{a} \\plus{} \\sqrt {3} \\plus{} \\frac {1}{2}a \\equal{} 2 \\iff a \\equal{} \\frac {2(3 \\minus{} \\sqrt {3})}{\\sqrt {3} \\plus{} 1} \\equal{} 4\\sqrt {3} \\minus{} 6$. Then $ DF^2 \\equal{} ( \\minus{} 2)^2 \\plus{} (4\\sqrt {3} \\minus{} 6)^2 \\equal{} 88 \\minus{} 48\\sqrt {3}$. The area of this triangle is then this times $ \\frac {\\sqrt {3}}{4}$, or $ 22\\sqrt {3} \\minus{} 36$.[/hide]",
  "Solution_2": "Your work was fine until the final answer.",
  "Solution_3": "Ah, found the error and edited it. Thanks.",
  "Solution_4": "[hide=\"Alternate Solution\"]\nLet $BF=f$, $DE=EF=FD=s$, and $EC=x$. Then, we get three equations, one from the Pythagorean Theorem and two from LoC: \\[ \\begin {eqnarray*} f^2 + 4 &=& s^2, \\\\ (4-f)^2 + (4\\sqrt{2}-x)^2 - 2(4-f)(4\\sqrt{2}-x)\\frac{1}{\\sqrt{2}} &=& s^2, \\\\ 4 + x^2 - \\frac {4x}{\\sqrt{2}} &=& s^2. \\]It is a bit of a pain to solve this, but not impossible. After much bashing, we see that $ s = \\pm 2 \\cdot \\sqrt {22 \\pm 12\\sqrt{3}} $. Obviously, the only non-extraneous solution is $ s = 2 \\cdot \\sqrt {22 - 12\\sqrt{3}} $, so we get \\[ [\\triangle DEF] = \\frac {\\sqrt{3}s^2}{4} = \\sqrt {3} \\cdot \\left( 22 - 12\\sqrt{3} \\right) = \\boxed {22\\sqrt{3} - 36}. \\]\n[/hide]",
  "Solution_5": "We could also use polar coordinates centered around D, noting that since $DE = DF$, we have in polar coordinates $E = (r, \\theta)$ and $F = (r,\\theta -60)$. Then use $r\\cos(\\theta)$ and $r\\sin(\\theta)$ along with the fact that for $E$, $y=-x+2$. "
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bd\u03b1 \u03bb\u03ad\u03b5\u03b9 \u03c0\u03c9\u03c2 \u03bb\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1\u03ba\u03ad\u03c2 \u03b5\u03be\u03b9\u03c3\u03ce\u03c3\u03b5\u03b9\u03c2??",
  "Solution_1": "[quote=\"Christiano89\"]\u03a5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03c4\u03bf \u03bf\u03c0\u03bf\u03af\u03bf \u03bd\u03b1 \u03bb\u03ad\u03b5\u03b9 \u03c0\u03c9\u03c2 \u03bb\u03cd\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1\u03ba\u03ad\u03c2 \u03b5\u03be\u03b9\u03c3\u03ce\u03c3\u03b5\u03b9\u03c2??[/quote]\r\n\r\n\u0394\u03b5\u03bd \u03b3\u03bd\u03c9\u03c1\u03b9\u03b6\u03c9 \u03b1\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c4\u03b5\u03c4\u03bf\u03b9\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf, \u03b7 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03c3\u03c4\u03b9\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1\u03ba\u03b5\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03b7, \u03c4\u03bf \u03c0\u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bd\u03b1 \u03c3\u03ba\u03b5\u03c5\u03c4\u03b5\u03c3\u03b5 \u03c4\u03b9\u03c2 \u03c3\u03c9\u03c3\u03c4\u03b5\u03c2 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b1\u03c3\u03b5\u03b9\u03c2... \u03c5\u03c0\u03b1\u03c1\u03c7\u03bf\u03c5\u03bd \u03ba\u03b1\u03c0\u03bf\u03b9\u03b1 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03b1 \u03b8\u03b5\u03c9\u03c1\u03b7\u03bc\u03b1\u03c4\u03b1 \u03bc\u03b5 \u03c4\u03b1 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03bb\u03c5\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1\u03ba\u03b5\u03c2 (\u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1\u03ba\u03b7 Cauchy, Jensen \u03ba\u03bb\u03c0) \u03ba\u03b1\u03b8\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03bc\u03b5\u03b8\u03bf\u03b4\u03bf\u03bb\u03bf\u03b3\u03b9\u03b5\u03c2 (\u03ba\u03c5\u03c1\u03b9\u03c9\u03c2 \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03b5\u03c2) \u03b3\u03b9\u03b1 \u03c4\u03b1 \u03bf\u03c0\u03bf\u03b9\u03b1 \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03b5\u03bd\u03b1 \u03ba\u03b5\u03c6\u03b1\u03bb\u03b1\u03b9\u03bf \u03c3\u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03b3\u03b9\u03b1 \u03bf\u03bb\u03c5\u03bc\u03c0\u03b9\u03b1\u03b4\u03b5\u03c2 \u03c4\u03bf\u03c5 Arthur Engel (Springer \u03b5\u03ba\u03b4\u03bf\u03c3\u03b5\u03b9\u03c2)",
  "Solution_2": "\u03bf\u03ba \u03c4\u03cc\u03c4\u03b5... \u0398\u03b1 \u03b2\u03c1\u03c9 \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03ba\u03b1\u03b9 \u03b8\u03b1 \u03c4\u03bf \u03c8\u03ac\u03be\u03c9. \u0395\u03c5\u03c7\u03b1\u03c1\u03b9\u03c3\u03c4\u03ce",
  "Solution_3": "\u0394\u03ce\u03c3\u03b5 \u03bc\u03bf\u03c5 Gmail \u03bd\u03b1 \u03c3\u03bf\u03c5 \u03c3\u03c4\u03b5\u03af\u03bb\u03c9 \u03b1\u03bd \u03bc\u03c0\u03bf\u03c1\u03ad\u03c3\u03c9 \u03b2\u03b9\u03b2\u03bb\u03af\u03b1 \u03bc\u03b5 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b9\u03b1\u03ba\u03ad\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2 \u03b7\u03bb\u03b5\u03ba\u03c4\u03c1\u03bf\u03bd\u03b9\u03ba\u03ac"
}
{
  "Tag": [
    "search",
    "algebra",
    "polynomial"
  ],
  "Problem": "This contest seemed harder than the last one.\r\n\r\n[hide=\"my results\"]\nbclgold     1  1  0  0  1  0  0  0  0  0\ngourmet     *  *  *  *  *  *  t  *  t  t\nsightsee    *  *  *  *  *  *  *  *  t  t\n[/hide]\n\n[hide=\"sightsee\"]\nBinary search on the answer, use Bellman-Ford to detect negative cycles.\n[/hide]\n\n[hide=\"gourmet\"]\nGreedy, use interval trees to make it NlogN.\n[/hide]\n\n[hide=\"bclgold\"]\nUse Rabin-Karp (polynomial) hashing, binary search on the agreement between prefix and suffix and then greedy.\n[/hide]",
  "Solution_1": "I did:\r\nbclgold     1  0  0  0  1  0  0  0  0  0\r\ngourmet     *  *  *  *  *  x  t  t  t  t\r\n\r\nbclgold I did the failed O(n) algo. Gourmet I did greedy, except I did it inefficiently [sort, then unnecessarily do #ofcows swaps, then start at the most expensive and work downwards, for guaranteed O(n^2) instead of more like O(nlogn).]",
  "Solution_2": "[code]\nbclgold     1  1  1  1  1  1  1  1  4  4\ngourmet     *  *  *  *  *  *  t  t  t  t\nsightsee    *  x  x  x  x  x  x  x  t  t\n[/code]\r\n\r\nMy sightsee isn't working correctly for some reason, and it's a bit slow in any case. 15 or so runs of O(VE) Bellman-Ford seems like it would time out, I think.",
  "Solution_3": "It might be because of double imprecision. Try adding epsilons (like 1e-9 or something) in your comparisons in Bellman-Ford, and see if that fixes it.",
  "Solution_4": "[code]\n           1  2  3  4  5  6  7  8  9 10\nbclgold     1  1  1  1  1  1  1  1  0  0\ngourmet     *  *  *  *  *  x  x  t  t  t\n[/code]\r\n\r\nO(N^2) on bcl. I got 6/8 of the cases in 9 and 10, but it seems I get no credit for that :( Can someone explain the method of suffix arrays used in the analysis?\r\n\r\nI had O(N log N) in gourmet, but I did not use long longs (64 bit) hence the two WA and coding errors (a comparison that is always true), made my set work in O(N) time instead of O(log N), hence my code was O(N^2). As of now, I have gotten my program to work through case 8, but I can't get 9 or 10.",
  "Solution_5": "I'd discuss problems, but I'm only in Bronze Division  :P \r\n\r\nDo you need to solve all the problems in one contest in order to be bumped up to Silver? I couldn't the details on moving between divisions.",
  "Solution_6": "To move up midcontest, you generally need all but one or two testcases at least. If you don't move up midcontest, 800 points generally will move you from bronze to silver and 700 points will generally move you from silver to gold (the cutoff was below 700 a lot last year).",
  "Solution_7": "[quote=\"nebula42\"]It might be because of double imprecision. Try adding epsilons (like 1e-9 or something) in your comparisons in Bellman-Ford, and see if that fixes it.[/quote]\r\n\r\nActually, it turns out the problem was that in my Bellman-Ford, I used index i instead of j when looping through the edges. I still can't get the last test case to work in time though. Did anybody else get it?",
  "Solution_8": "Double imprecision shouldn't be a problem on sightsee because they seem to be accepting answers within $ .01$ of the correct answer...\r\n\r\nI think the test data was a bit big. Richard Peng claims that if you stop updating when none of the values change, then it will run in time. Also, if that's too slow then try only updating edges coming out of vertices that were updated in the previous run. The largest data is random so this should be a significant improvement.\r\n\r\nMy guess is IOI scoring. (Maybe no one got the last couple cases on sightsee and they'll be weighted zero.)",
  "Solution_9": "[quote=\"Fraxia\"]I'd discuss problems, but I'm only in Bronze Division  :P \n\nDo you need to solve all the problems in one contest in order to be bumped up to Silver? I couldn't the details on moving between divisions.[/quote]\nSince you did bronze, did you think my analyses were clear enough?  :)  (If you haven't seen them yet, you can find them [url=http://ace.delos.com/DEC07]here[/url].)\n\n[quote=\"JSteinhardt\"]Double imprecision shouldn't be a problem on sightsee because they seem to be accepting answers within .01 of the correct answer...\n\nI think the test data was a bit big. Richard Peng claims that if you stop updating when none of the values change, then it will run in time. Also, if that's too slow then try only updating edges coming out of vertices that were updated in the previous run. The largest data is random so this should be a significant improvement.\n\nMy guess is IOI scoring. (Maybe no one got the last couple cases on sightsee and they'll be weighted zero.)[/quote]\r\nI meant double imprecision in comparisons, as in doing\r\n[code]if (a + b > c + epsilon)[/code]\ninstead of just\n[code]if (a + b > c)[/code]\r\nAlso, it's definitely possible to get a fast enough solution on 'sightsee', since I currently have a working solution that takes at most 0.15 seconds on the contest data. I do agree that the limits were a bit high though, especially since the problem was basically hard enough already.  :|",
  "Solution_10": "well.. in the gold1 sightseeing problem, how can u make the node weight into edge weight? are u making node an edge? if so, then how can u handle no path goes through a node multiple time? will it be automatically handled?",
  "Solution_11": "You can prove that the optimal cycle will simple, so you don't have to worry about repeated vertices."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "icosahedron"
  ],
  "Problem": "Show that the distance between opposite edges in an icosahedron is phi times the sidelength.",
  "Solution_1": "I thought that proofs weren't allowed in Getting Started?  :?\r\n\r\nEdit: Oh, I see you moved it.  :)"
}
{
  "Tag": [
    "vector",
    "linear algebra"
  ],
  "Problem": "hi...\r\nlet u v subspace in R^n and dimv=dimu=n-1 ,show it u+v=R^n",
  "Solution_1": "This is false if the two subspaces are equal, so you'd better include $ U\\ne V$ as a condition.\r\n\r\nIf they're not equal, then either there's an $ x\\in U\\setminus V$ or there's an $ x\\in V\\setminus U.$ What can you do with that?",
  "Solution_2": "u and v different vector i need complet proof pleas",
  "Solution_3": "[quote=\"npare\"]i need complet proof pleas[/quote]\r\nThis is not the recommended way to make the regular participants here want to answer you. We'd like to see some signs of effort on your part - including sometimes being satisfied with a hint."
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "real analysis",
    "derivative",
    "algebra",
    "functional equation"
  ],
  "Problem": "Find the continuous function $f(x)$ which satisfies for all real numbers $x$ the following equation.\r\n\r\n\\[\\int^x_0 f(t)dt+\\int^x_0 t f(x-t)dt=e^{-x}-1\\]",
  "Solution_1": "This is a good one for the use of Laplace transforms. If we let $F(s)=\\int_0^{\\infty}e^{-sx}f(x)dx$ be the Laplace transform of $f$, it isn't too hard to show that the Laplace transform of the first term (the indefinite integral of $f$) is $\\frac1sF(s).$ The second term is the convolution of $x$ with $f(x).$ The Laplace transform of the convolution is the product of the Laplace transforms, and the Laplace transform of $x$ is $\\frac1{s^2}.$ Thus if we take the Laplace transform of the whole equation, we get \r\n\r\n$\\frac1sF(s)+\\frac1{s^2}F(s)=\\frac1{s+1}-\\frac1s$\r\n\r\n$F(s)=-\\frac{s}{(s+1)^2}=-\\frac1{s+1}+\\frac1{(s+1)^2}$\r\n\r\nFrom this, $f(x)=-e^{-x}+xe^{-x}=e^{-x}(x-1).$",
  "Solution_2": "Hello! Kent Merryfield. \r\n\r\nThat's the very [b]Laplace Transform [/b].You are right.\r\nRegrettably,in Japan which you have been tired of hearing,(laghing!),senior high school student don't study Laplace Transform.I will post my solution later.\r\nPlease wait for a while.\r\n\r\nkunny",
  "Solution_3": "We could use integration by parts.. :)\r\n\r\nIf we take $F(x)=\\int^x_0 f(t)dt$ and $\\mathcal F(x)=\\int^x_0 F(t)dt$ we get, after changing the second intergal to $\\int^x_0 (x-t)f(t)$, $F(x)+\\mathcal F(x)=e^{-x}-1$, and left part is simply $\\frac{[e^x\\mathcal F(x)]'}{e^x}$",
  "Solution_4": "A second solution.\r\n\r\nNote as a preliminary that convolution is commutative, and that with a change of variables we could write that\r\n$\\int_0^xtf(x-t)dt=\\int_0^x(x-t)f(t)dt=x\\int_0^xf(t)dt-\\int_0^xtf(t)dt$\r\nDifferentiate this by the Fundamental Theorem of Calculus, using the product rule where appropriate. Its derivative is\r\n$\\int_0^xf(t)dt+xf(x)-xf(x)=\\int_0^xf(t)dt.$\r\n\r\nNow differentiate the original integral equation:\r\n\r\n$f(x)+\\int_0^xf(t)dt=-e^{-x}$\r\n\r\nPutting $x=0$ in this gives us $f(0)=-1.$\r\n\r\nDifferentiate a second time:\r\n\r\n$f'(x)+f(x)=e^{-x}.$\r\n\r\nThis is a first order linear differential equation; the technique of multiplying factors works easily for it. The solution becomes:\r\n\r\n$e^xf(x)=x+C$. The initial condition $f(0)=-1$ implies $C=-1.$\r\n\r\n$f(x)=e^{-x}(x-1).$\r\n\r\nI posted the Laplace transform solution first just because the problem is so beautifully suited to transform methods.",
  "Solution_5": "[quote=\"kunny\"]Regrettably,in Japan which you have been tired of hearing,(laghing!),senior highschool student don't study Laplace Transform.[/quote]\r\nIn the U.S., this would most likely be in a course intended for 2nd year (or sometimes 3rd year) college students in Differential Equations. It's not part of the AP Calculus syllabus that the top high school students follow; high school students typically would not have seen it. My second solution includes only topics that would be in the syllabus of AP Calculus BC.",
  "Solution_6": "Thank you for your solutions! grobber and Kent Merryfield.\r\n\r\nMy solution is exactly same as [b]Kent Merryfield's [/b]solution,which is the typical solution in Japanese Entrance Exam for university.To tell the truth,since 1997\r\ndifferential equation,functional equation have been eliminated in curicurum of senior high school's mathematics,but next year these are making revival.\r\nSo is linear transformation instead of dissapearing Complex plane (Gaussian plane).\r\n\r\nkunny"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Does anyone know how to solve the following problem, or the origin, known results or anything related?  Thank you!\r\n\r\nLet's call a triple of numbers good if one of them is the sum of the other two.  Call a number n admissible if the numbers 1,2,3,..., 3n can be partitioned into n good triples.  An obvious necessary condition that n be admissible is n to be 0 or 1 modulo 4.  Is the necessary condition also sufficient?",
  "Solution_1": "Some previous threads on this problem: \r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=218089\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?p=1054571#1054571\r\n\r\nI do not know if your conjecture is true. I think $ n\\equal{}24$ is the smallest value I'm not sure about (that is, can you partition $ \\{1,\\ldots,72\\}$ into 24 good triples?)."
}
{
  "Tag": [],
  "Problem": "I am somewhat confused, but was is the difference between a hedge fund and an investment bank?\r\nAlso is there anywhere online that will provide me with a list of hedge funds since I am having no luck finding one.\r\nThanks.",
  "Solution_1": "[url]http://en.wikipedia.org/wiki/Hedge_fund[/url] has a list.\r\n\r\nAlso: please start a new topic for a new question, rather than posting a somewhat off-topic question in a long-dormant thread.  Thanks.",
  "Solution_2": "Thanks for the list, and sorry for the off topic question."
}
{
  "Tag": [
    "inequalities theorems",
    "inequalities"
  ],
  "Problem": "Let $ a,b,c,d\\geq 0$ and $ \\frac{1}{p}\\plus{}\\frac{1}{q}\\equal{}1$, $ (p,q>1)$, and $ \\alpha\\plus{}\\beta \\equal{}1$, $ \\alpha , \\beta \\geq 0$ then\r\n\\[ (\\alpha a\\plus{}\\beta c)^{\\frac{1}{p}}(\\alpha b\\plus{}\\beta d)^{\\frac{1}{q}}\\geq \\alpha a^{\\frac{1}{p}}b^{\\frac{1}{q}}\\plus{}\\beta c^{\\frac{1}{p}}d^{\\frac{1}{q}}\\]",
  "Solution_1": "This is Holder's inequality..."
}
{
  "Tag": [],
  "Problem": "Let us play THE GAME. If you want to know the rules, ask at your own risk. If you know how to play THE GAME, I dare you to post a ddr emoticon telling me you know how the play THE GAME. You are allowed to annoy others. Begin![/i]",
  "Solution_1": "you made me lose, I lose",
  "Solution_2": "Don't lock, or else there will be no place to announce loss...\r\n\r\nI just lost...\r\n\r\n[Moderator: I propose PMing mathlete every loss]",
  "Solution_3": "Mods decide what gets locked, not the posters. Anyway, I won.\r\n[img]http://imgs.xkcd.com/comics/anti_mind_virus.png[/img]\r\nAt least this topic can be locked in peace now.\r\n\r\n[Moderator: Indeed!]\r\n[hide=\";)\"][Moderator: Well you may start one in Fun Factory][/hide]"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "there are 4 points on a plane :A,B,C,D\r\nprove that the length of all of AB,AC,AD,BC,...,CD (6 ones) are not  integer\r\nunfortunately my english is not good but i hope you understand what i mean...",
  "Solution_1": "We can understand your English. :lol:",
  "Solution_2": "I think you have to add no 3 are on one line cause\r\n$ (0|0) (3|0) (3|4) (6|0)$ would be a counterexample if i understood you right",
  "Solution_3": "[quote=\"Rafikafi\"]I think you have to add no 3 are on one line cause\n$ (0|0) (3|0) (3|4) (6|0)$ would be a counterexample if i understood you right[/quote]\r\nyes you are right \r\ni told this to my friend and he edited his problem \r\n[b] the number of the points is infinite and they are not all on a line [/b]"
}
{
  "Tag": [
    "calculus",
    "function"
  ],
  "Problem": "Find the point on hte graph of y=Square root of ( x+1) closet to the point (3,0)",
  "Solution_1": "This is a calculus problem again. You just want to minimize the distance between a point on the function and the given point. I don't think this kind of problem belongs in this forum.",
  "Solution_2": "Not necessarily calculus... but the same idea:\r\n\r\n$d^2 = (x-3)^2+(x+1) = x^2-5x+10 = \\left(x-\\frac{5}{2}\\right)^2+\\frac{15}{4}$ which has a minimum at $(x,y) = \\left(\\frac{5}{2}, \\frac{\\sqrt{14}}{2}\\right)$."
}
{
  "Tag": [],
  "Problem": "here is what i and Vihang(whose initiative this was) decide that why not we share our screen shots...the most beautiful.. i know there are lots of geeks out here....out of the 20 odd selected in INO! 4 visit AOPs/ML ...so why not \r\nlet me start with mine\r\n[url=http://ubuntuforums.org/attachment.php?attachmentid=35672&d=1182110755]this[/url] is how mine looks like :)",
  "Solution_1": "he would have got the idea from long back  :D  And how come it says that i cannot view the page like i have to be registered?  :huh:   :)",
  "Solution_2": "that is why I WROTE that IT WAS HIS INITIATIVE AND HE WANTED ME TO START :mad: \r\ntaht's because  u aren't registered [url=http://ubuntuforums.org/]here[/url]\r\nlet me find another way out",
  "Solution_3": "ok i am attaching a zipped version :D",
  "Solution_4": "[url=http://img120.imageshack.us/img120/4660/screenshotls7.png]Mine[/url]\r\n\r\nYeah that's XP ...\r\nMy FC6 looks like any other standard FC6 ... no point in putting that here.",
  "Solution_5": "so u have a mac leopard tranformation...gr8..\r\ntry compiz fusion if u can with FC6",
  "Solution_6": "nice idea vihang \r\n[url=http://img257.imageshack.us/my.php?image=ubuntufinaleww5.png]here[/url] is mine\r\n[url=http://img412.imageshack.us/my.php?image=screenshotyz8.png]here[/url] is  another"
}
{
  "Tag": [],
  "Problem": "Hello people!\r\n\r\nWelcome to [b]jain_harshi's one day tournament[/b]. This tournament's first round is made up of response games, 4 people each. Each game will have 10 questions. The winners move on to the semifinals, which is made up of countdowns. The final ends in a countdown. Please join!",
  "Solution_1": "uh sure jain",
  "Solution_2": "Ok, I'll participate.\r\n\r\nWhat day?",
  "Solution_3": "is this CTC-certified?",
  "Solution_4": "I don't believe it is; we're already having a CTC Tourney scheduled for mid-August - we wouldn't be having two tourneys - it defies the point of CTC.",
  "Solution_5": "[quote=\"myyellowducky82\"]is this CTC-certified?[/quote]\r\n\r\nNo, this tourney is illegal for all CTC members; it is not CTC-certified.",
  "Solution_6": "That pretty much makes it impossible, seeing as almost everyone is in CTC.",
  "Solution_7": "Does it have to be CTC Certified?",
  "Solution_8": "Sorry, but since this tournament is illegal, it has to canceled. I actually [b]forgot[/b] the CTC-certification rule.",
  "Solution_9": "It's not really illegal, it's just that no one in CTC is supposed to join it, which means few people will join it.",
  "Solution_10": "I am part of the CTC. It kind of like breaking a law.",
  "Solution_11": "It is breaking a law. Shut down this tournament or you are exiled from CTC.",
  "Solution_12": "I did. Can't you see? For sure, I WILL CLOSE IT AGAIN:\r\n\r\n[size=200]\n\n[b][i][u]This tournament is now closed, canceled, destroyed, etc.[/u][/i][/b][/size]",
  "Solution_13": "SURE I'LL JOIN. DON'T REALLY WANT TO BUT WHO CARES :rotfl:",
  "Solution_14": "But the [size=200]Tournament is closed[/size]. No one can now join",
  "Solution_15": "please lock this thread admins.",
  "Solution_16": "Admins rarely come onto this forum."
}
{
  "Tag": [
    "Miscellaneous Problems"
  ],
  "Problem": "Is there a $3 \\times 3$ magic square consisting of distinct Fibonacci numbers (both $f_{1}$ and $f_{2}$ may be used; thus two $1$s are allowed)?",
  "Solution_1": "no, it doesn't exist.\r\n\r\nwe have:\r\n\r\nA  B  C\r\nD  E  F\r\nG  H  I.\r\n\r\n$ f_n$ is the greatest number that we use.\r\n\r\nCASE 1)$ f_n$ is in A (or simmetrically in C,G,I)\r\nwe must have A+B+C=C+F+I so A+B=C+I but C+I $ \\le$ A < A+B, absurd.\r\n\r\nCASE 2)$ f_n$ is in B(or simmetrically in D,F,H)\r\nwe must have B+E+H=D+E+F  so B+H=D+F but D+F $ \\le$ B < B+H, absurd.\r\n\r\nLAST CASE)$ f_n$ is in E.\r\nwe must have G+H+I=A+B+C=D+E+F>E. but $ \\frac{(G\\plus{}H\\plus{}I)\\plus{}(A\\plus{}B\\plus{}C)}{2} \\le \\frac{\\sum_{i\\equal{}n\\minus{}6}^{n\\minus{}1}{f_i}}{2}\\equal{}\\frac{(f_{n\\minus{}1}\\plus{}f_{n\\minus{}2})\\plus{}(\\sum_{n\\minus{}6}^{n\\minus{}3}{f_i})}{2}< f_n$, absurde\r\n\r\nbye\r\npaolo",
  "Solution_2": "an easier approach in which we prove that even the sum of 3 different rows cannot be the same;\n\nLet f(n+4) be the largest occuring Fibonacci number. Then there must be a row in which the largest Fibonacci number is no larger than \nf(n+2). But the sum of that row is no larger than f(n) + f(n+1) + f(n+2) = f(n) + f(n+3) < f(n+2) + f(n+3) = f(n+4), impossible.",
  "Solution_3": "Note that the previous proof alsof easily generalizes to m*m magic squares; If we let f(m+n-1) be the largest occuring Fibonacci number, there must be a row in which the largest Fibonacci number is no larger than f(n). The sum in that row is smaller than or equal to: f(n-m+1) + f(n-m+2) + .. + f(n) < (m-1)f(n) < 1.5^(m-1)f(n) < f(n+m-1), contradiction. \n\nAnother approach; Let X = F(x) = the x-th fibonacci number. Let\n\nA B C\nD E F\nG H I\n\nbe our 3*3 magic square. Then A + I = 2E = C + G. This implies that max(A, C, G, I) is at least f(e+2). But this is already larger than 2E, contradiction."
}
{
  "Tag": [
    "calculus"
  ],
  "Problem": "Of course it is.\r\n\r\nDon't worry, another mod will respond to your problems (most of the time, anyway)  :D \r\n\r\nBTW, here is another, if you want.\r\n\r\nWhat are all possible solutions to $\\prod_{k=0}^{\\infty}x^{k}$?\r\n\r\n[size=75][color=red]b-flat: Post in a new thread  :D [/color][/size]",
  "Solution_1": "None, the exponent grows without bounds and $0^\\infty$ is an indeterminate form and therefore no real solutions.",
  "Solution_2": "Sorry, I meant to say, something more like what are all the things that $x\\times x\\times x\\cdots$ tends to (for non-negative $x$)?",
  "Solution_3": "Infinite product; the question of whether or under what circumstances it converges is an issue. I'd say it probably belongs in the \"calculus computations and tutorials\" forum, along with most other things involving the convergence of infinite sums or infinite products.\r\n\r\nThis infinite product tends to zero if $\\mid x\\mid <1.$ It converges to $1$ if $x=1.$ Otherwise, it diverges.\r\n\r\n(There's a convention with infinite products that would say tht even though the manner in which this tends to zero for $0<\\mid x\\mid <1$ is certainly a convergent sequence, we nonetheless say that the product diverges to zero.)",
  "Solution_4": "[quote=\"Kent Merryfield\"]Infinite product; the question of whether or under what circumstances it converges is an issue. I'd say it probably belongs in the \"calculus computations and tutorials\" forum, along with most other things involving the convergence of infinite sums or infinite products.\n\nThis infinite product tends to zero if $\\mid x\\mid <1.$ It converges to $1$ if $x=1.$ Otherwise, it diverges.\n\n(There's a convention with infinite products that would say tht even though the manner in which this tends to zero for $0<\\mid x\\mid <1$ is certainly a convergent sequence, we nonetheless say that the product diverges to zero.)[/quote]\r\nJust wondering, what happens if $x=-1$? Would you just say that it's undefined? (since it's either $1$ or $-1$, but it's impossible to tell which)",
  "Solution_5": "Sorry I didn't know that it was so complicated  :blush: \r\n\r\nI thought that it was just if $\\mid x\\mid <1$, then it is 0, if $x=1$, then it is 1, and if $x>1$, then it will keep getting bigger and bigger, approaching infinity.\r\n\r\nSorry again.  :oops:",
  "Solution_6": "[quote=\"nebula42\"][quote=\"Kent Merryfield\"]Infinite product; the question of whether or under what circumstances it converges is an issue. I'd say it probably belongs in the \"calculus computations and tutorials\" forum, along with most other things involving the convergence of infinite sums or infinite products.\n\nThis infinite product tends to zero if $\\mid x\\mid <1.$ It converges to $1$ if $x=1.$ Otherwise, it diverges.\n\n(There's a convention with infinite products that would say tht even though the manner in which this tends to zero for $0<\\mid x\\mid <1$ is certainly a convergent sequence, we nonetheless say that the product diverges to zero.)[/quote]\nJust wondering, what happens if $x=-1$? Would you just say that it's undefined? (since it's either $1$ or $-1$, but it's impossible to tell which)[/quote]\r\n\r\nWouldn't this be called indeterminate? Since we can't determine what the product is?"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Show that for any positive integer $n$, we can find a positive integer $f(n)$ such that $19^{f(n)}-97$ is divisible by $2^{n}$.",
  "Solution_1": "Let $k$ be natural number which satisfies $2^{n}\\parallel 19^{k}-1$. Then we can define $f(n+1)=f(n)$ if $2^{n+1}| 19^{f(n)}-97$ or otherwise $f(n+1)=f(n)+k$. Such a $k$ exists because of the lemma if $2^{\\alpha}\\parallel a^{2}-1$ then for any $\\beta \\geq 0$: $2^{\\alpha+\\beta}|a^{n}-1$ iff $2^{\\beta+1}|n$"
}
{
  "Tag": [],
  "Problem": "In 12 years Martha age will be two times of Bill's age now. When Bill's age was 1/2 of Martha's age then, Bill was 16 years younger than Martha is now. How old is Martha?",
  "Solution_1": "[quote=\"randomdragoon\"]In 12 years Martha age will be two times of Bill's age now. When Bill's age was 1/2 of Martha's age then, Bill was 16 years younger than Martha is now. How old is Martha?[/quote]\r\n[hide]\n$m+12=2b$\n$\\frac{m+12}{2}=m-16$\n$m+12=2m-32$\n$m=44$\nmartha was 44.\nWe don't need to find billy's age, but it is 28.[/hide]",
  "Solution_2": "[quote=\"bpms\"][quote=\"randomdragoon\"]In 12 years Martha age will be two times of Bill's age now. When Bill's age was 1/2 of Martha's age then, Bill was 16 years younger than Martha is now. How old is Martha?[/quote]\n[hide]\n$m+12=2b$\n$\\frac{m+12}{2}=m-16$\n$m+12=2m-32$\n$m=44$\nmartha was 44.\nWe don't need to find billy's age, but it is 28.[/hide][/quote]\r\nSorry, your solution isn't consistent with the last condition \"When Bill's age was 1/2 of Martha's age then, Bill was 16 years younger than Martha is now.\" Please note the convoluted use of present and past tense in the same sentence.\r\n\r\n(I'm not sure if I worded it properly -- what I mean is if Bill were 6 and Martha were 9, Bill was 1/2 of Martha's age then when Bill was 3 and Martha was 6 -- 3 years ago)."
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO",
    "summer program",
    "MathPath",
    "Mathcamp",
    "calculus"
  ],
  "Problem": "Hi, everyone, \r\n\r\nEver since Mathfanatic was thinking out loud about whether to go to Exeter (by the way, how are you, Mathfanatic?), I've been wondering about the prep school path to learning more mathematics. I recall that there are some other AoPSers in various prep schools. What do you think of the prep school experience? Would you recommend it to your best friend or your younger sibling? What are the advantages of prep school? Are there any disadvantages? How was the application process? What else should people who have never been to prep school know about prep school? \r\n\r\nCurious minds would like to know more. Please let us know.",
  "Solution_1": "Could you please define \"prep school?\" I am still quite unsure about the differences in schools. For instance, there are magnet schools, but there are also publicly funded arts schools... are these magnet schools as well? Are there private, non-parochial, non-prep schools, or are all private non-parochials considered prep schools? Or is there a standard for prep schools, such as offering a certain number of AP or IB courses? Does one, perhaps, get college credit for taking certain classes at a prep school (that are not part of the AP/IB systems)? And what about homeschooling? It seems like there might be some blurriness between a large homeschool system and a small private school... is there some sort of home-school registery? Ok, I realize I am now going overboard with questions about schooling, but if you could, perhaps, answer my first question (\"What is a prep school?\") I would be most grateful.",
  "Solution_2": "How about this? If you think your school is a prep school, please tell us about it. \r\n\r\nFor what it's worth, I don't think the homeschooling I'm doing now is the same thing as what I am asking about.",
  "Solution_3": "Yeah, I know it's kind of cheap to double-post to bump up a thread, but I'm really curious about what the math-liking people here who go to prep schools think of them. One kind of school I have in mind is [url=http://www.exeter.edu]Phillips Exeter Academy[/url], which is conspicuous for the AMC scores of some of its students. I'm sure that there are AoPSers who go to other schools, who can perhaps relate why they didn't stay in the local public school, and perhaps why they didn't go to some other prep school they could have gone to. \r\n\r\nThanks to anyone who comes out to represent alma mater and tell us like it is.",
  "Solution_4": "My kids (a high school freshman in IB program and one in 3rd grade) are not in prep schools. We visited one of our local prep schools and also looked into Exeter a year ago when my older one was applying for high school. The local prep school is a very good one with mostly rich kids there (very few on need based scholarship). The enrollment is on the competitive base. Good prep schools do ask for PSAT or SSAT in addition to GPA, recommendation, writing assays, and tuition, etc. In our region, we have private schools, two prep schools, two IBs (students in the same enrollment process divided into two IB schools by east or west of a specific road for transportation purpose) and other subject-specific magnet schools, such as music and art. So the choices are plenty. My older one chose IB over the prep school since she did not felt that the prep school would be able to provide better education for her. The IB class size per year (around 110/yr) is about the same as the prep school. Her IB graduating class has12 national merit scholar finalists, which at least double the number of the prep school. Our school district, including private and prep schools (a very large school district, about the 10th largest one in the country) did not have any USAMO qualifier for 2003 and 2004. She is the only qualifier from here for 2005. Math wise, she is probably better than anyone from that prep school. However, that prep school had one finalist in the US bio Olympiad last year (top 20 students in the country are finalists), and one this year too. That is easy to imagine though, considering large number of students in that prep school are children of lawyers and doctors. Teachers in the prep school drive cars in the level of Toyota Corolla. But many students in the upper class drive BMW and Mercedes. \r\n\r\nWe looked into Exeter since Exeter made half of the US IMO team last year (Tony Zhang, Oleg Golberg, TianKai Liu, all earned gold medals). Those three were not initially from Exeter but were attracted there probably because of Zuming Feng. Exeter is a prep but a boarding school. Going there is more like to start your kids with college life at an age of 14. Maybe it is truly worth of going Exeter if he or she makes the IMO team. Even making the team, in the long run, I don\u2019t know whether your children get more benefits or harm by starting college-like life that early. But we are still thinking about Exeter, probably for junior and senior years.",
  "Solution_5": "[quote=\"qvc122\"]In our region, we have private schools, two prep schools, two IBs (students in the same enrollment process divided into two IB schools by east or west of a specific road for transportation purpose) and other subject-specific magnet schools, such as music and art. So the choices are plenty.[/quote] \n\nThanks for your reply. It sounds like you have a good experience base for making comparisons among schools. Here in Minnesota we have public school open enrollment, which means in principle that students could attend any public school that they can reach by their own transportation. (There are a few exceptions, but that is the general rule.) Alas, none of the Minnesota public schools have consistently strong math teams, although there are always multiple USAMO qualifiers here. Nor do any of privately operated schools here place a high emphasis on math preparation--no matter what they say to the contrary. I have been trying to develop more mathematical culture in our local homeschooling network, but my own limitations as a math coach mean that I am looking EVERYWHERE for other possibilities for my son's education. It seems that most of Minnesota's math stars have parental influence on their education from math-educated parents, and I am not that kind of parent. \n\n[quote=\"qvc122\"]We looked into Exeter since Exeter made half of the US IMO team last year (Tony Zhang, Oleg Golberg, TianKai Liu, all earned gold medals). Those three were not initially from Exeter but were attracted there probably because of Zuming Feng.[/quote]\r\n\r\nYou aptly described Exeter is being much like going to college early in the part of your post that I didn't quote. Exeter is intriguing because of the strong math team there. My son could not find classmates like that anywhere in our state, or at least not that many. This is a lot to think about during the next year.",
  "Solution_6": "Well I'd actually like to comment on the other path, of going to a public high school, the path i did, and why i wouldnt recommend it and wish i had been less lazy in gr8 (wow, i must really have been lazy cuz im still really lazy and can still call the lazy). Anyway, I never thought of private or prep schools, thinking they were just where the rich kids went to stay away from the rest of us. Big Mistake! Here I am, at the end of my junior year and guess what, I've learned absolutely nothing. I've spent the past 3 years of my sleeping through classes and wondering why is wasnt on drugs. I never really get along with the people here becasue we just don't think the same. Sure, I have friends and have had quite a bit of fun, but I still feel that I can't tell people about my interest in advanced math, except for my physics teacher, but he's special. There are several good teachers, but also many who hate intelligent people and are intimidated by us so the best you can do with them is mid 80s even when you deserve well into the 90s (e.g. in my freshman year I got an 87 in science, now I have a 100 in chem, 96 in bio and 97 in physics). The rare intelligent students are not encouraged to pursue advanced learning and all of our schools funds go to the football team.  I was the first student in my school's history to write the COMC and will the first to try for any Olympiad. I have no competition in the classes i bother to wake up for and even in the ones i never wake up in i beat 95% of the people in it. I don't think I would have enjoyed the homeschooling experience because I would miss socializing and gambling (it really is an addiction:)). I wish i had been exposed to this level of knowledge before i was a sophmore and even then i was not prepared to realize or accept the quality of the individuals that were out there, including the vast number of elementary school kids who could teach me a lot about math. If you have the choice, i'd recommend going to a private/prep/gifted school where you will be challenged and exposed to the competition and learning that a young mind needs, don't kill your or your child's love of learning.",
  "Solution_7": "Yes, I think it is very important, at least, to do no harm. I had some school experiences that almost turned me off to learning, and I wouldn't want anyone in the current generation to have that kind of school experience.",
  "Solution_8": "Some of the public school programs, such as IBs are very similar to local prep school if not stronger, but most of IBs are not as strong as Exeter. In fact, IBs are public fund college prep programs, which have quite vigorous selection standard. Exeter is almost like a private national prep school. \r\n\r\nOur state has 13 USAMO qualifiers this year (and 3 MOPpers from our state this year), but our city only has one USAMO qualifier. \r\n\r\nIf A+ math is determined to prevail in math, maybe he should go to Exeter. In our situation, even my husband can coach my daughter; I still think she should go to Exeter if she is truly serious about any chance of making IMO team in the future, since Zuming Feng is so experienced in making national winners. \r\n\r\nTokenadult, I wish you and A+ math have a great time with Mathpath in Colorado.",
  "Solution_9": "i feel like i should post since i have gone to both public school and a prep school (exeter).  for 9th and 10th grade i went to a decent public school in colorado, but i didn't feel challenged academically or even encouraged.  my parents had to pay for me to take the amc, because my school wouldn't.  i felt like my interests and goals were fundamentally different than almost everyone in my school.\r\n\r\ni went to canada/usa mathcamp after 9th grade (and after 10th), and that's when i started thinking about changing schools.  several students there were going to boarding schools and seemed to like them, and i saw a prep school experience as being something like a year-round mathcamp.  \r\n\r\nlater that summer i received a letter from exeter that outlined some of the math opportunities there, and i immediately wanted\r\nto apply.  and after one year there, i know it was the right decision.  i have learned so much, about math and otherwise, from going to exeter.  fall term i had mr. feng for calculus, and i have been in math club the whole year.  i've had the opportunity to go to HMMT and ARML (we were third!!), which i never would have been able to do in colorado.  \r\n\r\nbut the prep school experience isn't for everyone, and i don't think i would have been ready to leave home as a freshman.  but i would definitely recommend applying to exeter to anyone who has a serious interest in academics (not just math) and feels capable of living on their own.  if IMO is the goal, exeter's definitely a great place to be.  i know i have improved a great deal.  but exeter academics are very demanding and that can make it difficult to dedicate as much time as one might want to math.  exeter wants its students to be somewhat well-rounded, which is important to remember.  but the opportunities for advancement in math are completely worth it, and there is definitely something to be said for being well-rounded. \r\n\r\ni'd be happy to try to answer other questions about the pros/cons of public/prep schools, admissions process, etc.  it was a big decision in my life and there's quite a bit more i could say about it.",
  "Solution_10": "[quote=\"dancer87\"]i feel like i should post since i have gone to both public school and a prep school (exeter).  for 9th and 10th grade i went to a decent public school in colorado, but i didn't feel challenged academically or even encouraged.[/quote]\r\n\r\nThanks for replying. I'm especially interested in your experiences because I know Colorado has an unusually large number of schools that purport to be academically advanced for highly gifted students (most private, some charter schools). And I liked hearing about how you learned about Exeter--by meeting classmates at summer programs. After all, the main issue is the overall experience at the school, and most of what I hear about Exeter so far sounds quite positive.",
  "Solution_11": "Thanks for the previous responses. Does anyone else have something to say about this issue? Current students?",
  "Solution_12": "..... what's a prep school?  :?  :D Is it like a private school?",
  "Solution_13": "[quote=\"chess64\"]..... what's a prep school?  :?  :D Is it like a private school?[/quote]\r\n\r\nI'd be happy to hear definitions from other people. The word \"prep\" comes from \"preparatory,\" and comes from the idea that some schools intentionally prepared students for college, while other schools were designed to be last stops before getting a job with only a high school diploma. I don't know if all prep schools are privately operated, or if some are government-operated (\"public\").",
  "Solution_14": "Like Chigr, I've never gone to prep/private school, but I can give a rough description of my experience (which was much better than Chigr's) in public school.  My school has a math culture, but it was definitely not as strong a culture as you would find even at other public schools.  The last year or two, it was a bit frustrating that the math culture here didn't match my desire to improve, but I was nowhere near the situation Chigr mentioned in which he was unable to share his mathematical interest.  People here definitely have an appreciation for math and mathematical talent, it's just that there are not as many people as I would like who really want to develop that talent to its full potential.  My biggest regret is that I didn't step up and take a leadership role early on and push to develop that mathematical culture until the last year or so.  As a result, when I was ready to become more of a leader, there wasn't time to do nearly as much as I would have liked.  Fortunately, it doesn't seem like you'll have any problem with lack of ability/willingness to try to start a math culture.\r\n\r\nIf you want to take advantage of open enrollment here, some schools you might want to start with are Eden Prairie, St. Paul Academy, Wayzata, and maybe Mounds View.  Those four have had the most consistently strong math teams in the state.  I would also recommend looking into UMTYMP.  I've never actually been in the program (I took the qualifying test once and managed to fail it), but from what I've heard, it seems to be a very good program (if A+MATH doesn't mind a lot of work).  I think there tends to be a good chunk of the ARML team each year that has taken classes there.",
  "Solution_15": "about schools in colorado....yes, there are a lot charter schools and other schools that are not quite open enrollment.  however, perhaps because of colorado's location, there doesn't seem to be a huge interest in extracurricular math.  in general though, i think colorado has a very good school system, but my school was not one of the better ones in the state.  going to exeter has definitely opened up so many more opportunities than i would have had at my old school, and i feel like my college options are also more open.",
  "Solution_16": "Um... call me stupid, but where [i]is[/i] Exeter? It's a prep school, right? Is it like a math/science oriented school?",
  "Solution_17": "Exeter is a private school in New Hampshire. It's one of the most popular for students strong in math and sciences. Its rival is Andover. Exeter (I think) has the largest library of all high schools around the country, but Andover has one of the biggest stuffed animal collections. You pick which is better.\r\n\r\nAlso, Exeter gets a huge endowment each year, one substantially bigger than almost all its sister schools.",
  "Solution_18": "[quote=\"yif man12\"]Exeter is a private school in New Hampshire. It's one of the most popular for students strong in math and sciences. [/quote]\r\n\r\nI'll note that this popularity among the math set is a relatively new phenomenon, and I think it almost entirely due to Zuming Feng's fantastic efforts there in attracting and educating top students.  (For those of you who don't know who Zuming is, he's the leader of the US IMO team.  He's also doing much of the work to build the tests for [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url].)",
  "Solution_19": "As a rising junior at [url=http://www.choate.edu]Choate Rosemary Hall[/url] in Connecticut, I'll add my two cents.\r\n\r\n\r\n[quote=\"tokenadult\"]What do you think of the prep school experience?[/quote]\nIt's changed my life. I started Choate interested in math and not much else--you could have fairly characterized me as one-dimensional. But I started taking advantage of opportunities around campus and found that I could maintain my interest in math in addition to developing new interests.\n\nFreshman year I was a day student, which means that I commuted to campus each day. I live about 45 minutes away from Choate, and I was lucky enough to be in a carpool which made life easier for my parents. I joined the math team early on and enjoyed it a lot, but competition math was still my sole interest.\n\nSophomore year I became a boarder and my life began to change. I joined several new clubs, played interscholastic squash, and was elected to the Student Council. The most valuable change that happened to me, however, was that I became a more relaxed and outgoing person. I'd been quiet and shy for much of my life, but living in a dorm helped me learn how important being social is to being happy. Prep school students who spend their time solely pursuing academics are missing the greatest opportunity that their schools offer--the opportunity to grow as a person, not only as an academic machine.\n\n\n[quote=\"tokenadult\"]Would you recommend it to your best friend or your younger sibling?[/quote]\nI would recommend Choate to almost anyone. The only exception I could think of is if a prospective student has only one interest and wants to spend all of his or her time on it and pursue nothing else. This is simply impossible at Choate (or any prep school) because the workload for all classes is too much to entirely focus on only one thing.\n\n\n[quote=\"tokenadult\"]What are the advantages of prep school?[/quote]\n\nOpportunities! Students at Choate can sign up for over thirty different non-academic clubs, from a cappella singing to the Spanish club to the debate team. In one day, a student can attend five or six interesting classes, practice with their sports team, have dinner, and go to a club meeting--all before 7:00! This is something that's hard to find anywhere else.\n\n\n[quote=\"tokenadult\"]Are there any disadvantages?[/quote]\nAt Choate, the workload is so intense that I haven't been able to spend as much time preparing for math competitions as I would have liked to. However, some of this was self-imposed--I chose to take six courses this year instead of the standard five, so I had more work than was necessary--and there have also been opportunities for me to weave my interest in competiton math into my life at Choate. I'm an active member of the math team (which recently took 1st place in the Connecticut championships for the [i]n[/i]th year in a row), and for my final project in BC Calculus I studied Jensen's Inequality. I'm also trying to set up a Directed Study for next year in which I would take the WOOT class on Art of Problem Solving and, under the guidance of the coach of our math team, solve as many problems as possible and develop my proof-writing skills. Overall, it's not easy to continue a thorough study of competition math, but with dedication it is possible.\n\n\n[quote=\"tokenadult\"]How was the application process?[/quote]\r\nIt wasn't particularly grueling. Students who want to apply to Choate take a student-guided tour of campus and have an interview with a member of the Admissions department. The application was pretty standard...if I remember correctly, it asked for teacher recommendations, a grade transcript, awards and extracurricular activities, and a few essays.\r\n\r\n\r\nI encourage you to visit [url=http://www.choate.edu]Choate's newly-designed website[/url] for more detailed information, and I'd also be happy to answer any other questions you may have. I wish you and your son the best of luck in whatever path you decide to take.",
  "Solution_20": "Thank you to JSRosen3 for the detailed description of Choate.",
  "Solution_21": "Could you attend Exeter if you lived in, say, Delaware? :? I don't think I want to do prep school, though... it's like boarding school...",
  "Solution_22": "If you lived in Deleware, you'd definitely have to board there. Most of the prep schools around here (New England) have programs for day students, who commute each day to and from school. In most of them, boarders make up about 60 to 80 percent of the students.\r\n\r\nI don't think admissions are very difficult at all. It's just standard application materials. Teacher reccomendations, grade transcripts, some information about yourself, what you like to do and what your hobbies are, what extracurriculars you do, and one or two essays. I believe that the belief that it's extraordinarily difficult to get in is false. True, you have to have pretty good grades and have a few distinctions, but you don't have to have been getting straight As since 4th grade. I got accepted to the schools I applied to despite being an A-/B+ student for pretty much all of 7th grade and even after getting a B- in English in my first term of 8th grade. My advice would be to apply to a couple schools instead of just one. Get to know the schools. You never know how much you might like one school. I first wanted to go to Exeter like a lot of other people but after visiting, Choate, I found that I liked it a lot. So when I got accepted to both, I ended up choosing Choate.",
  "Solution_23": "For Exeter upper class (11th 12th grades), you need SAT or PSAT as part of the application requirements. For lower class (9th and 10th), you need SSAT. Your parents also need to pay tuition and fees which is somewhere around $35,000 per year. Exeter does have need based scholarship, then you have to be outstanding for them to accept you and fund you.",
  "Solution_24": "[quote=\"qvc122\"]Exeter does have need based scholarship, then you have to be outstanding for them to accept you and fund you.[/quote]\r\n\r\nThe way Exeter puts it is that admissions is more competitive for applicants who request financial aid. That is because Exeter has not yet fully endowed its financial aid fund, so it thinks it can't afford to make financial aid offers to all the needy students who apply. On the other hand, there are definitely quite poor students who attend Exeter on full financial aid. \r\n\r\nOff this subject: have any of you ever seen the reference book about independent schools called the \"Blue Book\"? I saw it today in a public library, and I was astonished to notice that it made no mention of Exeter whatever.",
  "Solution_25": "[quote=\"Chigr\"]The rare intelligent students are not encouraged to pursue advanced learning and all of our schools funds go to the football team.[/quote]\r\n\r\n:/ that must suck especially if you enjoy math. not all high schools are like that though... tons of schools in California especially the Bay Area are really great about academic activities.",
  "Solution_26": "[quote=\"rrusczyk\"][Exeter's] popularity among the math set is a relatively new phenomenon, and I think it almost entirely due to Zuming Feng's fantastic efforts there in attracting and educating top students.[/quote]\r\n\r\nIt is new that people go there specifically for IMO-level preparation.  But before Z.F., there was certainly no shortage of elite (both in ability level and financial resources) math/science types who went there for at least the last 1-3 years of school as phase 1 of a sequence leading through high school competitions, Harvard/Princeton/MIT/etc, the Putnam, and a PhD at similar university.  There was another teacher there at the time who was an active organizer of math competitions, co-authored some books.  I don't remember the name.\r\n\r\nIt could also be that Exeter hiring Feng was an effect as well as a cause of this concentration of talent --- the school realizing it had high-level young mathematicians, and keeping an eye out for somebody who could meet their needs.",
  "Solution_27": "[quote=\"qvc122\"]For Exeter upper class (11th 12th grades), you need SAT or PSAT as part of the application requirements. For lower class (9th and 10th), you need SSAT. Your parents also need to pay tuition and fees which is somewhere around $35,000 per year. Exeter does have need based scholarship, then you have to be outstanding for them to accept you and fund you.[/quote]\r\n\r\nUnless a student is specifically on the IMO track, getting the full value of the 35K(arat) would require [i]not[/i] excelling academically, as it takes too much time from networking.  The future Bush II and Kerry types (scholastically mediocre, socially enthusiastic, ambitious, alcohol-tolerant) are the ones who get their money's worth from the fancy private schools.",
  "Solution_28": "[quote=\"tokenadult\"]Off this subject: have any of you ever seen the reference book about independent schools called the \"Blue Book\"? I saw it today in a public library, and I was astonished to notice that it made no mention of Exeter whatever.[/quote]\r\nWe have a book sent by TIP program (I could not find that book at this moment), which listed most of the private schools and boarding schools. Exeter is among them. Exeter was also mentioned in one of the bestseller books (Brining Down the House), written by Ben Mezrich. Exeter caught my attention from last year when I visited AMC site and found many USAMO qualifiers from there. \r\nMy understanding is some of the IMO members from Exeter paid full tuition and fees to Exeter (if their parents were capable of affording the cost). However, if a student is at that caliber and wants to attend Exeter and has a financial need, Exeter will support that student.\r\n\r\nRegarding Bush and Kerry, I always thought that Kerry was a scholar. However, not long time ago, NPR news said that his first year transcript at Yale showed 4 Ds, worse than Bush's. I was shocked.",
  "Solution_29": "[quote=\"qvc122\"]\nUnless a student is specifically on the IMO track, getting the full value of the 35K(arat) would require [i]not[/i] excelling academically, as it takes too much time from networking.  The future Bush II and Kerry types (scholastically mediocre, socially enthusiastic, ambitious, alcohol-tolerant) are the ones who get their money's worth from the fancy private schools.[/quote]\r\n\r\ni'll be a senior at exeter this coming year.  i've have not met anyone who went for the sake of networking.  (read dancer87's post.)\r\n\r\nmost people go there for a challenging academic environment.  we learn a lot from each other, especially since we use the harkness method.\r\n\r\nfrom the exeter website:\r\n[i]\"Since the arrival of 'Harkness tables' on campus, the principal mode of instruction at Exeter has been discussion around an oval table. The Harkness table is central to both the Exeter classroom and the Exeter curriculum. Though teaching and learning look different in different disciplines and at different levels of study, all Exeter teachers and students are committed to an ideal of active, participatory, student-centered learning which values teaching students not just a given course's content but the skills required to become their own and each others' teachers. As the physical table itself implies, learning at Exeter is a cooperative enterprise in which the students and teacher work together as partners.\n\nThe academic life of the school depends upon the quality of each student's preparation for active participation in learning. As one of the schools legendary teachers said, \"What happens in class depends upon what the students have done before the class begins.\" A place at the Harkness table requires students to exercise a high degree of self-discipline, and to engage eagerly and energetically with both peers and instructors. Exeter is a school for students who take pleasure in this distinctive mode of teaching and learning in which each member of each class is in some measure a teacher of all the others.\"[/i]\r\n\r\ni think that going to exeter for math is great, even if you're not at the olympiad level.  if i went to public school, i would have taken AB calc senior year.  but there are accelerated courses and more options.  and the department is pretty nice about letting you skip (not much public school bureaucratic nonsense).  so i got to take BC calc junior year.  we also have multivariable calculus and other courses (analysis, linear algebra, non-euclidean geometry this year... in the past they have offered combinatorics, topology, etc).\r\nfor the most part, my teachers have been outstanding.  maybe mr. parris was the teacher fleeing_guest was wondering about?  he is incredibly smart, as well as a very nice guy.  and in my opinion, mr feng cares about more than the success of future IMO teams.  he makes math club great with his elegant solutions and ridiculous jokes.  i hadn't even heard about the AMC competitions or anything before exeter, so mr feng was first introduced to me as \"the guy who does math club.\"  last year, there was a group of people who met every day to do olympiad problems.  any math clubber who was willing to put in the time and effort was welcome... even people who didn't make usamo went.\r\n\r\nand exeter also has many other fantastic departments!  we have so many talented musicians, and i have had a great time in the sciences.  i used to hate english; now i love it!\r\n\r\nthere are downsides, too.  6 ish hours of homework a night (?).  AP season was wiiiiild trying to prep on top of that.  it can be hard to keep up with extracurriculars.  but as long as you stay committed to everything you do, i think it's worth it.  you learn how to think, how to study... you can feel yourself improving.  except for selected high-stress times during junior year when everyone is a sleep-deprived zombie, it is a great environment.\r\n\r\ni'd be happy to answer any questions!\r\nemail: slcampbell@exeter.edu\r\nIM: saracampbellsoup",
  "Solution_30": "Sara, thanks for your comments. \r\n\r\nWhat do you think of Exeter's library? When I land on its online catalog, it doesn't seem to have many books on the subjects I search for. It is said to be, credibly, the world's largest high school library, but we have access to a very large research university library here and I'm still trying to figure out if going to Exeter means the improved classroom environment (e.g., smart classmates) and flexible, rigorous curriculum will make up for much more limited library resources. \r\n\r\nBTW, is mathfanatic just too busy these days to post much? I'm especially interested in his reaction to Exeter after his first year there, since he was thinking out loud about it on AoPS a year ago. I also appreciate hearing from AoPSers who toured Exeter but enrolled elsewhere--what was the difference between your school and Big Red that prompted you to enroll where you enrolled?",
  "Solution_31": "people don't seem to have a problem with the library... i don't feel limited.  what subjects were you searching for?\r\nmathfanatic had a very impressive first year; perhaps he is still catching up on sleep.\r\nmy parents wouldn't let me be a boarding student, so i only applied to exeter.  i was kind of torn between staying with my friends at public school and going to exeter, but only until i went to the preview day and saw everything in action.  i didn't know that there was a place where so many people could genuinely care about so many things.",
  "Solution_32": "Just to keep this a little less prep-school biased, there ARE some very, [i]very[/i] good public schools that provide many, if not all, of the resources the aforementioned prep schools provide, for nothing more than the matriculation fees. for those of you out there whose families can't afford you 12 years of college education (4 years of prep, 4 years of undergrad, and 4 years of grad school), you should REALLY look into public magnet schools in your state. some states, like my own, are blessed with superb public magnets (although im not really sure one would call our school a \"magnet\"), some of which even offer boarding (that, of course, costs extra money, but still nowhere near 35k). if i'm not mistaken, two of our favorite admins are graduates of public schooling, and look what became of them. you make your own opportunities.",
  "Solution_33": "[quote=\"saracampbellsoup\"][quote=\"fleeting_guest\"]Unless a student is specifically on the IMO track, getting the full value of the 35K(arat) would require [i]not[/i] excelling academically, as it takes too much time from networking.[/quote]\n\ni've have not met anyone who went for the sake of networking.  (read dancer87's post.)\nmost people go there for a challenging academic environment.[/quote]\r\n\r\nNetworking does not imply carrying around business cards, or a specific intention to form potentially exploitable relationships.   Most people don't have the purpose or even the concept of \"networking\" during their schooling or participation in activities such as summer camps and math competitions.  Nevertheless, years later that is one of the principal outcomes, and in many (economic and non-economic) ways the most valuable one. \r\n\r\nFor any given academic or intellectual goal of a prep school (including IMO preparation, though not including access to specific IMO coaches such as Mr. Feng), there is usually a surrogate or even a superior means of attaining it for far less than the cost of attendance.  What is actually purchased is rarefaction of one's peer group (which to some extent, and certainly in the hopes of some parents, will set the peer group for the student later in life).  To the extent that one concentrates on the books and doesn't make peer connections, one doesn't reap anywhere near the same advantages.  The intellectual benefits are easier to substitute than the social ones. \r\n\r\n Please note that there is no implied opinion of whether the attendance is \"worth it\" at 35K or any other price, and no evaluation for or against networking.  I am pointing out only that, whatever the value may be, the bulk of it ultimately lies in the peer relationships, and that people who operate in ways that develop the relationships are the ones getting their money's worth.   It may be counterintuitive, but studying to get top grades may actually be against your interests at (for example) Exeter.",
  "Solution_34": "[quote=\"fleeting_guest\"]  It may be counterintuitive, but studying to get top grades may actually be against your interests at (for example) Exeter.[/quote]\r\n\r\nDepends on who you are studying with ;)\r\n\r\nAs for Zuming as a cause or effect, check out the record of Exeter before and after Zuming (not to mention the degree to which they recruit top math students before and after).  Indeed, they had strong students before Zuming, but their track record post-Zuming is much more impressive, and not only at the very top - their depth is leading them to be competitive at Harvard-MIT and ARML, as well.",
  "Solution_35": "[quote=\"rrusczyk\"] As for Zuming as a cause or effect, check out the record of Exeter before and after Zuming (not to mention the degree to which they recruit top math students before and after).  Indeed, they had strong students before Zuming, but their track record post-Zuming is much more impressive, and not only at the very top - their depth is leading them to be competitive at Harvard-MIT and ARML, as well.[/quote]\r\n\r\nYes, apparently he had a big effect.  The speculation about the cause (of his hiring) was that Exeter may have found itself in a position similar to that of, say, Harvard (at undergraduate level) some years ago, with a large pool of talented students who were underserved by the existing options, which at that time consisted of advanced courses and participation in the standard competitions.  Other than Stuyvesant, I can't think of any other US schools that would be such a good match for an IMO expert.\r\n\r\nAs you mention, the talent pool grows through recruitment when people notice the AMC results.  This was also what happened for Harvard with the Putnam competitions, but Exeter is in an even better position in this respect as it has (as far as I know) no comparable national-level rivals analogous to MIT, Princeton, etc who might balance the scales a bit over time.",
  "Solution_36": "There are several other schools which get a similar level of talent as Exeter.  To name a few: Thomas Jefferson in VA (2 IMO team members this year, more USAMO participants than any other school for years), AAST in NJ, IMSA in IL, Montgomery-Blair in MD.  Even Saratoga HS, which seems to do very little with the talent it brings in, arguably has the same caliber of students coming into the school (3 students there were top 5 in National MATHCOUNTS - not even Exeter can say that right now).  Several of the other schools in the Bay Area also have outstanding talent, such as Gunn and Lynbrook (the latter benefits from receiving students from one of the top middle school programs in the country).",
  "Solution_37": "Those schools draw on a county or state population.  Exeter is, as far as I know, the only with both a national reach and the concentration of mathematical talent.  They can also attract some of the talent once it appears at other schools, either by reputation or by recruitment (offering scholarships).  If you live in Colorado, you can't go to TJHS, but you can attend Exeter; and if you happen to also have an IMO gold medal, they may hook you up with a scholarship. \r\n\r\nWith that said, I think it's easier to make the IMO from a regular public high school, as one has more free time.",
  "Solution_38": "I'm just noting that Stuyvesant (which also does not have a national reach) is not the only other great school for olympiad math students.  Indeed, Exeter has a big advantage in being able to recruit nationally, but these other schools are free, and many parents are not ready to send their kids off to a boarding school.  It's not so clear that Exeter has a natural advantage over TJ.\r\n\r\nAs for regular public school being your best bet, I would have agreed, but the past few years' teams suggest otherwise.  This year: 4 from magnet schools, 1 from Exeter, 1 from a regular public school.  Though I do agree in general that the highly self-motivated student is probably better off in the long run being at a school that doesn't make so many time demands.  One problem these days is that even the 'regular schools' have absurd AP/IB requirements, so these students have the same time pressure as those at a TJ or Exeter, without the excellent peer group or instruction to help them achieve their goals.",
  "Solution_39": "Stuyvesant and maybe a couple of the public magnet schools have some of the peer-group-rarefaction (or the potential for it), as well, though in a more haphazard way than at a place like Exeter, and with different (or more particular) demographics.  And you are right that Stuyvesant is not the only public school with some math culture.\r\n\r\nAbout the number of IMO students from magnet vs other type of schools, it makes sense, as most people do not have definite enough IMO ambitions (or even know what IMO is) at the age of high-school selection to actually opt for an easier regular school; but anyone able to make IMO would at that age be a likely candidate for the local magnet school, if one exists.  I suppose that families of IMO candidates are also more liable to move to communities that have special schools, or good private schools.",
  "Solution_40": "Exeter is a prep high school. It does not have middle school portion. So they do not participate in Mathcounts."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "If $a,b$ be natural numbers such that $ab|(a^2+b^2+1)$ then\r\nprove that $\\frac{a^2+b^2+1}{ab}=3$",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=71760\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=40207\r\n\r\nYimin",
  "Solution_2": "And one more http://www.mathlinks.ro/Forum/viewtopic.php?t=91590 :D"
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "I think someone has asked this but I cannot seem to find it.\r\nThere seems to be no more ... option...",
  "Solution_1": "In class, Mr. Vornicu told us that only mods and admins have it now :(",
  "Solution_2": "Yeah I saw.",
  "Solution_3": "Just wondering, but how come?",
  "Solution_4": "Well I know some people put really really big annoying pictures on...\r\n\r\nOr animated pics...",
  "Solution_5": "[quote=\"Valentin Vornicu\"]Some people were abusing the privelages[/quote]\r\n\r\nSo sad but so true",
  "Solution_6": "So we probably (regular users) won't be able to have this feature back?",
  "Solution_7": "Hmm, I hope it's not my fault.  :P \r\nAnimated avatars... (Never big though) and text in front of my name. :P\r\nIf so, sorry!",
  "Solution_8": "[quote=\"snee\"]Hmm, I hope it's not my fault.  :P \nAnimated avatars... (Never big though) and text in front of my name. :P\nIf so, sorry![/quote]\r\nNah probably not. It was probably a lot of different people.\r\nBut don't the avatars have a size limit? Then how can it be *too* big?",
  "Solution_9": "They don't have a size limit.  I forget who, but someone had a huge aops logo before...at least it was after class.",
  "Solution_10": "Can there at least be an option to have no avatar? I accidentally chose a default avatar and now I cant change it back to no avatar because it won't let me. (I despise those default avatars...)",
  "Solution_11": "[quote=\"AIME15\"]They don't have a size limit.  I forget who, but someone had a huge aops logo before...at least it was after class.[/quote]\r\nI think they should. Like 80 by 80 or so.",
  "Solution_12": "buzzer: you can. (1) log out and log back in. (2) open the avatar panel again and click on the default avatars again. Then hit OK.\r\n\r\nducky: It's actually 80x100",
  "Solution_13": "Well I think it would be helpful if you didn't have to log out completely to change your avatar to a blank one...",
  "Solution_14": "we should all have a huge voting rebellion to have the avatars back...",
  "Solution_15": "[quote=\"coolts\"]we should all have a huge voting rebellion to have the avatars back...[/quote]We would just punish the starter/leader of the rebellion :)",
  "Solution_16": "admins are evil and must be stopped... except when teach Intermediate NT",
  "Solution_17": "[quote=\"mathemonster\"]admins are evil and must be stopped... except when teach Intermediate NT[/quote]\r\n\r\nThat might get you warned or even banned. :)",
  "Solution_18": "But can we use our AoPS avatar in the classroom? I understand people were abusing the privilege of using URLs but how about our forum avatar?",
  "Solution_19": "[quote=\"ernie\"][quote=\"mathemonster\"]admins are evil and must be stopped... except when teach Intermediate NT[/quote]\n\nThat might get you warned or even banned. :)[/quote]He was making a subtle joke .. \r\n\r\nAnyway we've had enough discussion on this topic, it ends here."
}
{
  "Tag": [
    "algebra",
    "system of equations"
  ],
  "Problem": "quick, tell me how to solve this problem:\r\nLet a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7. Then a^2-b^2+c^2 is what?\r\n\r\nI know it's a problem involving factoring techinques. However, when I did the problem, I was unable to find a way to combine them nicely into factorable terms. I know the answer, but I don't know how to make such insight in combining equations in such ways in order to solve the problem. Can anyone please enlighten me?",
  "Solution_1": "This is how I did it...\r\n\r\nYou see that there are 3 unknowns and 2 equations so you can plug in anything for the variable that suits you.\r\n\r\nSo I plugged in 0 = b\r\n\r\na + 8c = 4\r\n\r\n8a - c = 7\r\n\r\nHere solve for a and c and find the sum of their squares\r\n\r\nc = 8a - 7 = (4 - a )/8 ...  64a - 56 = 4 - a so a = 60/65 = 12/13\r\n\r\nc = 8*12/13  - 7 = 5/13\r\n\r\nSo a :^2:  + c :^2: = 169/169 = 1",
  "Solution_2": "Thats how I first did it, and thats probably all you need for that type of competition (I think that ones multichoice).\n\n\n\nHowever, heres a nicer solution, and one that would also be enough for a proof.\n\n[hide]\n\nBasically I just tried a few things, adding the given equations, etc. I also wanted to get things with a-b and a+b in it. In any case, I just happened to notice that if you double the first equation and add the second, you get 10a - 10b + 15c = 15, that looks very nice so we simplify it to 2(a-b) = -3(c-1).\n\n\n\nNow double the second one and subtract the first. Then 15a + 15b - 10c = 10, so 3(a+b) = 2(c+1). Multiply our two equations together, and 6(a^2 - b^2) = -6(c^2 - 1), so a^2 - b^2 + c^2 = 1. \n\n[/hide]",
  "Solution_3": "System of equations:\r\n8a+4b-c=7\r\n8(a-7b+8c=4) = 8a - 56b + 64c = 32\r\n-60b + 65c = 25\r\n13c - 12b = 5,\r\nc, b = 5 is a possibility\r\nPlug in: a - 35 + 40 = 4\r\na + 5 = 4\r\na = -1\r\n\r\na^2-b^2+c^2 = 1 - 25 + 25\r\n= 1",
  "Solution_4": "Thanks for the helps!! Now I understand..."
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "probability"
  ],
  "Problem": "A point $ (x,y)$ is randomly picked from inside the rectangle with vertices\r\n $ (0,0)$, $ (3,0)$, $ (3,2)$, and $ (0,2)$.  What is the probability that\r\n $ x &lt; y$?",
  "Solution_1": "[quote=\"GameBot\"]A point $ (x,y)$ is randomly picked from inside the rectangle with vertices\n $ (0,0)$, $ (3,0)$, $ (3,2)$, and $ (0,2)$.  What is the probability that\n $ x \\& lt; y$?[/quote]\r\n\r\nDraw the line $ y = x$.  You want to compute the area less than that line in the rectangle with points $ (0,0) (3,0), (3,2), and (0,2)$.  This area is $ 4$, if you use the formula.  The total area is $ 6$, so I think the probability is $ \\boxed{\\frac{2}{3}}$.  \r\n\r\nSomeone check my work please  :wink:",
  "Solution_2": "Geometric probability. \r\n[asy]size(250);\nimport graph;\nxaxis(\"$x$\",-0.5,6,Arrows);\nyaxis(\"$y$\",-0.5,5,Arrows);\ndot(\"$(0,0)$\",(0,0),SE,black);\ndot(\"$(3,0)$\",(3,0),S,black);\ndot(\"$(0,2)$\",(0,2),W,black);\ndot(\"$(3,2)$\",(3,2),NE,black);\ndraw((0,0)--(3,0)--(3,2)--(0,2)--cycle);\ndraw(\"$x<y$\",(-0.5,-0.5)--(2.5,2.5),NW,dashed,Arrows);[/asy]\r\nThe small triangle is the set of points that satisfy. Its area is $ 2\\times2\\equal{}4$ and the area of the whole rectangle is $ 2\\times3\\equal{}6,$ so the probability is $ \\frac{4}{6}\\equal{}\\boxed{\\frac{2}{3}}.$\r\n\r\nYour work is correct :)",
  "Solution_3": "Actually, the answer should be $ \\boxed{\\frac{1}{3}}$ since the area of the triangle is 1/2bh, or 2."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "inequalities",
    "number theory",
    "perpendicular bisector",
    "Diophantine equation"
  ],
  "Problem": "Is it the right section or not? ;)\r\n\r\nThe convex quadrilateral $ABCD$ is inscribed in the circle with diameter $[AD]$.\r\nAssume that $AB=BC=a,CD=b$ and $AD=d$ where $a,b,d$ are integers and $a \\neq b$.\r\nDetermine the least possible value for $d$. \r\n\r\nPierre.",
  "Solution_1": "[color=red][Author edit: Corrected a calculation mistake.][/color]\r\n\r\nThanks, Pierre, I will remember this problem as one of the first nontrivial number theory problems I solved...\r\n\r\nThe least possible value for d is d = 8. Hereby, a = 2 and b = 7.\r\n\r\nIndeed we begin with some geometry. Since the points A, B, C, D are supposed to lie on the circle with diameter AD, we have OA = OB = OC = OD, where O is the midpoint of the segment AD.\r\n\r\nNow, from OA = OC and AB = BC, it follows that the two points O and B lie on the perpendicular bisector of the segment AC. In other words, the line OB is perpendicular to the line AC and passes through the midpoint M of the segment AC.\r\n\r\nNow, since the point O is the midpoint of the segment AD, we have $\\displaystyle OD = \\frac{AD}{2}$. Since OA = OB = OC = OD, this entails $\\displaystyle OA = OB = OC = OD = \\frac{AD}{2}$.\r\n\r\nOn the other hand, the point M is the midpoint of the segment AC, and thus $\\displaystyle AM = \\frac{AC}{2}$.\r\n\r\nFinally, $\\displaystyle MO = \\frac{CD}{2}$, since the points M and O are the midpoints of the sides AC and AD of the triangle ACD. Hence, $\\displaystyle BM = OB - MO = \\frac{AD}{2} - \\frac{CD}{2} = \\frac{AD-CD}{2}$.\r\n\r\nThe Pythagoras theorem, applied to the right-angled triangle AMB, yields\r\n\r\n$\\displaystyle AB^2 = BM^2 + AM^2 = \\left( \\frac{AD-CD}{2} \\right)^2 + \\left( \\frac{AC}{2} \\right)^2 = \\frac{\\left( AD-CD \\right)^2 + AC^2}{4}$\r\n$\\displaystyle = \\frac{AD^2 + CD^2 - 2 \\cdot AD \\cdot CD + AC^2}{4} = \\frac{AD^2 + \\left( CD^2 + AC^2 \\right) - 2 \\cdot AD \\cdot CD}{4}$.\r\n\r\nNow, by the Pythagoras theorem in the right-angled triangle ACD (this triangle is right-angled since < ACD = 90, what is because the point C lies on the circle with diameter AD), we have $CD^2 + AC^2 = AD^2$, and thus\r\n  \r\n$\\displaystyle AB^2 = \\frac{AD^2 + AD^2 - 2 \\cdot AD \\cdot CD}{4} = \\frac{AD^2 - AD \\cdot CD}{2} = \\frac{AD \\cdot \\left( AD-CD \\right) }{2}$.\r\n\r\nThus, $2 \\cdot AB^2 = AD \\cdot \\left( AD-CD\\right)$. In other words, $2a^2 = d \\left( d-b \\right)$. Hence we have proven:\r\n\r\n[b]Lemma 1.[/b] In the configuration of your problem, the equality $2a^2 = d \\left( d-b \\right)$ holds.\r\n\r\nNow we will show:\r\n\r\n[b]Lemma 2.[/b] Conversely, if a, b, d are any three positive real numbers satisfying the equation $2a^2 = d \\left( d-b \\right)$, then there exists a convex quadrilateral ABCD such that this quadrilateral is inscribed in the circle with diameter AD, and the equations AB = BC = a, CD = b and AD = d hold.\r\n\r\n[i]Proof.[/i] Let AD be an arbitrary segment in the plane such that the length of this segment is AD = d. Construct the circle with diameter AD. Now, since $2a^2 = d \\left( d-b \\right) = d^2 - db < d^2$, we have $\\displaystyle a < \\frac{\\sqrt2}{2} d$, and thus a < d. Hence, the circle with diameter AD (of length d) must have two common points with the circle with center A and radius a. Let B be one of these two common points. Then this point B lies on the circle with diameter AD, and AB = a. Let O be the midpoint of the segment AD, i. e. the center of the circle with diameter AD. If we reflect the point A in the line OB, we get a new point C which also lies on the circle with diameter AD (just because the circle with diameter AD is symmetric with respect to the line OB, since this line passes through the center of that circle). Moreover, from the reflection, we have AB = BC. Combined with AB = a, this gives AB = BC = a.\r\n\r\nHence we have obtained a quadrilateral ABCD inscribed in the circle with diameter AD, and satisfying the equations AB = BC = a and AD = d.\r\n\r\nNow, we can use our inequality $\\displaystyle a < \\frac{\\sqrt2}{2} d$ to show that the quadrilateral ABCD is convex: In fact, if we consider the arc ABD on the circle with diameter AD, and if K is the midpoint of this arc, then the triangle AOK is right-angled and isosceles, so that $\\displaystyle AK = \\sqrt2 \\cdot OA = \\sqrt2 \\cdot \\frac{AD}{2} = \\sqrt2 \\cdot \\frac{d}{2} = \\frac{\\sqrt2}{2} d$, so that $\\displaystyle a < \\frac{\\sqrt2}{2} d$ implies a < AK, so that AB < AK. Thus, arc AB < arc AK. But since the point C is the reflection of the point A in the line OB, we have arc AC = 2 arc AB, so that arc AC = 2 arc AB < 2 arc AK = 2 $\\cdot$ 90 = 180 = arc AD. Hence, the quadrilateral ABCD is convex.\r\n\r\nFinally, we have to show that CD = b. In fact, from Theorem 1, applied to the quadrilateral ABCD, we obtain $2a^2 = d \\left( d-CD \\right)$. On the other hand, we know that $2a^2 = d \\left( d-b \\right)$. This immediately yields CD = b. Hence Lemma 2 is proven.\r\n\r\nNow, Lemma 1 and Lemma 2 show that for four positive numbers a, b, c, d, the existence of a convex quadrilateral ABCD such that this quadrilateral is inscribed in the circle with diameter AD, and the equations AB = BC = a, CD = b and AD = d hold, is equivalent to the equation $2a^2 = d \\left( d-b \\right)$. Therefore, your problem reduces to the question of finding the least positive integer d such that there exist two positive integers a and b with $a \\neq b$ and satisfying this equation.\r\n\r\nWell, the equation $2a^2 = d \\left( d-b \\right)$ is a diophantine equation, and there are lots of methods for solving such ones. Actually, you should note that always $d \\geq b$ (otherwise, d (d - b) would be negative, but the equation $2a^2 = d \\left( d-b \\right)$ implies that d (d - b) is twice a perfect square, and perfect squares can't be negative!). Hence, if d is a fixed natural number, then the only positive integers b coming in question for the equation $2a^2 = d \\left( d-b \\right)$ are the numbers b = 1, b = 2, ..., b = d. These are finitely many numbers and we can check them through.\r\n\r\nBy this kind of checking, we see that for every of the numbers d = 1, d = 2, d = 3, d = 4, d = 5, d = 6 and d = 7, there do not exist two positive integers a and b with $a \\neq b$ and satisfying the diophantine equation $2a^2 = d \\left( d-b \\right)$. In the case d = 8, we find the solution a = 2, b = 7, d = 8. Hence, d = 8 is the required least positive value.\r\n\r\n  Darij",
  "Solution_2": "I did almost the same thing, getting $(a,b,d)=(2,7,8)$ (I missed this case too at first :)).",
  "Solution_3": "Nice job Darij but, unfortunately, it seems that you made some error in your checking, because for $d=8$, we may choose $a=2$ and $b=7$...\r\nThus, the minimal value is in fact $d=8$.\r\n\r\nMoreover, i think some computations can be done with less efforts. Recall that $AB=BC=a, CD=b$ and $DA=d$.\r\nThus, from Pythagore's theorem : $BD= \\sqrt {d^2 - a^2}$ and $AC = \\sqrt {d^2-b^2}$.\r\nSince $ABCD$ is inscriptible, we may use Ptolemee's theorem, which leads to :\r\n$(ab+ad)^2 = (d^2-b^2)(d^2-a^2)$ that is $2a^2 = d(d-b)$ as you noticed.\r\n\r\nAnd if you want to avoid some boring cases to be checked, and to prove an extra result, we may prove that in that case $d$ cannot be prime. Indeed, first note that (1) implies that $d^2 = 2a^2 + bd > 2a^2$ so that $d> a \\cdot \\sqrt {2}$.\r\nNow assumethat $d$ is a prime :\r\nClearly $d=2$ leads to $a=b=$ which contradicts that $a \\neq b$. Thus, $d \\geq 3$ and since $d$ divides $2a^2$, it follows that $d$ divides $a$.\r\nBut, in that case : $a \\geq d > a \\cdot \\sqrt {2}$ which is absurd.\r\n\r\nPierre.",
  "Solution_4": "[quote=\"grobber\"]I did almost the same thing, getting $(a,b,d)=(2,7,8)$ (I missed this case too at first :)).[/quote]\n\nOh, thanks Grobber, I have now corrected the bug.\n\nOther solutions are a = 3, b = 7 and c = 9, or also a = 6, b = 1 and c = 9. It could be interesting to parametrize all solutions, but I am too tired now...\n\n[quote=\"pbornsztein\"]Nice job Darij but, unfortunately, it seems that you made some error in your checking, because for $d=8$, we may choose $a=2$ and $b=7$...[/quote]\n\nThanks to you too... anyway, I have corrected it now.\n\n[quote=\"pbornsztein\"]Moreover, i think some computations can be done with less efforts. Recall that $AB=BC=a, CD=b$ and $DA=d$.\nThus, from Pythagore's theorem : $BD= \\sqrt {d^2 - a^2}$ and $AC = \\sqrt {d^2-b^2}$.\nSince $ABCD$ is inscriptible, we may use Ptolemee's theorem, which leads to :\n$(ab+ad)^2 = (d^2-b^2)(d^2-a^2)$ that is $2a^2 = d(d-b)$ as you noticed.[/quote]\n\nIndeed, this is shorter, but my proof doesn't use Ptolemy. I actually see from your post and from Grobber's post on another thread that I should learn solving problems with Ptolemy; I am currently not so good at that...\n\n[quote=\"pbornsztein\"]And if you want to avoid some boring cases to be checked, and to prove an extra result, we may prove that in that case $d$ cannot be prime.[/quote]\r\n\r\nThis is what I actually did! But then I didn't want to write it up...\r\n\r\n  Darij"
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "Given that both the roots of the quadratic equation $x^2+bx+c=0$ are greater than 1 . Show that $b+c+1>0$   :)",
  "Solution_1": "Since $a=1 >0$, so $f(x) >0 \\Longleftrightarrow x \\in (-\\infty, x_1) \\cup (x_2, \\infty)$. Since the root of the function's roots are all greater than $1$, so $f(1)>0$. So $1+b+c>0$.",
  "Solution_2": "Well done shobber , you got it right  :D . By the way , what program you use to plot the graph ?",
  "Solution_3": "[quote=\"shyong\"]Well done shobber , you got it right  :D . By the way , what program you use to plot the graph ?[/quote]\r\nIt is called the geometer's sketchpad. There is a topic in the Other Problem Solving Topic discussing it. I've just bumped it up."
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Suppose that\r\n$ f(x)\\equal{}c_{0}\\plus{}c_{1}x\\plus{}c_{2}x^{2}...$\r\nis an even function. Show that\r\n$ f(x)\\equal{}c_{1}x\\plus{}c_{3}x^{3}\\plus{}c_{5}x^{5}...$",
  "Solution_1": "Show what?  :maybe:",
  "Solution_2": "[quote=\"AstroPhys\"]Show what?  :maybe:[/quote]\r\n\r\nShow that the first function is congruent to the second if f(x) is even.",
  "Solution_3": "Really? That second function sure looks odd to me.\r\n\r\n... I guess I meant that in two ways.\r\n\r\nIf you meant even function and $ f(x) \\equal{} c_{0}\\plus{}c_{2}x^{2}\\plus{}c_{4}x^{4}$, then this is easy by noticing that $ f$ is even iff $ c_{1}\\equal{} 0$, so then just let $ c_{4}\\equal{} 0$ and you're done.\r\n\r\nIf you meant odd function and what you have above, then\r\n$ f(x) \\equal{}\\minus{}f(\\minus{}x)$\r\n$ c_{0}\\plus{}c_{1}x\\plus{}c_{2}x^{2}\\equal{}\\minus{}c_{0}\\plus{}c_{1}x\\minus{}c_{2}x^{2}$\r\n$ c_{0}\\plus{}c_{2}x^{2}\\equal{} 0$\r\nThus $ f(x) \\equal{} c_{1}x$, so let $ c_{3}\\equal{} 0$, $ c_{5}\\equal{} 0$, and you're done.",
  "Solution_4": "I'm very sure I meant even function, and yes, that is correct.",
  "Solution_5": "I'm afraid I don't follow.  That second function you have written is very clearly an odd function."
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Is anyone familiar with a program called Scilab (which I'm told is really similiar to Matlab), to write a program to compute the exponential function e^x using only the +,-,*,/ operations (ie. no logs or exponents)?\r\n\r\n The program is supposed to have the same accuracy and limitations as the built-in routine in Scilab or equivalent, so i am told.\r\n\r\nHow does one in general write a program to solve a math question? \r\n\r\nWhat about the function e^x? Can we approximate it using its Taylor series expansion? Does anyone know the accuracy or limitations that the built-in program of Scilab has?",
  "Solution_1": "I am not familiar with Scilab, but this link might help\r\n\r\nhttp://en.wikipedia.org/wiki/Exponential_function#Computing_exp.28x.29_for_real_x",
  "Solution_2": "There should be something in the command list: http://www.scilab.org/product/man-eng/index.html"
}
{
  "Tag": [
    "inequalities solved",
    "inequalities"
  ],
  "Problem": "Prove the ineq: $ \\sqrt[n]{n+ \\sqrt[n]{n}}+ \\sqrt[n]{n- \\sqrt[n]{n}} \\leq 2 \\sqrt[n]{n}$ , $ \\forall n \\in N$, $ n \\geq2$.\r\n\r\n\r\ncheers! :D :D :cool:",
  "Solution_1": "For x>0, (1+x)^(1/n) \\leq 1+x/n\r\n\r\nso\r\n \r\n(n+n^(1/n))^(1/n) = n^(1/n).(1+n^(1/n -1))^(1/n) \\leq n^(1/n)(1+1/n.n^(1/n -1))\r\n\r\n(1) (n+n^(1/n))^(1/n)  \\leq  n^(1/n) + n^(2/n -2)\r\n\r\nthe same idea =>\r\n\r\n(2) (n-n^(1/n))^(1/n)  \\leq  n^(1/n)  - n^(2/n -2)\r\n\r\n \r\nAdd (1) & (2)",
  "Solution_2": "The problem is trivial:\r\n  Just use the fact that x^1/n+y^1/n<=2*((x+y)/2)^1/n (Jensen).",
  "Solution_3": "You are too hard with Moubi, Harazi... :D \r\n\r\nPierre.",
  "Solution_4": "Not at all (or at least I really didn't have any intention), but I apologise if this is the impression. Anyway, you must admit that the problem is very simple. Moubi has found a beautiful solution, so why should I be hard with him?"
}
{
  "Tag": [
    "geometry",
    "trapezoid",
    "trigonometry",
    "geometry unsolved"
  ],
  "Problem": "M is the midpoint of side AB of triangle ABC.The inscribed circle touches AB and AC in P and Q. The excircle that lies in different halfplane from A with respect to BC touches AB and AC in R and S.Prove that the perpendicular distance from M to PQ equals the perpendicular distance from M to RS",
  "Solution_1": "Are you sure about this problem??? I think this statement doesn't hold",
  "Solution_2": "In my opinion, $M$ is the midpoint of the side $BC\\ !$",
  "Solution_3": "[hide=\"Here is the corrected enunciation and its solution.\"]\n\n[color=blue]Let $ABC$ be a triangle. [u]Denote:[/u] the distance $\\delta_{d}(X)$ of the point $X$ to the line $d$; the midpoint $M$ of the side $[BC]$; the points $P\\in AB$, $Q\\in AC$ where the incircle $C(I,r)$ touches the respective sides; the points $R\\in AB$, $S\\in AC$ where the exincircle $C(I_a,r_a)$ touches the respective sides. Prove that $\\delta _{PQ} (M)=\\delta_{RS} (M).$[/color]\n\n[b]A solution.[/b] The quadrilateral $PQSR$ is isosceles trapezoid ($PR=QS=a$,\n\n$PQ\\parallel RS$, i.e. $\\frac{PQ}{p-a}=\\frac{RS}{p}=2\\sin \\frac A2$) and $PB=SC=p-b$.\n\nSo that, the point $M$ is the middlepoint of the segment $[PS]$, i.e.\n\n $\\delta_{PQ}(M)=\\delta_{RS}(M)\\ \\left(\\ =\\frac 12\\cdot BC\\cdot \\cos \\frac A2\\ \\right).$\n\n[u]Remark.[/u] You can use and the following property:\n$D\\in BC\\cap C(I,r)$, $E\\in BC\\cap C(I_a,r_a)$, $\\{D,D'\\}=ID\\cap C(I,r)$, $\\{E,E'\\}=I_aE\\cap C(I_a,r_a)$, $\\Longrightarrow$ $A\\in ED'\\cap E'D.$ [/hide]",
  "Solution_4": "Let $D\\equiv AI \\cap \\odot(ABC)$; it is the midpoint of $II_a$, hence it it is projected onto $AB$ and $AC$ at the midpoints $K$ and $L$ of $PR$ and $QS$ respectively. $PRSQ$ is an isosceles trapezoid, so $KL \\parallel PQ \\parallel RS$. See that $M$ is the projection of $D$ onto $BC$, therefore $K, M$ and $L$ are collinear (Simson line), done.\n\nBest regards,\nsunken rock"
}
{
  "Tag": [
    "function",
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "(tan2x)^2 + 2(tan2x)(tan3x) = 1\r\n\r\nafter arranging, I get\r\n\r\n7T^3 - 35T^2 + 21T - 1 = 0  (T = (tanx)^2)\r\n\r\nNow..... how?\r\n\r\nIn my opinion, using cardano method & arctan function is not elegant.\r\n\r\nHelp me, plz~",
  "Solution_1": "$x=\\frac{\\pi}{14},\\cdots$",
  "Solution_2": "[quote=\"antinoise007\"](tan2x)^2 + 2(tan2x)(tan3x) = 1\n\nafter arranging, I get\n\n7T^3 - 35T^2 + 21T - 1 = 0  (T = (tanx)^2)\n\nNow..... how?\n\nIn my opinion, using cardano method & arctan function is not elegant.\n\nHelp me, plz~[/quote]\r\n\r\nthe equation gives \r\n\r\n$\\tan 2x \\frac{\\tan 2x+\\tan 3x}{1-\\tan2x \\tan 3x}= 1$\r\n\r\nprovided ,$1-\\tan2x \\tan 3x\\neq 0$\r\nwhich gives two cases \r\n\r\n$\\tan 2x = \\cot 5x$\r\nor \r\n$\\tan 2x = \\cot 3x$\r\n\r\nthen i suppose u can proceed ....."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find a group of $4$ positive integer numbers $(a;b;c;d)$ that satisfies the following equation $a^{4}+b^{7}+c^{9}=d^{11}$.",
  "Solution_1": "[quote=\"Tiks\"]Find a group of $4$ positive integer numbers $(a;b;c;d)$ that satisfies the following equation $a^{4}+b^{7}+c^{9}=d^{11}$.[/quote]\r\nIf $(a,b,c,d)$ is solution, $(at^{m/4},bt^{m/7},ct^{m/9},dt^{m/11})$ is solution too, were $m=4*7*9*11$.\r\nWe can find solution $a=3^{63},b=3^{36},c=3^{28},d=3^{23}$.\r\nWe can find solutions (3 typs), when all numbers are powers of 2. For example $a=2^{646},b=2^{369},c=2^{287},d=2^{235}$."
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "modular arithmetic",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "what is the number of triangles whose side lenghts are integers and whose perimeter is n???!!!!!!!!",
  "Solution_1": "Please, don't call problems like \"scary problem\", \"hard problem\" or something like that. Problem's title should describe its essence.\r\n\r\nRead http://www.mathlinks.ro/viewtopic.php?p=140098",
  "Solution_2": "ok !! so the answer is????????? :)",
  "Solution_3": "It depends on what your are calling 'triangles'.\r\nAre $(a,a,b)$ and $(a,b,a)$ supposed to be distincts or not?\r\n\r\nPierre.",
  "Solution_4": "no they aren't distinct",
  "Solution_5": "hey what's the matter? nobody can solve it or nobody tried?",
  "Solution_6": "Nobody tries :P",
  "Solution_7": "helllllllllllllllllllllllp , u r wrong myth , scary is the best name we can give to this problem , come on guys , grobber ,orl, myth , bornzstein , einstein????",
  "Solution_8": "[quote=\"wisemaniac\"]what is the number of triangles whose side lenghts are integers and whose perimeter is n???!!!!!!!![/quote]\r\n\r\nso you seek the number of solutions to $a+b+c = n$ with $a \\leq b \\leq c$ and $a+b > c$.  these conditions are together equivalent to $ [(n+2)/2] \\leq a+b \\leq [2n/3] $ and $ a \\leq b $.  in general, the number of solutions to $a+b = k$ with $ a \\leq b $ is $[k/2]$; and the problem is basically solved from there...",
  "Solution_9": "You count some solutions twice. Right? How do you distinguish $b\\leq c$ or $c\\leq b$?\r\nI am writing my own solution right now. But your approach is nice, though it needs some corrections.",
  "Solution_10": "Ok. Let's count the number of all triples $(a,b,c)$ with $a\\leq b\\leq c$, $a+b+c=n$, $c<a+b$.\r\n\r\nLet's count the number of triple with $a+b+c=n$ and $a\\leq b\\leq c$.\r\nThere are $C_{n-1}^2$ solutions to the $a+b+c=n$. Each such solution we can rearrange to $a\\leq b\\leq c$. If $a\\neq b\\neq c$, then there are 6 rearrangements, but if two or three of these numbers are equal then the number of rearrangemts is less than 6.\r\nThe number of triples with two equal numbers is $\\left[\\frac{n-1}{2}\\right]-[3\\mid n]$. They were counted in $C_{n-1}^2$ thrice. \r\nTherefore the number of solutions $a+b+c=n$ with $a\\leq b\\leq c$ is \\[\\frac{C_{n-1}^2-3\\cdot \\left[\\frac{n-1}{2}\\right]+3\\cdot[3\\mid n]}{6}+\\frac{\\left[\\frac{n-1}{2}\\right]-[3\\mid n]}{3}+[3\\mid n]=\\frac{C_{n-1}^2-\\left[\\frac{n-1}{2}\\right]+7\\cdot [3\\mid n]}{6}.\\]\r\nNow let's count the number of solutions of $a+b+c=n$ with $a\\leq b\\leq c$ and $c\\geq a+b$.\r\nActually we need to find the number of triple with $n-2>c\\geq \\left[\\frac{n+1}{2}\\right]$. For fixed $c$ we have $\\left[\\frac{n-c}{2}\\right]$ choices for pair $(a,b)$.\r\n\r\nThus final answer is $\\frac{C_{n-1}^2-\\left[\\frac{n-1}{2}\\right]+7\\cdot [3\\mid n]}{6}-\\sum_{c=[\\frac{n+1}{2}]}^{n-2} \\left[\\frac{n-c}{2}\\right]$.\r\n\r\nOf course, you can find closed form for it, but I am not sure it is nice looking (something depening on $n\\pmod{4}$ :? )",
  "Solution_11": "a friend have just told told me that it's for sure exactlythe integer part of:\r\n1/48(n\u00b2+3n+21+((-1)^n)*3n). what do u think?",
  "Solution_12": "the number of triples $(x,y,z)$ with all the entries positive is simply\r\n$t_n=\\binom{n-1}{2}$.\r\nnow among these we will have to discard the triples that don't form triangles;this simply has to throw away the triples with (exactly) one of $x,y,z$ being $\\geq n/2$; suppose all $x,y,z<n/2$ but say $x+y\\leq z$; then $x+y+z\\leq 2z \\Rightarrow 2z\\geq n$ and that is ruled out.\r\ncounting the number of triples with say $x\\geq n/2$ is the same as having the number of triples with $y+z\\leq n/2$ and that is the same as the number of triples satisfying $x+y+z'=[n/2],x,y>0$ and this number is $\\binom{[n/2]}{2}$.\r\n\r\ni see that many others have already written.....",
  "Solution_13": "[quote=\"Myth\"]You count some solutions twice. Right? How do you distinguish $b\\leq c$ or $c\\leq b$?\nI am writing my own solution right now. But your approach is nice, though it needs some corrections.[/quote]\r\n\r\nno... the conditions $[(n+2)/2] \\leq a+b \\leq [2n/3]$ and $a \\leq b$ guarantee that $a \\leq b \\leq c$",
  "Solution_14": "yep, you're right. :("
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "A King gathers 100 men from his country in a room and says, \"I will give 1,000,000 dollars to the Laziest Person here\". He then asks them to raise their hands if they're the laziest person here. Everybody except for one person raise their hands. When the King asks him why he didnt raise his hand, the persons response makes the King give him the 1,000,000 dollars. What was the mans response?\r\n\r\nI think this is a pretty cool problem, I think its well known though.",
  "Solution_1": "uhh... is it[hide=\"?\"]\"I'm too lazy to raise my hand.\"[/hide]",
  "Solution_2": "[quote=\"Hamster1800\"]uhh... is it[hide=\"?\"]\"I'm too lazy to raise my hand.\"[/hide][/quote]\r\n\r\nThat's exactly what I thought.",
  "Solution_3": "[quote=\"chess64\"][quote=\"Hamster1800\"]uhh... is it[hide=\"?\"]\"I'm too lazy to raise my hand.\"[/hide][/quote]\n\nThat's exactly what I thought.[/quote]\r\nThat's exactly what I was going to post.",
  "Solution_4": "hehe guess it was well known, i never heard it untill my dad told me  :blush:",
  "Solution_5": "actually i'd never heard it until you told it to us but the answer was really obvious. Unfortunately, the answer actually WOULDN\"T work as raising your hand takes less energy than to speak that response. Therefore, he also must be too lazy to answer so another answer might be he says nothing?",
  "Solution_6": "[quote=\"Hamster1800\"]actually i'd never heard it until you told it to us but the answer was really obvious. Unfortunately, the answer actually WOULDN\"T work as raising your hand takes less energy than to speak that response. Therefore, he also must be too lazy to answer so another answer might be he says nothing?[/quote]\r\n\r\nHahaha...so I guess the King will get angry at him for not answering, chop his head off.  Then he'll keep the money.",
  "Solution_7": "He was just to lazy to lift his hand",
  "Solution_8": "Thank you Davendra for so graciously posting the solution, after it has already been posted, and confirmed for the (something like...) 5th time.  :)",
  "Solution_9": "I think the guy said \"I'm too lazy to raise my hand!\""
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "quadratics",
    "LaTeX"
  ],
  "Problem": "Given that $x^3+3x^2+3x+1=(x+3)^3$, find all real $x$ such that $x^3+3x^2+3x=1$",
  "Solution_1": "But the first condition isn't true...",
  "Solution_2": "is there a way to factorize x^3+3x^2+3x=14? i can't bring it down to a quadratic equation.......",
  "Solution_3": "Looking at the $/LaTeX$ code, it says the thrid degree equation is equal to $(x+3)$ to the power of #\r\n\r\nI don't know what that means...  :?",
  "Solution_4": "is # a variable or something that didn't appear?",
  "Solution_5": "i';m pretty sure he meant that \\[\\text{given  } x^3+3x^2+ 3x +1 = {(x+1)}^3\\]",
  "Solution_6": "so, then its find all real x if (x+1)^3-1=14?",
  "Solution_7": "Well, that's a pretty nasty radical answer.",
  "Solution_8": "Sorry everybody, I fixed the typo. Pressing down the shift key too often is not a good thing.  :oops:",
  "Solution_9": "Oh, and one more thing. X isn't a nasty radical, it's a clean radical.  ;)",
  "Solution_10": "The first condition is still false  :?",
  "Solution_11": "I don't believe this problem is on the level of MOEMS.\r\n\r\nIf you are not certain of the MOEMS curriculum level, please do not post problems in the MOEMS forum."
}
{
  "Tag": [
    "function",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Everyone likes to study the asymptotics of the Riemann zeta function near the critical strip, because that's what's actually got deep meaning in analytic number theory. But what about in the \"boring\" parts of the plane? \r\n\r\nLet's just work on the real line. We know that $ \\lim_{x \\to \\infty} \\zeta(x) \\equal{} 1$. Following up on this, here's the problem I propose: \r\n\r\nFind (explicitly) an elementary function $ g(x)$ so that $ \\lim_{x \\to \\infty} \\frac{\\zeta(x) \\minus{} 1}{g(x)} \\equal{} 1$.",
  "Solution_1": "$ 2^{\\minus{}x}$  :rotfl:",
  "Solution_2": "Now what is $ \\sum_{n\\equal{}2}^\\infty (2^{n} (\\zeta(n) \\minus{} 1) \\minus{} 1)$ ? ;)",
  "Solution_3": "$ \\frac{2^n}{3^n}$, of course. We could go on like this forever.",
  "Solution_4": "Dang. I should put less scary words in my topics. Probably the only people who look at it are the ones for whom it's trivial...  :("
}
{
  "Tag": [
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Let $ i$, $ n \\in\\mathbb{N}^{*}$. And let $ a_{i}= \\min\\{k+\\frac{i}{k}| k \\in\\mathbb{N}^{*}\\}$. Find the value of $ \\sum_{i=1}^{n^{2}}[a_{i}]$, where $ [x]$ means the largest integer no larger than $ x.$",
  "Solution_1": "[quote=\"April\"]Let $ i$, $ n \\in\\mathbb{N}^{*}$. And let $ a_{i}= \\min\\{k+\\frac{i}{k}| k \\in\\mathbb{N}^{*}\\}$. Find the value of $ \\sum_{i=1}^{n^{2}}[a_{i}]$, where $ [x]$ means the largest integer no larger than $ x.$[/quote]\r\nFrom $ k+\\frac{i}{k}\\le k+1+\\frac{i}{k+1}$ we get $ i\\le k(k+1)$.\r\nTherefore $ a_{i}=k+\\frac{i}{k}=2k+\\frac{i-k^{2}}{k}$, were $ k(k-1)\\le i\\le k(k+1)$.\r\n$ [a_{i}]=2k-1$, when $ k(k-1)\\le <k^{2}$ and $ [a_{i}]=2k$, when $ k^{2}\\le i<k(k+1)$. Therefore\r\n\\[ \\sum_{i=1}^{n^{2}}[a_{i}]=2n+\\sum_{k=1}^{n-1}k*(2k)+\\sum_{k=2}^{n}k*(2k-1)=2n+\\frac{n(n-1)(2n-1)}{3}+\\frac{n(n+1)(4n-1)}{6}-1=\\frac{8n^{3}-3n^{2}+13n-12}{6}\\]."
}
{
  "Tag": [],
  "Problem": "Juanita paid $ \\$18$ for a book after a $ 10\\%$ discount. What was the number of dollars in the original price of the book?",
  "Solution_1": "$ x$ is original price\r\n$ 18 \\equal{} 90\\%x$\r\n\r\n$ 18 \\equal{} \\frac {9x}{10}$\r\n\r\n$ 2 \\equal{} \\frac {x}{10}$\r\n\r\n$ 20 \\equal{} x$\r\n20"
}
{
  "Tag": [
    "floor function",
    "geometry",
    "3D geometry",
    "pigeonhole principle",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Let $S$ be the set of $n$-uples $\\left( x_{1}, x_{2}, \\ldots, x_{n}\\right)$ such that $x_{i}\\in \\{ 0, 1 \\}$ for all $i \\in \\overline{1,n}$, where $n \\geq 3$. Let $M(n)$ be the smallest integer with the property that any subset of $S$ with at least $M(n)$ elements contains at least three $n$-uples \\[\\left( x_{1}, \\ldots, x_{n}\\right), \\, \\left( y_{1}, \\ldots, y_{n}\\right), \\, \\left( z_{1}, \\ldots, z_{n}\\right) \\] such that\r\n\\[\\sum_{i=1}^{n}\\left( x_{i}-y_{i}\\right)^{2}= \\sum_{i=1}^{n}\\left( y_{i}-z_{i}\\right)^{2}= \\sum_{i=1}^{n}\\left( z_{i}-x_{i}\\right)^{2}. \\]\r\n(a) Prove that $M(n) \\leq \\left\\lfloor \\frac{2^{n+1}}{n}\\right\\rfloor+1$.\r\n(b) Compute $M(3)$ and $M(4)$.",
  "Solution_1": "Call two n-tuples neighbors if they differ by exactly one element. If $M(n) > \\lfloor \\frac{2^{n+1}}{n}\\rfloor+1$, then for the corresponding S, each element of S has n neighbors. Now $nM(n) > n \\lfloor \\frac{2^{n+1}}{n}\\rfloor+1 \\ge 2^{n+1}$, so there is some n-tuple x such that three elements a, b, and c of S are neighbors of x. Then these three elements eacher differ from each other in two places, a contradiction.\r\n\r\nNow to find M(3) and M(4). For n = 3, we can have 000, 100, 010, 110. If we could have 5 elements, we can't have three elements with the same number of 1's. Also, either some two have exactly one 1, or some two have exactly one 0, and WLOG we'll take the latter. Then we can not have 000, so then we must have some pair having exactly one 1. But that means we can't have 111, so either some three have exactly one 1 or some three have exactly one 0, contradiction.\r\n\r\nFor n = 4, we can have 0000, 1000, 0100, 1010, 0101, 0111, 1011, 1111, and by the arguments in proving the general bound, we can show that this is optimal.",
  "Solution_2": "[quote=\"probability1.01\"]If $M(n) > \\lfloor \\frac{2^{n+1}}{n}\\rfloor+1$, then for the corresponding S, each element of S has n neighbors. Now $nM(n) > n \\lfloor \\frac{2^{n+1}}{n}\\rfloor+1 \\ge 2^{n+1}$, so there is some n-tuple x such that three elements a, b, and c of S are neighbors of x. Then these three elements eacher differ from each other in two places, a contradiction.[/quote]\r\n\r\nI don't get a single thing from it, could you elaborate on it?  maybe I'm just blind at the moment :maybe:",
  "Solution_3": "[quote=\"probability1.01\"]\nNow to find M(3) and M(4). For n = 3, we can have 000, 100, 010, 110. If we could have 5 elements, we can't have three elements with the same number of 1's. Also, either some two have exactly one 1, or some two have exactly one 0, and WLOG we'll take the latter. Then we can not have 000, so then we must have some pair having exactly one 1. But that means we can't have 111, so either some three have exactly one 1 or some three have exactly one 0, contradiction.\n[/quote]\r\n\r\nactually $M(3)=5$ is the sollution.  :wink: and a hint for finding this is by considering a cube and translating the problem geometrically.",
  "Solution_4": "I think I proved M(3) = 5. Except I thought M(n) was the largest integer such that you don't have three elements of equal distance to each other, so it looks like I have M(3) = 4. Unless you really meant M(3) = 6? To Megus, I just used the pigeonhole principle...  :maybe:",
  "Solution_5": "[quote=\"probability1.01\"]$ n \\lfloor \\frac {2^{n \\plus{} 1}}{n}\\rfloor \\plus{} 1 \\ge 2^{n \\plus{} 1}$, [/quote]\r\n\r\nThis is not true! Take $ n\\equal{}5$ as an example.",
  "Solution_6": "So, any corrections?",
  "Solution_7": "i think he wanted to write $ n(\\lfloor \\dfrac{2^{n \\plus{} 1}}{n} \\rfloor \\plus{} 1)$\r\nthen there is no error in the solution :lol:\r\n\r\nand maybe $ M(3)\\equal{}5$ $ M(4)\\equal{}9$",
  "Solution_8": "[quote=pohoatza][quote=\"probability1.01\"]\nNow to find M(3) and M(4). For n = 3, we can have 000, 100, 010, 110. If we could have 5 elements, we can't have three elements with the same number of 1's. Also, either some two have exactly one 1, or some two have exactly one 0, and WLOG we'll take the latter. Then we can not have 000, so then we must have some pair having exactly one 1. But that means we can't have 111, so either some three have exactly one 1 or some three have exactly one 0, contradiction.\n[/quote]\n\nactually $M(3)=5$ is the sollution.  :wink: and a hint for finding this is by considering a cube and translating the problem geometrically.[/quote]\nWhat are you talking about?How can you get 5?\n"
}
{
  "Tag": [
    "function",
    "geometry",
    "\\/closed"
  ],
  "Problem": "How do you take a screen shot of the page you are on and then put it in a message?\r\nI think this was asked before if so im sorry :(",
  "Solution_1": "Press Print Scr on your keep board to take a picture and ctrl-v to paste.",
  "Solution_2": "[url=http://www.google.com/search?q=how+to+screenshot&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]Are you new to the internet, by chance? Google is your friend[/url]",
  "Solution_3": "i mean putting a picture in an AoPS message.",
  "Solution_4": "You can upload it to another host (such as tinypic) and embed it using the Img tags and then place it in within your image wherever you like, or you can send it as an attachment and click \"Add an Attachment\" next to \"Message Body\" when composing a message.",
  "Solution_5": "so u can save it as a picture? how would u do that?",
  "Solution_6": "Did you even click that link? *sigh*\r\n\r\n1) Press PrintScreen\r\n2) Open Paint (or another iamge editing program)\r\n3) Right-click -> Paste or Press Ctrl-V\r\n4) Click Save As and save where you want",
  "Solution_7": "Don't forget to crop out the stuff you don't want. No point in uploading a massive PNG file onto AoPS.  :wink:",
  "Solution_8": "thanks!! (by the way that link didnt work Anaxerzia.)",
  "Solution_9": "..dude the link works :huh:\r\n\r\nFYI, on a Mac, it's Cmd+Shift+3 for a fullscreen capture, and Cmd+Shift+4 for a window. Pressing spacebar after Cmd+Shift+4 should enable you to screenshot a selection.\r\n\r\non Linux with GNOME, you can go to Accessories > Take Screenshot. dunno about KDE.\r\n\r\nI've also managed to tweak settings so that I can drag my cursor across the screen while holding the Command key to capture a portion.. forgot how. google it.",
  "Solution_10": "is cropping a pic like erasing the stuff u dont want?",
  "Solution_11": "Yes; when you go onto a graphics program, for example the GIMP, you use the crop function and select an area you want to keep. Everything else will be deleted.\r\n\r\nOr, on some computers, like mine, when you take a screen shot, you can just choose a option that only takes a picture of an open window.",
  "Solution_12": "i used paint."
}
{
  "Tag": [
    "calculus",
    "integration"
  ],
  "Problem": "The ring is a two dimensional figure. Given that the line charge density \"U\", what can we say about the Eletric field everywhere inside such a ring of charge?",
  "Solution_1": "have you heard of the shell theorem?",
  "Solution_2": "Yes i did. The goal of  shell theory is to describe the energy levels of nuclei and transitions between these levels. Ideally the theory is based on known nuclear forces.",
  "Solution_3": "I believe the shell theorem states that inside a uniform circular shell, the gravitational field is 0.  \r\nAnd electric fields work a lot like gravity...",
  "Solution_4": "Shell Theory and Shell Theorem are two different things, I believe... :)\r\n\r\nAnyway, daveed is right. You should try to prove it, it's not that hard.",
  "Solution_5": "Ooops, actually - daveed is not so right. The shell theorem works in 3D with all $F=\\frac{k}{x^2}$ forces but not in 2D! In 2D, it could probably work if the force was $F=\\frac{k}{x}$.",
  "Solution_6": "oof.  My apologies.  If that's so, maybe you could get the result from some integration?",
  "Solution_7": "The electric field inside any conductor, when a constant current is travelling through it, is zero. Now, that and any charge distribution usualy creates a field which is found by solving Poisson or d'Alembert equations with proper boundary conditions.",
  "Solution_8": "Ooops, i think the problem mentions no conductor.But it's true that the potential at every point on the ring is the same.Take it to be 0. Inside, the potential satisfies Laplace's equation."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "What is the square root of the cube root of 64?",
  "Solution_1": "This is the sixth root of 64=2^6, which is 2.",
  "Solution_2": "64's cube root is 4, because 4*4*4=64.  4's square root is 2, because 2*2=4, hence the answer is 2. :surf:"
}
{
  "Tag": [
    "trigonometry",
    "trig identities",
    "Law of Cosines",
    "geometry",
    "congruent triangles"
  ],
  "Problem": "Ten equally spaced points are plotted on a circle's circumference.  Connecting each point to each other point 3 points away, a regular convex decagon is formed in the center of the circle. Show the length of a side of the decagon is equal to the length of the radius.\r\n\r\ni dont think its possible...because that would mean each sector containing 2 radii and one side of the decagon must be equilateral, and its not.",
  "Solution_1": "Let the length of the radius and decagon's side be $ r$ and $ s$ respectively. Consider the triangle made by two radii and a side. Because the sides span 3 points, the central angle measures $ 3/10 \\cdot 360 \\equal{} 108$ degrees. By the law of cosines, $ 2r^2 \\minus{} r^2 \\cos 108 \\equal{} s^2$, so $ \\frac{s}{r} \\equal{} \\sqrt{2 \\minus{} \\cos 108}$. It's easy to see this is not $ 1$.",
  "Solution_2": "Obviously there was a transcription error somewhere. I tried hard to figure out what the right wording should be. This version works.\r\n[hide]Ten equally spaced points are plotted on a circle's circumference.  Connecting each point to the next, a regular convex decagon is formed. Connecting each point to each other point 3 points away, a smaller regular convex decagon is formed in the center of the circle. Show the length of a side of the larger decagon is equal to the length of the radius of the circle circumscribed about the smaller decagon.[/hide]\r\nThere may be a simpler, more straightforward wording that works too. Comments?",
  "Solution_3": "sorry it should read as follows:\r\n\r\nTen equally spaced points are plotted on a circle's circumference These points connect to form a regular convex decagon. Connecting each point to each other point 3 points away, a regular convex decagon is formed in the center of the circle. Show the difference between the lengths of the sides of the decagons is equal to the length of the radius.",
  "Solution_4": "any help please?",
  "Solution_5": "By the side lenght of the larger decagon, i mean the length of one point to another point 3 points away.\r\n\r\nThe larger decagon is a regular 10-pointed star",
  "Solution_6": "Maybe something is lost in translation. The way I understand the drawing it still does not add up.\r\nIf you draw it, and add segments going from one point of the 10-point star to the opposite one (diameters), you will find isosceles triangles all over.\r\nThe angles can be figured as 36, 72 and 108 degrees (pi/5, 2pi/5. and 3pi/5).\r\nThere are many congruent triangles and many congruent sides.\r\nConsider one side of the inner decagon. I find two segments of the same length along each of the segments drawn from one point in the circumference to another 3 points away (\"sides of the larger decagon\").\r\nConsider a radius drawn from one of the points of the star. There is a segment of equal length on the adjacent \"side of the larger decagon\".\r\nThe problem is that the two segments I find along the \"side of the larger decagon\" (one as long as the radius, the other as long as the side of the small decagon) do not add up to the \"side of the larger decagon\"."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "trigonometry"
  ],
  "Problem": "Prove: If Q is an orthogonal matrix, then each entry of Q is the same as its cofactor if det(Q)=1 and is the negative of its cofactor if det(Q)=-1.\r\n\r\nProve: Let $ u$ be a vecter in an inner product space V, and let $ \\{v_1,v_2,...,v_n\\}$ be an orthonormal basis for V, and if $ \\alpha_i$ is the angle between $ u$ and $ v_i$, then\r\n\r\n$ \\sum_{i\\equal{}1}^n\\cos^2 \\alpha_i\\equal{}1$.",
  "Solution_1": "[b]a)[/b]Let $ A$ be an invertible matrix. Then $ A\\text{cof}(A)^{t}\\equal{}\\det(A)I$. Now, compare it with the definition of an orthogonal matrix.\r\n\r\n[b]b)[/b]Let $ 0\\neq u\\in V$. $ u\\equal{}\\sum_{i\\equal{}1}^{n}c_{i}v_{i}$ for some scalars $ c_{i}$ in the ground field. Use $ \\cos(\\alpha_{i})\\equal{}u.v_{i}/|u|$ to conclude."
}
{
  "Tag": [
    "function",
    "modular arithmetic"
  ],
  "Problem": "All are questions from the path to combinatorics for undergraduates:\r\n\r\n1)Determine the no. of functions f: {1,2,3,...,1999) -> {2000,2001,2002,2003} satisfying the condition that f(1) + f(2) + f(3) ....+ f(1999) is odd.\r\n\r\n[hide]My answer :$ 1000. 2^{1999}$[/hide]\n\n2) In how many ways can COMPUTER be spelled by moving either down or diagonally to the right in fig:\n\nC \nO O \nM M M\nP  P  P  P\nU  U  U  U  U\nT  T  T  T  T  T \nE  E  E   E  E  E  E\nR  R  R   R  R  R  R  R \n\n\n[hide]I tried generating function[/hide]\n\n\n3) A basketball team consists of 12 pairs of twin brothers. On the first day of training, all 24 players stand in a circle in such a way that all pairs of twin brothers are neighbors. In how many ways can this be done?\n \n\n[hide]My answer :$ 11! . 2^{12}$[/hide]",
  "Solution_1": "[hide=\"Solution to 1\"]Instead consider functions $ g: \\{1,2,3,\\ldots ,1999\\}\\rightarrow\\{0,1\\}$.  We want the number of such functions with $ g(1) \\plus{} g(2) \\plus{} g(3) \\plus{} \\cdots \\plus{} g(1999) \\equal{} n$ for $ n$ odd.  For any $ n$ the number of functions is $ \\binom{1999}{n}$, so we want $ \\binom{1999}{1} \\plus{} \\binom{1999}{3} \\plus{} \\binom{1999}{5} \\plus{} \\cdots \\plus{} \\binom{1999}{1999}$, which is $ 2^{1998}$ (Try to prove this  :) ).  Now we want to looks at functions $ f: \\{1,2,3,\\ldots ,1999\\}\\rightarrow\\{2000,2001,2002,2003\\}$ such that the sum of function values is still odd.  There is a $ 2^{1999}$ to 1 correspondence between the number of functions $ f$ and the number of functions $ g$, since if $ f(i) \\equal{} 0$ then we make $ g(i) \\equal{} 2000,2002$ and if $ f(i) \\equal{} 1$ then we make $ g(i) \\equal{} 2001,2003$ (this is also clearly reversible).  Therefore, the answer is $ 2^{3997}$.[/hide]\nEDIT:\n[hide=\"Much nicer solution to 1\"]There are $ 4^{1999}$ such functions ignoring the sum condition.  Now given a function $ f$ consider the function $ g$ such that $ g(n) \\equal{} 4003 \\minus{} f(n)$ for each $ n$.  Then $ g(n)$ has different parity than $ f(n)$ for each $ n$, so $ g(n)\\equiv f(n) \\plus{} 1999\\pmod{2}$, so exactly one of $ f(1) \\plus{} \\cdots \\plus{} f(1999)$ and $ g(1) \\plus{} \\cdots \\plus{} g(1999)$ is odd.  We can pair up all the functions in this way, and exactly one of each pair of functions will satisfy the desired property.  Therefore, the answer is $ 4^{1999}/2 \\equal{} 2^{3997}$.[/hide]\n[hide=\"Hint to 2\"]Consider the number of choices at each step in spelling the word.  :wink: [/hide]\r\nYour answer to 3 is correct."
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Two players fill an $ n\\times n$ square board with $ 0$ and $ 1$. As soon as the board is filled, they calculate the sum $ r_i,c_i$ of numbers in $ i$th row and $ i$th column, respectively. Let\r\n\r\n$ R\\equal{}\\sum_{i\\equal{}1}^{n}ir_i, C\\equal{}\\sum_{i\\equal{}1}^{n}ic_i$. The first player wins if $ R<C$, and the second wins otherwise.\r\n\r\nWho has a winning strategy?",
  "Solution_1": "[b]Solution:[/b] Nobody has the winning strategy. Consider the gain matrix of the 1st player on a $ nXn$ board when every entry in the board is made 1. Let $ g(a,b)$ be the gain of the first player if 1 is entered in the square $ (a,b)$ (considering leftdown square to be $ (1,1)$).\r\n\\[ g(a,b) \\equal{} a \\minus{} b\r\n\\]\r\nHence the game is equally balanced.",
  "Solution_2": "I don't really get it... what is the gain matrix? Why the game is equally balanced?\r\n\r\nPlease explain a bit more  :)",
  "Solution_3": "The picture shows the $ 3X3$ gain matrix for first person. For the first person (who tries to maximize the column sum) 1 in (3,1) gives him the score of 3 while it gives the second person score 1. It is clear from the gain matrix that game is equally balanced.\r\n\r\nAssuming both players make intelligent moves (like in all game problems), [b]a[/b] strategy is as follows: \r\nIf the P1 puts 1 in (3,1) or a 0 in (1,3)then P2 has to put a 1 in (1,3) or 0 in (3,1)\r\n\r\nIf the P1 puts 1 in (2,1) or a 0 in (1,2)then P2 has to put a 1 in (1,2) or 0 in (2,1)\r\n\r\nIf the P1 puts 1 in (3,2) or a 0 in (2,3)then P2 has to put a 1 in (2,3) or 0 in (3,2)\r\n\r\nI hope now it is clear ?",
  "Solution_4": "Umm... it's not really clear. It doesn't seem rigorous at all.\r\n\r\nHow do you know those are the most intelligent moves?",
  "Solution_5": "[quote=\"easternlatincup\"]Two players fill an $ n\\times n$ square board with $ 0$ and $ 1$.[/quote]\r\n{Both players can use both $ 0$ and $ 1$} or {The first player puts $ 0$ and the 2nd player puts $ 1$}?",
  "Solution_6": "Both players can use both $ 0$ and $ 1$  :)",
  "Solution_7": "I think Srikanth is right.\r\n\r\nTo make his solution more understandable and more rigorous we need to show that each player has a not-loosing(winning or drawing) strategy.\r\n\r\nSuppose a player decides that he wont let the other player win and he is intelligent enough. In each move he might copy the other player's move in the centrally symmetric cell. It might be that there is no such centrally symmetric cell (For example, He is the 1st player making his 1st move, or the previous move had been in the center-cell etc.). In that case he may put a 1 in his most-gain cell. His most-gain cell will have a non-negative value because in such a situation the board would be completely centrally symmetric. Thus the game must end in a draw or a win in favor of the discussed player. (I am not sure if it can be a win but that doesn't matter.)\r\n\r\n(a,b) and (c,d) are centrally symmetric iff a+c=n+1 and b+d=n+1.",
  "Solution_8": "Well, I wrote my solution short and sweet in my first post (username: srikanth) as we typically write solutions to most game problems ( especially Russian olympiads).\r\nHere are more details which have could have been got from the gain table.\r\n\r\nThe gain table for the 1st and 2nd player look like\r\n\r\n-2 -1  0           \r\n-1  0  1            \r\n 0  1   2           \r\n\r\n2  1  0\r\n1  0 -1\r\n0 -1 -2\r\n\r\nBy intelligent move I mean operations like {p1 put 1 in (1,3) that gives him a gain of -2} are not done by p1.\r\n\r\nWe call (1,3) and (3,1) to be corresponding squares. If P1 puts 1 in (3,1) , p2 retaliates by putting 1 in (1,3). Like wise {(1,2) and (2,1)} {(2,3) and (3,2)} can be corresponding squares. The other possibility is {(1,2) and (3,2)} and {(2,1) and (2,3)}. This new possibility merely comes as the number in the squares are same(here 1).\r\n\r\nI have been as explicit as I could have been. Hope the solution is clear now. generalization to n is trivial."
}
{
  "Tag": [],
  "Problem": "Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate\nsix more bananas than on the previous day. How many bananas did Big Al eat\non May 5?",
  "Solution_1": "Let x equal the number of bananas he ate on May 3. Thus we have $ x\\minus{}12\\plus{}x\\minus{}6\\plus{}x\\plus{}x\\plus{}6\\plus{}x\\plus{}12\\equal{}100$, or $ x\\equal{}20$. Thus Big Al ate $ 20\\plus{}12\\equal{}\\boxed{32}$ bananas on May 5.",
  "Solution_2": "[quote=\"$LaTeX$\"]Let x equal the number of bananas he ate on May 3. Thus we have $ x\\minus{}12\\plus{}x\\minus{}6\\plus{}x\\plus{}x\\plus{}6\\plus{}x\\plus{}12\\equal{}100$, or $ x\\equal{}20$. Thus Big Al ate $ 20\\plus{}12\\equal{}\\boxed{32}$ bananas on May 5.[/quote]\n[hide=\"A Solution That May Be A Tiny Bit Quicker\"]We know that on the day in the middle (which is May 3), Big Al must have ate the mean of the number of banana's he ate on each day.  Since we are already given the total, we can just divide by the number of days and get $\\frac{100}{5}=20$ bananas.  Then we see that May 5 is only 2 days after May 3 so we just add $6\\times2=12$ to the number of bananas Big Al ate on May 3.  So Big Al ate $20+12=\\boxed{32}$ bananas on May 5.[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "calculus computations"
  ],
  "Problem": "Given $ f(x)$ is an odd function defined everywhere,periodic with period $ 2$ and integrable on every interval.\r\n\r\nLet $ g(x)\\equal{}\\int_{0}^x f(t) dt$.\r\n\r\nThen prove that $ g(2n)\\equal{}0$ for every integer $ n$.",
  "Solution_1": "Hi :) !!\r\n\r\n$ f$ is odd and $ 2$-periodic function in everywhere than $ \\int_{2(k \\minus{} 1)}^{2k} f(t)dt \\equal{} 0$  for every integr $ k$\r\n\r\nthan \r\n\r\n$ g(2n) \\equal{} \\int_0^{2n}f(t)dt \\equal{} \\sum_{k \\equal{} 1}^{n} \\int_{2(k \\minus{} 1)}^{2k}f(t)dt \\equal{} 0$\r\n\r\n...\r\ni taked that $ n \\geq 1$ because $ g$ is no-odd  function .\r\n\r\nPS: we can prove that by a lof of methode .\r\n\r\nand thanks",
  "Solution_2": "Thank you mathema.\r\nCan you show any of the other methods",
  "Solution_3": "$ \\int_0^{2n} f(x) \\ dx \\equal{} \\int_{\\minus{}n}^{n} f(x\\minus{}2n) \\ dx \\equal{}  \\int_{\\minus{}n}^{n} f(x) \\ dx \\equal{} 0 \\ (\\ \\because $ f(x) has period 2 and integration of an odd function over a symmetric interval about 0 gives 0)",
  "Solution_4": "[quote=\"pankajsinha\"]Thank you mathema.\nCan you show any of the other methods[/quote]\r\nOK :) !!\r\n\r\nafter the methode proposed by [b]hsbhatt[/b] i give one other nice :) !!\r\n\r\nif $ h$ is a odd and $ T$-periodic function than we have $ \\int_0^T h(t)dt \\equal{} 0$\r\n\r\nthan our exemple ($ T \\equal{} 2$) :\r\n\r\n$ g(x) \\equal{} \\int_0^x f(t)dt \\equal{} x \\int_0^1 f(tx)dt$\r\n\r\nthan $ g(2n) \\equal{} 2n \\int_0^1 f(2nt)dt$\r\n\r\nwe know that the fuction: $ h\\quad : \\quad t \\longrightarrow f(2nt)$ is odd and $ 1$-periodic fuction than $ g(2n) \\equal{} 0$ ...\r\n\r\nand thanks :)",
  "Solution_5": "For $ n\\in N$ and $ f(t)$ periodic with period $ 2$ \r\n$ g(2n) \\equal{}\\int_{0}^{2n}f(t)dt \\equal{}n\\int_{0}^{2}f(t)\\equal{}n\\int_{\\minus{}1}^{1}f(t)\\equal{}0  .$ (as $ f(x)$ is an odd function)\r\nNote for periodic function with period $ T$, we have $ \\int_{a}^{a\\plus{}T}f(t)dt \\equal{}\\int_{0}^{T}f(t)$\r\nI chosen $ a\\equal{}\\minus{}1$ in last integral"
}
{
  "Tag": [
    "function",
    "symmetry",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Prove or give a counterexample:\r\n\r\nIf an analytic function mapping the unit disk into itself has two fixed points in the unit disc, then it must be the identity function.",
  "Solution_1": "From Schwarz lemma it's clear that it's an automorphism of the unit disc (and in particular a M\u00f6bius transform). Such an automorphism maps the unit circle onto itself as well, and such a map cannot have two fixed points in the unit disc without being the identity function. One argument is that m\u00f6bius transforms preserve symmetry w.r.t. circles"
}
{
  "Tag": [
    "modular arithmetic",
    "calculus",
    "integration"
  ],
  "Problem": "If $ a,b,c\\in {\\rm Z}$ and\n\\[ \\begin{array}{l} {x\\equiv a\\, \\, \\, \\pmod{14}} \\\\\n{x\\equiv b\\, \\, \\, \\pmod {15}} \\\\\n{x\\equiv c\\, \\, \\, \\pmod {16}} \\end{array}\n\\]\n, the number of integral solutions of the congruence system on the interval $ 0\\le x < 2000$ cannot be\n\n$\\textbf{(A)}\\ 0  \\qquad\\textbf{(B)}\\ 1  \\qquad\\textbf{(C)}\\ 2  \\qquad\\textbf{(D)}\\ 3  \\qquad\\textbf{(E)}\\ \\text{None}$",
  "Solution_1": "[hide=\"Solution\"]if $ a$ is odd abd $ c$ is even, there is no solution. Solution set can be empty.\n\nLet $ a \\equal{} 2a'$ and $ c \\equal{} 2c'$, then the congruence\n\\[ \\begin{array}{l}{x\\equiv 2a'\\,\\,\\,\\left(mod\\, 7\\right)} \\\\\n{x\\equiv b\\,\\,\\,\\left(mod\\, 15\\right)} \\\\\n{x\\equiv 2c'\\,\\,\\,\\left(mod\\, 16\\right)}\\end{array}\n\\]\nhas a solution in the form $ x \\equal{} 7 \\cdot 15 \\cdot 16 \\cdot k \\plus{} m \\equal{} 1680k \\plus{} m$ (Chinese Remainder Theorem). The number of solution on the interval $ 0\\le x < 2000$ is $ 1$ or $ 2$.\nWe will show that the solution of the second congruence is also the solution of the first congruence. \nFrom the second congruence, $ x \\equal{} 1680k \\plus{} 7p \\plus{} 2a' \\equal{} 1680k \\plus{} 16q \\plus{} 2c' \\equal{} 1680k \\plus{} 15r \\plus{} b$.  $ RHS$ is even, so $ p$ must be even. $ x \\equal{} 1680k \\plus{} 14p' \\plus{} 2a' \\equal{} 1680k \\plus{} 16q \\plus{} 2c' \\equal{} 1680k \\plus{} 15r \\plus{} b$. If we rewrite the equation as congruence, we get\n\\[ \\begin{array}{l}{x\\equiv 2a'\\,\\,\\,\\left(mod\\, 14\\right)} \\\\\n{x\\equiv b\\,\\,\\,\\left(mod\\, 15\\right)} \\\\\n{x\\equiv 2c'\\,\\,\\,\\left(mod\\, 16\\right)}\\end{array}\n\\]\nSimilar approach for $ a$ and $ c$ are odd.\nSo the answer is $ C$ which is $ 3$.[/hide]"
}
{
  "Tag": [
    "geometry",
    "trapezoid",
    "symmetry",
    "rectangle",
    "AMC",
    "AMC 10"
  ],
  "Problem": "The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $ 9$ trapezoids, let $ x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $ x$?\r\n[asy]unitsize(4mm);\ndefaultpen(linewidth(.8pt));\nint i;\nreal r=5, R=6;\n\npath t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0);\nfor(i=0; i<9; ++i)\n{\ndraw(rotate(20*i)*t);\n}\ndraw((-r,0)--(R+1,0));\ndraw((-R,0)--(-R-1,0));[/asy]$ \\textbf{(A)}\\ 100 \\qquad\r\n\\textbf{(B)}\\ 102 \\qquad\r\n\\textbf{(C)}\\ 104 \\qquad\r\n\\textbf{(D)}\\ 106 \\qquad\r\n\\textbf{(E)}\\ 108$",
  "Solution_1": "[hide]Extend the legs of the trapezoids: by symmetry, they all meet.  There are 9 trapezoids, so there are 9 congruent angles in 180 degrees: hence the small angles are each 20 degrees.  By symmetry, the smaller angles of the trapezoids are equal and the sum of two of them and 20 is 180, so one of them is 80 degrees.  Thus the larger angle is this amount from 180, or 100 degrees.  $ \\boxed{A}$[/hide]",
  "Solution_2": "[hide]The arch is half of a regular $ 18$-gon. Thus the measure of each interior angle is $ 180 \\minus{} \\frac {360}{18} \\equal{} 160$ degrees. Because corresponding base angles of an isosceles trapezoid are congruent, each of the base angles on the outside of the arch is $ 160/2 \\equal{} 80$ degrees. Because opposite angles of an isosceles trapezoid are supplementary, the other two angles are $ 180 \\minus{} 80 \\equal{} 100$ degrees, or $ A$.\n\nWas it just me or was this problem ridiculously easy for #24. The answer choices were hardly even good distractors as far as I can tell. [/hide]",
  "Solution_3": "I was pretty annoyed with this problem. For some reason, I was forgetting to subtract 80 from 180. If they had put 80 as an answer, I bet less people would have gotton this right. How would someone get 106?!?!?!?!",
  "Solution_4": "@henry To stop clueless people from protractoring-it, of course :]. I didnt even realize I had a protractor until too late xD.\r\nHere's the stupid way that I did it\r\n[hide]\nSince the trapezoids were isosceles, drop their altitudes to cut them into rectangles and right triangles. Now, if we remove all the rectangles, the triangles will form a 180 degree 9-sided \"loop\" made up of 18 triangles. So, the smaller angle of each triangle is 10 degrees, and thus the larger angle of the trapezoid is 90+10=100 degrees.\n\nDidn't think to use my protractor.\nThen I looked back and realized that it was impossible to get 102 or 104 xD.\nAlso, I didn't even notice that the figure was half of a regular 18-gon.\n\n\n[/hide]",
  "Solution_5": "If you look closely at the diagram on the actual page it is slightly distorted - they tried to prevent people from protractoring it.",
  "Solution_6": "I was surprised how easy this problem was, as a #24 question. It's just isoceles triangles.",
  "Solution_7": "[quote=\"l Bob l\"]I was surprised how easy this problem was, as a #24 question. It's just isoceles triangles.[/quote]\r\n\r\nI agree. This was the easiest #24 on an AMC10 I'd seen in all my years. I literally solved this in 1 minute.",
  "Solution_8": "Same, there was nothing to this problem, :P #7,10 and others, while they were easy deffinitely much harder than #24, which any moronic sophomore can do...",
  "Solution_9": "[hide=Solution]Let the answer be $x.$ This means that $$\\frac{8(180)-2(180-x)}{8}=360-2x\\Leftrightarrow180-\\frac{180-x}{4}=360-2x\\Leftrightarrow135+\\frac{x}{4}=360-2x\\Leftrightarrow225=\\frac{7x}{4}\\Leftrightarrow x=\\frac{4\\cdot225}{7}=100\\Longrightarrow\\boxed{\\mathbf(A)}.$$[/hide]",
  "Solution_10": "$$180-\\frac{180-\\tfrac{180}{9}}{2}$$\n\n@below that is correct !",
  "Solution_11": "[quote=asdf334]$$180-\\frac{180-\\tfrac{180}{9}}{2}$$[/quote]\n\nand this is a problem 24",
  "Solution_12": "The shape made by the outer side lengths is half a 18 sided regular polygon so it has angle sum $\\frac{16 \\cdot 180}{2} = 1440$. The small interior angle is $180-x$, so $2(180-x) + 8(2(180-x))=18(180-x) = 1440 \\implies x = \\boxed{100}$.",
  "Solution_13": "[hide=Solution]Reflect the shape over the bottom line and notice that it is a regular $18$-gon. Each interior angle has measure $180-360/18=160,$ so $$2(180-x)=160\\implies 180-x=80\\implies x=\\boxed{100\\quad\\text{(A)}}.$$[/hide]",
  "Solution_14": "Because the interior polygon is a decagon, we know $$180(10 - 2) = 2(180 - x) + 8(360 - 2x)$$ which yields $x = \\boxed{100.}$"
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "Let's make a thread about good books that would be ideal Christmas presents.  Don't limit yourself to just mathematics books!",
  "Solution_1": "-America (The Book)\r\n-Anything Dilbert-related\r\n-Dave Barry stuff is good.\r\n-The Onion (Ad Nauseam, Our Dumb Century, etc.)\r\n-various math textbooks, although they are pretty expensive\r\n-Art of Problem Solving!\r\n\r\nProbably depends on people's interests (don't give anything Michael Savage-related to a liberal kid)",
  "Solution_2": "personally, I absolutely loved \"The Time Traveler's Wife\", by Audrewy Niffenegger\r\n\r\nWhat a great book.  I picked it up while walking around at the bookstore... read a paragraph and decided I had to have it.\r\n\r\nIt's really good  :)",
  "Solution_3": "I've been reading books by Steinbeck lately, my favorite so far is East of Eden.  I don't know if these are really great christmas ideas since the books aren't jolly and joyful.",
  "Solution_4": "[quote=\"SnowStorm\"]I've been reading books by Steinbeck lately, my favorite so far is East of Eden.  I don't know if these are really great christmas ideas since the books are jolly and joyful.[/quote] They are or they arent?",
  "Solution_5": "hmm, I don't think it's very jolly...",
  "Solution_6": "Whoops.. meant for it to say \"aren't\" sorry",
  "Solution_7": "It's about time this thread was made :lol: \r\n\r\nI'm one of those people that reads about 2 pages of a particular book, and then immediately decides whether it's boring or not.  Generally, I like books with an interesting concept, otherwise, I stick with Mathematical texts.\r\n\r\nOne of my favourite books is \"Battle Royale\", an English translated Japanese depiction of a modern fascist society (well, a particular event of that society).  There is a movie based on this, and I'd have to say that both movie and novel are well done.\r\n\r\nWhile I won't spoil the plot, I will say that YOU MUST READ THIS BOOK.  While not Christmas \"jolly\", the novel plot, ending, and concept are absolutely gratifying.",
  "Solution_8": "Anyone ever read any of Tom Clancy's books? I like Patriot Games and the Hunt for Red October\r\n\r\nAlso:\r\nLotR\r\nBartimaeus Trilogy\r\nHarry Potter\r\n\r\netc. etc. anything big that'll last.",
  "Solution_9": "The Revelation Space triolgy is great! Chasm city isn't in the triolgy, but it still takes place in the same universe. It's writing stlye is way different, an assasin's diary instead of a space opera.\r\n\r\nPink Floyd and White Stripes cd's are good. So are compilations of Queen, Led zepplin, and other bands.\r\n\r\nChess, math, programming, and electronics books are always good.\r\n\r\nMusic method books and song books for assorted instuments are great. I plan on getting a guitar  :) Some jazz lead sheets for pianists are great :lol: \r\n\r\nRubiks cubes, juggling balls, and other skills toys are great.\r\n\r\nI love giant puzzle compilations, especially computer programming contest compilations.\r\n\r\nIf you wanted to delve deeper into nerdery, MtG and D&D are options, I guess.]\r\n\r\nMicrochip programming is fun. So are burnable gameboy carts. Programming gaming consoles= fun. If you wanted to avoid specs or installing linux on to some things, the Xgamestation is fun.\r\n\r\nNerf and water guns are fun, a staining agent like kool-aid makes it even more fun.\r\n\r\nOh, an external hard drive, more RAM, and a new GPU are also high on my list.",
  "Solution_10": "Les Miserables is a really good book, its a depiction of French society of back then and some stuff, but really really long, although quite exciting....\r\n\r\nAOPS and AACOPS (art & craft of problem solving) are good books also, ith addition of USSR Math Olympiad Book. which has some pretty cool problems..."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "system of equations",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "\\[ \\sum\\limits_{k \\equal{} 0}^n {k^p } \r\n\\]\r\n\r\n\r\nbernoulli, but I can't get it...\r\nthx",
  "Solution_1": "better state it as $ \\sum_{k \\equal{} 1}^n \\; k^p,$ where $ p\\in\\mathbb Z$\r\n\r\n\r\nsee\r\nhttp://www.trans4mind.com/personal_development/mathematics/series/sumsBernoulliNumbers.htm\r\n\r\n\r\nalso see equation $ (13)$ \r\nhttp://mathworld.wolfram.com/PowerSum.html",
  "Solution_2": "The name for the general formula is [url=http://mathworld.wolfram.com/FaulhabersFormula.html]Faulhaber's formula[/url].",
  "Solution_3": "An outline of the proof:\r\n\r\nWrite down the polynomial formula $ \\sum_{k\\equal{}1}^n k^p\\equal{}\\frac1{p\\plus{}1}n^{p\\plus{}1}\\plus{}\\sum_{j\\equal{}0}^{p\\minus{}1}a_jn^{p\\minus{}j}$. Take its difference, and expand the $ (n\\plus{}1)^r\\minus{}n^r$ terms on one side with the binomial theorem. You get $ n^p$ from the $ \\frac1{p\\plus{}1}n^{p\\plus{}1}$ term, and the other terms should balance to zero. That leaves a system of equations in the $ a_j$; insert the appropriate factors to turn it into the defining equation of the Bernoulli numbers $ (e^x\\minus{}1)\\sum_{j\\equal{}0}^{\\infty}B_nx^n\\equal{}x$."
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "Vanessa has N pieces of wedding cake.  If she arranges them in groups of 5, she winds up with 4 left over.  If she arranges the pieces in groups of 8, she ends up with 7 left over.  If Vanessa decides to arranges the pieces of cake in groups of 20, how many pieces will be left over?",
  "Solution_1": "[color=red][quote]Vanessa has N pieces of wedding cake. If she arranges them in groups of 5, she winds up with 4 left over. If she arranges the pieces in groups of 8, she ends up with 7 left over. If Vanessa decides to arranges the pieces of cake in groups of 20, how many pieces will be left over?[/quote]\r\n\r\nANSWER: 19\r\n\r\nwell, since all multiples of 5 end with 0 or 5 in the units digit N must end with a 4 or 9. all multiples of 8 end in 8, 6,4,2, and 0. N can end in 5,3,1,9. or 7. one can see that  9 is shared between the two. so we can conclude that N must end in a 9. if N gives a remainder of 7 when divided by 8 one can tell that N is one less than a multiple of 8. as long as it ends in a 9. so 9+1=?, anyone, anyone, okay i can see that you dont know, the world these days, okay, 9+1=10. so the multiple of 8 must end in a 0. i just randomly picked 80 as my number. so 80-1=79. N can be many different numbers like 39. but 80 came to my head before 40. were not done with the question yet. we still have to find how many pieces will be left over if we Vanessa arranges the pieces of cake in groups of 20. its pretty straight forward from here. but ill tell anyhow. 79/20=3r19. the 19 is what we are looking for here, not the 3.  wow. that took some typing to explain. my way may not be the fastest, but it didnt take me more than 30 seconds to solve. it just took some time to type. lool.  i remember in this year's state mathcounts there was a question like this except it was like n^2 divided by a number gave a remainder of something and n^3 divided by a number gave a remainder of something else, find n. anyways. ummmm. hope that helped. i am wanting to know if there is a faster way. oh yeah, that question that i mentioned that came from the state competition was also a countdown round question in nationals.  ummmm. and if i dont see you again good morning good evening and good night. nuff said. wait. the answer is 19. just so you wont have to look through my passage again. now anwho. nuff said",
  "Solution_2": "[hide]\n\nn=5m + 4 = 5m - 1\nn also = 8x + 7 = 8x-1\n\nthus n+1=5m=8x.  the smallest integer n satisfying this is 39 (n+1)=40.\n\nupon division by 20, the remainder is 19.\n[/hide]"
}
{
  "Tag": [
    "modular arithmetic",
    "number theory",
    "number theory unsolved"
  ],
  "Problem": "p is a prime number\r\n$4^{p-1}-1$ is $0$ (mod $p$) is easy to prove by Fermat's little theorem\r\nbut I don't know how about in mod$p^{2}$",
  "Solution_1": "It's not 0...\r\n\r\np=3 => $4^{2}-1 = 15 \\equiv 6 \\pmod 9$"
}
{
  "Tag": [
    "trigonometry",
    "logarithms",
    "function",
    "limit",
    "calculus",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "can you help me out these problems\r\n\r\n1) $ \\tan{x}-\\sin{x}> \\frac{x^3}{2} (0<x<\\frac{\\pi}{2})$\r\n2) $ \\frac{1}{\\sin^{2}{x}}< \\frac{1}{x^2}+1-\\frac{4}{{\\pi}^2} (0<x<\\frac{\\pi}{2})$\r\n3) $ \\sin{2x}<\\frac{2}{3{x}-x^2}(0<x<\\frac{\\pi}{2})$\r\n4) $ \\frac{2}{{e}}<a^{\\frac{a}{1-a}}+a^{\\frac{1}{1-a}} (0<a<1) $\r\n5) $ \\frac{\\sin{x}}{\\sin{y}}<\\frac{x-\\frac{x^3}{6}}{ y-\\frac{y^3}{6}} (0<x<y<\\sqrt{6})$\r\n6) $ \\frac{1}{x}\\ln{(a^x+b^x)}> \\frac{1}{y}\\ln{(a^y+b^y)} (a,b=const; a>0;b>0;0<x<y) $",
  "Solution_1": "1) What about expanding [tex]\\mbox{tan} x - \\mbox{sin} x[/tex] into a serie :)",
  "Solution_2": "yeah,i have try it out but there is no result",
  "Solution_3": "hmm you are in radian right?",
  "Solution_4": "Hello monster, I can solve the ineq 5. \r\n Consider the function $ f(x)= \\frac{x-\\frac{x^3}{6}}{sin(x)}$\r\n $ f'(x)= (1/6)\\frac{ 6sin(x)-3sin(x)x^2-6xcos(x)+x^3cos(x)}{1-cos^2(x)}$\r\n Consider the function $ g(x)= 6sin(x)-3sin(x)x^2-6xcos(x)+x^3cos(x)$\r\n Actually if $ g(x) < 0$ then $ f'(x)<0$ give us the desired result.\r\n     Now prove $ g(x) <0$\r\n $ g'(x) = -x^3sin(x) < 0$ => $ g(x)<g(0)=0$ (q.e.d)     \r\n\r\nTh\u1ea5y hay ko?  :D  :D  :D",
  "Solution_5": "And this is the solution of problem 6. It's a nice problem.\r\nSince a and b are symmetric we can assume that $ b \\geq a$.\r\nBy putting $ c=\\frac{b}{a}$ we have that \r\n$ \\frac{1}{x}ln(a^x+b^x)= ln(a) + \\frac{1}{x}ln(1+c^x)$ with $ c \\geq 1$\r\nTo get the resulf, all we must do now is to show that $ f(x)= \\frac{1}{x}ln(1+c^x)$ is strictly decreasing.\r\n$ f'(x)= \\frac{xc^xln(c)-(1+c^x)ln(1+c^x)}{(1+c^x)x^2}$\r\nFor the decrease of f(x), we must show that $ xc^xln(c)-(1+c^x)ln(1+c^x)<0$\r\nEquivalent to $ (c^x)^{c^x}<(1+c^x)^{1+c^x}$.\r\nThis one is true since $ c^x>1$ ( c>=1,x>0)",
  "Solution_6": "For the problem 1 ( i think it's the hardest problem), see here :(the 4th post) \r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=273401#273401",
  "Solution_7": "This is my sol for the problem 4, it's a boring one.\r\n$ f(x)=\\frac{1}{x^2}-\\frac{1}{tan^2(x)}-\\frac{4}{\\pi^2}$\r\n$ f'(x)<0$ $\\Leftrightarrow$  $ g(x)=\\frac{tan(x)}{\\sqrt[3]{1+tan^2(x)}}-x>0$\r\nOf course if $ g'(x)>0$ then $ g(x)>g(0)=0$. I'd prove $ g'(x)>0$\r\n$ g'(x)>0 \\Leftrightarrow 3+tan^2(x)>3(1+tan^2(x))^{\\frac{1}{3}}$\r\nEquivalent to $ (3+tan^2(x))^3>27+27tan^2(x)$ ( this is true).\r\nThus, $ f'(x)<0$, so $ f(x)> \\lim_{x \\to \\frac{\\pi}{2}}f(x)=0$ (q.e.d)",
  "Solution_8": "Sorry, it's problem 3",
  "Solution_9": "try them using calculus\r\nI think this is an appropriate way!\r\nThey look good!!!! :)  :D",
  "Solution_10": "And finally the last is solved.\r\nBy putting $ x=a/(1-a)$ the ineq is equivalent to $ e>(1+\\frac{1}{x})^x(1+\\frac{1}{2x+1})$\r\nor $ 1> xln(1+\\frac{1}{x})+ln(1+\\frac{1}{2x+1})=f(x)$\r\nThis one can be easily proved by diferentiating f(x) to the $ 2^{th}$ derivative.\r\nCheer! All of them are defeated.",
  "Solution_11": "yeah, i have solution for problem 4\r\nby taking logarithm both handsides of the ineq, we have\r\n$ \\ln{2}-1<\\frac{a}{1-a}\\ln{a}+\\ln{a+1} $\r\nequivalent to \r\n$ a(\\ln{2}+\\ln{a})+(1-a)\\ln{a+1}-a>\\ln{2}-1 $\r\nput $ f(a)=a(\\ln{2}+\\ln{a})+(1-a)\\ln{a+1}-a $\r\nwe have ineq  $ f(x)>f(1) $\r\nneed to prove that $ f'(a)=g(a)=\\ln{2}+\\ln{a}-\\ln{a+1}-\\frac{1-a}{1+a}<0   (0<a<1) $\r\n we have $ g'(a)=\\frac{1}{a}-\\frac{1}{1+a}+\\frac{1}{{(1+a)}^2}$ \r\n                $     =\\frac{a+3}{a*(a+1)^2}>0$ (0<a<1)           \r\n         so $ f'(a)=g(a)<g(1) =0 $\r\n           $\\Longleftrightarrow f(x)>f(1) \\forall a\\in]0,1[$\r\nproblem solved. i feel very happy  :rotfl:  :lol:",
  "Solution_12": "yeah,i have some more problems\r\n*) $ \\frac{1}{x-y}(\\ln{\\frac{x}{1-x}}-\\ln{\\frac{y}{1-y}})>4         (0<x,y<1;x_is_not_equal_to_y) $\r\n*) $ \\frac{x^{2}y}{z}+\\frac{y^{2}z}{x}+\\frac{z^{2}x}{y}>x^2+y^2+z^2           (x>y>z>0) $\r\n*) $ \\cos({x+y})<\\frac{y\\sin{x}}{x\\sin{y}} (x,y>0; x+2y<\\frac{5\\pi}{4}) $\r\n   :P  :P  :P try out to solve it",
  "Solution_13": "Yes, now i try to solve the $ 2^{nd}$ problem :\r\n   For $ x>y>z>0$ we have $ (x-y)(y-z)(x-z)(xy+yz+zx) >0$ \r\n       $ %Error. \"Loghrightarrow\" is a bad command.\n\\sum_{cyc}\\frac{x^2y}{z}>\\sum_{cyc}\\frac{x^2z}{y}$(1)\r\nAnd by Cauchy-Schwarts: $ (x^2+y^2+z^2) = (\\sum_{cyc}x\\sqrt{\\frac{y}{z}}x\\sqrt{\\frac{z}{y}}) \\leq \\sqrt{(\\sum_{cyc}\\frac{x^2y}{z})(\\sum_{cyc}\\frac{x^2z}{y})}$(2)\r\n(1) and (2) lead us directly to the answer.  :lol:",
  "Solution_14": "Problem 1 is an easy Langrange exercise. Apply Langrange theorem for the function $ ln(\\frac{x}{1-x})$ and then use AM-GM to get the resulf.",
  "Solution_15": "I have solved the  problem 1 but not follow \"larange\", here is my solution\r\nAssume that $ x>y$ , so we have $ x-y>0 $\r\nMultiply both handside of the ineq with $ x-y$, we have\r\n$\\ln{\\frac{x}{1-x}}-\\ln{\\frac{y}{1-y}}>4(x-y)$\r\nequivalent to $\\ln{\\frac{x}{1-x}}-4x>\\ln{\\frac{y}{1-y}}-4y $\r\nput $ f(t)=\\ln{\\frac{t}{1-t}}-4t $\r\nwe need to prove that $ f(x)>f(y) (1>x>y>0) $\r\nso to get the result,we must show that $ f(t)=\\ln{\\frac{t}{1-t}}-4t $ is strictly increasing in the open interval $ ]0,1[$\r\nand it is nothing difficult to show that $ f'(t)>0 \\forall t\\in ]0,1[ $\r\n         --------------$ qua trum $-------------------",
  "Solution_16": "Ok, I think your \"quatrum\" means you need a one-line solution. It's here:\r\n$ \\frac{1}{x-y}(\\ln{\\frac{x}{1-x}}-\\ln{\\frac{y}{1-y}}) = \\frac{1}{c(1-c)}>4$ (0<c<1)",
  "Solution_17": "the last problem has been defeated\r\nwe have that $ x+2y<\\frac{5\\pi}{4} $ so $ y<\\frac{5\\pi}{8}$\r\n=>$\\sin{y}>0 $\r\nthe ineq equivalent to $x\\sin{y}\\cos({x+y})-y\\sin{x}<0 $\r\n                       <=> $ x(\\sin({x+2y})-\\sin{x})-2y\\sin{x}<0 $\r\n                       <=> $ x\\sin({x+2y})-\\sin{x}*(x+2y)<0 $\r\n                       <=> $ \\frac{\\sin({x+2y})}{(x+2y)}<\\frac{\\sin{x}}{x} $\r\nput $ f(t)=\\frac{\\sin{t}}{t}---- and ------k=x+2y $\r\nwe need to prove that \r\n               $ f(k)<f(x) -----(x,k>0;0<x<k<\\frac{5\\pi}{4}) $\r\nand it is not difficult to prove that. hehehehe :D  :D  :D\r\n-------------------quatrum-----------------------",
  "Solution_18": "Well done! This's a nice proof. I appreciate it!",
  "Solution_19": "Good thing Valentin didn't see this, otherwise he'd have said:\r\n\r\n\"monster_of_god, post them in separate topics! This is now locked.\"\r\n\r\n :lol:  :lol:  :lol:",
  "Solution_20": "yes,pray for god  :D  :D  :D",
  "Solution_21": "Our work is almost, but not completely, done, since we missed a problem ( the problem about $ sin(2x)$. We should try to solve it.",
  "Solution_22": "Finally I complete it. It's equivalent to $ f(x)=\\frac{1}{sin(2x)}-3x+x^2>0$.\r\n$ f'(x)=-2\\frac{cos(2x)}{sin^2(2x)}-3+2x$ (1)\r\n$ f\\\"(x)=8\\frac{cos^2(2x)}{sin^3(2x)}+\\frac{4}{sin(2x)}+2>0$ (2)\r\n(2) shows that f is a strictly convex function. So its graph is above its tangents.\r\nWe take two tangents $ D_1$,$ D_2$ at $ x=\\frac{\\pi}{4}$ and $ x=\\frac{\\pi}{6}$.\r\nThey meet each other at an intersection point. By computation we can show this point is above the x-axis. So the gragh of f is above the x-axis, and the resulf follows.\r\n\r\n\"Who's the god's monster killer? He's the master of darkness.\" :D  :D  :D[/b]"
}
{
  "Tag": [],
  "Problem": "$a>0,b>0,c>0, 0\\leq k\\leq8.$ Prove that \r\n$\\sqrt{1+\\frac{2ka}{b+c}}+\\sqrt{1+\\frac{2kb}{c+a}}+\\sqrt{1+\\frac{2kc}{a+b}} \\ge3\\sqrt{1+k}$",
  "Solution_1": "Could you give your solution . It is unsolved for four days. thanks",
  "Solution_2": "[quote=\"silouan\"]Could you give your solution . It is unsolved for four days. thanks[/quote]\r\nLet $a+b+c=3.$ then:\r\n$\\sum\\sqrt{1+\\frac{2ka}{b+c}}\\geq3\\sqrt{1+k}\\Leftrightarrow$\r\n$\\sum(\\sqrt{1+\\frac{2ka}{3-a}}-\\frac{3k}{4\\sqrt{1+k}}(a-1)-\\sqrt{1+k})\\geq0\\Leftrightarrow$\r\n$\\sum(4\\sqrt{(1+k)(1+\\frac{2ka}{3-a})}-(3ka+k+4))\\geq0\\Leftrightarrow$\r\n$\\sum\\frac{k(a-1)^2(3ka+8-k)}{(3-a)(4\\sqrt{(1+k)(1+\\frac{2ka}{3-a})}+3ka+k+4)}\\geq0.$ :)"
}
{
  "Tag": [],
  "Problem": "this is an easy problem form the first round of 1994 IRMO.\r\nhow many natural solutions does the following equation has?\r\n$2^x+1=3^y$",
  "Solution_1": "This can be solved by following method.\r\n\r\nIf $x \\geq 4$ from $ord_{16}3=4$,  We get $4|y$.\r\n\r\nso, $3^y = 1 (mod 5)$. This means that $2^x=0(mod 5)$. contradiction!\r\n\r\nTherefore $x< 4$ \r\n\r\nSo, possible solution is $(x,y)=(1,1),(3,2)$\r\n\r\n$Q.E.D.$",
  "Solution_2": "there's another way, too. :D",
  "Solution_3": "Maybe this solution is a bit more complicated...\r\n\r\nFirst of all, $2^x+1 \\equiv (-1)^x+1 \\equiv 0 (\\mod 3)$, so $x$ must be odd. Then we have $2^x+1=3^y$ and because $x$ is odd, we can write $(2+1) \\cdot (2^{x-1}-2^{x-2}+...-2+1)=3^y$, if $x \\neq 1$. (If $x=1$, we get one solution, that is $(1,1)$). This is equivalent to $2^{x-1} - 2^{x-2} + ... - 2 + 1 = 3^{y-1}$. That means $3^{y-1} \\equiv (-1)^{x-1} - (-1)^{x-2} + ... - (-1) + 1 \\equiv x \\equiv 0 (\\mod 3)$, so $3|x$. That means we can write $x=6k+3$. Let's put this in the first equation: $2^{3 \\cdot (2k+1)} + 1 = 3^y$. Again, we can write $2^{2k} - 2^{2k-1} + ... - 2 + 1 \\equiv 2k+1 \\equiv 0 (\\mod 3)$, because there is no solution for $y=1$. So $3|2k+1$. Thus we can write $2k+1=3l$, so $x=18l+9$. If $y=2$, we get a solution $(3,2)$. If $y \\neq 2$, we can repeat the process we were doing before, and get $2l+1=3t$, so $x=54t+27$. But that means $t<l$ and by the method of infinite descent, there is no solution.\r\nThus the only two solution are $(1,1)$ and $(3,2).$",
  "Solution_4": "[quote=\"Chang Woo-JIn\"]This can be solved by following method.\n\nIf $x \\geq 4$ from $ord_{16}3=4$,  We get $4|y$.\n\nso, $3^y = 1 (mod 5)$. This means that $2^x=0(mod 5)$. contradiction!\n\nTherefore $x< 4$ \n\nSo, possible solution is $(x,y)=(1,1),(3,2)$\n\n$Q.E.D.$[/quote]\r\n\r\nWhat is ord?",
  "Solution_5": "Maybe $ord$ is short for $order$, in which case $ord_{a}p$, where $p$ is prime, would be the the lowest $n$ for which ${a^{n} \\equiv 1 \\mod{p}}$.  But I think Chang means something different, because by this definition, $ord_{16}3=2$ not 4.",
  "Solution_6": "But isn't then $odr_{16}3=1$?",
  "Solution_7": "hmm.. Sorry for My Symbol. But This symbol is universally used.\r\n\r\n$ord_{m} a$ means the value $t$ s.t lowest power of $a$ satisfies $a^t =1 (mod m)$\r\n\r\n\r\n\r\n :D",
  "Solution_8": "Oh, sorry, I just had the arguments the other way around.  Thanks for clarifying."
}
{
  "Tag": [
    "limit",
    "calculus",
    "derivative",
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Find \r\n\\[ \\lim_{x\\to 0}\\lim_{n\\to \\infty} (e^{\\sqrt{x}}+e^{-\\sqrt{x}})^{(n)}  \\]\r\nWhere $f^{(n)}$ is the $n$-th derivative of $f$.\r\n\r\nThe answer was that the limit is 0. But I even cannot understand the problem.. :?",
  "Solution_1": "$f(x)=e^{\\sqrt{x}}+e^{-{\\sqrt{x}}}=2\\sum_{k=0}^{\\infty}\\frac{x^k}{2k!}\\Longrightarrow f^{(n)}(x)=2\\sum_{k=n}^{\\infty}\\frac{k(k-1)...(k-n+1)x^{k-n}}{2k!}\\Longrightarrow\\\\ \\\\ \\lim_{x\\to 0}\\lim_{n\\to \\infty} (e^{\\sqrt{x}}+e^{-\\sqrt{x}})^{(n)}= \\lim_{x\\to 0}\\lim_{n\\to \\infty}f^{(n)}(x)=\\\\ \\\\ \\lim_{x\\to 0}\\lim_{n\\to \\infty}2\\sum_{k=n}^{\\infty}\\frac{k(k-1)...(k-n+1)x^{k-n}}{2k!}=\\lim_{n\\to \\infty}\\lim_{x\\to 0}2\\sum_{k=n}^{\\infty}\\frac{k(k-1)...(k-n+1)x^{k-n}}{2k!}=\\\\ \\\\ \\lim_{n\\to \\infty}\\frac{k(k-1)...(k-n+1)}{2k!}=0$\r\n\r\n\r\n\r\n\r\n ????  :?",
  "Solution_2": "I do not understand what is the limit of thr n-th derivative$\\lim f^{(n)}$... :?  I think that, this limit is a function, name it $g$, and we are suposed to calculate $\\lim_{x\\to 0} g(x)$.. \r\nI also thought to your aproach but I wonder why can you interchange the limits...  :?",
  "Solution_3": "[quote=\"xirti\"]I do not understand what is the limit of thr n-th derivative$\\lim f^{(n)}$... :?  I think that, this limit is a function, name it $g$, and we are suposed to calculate $\\lim_{x\\to 0} g(x)$.. \nI also thought to your aproach but I wonder why can you interchange the limits...  :?[/quote]By Ramanujan's argument, for $x > 0$,\r\n$\\frac{1}{2}f^{(n)}(x) = \\sum_{k=n}^\\infty \\frac{k!}{(k-n)!(2k)!}x^{k-n} = \\sum_{j=0}^\\infty \\frac{(j+n)!}{j!(2n+2j)!}x^j$.\r\nYou are right in that you must first find the limit of these functions as $n \\to \\infty$. For fixed $x$ therefore \r\n$0 \\le \\frac{1}{2}f^{(n)}(x) \\le \\frac{1}{2n}\\sum_{j=0}^\\infty \\frac{1}{j!}x^j$\r\nwhich should point you in the right direction :)\r\n\r\nThe function $f$ is the [url=http://mathworld.wolfram.com/Mittag-LefflerFunction.html]Mittag-Leffler function[/url] $E_2$ up to a factor 2.",
  "Solution_4": "thank you for clarifying this problem :lol:"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "percent"
  ],
  "Problem": "If an edge of a cube is doubled, its surface area is increased by what percent?",
  "Solution_1": "[quote=\"236factorial\"]If an edge of a cube is doubled, its surface area is increased by what percent?[/quote]\r\n\r\n[hide]Say the original cube was a unit cube. Then the surface area would be $6$\n\nIf each edge is doubled, the surface area would turn to $2*2*6=24$. \n\nSo it increased by $300\\%$[/hide]",
  "Solution_2": "Are you sure it was [i]increased[/i] by 400%?  :P",
  "Solution_3": "Since it increased by 18, wouldn't it increase by $300\\%$?  :?",
  "Solution_4": "[quote=\"Drunken_Math\"]Since it increased by 18, wouldn't it increase by $300\\%$?  :?[/quote]\r\n\r\nYeah, but you put 400...."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let be $ a,b,c\\in \\left[\\frac{1}{3} , 3\\right]$ . Show that :\r\n\r\n$ \\frac {a}{a \\plus{} b} \\plus{} \\frac {b}{b \\plus{} c} \\plus{} \\frac {c}{c \\plus{} a}\\ge \\frac {7}{5}$",
  "Solution_1": "[quote=\"alex2008\"]Let be $ a,b,c\\in \\left[\\frac {1}{3},3\\right]$ . Show that :\n\n$ \\frac {a}{a \\plus{} b} \\plus{} \\frac {b}{b \\plus{} c} \\plus{} \\frac {c}{c \\plus{} 1}\\ge \\frac {7}{5}$[/quote]\r\nAre you sure that the last term is not $ \\frac{c}{c\\plus{}a}$? :huh:  :oops:",
  "Solution_2": "[quote=\"alex2008\"]Let be $ a,b,c\\in \\left[\\frac {1}{3} , 3\\right]$ . Show that :\n\n$ \\frac {a}{a \\plus{} b} \\plus{} \\frac {b}{b \\plus{} c} \\plus{} \\frac {c}{c \\plus{} a}\\ge \\frac {7}{5}$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=175423"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra unsolved"
  ],
  "Problem": "Show that the only polynomial of odd degree satisfying $p(x^2-1) = p(x)^2 - 1$ for all $x$ is $p(x) = x$",
  "Solution_1": "The generalization is here:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=247124#p247124",
  "Solution_2": "[incomplete solution]\nif $P(x)\\in \\mathbb C[x]$\n              $P(x^2-1)=P(x)^2-1$\nset $-x$:$P(x^2-1)=P(-x)^2-1$ \nhence$P(x)^2=P(-x)^2$\nthen $P(x)=-P(-x)$ or $P(x)=P(-x)$  .\nThe second result isn't true (because $P$ has odd degree)\nand so $P(-x)=-P(x)$ \nhence there is $ Q(x)\\in \\mathbb C[x]$ that $P(x)=xQ(x^2)$\n.\n.\n.\n.\n.\n.",
  "Solution_3": "First, $P(x)^2=P(-x)^2$ for all $x$. So, either $P(x)=P(-x)$ infinitely often (io) or $P(x)=-P(-x)$ io. The former case is impossible: as the degree of $P$ is odd, then the sign of $P(x)$ and $P(-x)$ are different for sufficiently large $x$. So, the latter case holds; in particular $P(x)+P(-x)=0$. Thus $P(0)=0$, $P(1)=1$ and $P(-1)=-1$. Next, define $a_n$ such that $a_0=0$, $a_{n+1}=\\sqrt{a_n+1}$, $n\\ge 0$. Clearly $a_n>1$ for all $n\\ge 2$. We show $P(a_n)=a_n$ for all $n$. The case $n=0,1$ is clear, let $P(a_n)=a_n$. Taking $x=a_{n+1}$, we find $P(a_{n+1})^2 = a_n+1$, so $P(a_{n+1})=\\pm a_{n+1}$. Let $P(a_{n+1})=-a_{n+1}$. Then, taking $x=a_{n+2}$, we get $P(a_{n+2})^2 = 1-a_{n+1}<0$, a contradiction. Hence, $P(a_n)=a_n$ for all $n$. From here, we get $P(x)-x$ has infinitely many roots (any $a_i$), so $P(x)=x$ for all $x$."
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "look, i understand that the thing im trying to prove is false but for some reason i can't find the mistake in my proof, and im really confused by mods.\r\n\r\ni was just sitting one day and i consider if all pythagorean triples were thought of in mod 5. i then came to this weird conclusion that why couldn't 4,5,and 6 be a pythagorean triple? i understand that numerically this is preposterous, but this is my reasoning:\r\n\r\n$ 4^{2}\\plus{}5^{2}\\equal{}6^{2}$\r\ntake the mod 5 of each part of the equation, you get \r\n$ 4^{2}\\equiv 1\\pmod{5}$\r\n$ 5^{2}\\equiv 0\\pmod{5}$\r\n$ 6^{2}\\equiv 1\\pmod{5}$\r\n\r\nwe then get\r\n$ 1\\plus{}0\\equal{}1\\pmod{5}$...which is true?\r\n\r\ni understand im wrong, but can anybody point me out where my proof goes askew? \r\n\r\nthanks!",
  "Solution_1": "You have proved that $ 4^{2}\\plus{}5^{2}\\equiv{}6^{2}\\pmod{5}$, but not $ 4^{2}\\plus{}5^{2}\\equal{}6^{2}$\r\n\r\nThink of mods more as useful restriction that allow us to pinpoint things such that all Pythagorean Triples are multiples of 5, but cant all ways assume the opposite because just because something follows [i]a[/i] restriction doesn't mean it follows [i]all[/i].",
  "Solution_2": "Basically:\r\n\r\n$ A\\equal{}B\\Rightarrow A\\equiv B (modn)$ is true\r\n\r\nbut\r\n\r\n$ A\\equiv B ( modn)\\Rightarrow A\\equal{}B$ is not.\r\n\r\nLike $ 1$ and $ 4$ they both have residue class $ 1$ modulo $ 3$ but they are not equal.",
  "Solution_3": "thanks guys!\r\ni see my mistake. this was just a failed attempt of mine at solving a pythagorean triple proof using mods. :blush:  :lol:"
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "Let $f(x)$ be a monic polynomial of degree $n$ with real coefficients. Prove that \r\n\\[ \\max_{0\\le i \\le n, i\\in {\\mathbb Z}} |f(i)| \\geq \\frac{n!}{2^n}.  \\]",
  "Solution_1": "I think this math can be solved by using lagrange",
  "Solution_2": "I think you need to show how you would do this using Lagrange :)",
  "Solution_3": "[hide=\"hint\"]\nThis might help you http://en.wikipedia.org/wiki/Forward_difference_operator\n\nIn particular, what's the n-th forward difference of a polynomial of degree n?\n[/hide]"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "search"
  ],
  "Problem": "Prove that there exists no non-constant polynomial $ f(x)$ with integer coefficients with $ f(x)$ prime for all $ x$.",
  "Solution_1": "This question has been asked multiple times in various forums. As JBL would say, \"Search the forums.\"\r\n\r\n[hide=\"Though to answer\"]\n[quote=\"[i]Excursions in Number Theory[/i] by Ogilvy and Anderson\"]\nSuppose such a polynomial exists:\n$ Y(x) \\equal{} a_{0} \\plus{} a_{1}x \\plus{} a_{2}x^{2} \\plus{} ... \\plus{} a_{n}x^{n}$.\n$ \\dots$\nSince it is assumed to produce primes for all $ x$, select some $ x$, say $ x \\equal{} b$, that yeilds prime $ p$:\n$ Y(b) \\equal{} p$\n[/quote]\nThen to paraphrase:\nFollowing from the assumption $ Y(b \\plus{} mp) \\ , \\ m \\in \\mathbb{Z}$ must also be prime.\nFinding its value\n$ Y(b \\plus{} mp) \\equal{} a_{0} \\plus{} a_{1}(b \\plus{} mp) \\plus{} a_{2}(b \\plus{} mp)^{2} \\plus{} ... \\plus{} a_{n}(b \\plus{} mp)^{n}$.\nMultiplying that shows $ p | Y(b \\plus{} mp)$. Contradiction. $ \\square$.\n[/hide]",
  "Solution_2": "If you know greek http://www.mathlinks.ro/viewtopic.php?t=126749\r\nIf you dont http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html",
  "Solution_3": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=88190084&t=180433]A topic where this was recently discussed[/url]"
}
{
  "Tag": [
    "factorial",
    "induction"
  ],
  "Problem": "Find the sum of $1\\cdot 1!+2\\cdot 2!+3\\cdot 3!+\\cdots+(n-1)(n-1)!+n\\cdot n!$, where $n!=n(n-1)(n-2)\\cdots2\\cdot1$.",
  "Solution_1": "[hide]\nnote that $(n+1)!=(n!)(n+1)=(n)(n!)+n!$\nso it telescopes to...\n\n$%Error. \"dsiplastyle\" is a bad command.\n\\sum_{k=1}^{n}k*k!=(n+1)!-n!$\n[/hide]",
  "Solution_2": "maokid7\r\n[hide]Don't you mean $(n+1)!-1$?[/hide]",
  "Solution_3": "[hide=\"Kindda obvious hint\"]$1\\cdot 1!+2\\cdot 2!+3\\cdot 3!+\\cdots+(n-1)(n-1)!+n\\cdot n!=$\n$1^2 \\cdot 0!+2^2 \\cdot 1! + 3^2 \\cdot 2! + \\cdots +(n-1)^2 \\cdot (n-2)!+n^2 \\cdot (n-1)!$\nWhich telescopes to...\n$\\sum_{k=1} ^ {n} = k^2 \\cdot (k-1)!$[/hide]\r\n\r\nI'm not sure if it helps a lot but I think its an easier way to look at it..........",
  "Solution_4": "[quote=\"sthebedagain\"]maokid7\n[hide]Don't you mean $(n+1)!-1$?[/hide][/quote]\r\n\r\nmaokid told us that it telescopes like this; $(n+1)! - n!$ so the final answer would be $(n+1)!-1$ like you said. :lol:",
  "Solution_5": "Checking your work is a problem solving tool that should not be neglected. :D\r\n\r\n[hide=\"Solution\"]Note that for any positive integer $n,$ \n\\begin{eqnarray*}n\\cdot n!+(n-1)\\cdot(n-1)!&=&(n^{2}+n-1)(n-1)!\\\\ &=& (n+1)!-(n-1)!.\\end{eqnarray*}\n\nWe must now consider when $n$ is odd or even.\n\nIf $n$ is odd, \\[(n+1)!-(n-1)!+(n-1)!-(n-3)!\\cdots-2!+2!-0!.\\] If $n$ is even, \\[(n+1)!-(n-1)!+(n-1)!-(n-3)!\\cdots-3!+3!-1!.\\] In both cases, the expression telescopes into $(n+1)!-1.$[/hide]",
  "Solution_6": "$(n+1)!-1$ is what I got for the final answer. So I agree with everyone else.",
  "Solution_7": "you could also use induction?\r\n\r\n[hide]\n\nlooking at $n=1,2,3,4$ we get $1,5,23,119$ respectively so we see a pattern\n\n\nwe conjecture $(n+1)!-1$ is the sum, and now we use induction\n\nbase case is true and we assume $n$ case is true, now we look at $n+1$ case, so we have:\n\n$1\\cdot 1!+2\\cdot 2!+\\cdots+n\\cdot n!+(n+1)\\cdot (n+1)! = (n+1)!-1+(n+1)\\cdot (n+1)! = (n+2)!-1$\n\nso our induction is complete\n\n[/hide]",
  "Solution_8": "Yeah, solving directly and trought induction will both bring the same solution. I think induction is harder to see though.",
  "Solution_9": "Induction is a lot easier to see, in my opinion, especially if you approached this problem by doing numerical examples.",
  "Solution_10": "Hello! My solution isn`t nice.",
  "Solution_11": "If I am not wrong, your download does not give us a solution. All it says is that by induction .... is true.",
  "Solution_12": "Good catch Varun; his download is not a proof, merely two lines stating that the sum is ${ (n + 1)! - 1}$.\r\n[hide=\" A real proof\"]\nWe notice that $ n \\cdot n! = (n + 1)! - n!$\n$ \\implies \\displaystyle\\sum_{k = 1}^{n}{k \\cdot k!} = \\displaystyle\\sum_{k = 1}^{n}{(k + 1)! - k!} = (n + 1)! - 1! = \\boxed{(n + 1)! - 1}$[/hide]\r\n\r\nNote: Why does the (n+1)!-1 in the box look so weird?\r\nEdit: Thanks TachyonPulse.",
  "Solution_13": "Use \\boxed{} instead of \\fbox{}.",
  "Solution_14": "(n+1)!-1 by telescoping, or just by adding an extra 1x1! in there."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "Evaluate $\\big(\\root 3 \\of {0.000343}\\times\\root 3 \\of {0.008}\\big)^2$",
  "Solution_1": "[hide]If you know you square number, then you know that 7^3=343, 10^-6=.000001, 2^3=8, and 10^-3=.001.\nif you cube root them, you get:\n(7*10^-2*2*10^-1)^2\n(14*10^-3)^2\n196*10^-6\n>>.000196<<<[/hide]",
  "Solution_2": "it's 0.07  times 0.2 , squared.\r\nwhich is .014 squared\r\n\r\n0.000196. right?",
  "Solution_3": "I also got .000196 (it's just common squares and cube roots (Ex: 343 = $7^{3}$) in decimal form...",
  "Solution_4": "[quote=\"rcv\"]Evaluate $\\big(\\root 3 \\of {0.000343}\\times\\root 3 \\of {0.008}\\big)^2$[/quote]\r\n\r\n[hide=\"solution\"]$(.07\\cdot .2)^2$\n\n$=(.014)^2=..000196$[/hide]",
  "Solution_5": "0.07* 0.2= .014\r\n\r\n.014^2= .000196.\r\n\r\nThe answer, I hope.  :lol:"
}
{
  "Tag": [],
  "Problem": "when the compound is treated with\r\na.$ \\text{NH}_{2}\\text{NH}_{2}(\\text{KOH})$\r\nb.$ \\text{Zn\\minus{}Hg(Conc. HCl)}$",
  "Solution_1": "As you might already know, the first reaction, (a), is the Wolff-Kishner reduction which is used to reduce C=O bonds to -CH2-, under alcaline conditons; the second, (b), is the Clemmensen reduction which does the same but under acidic conditions. I was unable to open your picture, so I don't know what is the functional group in lowe left corner of your molecule, but it is expected that the right hand carbonyl will be reduced in both cases.",
  "Solution_2": "the lower left corner is $ \\text{OCH}_{3}$\r\nnow answer it.also whta picture formats does ur computer open\r\nalso what about the mail i PMed u",
  "Solution_3": "The methoxy group, as all ethers, is stable under alcaline conditions. Under acidic conditions, it could, at most, be converted into an hydroxi group.",
  "Solution_4": "well the actual answer is",
  "Solution_5": "That's what I said: the ether group was converted into an hydroxyl group under Clemmensen conditions, in this case methanol and the hydroxy substituted steroid. However, under the strongly acidic conditions, the sterol was converted into the chloride via a SN reaction. The protonated ether could also be directly converted into the chloride.",
  "Solution_6": "yes thanks  :oops:"
}
{
  "Tag": [],
  "Problem": "Show that the number \r\n\\[ \\underbrace{11...1}_{n-1}\\underbrace{222...25}_{n} \\]\r\n\r\nis a perfect square",
  "Solution_1": "[quote=\"felipesa\"]Show that the number \n\\[ \\underbrace{11...1}_{n-1}\\underbrace{222...25}_{n}  \\]\n\nis a perfect square[/quote]\r\n\r\nassuming that you mean $\\underbrace{11...1}_{n-1}\\underbrace{222...2}_{n}5$\r\n\r\nthis is just $(\\underbrace{3...3}_{n-1}5)^2$",
  "Solution_2": "[hide][quote=\"Altheman\"][quote=\"felipesa\"]Show that the number \n\\[ \\underbrace{11...1}_{n-1}\\underbrace{222...25}_{n}  \\]\n\nis a perfect square[/quote]\n\nassuming that you mean $\\underbrace{11...1}_{n-1}\\underbrace{222...2}_{n}5$\n\nthis is just $(\\underbrace{3...3}_{n-1}5)^2$[/quote]\n\nYes you must consider \\[ \\underbrace{11...1}_{n-1}\\underbrace{222...2}_{n}5  \\]\n\nbut the  answer should be a fraction[/hide]",
  "Solution_3": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58633",
  "Solution_4": "thanks for the link chess64!! ;) \r\n[hide]\nPS:to altheman\nWhen i said fraction i was talking about something like\n\\[ \\underbrace{111...1}_{n-1}\\underbrace{222...2}_{n}5=\\underbrace{111...1}_{n-1}10^{n+1}+\\underbrace{222...2}_{n}10+5 \\]\n\\[ \\frac{1}{9}\\underbrace{999...9}_{n-1}.10^{n+1}+\\frac{1}{9}.2.\\underbrace{999...9}_{n-1}.10+5 \\]\n\\[ \\frac{1}{9}.(10^{n-1}-1).10^{n+1}+\\frac{1}{9}.2.(10^n-1).10+5 \\]\n\\[ \\frac{1}{9}.(10^{2n}-10^{n+1}+2.10^{n+1}-20+45)=\\boxed{\\left(\\frac{10^n+5}{3}\\right)^2} \\] ;)[/hide]"
}
{
  "Tag": [
    "absolute value"
  ],
  "Problem": "If $ |x^2 \\minus{} 4| < n$ for all $ x$ such that $ |x \\minus{} 2| < 0,01,$ the least value what we can use for $ n$ is:\r\n\r\na) $ 0,0301$\r\nb) $ 0,0349$\r\nc) $ 0,0399$\r\nd) $ 0,0401$\r\ne) $ 0,0499$",
  "Solution_1": "I'm guessing by 0,01 you mean 0.01, but:\r\nwe have that, from the first absolute value, that:\r\n2.01>x>1.99\r\nFrom the second we get that:\r\n-n<(x-2)(x+2)<n\r\nThe least value of x we can possibly guess has to be greater than 1.99, so we \r\nknow that if we plug in 1.99 for x we'll be able to find the least value of n. \r\nSo, plugging this in, we get:\r\n-n<(-.01)(3.99)<n or\r\nn>.0399>-n\r\nHence, the least value of n is[u] .0399[/u] or [u]C[/u]",
  "Solution_2": "[quote=\"Z-coli\"]I'm guessing by 0,01 you mean 0.01[/quote]\r\n\r\nIn Brazil we use 0,01 and not 0.01.\r\n\r\nForgot to use USA notation.\r\n\r\n[hide=\"Another Solution\"]\n\n[list]$ |x-2|<0.01\\Longrightarrow 1.99<x<2.01$[/list]Hence\n\n[list]$ 1.99^2<x^2<2.01^2\\\\\n1.99^2-4<x^2-4< 2.01^2-4\\\\\n-0.0399<x^2-4<0.0401$[/list]\n\nAnything wrong?[/hide]"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "I'm making a list, similar the AoPS resources page, of physics book that would make the average AoPSer stand a good chance of succeeding internationally on physics contests.\r\n\r\nI have no experience with physics or physics contests. So I was wondering if one of the more experianced members could help me with my compilation.\r\n\r\nSummary of the Request: \r\nReccomend a list of books, that read sequentially and attentively, would give a good student a complete theorectical backbone for IPho physics. Problem books are easy to come by...\r\n\r\nPlease help :(",
  "Solution_1": "Probably [url=http://www.amazon.com/gp/product/0471332364/102-7962144-6561747?v=glance&n=283155]Fundamentals of Physics[/url] contains a good deal of the core concepts.\r\n\r\nBy the way, when are the tests?",
  "Solution_2": "The Physics Resources topic in this forum already has a very similar discussion? You might also consider searching for it on the rest of the Physics forum....",
  "Solution_3": "[quote=\"Rushil\"]The Physics Resources topic in this forum already has a very similar discussion? You might also consider searching for it on the rest of the Physics forum....[/quote]\r\nThe physics resources doesn't really contain any detail or pedalogical breakdown.\r\nIt doesn't even consider the relative difficulty.\r\n\r\nThe syllabus is available [url=http://www.ipho2005.com/]here.[/url]",
  "Solution_4": "If you create a resources page for physics,i think you can search on google for Indian,australian,etc physics olympiads at all levels and integrate them all at one place.I'll search for some...\r\n\r\nSome books-1)Resnick and Halliday (Walker and Krane edition)\r\n2)200 puzzling physics problems\r\n3)Richard Feynman Lectures",
  "Solution_5": "[quote=\"shadysaysurspammed\"]\n200 puzzling physics problems\n[/quote]\r\n\r\nThis is one of the best books for preparing for the IPHO...my favourite physics question all come from there!"
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "If the group $ G$ has order $ 4n\\plus{}2$, then there exists a non-trivial normal subgroup $ G^{\\prime}$, where $ G^{\\prime}\\neq G$.",
  "Solution_1": "If we let $ G$ act on itself by right multiplication, we notice that the kernel is trivial, hence $ G$ is isomorphic to a subgroup $ H$ of $ S_{\\left|G\\right|}$. \r\n\r\nFurthermore, $ H \\not \\subseteq A_{\\left|G\\right|}$, since an element $ g$ of order $ 2$ induces an odd permutation on $ G$. Indeed, the elements of $ G$ can be divided in pairs, each of them being interchanged by letting $ g$ work on $ G$. Since there are an odd number of such pairs, the cycle decomposition of the permutation induced by $ g$ has an odd number of cycles. \r\n\r\nNow it follows that $ [H : A_{\\left|G\\right|} \\cap H] \\equal{} 2$, hence $ G$ has a normal subgroup of index $ 2$.",
  "Solution_2": "Yes, that is the same solution as mine.\r\n\r\n$ \\underline{G \\hookrightarrow S_{|G|}: }$ $ \\varphi: G\\rightarrow S_{|G|}, \\varphi: g\\mapsto L_g,$ where $ L_g: S_{|G|}\\rightarrow S_{|G|}, L_g: g\\mapsto ga.$\r\n\r\n$ \\underline{L_g\\in S_{|G|}: }$ \r\n$ L_g(a)\\equal{}L_g(b)\\Leftrightarrow ga\\equal{}gb \\Leftrightarrow a\\equal{}b$ (Cancelation law). So $ L_g$ is injective.\r\nBecause $ |L_g(S_{|G|})|\\equal{}|S_{|G|}|$, it is a bijection.\r\n\r\n$ \\underline{\\varphi\\; \\mathrm{homomorphism: }}$\r\n$ \\varphi(ab)\\equal{}\\varphi(a)\\varphi(b)\\Leftrightarrow L_{ab}\\equal{}L_aL_b \\Leftrightarrow L_{ab} (g) \\equal{}L_aL_b (g), \\forall g\\in S_{|G|}\\Leftrightarrow \\newline\r\n\\Leftrightarrow L_{ab}(g)\\equal{}abg\\equal{} L_a(bg)\\equal{}L_a L_b (g)$.\r\n\r\n$ \\underline{\\varphi\\; \\mathrm{injective}: }$ $ \\varphi(a)\\equal{}\\varphi(b)\\Leftrightarrow L_a\\equal{}L_b \\Leftrightarrow L_a(g)\\equal{}L_b(g), \\forall g\\in S_{|G|} \\Leftrightarrow ag\\equal{}bg \\Leftrightarrow a\\equal{}b.$\r\n\r\nIn our problem $ \\varphi: G \\rightarrow S_{4n\\plus{}2}, \\varphi: g \\mapsto L_g$.\r\n\r\nSince $ G$ is even, there exists an element $ g$ of order $ 2$ in $ G$. Then also $ \\mathrm{order}\\varphi(g)\\equal{}2$, since $ \\mathrm{order}(\\varphi(g))\\leq \\mathrm{order}(g)$, but $ \\varphi$ is an injection, therefore the order of $ \\varphi(g)$ is not $ 1$.\r\n\r\nTherefore $ \\varphi(g)$ is the product of disjunct transpositions. Except that $ \\varphi(g)$ does not have a fixed point, otherwise $ L_g(a)\\equal{}ga\\equal{}a$ for some element $ a\\in G$. Then $ g$ is unity. Contradiction. \r\n\r\nThat means $ \\varphi(g)$ is  a product of $ 2n\\plus{}1$ transpositions, therefore an odd permutaon. That means $ \\varphi(G)\\cap A_{4n\\plus{}2}\\neq \\varphi(G)$ is normal and nontrivial in  $ \\varphi(G)$. Then $ \\varphi^{\\minus{}1}({\\varphi(G)\\cap A_{4n\\plus{}2}})\\neq G$ is normal and nontrivial in $ G$. (We have to narrow $ S_{4n\\plus{}2}$ to $ \\varphi(G)$ first.)"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Enuntul acestei probleme este urmatorul:\r\n\r\n\"Diagonalele patrulaterului inscriptibil ABCD se intersecteaza in punctul K. \r\nCercul circumscris triunghiului ABK intersecteaza a doua oara dreptele BC si AD\r\nin punctele M si N. Demonstrati ca: |KM|=|KN|.\"\r\n\r\nSOLUTIE(MM): [Notez cu m(XYZ) masura unghiului XYZ]\r\nPe o figura in care B este pe [MC] si N pe [AD], avem:\r\nBKNM-inscriptibil--->m(MNK)=m(KBC);     (1)\r\nABCD-inscriptibil--->m(KBC)=m(KAN);     (2)\r\n  m(KAN)=m(NMK)(subantin arcul NK).      (3)\r\nDin (1), (2) si (3)--->m(MNK)=m(NMK)--->|KM|=|KN|.",
  "Solution_1": "Interesante problemele, dar ar fi bine sa invatati sa folositi LaTeX, totul ar arata mult mai bine si ar fi mai usor de citit. Si nici nu e greu."
}
{
  "Tag": [
    "inequalities",
    "Cauchy Inequality",
    "inequalities unsolved"
  ],
  "Problem": "Let $ a,b,c\\in R^ \\plus{}$. Prove that\r\n\r\n$ \\sum_{cyc} \\frac {ab(a^3 \\plus{} b^3)}{a^2 \\plus{} b^2}\\geq \\sqrt {3abc(a^3 \\plus{} b^3 \\plus{} c^3)}$",
  "Solution_1": "[quote=\"easternlatincup\"]Let $ a,b,c\\in R^ \\plus{}$. Prove that\n\n$ \\sum_{cyc} \\frac {ab(a^3 \\plus{} b^3)}{a^2 \\plus{} b^2}\\geq \\sqrt {3abc(a^3 \\plus{} b^3 \\plus{} c^3)}$[/quote]\r\n\r\n$ x,y,z,a,b,c$ positiv real number \r\n$ \\frac{x(b\\plus{}c)}{y\\plus{}z}\\plus{}\\frac{y(c\\plus{}a)}{x\\plus{}z}\\plus{}\\frac{z(a\\plus{}b)}{y\\plus{}x}\\geq\\sqrt{3(ab\\plus{}bc\\plus{}ac)}$\r\n\r\n----> \r\n\r\n$ \\frac{xy(x^{3}\\plus{}y^{3})}{x^{2}\\plus{}y^{2}}\\equal{}\\frac{(xy)^{2}}{(xz)^{2}\\plus{}(yz)^{2}}*(\\frac{x^{3}z^{2}}{xy}\\plus{}\\frac{y^{3}z^{2}}{xy}))\\equal{}\\frac{(xy)^{2}}{(xz)^{2}\\plus{}(yz)^{2}}*(\\frac{x^{2}z^{2}}{y}\\plus{}\\frac{y^{2}z^{2}}{x}))$---->\r\n\r\n\r\n$ \\sum_{cyc} \\frac {ab(a^3 \\plus{} b^3)}{a^2 \\plus{} b^2}\\equal{}\\sum_{cyc} \\frac{(ab)^{2}}{(ac)^{2}\\plus{}(bc)^{2}}*(\\frac{a^{2}c^{2}}{b}\\plus{}\\frac{b^{2}c^{2}}{a}))\\geq\\sqrt{3(\\sum_{cyc} \\frac{a^{2}c^{2}}{b}*\\frac{b^{2}c^{2}}{a})}\\equal{}\\sqrt{3(a^{4}abc\\plus{}ab^{4}c\\plus{}abc^{4})}\\equal{}\\sqrt{3abc(a^{3}\\plus{}b^{3}\\plus{}c^{3})}$",
  "Solution_2": "Can you post a proof for the lemma,\r\n\r\n$ \\frac{x(b\\plus{}c)}{y\\plus{}z} \\plus{} \\frac{y(c\\plus{}a)}{z\\plus{}x} \\plus{} \\frac{z(a\\plus{}b)}{x\\plus{}y} \\geq \\sqrt{3(ab\\plus{}bc\\plus{}ca)}$ ?\r\n\r\nI've had no success in proving the lemma..",
  "Solution_3": "[quote=\"qwerty414\"]Can you post a proof for the lemma,\n\n$ \\frac {x(b \\plus{} c)}{y \\plus{} z} \\plus{} \\frac {y(c \\plus{} a)}{z \\plus{} x} \\plus{} \\frac {z(a \\plus{} b)}{x \\plus{} y} \\geq \\sqrt {3(ab \\plus{} bc \\plus{} ca)}$ ?\n\nI've had no success in proving the lemma..[/quote]\r\n suppose that a+b+c=1 and after that use cauchy inequality"
}
{
  "Tag": [],
  "Problem": "I've got two questions, but I'll post the second after I get the first one cleared up.  I got the answer to this, but I just listed the first 7 terms of the sequence and saw the pattern.  I can't figure out a logical way to think about why this is, though.  Help?\r\n\r\nAn elf can climb a staircase by jumping either 1 or 2 stairs at a time.  If [tex]e_n[/tex] represents the number of different ways the elf can reach the [tex]n[/tex]th stair of the staircase, define the sequence {[tex]e_n[/tex]} recursively.",
  "Solution_1": "[tex]e_n = e_{n-1} + e_{n-2}[/tex]\r\n\r\nbecause if the last jump is 2 steps, then there are [tex]e_{n-2}[/tex] ways, and if the last jump is 1 step, then there are [tex]e_{n-1}[/tex] ways.",
  "Solution_2": "Ohhh, I get it :)  Thanks.  Now this one... I don't even know where to start :(\r\n\r\nSuppose that the binary \"alphabet\" contains only two letters, A and B.  Then there would be two \"words\" of length 1: A, B; four \"words\" of length 2: AA, AB, BA, BB; and eight \"words\" of length 3: AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB.  Define {[tex]w_n[/tex]} recursively if [tex]w_n[/tex] represents the number of \"words\" of length n that do [u]not[/u] contain the BB pattern.",
  "Solution_3": "[tex]w_n = w_{n-1} + w_{n-2}[/tex]\r\n\r\nbecause for every sequence of length n-1, you just add an A.  You can also add a B for every sequence of length n-1 ending in A.  The number of sequences of length n-1 ending in A is equal to however many sequences of length n-2 ending in anything."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "reflection",
    "similar triangles",
    "geometry proposed"
  ],
  "Problem": "Let $ \\omega$ be the circumcircle of triangle $ ABC$ and $ P,Q$ be the points on $ \\omega$.Let $ P_{1},P_{2},P_{3}$ be a symmetric points to $ P$ w.r.t $ BC,AC,AB$ respectively.Denote $ A'\\equal{}QP_{1}\\cap BC$,$ B'\\equal{}QP_{2}\\cap AC$,$ C'\\equal{}QP_{3}\\cap AB$.Prove that $ A',B',C'$ are collinear.",
  "Solution_1": "The points $ A', B', C'$ are the points such that if a ray of light is fired in the direction from $ Q$ to any of them, it will \"bounce off\" the corresponding side to form a line containing $ P$.  So then it seems that, based on the similar triangles formed, we can draw the Simson line wrt Q and use Menelaus a lot.  Just an idea.",
  "Solution_2": "Here is an image to this unsolved problem:\r\n[img]http://img411.imageshack.us/img411/830/asan3kw6.png[/img]",
  "Solution_3": "Could you K81o7 continue your proof,or maybe somebody else will post his own solution?\r\nI don't know why,but i think that this is not hard problem,and there must be easy solution :roll: \r\nThank you.",
  "Solution_4": "This problem is unsolved for about 3 months.Please,let me know if someone trying to solve it,or tried to solve it,and prove or find out some useful properties to solve this problem. :( \r\nI hope we can find synthetic solution,especially without using complex numbers.\r\nBest regards,Erken.",
  "Solution_5": "Almost one year left since the time when I've posted the problem,but it is still unsolved.\r\nSometime ago I've finally managed to solve this problem,but I feel that my solution is still too messy.I would like to post the moments which were used by me to start that solution and to end it.\r\n[hide=\"How to Start:\"]Try to reflect the point $ Q$ w.r.t the sides of triangle $ \\triangle ABC$ \n[/hide]\n[hide=\"How to End:\"]Use the well-known fact which tell us that $ \\overline{P_1P_2P_3}$ and $ \\overline{Q_1Q_2Q_3}$ intersect at $ H$,orthocenter of $ \\triangle ABC$.\n[/hide]",
  "Solution_6": "Dear Mathlinkers,\r\nthe line A'B'C' is the Seimiya's line od P and Q wrt ABC.\r\nYou can see my synthetic proof and more on my website\r\nhttp://perso.orange.fr/jl.ayme  (vol.3 Un orthop\u00f4le sur la droite de Seimiya  and La droite se Seimiya-Ayme-Turner)\r\nSincerely\r\nJean-Louis"
}
{
  "Tag": [],
  "Problem": "I can't get my head around trying to solve this particular surd simultaneous equation so could someone please help me by showing some methods to solving this. Thanks.\r\n\r\n$ 5x\\minus{}3y \\equal{} 41$ [1]\r\n$ (7\\surd2)x \\equal{} (4\\surd2)y \\equal{} 82$ [2]",
  "Solution_1": "[quote=\"pizza1512\"]I can't get my head around trying to solve this particular surd simultaneous equation so could someone please help me by showing some methods to solving this. Thanks.\n\n$ 5x\\minus{}3y \\equal{} 41$ [1]\n$ (7\\surd2)x \\equal{} (4\\surd2)y \\equal{} 82$ [2][/quote]\r\n[hide=\"Solution\"]Take the second equation and divide it by $ \\sqrt{2}$. Now you have $ 7x\\plus{}4y \\equal{} 41\\sqrt{2}$.\n\nNow change them to $ 20x\\minus{}12y \\equal{} 164$ and $ 21x\\plus{}12y \\equal{} 123\\sqrt{2}$.\n\nAdd them to get $ 41x \\equal{} 164\\plus{}123\\sqrt{2}$, and solve to get $ x \\equal{} 4\\plus{}3\\sqrt{2}$. Plug $ x$ into one of the equations to get $ y \\equal{}\\minus{}7\\plus{}5\\sqrt{2}$.\n\nThe solution is $ \\boxed{\\left(4\\plus{}3\\sqrt{2},\\minus{}7\\plus{}5\\sqrt{2}\\right)}$.[/hide]",
  "Solution_2": "[quote=\"pizza1512\"]I can't get my head around trying to solve this particular surd simultaneous equation so could someone please help me by showing some methods to solving this. Thanks.\n\n$ 5x\\minus{}3y \\equal{} 41$ [1]\n$ (7\\surd2)x \\equal{} (4\\surd2)y \\equal{} 82$ [2][/quote]\r\nDo you mean $ (7\\surd2)x\\plus{}(4\\surd2)y \\equal{} 82$ ?",
  "Solution_3": "[quote=\"Quevvy\"][quote=\"pizza1512\"]I can't get my head around trying to solve this particular surd simultaneous equation so could someone please help me by showing some methods to solving this. Thanks.\n\n$ 5x\\minus{}3y \\equal{} 41$ [1]\n$ (7\\surd2)x \\equal{} (4\\surd2)y \\equal{} 82$ [2][/quote]\nDo you mean $ (7\\surd2)x\\plus{}(4\\surd2)y \\equal{} 82$ ?[/quote]\r\n\r\nOops... I typed it incorrectly... yep it should be $ (7\\surd2)x\\plus{}(4\\surd2)y \\equal{} 82$",
  "Solution_4": "You can use \"\\sqrt{}\" instead of \"\\surd\" for better-looking square roots. Other commands can also be learned [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url].",
  "Solution_5": "lol , I was thinking, what? It shouldn't be =. Now the + makes complete sense. :p\r\n\r\n$ (7\\sqrt{2})x\\plus{}(4\\sqrt{2})y \\equal{} 82$ :)"
}
{
  "Tag": [],
  "Problem": "Find all pairs $(k,m)$ of positive integers such that $k^{2}+4m$ and $m^{2}+5k$ are both perfect squares.",
  "Solution_1": "[hide]Let's put $k^{2}+4m=(k+p)^{2},\\; m^{2}+5k=(m+q)^{2}$. Then $p,q\\in\\mathbb{N}^{+}$. Transforming these equations we get:\n$4m=2kp+p^{2}$\n$5k=2mq+q^{2}$. \nPlugging the first into the second and then the second into the first gives\n$5k=2q\\cdot \\frac{2kp+p^{2}}{4}+q^{2}$\n$4m=2p\\cdot\\frac{2mq+q^{2}}{5}+p^{2}$\nwhich is equivalent to\n$(5-pq)k= \\frac{p^{2}q}{2}+q^{2}$. \n$\\left(4-\\frac{4}{5}pq\\right)m= \\frac{2pq^{2}}{5}+p^{2}$\nSince $k$ and $\\frac{p^{2}q}{2}+q^{2}$ are positive numbers, also $5-pq$ is positive, i.e. $pq\\leqslant 4$. Thus $pq\\in\\{1,2,3,4\\}$. From the second equation: $m=\\frac{2pq^{2}+5p^{2}}{20-4pq}$, so $2|5p^{2}$. Hence $2|p$. The only possible pairs $(p,q)$ are $\\{2,1\\}$, $\\{2,2\\}$, $\\{4,1\\}$. After evaluating $(k,m)$ for each pair we see that the pairs $(k,m)$ we got satisfy the given conditions. Therefore the final result is $(k,m)\\in \\{(1,2),(8,9),(9,22)\\}$. [/hide]",
  "Solution_2": "[hide]My solution: :blush:\n\n[b]1.[/b] $k\\geq m$\n\n$k^{2}<k^{2}+4m\\leq k^{2}+4k<k^{2}+4k+4=(k+2)^{2}$\n\n$\\Rightarrow k^{2}<k^{2}+4m<(k+2)^{2}\\Rightarrow k^{2}+4m=(k+1)^{2}$\n\n$\\Rightarrow k=\\frac{4m-1}{2}=2m-\\frac{1}{2}$, so $k$ can't be intger.\n\n[b]2.[/b] $k<m$\n\n$m^{2}<m^{2}+5k<m^{2}+5m<m^{2}+5m+6.25 = (m+2.5)^{2}$\n\n$\\Rightarrow m^{2}+5k=(m+1)^{2}$ or $m^{2}+5k=(m+2)^{2}$. $\\Rightarrow 5k=2m+1$ or $5k=4m+4$.\n\n[b]2.1.[/b] $5k=2m+1$\n\nSo $k^{2}+4m=k^{2}+10k-2$ and\n\n$k^{2}<k^{2}+10k-2< k^{2}+10k+25=(k+5)^{2}$\n$\\Rightarrow k^{2}<k^{2}+10k-2<(k+5)$ and we have to consider 4 cases, but only two of them gives us a solution: for $k^{2}+10k-2=(k+2)^{2}$ $k=1; m=2$ and for $k^{2}+10k-2=(k+4)^{2}$ $k=9; m=22$\n\n[b]2.2.[/b] $5k=4m+4$\n\nSo $k^{2}+4m=k^{2}+5k-4$ and\n\n$k^{2}<k^{2}+5k-4<k^{2}+5k+6.25=(k+2.5)$\n$\\Rightarrow k^{2}<k^{2}+5k-4<(k+2.5)^{2}$ and we have to consider 2 cases, but only one of them gives us a solution: for $k^{2}+5k-4=(k+2)^{2}$ $k=8; m=9$.[/hide]"
}
{
  "Tag": [
    "symmetry",
    "function",
    "geometry",
    "geometric transformation",
    "reflection",
    "probability",
    "expected value"
  ],
  "Problem": "Prove that $ \\sum_{t\\equal{}1}^{n} 2t {n \\choose t}^2 \\equal{} n {2n \\choose n}$ .",
  "Solution_1": "[hide=\"Solution\"] Consider the set of lattice paths from $ (0, 0)$ to $ (2n, 0)$ in steps of the form $ (1, 1)$ and $ (1, \\minus{} 1)$.  There are $ {a \\choose b}$ paths from $ (0, 0)$ to $ (a, a \\minus{} 2b)$, so partition the paths according to which point of the form $ (n, n \\minus{} 2t)$ they pass through; this is the familiar combinatorial proof that\n\n$ \\sum_{t \\equal{} 0}^{n} {n \\choose t}^2 \\equal{} {2n \\choose n}$.\n\nSince the paths have a reflectional symmetry across the $ x$-axis, it then follows that the expected value of $ n \\minus{} 2t$ is $ 0$, hence the expected value of $ t$ is $ \\frac {n}{2}$, which is equivalent to the desired result. [/hide]\r\n\r\nA generalization was recently posted on the xkcd forums [url=http://echochamber.me/viewtopic.php?f=3&t=39539]here[/url].",
  "Solution_2": "t0r, you have a 1 in a lower index of summation that really wants to be a 0.\r\n\r\nThis is almost not worth mentioning, but the desired result is easily seen to be an algebraic consequence to Vandermonde's identity:\r\n\\begin{align*} \\sum_{t \\geq 1} 2t \\binom{n}{t}^2 & = \\sum_{t \\geq 1} 2n \\binom{n - 1}{t - 1}\\binom{n}{t} \\\\\r\n& = 2n \\binom{2n - 1}{n - 1} \\\\\r\n& = n \\binom{2n}{n}. \\end{align*}\r\n(And of course we could do it with generating functions, as well.)",
  "Solution_3": "I see.. thanks. Is there a way, though, to prove it using binomial expressions? For example, we can prove that $ \\sum_{k\\equal{}0}^n {n \\choose k}^2 \\equal{} {2n \\choose n}$ by considering $ (1\\plus{}x)^n (1\\plus{}x)^n \\equiv (1 \\plus{} x)^{2n}$ and looking for the coefficient of $ x^n$.",
  "Solution_4": "Right, with generating functions ;)  I think the thread t0r links to has one such solution, but here it is anyhow: $ \\sum_{t \\equal{} 0}^n \\left(2t \\binom{n}{t}\\right) \\cdot \\binom{n}{t}$ is the coefficient of $ x^{n \\minus{} 1}$ in $ \\left(2\\cdot \\frac{d}{dx}(1 \\plus{} x)^n\\right) \\cdot (1 \\plus{} x)^n$.  But that's just $ 2n (1 \\plus{} x)^{2n \\minus{} 1}$, so it's also equal to $ 2n \\binom{2n \\minus{} 1}{n} \\equal{} n \\binom{2n}{n}$.  (The manipulations of this solution are pretty much identical to the manipulations in my solution.)",
  "Solution_5": "Nice! I couldn't find the right generating function.",
  "Solution_6": "Hmm... I tried to find a clean committee-forming-style proof, but ended up down JBL's Vandermonde route. However, I did find a combinatorial proof that $ k\\binom{n}{k}\\equal{}n\\binom{n\\minus{}1}{k\\minus{}1}$  :dry: \r\n\r\n[hide]$ k\\binom{n}{k}$ counts the number of ways to choose $ k$ of $ n$ objects, disregarding order, then picking one of those $ k$ objects (the objects here are distinct). We could also fix our chosen object in our resulting set, and fill in the gaps around it with other objects in $ \\binom{n\\minus{}1}{k\\minus{}1}$ ways, where there are $ n$ ways to choose that object. This means that $ k\\binom{n}{k}\\equal{}n\\binom{n\\minus{}1}{k\\minus{}1}$.\n\nThen our sum is $ \\sum_{k\\equal{}0}^{n}2k\\binom{n}{k}^2\\equal{}2n\\sum_{k\\equal{}0}^{n}\\binom{n\\minus{}1}{k\\minus{}1}\\binom{n}{k}$, blah blah blah Vandermonde.[/hide]",
  "Solution_7": "[quote=\"grn_trtle\"]Hmm... I tried to find a clean committee-forming-style proof[/quote]\r\nI gave a proof of the generalization on the xkcd thread I linked to using linearity of expectation.  I think a committee-forming proof would be essentially the same.\r\n\r\nEdit:  [hide=\"Here it is, basically\"] Pick two committees $ A, B$ of the same size out of a group of $ n$ people and pick a president in $ A$.  The total number of ways to do this is as follows:  line up two copies of the group of $ n$ people and pick a subset of size $ n$; there are $ {2n \\choose n}$ ways to do this.  Declare the people picked in the first copy committee $ A$ and declare the complement of the people picked in the second copy committee $ B$.  There are, on average, $ \\frac {n}{2}$ possible presidents in $ A$; the rigorous way to prove this is essentially the reflection argument I gave. [/hide]",
  "Solution_8": "We can use pairing to reduce the problem to a known one...\r\n\r\nLet $ S\\equal{}\\sum_{k\\equal{}0}^n2t\\binom{n}{t}^2\\equal{}\\sum_{k\\equal{}0}^n2(n\\minus{}t)\\binom{n}{n\\minus{}t}^2$\r\n\r\nThen $ 2S\\equal{}\\sum_{k\\equal{}0}^n2n\\binom{n}{t}^2$\r\n\r\n$ S\\equal{}n\\cdot\\sum_{k\\equal{}0}^n\\binom{n}{t}^2$\r\n\r\nAs a comment, can we evaluate the sum where $ t$ is replaced by $ t^a$ where $ a$ is a natural number (of course the answer is that it is not that hard, but i'm curious what methods work best here, recursion, generating functions etc).",
  "Solution_9": "Small typo: Your sums should be $ \\sum_{t\\equal{}0}^{n}$. I do the same thing all the time ;)",
  "Solution_10": "[quote=\"Altheman\"]can we evaluate the sum where $ t$ is replaced by $ t^a$ where $ a$ is a natural number (of course the answer is that it is not that hard, but i'm curious what methods work best here, recursion, generating functions etc).[/quote]\r\nAs is usually the case in these situations, it's much easier to replace $ t$ with $ {t \\choose a}$ and interpolate from there.  This is true of the generating functions approach and the recursive approach and the probabilistic approach (although I don't understand the probabilistic approach well for higher moments).",
  "Solution_11": "Here is a direct [hide=\"combinatorial proof\"]\n\nSuppose you have a class of $ n$ boys and $ n$ girls. Let's say you want to choose a group of $ n$ students with one boy among them as their leader. \nTo count the no. of ways in which you can do this:\nChoose $ i$ boys and $ n\\minus{}i$ girls with one of the $ i$ boys as their leader. This can be done in $ i \\binom n i\\binom n {n\\minus{}i}$ ways.\nBut $ \\binom n {n\\minus{}i} \\equal{} \\binom n i$, so the no. ways to make a selection is $ \\sum_{i\\equal{}1}^n i {\\binom n i}^2$\n\nCounting in another way, you can choose a leader from one of the $ n$ boys and then choose $ n\\minus{}1$ kids from the remaining $ 2n\\minus{}1$. So, $ n \\binom {2n\\minus{}1} {n\\minus{}1}$ ways.\n\nSo, $ 2 \\sum_{i\\equal{}1}^n i {\\binom n i}^2 \\equal{} 2n \\binom {2n\\minus{}1} {n\\minus{}1} \\equal{} n{2n\\choose n}$\n\n\nI didn't read the other proofs so forgive me if I have repeated someone's solution.[/hide]"
}
{
  "Tag": [
    "vector",
    "linear algebra"
  ],
  "Problem": "Let $ \\mathbb U$ and $ \\mathbb W$ be the linear subspaces of $ \\mathbb R^4$ such that $ \\mathbb U=<\\{(1,0,1,2),(0,-1,0,1)\\}>$ and $ \\mathbb W=\\{(a,b,c,d)\\in\\mathbb R^4|3d-a-b=0\\,\\wedge\\,4c+d-b=0\\}.$ Find $ (\\mathbb U+\\mathbb W)^\\perp.$\r\n\r\n[hide=\"My work\"]\n\\begin{eqnarray*}\n  \\mathbb W&=&\\left\\{ \\left( a,\\frac{a+12c}{2},c,\\frac{a+4c}{2} \\right)\\in \\mathbb{R}^{4},\\forall \\,a,c\\in \\mathbb{R} \\right\\} \\\\ \n & =&\\left\\{ a\\left( 1,\\frac{1}{2},0,\\frac{1}{2} \\right)+c\\left( 0,6,1,2 \\right),\\forall \\,a,c\\in \\mathbb{R} \\right\\} \\\\ \n & =&<\\left\\{ \\left( 1,\\frac{1}{2},0,\\frac{1}{2} \\right),\\left( 0,6,1,2 \\right) \\right\\}> \\\\ \n & =&<\\left\\{ (2,1,0,1,),(0,6,1,2) \\right\\}>. \n\\end{eqnarray*}\n\nThus we get $ \\mathbb U+\\mathbb W=\\{(1,0,1,2),(0,-1,0,1),(2,1,0,1,),(0,6,1,2)\\},$ provided that these are linearly independent.\n\nHence $ (\\mathbb U+\\mathbb W)^\\perp=\\left\\{ (x,y,z,u)\\in \\mathbb{R}^{4}\\left| \\begin{aligned}\n   \\left\\langle (x,y,z,u),(1,0,1,2) \\right\\rangle &=0 \\\\ \n  \\left\\langle (x,y,z,u),(0,-1,0,1) \\right\\rangle &=0 \\\\ \n  \\left\\langle (x,y,z,u),(2,1,0,1) \\right\\rangle &=0 \\\\ \n  \\left\\langle (x,y,z,u),(0,6,1,2) \\right\\rangle &=0 \n\\end{aligned} \\right. \\right\\}$\n\nIs this right?\n[/hide]",
  "Solution_1": "You are on the right track but your answer can be simplified a lot.",
  "Solution_2": "Thanks, but what would you do then?",
  "Solution_3": "[hide=\"Quick Answer\"]\nIndeed, the 4 vectors in $ \\mathbb{U}\\plus{}\\mathbb{W}$ are linearly independent since their determinant is nonzero (7, I believe), but doesn't this mean that $ \\mathbb{U}\\plus{}\\mathbb{W} \\equal{} \\mathbb{R}^4$? Therefore, $ (\\mathbb{U}\\plus{}\\mathbb{W})^\\perp \\equal{} \\{(0,0,0,0)\\}$.[/hide]"
}
{
  "Tag": [
    "trigonometry",
    "induction",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "We have cosx + cosy + cosz = sinx + siny + sinz = 0 ... and we must prove:\r\ncos(n*x) + cos(n*y) + cos(n*z) = sin(n*x) + sin(n*y) + sin(n*z) = 0\r\n\r\nI tried hard to solve it but i didn't  :(  .. help me please! .. i think we must start from cosx<0 , cosy>0, cosz>0.  :help:",
  "Solution_1": "complex numbers(de moivre's formula) maybe.\r\n\r\nLet $z_{1}=\\cos x+i\\sin x$\r\n$z_{2}=\\cos y+i\\sin y$\r\n$z_{3}=\\cos z+i\\sin z$\r\nwe have $z_{1}+z_{2}+z_{3}=0$\r\n\r\nnow someone can complete? :wink:",
  "Solution_2": "Let $A(\\cos nx,\\ \\sin nx), \\ B(\\cos ny,\\ \\sin ny),\\ C(\\cos nz,\\ \\sin nz)$, consider the barycentre of $\\triangle{ABC}$.",
  "Solution_3": "[quote=\"kunny\"]Let $A(\\cos nx,\\ \\sin nx), \\ B(\\cos ny,\\ \\sin ny),\\ C(\\cos nz,\\ \\sin nz)$, consider the barycentre of $\\triangle{ABC}$.[/quote]\r\n\r\nCan you be more explicit?",
  "Solution_4": "[quote=\"St3faN\"][quote=\"kunny\"]Let $A(\\cos nx,\\ \\sin nx), \\ B(\\cos ny,\\ \\sin ny),\\ C(\\cos nz,\\ \\sin nz)$, consider the barycentre of $\\triangle{ABC}$.[/quote]\n\nCan you be more explicit?[/quote]\r\n\r\nSorry that this didn't make sence.",
  "Solution_5": "if $x=\\frac{1}{6}\\pi$, $y=\\frac{5}{6}\\pi$, $z=\\frac{9}{6}\\pi$, $n=3$\r\nthen,\r\n$\\cos x+\\cos y+\\cos z = \\frac{\\sqrt{3}}{2}+\\left (-\\frac{\\sqrt{3}}{2}\\right )+0 = 0$ :) \r\n$\\sin x+\\sin y+\\sin z = \\frac{1}{2}+\\frac{1}{2}+(-1) = 0$ :) \r\n$\\cos nx+\\cos ny+\\cos nz = 0+0+0 = 0$ :) \r\n$\\sin nx+\\sin ny+\\sin nz = 1+1+1 = 3$ :(\r\n\r\nIs a condition of a problem right?",
  "Solution_6": "[quote=\"Kouichi Nakagawa\"] \nIs a condition of a problem right?[/quote]\r\n\r\noh :wallbash: .. I forgot about it : n>3. and n isn't a multiple of 3.",
  "Solution_7": "Good!  Now consider\r\n\\[\\frac{1}{z_{1}}+\\frac{1}{z_{2}}+\\frac{1}{z_{3}}.\\]\r\n\r\n[quote=\"nayel\"]complex numbers(de moivre's formula) maybe.\n\nLet $z_{1}=\\cos x+i\\sin x$\n$z_{2}=\\cos y+i\\sin y$\n$z_{3}=\\cos z+i\\sin z$\nwe have $z_{1}+z_{2}+z_{3}=0$\n\nnow someone can complete? :wink:[/quote]",
  "Solution_8": "well, $\\frac{1}{z_{1}}+\\frac{1}{z_{2}}+\\frac{1}{z_{3}}=\\cos x+\\cos y+\\cos z-i(\\sin x+\\sin y+\\sin z)=0$\r\n\r\nnow can it be shown by induction that $z_{1}^{n}+z_{2}^{n}+z_{3}^{n}=0$? :?: \r\nif so then $\\frac{1}{z_{1}^{n}}+\\frac{1}{z_{2}^{n}}+\\frac{1}{z_{3}^{n}}=0$\r\n\r\nso $\\cos nx+\\cos ny+\\cos nz=0=\\sin nx+\\sin ny+\\sin nz$ and we will be done.\r\n\r\nthanks for your help, nsato.\r\n\r\nEDIT: sorry i found some mistakes in my induction.",
  "Solution_9": "but if\nWe have sinx + siny + sinz =  sin3x + sin3y + sin3z = 0  prove\ncos2xcos2ycos2z>=0",
  "Solution_10": "any ideea?"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "For each positive integer $n$, find positive integer solutions $x_{1}, x_{2}, ..., x_{n}$ to the equation\\[\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+...+\\frac{1}{x_{n}}+\\frac{1}{x_{1}x_{2}...x_{n}}= 1.\\]\r\n\r\nHow are you supposed to make eclipses?",
  "Solution_1": "[code] \\ldots  \\text{ or } \\cdots [/code] gives $\\ldots \\text{ or }\\cdots$\r\n\r\n[hide=\"For the problem:\"]\nI remember seeing this before. Start with smaller cases, and once you find $x_{1}$ and $x_{2}$, I think you can define a recursion.  :) [/hide]",
  "Solution_2": "[hide=\"Hint\"]Induction[/hide]",
  "Solution_3": "There might be more to it than just \"induction\"",
  "Solution_4": "Use induction to prove that $x_{n+1}=\\left(\\prod_{i=1}^{n}x_{i}\\right)+1$."
}
{
  "Tag": [
    "calculus",
    "integration",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Find integral closure of $ \\mathbb{C}[x]$ in $ \\mathbb{C}(x)(\\sqrt[d]{1\\minus{}x^d})$. I suspect that it must be $ \\mathbb{C}[x](\\sqrt[d]{1\\minus{}x^d})$. How do you prove this?",
  "Solution_1": "I didn't check the answer, but you can use Serre's criterion for normal."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "geometric transformation",
    "trigonometry",
    "modular arithmetic",
    "Gauss",
    "cyclic quadrilateral"
  ],
  "Problem": "Given triangle $ ABC$. Let $ M$ be an arbitray point on A-altitude. $ CM\\cap (BC) \\equal{} \\{K\\}, BM\\cap (BC) \\equal{} \\{L\\}$, $ CM\\cap (AMB) \\equal{} \\{X\\}, BM\\cap (ACM) \\equal{} \\{Y\\}$.\r\n Prove that the midpoints of $ BC, XY$ and $ KL$ are collinear.",
  "Solution_1": "Could you please watch the problem? I think there is a typo, since $ CM\\cap (BC) \\equal{} \\{C\\},BM\\cap (BC) \\equal{} \\{B\\}$, or did you use the notation $ (XY)$ to say the circle with diameter $ XY$ :oops:",
  "Solution_2": "Yes, $ (BC)$ is a circle with diameter $ BC$.",
  "Solution_3": "$ O,F,P$ is collinear\r\n$ \\Leftrightarrow FP\\perp KL$\r\n$ \\Leftrightarrow \\vec {FP}.\\vec {KL}\\equal{}0$\r\n$ \\Leftrightarrow (\\vec {LX}\\plus{}\\vec {KY})(\\vec {LH}\\plus{}\\vec {HK})\\equal{}0$\r\n$ \\Leftrightarrow \\vec {LX}.\\vec {HK}\\plus{}\\vec {KY}.\\vec {LH}\\equal{}0$ $ (1)$\r\nfrom $ (\\vec {LX},\\vec {HK})\\equiv \\minus{}(\\vec {KY},\\vec {LH})(mod 2\\pi)$\r\n$ (1)\\Leftrightarrow LX.HK\\equal{}KY.LH$\r\n$ \\Leftrightarrow LX.HB\\equal{}KY.HC$\r\n$ \\Leftrightarrow HB.CX\\equal{}HC.BY$\r\n$ \\Leftrightarrow \\frac {HB}{HC}\\equal{}\\frac {AX}{AB}$\r\nTrue.\r\nDone.",
  "Solution_4": "$ \\bullet$ Let be the point $ E\\equiv BK\\cap CL$, which lies on the line of the $ A$-altitude $ AD$ of the given triangle $ \\bigtriangleup ABC$, as well.\r\n\r\nSo, based on the below [b][size=100]Lemma[/size][/b], we have that $ \\angle BEX = \\angle CEY$ $ ,(1)$\r\n\r\nFrom $ (1)$ $ \\Longrightarrow$ $ \\angle BEY = \\angle CEX$ $ ,(2)$\r\n\r\nFrom $ (2)$ and because of $ \\angle EBY = \\angle ECX$, we have that the triangles $ \\bigtriangleup EBY$ and $ \\bigtriangleup ECX$ are similar.\r\n\r\nFrom this similarity and because of $ (1)$ we have that $ \\frac {KX}{KC} = \\frac {LY}{LB}$ $ ,(3)$\r\n\r\n$ \\bullet$ We consider now the non-convex quadrilateral $ BCXY$ and based on the [b][size=100]ERIQ theorem[/size][/b] ( = [b]E[/b]qual [b]R[/b]atios [b]I[/b]n [b]Q[/b]uadrilateral ), because of $ (3)$, we conclude that the midpoints of the segments $ BC,\\ KL,\\ XY$ are collinear and the proof is completed.\r\n\r\n[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \\bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on its $ A$-altitude $ AD.$ We denote as $ M,\\ N,$ the points of intersections of the circumcircles $ (O_{1}),\\ (O_{2})$, of the triangles $ \\bigtriangleup BHP,\\ \\bigtriangleup CHP$ respectively, where $ H$ is the orthocenter of $ \\bigtriangleup ABC.$ Prove that $ \\angle BAM = \\angle CAN$.[/color][/size][/b]\r\n\r\nKostas Vittas.\r\n\r\nPS. I will post here later, the proof of the above [b]Lemma[/b] I have in mind.",
  "Solution_5": "[quote=\"vittasko\"][b][size=100][color=DarkBlue]LEMMA. - A triangle $ \\bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on its $ A$-altitude $ AD.$ We denote as $ M,\\ N,$ the points of intersections of the circumcircles $ (O_{1}),\\ (O_{2})$, of the triangles $ \\bigtriangleup BHP,\\ \\bigtriangleup CHP$ respectively, where $ H$ is the orthocenter of $ \\bigtriangleup ABC.$ Prove that $ \\angle BAM = \\angle CAN$.[/color][/size][/b][/quote]\r\n[b][size=100]PROOF OF THE LEMMA.[/size][/b] - We draw the circumcircles $ (K),\\ (L),$ of the triangles $ \\bigtriangleup AHB$ and $ \\bigtriangleup AHC$ respectively and let be the points $ X\\equiv (K)\\cap CM$ and $ Z\\equiv (L)\\cap BN.$\r\n\r\nIt is easy to show that $ \\angle XAB = \\angle XHB = \\angle ZHC = \\angle ZAC$ $ \\Longrightarrow$ $ \\angle XAB = \\angle ZAC$ $ ,(1)$\r\n\r\n$ \\bullet$ Let be the points $ Q\\equiv AC\\cap (O_{2})$ and $ R\\equiv AB\\cap (O_{1})$ and from $ (AQ)\\cdot (AC) = (AP)\\cdot (AH) = (AR)\\cdot (AB)$, we conclude that the quadrilateral $ BCQR$ is cyclic and so, we have that $ \\angle AQR = \\angle B$ $ ,(2)$\r\n\r\nAlso, from the cyclic quadrilateral $ BCEF,$ we have that $ \\angle AEF = \\angle B$ $ ,(3)$\r\n\r\nFrom $ (2),\\ (3)$ $ \\Longrightarrow$ $ \\angle AQR = \\angle AEF$ $ \\Longrightarrow$ $ QR\\parallel EF$ $ ,(4)$\r\n\r\nIt is easy now to show that $ QN\\parallel AZ$ $ ,(5)$ from $ \\angle QNH = \\angle QCH = \\angle AZH$ and similarly, we have that $ RM\\parallel AX$ $ ,(6)$\r\n\r\nFrom $ (4),\\ (5),\\ (6)$ $ \\Longrightarrow$ $ \\frac {FM}{MK} = \\frac {FR}{RA} = \\frac {EQ}{QA} = \\frac {EN}{NZ}$ $ \\Longrightarrow$ $ \\frac {KM}{MX} = \\frac {EN}{NZ}$ $ ,(7)$\r\n\r\nHence, because of $ (7)$, from the similar right triangles $ \\bigtriangleup FAX,\\ \\bigtriangleup EAZ,$ we conclude that $ \\angle MAB = \\angle NAC$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.\r\n\r\nKostas Vittas.",
  "Solution_6": "Thanks you dear [b]Elementarry[/b] and [b]Kostas Vittas[/b] for your nice solutions.\r\nHere is my solution:\r\n$ \\vec{EF} \\equal{} \\frac {1}{2}(\\vec{KX} \\plus{} \\vec{LY}), \\vec{OE} \\equal{} \\frac {1}{2}(\\vec{BK} \\plus{} \\vec{CL})$. So we need to prove that $ \\vec{KX} \\plus{} \\vec{LY} \\equal{} k(\\vec{BK} \\plus{} \\vec{CL})$. \r\nA translation $ T_{\\vec{LK}}: Y\\rightarrow P, C\\rightarrow Q$. Denote $ R, S$ the midpoints of $ XP, BQ$\r\nSo we only need to show that $ R,K,S$ are collinear.\r\nNote that $ \\angle BKX \\equal{} \\angle QKP \\equal{} 90^o$. Let $ SK\\cap PX \\equal{} \\{R'\\}, BK\\cap CL \\equal{} \\{J\\}$\r\n$ \\frac {\\sin \\angle SKQ}{\\sin \\angle SKB} \\equal{} \\frac {BK}{KQ} \\equal{} \\frac {BK}{LC} \\equal{} \\frac {\\sin \\angle BJA}{\\sin \\angle CJA}$\r\nBut $ \\angle SKQ \\plus{} \\angle SKB \\equal{} \\angle BJC \\equal{} \\angle BJA \\plus{} \\angle CJA$ so $ \\angle SKQ \\equal{} \\angle BJA, \\angle SKB \\equal{} \\angle CJA (1)$\r\n$ R'\\equiv R \\Leftrightarrow \\frac {KX}{KP} \\equal{} \\frac {\\sin \\angle R'KP}{\\sin \\angle R'KX} \\equal{} \\frac {\\cos \\angle SKQ}{\\cos \\angle RKS}$\r\n$ \\Leftrightarrow \\frac {\\cos \\angle SKQ}{\\cos \\angle RKS} \\equal{} \\frac {KX}{LY} \\equal{} \\frac {BK.\\cot \\angle BXK}{CL.\\cot \\angle CYM} \\equal{} \\frac {BK}{CL}.\\frac {\\cot \\angle BAM}{\\cot CAM}$\r\n$ \\equal{} \\frac {\\sin \\angle BJM}{\\sin \\angle CJM}.\\frac {\\cot \\angle BJM}{\\cot \\angle CJM} \\equal{} \\frac {\\cos \\angle BJM}{\\cos \\angle CJM}$\r\nIt's right from $ (1)$.\r\nWe are done.",
  "Solution_7": "Dear Livetolove and Mathlinkers,\r\nin order to have perhaps another approach of your nice problem consider the circle going through B, M, C and noted U his center.\r\nProuve: UM is parallel to OFP.\r\nSincerely\r\nJean-Louis",
  "Solution_8": "Based on the [b][size=100]Lemma[/size][/b] I mentioned before, we can have another more elementary approach as follows :\r\n\r\nWe consider ( see the [b]schema t=306006[/b] ) the similar right triangles $ \\bigtriangleup KEX,\\ \\bigtriangleup LEY,$ erected on the side-segments $ EK,\\ EL$ respectively, of the triangle $ \\bigtriangleup EKL,$ outwardly to it.\r\n\r\nSo, it is well known that the line segment $ OQ$ as the midperpendicular of $ KL,$ bisects the segment $ XY$ and the proof is completed.\r\n\r\nKostas Vittas.",
  "Solution_9": "This is my solution. It is based on the same idea with Mr.[b]vittasko[/b]\r\n\r\nDenoted by $ E\\equiv BK\\cap CL, P\\equiv EB\\cap (AMB), Q\\equiv EC\\cap (AMC)$. It is clearly that $ \\overline{EP}\\cdot\\overline{EB}\\equal{}\\overline{EA}\\cdot\\overline{EM}\\equal{}\\overline{EQ}\\cdot\\overline{EC}$ so $ B,C,P,Q$ are concyclic. Hence $ (BL,BQ)\\equiv (CP,CK) \\pmod{\\pi}$. Moreover, $ (BP,BL)\\equiv (CP,CL) \\pmod{\\pi}$ so $ (BL,BQ)\\equiv (CP,CK) \\pmod{\\pi}$. On the other side $ (XP,XC) \\equiv (BK,BL) \\equiv (CK,CL) \\equiv (YB,YQ) \\pmod{\\pi}$. Therefore the triangles $ \\triangle PXC$ and $ \\triangle QYB$ are similar. This yields $ \\frac{KX}{KC} \\equal{} \\frac{LY}{LB}$ so the midpoints of $ BC,KL,XY$ are collinear. Q.E.D.",
  "Solution_10": "Dear Mathlinkers,\r\nanother idea : the line OFP goes also through the midpoints of the segment joining the centers of (AMB) and (ACM).\r\nSincerely\r\nJean-Louis",
  "Solution_11": "Use [b]Lemma[/b]:\r\n$ \\frac {AE}{AB} \\equal{} \\frac {CK}{CD}$ and $ \\frac {LA}{LC} \\equal{} \\frac {NE}{NK} \\equal{} \\frac {MB}{MD}$\r\nthen $ L,N,M$ are collinear.\r\nApply we $ O,F,P$ are collinear.",
  "Solution_12": "Dear Mathlinkers,\r\nsorry my last result is wrong!!!\r\nSincerely\r\nJean-Louis",
  "Solution_13": "Another problem:\r\nGiven triangle $ ABC$ and altitude $ AA_1$. Let $ P$ be an arbitrary point on $ AA_1. BP\\cap AC\\equal{}\\{B_1\\}, CP\\cap AB\\equal{}\\{C_1\\}$\r\n$ A_1B_1\\cap (AB)\\equal{}X, A_1C_1\\cap (AC)\\equal{}Y$. Prove that $ A$, the midpoint of $ B_1C_1$ and the midpoint of $ XY$ are collinear.",
  "Solution_14": "It is not true.",
  "Solution_15": "[quote=\"livetolove212\"]Another problem:\nGiven triangle $ ABC$ and altitude $ AA_1$. Let $ P$ be an arbitrary point on $ AA_1. BP\\cap AC \\equal{} \\{B_1\\}, CP\\cap AB \\equal{} \\{C_1\\}$\n$ A_1B_1\\cap (AB) \\equal{} X, A_1C_1\\cap (AC) \\equal{} Y$. Prove that $ A$, the midpoint of $ B_1C_1$ and the midpoint of $ XY$ are collinear.[/quote]\n[quote=\"Elementaryyy\"]It is not true.[/quote]\r\nI think that [b]livetolove212[/b] thought about a different result.\r\n\r\nSurely, it is well known that the line connecting the midpoints of the segments $ XY,\\ B_{1}C_{1}$ he mentioned, passes through the midpoint of the segment $ AA_{1}.$\r\n\r\nBest regards, Kostas Vittas.",
  "Solution_16": "[quote=\"Elementaryyy\"]It is not true.[/quote]\n\nWhy is it not true?\nActually, this problem is [url=http://www.mathlinks.ro/viewtopic.php?p=1474889#1474889]Vietnam TST 2009-problem 5[/url] and I see that it is also an application of [b]ERIQ theorem[/b] [b]Mr. Kostas Vittas[/b] has mentioned.\n\n[quote=\"vittasko\"]I think that livetolove212 thought about a different result. \n\nSurely, it is well known that the line connecting the midpoints of the segments $ XY, B_1C_1$ he mentioned, passes through the midpoint of the segment  $ AA_1$\n\nBest regards, Kostas Vittas.[/quote]\r\n\r\nDear [b]Mr. Kostas Vittas[/b], I think it is not true.",
  "Solution_17": "[quote=\"livetolove212\"]\nWhy is it not true?\nActually, this problem is [url=http://www.mathlinks.ro/viewtopic.php?p=1474889#1474889]Vietnam TST 2009-problem 5[/url] and I see that it is also an application of [b]ERIQ theorem[/b] [b]Mr. Kostas Vittas[/b] has mentioned.\n\n[quote=\"vittasko\"]I think that livetolove212 thought about a different result. \n\nSurely, it is well known that the line connecting the midpoints of the segments $ XY, B_1C_1$ he mentioned, passes through the midpoint of the segment  $ AA_1$\n\nBest regards, Kostas Vittas.[/quote]\n\nDear [b]Mr. Kostas Vittas[/b], I think it is not true.[/quote]\r\nSee fig:$ A,J,M$ is not collinear.\r\nDear  [b]Mr. Kostas Vittas[/b] $ J,K,M$ is on $ Gauss$ line of $ AC_1HB_1$.",
  "Solution_18": "$ (AB)$ is the circle with diameter $ AB$.",
  "Solution_19": "[quote=\"livetolove212\"]$ (AB)$ is the circle with diameter $ AB$.[/quote]\r\n :ewpu: Sorry for my blindness dear [b]livetolove212[/b]. I was sure that you thought something different ( than the [b]Gauss line[/b] ) but, I didn't see your notation of $ (AB)$ as the circle with diameter $ AB$.\r\n\r\nBest regards, Kostas Vittas.",
  "Solution_20": "Dear Mr. [b]Kostas Vittas, Elementarry[/b] and Mathlinkers,\r\nSorry for my careless style.\r\nBut, any idea for this problem? With another expression of [b]Vietnam TST 2009-problem 5[/b], I think we would have a shorter solution for it.",
  "Solution_21": "Use [b]Lemma[/b] I posted:\r\nA line through $ A$ parallel $ BC$ cut $ A_1C_1,A_1B_1$ at $ R,S$.\r\nFrom $ A_1(R,A,S,C)\\equal{}\\minus{}1$ we have $ AR\\equal{}AS$.\r\nAnd not hard to prove that $ \\frac {YC_1}{YR}\\equal{}\\frac {XB_1}{XS}$ then from [b]Lemma[/b]\r\nwe have $ A,I,J$ are collinear."
}
{
  "Tag": [
    "blogs",
    "search"
  ],
  "Problem": "I'm a total beginner at Linux, and I want to install it one a brand new slave hard drive.  I have a beginners book, and they recommend [url=http://www.mandriva.com/en/download/mandrivaone]Mandriva[/url] but the problem is, I don't know which version to download.  I'm a total noob, so I want a graphical interface.  The book talks about KDE, so I think I should download KDE, except it talks about \"non-free drivers.\"  What exactly does this mean?\r\n\r\nSince I wasn't sure, I just downloaded the Purely Free version.  Will I still be able to use the KDE window manager?\r\n\r\nAgain, I'm totally new, so any beginner's tips would be useful.  Also, I may switch to Ubuntu, but I want to get good at Linux first, and everyone tells me Mandriva is the way to go for beginners.",
  "Solution_1": "If I remember correctly KDE comes with Mandriva, so you won't need to download anything. Non-free drivers are just the official Nvidia or ATI drivers, which should automatically be installed. Go on and jump in; the best way to learn isn't from the book but from trying.",
  "Solution_2": "I actually started with Ubuntu and it's rather easy. Once you have the disk (LOOONG download), the installation is fairly easy. It should be able to automatically resize your windows partition (if you have one) to make space for its partitions.\r\n\r\nI don't know anything about Mandriva, and the link you provided didn't really help, but Ubuntu has a nice GUI and applications that can do a lot of the administrative stuff that you'd often do in a terminal, but to get good with the terminal you have to use the terminal. Just don't play around with superuser too much ;). Personally, I've always been told that Ubuntu is for beginners and Gentoo is for the more experienced.",
  "Solution_3": "Mandriva is a pain.  I can use the Live CD, but it causes problems when I try to install to the external hard drive.  I'm gonna go ahead and format the hard drive to kill Mandriva, and then try again with Ubuntu.  If I do that, will it automatically kill the bootloader Mandriva installed on my Windows drive?\r\n\r\nAlso, I think I am lucky enough to have a modem that works with Linux!",
  "Solution_4": "[quote=\"mysmartmouth\"]Mandriva is a pain.  I can use the Live CD, but it causes problems when I try to install to the external hard drive.  I'm gonna go ahead and format the hard drive to kill Mandriva, and then try again with Ubuntu.  If I do that, will it automatically kill the bootloader Mandriva installed on my Windows drive?\n\nAlso, I think I am lucky enough to have a modem that works with Linux![/quote]\r\n\r\nIt'll update the bootloader, but you'll still have a (non-working) Mandriva entry in the bootloader. (Incidentally, Ubuntu will automatically format for you :))\r\n\r\nIf you have to have dialup, congrats on getting a good modem!\r\n\r\nGentoo is for people who have lots of time and/or have specialized hardware or want to learn more than they ever wanted to know about their computer. Ubuntu is kinda like macs in that it just works, but Gentoo has more stuff you can easily install. Further, Gentoo is faster once compiled.",
  "Solution_5": "Hm, Ubuntu wouldn't install.  I totally uninstalled everything Linux after much trial and tribulation.  I am going to buy a new computer when Vista comes out, and then convert this to exclusively Linux.\r\n\r\nBut I came out of this with some Linux experience, as well as extensive troubleshooting experience (see [url=http://seansoni.blogspot.com]my blog[/url]).  Also, I learned how to use the command line more effectively.",
  "Solution_6": "Getting started with Ubuntu was very easy for me...their forums(http://www.ubuntuforums.org) were a huge help in setting it up and stuff.  All you have to do is enter a search for your specific problem and they usually have a few threads on how to help you out with it.  \r\n\r\nNow, I use strictly Ubuntu and I don't think I'll be going back to Windows again.  I'd recommend Ubuntu as a beginner distro because of its friendly GUI interface that helps you adjust to linux.",
  "Solution_7": "I like both Ubuntu and SuSE, but I personally think Mandriva sucks and it's way too hard to install anything. (K)ubuntu is the best distro I've used so far (I've used ubuntu, suse, mandriva, gentoo, and maybe others that I don't remember right now).",
  "Solution_8": "I agree with Chess64. I started with SuSE several years ago, and have since used Gentoo, CentOS, Slackware and Kubuntu. Kubuntu is easily the nicest distro I've used - really painless installation (especially given that up until that point I'd been using Gentoo everywhere, which has an install process that can last literally days), stable system and nice package management. \r\n\r\nEach distro has its strong points, so I suppose it's really a personal choice. SuSE has a nice config tool (YaST), Gentoo (despite what a lot of people say) does indeed help you learn more about the workings of the OS, as long as you're willing to do more than mindlessly follow the install process (and if you're very patient :P) and CentOS is nice if you want to play around with server-side stuff. Kubuntu (and Ubuntu I suppose) is generally easy to use, and lets you get on with whatever you want to do, without much hassle. \r\n\r\nGood luck whichever you choose :)"
}
{
  "Tag": [
    "floor function",
    "Olimpiada de matematicas"
  ],
  "Problem": "sean $a_{1},a_{2},a_{3},...,a_{100}$ son numeros $R+$, \r\nsi: $a_{1}\\geq a_{2}\\geq a_{3}\\geq ...\\geq a_{100}\\geq 0$, \r\n     $a_{1}+a_{2}\\leq 100$, $a_{3}+a_{4}+a_{5}+...+a_{100}\\leq 100$, \r\nhallar el maximo valor de $a_{1}^{2}+a_{2}^{2}+...+a_{100}^{2}$, y indicar para que configuraciones de $(a_{1}, a_{2},...,a_{100})$ se da este maximo valor.",
  "Solution_1": "Este problema esta horrible. O por lo menos asi esta mi solucion.\r\nEl m\u00e1ximo valor que toma la suma es $100^{2}$.\r\n\r\n[b]Demostracion:[/b]\r\nSea $s = \\sum_{i=1}^{n}a_{i}^{2}$\r\nSea $t=\\sum_{i=3}^{n}a_{i}$\r\nSea $u=\\sum_{i=3}^{n}a_{i}^{2}$\r\n\r\nLo primero es que $a_{2}\\le 50$ pues es menor o igual que $a_{1}$.\r\nFijemos primero $a_{2}$. Notemos que la suma $a_{1}^{2}+a_{2}^{2}$ se maximiza si hacemos $a_{1}$ maximo es decir $a_{1}= 100-a_{2}$.\r\n\r\nAhora si $98a_{2}\\le 100$ es decir $a_{2}\\le \\frac{50}{49}$ tenemos que:\r\n$s\\le (100-a_{2})^{2}+99a_{2}^{2}= 100^{2}-200a_{2}+100a_{2}^{2}\\le 100^{2}$ con igualdad si $a_{2}=2$ que no se puede pues $2> \\frac{50}{49}$ o si $a_{2}=0$\r\n\r\nEntonces podemos asumir que $a_{2}> \\frac{50}{49}$.\r\nSea $k=\\lfloor \\frac{100}{a_{2}}\\rfloor$\r\nSabemos que $t=100$, de lo contrario podemos empezar a sumarle valores a los $a_{i}$ ($i\\ge3$) y obtenemos sumas mayores de los cuadrados(hasta que la suma de los $a_{i}$ sea $100$). \r\nAhora probaremos que:\r\n$u \\le ka_{2}^{2}+(100-ka_{2})^{2}$\r\n\r\nSabemos que:\r\n$a_{i}\\le a_{2}$ para $i\\ge 3$ (de ahora en adelante $i,j \\ge 3$)\r\nAhora haremos smoothing para probar lo que queremos.\r\nSi $a_{i}+a_{j}\\le a_{2}$ tomamos $a'_{i}= a_{i}+a_{j}$ y $a'_{j}=0$ y $a'_{i}^{2}+a'_{j}^{2}=(a_{i}+a_{j})^{2}\\ge a_{1}^{2}+a_{2}^{2}$(notemos que el orden de los $a_{i}$ para $3\\le i\\le 100$ no importa, lo unico que importa es que $a_{2}\\ge a_{i}\\ge 0$. Al final se pueden ordenar).\r\nSi $a_{i}+a_{j}= a_{2}+\\alpha$ con $a_{2}> \\alpha \\ge 0$ ponemos:\r\n$a'_{i}=a_{2}$ y $a'_{j}=\\alpha$ y tenemos que:\r\n$a_{2}^{2}+\\alpha^{2}= (a_{2}+\\alpha)^{2}-2a_{2}\\cdot \\alpha \\ge (a_{i}+a_{j})^{2}-2a_{i}\\cdot a_{j}= a_{i}^{2}+a_{j}^{2}$ pues $2a_{i}\\cdot a_{j}\\ge 2a_{2}\\cdot\\alpha$ pues suman lo mismo pero $a_{i}, a_{j}$ se acercan mas al promedio.\r\nCon esto queda probado lo que queriamos, que es tener la maxima cantidad de $a_{i}= a_{2}$ (que son $k$) y el resto en un solo $a_{j}$.\r\nLlamemos $a_{2}= a$ para que sea mas corto de escribir. \r\nLo que queremos probar ahora es: \r\n$(100-a)^{2}+a^{2}+ka^{2}+(100-ak)^{2}\\le 100^{2}$ $\\iff$\r\n$50+\\frac{(k^{2}+k+2)a^{2}}{200}\\le a(k+1)$ \r\nPero sabemos que $k=\\frac{100}a-r$ con $0\\le r < 1$\r\nReemplazando obtenemos que lo anterior se cumple si y solo si:\r\n$a(r^{2}-r+2) \\le 100$ despues de expandir y cancelar terminos.\r\nPero $r^{2}-r \\le 0$ pues $0\\le r <1$ \r\nEntonces tenemos que:\r\n$a(r^{2}-r+2) \\le 2a \\le 100$ que es cierto pues $a\\le 50$ \r\n\r\nPor lo que tenemos que el minimo $s$ es en efecto $100^{2}$ y la igualdad se da solamente en los siguientes dos casos:\r\n$a_{1}=100$ y $a_{2}=a_{3}=...=a_{100}=0$\r\n$a_{1}=a_{2}=a_{3}=a_{4}=50$ y $a_{5}=a_{6}=...=a_{100}=0$\r\nQED",
  "Solution_2": "Nadie tiene una forma mas bonita de solucionar este problema? He estado pensandolo y no se me a ocurrido nada. :mad:"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "rectangle",
    "calculus",
    "calculus computations"
  ],
  "Problem": "I need help with another EVP.\r\n\r\nA window has perimeter 10 m and is the shape of a rectangle with the top edge replaced by a semicircle. Find the dimensions of the rectangle if the window admits the greatest amount of light.\r\n\r\nThanks!",
  "Solution_1": "$ P\\equal{}(\\pi \\plus{}2)r\\plus{}2h$\r\n\r\n$ A\\equal{} \\frac{\\pi}{2} r^2 \\plus{}2rh$\r\n\r\n$ 2h\\equal{}10\\minus{}(\\pi\\plus{}2)r$\r\n\r\n$ A(r)\\equal{}\\frac{\\pi}{2}r^2\\plus{}10r\\minus{}(\\pi \\plus{}2)r^2\\equal{}\\minus{}(\\frac{\\pi}{2}\\plus{}2)r^2\\plus{}10r$\r\n\r\n$ A'(r)\\equal{}\\minus{}(\\pi\\plus{}4)r\\plus{}10$, $ A'(r)\\equal{}0 \\to r\\equal{}\\frac{10}{\\pi\\plus{}4}\\equal{}h$\r\n\r\nso, we get height above with width twice that. (for the rectangular dimensions)"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "In how many ways can you locate 17 different books in 3 different boxes?\r\n\r\nI dont understand how to resolve this exercise. I have the solution 129140163",
  "Solution_1": "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51376&search_id=837198339]Take a look here[/url]",
  "Solution_2": "thanks kunny, so for the first part is variations with repetition.",
  "Solution_3": "Don't mention it. :) You are right.",
  "Solution_4": "What about the second part:\r\n\r\nHow many ways if all the boxes must be full?\r\n\r\nThe solution: 128746950\r\n\r\nApparently is 3^17 - 3(2^17 - 1)\r\n\r\nI dont understand that number 3 multiplying (2^17 - 1) ?? why??"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "This problem has been posted before on the forum, but I would like to see a different proof, preferably without substitutions.\r\n\r\nGiven $a,b,c,x,y,z\\geq 0$ and $x+y+z=1$, prove that:\r\n\r\n$ax+by+cz+2\\sqrt {(ab+bc+ca)(xy+yz+zx)}\\leq a+b+c$\r\n\r\nThanks!",
  "Solution_1": "Any ideas?  I am finding this problem quite frustrating.",
  "Solution_2": "$a+b+c=\\sqrt{\\left(a^2+b^2+c^2+2(ab+bc+ca)\\right)\\left(x^2+y^2+z^2+2(xy+yz+zx)\\right)} \\ge ax + by + cz + 2\\sqrt{(ab+bc+ca)(xy+yz+zx)}$\r\nCauchy!",
  "Solution_3": "very cool!",
  "Solution_4": "I don't know whether to laugh at myself or thank you...so I'll just do both.  Thanks."
}
{
  "Tag": [
    "logarithms",
    "algebra",
    "functional equation",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Show that the functional equation\r\n\r\n$ f\\left(\\frac{2x}{1+x^{2}}\\right) = (1+x^{2})f(x)$\r\n\r\nis satisfied by\r\n\r\n$f(x) = 1+\\frac{1}{3}x^{2}+\\frac{1}{5}x^{4}+\\frac{1}{7}x^{6}+\\cdots$ with $|x| < 1$.",
  "Solution_1": "Simply check that $\\frac{\\arctan(iz)}{iz}=\\frac{-1}{2z}\\ln \\frac{1-z}{1+z}$, verifies the functional equation."
}
{
  "Tag": [
    "graph theory",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Let a set $X$ be given which consists of $2 \\cdot n$ distinct real numbers ($n \\geq 3$). Consider a set $K$ consisting of some pairs $(x, y)$ of distinct numbers $x, y \\in X$, satisfying the two conditions:\r\n\r\n[b]I.[/b] If $(x, y) \\in K$ then $(y, x) \\not \\in K$.\r\n[b]II.[/b] Every number $x \\in X$ belongs to at most 19 pairs of $K$.\r\n\r\nShow that we can divide the set $X$ into 5 non-empty disjoint sets $X_1, X_2, X_3, X_4, X_5$ in such a way that for each $i = 1, 2, 3, 4, 5$ the number of pairs $(x, y) \\in K$ where $x, y$ both belong to $X_i$ is not greater than $3 \\cdot n$.",
  "Solution_1": "nobody?   ",
  "Solution_2": "This is a graph theory problem, the numbers are vertices and the pairs are edges. For each $x\\in X$ let $d(x)$ be the degree (number of edges) of $x$. If $U,V$ are subsets of $X$ let $d(U,V)$ be the number of edges between $U$ and $V$. For each decomposition of $X$ into components, an edge $xy$ is called intercomponent if $x$ and $y$ belongs to different components. Let $X = (A,B,C,D,E)$ be the decomposition such that the number of intercomponent edges are maximal. For each $a \\in A$ we claim that $d(x,A) \\leq d(x,B)$. If not, moving $a$ to $B$ would increase the number of intercomponents. Similar for $B,C,D,E$. Thus $5d(x,A) \\leq d(x,A) +d(x,B)+d(x,C)+d(x,D)+d(x,E) \\leq 19$ hence $d(x,A) \\leq 3$ for each $a \\in A$.\nTherefore, $d(A,A) \\leq 3n$. Same argument for other components.",
  "Solution_3": "This is basically the same argument that shows up more commonly when we want only 2 disjoint sets (the max-cut problem):\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=392464"
}
{
  "Tag": [
    "pigeonhole principle"
  ],
  "Problem": "Prove: If there are 6 people at a party, then either 3 of them knew each other before the party or 3 of them were complete strangers before the party.",
  "Solution_1": "In fact you can do better: there are at least 2 such triples.",
  "Solution_2": "Wouldn't that just be common sense?  Since there are a total of six people, if 3 people didn't know each other, then 4 people would have know each other.  The sum of a+b=6.  Either a or b must be 3 or greater.\r\n\r\nOr is it something way different?",
  "Solution_3": "[quote=\"hoshattack\"]The sum of a+b=6.[/quote]\r\n\r\nConsider a party with three pairs of friends.  If by \"a\" you mean \"the largest group of people that know each other\" and by \"b\" you mean \"the largest group of people that don't know each other,\" then $ a \\equal{} 2, b \\equal{} 3, a \\plus{} b \\equal{} 5$.\r\n\r\nThe general problem is far from common sense, and is in fact an outstanding problem in graph theory.  See [url=http://mathworld.wolfram.com/RamseyNumber.html]Ramsey numbers[/url].",
  "Solution_4": "This is equivalent to saying that $ K_{6}$ has a monochromatic triangle, right?",
  "Solution_5": "[quote=\"diophantient\"]This is equivalent to saying that $ K_{6}$ has a monochromatic triangle, right?[/quote]\r\n\r\nExactly.  Assuming that you mean each edge in $ K_6$ is colored one of two colors and a monochromatic triangle is three vertices where each edge is the same color.  Or with pleurestique's addition, that there are two such triangles.",
  "Solution_6": "[hide=\"solution\"]put it on a graph, red=friends, blue=non-friends. consider any vertex A, by pigeonhole some 3 edges, say AB,AC,AD going out of it have to be the same color, say red. if any of BC,CD,DB are red then we get a red triangle so they're all blue, but then we get a blue triangle.[/hide]",
  "Solution_7": "[quote=\"hoshattack\"]Wouldn't that just be common sense?  Since there are a total of six people, if 3 people didn't know each other, then 4 people would have know each other.  The sum of a+b=6.  Either a or b must be 3 or greater.\n\nOr is it something way different?[/quote]\r\n\r\nPigeon hole principle:\r\n\r\nIf we have n+1 pigeons and want to keep them in n boxes(jaja), no matter how we do this there will always at least 2 pigeons in one box.\r\n\r\nTraduced: common sense. \r\n\r\n :rotfl: \r\n\r\n\r\nThis is just a ver specif way to define it, and I'm very far to be saying this is useless.\r\n\r\n\r\n :)"
}
{
  "Tag": [
    "linear algebra"
  ],
  "Problem": "Do there exist 10 distinct integers, the sum of any 9 of which is a perfect square?",
  "Solution_1": "does integer mean >= 0 ?\n\n\n\n\n\nIf not : \n\n-3050936351, 48427561, 392801329, 431065081, 435316609, 435789001, 435841489, 435847321, 435847969, 435848041 \n\nis a solution !\n\n\n\n[hide]\n\nUsing linear algebra :\n\n\n\nA*X = Y where \n\n\n\nA = \n\n(1,1,...,1,0)\n\n(1,1,...,0,1)\n\n..\n\n(0,1,...,1,1)\n\n\n\nA is inversible (is it the right word ? = not singular, det != 0), and B=A^1 is \"routine\" to evaluate. Also det(A)=9.\n\nThen take V = (3^2,3^4, ..., 3^20)\n\n\n\nX = B*V is then a solution.\n\n\n\n\n\nIf it's an olympiad level problem, no doubt there is an easy (and less ugly   ) solution. At least we now know a solution exists.[/hide]"
}
{
  "Tag": [],
  "Problem": "How to do this ?\r\n\r\nA plane has been partitioned into $n$ regions by 3 sets of parallel lines such that no three meet in a point. What is the least number of lines required such that $n>2001$  ?",
  "Solution_1": "Hint: Consider what happens when you add one new line to an arbitrary partition."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Given $ a, b, c \\geq\\ 0.$ Prove that:\r\n$ (a\\plus{}b\\plus{}c)^3 \\geq\\ 27abc \\plus{} 6\\sqrt[]{3}(a\\minus{}b)(b\\minus{}c)(c\\minus{}a)$",
  "Solution_1": "Try $ a \\equal{} \\min\\{a,b,c\\},$ $ b \\equal{} a \\plus{} x$ and $ c \\equal{} a \\plus{} y.$\r\n If your inequality is true then this way helps to you. :wink:",
  "Solution_2": "I know. But can U check for me it is trues or not. If it's trues, it's very nice althought easy  :)",
  "Solution_3": "Equality occurs when $ a \\equal{} b \\equal{} c > 0$ or $ a \\equal{} k(3 \\minus{} \\sqrt []{3}); b \\equal{} k(3 \\plus{} \\sqrt []{3}); c \\equal{} 0$ and any cyclic."
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "I know this must be an easy problem to be in a BC calculus exam, but I'm just drawing a blank:\r\n\r\nA particle moves along the x-axis so thats its acceleration at any time t is a(t) = 2t - 7. If the initial velocity of the particle is 6, at what time t during the interval 0<= t <= 4 is the particle farthest to the right?",
  "Solution_1": "$v(t)=\\int a(t) dt=t^2-7t+Const.$, from $v(0)=6$, we have $v(t)=t^2-7t+6=(t-1)(t-6)$.\r\nLet $x(t)$ be the dislocation at the time $t$, $x'(t)=v(t)$, for $0\\leqq t\\leqq 4$,yielding the answer $t=1$.",
  "Solution_2": "thanks  :)"
}
{
  "Tag": [
    "modular arithmetic",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove that there are infinite prime numbers of the form $ 4n\\minus{}1$, where $ n \\in N$.",
  "Solution_1": "$ P_{1}\\equal{}7,...,P_{n}\\equal{}4k\\plus{}3$,$ i>j$ $ \\implies$  $ P_{i}>P_{j}$, $ (P_{i}\\equiv \\minus{}1\\pmod{4})$\r\n$ \\implies$  $ 4.(P_{1}.P_{2}...P_{n})\\plus{}3\\equal{}q$\r\nif  $ q$  is prime  $ \\implies$  $ q>P_{n}$  contradiction!  \r\nif  $ q$  is coprime $ \\implies$  ,$ q$ has divisor same form.But $ P{i}\\nmid q$  contradiction!",
  "Solution_2": "[quote=\"Obel1x\"]Prove that there are infinite prime numbers of the form $ 4n \\minus{} 1$, where $ n \\in N$.[/quote]\r\n\r\nLet $ p_1 ,p_2 , ...,p_n$ be all prime numbers.\r\n\r\nSuppose that number $ 4 p_1 p_2 ...pn \\minus{} 1$  is composite \r\n\r\nThen it has at least one prime divisor $ 1 < p_i \\leq n$ , but number $ 4p_1 p_2 ...p_n \\minus{} 1$ is not dividable by any of $ p_i$ ,contradiction"
}
{
  "Tag": [
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "I've posted this problem in computational calculus... section, but nobody solved it. I consider this is hard.\r\nfind $y(x)$ so that\r\n\\[ (2+2x \\sqrt{y})x dx + (2+x^2 \\sqrt{y})y dy =0  \\]",
  "Solution_1": "[quote=\"Beginner\"]I've posted this problem in computational calculus... section, but nobody solved it. I consider this is hard.\nfind $y(x)$ so that\n\\[ (2+2x \\sqrt{y})x dx + (2+x^2 \\sqrt{y})y dy =0  \\][/quote]\r\n\r\nMaybe try to find an integrating factor to make it exact.",
  "Solution_2": "I don't think that help.\r\nlet $k(x)$ be the integrating factor\r\nthen we need to find $k(x)$ so that\r\n\\[ \\frac{\\partial (2xk(x)+2x^2 k(x) \\sqrt{y})}{\\partial y} = \\frac{\\partial (2yk(x)+x^2 y k(x) \\sqrt{y})}{\\partial x}  \\]\r\nthis yield\r\n\\[ x^2 k(x) \\sqrt{y} = k'(x) y^2 (2+x^2 \\sqrt{y}) + 2x k(x) y^2 \\sqrt{y}  \\]  :ninja: \r\n :(  :? \r\ni also try using subtitutioun $u=x^2$ and $v=y^2$ but i don't know how to continue\r\nanybody solve it?\r\n\r\nNOTE: My friend from other faculty ask me to compute this ODE, maybe this appears in his modeling of problem, and maybe there is initial value for this ODE so that it can solve using numerical method."
}
{
  "Tag": [
    "ratio"
  ],
  "Problem": "There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?",
  "Solution_1": "[quote=\"Jos\u00e9\"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]\r\nWhat is the ratio of boys to girls?",
  "Solution_2": "[hide]$b+g=210\\\\ \\frac{1}{2}g+\\frac{1}{3}b=85\\\\ \\frac{1}{2}g+\\frac{1}{3}(210-g)=85\\\\ \\frac{1}{6}g=15\\\\ g=90\\\\ b=120$\n[/hide]",
  "Solution_3": "[quote=\"Jos\u00e9\"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]\r\n\r\n[hide]Call the number of girls $x$ and the number of boys $y$\n.Thus, $x+y=210$. Half the girls and a third of the boys on bicycles can be represented as $\\frac{x}{2}+\\frac{y}{3}=85$.\nSolving both as yields x=90, y=120.[/hide]",
  "Solution_4": "[quote=\"mtms5467\"][quote=\"Jos\u00e9\"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]\nWhat is the ratio of boys to girls?[/quote]\r\n\r\nyou don't need to know the ratio, otherwise it would be trivial  :wink:",
  "Solution_5": "[quote=\"daermon\"][quote=\"mtms5467\"][quote=\"Jos\u00e9\"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]\nWhat is the ratio of boys to girls?[/quote]\n\nyou don't need to know the ratio, otherwise it would be trivial  :wink:[/quote]\r\nOh. Whoops. :oops: \r\n\r\n[hide=\"answer\"]Call the number of boys $a$ and the number of girls $b$. We can set up the equations $\\frac{1}{2}b+\\frac{1}{3}a=85$ and $a+b=210$. Solving we find $a=120$ and $b=90$.[/hide]",
  "Solution_6": "correct!  :thumbup:",
  "Solution_7": "[hide]$a+b=210$\n$\\frac a2+\\frac b3=85$\n$3a+2b=510$\n$2a+2b=420$\n$a=90$\n$b=120$\n[/hide]",
  "Solution_8": "[quote=\"Jos\u00e9\"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]\r\n\r\nok so w=women and b=boys\r\n\r\n1/2 w + 1/3 b = 85\r\nw+b=210\r\nw= 210-b\r\n\r\n(210-b)/2 + 1/3 b =85\r\n105- 1/2b + 1/3 b= 85 \r\n1/6b=20 \r\nb=120\r\ng= 90\r\n\r\n( Is this right :?: )",
  "Solution_9": "210(you said bluntly boys AND girls)\r\n\r\nor if you meant the num. of boys and the num. of girls....[hide]120 boys, 90 girls.[/hide]",
  "Solution_10": "[quote=\"Jos\u00e9\"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]\r\n\r\n[hide=\"...\"]\nThere are 90 girls, and 120 boys, but you didn't exactly ask a question...\n\nmtms5467, the ratio is 4:3...[/hide]",
  "Solution_11": "easy:90 girls and 120 boys",
  "Solution_12": "I think there are 90 girls and 120 boys. Is that right? :?:",
  "Solution_13": "Yes, it's right. I think everybody knows the answer, so it's useless to post another solution",
  "Solution_14": "[quote=\"Jos\u00e9\"]There are $210$ people in a gym. Half of the women and the third part of the boys are each at a bicycle. If there are $85$ bicycles occupied, how many boys and girls are there at the gym?[/quote]\r\n\r\n\r\n[hide]\n$b+g=210$\n$b/3+g/2=85$\n$b+3g/2=255$\n$b+g=210$\n$g/2=45$\n$g=90$\n$b=210-90$\n$b=120$[/hide]"
}
{
  "Tag": [
    "number theory",
    "prime numbers",
    "relatively prime"
  ],
  "Problem": "If the quotient of a/b has a period of \"c\" figures and (b-1)/c results to be a non-integer number, then \"b\" is NOT a prime number.  \"a\" and \"b\" are two integers, primes between them.   \r\n  \r\nFor instance, from 2/57 results the quotient 0,035087719298245614 which has a period of 18 figures and because the result of (57-1)/18 is 3,1 which is a non-integer, then 57 is not prime.  \r\n\r\nI have to mention that this conjecture doesn't determine if \u201cb\u201d is prime but if \u201cb\u201d is not prime, since it does not succeed in finding all numbers that are not prime.  \r\nThe non-prime numbers that the conjecture does not find on an interval between 3 and 5000 (from which all multiples of 2 and 5 have been excluded) are: 9 - 33 - 91 - 99 - 259 - 451 - 481 - 561 - 657 - 703 - 909 - 1233 - 1729 -  2409 - 2821 - 2981 - 3333 - 3367 - 4141 - 4187 - 4521 -  \u2026\r\n  \r\nThe Carmichael numbers are a subset of the whole non-prime numbers which the conjecture can not identify.\r\nI would appreciate any suggestions or demonstrations regarding this conjecture. Thank you all in advance.",
  "Solution_1": "This conjecture is true- it's a special case of Fermat's \"little\" theorem. As a computational method, it's entirely contained in the idea of pseudoprimes (base 10 in this case).\r\n\r\nAssume that $ b$ is relatively prime to $ 10$ and $ a$ is relatively prime to $ b$; the alternative isn't interesting. The period length $ c$ is the smallest positive $ k$ for which $ 10^k\\equiv 1\\mod b$. If $ b$ is prime, $ 10^{b\\minus{}1}\\equiv 1\\mod b$. Since the only $ k$ with $ 10^k\\equiv 1$ are the multiples of $ c$, $ c|b\\minus{}1$.",
  "Solution_2": "[quote=\"jmerry\"]This conjecture is true- it's a special case of Fermat's \"little\" theorem. As a computational method, it's entirely contained in the idea of pseudoprimes (base 10 in this case).\n\nAssume that $ b$ is relatively prime to $ 10$ and $ a$ is relatively prime to $ b$; the alternative isn't interesting. The period length $ c$ is the smallest positive $ k$ for which $ 10^k\\equiv 1\\mod b$. If $ b$ is prime, $ 10^{b \\minus{} 1}\\equiv 1\\mod b$. Since the only $ k$ with $ 10^k\\equiv 1$ are the multiples of $ c$, $ c|b \\minus{} 1$.[/quote]\r\n\r\n\r\nCould you show me with a little example how can you find the unknown digit \"x\" of 2/7=0,285X14, developing all the operations? Thank you.",
  "Solution_3": "[quote=\"Explorer\"]Could you show me with a little example how can you find the unknown digit \"x\" of 2/7=0,285X14, developing all the operations? Thank you.[/quote]\r\n\r\nI already answered this question in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201696]other thread[/url].  Here is a computational demonstration:\r\n\r\nWe want to examine $ k \\equal{} \\frac{10^6 \\minus{} 1}{7}$.  We can easily compute that $ k \\equal{} 142857, ak \\equal{} 285714$ which gives your answer directly, or we can do the following:\r\n\r\nSince $ 10^6 \\equiv 1 \\bmod 11$, we know that $ 11 | k$.  Then $ 11 | ak \\equal{} 285X14$, which leaves remainder\r\n\r\n$ (4 \\plus{} X \\plus{} 8) \\minus{} (1 \\plus{} 5 \\plus{} 2) \\equiv 4 \\plus{} X \\bmod 11$\r\n\r\nwhich gives $ X \\equal{} 7$.",
  "Solution_4": "[quote=\"t0rajir0u\"][quote=\"Explorer\"]Could you show me with a little example how can you find the unknown digit \"x\" of 2/7=0,285X14, developing all the operations? Thank you.[/quote]\n\nI already answered this question in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201696]other thread[/url].  Here is a computational demonstration:\n\nWe want to examine $ k \\equal{} \\frac {10^6 \\minus{} 1}{7}$.  We can easily compute that $ k \\equal{} 142857, ak \\equal{} 285714$ which gives your answer directly, or we can do the following:\n\nSince $ 10^6 \\equiv 1 \\bmod 11$, we know that $ 11 | k$.  Then $ 11 | ak \\equal{} 285X14$, which leaves remainder\n\n$ (4 \\plus{} X \\plus{} 8) \\minus{} (1 \\plus{} 5 \\plus{} 2) \\equiv 4 \\plus{} X \\bmod 11$\n\nwhich gives $ X \\equal{} 7$.[/quote]\r\n\r\n\r\nSorry, I have posted in the wrong thread. Thanks a lot for making me notice that and also for your answer."
}
{
  "Tag": [
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "It is common to develop the real numbers axiomatically. That is, we take $\\mathbb{R}$ to be a set which satisfies the axioms of an ordered field, and which also satifies one additional axiom. What should that one additional axiom be? I offer seven possible choices:\r\n\r\n1. Every nonempty subset that is bounded above has a least upper bound.\r\n\r\n2. Suppose $A\\cup B=\\mathbb{R}, A\\cap B=\\emptyset, A\\ne\\emptyset, B\\ne\\emptyset$, and for all $a\\in A,b\\in B, a<b$, then either $A$ has a greatest element or $B$ has a least element.\r\n\r\n3. Monotone bounded sequences converge.\r\n\r\n4. Cauchy sequences converge.\r\n\r\n5. If $I_1\\supset I_2\\supset I_3\\supset\\cdots$ are nested closed bounded intervals, then $\\bigcap_{n=1}^{\\infty}I_n\\ne\\emptyset$.\r\n\r\n6. Bounded sequences have convergent subsequences.\r\n\r\n7. Every decimal expansion converges to a real number, and every real number has at least one decimal expansion.\r\n\r\n(Note: the second phrase in #7 doesn't look like completeness, but without it, we wouldn't have the Archimedean property.)\r\n\r\nThe most common textbook path through this would start by assuming 1 as an axiom. I would consider the following implications \"textbook obvious\":\r\n$1\\implies 3\\implies 5\\implies 6\\implies 4, 1\\implies 2, 3\\implies 7.$\r\n\r\nI have two challenges:\r\n\r\nFirst, to show that all seven of these statements are equivalent; that we could start with any one as the axiom and prove the other six as theorems.\r\n\r\nSecond, to prove as many of the 42 direct implications as possible without going through one of the other statements. Clearly, you can't do all of these, but which ones can you do?\r\n\r\nAny construction of the real numbers should lead fairly quickly to one of these properties.",
  "Solution_1": "What about the Dedekind method  ( Dedekind slice )of constructing real numbers??\r\n\r\n ( A real number is a set of rational numbers with some properties )",
  "Solution_2": "The dedekind methode is very similar to/exactly the same as the second axiom.\r\n(Dedekind axioms: a dedekindian slice is a (ordered) pair $(A,B)$ of two sets of rational numbers that fulfil $A\\cup B=\\mathbb{Q}, A\\cap B=\\emptyset, A\\ne\\emptyset, B\\ne\\emptyset, \\forall a \\in A, b \\in B : a<b$ and $B$ has no least element).\r\nBut I think there is an other problem: not all these axioms guarant the Archimedean property (for example the 4.)"
}
{
  "Tag": [],
  "Problem": "If in the formula $ C \\equal{} \\frac {en}{R\\plus{}nr}$, $n$ is increased while $ e$, $R$ and $r$ are kept constant, then $C$:\n\n$\\textbf{(A)}\\ \\text{Increases} \\qquad\n\\textbf{(B)}\\ \\text{Decreases} \\qquad\n\\textbf{(C)}\\ \\text{Remains constant} \\qquad\n\\textbf{(D)}\\ \\text{Increases and then decreases} \\qquad\\\\\n\\textbf{(E)}\\ \\text{Decreases and then increases}$",
  "Solution_1": "[hide=\"solution\"]e, R, and r are constants, so we arbitrarily assign values to them.  To keep it simple, we make all equal to 1.  \n\n$ C = \\frac{1*n}{1+n*1}$\n\n$ C = \\frac{n}{1+n}$\ntesting n=1, 2, 3, we get 1/2, 2/3, 3/4, respectively, or 6/12, 8/12, 9/12.  \nWith the numerator just 1 less than the denominator, C is progressing towards 1.  \n\nThus, C increases, and the answer is A.  [/hide]",
  "Solution_2": "[quote=\"Tecnogram888\"][hide=\"solution\"]e, R, and r are constants, so we arbitrarily assign values to them.  To keep it simple, we make all equal to 1.  \n\n$ C = \\frac{1*n}{1 + n*1}$\n\n$ C = \\frac{n}{1 + n}$\ntesting n=1, 2, 3, we get 1/2, 2/3, 3/4, respectively, or 6/12, 8/12, 9/12.  \nWith the numerator just 1 less than the denominator, C is progressing towards 1.  \n\nThus, C increases, and the answer is A.  [/hide][/quote]\r\nWhat if $ R=0$? Or what if $ R<0$?",
  "Solution_3": "[hide=\"sol.\"]en/R+nr-en+e/R+nr+r=enR+ern^2+enr-enR-ern^2-er-enr/(R+nr)(R+nr+r)\n=-er/(R+nr)(R+nr+r).\nSo, it increases if er is pos., decreases if er is neg.[/hide]",
  "Solution_4": "Sorry for revival, but I'm not quite sure I understand the answer. Can anyone clarify?",
  "Solution_5": "$C=\\frac{e}{r+\\frac{R}{n}}$ which, clearly, is increasing with $n$.",
  "Solution_6": "Not if R is negative, e is positive, etc.",
  "Solution_7": "How would you solve this without testing the cases of\n\nNegative.  Negative\n\nPositive.  Positive\n\nNegative positive \n\nPositive negative\n\n\n\nAnd so what exactly is the correct answer?",
  "Solution_8": "The official answer key says B, but i think the problem isn't clearly stated as it doesnt say where n starts",
  "Solution_9": "$C(R+nr) = en$\n$CR = en-Cnr$\n$CR = n(e-Cr)$\n$\\frac{CR}{e-Cr} = n$\nAssume R is positive for now.\n$\\frac{e-Cr}{C} \\cong \\frac1n$\nI use congruent to mean in/decreases in the same direction.\nNote as C increases, the numerator decreases, and the denominator increases, so $\\frac1n$ decreases, so n increases.\nHowever, n decreases if R is negative.",
  "Solution_10": "One could argue that C is correct if e=0 or R=0"
}
{
  "Tag": [
    "trigonometry",
    "blogs",
    "complex numbers"
  ],
  "Problem": "Does anyone know of a good trigonometry/complex number book?",
  "Solution_1": "I used a regular precal book when I learned precal a few years ago (it was Larson's). Most books do not properly cover problem solving so I reccomend ACoPS and the USSR problem book for good introductory chapters into easy trig/complex numbers problems. Or you could wait until the AoPS book. :)",
  "Solution_2": "103 trigonometry problems by titu andreescu",
  "Solution_3": "Depends what you want- if its a problem book you're after, 103 Trigonometry Problems is a gem. For a challenging introductory text however, good trigonometry-centric books are far and few these days. You could go with the classics which are as good as ever- like the Michigan University reprint of S L Loney's Trigonometry (volumes 1 and 2, where 2 introduces complex numbers); or Dover's edition of E W Hobson's book.",
  "Solution_4": "book S L Loney's Trigonometry  link: http://rapidshare.com/files/141478225/Loney-Plane_trigonometry.rar.html",
  "Solution_5": "there is also a blog called IIT dreams where you can find like a whole bunch of stuff."
}
{
  "Tag": [
    "probability",
    "combinatorics proposed",
    "combinatorics",
    "TST",
    "China TST"
  ],
  "Problem": "Let $ S$ be a set that contains $ n$ elements. Let $ A_{1},A_{2},\\cdots,A_{k}$ be $ k$ distinct subsets of $ S$, where $ k\\geq 2, |A_{i}| \\equal{} a_{i}\\geq 1 ( 1\\leq i\\leq k)$. Prove that the number of subsets of $ S$ that don't contain any $ A_{i} (1\\leq i\\leq k)$ is greater than or equal to $ 2^n\\prod_{i \\equal{} 1}^k(1 \\minus{} \\frac {1}{2^{a_{i}}}).$",
  "Solution_1": "Fang-jh: Which year's TST is this?\r\n\r\nI don't see any reason to require the sets $ A_i$ to be different, by the way. The condition $ a_i\\geq 1$ is equally useless.\r\n\r\n  darij",
  "Solution_2": "[quote=\"darij grinberg\"]The proof is flawed. $ H_{A\\cup B} \\equal{} H_A\\cup H_B$, not $ H_{A\\cup B} \\equal{} H_A\\cap H_B$. Mind the difference between \"do not contain $ A$\" and \"do not intersect $ A$\".\n\nFang-jh: Which year's TST is this?\n\nI don't see any reason to require the sets $ A_i$ to be different, by the way. The condition $ a_i\\geq 1$ is equally useless.\n\n  darij[/quote]\r\nit is problem 3 of the fourth quiz of Chinese TST 2008  (March 22,2008)",
  "Solution_3": "Sorry for a potentially stupid question, but what is a \"quiz\"? Is it just a round of the TST, or is it something easier than the usual TST's?\r\n\r\n  darij",
  "Solution_4": "[quote=\"darij grinberg\"]Sorry for a potentially stupid question, but what is a \"quiz\"? Is it just a round of the TST, or is it something easier than the usual TST's?\n\n  darij[/quote] \r\nThere are 6 quizzes in total, it is easier than the usual TST.",
  "Solution_5": "It directly follows from Kleitman lemma (which is itself quite subtle):\r\n\r\nIf $ U$ and $ V$ are two collections of subsets of a set $ S$, each of them closed under taking subsets (i.e. $ A\\in U$, $ B\\subset A$ implies $ B\\in A$) then $ |U\\cap V|\\ge |U|\\cdot |V|/2^{|S|}$.\r\n\r\nThat is, events ``to belong to $ U$'' and ``to belong to $ V$' for randomly chosen subset of $ S$ correlate positively (or how to say it?)",
  "Solution_6": "Incorrect, sorry:\n\nSo let us consider a random subset $X$ of $S$ that is constructed by placing each element of $S$ in $X$ with probability $\\frac{1}{2}$. Now, consider the probability that some specific $A_i$ is completely contained in $X$. That means for each $x \\in A_i$, we must have placed $x$ in $X$, and this happens with probability $\\frac{1}{2^{a_i}}$. Thus, the probability that $A_i$ is not completely contained in $X$ is $1 - \\frac{1}{2^{a_i}}$. Thus, the probability that no $A_i$ is contained in $X$ is simply $ \\prod_{i = 1}^k(1-\\frac{1}{2^{a_{i}}}). $ But there are $2^n$ total subsets of $S$, and then the result follows. $\\blacksquare$",
  "Solution_7": "[quote=\"TheStrangeCharm\"]Thus[/quote] These are not independent events!",
  "Solution_8": "[quote=JBL][quote=\"TheStrangeCharm\"]Thus[/quote] These are not independent events![/quote]\n\nSure,you can't multiply them like this!",
  "Solution_9": "Sorry for reviving \nBut is there any solution with expected value?",
  "Solution_10": "Take a random subset $U$ of $S$. Let $E_i\\,,\\,i=1,2,\\dots,k$ be the event that $U$ doesn't contain $A_i$. We will prove \n$$P(E_1\\cap E_2\\cap\\dots\\cap E_k)\\geq \\prod_{i = 1}^k(1 - \\frac {1}{2^{a_{i}}}).$$\nThe proof goes by induction. We start with the equality:\n\\[P(E_2\\cap\\dots\\cap E_k)=P(E_1)P_{E_1}(E_2\\cap\\dots\\cap E_k)+P(\\overline{E_1})P_{\\overline{E_1}}(E_2\\cap\\dots\\cap E_k).\\]\nIt follows:\n\\begin{align*}(*)\\,\\,\\,\\,\\,\\,\\,\\,\\, P(E_1\\cap\\dots\\cap E_k) &= P(E_1)P_{E_1}(E_2\\cap\\dots\\cap E_k) = \\\\\n&= P(E_2\\cap\\dots\\cap E_k) - P(\\overline{E_1})P_{\\overline{E_1}}(E_2\\cap\\dots\\cap E_k)\\\\ \n&\\geq P(E_2\\cap\\dots\\cap E_k)- P_{\\overline{E_1}}(E_2\\cap\\dots\\cap E_k). \\end{align*}\n\nLet's consider the probability $P_{\\overline{E_1}}(E_2\\cap\\dots\\cap E_k)$. Let us denote $A'_i := A_i\\setminus A_1\\,,\\,i=2,3,\\dots,k$. We want to estimate the probability the random set $U$ to contain $A_1$ and $U_1 :=U\\setminus A_1$ not to contain any $A'_i\\,,\\,i=2,\\dots,k$. To each such $U_1$ corresponds $2^{a_1}$ sets $U :=U_1\\cup U'$, where $U'\\subset A_1$ not containing any $A'_i\\,,\\,i=2,\\dots,k$. \nIt means: \n$$P_{\\overline{E_1}}(E_2\\cap\\dots\\cap E_k) =2^{-a_1}P(E'_2\\cap\\dots\\cap E'_k) \\leq  2^{-a_1}P(E_2\\cap\\dots\\cap E_k) $$\nHere the event $E'_i,i=2,\\dots,k$ means the random set $U$ not to contain $A'_i$.\n\nNow, we use the induction hypothesis, so $P(A_2\\cap\\dots\\cap A_k)\\geq \\prod_{i = 2}^k(1 - \\frac {1}{2^{a_{i}}})$. Then $(*)$ yields:\n\\[P(A_1\\cap\\dots\\cap A_k)\\geq \\prod_{i = 2}^k(1 - \\frac {1}{2^{a_{i}}}) (1-\\frac{1}{2^{a_1}})=\\prod_{i = 1}^k(1 - \\frac {1}{2^{a_{i}}}).\\] \nwhich completes the induction step.",
  "Solution_11": "$\\textbf{Solution :}$\n\nInduction on $\\sum_{i\\not= j} |A_i\\cap A_j|$. Consider larger set $S':= S\\cup\\{n+1\\} = [1, n+1]$ and same subsets $A_1,\\ldots , A_k$ of it. So it's enough to prove that number of subsets of $S'$ that don't contain any $A_i$ is greater than or equal to $2^{n+1}\\prod_{i=1}^k(1-\\frac{1}{2^{a_i}})$. We can assume that $\\exists a\\in A_1\\cap A_2$. Consider new subsets of $S'$ : $\\{n+1\\}\\cup A_1\\setminus\\{a\\}, A_2, A_3, \\ldots , A_k$. Consider next numbers :\n\n1) Let $X$ be number of subsets of $S'$ that don't contain any $A_i$.\n\n2) Let $X_{n+1}^a$ be number of subsets of $S'$ that don't contain any $A_i$ and also don't contain element $n+1$ but contain element $a$.\n\n3) Let $X_{a}^{n+1}$ be number of subsets of $S'$ that don't contain any $A_i$ and also don't contain element $a$ but contain element $n+1$.\n\n4) Let $X_{n+1, a}$ be number of subsets of $S'$ that don't contain any $A_i$ and also don't contain elements $n+1, a$.\n\n5) Let $X^{n+1, a}$ be number of subsets of $S'$ that don't contain any $A_i$, but contain elements $n+1, a$.\n\n6) Let $Y$ be number of subsets of $S'$ that don't contain any $A_i$ for $i\\not= 1$ and don't contain set $\\{n+1\\}\\cup A_1\\setminus\\{a\\}$.\n\n7) Similarly define numbers $Y_{n+1}^a, Y_{a}^{n+1}, Y_{n+1, a}, Y^{n+1, a}$.\n\nBy induction hypothesis we know that $Y \\geq 2^{n+1}\\prod_{i=1}^k(1-\\frac{1}{2^{a_i}})$, so it's enough to prove that $X\\geq Y$.\n\n$\\textbf{Lemma 1 :}$\n\nWe get that $X_{a}^{n+1} + X_{n+1}^a\\geq Y_{a}^{n+1} + Y_{n+1}^a$.\n\n$\\textbf{Lemma 2 :}$\n\nWe get that $X_{n+1, a} = Y_{n+1, a}$.\n\n$\\textbf{Lemma 3 :}$\n\nWe get that $X^{n+1, a} = Y^{n+1, a}$.\n\nProofs of lemmas 1-3 is simple exercise. So, finally, we get that $X = (X_{a}^{n+1} + X_{n+1}^a) + X_{n+1, a} + X^{n+1, a}\\geq (Y_{a}^{n+1} + Y_{n+1}^a) + Y_{n+1, a} + Y^{n+1, a}\\geq 2^{n+1}\\prod_{i=1}^k(1-\\frac{1}{2^{a_i}})$. $\\Box$",
  "Solution_12": "Here's a solution using a nuke.\n\n[b] Theorem: [/b] ([url=\"https://en.wikipedia.org/wiki/FKG_inequality\"]FKG Inequality[/url]) Let $X_1, \\dots, X_n$ be decreasing functions on $\\{0, 1\\}^n$ (or any lattice). Then $\\mathbb{E}[\\prod X_i] \\ge \\prod \\mathbb{E}[X_i].$\n\nNow to finish use this inequality for the variables $X_i(S) = 0$ if $A_i \\subseteq S$ and $1$ otherwise.",
  "Solution_13": "Keeping stuff low-powered, I guess.\n-------\nWe begin with the key lemma.\n\n[color=#c00][b]Lemma.[/b][/color] If we replace any element $x$ in one of the $X_i$s with an element $y$ that does not appear in any of the other $X_i$s, then the number of subsets of $\\{1,2,\\cdots,n\\}$ not containing any of the other $X_i$s does not increase.\n\n[color=#c00][i]Proof.[/i][/color] WLOG the $X_i$ having $x$ replaced with $y$ has index $1$. Consider the $X_i$: $X_1\\setminus \\{x\\}, X_2, X_3, \\ldots$. There are two classes of subsets of $\\{1,2,\\ldots,n\\}$ that work for $(X_1\\setminus \\{x\\})\\sqcup \\{y\\}, X_2, \\cdots, X_m$: Ones that also work $X_1\\setminus \\{x\\}, X_2, X_3, \\ldots$, and ones that do not.\n\nNotice that all of the subsets that work for $X_1\\setminus \\{x\\}, X_2, X_3, \\ldots$ also work for both $(X_1\\setminus \\{x\\})\\sqcup \\{y\\}, X_2, \\cdots, X_m$ and $X_1,X_2,\\cdots X_m$. \n\nNow, we look at the subsets that do not work for \\[(X_1\\setminus \\{x\\}), X_2, X_3\\cdots, X_m\\] but do work for $X_1, X_2, \\cdots, X_m$. These subsets are exactly those which contain $X_1\\setminus \\{x\\}$ that do not contain $x$ and $X_2, X_3, \\cdots, X_m$.\n\nSimilarly, we look at the subsets that do not work for \\[(X_1\\setminus \\{x\\}), X_2, X_3\\cdots, X_m\\] but do work for $(X_1\\setminus \\{x\\})\\sqcup \\{y\\}, X_2, X_3,\\cdots, X_m$. These subsets are exactly those which contain $X_1\\setminus \\{x\\}$ that do not contain $y$ and $X_2, X_3, \\cdots, X_m$.\n\nThus, it suffices to show that there are at least as many subsets that contain $X_1\\setminus \\{x\\}$ that do not contain $x$ and $X_2, X_3, \\cdots, X_m$ than those that contain $X_1\\setminus \\{x\\}$ that do not contain $y$ and $X_2, X_3, \\cdots, X_m$. But notice that if a subset contains both $x$ and $y$ or neither $x$ nor $y$ and works for the second case, it also works for the first case. If a subset $S$ contains only $x$ and works for the second case, then $(S\\setminus \\{x\\})\\sqcup \\{y\\}$ also works for the first case. Thus, the first is at least as large, and the proof of the Lemma is complete.\n---------\nNow, we show the problem for $\\sum_i |X_i|\\leq n$ and $|X_i|$ fixed for each $X_i$.\n[color=#c00][i]Proof.[/i][/color] If we have two equal numbers in two of the $X_i$, then we know that we can replace one of them with a number that does not appear in any of the $X_i$. By the size assumption, we won't run out of numbers, so the minimum number of subsets not containing any of the $X_i$ is attained when all of the $X_i$ are disjoint. Then, by a simple counting argument (e.g. using proportions, similar to the proof of the formula of Euler's Totient Function), we have the desired bound of \\[2^n\\cdot\\prod_{i=1}^{m}\\left(1-\\frac{1}{2^{|X_i|}}\\right).\\]\n------\nFinally, we show the problem for arbitrary fixed $|X_i|$. Let $N$ be a number larger than $\\sum_{i}|X_i|$. Then, we have, for any choice of $X_i$,\n\\begin{align*}\n\\text{[\\# subsets of $[n]$ not containing any $X_i$]} &= \\frac{1}{2^{N-n}}\\text{[\\# subsets of $[N]$ not containing any $X_i$]}\\\\\n&\\geq \\frac{1}{2^{N-n}}\\cdot 2^N\\cdot\\prod_{i=1}^{m}\\left(1-\\frac{1}{2^{|X_i|}}\\right)\\\\\n& =2^n\\cdot\\prod_{i=1}^{m}\\left(1-\\frac{1}{2^{|X_i|}}\\right)\n\\end{align*}\nas requested. $\\blacksquare$",
  "Solution_14": "Ha! This will be a good example to illustrate a basic spirit of probablistic method - making counting simpler!"
}
{
  "Tag": [
    "number theory",
    "least common multiple",
    "algebra",
    "polynomial",
    "IMO",
    "IMO 1981"
  ],
  "Problem": "[b]a.)[/b] For which $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers?\r\n\r\n[b]b.)[/b] For which $n>2$ is there exactly one set having this property?",
  "Solution_1": "[hide]Let $k = \\prod p_{i}^{e_{i}}$ be the largest element of the set.  Then $k$ divides the least common multiple of the other elements of the set iff the set has cardinality of at least $\\max \\{ p_{i}^{e_{i}}\\}+1$, since for any of the $p_{i}^{e_{i}}$, we must go down at least to $k-p_{i}^{e_{i}}$ to obtain another multiple of $p_{i}^{e_{i}}$.  In particular, there is no set of cardinality $3$ satisfying our conditions, because each number larger than or equal to $3$ must be divisible by a number that is larger than two and is a power of a prime.\n\nFor $n > 3$, we may let $k = \\operatorname{lcm}[n-1, n-2] = (n-1)(n-2)$, since all the $p_{i}^{e_{i}}$ must clearly be less than $n$ and this product must also be larger than $ n$ if $n$ is at least $4$.  For $n > 4$, we may also let $k = \\operatorname{lcm}[n-2, n-3] = (n-2)(n-3)$, for the same reasons.  However, for $ n = 4$, this does not work, and indeed no set works other than $\\{ 3,4,5,6 \\}$.  To prove this, we simply note that for any integer not equal to $6$ and larger than $4$ must have some power-of-a-prime factor larger than $3$.\n\nQ.E.D.[/hide]",
  "Solution_2": "Let $A: (a, a+1, a+2, \\cdots, a+n-1)$ such that $a, n \\in \\mathbb{Z}^{+}.$ In other words, this is our desired set.\r\n\r\n[b]Lemma: $\\exists A$ such that $A:{a, a+1, a+2, \\cdots, a+n-1}$ where $a, n \\in \\mathbb{Z}^{+}$ and $a+n-1|lcm(a, a+1, \\cdots, a+n-2)$ if and only if $(n-1)! | a+n-1.$[/b]\r\n\r\nIf we take the polynomial $P(a) = a(a+1)(a+2)\\cdots(a+n-2)$ and find the remainder when divided by $a+n-1,$ we simply find $\\frac{a+n-1}{P(-n+1)}.$ If this remainder is equal to zero, then we can solve for $a.$ If $a$ is an acceptable value, we can then construct an appropriate set $A.$ Note that\r\n\r\n$P(-n+1) = (-n+1)(-n+2)(-n+3)\\cdots(2)(1) = (-1)^{n-1}(n-1)!$\r\n$\\Leftrightarrow \\frac{a+n-1}{P(-n+1)}= \\frac{a+n-1}{(-1)^{n-1}(n-1)!}$\r\n\r\nThus, if $(n-1)!|a+n-1,$ we have that there exists a valid set $A$ with cardinality $n.$ $\\blacksquare$\r\n\r\nWe will now commence proving parts A and B; this lemma will be used in both parts.\r\n\r\n[hide=\"Part A\"]\n[b]Claim: For $\\all n \\ge 4,$ there exists such a set $A.$[/b]\n[b]Proof:[/b]\n\nIf $a = (n-1)!-(n-1)$ and $a > 0,$ or\n\n$(n-1)!-(n-1) > 0 \\Leftrightarrow (n-1)! > (n-1) \\Leftrightarrow n \\ge 4,$\n\nthere exists such a sequence, according to our lemma. Such a sequence would be: $A: ((n-1)!-n+1, (n-1)!-n+2, (n-1)!-n+3, \\cdots, (n-1)!+1).$\n\nThus, we can say that a valid sequence exists for $\\all n \\ge 4.$ $\\blacksquare$\n\n[/hide]\n[hide=\"Part B\"]\n[b]Claim: Only for $n = 4$ does a unique $A$ exist.[/b]\n[b]Proof:[/b]\n\nAccording to our lemma, we have that $(n-1)! | a+n-1.$ The largest such $a$ that satisfies this is $(n-1)!-(n-1).$ The next largest that satisfies this is $(n-2)!-(n-1).$ However, since we only want one solution, we let $(n-2)!-(n-1) \\le 0 \\Leftrightarrow n \\le 4.$\n\nSince $2 < n \\le 4$ and $n = 3$ doesn't yield any valid $A,$ the only such $n$ is $n = 4.$ The unique sequence $A$ for $n=4$ is $A: (3, 4, 5, 6).$ $\\blacksquare$\n\n[/hide]\r\n\r\nCould someone grade me out of seven points and give me style points as well? I would really appreciate it  :)",
  "Solution_3": "Basically for a set to satisfy the property, the largest number must not have any prime powers that are not divisible by any of the other $n-1$ numbers. \n\nFor a given $n$, let $k_n$ be the largest element. \n\nThe answer for part $a)$ is $\\boxed{n\\ge 4}$. \n\nPart A1: Show that there is no such set for $n=3$. \nSuppose there was. Then the largest element must not have a prime power that is not in the first two elements. So all of it's prime factors are either divisible by the first or second element, which implies they must all be $2$. But $\\nu_2 (k_n)>\\nu_2 (k_n-2)$ in this case, a contradiction. $\\blacksquare$\n\n\n\nPart A2: Show that there always exists a set for $n\\ge 4$. \nWe let $k_n=\\prod_{i=1}^{x} p_i^{a_i}$ for nonnegative integers $a_i$ and $p_i$ is the $i$th prime. We let $p_i^{a_i}<n$ for all $i$. It's clear that this works because any $p_i^{a_i}$ divides some other element as $p_i^{a_i}<n$. Such a product of prime powers  exists for $n\\ge 4$ because the product of all distinct primes less than $n$ is always greater than $n$ (not that this is not the case for $n=3$). $\\blacksquare$\n\nThe answer for part $b)$ is $\\boxed{n=4}$. \n\nPart B1: Show that $\\{3,4,5,6\\}$ is the unique set for $n=4$. \nAgain let $k_4=\\prod_{i=1}^{x} p_i^{a_i}$. Clearly all the $p_i$ are either $2$ or $3$. The highest possible power of $2$ is $1$ and the highest possible power of $3$ is also $1$. But we also need $k_4\\ge 4$, for the set to contain only positive integers. So we must have $a_1=a_2=1$. Thus, $6$ is the only possible value of $k_4$. $\\blacksquare$\n\nPart B2: Show that there exist more than $1$ set for $n>4$. \nLet $k_n=\\prod_{i=1}^{x} p_i^{a_i}$. Let $p_x$ be the largest prime less than $n$. Then set all $a_i$ to $1$. This works as $k_n=\\prod_{i=1}^x p_i>n$. Now since $n>4$, we can also set $a_1=2$, which also works. $\\blacksquare$"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $a,b,c>0$. Prove or disprove that\r\n$a^6b^3+b^6c^3+c^6a^3 \\geq a^5b^2c^2+a^2b^5c^2+a^2b^2c^5$",
  "Solution_1": "It's true by Muirhead ($(6,3,0) \\succeq (5,2,2)$) or more simply by rearrangement ?",
  "Solution_2": "I'm wondering somewhat about that. Can Muirhead be used with non-symmetric sum?",
  "Solution_3": "Indeed you're right, Muirhead can't be used here...",
  "Solution_4": "Then can we use weight AM-GM to help?",
  "Solution_5": "Rather use H\u00f6lder: \\[ \\begin{aligned} & \\left(a^6b^3 + b^6c^3 + c^6a^3\\right)^3 \\\\ = & \\left(a^6b^3 + b^6c^3 + c^6a^3\\right)\\left(a^6b^3 + b^6c^3 + c^6a^3\\right)\\left(c^6a^3 + a^6b^3 + b^6c^3\\right) \\\\ \\geq & \\left(\\sqrt[3]{a^6b^3a^6b^3c^6a^3} + \\sqrt[3]{b^6c^3b^6c^3a^6b^3} + \\sqrt[3]{c^6a^3c^6a^3b^6c^3}\\right)^3 \\\\ = & \\left(a^5b^2c^2 + b^5c^2a^2 + c^5a^2b^2\\right)^3. \\end{aligned}  \\]"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "determine if $ n!\\minus{}1$ is composite for infinitely many $ n$'s, where $ n$ is a positive integer",
  "Solution_1": "$ (p \\minus{} 2)! \\equiv 1 \\mod{ p }$.\r\n\r\nIs it an open question if $ n! \\minus{} 1$ or $ n! \\plus{} 1$ is prime for infinitely many $ n$'s?"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let $ a,b,c \\ge 0, ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that:\r\n$ \\frac {ab \\plus{} 1}{a \\plus{} b} \\plus{} \\frac {bc \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {ca \\plus{} 1}{c \\plus{} a} \\ge 3$\r\n\r\n[b]Vietnam Inequality Forum:[/b] http://www.batdangthuc.net/forum/index.php\r\n[b]Welcome to VI![/b]",
  "Solution_1": "[quote=\"NguyenDungTN\"]Let $ a,b,c \\ge 0, ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that:\n$ \\frac {ab \\plus{} 1}{a \\plus{} b} \\plus{} \\frac {bc \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {ca \\plus{} 1}{c \\plus{} a} \\ge 3$\n\n[b]Vietnam Inequality Forum:[/b] http://www.batdangthuc.net/forum/index.php\n[b]Welcome to VI![/b][/quote]\r\nineq <=> $ p^2 \\plus{} (r \\minus{} 3)p \\plus{} 2r \\plus{} 3\\ge 0$\r\n$ \\delta \\equal{} r^2 \\minus{} 18r \\plus{} 1\\le 0$ (with $ r\\le \\frac {1}{3.\\sqrt {3}}$) \r\nso ineq was proved  :D",
  "Solution_2": "We also have the following result\r\n\\[ \\sum\\frac{a\\plus{}bc}{1\\plus{}ab} \\ge \\frac{3(\\sqrt{3}\\plus{}1)}{4}\r\n\\]\r\nfor any $ a,b,c \\ge 0,ab\\plus{}bc\\plus{}ca\\equal{}1$ :)",
  "Solution_3": "[quote=\"vodanh_tieutot\"][quote=\"NguyenDungTN\"]Let $ a,b,c \\ge 0, ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that:\n$ \\frac {ab \\plus{} 1}{a \\plus{} b} \\plus{} \\frac {bc \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {ca \\plus{} 1}{c \\plus{} a} \\ge 3$\n\n[b]Vietnam Inequality Forum:[/b] http://www.batdangthuc.net/forum/index.php\n[b]Welcome to VI![/b][/quote]\nineq <=> $ p^2 \\plus{} (r \\minus{} 3)p \\plus{} 2r \\plus{} 3\\ge 0$\n$ \\delta \\equal{} r^2 \\minus{} 18r \\plus{} 1\\le 0$ (with $ r\\le \\frac {1}{3.\\sqrt {3}}$) \nso ineq was proved  :D[/quote]\nWhen equality holds?\n[quote=\"can_hang2007\"]We also have the following result\n\\[ \\sum\\frac {a \\plus{} bc}{1 \\plus{} ab} \\ge \\frac {3(\\sqrt {3} \\plus{} 1)}{4}\n\\]\nfor any $ a,b,c \\ge 0,ab \\plus{} bc \\plus{} ca \\equal{} 1$ :)[/quote]\r\nNice problem, Vo Quoc Ba Can!",
  "Solution_4": "[quote=\"vodanh_tieutot\"][quote=\"NguyenDungTN\"]Let $ a,b,c \\ge 0, ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that:\n$ \\frac {ab \\plus{} 1}{a \\plus{} b} \\plus{} \\frac {bc \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {ca \\plus{} 1}{c \\plus{} a} \\ge 3$\n\n[b]Vietnam Inequality Forum:[/b] http://www.batdangthuc.net/forum/index.php\n[b]Welcome to VI![/b][/quote]\nineq <=> $ p^2 \\plus{} (r \\minus{} 3)p \\plus{} 2r \\plus{} 3\\ge 0$\n$ \\delta \\equal{} r^2 \\minus{} 18r \\plus{} 1\\le 0$ (with $ r\\le \\frac {1}{3.\\sqrt {3}}$) \nso ineq was proved  :D[/quote]\r\n\r\nWhy you can say that $ p^2 \\plus{} (r \\minus{} 3)p \\plus{} 2r \\plus{} 3\\ge 0$\r\n\r\nCan you please explain it to me?\r\n\r\nThank you very much.",
  "Solution_5": "[quote=\"vodanh_tieutot\"]Let $ a,b,c \\ge 0, ab \\plus{} bc \\plus{} ca \\equal{} 1$. Prove that:\n$ \\frac {ab \\plus{} 1}{a \\plus{} b} \\plus{} \\frac {bc \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {ca \\plus{} 1}{c \\plus{} a} \\ge 3$\n\n\nineq <=> $ p^2 \\plus{} (r \\minus{} 3)p \\plus{} 2r \\plus{} 3\\ge 0$\n$ \\delta \\equal{} r^2 \\minus{} 18r \\plus{} 1\\le 0$ (with $ r\\le \\frac {1}{3.\\sqrt {3}}$) \nso ineq was proved  :D[/quote]\r\n\r\n\r\nWhy $ \\delta \\equal{} r^2 \\minus{} 18r \\plus{} 1\\le 0$  with $ r\\le \\frac {1}{3.\\sqrt {3}}$????. \r\n\r\nSorry, but I don't see it.  :maybe:  \r\n\r\nCan anyone,please,  explain it to me.  :wink: \r\n\r\nThank you very much.  :)",
  "Solution_6": "NguyenDungTN, can you please post your solution?\r\n\r\nI cannot solve it :( \r\n\r\nThank you very much.",
  "Solution_7": "[quote=\"manlio\"]NguyenDungTN, can you please post your solution?\n\nI cannot solve it :( \n\nThank you very much.[/quote]\r\n[hide=\"Hint\"]We have:\n$ \\frac {ab \\plus{} 1}{a \\plus{} b} \\plus{} \\frac {bc \\plus{} 1}{b \\plus{} c} \\plus{} \\frac {ca \\plus{} 1}{c \\plus{} a} \\ge 3$\n$ \\Leftrightarrow \\sum_{cyc} (ab\\plus{}1)(c\\plus{}a)(c\\plus{}b) \\ge 3(a\\plus{}b)(b\\plus{}c)(c\\plus{}a)$\n$ \\Leftrightarrow \\sum_{cyc} (ab\\plus{}1)(c^2\\plus{}1) \\ge 3[(a\\plus{}b\\plus{}c)(ab\\plus{}bc\\plus{}ca)\\minus{}abc]$\n$ \\Leftrightarrow (a^2\\plus{}b^2\\plus{}c^2)\\plus{}ab\\plus{}bc\\plus{}ca\\plus{}abc(a\\plus{}b\\plus{}c)\\plus{}3\\plus{}3abc \\ge 3(a\\plus{}b\\plus{}c)$\n$ \\Leftrightarrow (a\\plus{}b\\plus{}c)^2\\plus{}abc(a\\plus{}b\\plus{}c\\plus{}3)\\plus{}2 \\ge 3(a\\plus{}b\\plus{}c)$\nNow let $ p\\equal{}a\\plus{}b\\plus{}c,q\\equal{}ab\\plus{}bc\\plus{}ca\\equal{}1,r\\equal{}abc$\nThe inequality becomes:\n$ p^2\\plus{}r(p\\plus{}3)\\minus{}3p\\plus{}2 \\ge 0$\n$ \\star p>2\\Rightarrow p^2\\minus{}3p\\plus{}2\\plus{}r(p\\plus{}3)\\equal{}(p\\minus{}1)(p\\minus{}2)\\plus{}r(p\\plus{}3)>0$\n$ \\star p \\le 2$\nBy Schur inequality, we have:\n$ p^3\\plus{}9r \\ge 4pq$ or $ r \\ge \\frac{4p\\minus{}p^3}{9}$\nSo we need to prove that:\n$ p^2\\minus{}3p\\plus{}2\\plus{}(p\\plus{}3)\\frac{4p\\minus{}p^3}{9} \\ge 0$\n$ \\Leftrightarrow p^4\\plus{}3p^3\\minus{}13p^2\\plus{}15p\\minus{}18 \\le 0$\n$ \\Leftrightarrow (p\\minus{}2)(p^3\\plus{}5p^2\\minus{}3p\\plus{}9)$\nBecause $ p \\le 2$ and $ p^3\\plus{}5p^2\\minus{}3p\\plus{}9\\equal{}p^3\\plus{}4p^2\\plus{}\\left(p\\minus{}\\frac{3}{2}\\right)^2\\plus{}\\frac{27}{4}>0$\nso we are done.\nEquality hold if and only if $ a\\equal{}b\\equal{}1,c\\equal{}0$ and permutations.[/hide]",
  "Solution_8": "Dear NguyenDungTN, \r\n\r\nthank you very much for your nice solution and inequality.  :)",
  "Solution_9": "[quote=\"can_hang2007\"]We also have the following result\n\\[ \\sum\\frac {a \\plus{} bc}{1 \\plus{} ab} \\ge \\frac {3(\\sqrt {3} \\plus{} 1)}{4}\n\\]\nfor any $ a,b,c \\ge 0,ab \\plus{} bc \\plus{} ca \\equal{} 1$ :)[/quote]\r\n\r\n\r\nDear can_hang2007 have you a solution for this very hard inequality? I tried with full expansion by applying p,q,r method but it is almost impossible for me. :oops: \r\n\r\nThank you very much.",
  "Solution_10": "Can't we see the solution of the first problem without using $ pqr$ method?",
  "Solution_11": "[quote=\"can_hang2007\"]We also have the following result\n\\[ \\sum\\frac {a \\plus{} bc}{1 \\plus{} ab} \\ge \\frac {3(\\sqrt {3} \\plus{} 1)}{4}\n\\]\nfor any $ a,b,c \\ge 0,ab \\plus{} bc \\plus{} ca \\equal{} 1$ :)[/quote]\r\nMy urgly solution.",
  "Solution_12": "Dear can_hang2007, \r\n\r\nthank you very much for your solution at this very hard inequality. :)"
}
{
  "Tag": [
    "LaTeX",
    "function",
    "USAMTS"
  ],
  "Problem": "Hi, \r\nI looked on the latex guide and couldn't find much. \r\nIs there any way to page break?\r\nFore xample, say im doing texniccenter latex, and have 1.5 pages for problem 1. And i want problem 2 to be on the third page. How do i do that without doing \"\\linbreak\" the needed number of times?\r\n\r\nIs there a \\pagebreakuntilendofpage type function?\r\n\r\nThanks",
  "Solution_1": "If you use the USAMTS template (which can be found on the USAMTS website), page breaks will be added automatically between solutions.",
  "Solution_2": "If you don't want to do that, you can use \\newpage",
  "Solution_3": "yep, thanks!",
  "Solution_4": "Also, it says \"You only need to submit this file and the PDF file with your solutions (and please attach any images you include).\" \r\n\r\nWhy can't I just submit the pdf file?",
  "Solution_5": "That's outdated...you can just upload the PDF file.",
  "Solution_6": "Hmmm I've always used \\begin{newpage} to make a new page.  It typically produces a few bad box errors and stuff though.  Is there another \\lpagebreakish command?"
}
{
  "Tag": [],
  "Problem": "Given $ 2x^{2} \\plus{} y^{2} \\plus{}2xy \\minus{}12x \\minus{}8y \\minus{}5\\equal{}0$, find the maximum and minimum values of $ 7x\\plus{}4y$.",
  "Solution_1": "hello, the maximum ist $ 47$ and will be attained for $ x\\equal{}5,y\\equal{}3$ , the minimum will reached for $ x\\equal{}\\minus{}1,y\\equal{}1$ and is $ \\minus{}3$.\r\nSonnhard.",
  "Solution_2": "This problem is readily solved by geometric means after [hide=\"a bit of algebra\"]\nWrite as:\n[color=darkblue][i]Given $ x^2 \\plus{} (x \\plus{} y)^2 \\minus{} 4x \\minus{} 8(x \\plus{} y) \\plus{} 5 \\equal{} 0$, find the bounds on $ 3x \\plus{} 4(x \\plus{} y)$[/i][/color]\n\nPerform the change of variables $ z \\equal{} x \\plus{} y$ to rewrite as:\n[color=darkblue][i]Given $ x^2 \\plus{} z^2 \\minus{} 4x \\minus{} 8z \\plus{} 5 \\equal{} 0$, find the bounds on $ 3x \\plus{} 4z$[/i][/color]\n\nComplete squares and define an auxiliary variable $ M$ to rewrite as:\n[color=darkblue][i]Given $ (x \\minus{} 2)^2 \\plus{} (z \\minus{} 4)^2 \\equal{} 15$ and $ 3x \\plus{} 4y \\equal{} M$, find the bounds on $ M$.[/i][/color][/hide]",
  "Solution_3": "Thanks Sonnhard and TZF."
}
{
  "Tag": [
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ (G,.)$ is a group . \r\n\r\n$ \\exists n \\in N \\ ; \\ n >1 : (a.b)^n\\equal{}a^n.b^n \\ , \\forall a,b \\in G$ .\r\n\r\nProve that : \r\n\r\ni ) $ (a.b)^{n\\minus{}1}\\equal{}b^{n\\minus{}1}.a^{n\\minus{}1}$\r\n\r\nii) $ a^{n}.b^{n\\minus{}1}\\equal{}b^{n\\minus{}1}.a^{n}$\r\n\r\n$ \\forall a,b \\in G$ .",
  "Solution_1": "1) $ a^nb^n\\equal{}(ab)^n\\equal{} a(ba)^{n\\minus{}1}b\\Rightarrow a^{n\\minus{}1}b^{n\\minus{}1}\\equal{}(ba)^{n\\minus{}1}$\r\n2) $ a^nb^{n\\minus{}1}\\equal{}aa^{n\\minus{}1}b^{n\\minus{}1}\\equal{}a(ba)^{n\\minus{}1}\\equal{}(ab)^{n\\minus{}1}a\\equal{}b^{n\\minus{}1}a^{n\\minus{}1}a\\equal{}b^{n\\minus{}1}a^n$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "limit",
    "Functional Analysis",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $A$ be a contraction ($\\|A\\|\\le 1$) on a complex Hilbert space, and let $D$ be the unit disc in the plane. Then, for every complex polynomial $p$, the following equality holds: $\\|p(A)\\|\\le\\sup_{z\\in D}|p(z)|$.",
  "Solution_1": "I think this was used in my analysis class as a step in the proof that every linear constant coefficient PDE has a weak solution. However my professor went over the proof quite quickly and I didn't understand very well... I will think on it though :)",
  "Solution_2": "Spectral radius $r(A)=\\lim\\|A^{n}\\|^{1/n}\\le1$ so $\\mathrm{Sp}\\,A\\subset D$ and thus $\\|p(A)\\|=\\max|p\\,(\\mathrm{Sp}\\,A)|\\le\\max|p\\,(D)|$.",
  "Solution_3": "[quote=\"Vladimir Lubyshev\"]$\\|p(A)\\|=\\max|p\\,(\\mathrm{Sp}\\,A)|$[/quote]\r\n\r\nIn other words, the norm of $p(A)$ is equal to its spectral radius? This certainly isn't true for all operators.",
  "Solution_4": "Oh... yes it is true for $A$ self-adjoint and $p$ having no poles inside $\\mathrm{Sp}\\,A$. I forgot that.\r\n\r\nI'll think on the problem when I have time. But do you know for sure the statement is true?",
  "Solution_5": "I even have a proof. It's only here so that you guys can enjoy it :)."
}
{
  "Tag": [
    "pigeonhole principle",
    "number theory solved",
    "number theory"
  ],
  "Problem": "The real number x between 0 and 1 has decimal representation .a<sub>1</sub>a<sub>2</sub>a<sub>3</sub>a<sub>4</sub>... with the following property: the number of distinct blocks of the form a<sub>k</sub>a<sub>k + 1</sub>a<sub>k + 2</sub>...a<sub>k + 2003</sub> as k ranges through all positive integers, is less than or equal to 2004. Prove that x is rational.",
  "Solution_1": "Did anyone think of this yet? Quite a nice question, but not so easy I think... I got somewhere using Pigeonhole Principle but I didn't manage to finish it off so far."
}
{
  "Tag": [
    "articles"
  ],
  "Problem": "I do. I've found a lot of tension between a few countries. I've also found a few illegal things in there.",
  "Solution_1": "What is it?\r\n\r\nA magazine?",
  "Solution_2": "Yup, I do suscribe to it.  How the hell do you find tension in a magazine?  And what illegal stuff are you referring to?",
  "Solution_3": "Using preformance enhancing drugs to improve your preformance. One guy hit farther than Babe Ruth, all thanks to drugs.",
  "Solution_4": "[quote=\"nutz_for2.718281828\"]Using preformance enhancing drugs to improve your preformance. One guy hit farther than Babe Ruth, all thanks to drugs.[/quote]\r\nThat was an article a couple of issues back, wasn't it?  A magazine reporting about illegal stuff, I don't see how that's particularly unusual...",
  "Solution_5": "its ok, but it has too much stuff.  i like The Week and Time better",
  "Solution_6": "[quote=\"nutz_for2.718281828\"]I do. I've found a lot of tension between a few countries. I've also found a few illegal things in there.[/quote]\r\nWhat sort of tensions,and what sort of illegal things are we speaking about?",
  "Solution_7": "[quote=\"star99\"]its ok, but it has too much stuff.  i like The Week and Time better[/quote]\r\n\r\nI think it is a great magazine and extremely insightful when compared to those two magazines (the quality of Time has decreased dramatically, IMO, it is now nothing more than a pile of cultural biases plopped on top of several events and alleged \"trends\").  I do not usually read as much of it as I'd like though, just cause it's kinda dense."
}
{
  "Tag": [
    "geometry",
    "Pythagorean Theorem"
  ],
  "Problem": "The baseball catcher needs to stop the player on first base from stealing second base. Approximately how far must the catcher throw the ball to get to second base? A regular baseball diamond is a square 90 feet on each side.\r\n\r\nA) 90 ft B) 127 ft C) 180 ft D) 64 ft E)100 ft",
  "Solution_1": "Off the top of my head, I know the answer is 127.  The distance the catcher has to throw is the hypotenuse of the right triangle formed by the first and second base paths.  So it's a triangle, 90 feet on two sides.  The third side is the desired distance.  The Pythagorean Theorem works."
}
{
  "Tag": [
    "group theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "I just started my group theory class:\r\n\r\n$ \\mathbb{Q}'$ of $ \\mathbb{Q}$ is the collection of cauchy sequences in $ \\mathbb{Q}$ mod the equivalence relation of being co-Cauchy. The arithmetic is defined by:\r\n\r\n$ A\\plus{}B \\equal{} (a_{n} \\plus{}b_{n})$\r\n$ A\\minus{}B \\equal{} (a_{n} \\minus{}b_{n})$\r\n$ AB \\equal{} (a_{n} *b_{n})$\r\n$ A/B \\equal{} (a_{n} /b_{n})$\r\n\r\nHow do you check if this is a field? Can someone show an example? Thanks!",
  "Solution_1": "Presumbably you have the definition of \"field\" ... check it.\r\nPart of this will be to check that that your operations are well-defined.  For example, to check that + is well defined, you need:  if $ (a_n)$ and $ (u_n)$ are co-Cauchy and $ (b_n)$ and $ (v_n)$ are co-Cauchy, then show $ (a_n\\plus{}b_n)$ and $ (u_n\\plus{}v_n)$ are co-Cauchy.",
  "Solution_2": "How come \"well definedness\" is not included amongst the other definitions of the field that I see such as closure, associativity, distributivity, and commutativity?\r\n\r\nHow do you show co-Cauchy with the above? Do you just add each element together and just show that each resulting sum still forms a Cauchy sequence?",
  "Solution_3": "Well-definedness is more likely a part of \"the addition/etc. can be defined that way\". That is, you need to show that the result is again a Cauchy sequence and that it doesn't depend on the equivalence class."
}
{
  "Tag": [
    "ratio",
    "geometry",
    "circumcircle",
    "conics",
    "geometric transformation",
    "reflection",
    "homothety"
  ],
  "Problem": "Hi, everyone  :roll: \r\n\r\nProve the following:\r\nif R be any point on a circle, A and B fixed points on a diameter and equidistant from the center,\r\nthe envelope of a line which cuts harmonically the two circles with A, B as centers and AR, BR as radii\r\nis independent of the position of R on the circle.\r\n\r\nThanks  :)\r\nPlease your answers. I am studying materials related to this post nowadays.",
  "Solution_1": "Let $ (O)$ be the circle with $ R$ on it. Let a line $ l$ cut the circle $ (A)$ at $ X, Y$ and the circle $ (B)$ at $ Z, T.$ Assume that the cross ratio $ \\frac {\\overline{XZ}}{\\overline{XT}} \\cdot \\frac {\\overline{YZ}}{\\overline{YT}} \\equal{} \\minus{} 1$ is harmonic. Let $ M, N$ be midpoints of $ XY, ZT$ and $ (M), (N)$ circles with diameters $ XY, ZT,$ intersecting at $ P, P'.$ Since $ PX \\perp PY,$ these lines bisect $ \\angle ZPT$ internally and externally, $ (M)$ is P-Apollonius circle of $ \\triangle PZT$ with circumcircle $ (N)$ $ \\Longrightarrow$ $ (M) \\perp (N).$ Inversion in $ (M)$ takes $ (N)$ to itself and $ Z$ to $ T$ $ \\Longrightarrow$ $ \\overline{MZ} \\cdot \\overline{MT} \\equal{} \\minus{} \\overline{MX} \\cdot \\overline{MY}$ $ \\Longrightarrow$ $ \\frac {p(M, (B))}{p(M, (A))} \\equal{} \\minus{} 1.$ In exactly the same way, $ \\frac {p(N, (B))}{p(N, (A))} \\equal{} \\minus{} 1.$ Since powers of $ M, N$ to $ (A), (B)$ are in the same ratio $ r \\equal{} \\minus{} 1,$ $ M, N$ are on a circle coaxal with $ (A), (B),$ centered on $ AB$ and passing through $ R.$ As $ \\frac {\\overline{AO}}{\\overline{BO}} \\equal{} r \\equal{} \\minus{} 1,$ $ \\odot(MNR) \\equiv (O).$ The feet $ M, N$ of perpendiculars $ AM \\perp l, BN \\perp l$ from the fixed points $ A, B$ to the arbitrary line $ l$ are therefore on the fixed circle $ (O)$ $ \\Longrightarrow$ $ l$ is tangent to a conic with foci $ A, B$ and pedal circle $ (O).$",
  "Solution_2": "Suppose that $ A,B$ lie inside the circle $ (O)$ centered at the midpoint of $ AB$ with radius $ OR \\equal{} \\varrho.$ Let the variable line $ \\ell$ cut $ (A)$ at $ P,Q$ and $ (B)$ at $ X,Y.$ Let $ M$ denote the orthogonal projection of $ A$ on $ \\ell,$ which is the midpoint of the chord $ PQ.$ Since $ (P,Q,X,Y) \\equal{} \\minus{} 1,$ by Newton's theorem we have $ MP^2 \\equal{} MX \\cdot MY$ $ \\Longrightarrow$ $ p(M,(A)) \\equal{} p(M,(B)).$ Hence $ M$ lies on the circle centered at the midpoint of $ AB$ and coaxal with $ (A)$ and $ (B),$ i.e. $ M \\in (O).$ Let $ A'$ be the reflection of $ A$ across $ \\ell$ and let $ N \\equiv \\ell \\cap BA'.$ $ A'$ moves on the homothetic circle $ (B, 2\\varrho)$ of $ (O)$ through the homothety centered at $ A$ with coefficient $ 2.$ Therefore $ NA \\plus{} NB \\equal{} NA' \\plus{} NB \\equal{} 2\\varrho \\equal{} \\text{const}$ $ \\Longrightarrow$ Locus of $ N$ is the ellipse $ \\mathcal{E}$ with foci $ A,B$ and major axis $ 2\\varrho.$  Since $ \\ell$ is the external bisector of $ \\angle ANB,$ it follows that $ \\ell$ is tangent to $ \\mathcal{E}.$ When $ A,B$ lie outside $ (O),$ analogous reasoning yields that $ \\ell$ is tangent to the hyperbola with foci $ A,B$ and major axis $ 2\\varrho.$"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "$ a,b,c,p,q\\in R^\\plus{}$\r\n$ abc\\equal{}1$ and $ p\\ge q$\r\nfind minimum value of $ \\dfrac{a^p}{b^q\\plus{}c^q}\\plus{}\\dfrac{b^p}{c^q\\plus{}a^q}\\plus{}\\dfrac{c^p}{a^q\\plus{}b^q}$",
  "Solution_1": "I think that p and q are in IN and $ p > q$\r\nChebychev:\r\n$ S \\equal{} \\dfrac{a^p}{b^q \\plus{} c^q} \\plus{} \\dfrac{b^p}{c^q \\plus{} a^q} \\plus{} \\dfrac{c^p}{a^q \\plus{} b^q}\\geq \\frac {1}{3}(a^p \\plus{} b^p \\plus{} c^p)\\left(\\sum \\frac {1}{a^q \\plus{} b^q}\\right)$\r\ncaushy:\r\n$ \\left(\\sum \\frac {1}{a^q \\plus{} b^q}\\right)(2(a^q \\plus{} b^q \\plus{} c^q))\\geq9$\r\n$ \\Leftrightarrow \\left(\\sum \\frac {1}{a^q \\plus{} b^q}\\right)\\geq\\frac {9}{2(a^q \\plus{} b^q \\plus{} c^q)}$\r\ntherefor:\r\n$ S \\geq\\frac {3}{2}\\times\\frac {a^p \\plus{} b^p \\plus{} c^p}{a^q \\plus{} b^q \\plus{} c^q}$\r\nwith Chebychev:\r\n$ S \\geq \\frac {3}{2}\\times\\frac {a^p \\plus{} b^p \\plus{} c^p}{a^q \\plus{} b^q \\plus{} c^q} \\geq \\frac {3}{2}\\times\\frac {1}{3}\\times\\frac {(a \\plus{} b \\plus{} c)(a^{p \\minus{} 1} \\plus{} b^{p \\minus{} 1} \\plus{} c^{p \\minus{} 1})}{a^q \\plus{} b^q \\plus{} c^q}\\geq \\frac {3}{2}$\r\nbecause with AM-GM:\r\n$ a \\plus{} b \\plus{} c \\geq3$\r\nthen:\r\n$ S\\geq \\frac {3}{2}$\r\nwith equality only if $ a \\equal{} b \\equal{} c$.",
  "Solution_2": "thanks  :)",
  "Solution_3": "[quote=\"pqrs\"]$ a,b,c,p,q\\in R^ \\plus{}$\n$ abc \\equal{} 1$ and $ p\\ge q$\nfind minimum value of $ \\dfrac{a^p}{b^q \\plus{} c^q} \\plus{} \\dfrac{b^p}{c^q \\plus{} a^q} \\plus{} \\dfrac{c^p}{a^q \\plus{} b^q}$[/quote]\r\nLet $ r \\equal{} \\frac{p}{q}$. And $ x \\equal{} a^q$, $ y \\equal{} b^q$, $ z \\equal{} c^q$.\r\nThen we have to find the minimum of $ \\frac{x^r}{y\\plus{}z} \\plus{} \\frac{y^r}{z\\plus{}x} \\plus{} \\frac{z^r}{y\\plus{}x}$ when $ xyz\\equal{}1$.\r\nWe prove that $ \\frac{x^r}{y\\plus{}z} \\plus{} \\frac{y^r}{z\\plus{}x} \\plus{} \\frac{z^r}{y\\plus{}x} \\ge \\frac{3}{2}$ $ \\iff$\r\n$ \\sum_{cyc} 2x^r(x\\plus{}y)(x\\plus{}z) \\equal{} \\sum_{cyc} 2x^{r\\plus{}2} \\plus{} 2x^{r\\plus{}1}y \\plus{} 2x^{r\\plus{}1}z \\plus{} 2x^ryz \\ge 3(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)$.\r\nWritten by muirhead means:\r\n$ 6[r\\plus{}2,0,0] \\plus{} 12[r\\plus{}1,1,0] \\plus{} 6[r,1,1] \\ge 12[1,1,1] \\plus{} 18[2,1,0]$. Which is obviously true when homogenizing since $ r \\ge 1$.",
  "Solution_4": "More directly, by Chebyschev\r\n\r\n$ \\sum \\frac {a^{p \\minus{} q \\plus{} q}}{b^q \\plus{} c^q} \\ge (\\sum \\frac {a^{p \\minus{} q}}{3})(\\sum \\frac {a^q}{b^q \\plus{} c^q})$\r\n\r\nBy AM - GM,\r\n\r\n$ \\sum \\frac {a^{p \\minus{} q}}{3} \\ge 1$.\r\n\r\nBut $ \\sum \\frac {a^q}{b^q \\plus{} c^q} \\ge \\frac {3}{2}$ is Nesbitt."
}
{
  "Tag": [],
  "Problem": "$a,b,c \\in \\mathbb{R}$\r\nFor all $n \\in \\mathbb{N}$ , we have:\r\n$[na] + [nb] = [nc]$ \r\nProve that at least one of those numbers (a,b,c) is an integer.",
  "Solution_1": "does anyone know the solution ?",
  "Solution_2": "Hmm... I have solution but it is messy, I have another idea, so I'll try to solve it in easier way and then post the solution.",
  "Solution_3": "Have a look here :\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?highlight=density&t=4568\r\n\r\nPierre."
}
{
  "Tag": [
    "complex analysis",
    "function",
    "integration",
    "calculus",
    "derivative",
    "complex analysis unsolved"
  ],
  "Problem": "Hello to all!\r\n\r\nI'm reading an article about series expansions and I found, what I think is a really strange result:\r\n\r\nWe have\r\n\r\n$ \\Theta(x)$ an holomorphic function inside a contour S, having only a root $ a$ inside S;\r\n\r\n$ x$ the affixe of a point inside S;\r\n\r\nFor every point $ z$ of the contour S:\r\n\r\n\\[ |\\Theta(x)|<|\\Theta(z)|\\]\r\n\r\nand the author states that the following:\r\n\r\n[i]The equation [/i]\r\n\r\n\\[ \\Theta(z)\\minus{}\\Theta(x)\\equal{}0\\]\r\n\r\n[i]has only one root $ z\\equal{}x$ inside $ S$, as we can see in this equality[/i]\r\n\\[ \\frac{1}{2i\\pi}\\int_{S}\\frac{\\Theta'(z)dz}{\\Theta (z)\\minus{}\\Theta(x)}\\equal{}\\frac{1}{2i\\pi}\\left[\\int_{S}\\frac{\\Theta'(z)dz}{\\Theta (z)}\\plus{}\\Theta(x)\\int_{S}\\frac{\\Theta'(z)dz}{\\Theta^{2}(z)}\\plus{}\\ldots\\right]\\equal{}\\frac{1}{2i\\pi}\\int_{S}\\frac{\\Theta'(z)dz}{\\Theta(z)}\\]\r\n\r\n[i]where the first and last members represent respectively, the number of roots of the considered equation and the number of roots of $ \\Theta(z)\\equal{}0$.[/i]\r\n\r\n\r\n\r\nI understand the first step of the equality (he used the same expansion as $ \\frac{1}{1\\minus{}x}$), but not the second... it looks as if he just made $ \\Theta(x)\\equal{}0$\r\n\r\nDoes anyone know what is happening there?\r\n\r\nThank you",
  "Solution_1": "Key words to look for: [url=http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem]Rouche's theorem[/url], [url=http://en.wikipedia.org/wiki/Argument_principle]argument principle[/url].\r\n\r\nIn the second term, $ \\frac{\\Theta'}{\\Theta^{2}}$ is the derivative of $ \\minus{}\\frac{1}{\\Theta}$, and we're integrating around a closed curve. It evaluates to zero by the Fundamental Theorem.",
  "Solution_2": "Thanks a lot!!!   :) \r\n\r\nI thought I was never going to figure it out!! :oops:"
}
{
  "Tag": [
    "number theory",
    "prime numbers"
  ],
  "Problem": "The sequence of the Human Genome was published 2003. Some people are using this code to generate music or to cipher messages or use DNA to do computing.\r\n\r\nI was wondering if there is any correlation between the DNA code and the sequence of prime numbers. I am not a mathematician but generally interested in Science, so forgive me if this  a \"weired\" idea.\r\nBut would it be not funny to be able to predict a DNA motif with number, especially prime number, theory?\r\n\r\nThanks!",
  "Solution_1": "Why primes and not composite, or Fermat numbers ?\r\nDo you have any reasonable reason to believe there is any link with [b]any[/b] known number ?\r\nTo give my sceptical answer:  You can \"find\" prime numbers anywhere in nature, but that's mostly (and most probably) numerology not mathematics.\r\n\r\nP.S: This post does not belong to the number theory section (tough it's \"related\" to prime) it should go in the Round Table section.\r\nP.S.S: Why do you need the human genome to generate music ? You could aswell (and with the same sucess) use the inscription on the butter you bought (let's suppose) today :)."
}
{
  "Tag": [],
  "Problem": "I was thinking to put this in SC competition but I decided not to because it might be hard for some beginners. Anyway, this is quite good thinking question so here it goes:\r\n\r\nd, a positive integer, is a nice number that can be resulted from either of expression a+b+c or a*b*c where a,b, and c are each distinct positive number.  Prove that (a+b+c)*d equals to a*b*c*d by defining each value for a,b,c, and d (a,b,c, and d are each different numbers) if a,b,c are consecutive numbers while a,b,c,d aren't.\r\n\r\nWording might be pretty confusing but you will find this easy.",
  "Solution_1": "But I thought:\n\n\n\n[hide]a+b+c = d\n\nd = a*b*c\n\n\n\nBy transitivity, \n\n\n\na+b+c = a*b*c\n\n\n\nMultiplying both sides by d,\n\n\n\n(a+b+c)*d = a*b*c*d[/hide]",
  "Solution_2": "[quote]Wording might be pretty confusing but you will find this easy.[/quote]\r\n\r\nNo offense, but I'll say.",
  "Solution_3": "Remember, you have to define the value for a,b,c, and d.",
  "Solution_4": "Answer: [hide]{a, b, c, d} = {1, 2, 3, 6}[/hide]",
  "Solution_5": "Excellent. Now, prove that it is only positive pair that's possible. Also, state the another pair that isn't all positive but satisfies the requirements I asked.",
  "Solution_6": "[hide]{a,b,c,d)={-1,-2,-3,-6}\n\nor it also could be {-1,0,1,0}[/hide]"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "For $0< a,b,c < \\frac{1}{2}$ prove that\r\n\r\n$(\\frac{1}{a}-1)(\\frac{1}{b}-1)(\\frac{1}{c}-1)\\geq (\\frac{3}{a+b+c}-1)^3$",
  "Solution_1": "Probably there is a more astute solution but log (1/x-1) is convex on ]0..1/2] + Jensen."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Let $O =(0,0), A = (1,3), B = (4,1)$ $D$ is on $OB$ such that $AD$ is perpendicular to $OB$. Find the area of triangle $OAD$.",
  "Solution_1": "The line through $O$ and $B$ is $y=\\frac{1}{4}x$\r\n\r\nSo let $D=(t, \\frac{1}{4})t$ $\\vec{AD}=[t-1 ; \\frac{1}{4}t-3]$ $\\vec{OB}=[4;1]$\r\n\r\n$AD$ is perpendicular to $OB$ so : $4t-4+\\frac{1}{4}t-3=0$ $t=\\frac{28}{17}$\r\n\r\nNow we know coordinations of $A,O,D$ so it's easy to calculate $AD$ and $DO$, \r\nArea of $\\triangle{ADO}= \\frac{1}{2}AD \\cdot DO$"
}
{
  "Tag": [
    "probability",
    "parameterization",
    "limit",
    "integration",
    "calculus",
    "probability and stats"
  ],
  "Problem": "$ X_1, ..., X_5$ are iid random variables having exponential distribution with parameter $ \\lambda$. Find $ P(X_1 \\geq 3|X_1 \\plus{} ... \\plus{} X_5\\equal{}5)$",
  "Solution_1": "Can anyone help me:?:",
  "Solution_2": "You may think about $ X_{1}$ as about minimum of five uniformly selected on $ [0;5]$ points.",
  "Solution_3": "$ (1 \\minus{} e^{ \\minus{} 3\\lambda})^5$ ?",
  "Solution_4": "[quote=\"roah\"]You may think about $ X_{1}$ as about minimum of five uniformly selected on $ [0;5]$ points.[/quote]\r\nCorrect answer is $ 0,0256$, but I don't know how to find it  :wink: If we think like You said we get $ 0,01024$ which is wrong. When I think about $ X_{1}$ as about minimum of FOUR uniformly selected on $ [0;5]$ points we get $ 0,0256$. But, could You explain this thinking? Why should I think about $ X_1$ like You wrote? I would be appreciate :)",
  "Solution_5": "it should be four, since the FIFTH point is AT 5 really, therefore we dont need to think about it.",
  "Solution_6": "But I still don't understand why these point are uniformly selected from $ [0,5]$, I wrote in my first post that $ X_1,...,X_5$ are exponential distributed random variables  :wink:. Any ideas to explain this fact?  :blush:",
  "Solution_7": "thinking about it today i came across the same issue you just asked...\r\n\r\nthe exponential thing doesnt seem to play a role if we can generalize like this.. as long as its a continuous distribution and $ X_n \\ge 0$ i think we can use the same argument..?",
  "Solution_8": "why don't we try to compute for example the exact conditional law of $ X_1$, give that $ X_1 + \\ldots, X_n = A$:\r\n$ X = X_1$ is independent of $ Y = X_2 + \\ldots + X_n$, and $ Y$ has a Gamma probability distribution with parameter $ (\\lambda, n - 1)$:\r\n\\[ P(Y \\in [y; y + dy]) = G(y) dy = \\frac {\\lambda^{n - 1}}{\\Gamma(n - 1)} y^{n - 2} \\exp( - \\lambda y) dy\r\n\\]\r\nNow, what we are trying to compute is:\r\n\\[ F(b,a) = \\lim_{\\epsilon \\to 0} \\frac {P[ (X \\leq b) \\cap (X + Y \\in [a; a + \\epsilon])] }{ P[ (X + Y \\in [a; a + \\epsilon])] } = P(X \\leq b | X_1 + \\ldots + X_n = a)\r\n\\]\r\nwhere\r\n\\[ P[ (X \\leq b) \\cap (X + Y \\in [a; a + \\epsilon]) = A(b,a + \\epsilon) - A(b,A)\r\n\\]\r\n\r\n\\[ P[ (X + Y \\in [a; a + \\epsilon]) = A(\\infty,a + \\epsilon) - A(\\infty,A)\r\n\\]\r\n\r\n\\[ A(b,a) = P[(X \\leq b) \\cap (X + Y \\leq a)] = \\int_{x = 0}^b \\lambda \\exp( - \\lambda x) [\\int_{y = 0}^{a - x} G(y)dy] dx\r\n\\]\r\nhence\r\n\\[ F(b,a) = \\frac {\\partial_{a} A(b,a) }{\\partial_a A(\\infty, a)} = \\frac {\\int_{x = 0}^b \\lambda \\exp( - \\lambda x) G(a - x) dx }{\\int_{x = 0}^{a} \\lambda \\exp( - \\lambda x) G(a - x) dx } = \\frac {\\int_{x = 0}^b x(a - x)^{n - 2} dx }{\\int_{x = 0}^{a} x(a - x)^{n - 2} dx }\r\n\\]\r\nwhence $ P(X \\leq b | X_1 + \\ldots + X_n = a) = (\\frac {b}{a})^n \\frac {\\int_{u = 0}^1 u(\\frac {a}{b} - u)^{n - 2} du }{\\int_{u = 0}^{1} u(1 - u)^{n - 2} du } = \\phi(\\frac {b}{a})$.\r\nThis shows that the conditional density of $ X_1$ given that $ X_1 + \\ldots + X_n = a$ is something like $ Cx(a-x)^{n-2} dx$\r\n(which is a little weird for $ n=2$, so I may have made a computation mistake somewhere  :blush: )",
  "Solution_9": "Let me try to pick up some loose ends and give some more details to the above posts. \r\n\r\nLook at the more general problem $ P(X_1 \\le b| X_1 \\plus{} \\ldots \\plus{} X_n \\equal{} a) \\equal{} ?$, where $ n \\ge 2$, the $ X_i$ are iid exponential random variables (parameter $ \\lambda >0$) and $ 0 \\le b \\le a$ (as considered by alekk). \r\n\r\nThe connection to the uniform distribution is the following: Because of the exponential distribution (which is crucial for this) the points $ Y_n : \\equal{} X_1 \\plus{} \\ldots \\plus{} X_n$ form a Poisson point process on the positive part of the real line. The random set of points $ Y \\equal{} \\{ Y_n: n \\ge 1\\}$ has the following properties: \r\nFor every interval $ I$ of length $ L$ the number of points of $ Y$ in $ I$ is Poisson-distributed (with parameter $ \\lambda L$) and for a fixed number $ n$ of points the conditional distribution of the positions of these $ n$ points in $ I$ is: The positions are independent and uniformly distributed.  The condition $ X_1\\plus{} \\ldots \\plus{} X_n \\equal{} a$ is the same as \"in the interval $ [0,a)$ there are exactly $ n\\minus{}1$ points of $ Y$\". The condition $ X_1\\plus{} \\ldots \\plus{} X_n \\equal{} a$ and $ X_1 \\le b$ is the same as \"in the interval $ [0,a)$ there are exactly $ n\\minus{}1$ points of $ Y$ and at least one of them is in $ [0,b]$\". (This elaborates the idea of roah, and the comment of kenn4000 that really only $ n\\minus{}1$ points are uniformly chosen, and $ X_1$ is indeed the minimum of $ n\\minus{}1$ independent uniformly distributed random variables.)\r\n\r\n\r\nTo apply this to the above problem: $ P(X_1 \\le b| X_2 \\plus{} \\ldots \\plus{} X_n \\equal{} a) \\equal{}$\r\n$ 1 \\minus{}P(Y$ has no point in $ [0,b]$ | $ Y$ has exactly $ n\\minus{}1$ points in $ [0,a)$) \r\n$ \\equal{} 1 \\minus{} P($of $ n\\minus{}1$ uniformly chosen independent points in $ [0,a)$ no point is in $ [0,b]) \\equal{} 1 \\minus{} (1 \\minus{} \\frac{b}{a})^{n\\minus{}1}$.\r\n\r\nThe calculation of alekk gives the same formula (small mistake in the last step of the calculation of $ F$):\r\n$ F(b,a) \\equal{} \\frac {\\partial_{a} A(b,a) }{\\partial_a A(\\infty, a)} \\equal{} \\frac {\\int_{x \\equal{} 0}^b (a \\minus{} x)^{n \\minus{} 2} dx }{\\int_{x \\equal{} 0}^{a} (a \\minus{} x)^{n \\minus{} 2} dx } \\equal{} 1 \\minus{} (1 \\minus{} \\frac{b}{a})^{n\\minus{}1}$,\r\n\r\nApplied to the original problem this gives $ P(X_1 \\geq 3|X_1 \\plus{} ... \\plus{} X_5\\equal{}5) \\equal{} (1 \\minus{} \\frac{3}{5})^4 \\equal{} 0.0256$.",
  "Solution_10": "Thank You very much. I have only one question. [b]alekk[/b] wrote \r\n$ A(b,a) \\equal{} P[(X \\leq b) \\cap (X \\plus{} Y \\leq a)] \\equal{} \\int_{x \\equal{} 0}^b \\lambda \\exp( \\minus{} \\lambda x) [\\int_{y \\equal{} 0}^{a \\minus{} x} G(y)dy] dx$\r\nso, how should I understand $ A(\\infty ,a)$ ?\r\nI think it is $ P[(X \\leq \\infty) \\cap (X \\plus{} Y \\leq a)]\\equal{}P(X \\plus{} Y \\leq a)$?  But in integral form? I guess it isn't \r\n$ \\int_{x \\equal{} 0}^\\infty \\lambda \\exp( \\minus{} \\lambda x) [\\int_{y \\equal{} 0}^{a \\minus{} x} G(y)dy] dx$ because this integral doesn't coverage on $ (0,\\infty)$. So my question is, how should I understand $ A(\\infty ,a)$? How to write this expresion using integrals?",
  "Solution_11": "[quote=\"zuraw\"]\n$ \\int_{x \\equal{} 0}^\\infty \\lambda \\exp( \\minus{} \\lambda x) [\\int_{y \\equal{} 0}^{a \\minus{} x} G(y)dy] dx$ because this integral doesn't coverage on $ (0,\\infty)$ So my question is, how should I understand $ A(\\infty ,a)$? How to write this expresion using integrals?[/quote]\r\n\r\nyes, as you said:\r\n$ A(\\infty,a)\\equal{}P[X\\plus{}Y \\leq a]\\equal{}\\int_{x \\equal{} 0}^\\infty \\lambda \\exp( \\minus{} \\lambda x) [\\int_{y \\equal{} 0}^{a \\minus{} x} G(y)dy] dx$ \r\nwhich is equal to:\r\n$ \\int_{x \\equal{} 0}^a \\lambda \\exp( \\minus{} \\lambda x) [\\int_{y \\equal{} 0}^{a \\minus{} x} G(y)dy] dx$"
}
{
  "Tag": [
    "MIT",
    "college",
    "Harvard",
    "Princeton",
    "parameterization",
    "topology",
    "Yale"
  ],
  "Problem": "Hi,\r\n\r\nWhat are the differences between top US Universities such as Harvard, Princeton, MIT with top UK Universities such as Cambridge and Oxford in terms of quality/character of students, environment, community, overall college experience (academically and non-academically), quality of teaching and X, \r\n\r\nwhere X is any parameter you may choose and any operation may be enclosed (If you know what I mean. I don't even sure whether I'm using the correct words! Haha! Never mind. You get the point. Just having a little fun here. Btw, it is not profane!)?\r\n\r\nPlease give your opinions. Thanks!",
  "Solution_1": "Honestly, you can't go wrong choosing any of those.  Do you want to have a chance to be around Stephen Hawking?  Go to Cambridge.  Do you want to be taught by the great Andrew Wiles?  Go to Princeton.  Do you want to say that you graduated from Harvard?  Go to Harvard.  You can't go wrong.",
  "Solution_2": "Oxford isn't really advisable - it's a reasonable amount worse than Cambridge for maths and sciences in general. (This is pretty widely accepted; I'm not just being all anti-Oxford because I'm at Cambridge.)\r\nThere are loads of great mathematicians here - Tim Gowers, Alan Baker, Ben Green, Peter Swinnington Dyer etc. - I imagine this is a trait shared with all the major U.S. ones mentioned.\r\n(A nice thing about Oxbridge - supervisions/tutorials - you could reasonably end up having a Field's medallist go over your solutions to example sheets once a week.)\r\nIn truth I'm just saying all that famous people stuff just to wow you - it really matters what they're like as teachers and I can't really comment on that.\r\nOne more advantage about the U.K; you only have to do maths, so none of all those other classes you might end up going to at a U.S. school.\r\n(BTW - Cambridge and Oxford are stunning towns - I don't want to offend anyone but I'm willing to go ahead and say they are nicer places than the locations of the U.S. ones mentioned.)",
  "Solution_3": "Don't make your decisions about what famous people are at the university. Take note of these key factors: \r\n\r\n1) Famous mathematicians and physicists might not actually have to teach undergrads anyway (Wiles rarely teaches undergrad classes here). If you're thinking grad school, it may be really competitive to get one of them for an advisor. Also some of the really famous people are also ancient and aren't doing much (like someone mentioned Swinnerton-Dyer before -- isn't he like 80 yrs old? and Stein at Princeton is almost that old, and he might stop teaching after this year.) \r\n\r\n2) Famous mathematicians and physicists aren't necessarily good teachers. For example, earlier this year I went to a lecture by Endre Szemeredi, who is one of the most brilliant combinatorialists alive. But his lecture was not good at all. And Chris Skinner (a great algebraic NT guy at princeton) is known for giving really tedious and painful and annoying assignments. And Stein is kinda boring and basically says exactly what's in the book (not surprising because he wrote it, but still). \r\n\r\nSo basically, my point is that you shouldn't base your decision on celebrities. That's not to say you won't benefit from them at all -- my friend at MIT has Munkres for topology and Artin for algebra and he says they're really good. And last year I did a reading course with Fefferman (fields medalist) and I got to have one-on-one discussions with him every week about geometric measure theory. \r\n\r\nNow I haven't been to Oxford or Cambridge and I don't know many people there, but I can say that the top USA universities are all pretty similar in overall quality, and you should choose between them based on your personal preferences. Do you want to live in a city? Then MIT or Harvard might be preferable over Yale and especially Princeton. If you prefer a more quiet setting with actual trees and green stuff and old buildings, reverse that order (and probably Oxbridge go ahead of all of the USA places). \r\n\r\nYou might also want to wait and find out what kind of financial aid you get (the USA places are all pretty generous but Harvard and Princeton are the best). \r\n\r\nOther factors might be the specific topics you want to study (for example, Princeton has a lot of people in analysis, number theory, and low dimensional topology, and I don't know the reputations of other schools' departments that well but you can find out easily enough), what options the schools offer (e.g. Princeton has a lot of interesting certificates you can get, but they won't let you double major), and what kind of classroom environment you want. Or extracurriculars -- do you want to do sports, join a performance group, join a math or physics club? Are you interested in frats and partying? You can ask a student at whichever university you're interested in to find out what the extracurriculars are like there. \r\n\r\nThere are tons of other factors too. But don't make 'what big name professor does the school have' one of them.",
  "Solution_4": "Hello everyone!\r\n\r\n  Im a student about 16 years old and im very interested about maths. What i mean is that i participate in mathematical competitions study hard maths especially number theory< i love Andreescu> but i think i dont have the <big> brain to go to I.M.O. My dream is to join a high college like Princeton, M.I.T and others. Can i do that withoiut taking place in an I.M.O??",
  "Solution_5": "You don't need to go to IMO in order to get into that kind of university. You just need to be a good student. In your application, you should stress the following points:\r\n\r\n(1) You have the necessary English skills and a good enough secondary school education to succeed in the university you're applying to. \r\n\r\n(2) You fall into one of the three major categories of students that are generally accepted to this type of university. They are: (i) Students who are amazingly good at one particular thing and not much else (you would probably need to go to IMO at least in order to qualify as this for math). (ii) Students who are well-rounded, i.e. good at a variety of things (and I don't mean maths, physics, and chemistry). If you do stuff like organizing a group, writing a lot, playing music, etc -- that will help in this category. (iii) Students who contribute significantly to the diversity of the community (you are Greek? Maybe you should emphasize Greek culture in your application. Do you play an instrument -- can you play Greek folk music? Would you want to get involved with (or create) a Greek culture student group at your university? etc...) \r\n\r\n(3) You have reasons for going to the university besides the name recognition. Learn some specific things about the university you're applying to and explain why those things make you want to go there. They don't want to be filled by a bunch of students who just want the name MIT or Princeton on their resumes (although it's inevitable that most people really are there for that reason). If you like number theory, in your Princeton application you could talk about the great number theorists in the Math Dept, like Sarnak and Skinner and Wiles and Katz.",
  "Solution_6": "[quote=\"Xevarion\"] (i) Students who are amazingly good at one particular thing and not much else (you would probably need to go to IMO at least in order to qualify as this for math).  [/quote]\r\n\r\nAren't there other ways of showing this?  :(",
  "Solution_7": "[quote=\"BanishedTraitor\"][quote=\"Xevarion\"] (i) Students who are amazingly good at one particular thing and not much else (you would probably need to go to IMO at least in order to qualify as this for math).  [/quote]\n\nAren't there other ways of showing this?  :([/quote]\r\nPublish something? My impression about math is that there are so many international students who are extremely good at math (but not particularly special in any other way) that you have to basically be one of the best in order to get in to a top university for that reason alone. Maybe if you don't go to IMO but show knowledge of advanced math and very high potential, that would be enough. Who knows, I'm not an admissions officer.",
  "Solution_8": "Well, I don't really mean Mathematics (too many good people  :D ) but what about Sciences?",
  "Solution_9": "[quote=\"BanishedTraitor\"]Well, I don't really mean Mathematics (too many good people  :D ) but what about Sciences?[/quote]\r\nI don't know enough about it. Sorry. I imagine anything beyond what a typical \"smart student\" would do in school in your country is at least a useful step in that direction."
}
{
  "Tag": [],
  "Problem": "Anyone interested in a chess tournament between mathlinks.ro members?",
  "Solution_1": "I would be if I had a clue besides the basic rules of how to play that game :)",
  "Solution_2": "the basic rules are everything you need  :D",
  "Solution_3": "I'm up for it, altough it has been a long time since I played \"seriously\" -- playing Go now instead.",
  "Solution_4": "Maybe someone could put up a poll regarding this.",
  "Solution_5": "Done.\r\nAlso,post which chess network would you like to play on.",
  "Solution_6": "im up for it, but i dont have a mathlinks account, so if you go to funorb.com, my username is hyperhavoc5. funorb has many different games"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let $a,b,c,d > 1$ natural numbers which satisfy $a^{b^{c^d}}=d^{c^{b^a}}$ . Prove that $a = d$ and $b=c$.",
  "Solution_1": "Who can post solution of this problem? Please post it!",
  "Solution_2": "I'm not sure if I'm right anway I'll post my solution since nobody else does...\r\n\r\nSince LHS=RHS they must contain the same prime factors, especially being $ a,b,c,d \\in \\mathbb N$ then $ a | d , d| a$\r\n\r\nwhich means $ a=d$. With substitutions we get $ {a}^{b^{{c}^{a}}}=a^{c^{b^{a}}}$$ \\Rightarrow  b^{c}=c^{b}$ $ \\Rightarrow b|c,c|b\\Rightarrow b=c$.\r\n\r\nPlease tell me if I'm wrong  :)",
  "Solution_3": "Your solution is based on the argument that, in general $ x^{y}=y^{x}\\Longleftrightarrow x|y\\wedge y|x\\Longleftrightarrow x=y$. The first equivalence is wrong. We have indeed $ 2^{4}=4^{2}$ and $ 4\\not=2$.",
  "Solution_4": "I see.. :blush:",
  "Solution_5": "however, i've been trying to find a second counter-example, but i can't find one. we can indeed use the following argument:\r\n\r\nIt is evident that (or quickly established), supposing $ x\\geq y$ by symetry, $ x^{y}=y^{x}\\Longleftrightarrow x=y^{k}$, $ k$ being a natural number. Now, we may say that the relation is equivalent to $ \\left(y^{k}\\right)^{y}=y^{y^{k}}\\Longleftrightarrow y^{k\\cdot y}=y^{y^{k}}\\Longleftrightarrow k\\cdot y=y^{k}$ which implies either $ k=1\\Longleftrightarrow x=y$ or $ y=2\\wedge x=4$. This lemma shows that, supposing we have established the relation $ a=d$, we also have the relation $ b=c$ (or the special case stated above)."
}
{
  "Tag": [],
  "Problem": "If $ 60$ miles per hour is $ 88$ feet per second, how many feet per second is $ 66$ miles per hour? Express your answer as a decimal to the nearest tenth.",
  "Solution_1": "$ 88 \\times \\frac{11}{10}\\equal{}\\frac{88\\plus{}880}{10}\\equal{}\\boxed{96.8}$."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "1)Given sequence $ x_n$ such that:$ x_n  \\in R;x_{n + 1}  = x_n  + \\frac{1}{2}(c{\\rm{os}}\\,x_n  + \\sin \\,x_n )(\\forall n \\in N^* )$.Find limited of sequence depending on $ x_1$.\r\n2)Given p be a prime and $ p \\ge 5$.Prove that exist $ q_1$,$ q_2$ such that $ 1 < q_1  < q_2  < p$ simultaneously $ q_1 ^{p - 1}  - 1$ and $ \\q_2 ^{p - 1}  - 1$ non divisible $ p^2$",
  "Solution_1": "some body help me !"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "If $a$, $b$, $c$ are positive real numbers, prove that\r\n$\\sqrt{\\frac{a}{b+c}}+\\sqrt{\\frac{b}{c+a}}+\\sqrt{\\frac{c}{a+b}}> 2$.",
  "Solution_1": "Using Cauchy:\r\n$a+(b+c) \\geq 2\\sqrt{a(b+c))}$\r\n=> $\\sqrt{\\frac{a}{b+c}} \\geq \\frac{2a}{a+b+c}.$\r\nSimilar:\r\n     $ \\sqrt{\\frac{b}{a+c}} \\geq \\frac{2b}{a+b+c}.$\r\n     $ \\sqrt{\\frac{c}{a+b}} \\geq \\frac{2c}{a+b+c}.$\r\n=> $L.H.S > \\frac{2a+2b+2c}{a+b+c} = 2.$\r\n(Because a,b,c>0).\r\n\r\nThe inequality very easy.",
  "Solution_2": "It was posted but i will give a solution\r\n$(a+b+c)^2\\geq4a(b+c)$ and equivalently $\\sqrt\\frac{a}{b+c}\\geq\\frac{2a}{a+b+c}$\r\nSame $\\sqrt\\frac{b}{a+c}\\geq\\frac{2b}{a+b+c}$ and\r\n$\\sqrt\\frac{c}{b+a}\\geq\\frac{2c}{a+b+c}$\r\nwe add the three ineqs and we have\r\n$\\sqrt\\frac{a}{b+c}+\\sqrt\\frac{b}{a+c}+\\sqrt\\frac{c}{b+a}\\geq2$\r\nthe equality holds when a=b+c ,b=c+a, c=a+b\r\nIf we add these we have that a+b+c=0 contradiction.\r\nSo $\\sqrt\\frac{a}{b+c}+\\sqrt\\frac{b}{a+c}+\\sqrt\\frac{c}{b+a}>2$",
  "Solution_3": "Metru, excuse me for saying this, but this inequality has been posted for many times and solved as well. ;)",
  "Solution_4": "Disprove or Prove that:\r\nIf $a,b,c>0$ and $k \\in N$ then:\r\n\r\n$ \\large \\sqrt[2k]{\\frac{a}{b+c}} + \\sqrt[2k]{\\frac{b}{a+c}} + \\sqrt[2k]{\\frac{c}{a+b}} > 2.$",
  "Solution_5": "[quote=\"tranvinhphuctk14\"]Using Cauchy:\n$a+(b+c) \\geq 2\\sqrt{a(b+c))}$\n=> $\\sqrt{\\frac{a}{b+c}} \\geq \\frac{2a}{a+b+c}.$\n[/quote]\r\ncan you explain this step in more details ... i dont know this cauchy ineq ;<",
  "Solution_6": "[quote=\"Gohan\"][quote=\"tranvinhphuctk14\"]Using Cauchy:\n$a+(b+c) \\geq 2\\sqrt{a(b+c))}$\n=> $\\sqrt{\\frac{a}{b+c}} \\geq \\frac{2a}{a+b+c}.$\n[/quote]\ncan you explain this step in more details ... i dont know this cauchy ineq ;<[/quote]\r\n\r\nyou can treat it as AM-GM by noticing that $x+y\\geq 2\\sqrt{xy}$",
  "Solution_7": "in vietnam, amgm is called cauchy, cauchy is called bun???ski",
  "Solution_8": "cezar lupu\r\n[url=http://kvant.mirror0.mccme.ru/pdf/2002/06/18.pdf]in another way:\r\n[/url]",
  "Solution_9": "[quote=\"siuhochung\"]in vietnam, amgm is called cauchy, cauchy is called bun???ski[/quote]\r\nYes, i'am sorry for mistake."
}
{
  "Tag": [
    "function",
    "inequalities",
    "calculus",
    "derivative",
    "algebra",
    "polynomial",
    "inequalities unsolved"
  ],
  "Problem": "Hi :D ! $ x,y,z$ are positive real numbers and $ 3x\\plus{}2y\\plus{}z\\equal{}6$ . prove that :\r\n$ x\\plus{}xy\\plus{}xyz\\leq 3$",
  "Solution_1": "i have an ugly soloution for this one but it seems too easy !! plz help me ?!?!",
  "Solution_2": "Let's even allow zero values, and try Lagrange multipliers. Build function $ L(x,y,z,\\lambda) \\equal{} x \\plus{} xy \\plus{} xyz \\minus{} \\lambda(3x\\plus{}2y\\plus{}z\\minus{}6)$. Then we get the system $ \\frac {\\textrm{d}} {\\textrm{d}x} L(x,y,z,\\lambda) \\equal{} 1\\plus{}y\\plus{}yz\\minus{}3\\lambda \\equal{} 0$, $ \\frac {\\textrm{d}} {\\textrm{d}y} L(x,y,z,\\lambda) \\equal{} x\\plus{}xz\\minus{}2\\lambda \\equal{} 0$, $ \\frac {\\textrm{d}} {\\textrm{d}z} L(x,y,z,\\lambda) \\equal{} xy\\minus{}\\lambda \\equal{} 0$. Plug $ \\lambda \\equal{} xy$ in the second equation to get $ x(1\\plus{}z\\minus{}2y) \\equal{} 0$, so $ z\\equal{}2y\\minus{}1$ ($ x\\equal{}0$ is on the contour, and provides a minimum, not a maximum). Plug again in the first equation to get $ 1\\plus{}y\\plus{}y(2y\\minus{}1)\\minus{}3xy \\equal{} 0$, so $ x \\equal{} \\frac {2y^2\\plus{}1} {3y}$. Back into the restriction, all these yield $ \\frac {2y^2\\plus{}1} {y} \\plus{} 2y \\plus{} (2y\\minus{}1) \\equal{} 6$, hence $ 6y^2 \\minus{}7y \\plus{} 1 \\equal{} 0$, so $ y\\equal{}1$, when $ x\\equal{}y\\equal{}1$, and $ y\\equal{} \\frac {1} {6}$, when $ z\\equal{}\\minus{}\\frac {2} {3}$ is not convenient. One can check that for $ x\\equal{}y\\equal{}z\\equal{}1$ the critical value of $ L \\equal{} 3$ is a maximum (on the contour, we have $ x\\equal{}0$, with $ L\\equal{}0$, or $ y\\equal{}0$, with $ L \\equal{} x \\leq 2 < 3$, or $ z\\equal{}0$, with $ L \\equal{} x\\plus{}xy \\leq \\frac {8} {3} < 3$).",
  "Solution_3": "Thanks a lot mavropnevma !! but is there any easier soloution (by AM-GM , Couchy ,...) ?",
  "Solution_4": "Ok, here is an elementary solution. Let's denote $ x \\equal{} a \\plus{} 1, y \\equal{} b \\plus{} 1, z \\equal{} c \\plus{} 1$. Then you can rewrite the condition of the problem as follows:\r\n$ 3a \\plus{} 2b \\plus{} c \\equal{} 0$ and you want to prove $ (a \\plus{} 1) \\plus{} (a \\plus{} 1)(b \\plus{} 1) \\plus{} (a \\plus{} 1)(b \\plus{} 1)(c \\plus{} 1)\\le 3$ or (after simplification) it's enough to verify\r\n$ 2ab \\plus{} bc \\plus{} ac\\plus{}abc\\le 0$. But that's almost obvious. Just write $ c \\equal{} \\minus{} (3a \\plus{} 2b)$ and you will get the inequality $ 3a^2 \\plus{} 2b^2\\plus{}ab(3a\\plus{}2b)\\plus{}3ab\\equal{}3a^2(b\\plus{}1)\\plus{}2b^2 (a\\plus{}1)\\plus{}3ab\\ge0$ for $ a,b,c\\ge \\minus{}1$ and $ 3a\\plus{}2b\\le 1$. Indeed, if $ ab\\ge 0$ we are done, and if not, then $ ab(3a\\plus{}2b)\\ge ab$ and we get that $ RHS\\ge 3a^2\\plus{}2b^2\\plus{}4ab\\ge 0$.",
  "Solution_5": "More elementary solution (but, yes, ugly).\r\n\r\nKeep $ x$ constant, then $ L \\equal{} x(1\\plus{}y\\plus{}yz)$, with $ z \\equal{} 6\\minus{}3x\\minus{}2y$, so $ L\\equal{} x(1\\plus{}y\\plus{}6y\\minus{}3xy\\minus{}2y^2) \\equal{} \\minus{}x(2y^2 \\plus{}y(3x\\minus{}7)\\minus{}1)$. The maximum for this occurs for $ y \\equal{} \\frac {7\\minus{}3x} {4}$, and amounts to $ \\frac {1} {8} x(9x^2 \\minus{}42x \\plus{} 57)$. Now revert to considering $ x$ variable in the interval $ [0,2]$. The derivative of the above polynomial annuls in $ x\\equal{}1$ and $ x\\equal{}\\frac {57} {27} > 2$, so its maximum is obtained for $ x\\equal{}1$, when we get value $ 3$. Then $ y$ is also equal to $ 1$, and finally $ z\\equal{}1$.",
  "Solution_6": "leshik, you wrote \"enough to verify $ 2ab \\plus{} bc \\plus{} ac\\le 0$\", but you forgot the term $ abc$, so one needs $ abc \\plus{} 2ab \\plus{} bc \\plus{} ac\\le 0$, with more work...",
  "Solution_7": "Thanks for the corrections.I've edited :blush:"
}
{
  "Tag": [],
  "Problem": "I was wondering where everybody in AoPS lived.  Because there are 195 countries in the world, I'm just going to do continents on the poll(but include a few big countries).  Post where you live here and answer the poll.\r\n\r\nNote: I have to delete a few poll options because the settings won't allow it, so if your country doesn't get separately represented and you think it should, that's probably why.",
  "Solution_1": "I live in a pineapple under the sea.",
  "Solution_2": "Similarly, I live under a rock under the sea, and I am quite stupid when compared to mewto's spongy intellect.",
  "Solution_3": "And I live in the middle of mewto and battered in a house that resembles my face. I am very grouchy compared to mewto and battered's two-legged kind....",
  "Solution_4": "hee hee.  THis isn't what I was thinking what your posts are like, but this is more fun than what I had in mind.  of course, no being silly in the polls!",
  "Solution_5": "If you would like me to be more formal...\r\nmy female clone lives in Washington... one of the states with the most rain. Sad part is we haven't had rain in... I don't know a month? And our temperatures are some what close to Nevada, :( I am NOT used to this type of weather.",
  "Solution_6": "I live in an igloo, and I hunt caribou for dinner. It should be obvious what country that is.",
  "Solution_7": "I live in an anchor under the sea, and I'm very cheap. I'm a big miser. I would even sell my soul for money :P . I make mewto work for me :P",
  "Solution_8": "So far, 75% of the people here(6 out of 8) live in the U.S.A.  I'm surprised there aren't any more from say, Eurpope and Asia.",
  "Solution_9": "[quote=\"meewhee009\"]And I live in the middle of mewto and battered in a house that resembles my face. I am very grouchy compared to mewto and battered's two-legged kind....[/quote]\r\n\r\nI live in a dome and i am a scientist with a white suit.",
  "Solution_10": "I live close to my computer.",
  "Solution_11": "I live in a billion-dollar box....",
  "Solution_12": "this is getting randomly SCARY.",
  "Solution_13": "I live in your wildest dreams.\r\nI am stuck in a temporal tunnel with my computer.",
  "Solution_14": "What the hell, am I the only Canadian here? (Besides dnkywin, of course. But then again, he's also American.)",
  "Solution_15": "I am Canadian, but I don't live in Canada.",
  "Solution_16": "[quote=\"PowerOfPi\"]I am Canadian, but I don't live in Canada.[/quote]\r\n\r\nSame deal with dnkywin. Why do people claim nationalities where they aren't a resident...? I mean, I'm Chinese both by origin and ethnicity, but I consider myself Canadian, because I've lived here for 12 years...",
  "Solution_17": "Looking at the polls, the AoPS population is no where near as diverse as I thought(at least in the FF).  I'm incredibly surprised I don't see any \"China's\" or \"India's\" considering the huge populations of the countries.",
  "Solution_18": "Maybe it is because people in those countries study harder and don't have the time for this stuff.\r\n\r\n@sudoku dragon: I consider myself Canadian because I was born in Canada. I moved a year ago, and not very far either. I still visit every few months.",
  "Solution_19": "[quote=\"meewhee009\"]And[b] I live in the middle of mewto[/b] and battered in a house that resembles my face. I am very grouchy compared to mewto and battered's two-legged kind....[/quote]\r\n\r\n*loud coughing*\r\n\r\nWell, maybe we have so many people living in the US because, hmm, well, this is a forum made for Americans and written in English.",
  "Solution_20": "And those asians do study all day.\r\nThey don't have time for all this stuff.\r\n\r\nNo wonder the Chinese come first every year. :huh: \r\nDon't know about Indians, they rarely post outside their national forum on Aops :rotfl:\r\n\r\n\r\n\r\n[b][color=red]Edited to avoid any misunderstanding[/color][/b]\r\n :wink:  :)",
  "Solution_21": "[quote=\"Xaenir\"]And those asian nerds do study all day.[/quote]\r\nConsidering AoPS is a worldwide forum, and this forum is frequented by people internationally, that's not a very smart thing to say.  :wink:",
  "Solution_22": "Xaenir, that's the second time you insult somebody. Do it once more, this goes to mods.\r\n\r\nI am the only one voted to live in Europe...",
  "Solution_23": "as you may have seen , I have long edited my post, so your post qualifies as spam :D\r\nBesides i didn't mean to insult anybody, just implied that asians work harder and are intellectually superior, you got it the wrong way. :wink:",
  "Solution_24": "No, wasn't it ''asian nerds''? Maybach's post tells it all. And that ''girls are lazy, girls are dumb''? \r\n\r\nI say again, you can say anything without insults. If you want to continue the discussion, PM me. I'm not against you, just giving you a little piece of advice :)",
  "Solution_25": "[b]nerd[/b] refers to a person who passionately pursues intellectual activities, esoteric knowledge, or other obscure interests rather than engaging in more social or popular activities.\r\n\r\nThat isn't an insult :!:",
  "Solution_26": "um.........look at this(Source:  Merriam Webster Dictionary&Thesaurus):\r\nnerd \\'n\u0259rd\\  n [perh. fr. nerd, a creature in the children's book If I Ran the Zoo (1950) by Dr. Seuss (Theodor Geisel)] (1951) : an unstylish, unattractive, or socially inept person ; esp: one slavishly devoted to intellectual or academic pursuits \u2039computer ~s\u203a\r\n\u2014 nerd\u00b7i\u00b7ness \\'n\u0259r-d\u0113-n\u0259s\\ n\r\n\u2014 nerd\u00b7ish \\'n\u0259r-dish\\ adj\r\n\u2014 nerdy \\-d\u0113\\ adj\r\n\r\nYour definition doesn't tell it all.  I'm not saying \"nerd\" is a terrible word to say, but it usually has a very negative connotation, which is why you certainly shouldn't call someone that, let alone steryotype billions of people considred \"Asians\".  Bottom line:  This is your second strike.  Don't do it again.",
  "Solution_27": "Well I didn't mean it as an insult.I apologize if I offended someone, is it alright now?\r\nAnd I think I would be more careful in the future",
  "Solution_28": "Japan!\r\n\r\n :lol:",
  "Solution_29": "FANTASY LAND!!!! :D",
  "Solution_30": "@Isabella,\r\nHa. Grammar gone wrong. I didn't think anyone would find that so I left it be. You proved me wrong  :lol:",
  "Solution_31": "to me, nerds are a type of candy\r\ni like nerds\r\nthey do taste quite sweet\r\ni live in a Gullible"
}
{
  "Tag": [],
  "Problem": "I want a recent national competition.\r\nI don't have anything to trade.\r\nWell, my teacher has 2005 (maybe?) state competition. \r\nAlso A few of the recent work books.\r\nPlease(If there is any chance of avoiding the actual trade, and I could just get the competition, It would be highly appreciated. I don't know if just giving away a nat competition is legal, though.) :help:",
  "Solution_1": "I have 2008 or 2006 nats, if you could trade anything it would be most helpful."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "factorial",
    "function",
    "domain",
    "Ring Theory",
    "superior algebra"
  ],
  "Problem": "Let $ f_1 ,f_2 \\in k[x_1,...,x_n]$ ($ k$ is a field) be polynomials with no common irreducible factors. Show That\r\n$ Syz(f_1 ,f_2)\\equal{}(f_2 ,\\minus{}f_1)k[x_1,...,x_n] \\subset k[x_1,...,x_n]^2$.\r\nWhat happens if $ f_1$ and $ f_2$ have common irreducible factors?",
  "Solution_1": "This sounds very much like homework, so I'd just give a few hints, but there is no much to hint here - if you can't solve the problem, you have either not understood the meaning of \"syzygy\", or you don't know some basic facts about factorial domains (also called unique factorization domains, UFDs, ...). If any of these is not familiar to you, read them up first!\r\n\r\nDo you agree that $ \\left(g_1,g_2,...,g_n\\right)\\in\\text{Syz}\\left(f_1,f_2,...,f_n\\right)$ if and only if $ g_1f_1 \\plus{} g_2f_2 \\plus{} ... \\plus{} g_nf_n \\equal{} 0$ ? Now think about what this means for $ n \\equal{} 2$. That is, $ g_1f_1 \\plus{} g_2f_2 \\equal{} 0$. That is, $ g_1f_1 \\equal{} \\minus{} g_2f_2$. If $ f_1$, $ f_2$, $ g_1$, $ g_2$ were integers, this would yield some divisibility... now, $ f_1$, $ f_2$, $ g_1$, $ g_2$ are not integers, but polynomials. What can you say about divisibility among polynomials?\r\n\r\nIf $ f_1$ and $ f_2$ have common irreducible factors - worst case: $ f_1 \\equal{} f_2$, there are many more syzygies. Do you see a very obvious one?\r\n\r\n  darij"
}
{
  "Tag": [
    "inequalities",
    "induction",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Show that:\r\n$F_{2n}\\leq \\frac{(2n+2)^n}{n!}$\r\nDenote :$F_{2n}$ is $2n$-th Fibolnaxi number\r\nIt's itshelf Mathlinks Contest Round 4 ,after some conclusions ,of course !  \r\nBut I have no final solution.Anybody has one??",
  "Solution_1": "It is trivial. Let Z(n) is right expression. We have Z(1)=4, Z(n)/Z(n-1)=2*(1+1/n)^n>4 if n>=2. Therefore Z(n)>=4^n.\r\nIf F(0)=a>=0,F(1)=b>=0,F(n+1)=F(n)+F(n-1) we have F(2n+2)=3F(2n)-F(2n-2)<=3F(2n+2). Therefore F(2n)<=F(2)*3^(n-1)<4^n if F(2)=a+b<4.",
  "Solution_2": "[quote=\"Rust\"]It is trivial. Let Z(n) is right expression. We have Z(1)=4, Z(n)/Z(n-1)=2*(1+1/n)^n>4 if n>=2. Therefore Z(n)>=4^n.\nIf F(0)=a>=0,F(1)=b>=0,F(n+1)=F(n)+F(n-1) we have F(2n+2)=3F(2n)-F(2n-2)<=3F(2n+2). Therefore F(2n)<=F(2)*3^(n-1)<4^n if F(2)=a+b<4.[/quote]\r\nPlease, use LaTeX...",
  "Solution_3": "It's easy to know $F_m\\geq F_{m-1}(m=1,2,\\ldots)$.So $F_{m+1}=F_m+F_{m-1}\\leq2F_m$.\r\nSo $F_{2n}=F_{2n-1}+F_{2n-2}\\leq2F_{2n-2}$.\r\nFirst,$F_2<4$,so the inequality is correct for $n=1$.\r\nBy induction,suppose $F_{2n}\\leq\\frac{(2n+2)^n}{n!}$\r\nSo $F_{2n+2}\\leq3F_{2n}\\leq3\\frac{(2n+2)^n}{n!}$.\r\nWe only need to show $3\\frac{(2n+2)^n}{n!}\\leq\\frac{(2n+4)^{n+1}}{(n+1)!}$.\r\nIt can be writed like this:$3(n+1)\\leq(2n+2)(1+\\frac{1}{n+1})^{n+1}$.\r\n$\\frac{3}{2}\\leq(1+\\frac{1}{n+1})^{n+1}$.\r\nAnd we can easily know $(1+\\frac{1}{n+1})^{n+1}\\geq1+(n+1)\\frac{1}{n+1}=2>\\frac{3}{2}$.\r\nSo the original inequality is correct. ;)"
}
{
  "Tag": [],
  "Problem": "There are 20 people in my club.  8 of them are left-handed.  15 of them like jazz music.  2 of them are right-handed and dislike jazz music.  How many club members are left-handed and like jazz music?",
  "Solution_1": "Draw a Venn diagram: one category is for left-handed, the other for likes jazz.  We are told 2 are right-handed and don't like jazz, so they are outside, meaning there are 18 inside.  Let $ x\\equal{}\\text{number of left\\minus{}handed people}$, $ y\\equal{}\\text{left\\minus{}handed and like jazz}$ and $ z\\equal{}\\text{like jazz}$.  Then $ x\\plus{}y\\plus{}z\\equal{}18$, $ x\\plus{}y\\equal{}8$, and $ y\\plus{}z\\equal{}15$.  Then $ y\\equal{}15\\minus{}z\\Rightarrow x\\plus{}15\\minus{}z\\equal{}8\\Rightarrow x\\equal{}z\\minus{}7$.  This means that $ z\\minus{}7\\plus{}15\\minus{}z\\plus{}z\\equal{}18\\Rightarrow z\\equal{}10\\Rightarrow x\\equal{}3,y\\equal{}5$.  So there are 5 people who like jazz and are left-handed.",
  "Solution_2": "Or for a faster way:\r\n\r\nWe want the lefties and those who like jazz, so we can disregard those 2 righties who don't like jazz. That leaves 18. Since 15 like jazz and 8 are leftie, our answer is simply $ (15 \\plus{} 8) \\minus{} 18 \\equal{} \\boxed{5}$.\r\n\r\nEDIT: Err...I knew that... :P Thanks, Coptie.",
  "Solution_3": "Uh izzy...$ (15\\plus{}8)\\minus{}18\\equal{}5\\neq3$"
}
{
  "Tag": [],
  "Problem": "The fraction $ \\frac{x}{5}$ does not change its value when 3 is added to both its numerator and its denominator. What is the value of $ x$?",
  "Solution_1": "hello \r\nu can simply say it should be $ 5$ \r\n$ \\frac{x\\plus{}3}{8 }\\equal{} \\frac{x}{5}$ \r\n$ x \\equal{}5$\r\nthank u",
  "Solution_2": "[hide=\"Clearer solution\"]\nWe have the equation \\[ \\frac{x}{5}\\equal{}\\frac{x\\plus{}3}{5\\plus{}3}.\\] Cross multiplying gives \\[ 8x\\equal{}5x\\plus{}15\\Rightarrow3x\\equal{}15\\Rightarrow x\\equal{}\\boxed{5}.\\][/hide]"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "The circumference of a circle is $ 24\\pi$ cm. What is the number of square centimeters in the area of the largest square that can fit inside the circle?",
  "Solution_1": "[asy]draw((1,0)--(0,1)--(-1,0)--(0,-1)--cycle);\ndraw((1,0)..(0,1)..(-1,0)..(0,-1)..cycle);\n\ndraw((1,0)--(-1,0));\ndraw((0,1)--(0,-1));\n\nlabel(\"$r$\",(0.5,0),S);\nlabel(\"$r$\",(0,0.5),W);\nlabel(\"$\\sqrt{2}r$\",(0.5,0.5),N);[/asy]\r\n\r\n$ 2\\pi r\\equal{}24\\pi \\iff r \\equal{} 12$.\r\nThus, $ (12\\sqrt{2})^2 \\equal{} \\boxed{288}$."
}
{
  "Tag": [
    "AMC"
  ],
  "Problem": "I need help on this problem. \r\n\r\nLet $a_1 = 1$ and for $n>1$, $a_n = 2a_{n-1}$ if $a_{n-1}$ is odd and equals $a_{n-1}+1$ if it's otherwise. Find $a_{2004}$.\r\n\r\nThe problem I'm facing is writing $a_n$ without using $a_{n-1}$. Like I see that:\r\n\r\n$a_4 = 2(2a_1+1)$\r\n$a_6 = 2[2(2a_1+1)+1]$\r\n\r\nand so forth but I'm having trouble writing $a_n$. Can anyone help me on how to do this? \r\n\r\nI posted in AMC forum because this type of problem appears quite often in AMC.",
  "Solution_1": "[hide=\"Hint\"]\\[a_2=2a_1=2\\]\n\\[a_4=2(a_2+1)=4+2\\]\n\\[a_6=2(a_4+1)=8+4+2\\]\n\\[a_8=2(a_6+1)=16+8+4+2\\]\nSee it now? ;) [/hide]",
  "Solution_2": "Oh..\r\n\r\nI was writing $a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 6, a_5 = 7, a_6 = 14, a_7 = 15, a_8 = 30$ and so forth. I didn't see it that way.\r\n\r\nSo $a_{2004}$ would be:\r\n\r\n[hide=\"Then..\"]\n\n$a_{2004} = 2^1+2^2+2^3+2^4+ \\cdots 2^{1002}$? [/hide]",
  "Solution_3": "Yup.\r\n\r\n[hide=\"That expression collapses...\"]\\[\\sum_{n=1}^{1002}2^n=\\sum_{n=0}^{1002}2^n-1=2^{1003}-2\\][/hide]"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Inscribed in a circle is a quadrilateral having side lengths 25,39,52 and 60 taken consecutively. What is the diameter of this circle?",
  "Solution_1": "[hide]just by luck, if you draw the diagonal, they're right triangles (and Pythagorean tripleets at that!) so the hypotenuse (and the diameter) is [b]65.[/b][/hide]",
  "Solution_2": "Or...\r\nThe magic of....\r\n[hide=\"Super Awesome Theorem of pwnage\"]\nPtolemy's Theorem!!!\n[/hide]\r\n\r\nOf course, it takes some more manipulation, because the diagonals of the quad aren't actually the diameters...",
  "Solution_3": "How would you use Ptolemy's Theorem on this problem?",
  "Solution_4": "[hide]Bhramagupta's formula says that the answer is $\\sqrt{63 \\cdot 49 \\cdot 36 \\cdot 28}$\n$84 \\cdot 21$\n$1764$[/hide]",
  "Solution_5": "Random trivia:\r\nWe can generalize Brhamagupta's Formula (or however you spell his name) to any quadrilateral, cyclic or non-cyclic. This formula is sometimes called Bretschneider's formula:\r\n$\\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\\cos^{2}{\\frac{A+C}{2}}}$, where $s$ is the semiperimeter, $a, b, c, d$ are the sides of the quadrilateral, and $A, C$ are opposite angles. In fact, any set of opposite angles will do."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "search",
    "algebra unsolved"
  ],
  "Problem": "Hi all!!!!!!!!! :wink: \r\nHere is a problem from VIETNAM 2003 olimpiad.\r\n$ B2.$  Define $ p(x) = 4x^{3}-2x^{2}-15x+9, q(x) = 12x^{3}+6x^{2}-7x+1$. Show that each polynomial has just three distinct real roots. Let $ A$ be the largest root of $ p(x)$ and $ B$ the largest root of $ q(x)$. Show that $ A^{2}+3B^{2}= 4$.",
  "Solution_1": "Please search it on this forum. It is posted before!"
}
{
  "Tag": [
    "function",
    "inequalities",
    "integration",
    "triangle inequality",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "let $ f_n$ be a sequence of non-negative measurable functions in $ L^p(\\Bbb R)$ for some $ 1 <p<\\infty$. show that $ f_n \\to f$ in $ L^p$ iff $ f_n^p \\to f^p$ in $ L^1(\\Bbb R)$.",
  "Solution_1": "Here's one direction:\r\n\r\nWe can show that for $ p\\ge 1$ and $ x\\ge 0,$ $ (1 \\plus{} x)^p\\ge 1 \\plus{} x^p.$\r\n\r\nFor $ a\\ge b > 0,$ we can let $ x \\equal{} \\frac ab \\minus{} 1$ and rearrange this to \r\n\r\n$ \\left(\\frac ab\\right)^p \\minus{} 1\\ge\\left(\\frac1b \\minus{} 1\\right)^p$ or $ a^p \\minus{} b^p\\ge(a \\minus{} b)^p$\r\n\r\nSo we have that $ |f_n \\minus{} f|^p\\le |f_n^p \\minus{} f^p|$ and hence $ \\|f_n \\minus{} f\\|_p^p\\le \\|f_n^p \\minus{} f^p\\|_1.$\r\n\r\nSo this gives us $ f_n^p\\to f^p$ in $ L^1$ implies $ f_n\\to f$ in $ L^p.$",
  "Solution_2": "Here is a somewhat weaker (I think, I very well could be wrong) version of the converse:\r\n\r\nSince $ f_{n}\\rightarrow f$ in $ L^{p}$, then $ \\left\\|f_{n}\\minus{}f\\right\\|_{p}\\rightarrow 0$ as $ n\\rightarrow\\infty$.  By the triangle inequality, $ \\left\\|f_{n}\\right\\|_{p}\\leq\\left\\|f_{n}\\minus{}f\\right\\|_{p}\\plus{}\\left\\|f\\right\\|_{p}\\Rightarrow\\left\\|f_{n}\\right\\|_{p}\\minus{}\\left\\|f\\right\\|_{p}\\leq\\left\\|f_{n}\\minus{}f\\right\\|_{p}$ and $ \\left\\|f\\right\\|_{p}\\leq\\left\\|f_{n}\\minus{}f\\right\\|_{p}\\plus{}\\left\\|f_{n}\\right\\|_{p}\\Rightarrow\\left\\|f\\right\\|_{p}\\minus{}\\left\\|f_{n}\\right\\|_{p}\\leq\\left\\|f_{n}\\minus{}f\\right\\|_{p}$.  Hence $ (\\left|f_{n}\\right|^{p})$ converges to $ \\left|f\\right|^{p}$ in $ L^{1}$.",
  "Solution_3": "[quote=\"JRav\"]Hence $ (\\left|f_{n}\\right|^{p})$ converges to $ \\left|f\\right|^{p}$ in $ L^{1}$.[/quote]\r\n\r\nare you sure that this is what you've shown?",
  "Solution_4": "By Minkowski's we have:\r\n$ \\left\\parallel{}f^{p}_{n}\\right\\parallel{}_{1} \\equal{} \\parallel{}f^{p}_{n} \\minus{} f^{p} \\plus{} f^{p}\\parallel{}_{1} \\leq \\parallel{}f^{p}_{n} \\minus{} f^{p}\\parallel{}_{1} \\plus{} \\parallel{}f^{p}\\parallel{}_{1}$\r\n\r\nThen we get:\r\n$ \\left| \\parallel{}f^{p}_{n}\\parallel{}_{1} \\minus{} \\parallel{}f^{p}\\parallel{}_{1} \\right| \\leq \\parallel{}f^{p}_{n} \\minus{} f^{p}\\parallel{}_{1} \\equal{} \\int \\left|\\left(f^{p}_{n} \\minus{} f^{p}\\right)^{\\frac{1}{p}}\\right|^{p} \\leq \\left\\parallel{}f_{n} \\minus{} f \\right\\parallel{}^{p}_{p} \\plus{} \\parallel{}f_{n}\\parallel{}^{p}_{p}$",
  "Solution_5": "for other direction : let $ M \\equal{} max\\{\\parallel{}f\\parallel{}_{p}, \\parallel{}f_{n}\\parallel{}_{p}: n\\in\\mathbb{N}\\}$, then $ \\int\\parallel{}f|^{p} \\minus{} |f_{n}|^{p}|d\\nu\\leq 2pM^{p \\minus{} 1}\\parallel{}f \\minus{} f_{n}\\parallel{}_{p}$\r\n\r\nProf. Kent proposed very nice solution but I think we can apply Fatou's lemma : \r\n\r\nwe have $ |f \\minus{} f_{n}|^{p}\\leq 2^{p}(|f|^{p} \\plus{} |f|_{n}^{p})$, so consider : $ 2^{p}(|f|^{p} \\plus{} |f|_{n}^{p}) \\minus{} |f \\minus{} f_{n}|^{p}\\to 2^{p \\plus{} 1}|f|^{p}$\r\n\r\nand apply the Fatou's lemma we have $ \\overline{lim}\\int|f \\minus{} f_{n}|^{p}\\leq 0$ .",
  "Solution_6": "1234567a: You have to pass to a subsequence first, though. So you get that every subsequence of $ f_n$ has a further subsequence that converges to $ f$ in $ L^p$, which is good enough.",
  "Solution_7": "[quote=\"blahblahblah\"]are you sure that this is what you've shown?[/quote]\r\nI showed that $ \\left|\\left\\|f\\right\\|_{p}\\minus{}\\left\\|f_{n}\\right\\|_{p}\\right|\\leq\\left\\|f_{n}\\minus{}f\\right\\|_{p}$, which means that $ \\left|\\int\\left|f_{n}\\right|^{p}\\minus{}\\int\\left|f\\right|^{p}\\right|\\rightarrow 0$, which I thought meant that $ \\left|f_{n}\\right|^{p}$ converged to $ \\left|f\\right|^{p}$ in the $ L^{1}$ norm but apparently I am mistaken.  Could you elaborate?",
  "Solution_8": "[quote=\"ficsur\"]let $ f_n$ be a sequence of non-negative measurable functions in $ L^p(\\Bbb R)$ for some $ 1 < p < \\infty$. show that $ f_n \\to f$ in $ L^p$ iff $ f_n^p \\to f^p$ in $ L^1(\\Bbb R)$.[/quote]\r\n\r\nI will prove that \" if $ f_n\\to f$ in $ L^p$, then $ f_n^p\\to f^p$ in $ L^1$\".\r\nIndeed, bacause the function $ \\varphi(x)\\equal{}x^p$ is convex with $ p>1$. By the convexity we have \r\n$ |x^p\\minus{}y^p|\\leq p(\\max\\{x,y\\})^{p\\minus{}1}|x\\minus{}y|$ with $ x,y\\geq 0$. \r\nFrom inequality $ \\max\\{x,y\\}\\leq x\\plus{}y$, we have $ |x^p\\minus{}y^p|\\leq p(x\\plus{}y)^{p\\minus{}1}|x\\minus{}y|$ (by $ p\\minus{}1>0$). Apply this inequality for $ f_n, f$, we obtain \r\n$ |f_n^p\\minus{}f^p|\\leq p(f_n\\plus{}f)^{p\\minus{}1}|f_n\\minus{}f|$\r\nHence, by using the Holder's inequality, we have \r\n$ \\parallel{}f_n^p\\minus{}f^p\\parallel{}_1\\leq p\\parallel{}(f_n\\plus{}f)^{p\\minus{}1}\\parallel{}_q\\parallel{}f_n\\minus{}f\\parallel{}_p(*)$\r\nhere, $ 1/p \\plus{} 1/q\\equal{}1$ therefor $ q(p\\minus{}1)\\equal{}p$.\r\n Thus, $ \\parallel{}(f_n\\plus{}f)^{p\\minus{}1}\\parallel{}_q\\equal{}\\parallel{}f_n\\plus{}f\\parallel{}_p^{p/q}\\leq 2^{p/q}(\\parallel{}f_n\\parallel{}_p\\plus{}\\parallel{}f\\parallel{}_p)^{p/q}$, this term is bounded, by the hypothesis $ f_n\\to f$ in $ L_p$\r\nThe proof is completed by the inequality (*).",
  "Solution_9": "[quote=\"JRav\"][quote=\"blahblahblah\"]are you sure that this is what you've shown?[/quote]\nI showed that $ \\left|\\left\\|f\\right\\|_{p} \\minus{} \\left\\|f_{n}\\right\\|_{p}\\right|\\leq\\left\\|f_{n} \\minus{} f\\right\\|_{p}$, which means that $ \\left|\\int\\left|f_{n}\\right|^{p} \\minus{} \\int\\left|f\\right|^{p}\\right|\\rightarrow 0$, which I thought meant that $ \\left|f_{n}\\right|^{p}$ converged to $ \\left|f\\right|^{p}$ in the $ L^{1}$ norm but apparently I am mistaken.  Could you elaborate?[/quote]\r\n\r\nwell, first of all, you've left out the $ 1/p \\minus{}$th roots.  Second of all, showing that $ |f_n|^p\\to |f|^p$ in $ L^1$ is the same as showing\r\n\\[ \\int \\parallel{}f_n|^p \\minus{} |f|^p|dx\\to 0,\\]\r\nwhich is harder, since $ \\int |u_n|dx \\to 0$ is stronger than $ |\\int u_ndx|\\to 0$.",
  "Solution_10": "Indeed it does.  Stupid mistake. :)"
}
{
  "Tag": [
    "trigonometry",
    "trig identities",
    "Law of Cosines",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Let \u03b1+\u03b2= Fixed Value\r\na , b is Fixed Value respectively\r\n\r\nFind  R , \u03b1 , \u03b2",
  "Solution_1": "Let $O$ be the center of the circle, and the segments $b=BC,a=CA$.\r\nAlso let the given angle $x=\\angle AOB$\r\n\r\nAt first, we can notice that we can easy construct this figure geometricaly.\r\n\r\nThe angle $\\phi=\\angle ACB$ is such that $\\phi+\\frac{x}{2}=\\pi\\Rightarrow \\phi = \\pi -\\frac{x}{2}$\r\n\r\nfrom the triangle $ABC$ we can get that\r\n\r\n$AB = 2Rsin \\phi\\Rightarrow AB^2 = 4R^2 sin^2 \\phi$\r\n\r\nLaw of cosines in $\\triangle ABC: AB^2 = a^2+b^2-2ab\\cdot cos \\phi$\r\n\r\nSo we have \r\n\r\n$4R^2 sin^2 \\phi = a^2+b^2-2ab\\cdot cos \\phi \\Rightarrow$\r\n\r\n$R^2 = \\frac{a^2+b^2-2ab\\cdot cos \\phi}{4sin^2 \\phi}$\r\n\r\n$R = \\frac{\\sqrt{a^2+b^2-2ab\\cdot cos \\phi}}{2sin \\phi}$\r\n\r\nOnce we have found $R$, we can find the angles on the center using the formulas\r\n\r\n$\\boxed{ sin \\frac{\\widehat{\\alpha}}{2} = \\frac{a}{2R} } \\boxed{sin \\frac{\\widehat{\\beta}}{2} = \\frac{b}{2R}}$\r\n\r\nWe can use sine and cosine of $\\ \\frac{x}{2}$ instead of $\\phi$:\r\n\r\n$\\phi = \\pi -\\frac{x}{2} \\Rightarrow$\r\n\r\n$sin \\phi = sin \\frac{x}{2}$\r\n\r\n$cos \\phi = -cos \\frac{x}{2}$"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Let an invertible square complex matrix A have Jordan form X. Find the Jordan form of the inverse of A.",
  "Solution_1": "This comes down to this question: can you find the inverse of one Jordan block? \r\n\r\nLet's do this in a slightly more general way. Suppose $ A\\equal{}\\lambda I\\plus{}N$ where $ N$ is nilpotent and $ \\lambda\\ne 0.$ What is $ A^{\\minus{}1}?$\r\n\r\nWrite $ A\\equal{}\\lambda(I\\plus{}\\lambda^{\\minus{}1}N).$ I claim that\r\n\\[ A^{\\minus{}1}\\equal{}\\lambda^{\\minus{}1}\\sum_{k\\equal{}0}^{\\infty}(\\minus{}1)^k\\lambda^{\\minus{}k}N^k\\equal{}\\sum_{k\\equal{}0}^{\\infty}(\\minus{}1)^k\\lambda^{\\minus{}k\\minus{}1}N^k.\\]\r\nNote that this apparent infinite series is in fact a finite sum, since $ N$ is nilpotent.\r\n\r\nAn example of this theorem:\r\n\\[ \\begin{bmatrix}\\lambda&1&0&0\\\\0&\\lambda&1&0\\\\0&0&\\lambda&1\\\\0&0&0&\\lambda\\end{bmatrix}^{\\minus{}1}\\equal{}\\begin{bmatrix}\\lambda^{\\minus{}1}&\\minus{}\\lambda^{\\minus{}2}&\\lambda^{\\minus{}3}&\\minus{}\\lambda^{\\minus{}4}\\\\0&\\lambda^{\\minus{}1}&\\minus{}\\lambda^{\\minus{}2}&\\lambda^{\\minus{}3}\\\\0&0&\\lambda^{\\minus{}1}&\\minus{}\\lambda^{\\minus{}2}\\\\0&0&0&\\lambda^{\\minus{}1}\\end{bmatrix}\\]\r\nOne you figure out how to do that to any Jordan block, you should be able to figure out how to handle several blocks put together.",
  "Solution_2": "Hi Kent. Your answer is not complete.\r\n1) $ \\lambda,\\mu\\not\\equal{}0,\\lambda\\not\\equal{}\\mu$ imply $ 1/\\lambda\\not\\equal{}1/\\mu$.\r\n 2) Let $ A\\equal{}\\lambda{I}\\plus{}J$ where $ J$ is a nilpotent Jordan block ($ J^n\\equal{}0,J^{n\\minus{}1}\\not\\equal{}0$). $ A^{\\minus{}1}\\equal{}(1/\\lambda)I\\plus{}N$ where $ N^n\\equal{}0,N^{n\\minus{}1}\\not\\equal{}0$. Then $ A^{\\minus{}1}$ is similar to $ (1/\\lambda)I\\plus{}J$, its Jordan form.\r\nConclusion: Let $ A\\in\\mathcal{M}_n(\\mathbb{C})$. If in the Jordan form of $ A$ , we change, for all $ i$, $ \\lambda_i$ with $ 1/\\lambda_i$ then we obtain the Jordan form of $ A^{\\minus{}1}$."
}
{
  "Tag": [
    "geometry",
    "ratio"
  ],
  "Problem": "Sorry this is a pretty standard question but when trying to come up with the formula for the area ratio asked in the following:\r\n\r\nSuppose $ ABC$ is a triangle and let $ D$, $ E$, $ F$ are on $ BC$, $ CA$, and $ AB$ respectively s.t.\r\n$ \\frac {BD}{DC} \\equal{} \\frac {1}{x}$\r\n\r\n$ \\frac {CE}{EA} \\equal{} \\frac {y}{1}$\r\n\r\n$ \\frac {AF}{FB} \\equal{} \\frac {1}{z}$\r\n\r\nIf $ AD$, $ BE$, $ CF$ are drawn then you get a a smaller triangle $ PQR$ where $ P$ is intersection of $ AD$ and $ BE$, $ Q$ is intersection of $ CF$ and $ BE$, and $ R$ is intersection of $ AD$ and $ CF$.\r\n\r\nCompute $ \\frac {[PQR]}{[ABC]}$\r\n\r\nI know this uses mass-point and is not pretty,as it looks at only two cevians at a time, etc (i've done it before, but can't seem to remember the sequence of steps i used) could somebody help?",
  "Solution_1": "Check out [url=http://www.cut-the-knot.org/triangle/RouthTheorem.shtml]Routh's Theorem[/url]."
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Given $\\sum^{n}_{i=1} x_i=n$ for $x_i\\in \\mathbb{R}$ and\r\n\r\n$\\sum^{n}_{i=1}x_i^4=\\sum^{n}_{i=1}x_i^3$\r\n\r\nSolve the system of equation for $x_i$ .\r\n\r\n :)",
  "Solution_1": "Two of the power means turn out to be equal, so we must have equality everywhere, and hence $x_1 = x_2 = \\dots = x_n = 1$.\r\n\r\nYou could also notice that $\\sum_{i = 1}^n\\left(x_i - 1\\right) = 0$ and $\\sum_{i = 1}^nx_i^3\\left(x_i - 1\\right) = 0$.\r\n\r\nSubtract the first equation from the second to get $\\sum_{i = 1}^n\\left(x_i - 1\\right)^2\\left(x_i^2 + x_i + 1\\right) = 0$. Hence $x_i = 1$ for all $i$.",
  "Solution_2": "Arne , I dont undestand what you mean by\r\n\r\n[quote]\nTwo of the power means turn out to be equal, so we must have equality everywhere, \n[/quote]\r\n\r\nCan you explain more detail ?\r\n :)",
  "Solution_3": "Let's see... We have \\[\\sqrt[4]{\\frac{\\sum_{i = 1}^nx_i^4}{n}} \\geq \\sqrt[3]{\\frac{\\sum_{i = 1}^nx_i^3}{n}}\\] by Power means, which gives \\[{{n \\cdot (\\sum_{i = 1}^nx_i^4})^3 \\geq (\\sum_{i = 1}^nx_i^3})^4\\] or \\[n \\geq \\sum_{i = 1}^nx_i^3.\\] So, \\[1 = \\frac{\\sum_{i = 1}^nx_i}{n} \\geq \\sqrt[3]{\\frac{\\sum_{i = 1}^nx_i^3}{n}}\\] which is the power mean inequality \"the wrong way around\".\r\n\r\nSo, equality must hold, and all variables must be equal.",
  "Solution_4": "Thats great !  Arne , thank you .   :)"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Prove that:$\\frac{1}{2+a}+\\frac{1}{2+b}+\\frac{1}{2+c}\\leq 1$\r\nwhere $a;b;c$ are non-negative and satisfying $abc=1$",
  "Solution_1": "[quote=\"chien than\"]Prove that:$\\frac{1}{2+a}+\\frac{1}{2+b}+\\frac{1}{2+c}\\leq 1$\nwhere $a;b;c$ are non-negative and satisfying $abc=1$[/quote]\r\n\r\nThis is a particular case (n = 3) of the following fact:\r\n\r\nIf $x_{1}$, $x_{2}$, ..., $x_{n}$ are n nonnegative reals that satisfy $x_{1}x_{2}...x_{n}=1$, then $\\sum_{i=1}^{n}\\frac{1}{n-1+x_{i}}\\leq 1$.\r\n\r\nSee\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=4951\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=27900\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=164\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=64349\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=116016\r\n\r\n(Someone should really moderate this forum...)\r\n\r\n  darij",
  "Solution_2": "It was also discussed here http://www.mathlinks.ro/Forum/viewtopic.php?t=131882"
}
{
  "Tag": [
    "logarithms"
  ],
  "Problem": "Source: MAML 2006 States!!\r\n\r\nIf $\\log_4{36}=a$ and $\\log_4{9}=b$, find $\\log_8{\\frac{216}{81}}$ in terms of $a$ and $b$.\r\n\r\nOh yea, you should have 2.5 minutes to do this question.",
  "Solution_1": "[hide=\"Took me around 60 - 80 seconds.\"]Break up the original $\\log_8 \\frac{216}{81} = 3 \\log_8 6 - 4 \\log_8 3$. Now change the a and b expressions to base 8, so $a= \\log_4 36 = \\frac{\\log_8 36}{\\log_8 4} = \\frac{\\log_8 36}{\\frac{2}{3}} = \\frac{3}{2} 2 \\log_8 6 = 3 \\log_8 6$ then doing the same for b yields $b=3 \\log_8 3$. Plugging back in appropriate a and b values into the simplified original, $3 \\log_8 6 - 4 \\log_8 3 = a- \\frac43 b$.[/hide]",
  "Solution_2": "[quote=\"weatherbottle\"]Source: MAML 2006 States!!\n\nIf $\\log_4{36}=a$ and $\\log_4{9}=b$, find $\\log_8{\\frac{216}{81}}$ in terms of $a$ and $b$.\n\nOh yea, you should have 2.5 minutes to do this question.[/quote]\r\n[hide]$\\log_8{\\frac{216}{81}}=\\log_2{\\frac{6}{3^{\\frac{4}{3}}}=\\log_2{6}-\\log_2{3^{\\frac{4}{3}}}=\\log_2{6}-\\frac{4}{3}\\log_2{3}}$.\n$a=\\log_2{6}$, $b=\\log_2{3}$, so $\\log_2{6}-\\frac{4}{3}\\log_2{3}=\\boxed{a-\\frac{4}{3}b}$.[/hide]",
  "Solution_3": "[quote=\"weatherbottle\"]Source: MAML 2006 States!!\n\nIf $\\log_4{36}=a$ and $\\log_4{9}=b$, find $\\log_8{\\frac{216}{81}}$ in terms of $a$ and $b$.\n\nOh yea, you should have 2.5 minutes to do this question.[/quote]\r\n[hide=\"logs are fun\"]\nLets start with $\\log_4{36}$\n$\\log_4 {36}=\\log_4 {4\\cdot9}=\\log_4 {4}+\\log_4 {9}$\n$a=1+\\log_4{9}$\n$\\log_4 {9}=a-1$\n$a-1=b$\nSo now we have to break down $\\log_8{\\frac{216}{81}}$.\nImmediately we see that that equals $\\log_8{216}-\\log_8 {81}$\nUsing the exponent rules, we get $3\\log_8 {2}-4\\log_8 {2}$\nAfter converting to base 4, we get $a-4b/3$\n[/hide]"
}
{
  "Tag": [
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "[b](1) Is there is any Geometrical method to find the value of cos(18 degree), cos(36 degree) ,cos(54 degree).[/b]",
  "Solution_1": "hello, maybe this will help you\r\nhttp://2000clicks.com/mathhelp/GeometryTrigSpecialAngles.htm\r\nSonnhard.",
  "Solution_2": "Sir have found no. idea how to find the value of sin(18 degree) or tan(18degree).\r\nso sir plz explain me.",
  "Solution_3": "we have $ \\sin 36^o \\equal{} \\cos 54^o$\r\nuse Triple Angle Formula  on $ RHS$ and Double Angle Formula on $ LHS$\r\nand it will solve the problem"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "incenter",
    "angle bisector"
  ],
  "Problem": "Right triangle $ ABC,$ with $ \\angle A\\equal{}90^{\\circ},$ is inscribed in circle $ \\Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \\angle EAC \\equal{} \\angle CAF.$ Segment $ BF$  meets $ \\Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.",
  "Solution_1": "[hide=\"click text\"]Let I be the incenter of AEF. you should prove <EIF=<EDF. or in other words <EDF=90\u00b0+(1/2)(<EAF).\nJust see that <EDB=<EAB=90\u00b0-(1/2)(<EAF) => <EDF=90\u00b0+(1/2)(<EAF).DONE[/hide]",
  "Solution_2": "let angle CAE=y, ACB=x\r\nthen ACF=90+y-x, EDF=90+y, AFE=90+x-2y, AEF=90-x\r\nhence OEC=y\r\nthen EC/ CF=AE/AF=sin(90+x-2y)/cos x\r\nwe get EM/MC=(cos x +sin(90+x-2y))/(sin(90+x-2y)-cosx) = tan ECO / tan y\r\nso tan ECO = tany( (cos x +sin(90+x-2y))/(sin(90+x-2y)-cosx)) = tan(90+y-x)\r\nso ECO = 90+y-x hence ACF,ECO are vertical angles which means A,C,O collinear.",
  "Solution_3": "this is fun, better than number 6.",
  "Solution_4": "[hide]\nWe can deduce that \\[ \\angle EOF \\equal{} 2(180 \\minus{} \\angle EDF) \\equal{} 2 \\angle EDB \\equal{} 2 \\angle EAB \\equal{} 2(90 \\minus{} \\angle EAD) \\equal{} 180 \\minus{} \\angle EAF\\], hence $ AFDE$ is cyclic. Since $ \\angle OEF \\equal{} \\angle OFE$, we can deduce that $ AO$ is the angle bisector of $ \\angle EAF$, hence $ A, C, O$ are collinear.\n[/hide]",
  "Solution_5": "Let $ AF$ meet the circuncircle of $ EDF$ again at $ E'$. Angle chasing shows that $ AEE'$ is isosceles, that is, $ E$ and $ E'$ are symmetrical wrt to $ AC$ and the result follows.",
  "Solution_6": "We solve it just by angles chasing : denote $ {\\angle ABC \\equal{} N}$\uff1b$ {\\angle BCA\\equal{}90\\minus{}N}$ ;$ {\\angle ACE\\equal{} M}$; then we get :$ {\\angle EAC\\equal{}180\\minus{}M\\minus{} N}$ $ {\\longrightarrow}$ $ {\\angle EAF \\equal{}360\\minus{}2M\\minus{}2N}$;and $ {\\angle BCE\\equal{}\\angle BCE\\equal{}M\\minus{}(90\\minus{}N)\\equal{}M\\plus{}N\\minus{}90}$; $ {\\longrightarrow}$  $ {\\angle EDF \\equal{}270\\minus{}M\\minus{}N}$ ;$ {\\angle EOF\\equal{}[180\\minus{}(270\\minus{}M\\minus{}N)]*2\\equal{}2M\\plus{}2N\\minus{}180}$;\r\n thus $ {\\angle EAF\\plus{}\\angle EOF\\equal{}180}$; $ {\\longrightarrow}$  $ {A,E,F,O}$ are con-cyclic; since $ {EO\\equal{}FO}$ ;$ {O}$ is on the angle bisector of $ {\\angle EAF}$; that is $ {A,C,O}$ are collinear.....$ {Q.E.D.}$[ :) ]",
  "Solution_7": "[quote]Right triangle $ ABC,$ with $ \\angle A \\equal{} 90^{\\circ},$ is inscribed in circle $ \\Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA > EC.$ Point $ F$ lies on ray $ EC$ with $ \\angle EAC \\equal{} \\angle CAF.$ Segment $ BF$  meets $ \\Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.[/quote]\r\nHere is my solution\r\n\r\nSince $ \\angle OED$ $ \\equal{}90^{\\circ}\\minus{}$ $ \\frac {1}{2}$ $ \\angle EOD$ $ \\equal{}90^{\\circ}\\minus{}$ $ \\angle DFE$ $ \\equal{}\\angle DCF$ $ \\equal{}\\angle DBE$, implying $ OE$ is a tangent from $ O$ to $ (\\Gamma)$. The same argument holds for $ OD$, which yields that $ (O)$ and $ (\\Gamma)$ are orthogonal. But in the other hand, due to the fact that $ (AB,AC,AE,AS)\\equal{}\\minus{}1$, then if $ W$ is the intersection of $ AB$ with $ EF$ then $ (EFCW)\\equal{}\\minus{}1$, which implies $ (BE,BF,BC,BW)\\equal{}\\minus{}1$ or we can say $ (BE,BD,BC,BA)\\equal{}\\minus{}1$ $ \\Longrightarrow$ $ ADCE$ is a [i]harmonic quadrilateral[/i]. As the result, $ AC$ with then tangents at $ D$ and $ E$ will concur, hence, $ AC,$ $ OD,$ $ OE$ will concur at $ O$, which implies $ A,$ $ C,$ $ O$ are collinear. $ \\square$",
  "Solution_8": "Notice $$-1=(E,F;C,\\overline{AB}\\cap\\overline{EF})\\stackrel{B}=(E,D;C,A)$$ by the Right Angles and Bisectors Lemma. Since $$\\angle CEO=90-\\tfrac{1}{2}\\angle FOE=90-\\angle EDB=90-\\angle EAB=\\angle CAE,$$ we know $\\overline{OE}$ is tangent to $\\Gamma$ at $E.$ Similarly, $O=\\overline{DD}\\cap\\overline{EE}$ so if $A'=\\overline{OC}\\cap\\Gamma,$ we see $(A',C;DE)=-1.$ Hence, $A=A'.$ $\\square$"
}
{
  "Tag": [
    "trigonometry",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Here is,  I think, a nice product.\r\n\r\n\r\n$\\prod_{n=0}^{\\infty}\\frac{(2n+1)^{4}}{(2n+1)^{4}-(2/\\pi)^{4}}=\\frac{2e\\sec(1)}{e^{2}+1}$.",
  "Solution_1": "[quote=\"didilica\"]Here is,  I think, a nice product.\n\n\n$\\prod_{n=0}^{\\infty}\\frac{(2n+1)^{4}}{(2n+1)^{4}-(2/\\pi)^{4}}=\\frac{2e\\sec(1)}{e^{2}+1}$.[/quote]\r\n\r\nLet consider a generalization: for $z\\in\\mathbb{C},$ put $f(z)=\\prod_{n=0}^{\\infty}\\frac{(2n+1)^{4}}{(2n+1)^{4}-(2z/\\pi)^{4}}$. Then,\r\n\r\n\\[f(z)=\\prod_{n=0}^{\\infty}\\left[\\frac{(2n+1)^{2}}{(2n+1)^{2}-(2z/\\pi)^{2}}\\cdot\\frac{(2n+1)^{2}}{(2n+1)^{2}+(2z/\\pi)^{2}}\\right] =\\left[\\prod_{n=0}^{\\infty}\\left(1-\\frac{4z^{2}}{(2n+1)^{2}\\pi^{2}}\\right)\\cdot\\prod_{k=0}^{\\infty}\\left(1+\\frac{4z^{2}}{(2k+1)^{2}\\pi^{2}}\\right)\\right]^{-1}\\]\r\n\r\nbut since we know that for all $z\\in\\mathbb{C},$\r\n\r\n\\[\\cos z =\\prod_{n=0}^{\\infty}\\left(1-\\frac{4z^{2}}{(2n+1)^{2}\\pi^{2}}\\right)\\ ,\\qquad\\qquad \\mbox{ and }\\qquad\\qquad\\mbox{cosh}z =\\prod_{k=0}^{\\infty}\\left(1+\\frac{4z^{2}}{(2k+1)^{2}\\pi^{2}}\\right)\\ ,\\] \r\n\r\nwe may write\r\n\r\n\\[f(z)=\\left(\\cos z \\cdot\\mbox{cosh}z\\right)^{-1}= \\sec z \\cdot\\mbox{sech}z =\\frac{2\\sec z}{e^{z}+e^{-z}}= \\frac{2e^{z}\\sec z}{e^{z}+1}\\]\r\n\r\nClearly the given product is then $f(1) = \\frac{2e\\sec(1)}{e^{2}+1}.$"
}
{
  "Tag": [
    "inequalities",
    "logarithms",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Let $a,\\ b$ be real numbers such that $\\log_{a}8=\\log_{\\sqrt{b}}2.$\r\n\r\n(1) Find the values of $\\log_{ab}a+\\log_{b}{\\frac{1}{\\sqrt{a}}}+(\\log_{a}b)^{2}.$\r\n\r\n(2) Solve the following inequality with respect to $x.$ \\[a^{3x}+a^{x+2}+a^{x}+a^{x-2}<b^{3x+3}+b^{3x}+b^{3x-3}+1\\]",
  "Solution_1": "$\\log_{a}8 = \\log_{\\sqrt{b}}2 \\Leftrightarrow \\frac{\\ln 8}{\\ln a}= \\frac{\\ln2}{\\ln \\sqrt{b}}\\Leftrightarrow 3 = \\frac{\\ln 8}{\\ln 2}= \\frac{\\ln a}{\\ln \\sqrt{b}}\\Leftrightarrow a^{2}= b^{3}$\r\n\r\n[b](1)[/b]  Hence the term is equal to $\\frac{3}{5}-\\frac{3}{4}+\\frac{4}{9}= \\frac{53}{180}$",
  "Solution_2": "O.K. :)"
}
{
  "Tag": [
    "number theory solved",
    "number theory"
  ],
  "Problem": "Find all positive integers n such that 2^n-1 is a multiple of 3 and 1/3*(2^n-1) is a divisor of 4m^2+1 for some integer m.",
  "Solution_1": "I think it is not hard !\r\nn=2^k.\r\nFirst, We prove that n is power of 2 :\r\n  from 2^n -1 is a mutilple of 3 hence n is even \r\n  assume n has an odd prime divisor d \r\n  2^d -4 is a mutilple of 4 that is 2^d-1 has prime divisor p is of the from  4k+3 .\r\n  then p divides 2^d-1 , which divides 2^n-1,which divides 4m^2+1\r\n  hence impossible\r\n  so n=2^k\r\n  consider f_k = 2^2^k +1\r\n  by the chinese remainder theorem we will see n=2^k satisfies .",
  "Solution_2": "nice problem\r\ni reached the same solution but once agen sum1 beat me to it\r\n :)",
  "Solution_3": "i can't understand that why p=3(mod4) can't divide 4m^2+1",
  "Solution_4": "think of Fermat's little theorem",
  "Solution_5": "sorry but could anyone explain why  numbers of the form $ n\\equal{}2^k$ satisfy conditions of the problem? J understand why other cannot be solutions but I can't understand why result follows from chinese remainder theorem.\r\n\r\nThanks for Your explanation.",
  "Solution_6": "I also think it is not very clear (actually, I think it is wrong)...\r\n\r\nLet $ n \\equal{} 2^k$, with $ k \\ge 1$. To prove that $ 4m^2 \\equiv \\minus{} 1 mod \\frac {2^n \\minus{} 1}{3}$ has a solution, we need to show that, for prime $ p$,\r\n\r\n$ (\\frac { \\minus{} 1}{p}) \\equal{} 1 \\forall p: p|\\frac {2^n \\minus{} 1}{3}$,\r\n\r\ni.e., $ p \\equiv 1 mod 4$ if $ p|\\frac {2^n \\minus{} 1}{3}$. It is easy to see with congruences that $ 2^n \\not\\equiv 1 mod 9$ (remembering that $ n \\equal{} 2^k$), so $ p \\not \\equal{} 3$. Now, if $ p|2^n \\minus{} 1$, we have $ ord_p2|n$. Thus, $ ord_p2$ is a power of $ 2$, say $ 2^t$. Since $ 2^{2^t} \\equiv 1 mod p$ and $ p > 3$, we have $ t > 1$. Then, as $ 2^{p \\minus{} 1} \\equiv 1 mod p$, $ 2^t|p \\minus{} 1 \\Rightarrow 4|p \\minus{} 1$, q.e.d."
}
{
  "Tag": [
    "greatest common divisor",
    "number theory",
    "least common multiple"
  ],
  "Problem": "the least common multiple of two nmbers is $2^{2}*3^{4}*5*7$. The greatest common divisor of the same two numbers is $2*3*5$. One of the numbers is 210. What's the other?[color=cyan]\n\n[size=50][hide=\"answer\"]3240[/hide][/size][/color]",
  "Solution_1": "[hide] The product of the GCF and LCM is the product of the two numbers.  When you multiply the GCF and LCM, you get $2^{3}*3^{5}*5^{2}*7$.  Call the other number n.  This means that 210n=8*243*25*7, so n=1620.  [/hide]",
  "Solution_2": "[quote=\"espark52\"]the least common multiple of two nmbers is $2^{2}*3^{4}*5*7$. The greatest common divisor of the same two numbers is $2*3*5$. One of the numbers is 210. What's the other?[color=cyan]\n\n[size=50][hide=\"answer\"]3240[/hide][/size][/color][/quote]\n[hide]\nLCM*GCF=Product\nLCM*GCF/1number=theother number\n$210=2*3*5*7$\nLCM*GCF$=2^{3}*3^{5}*5^{2}*7$\nLCM*GCF/210$=2^{2}*3^{4}*5$\n$=1620$\n[/hide]",
  "Solution_3": "[quote=\"espark52\"]the least common multiple of two nmbers is $2^{2}*3^{4}*5*7$. The greatest common divisor of the same two numbers is $2*3*5$. One of the numbers is 210. What's the other?[color=cyan]\n\n[size=50][hide=\"answer\"]3240[/hide][/size][/color][/quote]\r\n\r\nuhh... espark, i think the answer is 1620, not 3240 as you have posted.",
  "Solution_4": "[quote=\"davidlizeng\"][quote=\"espark52\"]the least common multiple of two nmbers is $2^{2}*3^{4}*5*7$. The greatest common divisor of the same two numbers is $2*3*5$. One of the numbers is 210. What's the other?[color=cyan]\n\n[size=50][hide=\"answer\"]3240[/hide][/size][/color][/quote]\n\nuhh... espark, i think the answer is 1620, not 3240 as you have posted.[/quote]\r\n\r\noh woops, ur right :blush: the reason i put 3240 is cause i mis-typed the problem it's supposed to be $2^{3}*3^{4}*5*7$"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "inradius",
    "incenter",
    "geometric transformation",
    "reflection",
    "homothety"
  ],
  "Problem": "[color=darkblue]Given are two parallel lines $d_{1}$ , $d_{2}$ and the circle $w$ which touches the line $d_{1}$ in the point $A$ and $d_{2}\\cap w=\\emptyset\\ .$\nFor a mobile point $M\\in d_{2}$ denote the intersections $X$ , $Y$ between the line $d_{1}$ and the tangents from the point $M$ to the circle $w\\ .$ \nProve that exist two constants $k$ , $l$ so that the product $AX\\cdot AY\\in \\{k,l\\}\\ .$\n\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=115606 (seek and will find ... the hint !).[/color]",
  "Solution_1": "[hide][url]http://www.mathlinks.ro/Forum/viewtopic.php?t=44854[/url][/hide]",
  "Solution_2": "The problem was also posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=85336[/url].\r\n\r\nAssume the circle center [color=blue][b]is[/b][/color] between the lines $d_{1},\\ d_{2}.$ Then $w$ is the incircle of $\\triangle MXY,$ so we denote its center I.  Let (O) be the circumcircle of this triangle, $(I_{m})$ its M-excircle, T the tangency point of $(I_{m})$ with $d_{1}\\equiv XY,$ and let $r,\\ r_{m},\\ h_{m}$ be its inradius, M-exradius, and M-altitude. Since $\\frac{r_{m}}{r}= \\frac{h_{m}}{h_{m}-2r}$ and since $r,\\ h_{m}$ are constant, $r_{m}$ is also constant. The bisector $MI \\equiv MI_{m}$ of $\\angle XMY$ meets (O) again at the midpoint P of the arc XY of (O) opposite to M. The incenter I and the M-excenter $I_{m}$ are both on a circle (P) centered at P and passing through the triangle vertices X, Y. Since this circle is centered on the perpendicular bisector p of XY and $I_{m}\\in (P),$ the reflection $I_{m}'$ of $I_{m}$ in p is also on (P). In addition, the reflection of T in p is A, the incircle tangency point with XY. Thus $I_{m}'A \\perp XY,\\ I_{m}'A = r_{m},$ and the points $I,\\ I_{m}' \\in P$ are both fixed. Hence the power of A to (P) is $AX \\cdot AY = AI \\cdot AI_{m}' = rr_{m}= \\frac{h_{m}r^{2}}{h_{m}-2r}$ is constant. If we substitute the distance  $d = h_{m}-r$ of the line $d_{2}$ from the center I of the circle $w \\equiv (I),$ we get\r\n\r\n$\\frac{AX \\cdot AY}{r^{2}}= \\frac{d+r}{d-r}$\r\n\r\nSimilarly, assuming the circle center [color=blue][b]is not[/b][/color] between the lines $d_{1},\\ d_{2},$ $w$ is the M-excircle of $\\triangle MXY$ and can get \r\n\r\n$\\frac{AX \\cdot AY}{r^{2}}= \\frac{d-r}{d+r}$\r\n\r\nThis proof is easily modified for the case $w \\cap d_{2}\\neq \\emptyset$ and M is still outside of $w.$ WLOG assuming AX > AY, $w$ is the X-excircle of $\\triangle MXY.$ For the proof, we use the Y-excircle, the radius of which is fixed. For example, if $r_{x}> h_{m}$ are fixed, since M is the internal homothety center of $(I_{x}),\\ (I_{y})$, $\\frac{r_{x}-h_{m}}{r_{x}}= \\frac{h_{m}-r_{y}}{r_{y}}$ and $r_{y}= \\frac{r_{x}h_{m}}{2r_{x}-h_{m}}$ is also fixed. The X, Y-excenters are both on a circle (P) centered at the midpoint P of the arc XY of (O) containing M. Similarly as before, (P) passes through the reflection of the Y-excenter in the perpendicular bisector of XY, which is fixed, etc.",
  "Solution_3": "[quote=\"Virgil Nicula\"][color=darkred]Given are two parallel lines $d_{1}$ , $d_{2}$ and the circle $w$ which touches the line $d_{1}$ in the point $T$ and $d_{2}\\cap w=\\emptyset\\ .$\nFor a mobile point $A\\in d_{2}$ denote the intersections $B$ , $C$ between the line $d_{1}$ and the tangents from the point $A$ to the circle $w\\ .$ \nProve that exist two constants $k$ , $l$ so that the product $TB\\cdot TC\\in \\{k,l\\}\\ .$[/color][/quote]\r\n[color=darkblue][b]Proof.[/b] I\"ll use the standard notations for the triangle $ABC\\ .$ Prove easily that the relations $(p-b)(p-c)=\\frac{r^{2}h_{a}}{h_{a}-2r}=\\frac{r_{a}^{2}h_{a}}{h_{a}+2r_{a}}\\ .$\n[hide=\"Here is prove.\"]\n$\\{\\begin{array}{cc}|\\begin{array}{c}h_{a}=\\frac{2pr}{a}\\\\\\\\ (p-a)(p-b)(p-c)=pr^{2}\\end{array}& (p-b)(p-c)(h_{a}-2r)=(p-b)(p-c)(\\frac{2pr}{a}-2r)=(p-a)(p-b)(p-c)\\cdot \\frac{2r}{a}=r^{2}h_{a}\\\\\\\\ |\\begin{array}{c}h_{a}=\\frac{2(p-a)r_{a}}{a}\\\\\\\\ p(p-b)(p-c)=(p-a)r^{2}_{a}\\end{array}& (p-b)(p-c)(h_{a}+2r_{a})=(p-b)(p-c)[\\frac{2(p-a)r_{a}}{a}+2r_{a}]=p(p-b)(p-c)\\cdot\\frac{2r_{a}}{a}=r_{a}^{2}h_{a}\\end{array}$[/hide] \n $TB\\cdot TC=(p-b)(p-c)\\Longrightarrow$ $\\{\\begin{array}{cccccc}\\mathrm{Case\\ 1.\\ : \\ }& w=C(I,r) & \\Longrightarrow & (p-b)(p-c)=\\frac{r^{2}h_{a}}{h_{a}-2r}& \\Longrightarrow &{ k=\\frac{r^{2}h_{a}}{h_{a}-2r}\\ }\\\\\\\\ \\mathrm{Case\\ 2.\\ : \\ }& w=C(I_{a},r_{a}) & \\Longrightarrow & (p-b)(p-c)=\\frac{r_{a}^{2}h_{a}}{h_{a}+2r_{a}}& \\Longrightarrow & l=\\frac{r_{a}^{2}h_{a}}{h_{a}+2r_{a}}\\ . \\end{array}$[/color]",
  "Solution_4": "that's a nice problem I see follow other way\r\nGiven $(I)$ touch line $d$ at $T$, find locus $A$ such that $(I)$ is incenter of $\\triangle ABC$ with $B,C$ move on $d$ and $TB.TC=const$.\r\n[hide=\"Solution\"]\na line $// BC$:D :D\n[/hide]"
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "algebra",
    "polynomial"
  ],
  "Problem": "I'll post COPSD 3rd B soon to encourage more middle schoolers to attend my competition.\r\n\r\nAnyone interested? (Even you qualified on COPSD 3rd A, you're welcome to just take it again for fun --> no prize for double qualifying though  :P )",
  "Solution_1": "When do you get when you qualify for the A one? (I'll do this one too  :D )",
  "Solution_2": "What's COPSD 3rd A?  :huh:",
  "Solution_3": "I'm interested, but could you please explain what COPSD 3rd A is? :? And what COSPD 3rd B is?",
  "Solution_4": "COPSD 3rd A was this thread 236factorial and Drunken_Math did:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=50030\r\n\r\nCOPSD 3rd B is just like COPSD 3rd A but with different set of problems that are in same level. The reason that I have A and B is because on the last time, some of people couldn't compete because of AP tests.\r\n\r\nAnyway, if you guys know AMC, this is same. Same level problems, just in different date.\r\n\r\n(Although you can do both  :P )\r\n\r\nTo qualify, you must send at least 7 SOLUTIONS (answers + explanation) to the problems that are correct. In another word, if you get 7 right, you're qualified. If you get more, good to you. If you get less,  :( .\r\n\r\nAnyone who qualifies from COPSD will go to COP Round 2, along with people from COP Round 1 (who are mostly high schoolers ---> This means you can go against the high school students!!).\r\n\r\nIf you guys have any more questions, please post!!",
  "Solution_5": "Oh, so now I get to participate in COP 5th now?  :D And can I do the COPSD B round at the same time?",
  "Solution_6": "I guess I will do it, it sounds fun.  Are the problems like AMC 10 level?  When are you going to post the questions?",
  "Solution_7": "[quote=\"236factorial\"]Oh, so now I get to participate in COP 5th now?  :D And can I do the COPSD B round at the same time?[/quote] \n\nYes. You're ALREADY qualified to COP Round 2 which is eh.. about level of later AMC-12 and possibly easy AIME.\n\n[quote=\"b-flat\"]Are the problems like AMC 10 level?[/quote]\n\nNo. They're level of mathcounts varying from chapter to national level.\n\n[quote=\"b-flat]When are you going to post the questions?[/quote]\r\n\r\nSoon. But I can't give you an exact date since I'm very busy this year (about 3 times more than last year). I'll announce it when it's ready. \r\n\r\nWe need more people to sign up!! Come on, guys! This is a VERY good practice for mathcounts!",
  "Solution_8": "I'm signing up!  :)",
  "Solution_9": "sign me up",
  "Solution_10": "me 2",
  "Solution_11": "Alright I think I'll have time to participate in this one. I'm sorry i didn't do 3rd A... I didn't have the time.",
  "Solution_12": "Here are the problems, folks! Enjoy!\r\n\r\nNote: These are VERY good practices for local, state, national mathcounts. Everybody who visit MC forum should give a shot. Also, some problems may require brute forcing and use of calculator so don't be afraid on this fact.\r\n\r\n[b]Problems[/b]\r\n\r\n1. Find the maximum or minimum height defined in equation $y = x^2-12x+2$.\r\n\r\n2. Find $a+b+c$.\r\n\r\n\\[2a+b+c = 10\\\\\r\na+2b+c = 12\\\\\r\na+b+2c = 14\\]\r\n\r\n3. In a 10-questioned test, the student can answer either true or false but must answer all of the 10 questions (no omitting!). How many ways can the student do this?\r\n\r\n4. Find $a$ and $b$ in:\r\n\r\n$\\sqrt{4 + \\sqrt{4 + \\sqrt{4 + \\sqrt{4 + \\cdots}}}} = \\frac{a+ \\sqrt b}{2}$\r\n\r\n5. The equilateral triangle $ABC$ has side length of 10. Connecting triangle $ABC$'s midpoints, we form triangle $A'B'C'$ with side length 5. In same manner, we form $A''B''C'', A'''B'''C'''$, and so on. Find the area of all equilateral triangles formed this way (including triangle $ABC$). In another word, what is $[ABC]+[A'B'C']+[A''B''C'']+[A'''B'''C''']+[A''''B''''+C''''] + \\cdots $? ($[z]$ denotes the area)\r\n\r\n6. Find the EXACT value of\r\n\r\n\\[\\sqrt{4 + 2 \\sqrt 3} + \\sqrt{4 - 2 \\sqrt 3}\\]\r\n\r\n7. The polynomial below has four roots. One of them is $4i$. Find the three other roots.\r\n\r\n\\[x^4-5x^3+22x^2-80x+96\\]\r\n\r\nNote on #7. This type of problem won't appear on Mathcounts history because $i$ indicates $\\sqrt{-1}$ which is a complex number. But here is a big hint on this problem. If a polynomial contains the $a+bi$ as one of root, it also contains $a-bi$ as another root.\r\n\r\n8. For non-real number $r,s$, we have:\r\n\r\n\\[r+s = 5\\\\\r\nrs = 10\\]\r\n\r\nFind $\\frac{1}{r^2} + \\frac{1}{s^2}$.\r\n\r\n9. How many number less than 500 satisfy:\r\n\r\n(i) has perfect square as one of its factors\r\n(ii) divisible by 7\r\n(iii) sum of its digit is 18\r\n\r\n10. In how many ways can you pick two numbers in a set $\\{1,2,3,4,5 \\}$ such that their sum is even?",
  "Solution_13": "All SOLUTIONS must be turned to ME by PM today (11:59:59).\r\n\r\nThank you and good luck!",
  "Solution_14": "I'm too late :(",
  "Solution_15": "[quote=\"nonie\"]I'm too late :([/quote]\r\n\r\nHere's a very good news for you.\r\n\r\nI decided to extend the deadline since I only received solution from only one contestant (can't believe only one in COP and only one in COPSD  :( ).\r\n\r\nSo, the new deadline is 10/3/05. Enjoy problem solving till then!\r\n\r\nRemember: I need the solutions to at least 7 of the problems."
}
{
  "Tag": [
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "li m x----8           sin{ x-10}/{10-x}\r\n\r\n{} denotes fractional part function\r\npl also explain little bit about fractional part function.i dont  know any thing about it\r\n\r\nQ2 lim X---0   Sec-(x/sin x)=l and limx---0 sec-(x/tan x)=m then which limit exist and which will not pl explain",
  "Solution_1": "The [url=http://mathworld.wolfram.com/FractionalPart.html]fractional part of a function[/url]."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "This Problem is marvelous ,i shall not say the source ,cause then you will not think on it ,but here it is [hide=\"Dont look\"]IMO 1991,Problem 6[/hide]\n\n[color=blue]Problem[/color]\nThe real number $a>1$ is given .\nA sequence like  \\[ \\{x_i \\}_{i=0}^{ \\infty}  \\]\nis called $LHTB(\\ little \\ hard \\ to \\ build )$ if for every two not-negative  integers like $i,j,that, i \\neq j$ \\[ |x_i-x_j| \\geq \\frac{1}{|i-j|^a}  \\] \n\n1.Prove that there exists a[b] bounded[/b] $LHTB$ sequence.\n\n2.[hide=\"The Original problem,\"]IMO1991,6[/hide] Built a[b] bounded[/b] $LHTB$ sequence.",
  "Solution_1": "$x_i=i$.",
  "Solution_2": "@lomos_lupin: Isn't this sequence supposed to be bounded in the original statement ?",
  "Solution_3": "If you want bounded sequence, then it can be build:\r\nLet $n=a_0+a_1*2+\\dots+a_k*2^k, a_i\\in \\{0,1\\}$.\r\n$x_n=2(a_0+a_1*2^{-1}+\\dots+a_k*2^{-k}).$",
  "Solution_4": ":blush:  Yes its bounded ,All of its fun was for being bounded.My bad\r\n\r\nA prove for existance of such a sequence:\r\n\r\n[color=blue]Solution[/color]Assume that we have built $x_0,...,x_{n-1}$ and we want to built the $x_n$ . Let $I$ be a bound for the sequence .for satisfying the condition of the problem ,we should have :\r\n\r\n \\[ |x_n-x_i| \\geq \\frac{1}{|n-i|^a} ,0 \\leq i \\leq n-1  \\]\r\n\r\nSo for existance of $x_n$ we should have :\r\n\r\n\\[ \\sum_{i=0}^{n-1} \\frac{1}{|n-i|^a} < I  \\]\r\n\r\nbut we know that  :\r\n\\[ \\sum_{i=1}^{n} \\frac{1}{i^a} < M= \\sum_{i=1}^{\\infty} \\frac{1}{i^a} < \\infty  \\]\r\n(Actually $M$ is a rational product of $\\pi^{a})$\r\nBy choosing $I=2M$ ,such a $x_n$ satisfying the condition of the problem exists.",
  "Solution_5": "To Rust : NO that sequance doesnt work ,Your sequance is not related to $a$,but it should be related.\r\nA book that i have present this sequence :\r\n\r\nLet the $x_n$ in the Rust sequence be $y_n$ then define your sequence as following \r\n\\[ x_n=\\frac{2^a-2}{2^a-1}y_n  \\]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "complex analysis"
  ],
  "Problem": "How would I use Residue Theorem and Laurent expansions to do the following problem?\r\n\r\n[img]http://i3.photobucket.com/albums/y57/lilshantypanty/prob1-1.jpg[/img]\r\n\r\nThanks!",
  "Solution_1": "For the first, how about first just expanding $ e^{1/z}$ in a Laurent expansion:\r\n\r\n$ e^{1/z}\\equal{}1\\plus{}1/z\\plus{}1/2(1/z)^2\\plus{}1/6(1/z)^3\\plus{}...$\r\n\r\nAlright, now expand $ Exp[1\\plus{}1/z\\plus{}1/2(1/z)^2\\plus{}1/6(1/z)^3\\plus{}...]$\r\n\r\nCan you show that I get for the residue:\r\n\r\n$ \\mathop\\text{Res}\\limits_{z\\equal{}0}\\left\\{e^{e^{1/z}}\\right\\}\\equal{}\r\n\\sum_{n\\equal{}1}^{\\infty}\\frac{n}{n!}$",
  "Solution_2": "[quote=\"sailorsenshi\"]How would I use Residue Theorem and Laurent expansions to do the following problem?\n\n[img]http://i3.photobucket.com/albums/y57/lilshantypanty/prob1-1.jpg[/img]\n\nThanks![/quote]\r\nI don't known. \r\nIs it right?\r\n${ e^{ix}=\\cos{x}+i\\sin{x}}$\r\n${ e^{x}=\\cos{\\frac{x}{i}}+i\\sin{\\frac{x}{i}}}$\r\nUse it , we  will ..."
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Hi,\r\nI need help in solving the equation.\r\n\r\n5x-7y=31\r\n-4x+2y=-14\r\n\r\nPlease solve by substitution and show steps.  Thanks",
  "Solution_1": "[quote=\"Trisha\"]Hi,\nI need help in solving the equation.\n\n5x-7y=31\n-4x+2y=-14\n\nPlease solve by substitution and show steps.  Thanks[/quote]\r\n\r\n[hide=\"hint\"]\n\nsolve for either variable in either equation.\n\nthen substitute that equation into the corresponding variable in the other equation\n\nsolve for the other variable\n\nplug the integer into an equation and solve for the original variable. \n[/hide]",
  "Solution_2": "[hide]\n1.$5x-7y=31$\n2.$-4x+2y=-14\\implies-2x+y=-7\\implies y=2x-7$\n\nSubstitute into 1\n\n$\\Rightarrow 5x-7(2x-7)=31\\implies x=2$\n\nsubstitute $x=2$ into 2\n\n$\\Rightarrow y=2(2)-7=-3$\n\n$\\boxed{x=2,y=-3}$\n[/hide]",
  "Solution_3": "Thanks SM4RT this is exactly what I wanted to see. :lol:",
  "Solution_4": "Another option is to convert into the coefficient matrix into reduced row echelon form...\r\n\r\nOr use cramer's rule, but both of these solutions are overkill on a two equation system.\r\n\r\nThe substitution method in general is the fastest way of solving small (2x2 or 3x3) systems.\r\n\r\nAnything larger, and matrix transforms are your friend."
}
{
  "Tag": [
    "real analysis",
    "topology",
    "real analysis unsolved"
  ],
  "Problem": "This question is from royden's real analysis from the chapter on lebesgue measure. Should be quite easy to solve. But i cant  :( \r\n\r\nLet $E$ be a set. Denote $m(E)$ to be the outer measure of $E$. Assume further that $m(E)$ is finite. Prove that the following are equivalent.\r\n\r\n(i) For all $\\epsilon>0$, there exists an open set $O\\supset E$ s.t. $m(O\\backslash E)<\\epsilon$.\r\n(ii) For all $\\epsilon>0$, there exists a finite union of open intervals $U$ s.t. $m(U\\Delta E)<\\epsilon$.\r\n\r\nNB: $U\\Delta E= (U\\backslash E) \\cup (E\\backslash U)$.",
  "Solution_1": "You missed a condition in (i). In royden's book, we also have $O\\supseteq E$.\r\n\r\n(i) => (ii)\r\n\r\nLet $\\epsilon>0$, by (i) there exists $O=\\bigcup I_{n}$ such that $m^\\ast(O\\setminus E)<\\epsilon/2$, where $I_{n}$ are open intervals.\r\nThen\r\n$\\sum|I_{n}| = m(O) \\leq m^\\ast(O\\setminus E)+m^\\ast(E) < \\frac \\epsilon2+m^\\ast E$ is finite.\r\n\r\nHence $\\sum |I_{n}|$ converges, so there exist $N$ such that $\\sum_{n=N}^\\infty |I_{n}| < \\frac\\epsilon 2$. Let $U = \\sum_{i=1}^{N}I_{n}$ then $m^\\ast(U\\Delta E) \\leq m^\\ast(U\\setminus E)+m^\\ast(E\\setminus U)<\\frac \\epsilon 2+\\frac\\epsilon 2 = \\epsilon$. (Note that $E\\setminus U$ is a subset of $\\sum_{i=N}^\\infty I_{n}$)\r\n\r\n(ii) => (i)\r\n\r\nLet $\\epsilon>0$, by (ii) we have a finite union of open intervals $U$ such that $m^\\ast(U\\Delta E)<\\frac\\epsilon 3$. By definite of outer measure, we have an open covering of $U\\Delta E$ thus an open set $G\\supseteq G\\Delta E$ such that $m(G)<m^\\ast(U\\Delta E)+\\frac\\epsilon 3 < \\frac{2}{3}\\epsilon$.\r\n\r\nLet $O=G\\cup U$, it is clear that $E\\subseteq O$, and\r\n\r\n$m^\\ast(O\\setminus E)\\leq m^\\ast(U\\setminus E)+m^\\ast(G\\setminus E)<\\frac\\epsilon3+\\frac23\\epsilon = \\epsilon$.",
  "Solution_2": "[quote=\"liyi\"]You missed a condition in (i). In royden's book, we also have $O\\supseteq E$.\n[/quote]\n\nooops.  :blush: i've just edited the post.\n\n[quote=\"liyi\"]\nLet $\\epsilon>0$, by (i) there exists $O=\\bigcup I_{n}$ such that $m^\\ast(O\\setminus E)<\\epsilon/2$, where $I_{n}$ are open intervals.\n[/quote]\r\n\r\nthink i'm forgetting some facts. But is it always true that any open set $O$ can be written as a countable disjoint union of open intervals? (i think you made use of the fact that it's disjoint when you assumed $\\sum|I_{n}|=m^{*}(O)$.)",
  "Solution_3": "Consider an open subset of the reals. Its connected components must be open, as given any element of a particular connected component, there exists a neighborhood of that element contained in the set, and the union of this neighborhood with the connected component is readily seen to still be connected. Thus, as open, connected subsets of the reals are simply open intervals, any open subset is a union of open intervals; they are disjoint as connected components are disjoint and there are a countable number as we can choose a rational in each of the open intervals.",
  "Solution_4": "[quote]think i'm forgetting some facts. But is it always true that any open set $O$ can be written as a countable disjoint union of open intervals? (i think you made use of the fact that it's disjoint when you assumed $\\sum|I_{n}|=m^{*}(O)$.)[/quote]\r\nYes, i think you can find the standard proof in Royden's book. But it is for sets on the real line only.",
  "Solution_5": "Ok. QED. Thanks for the help!"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "a,b,c>0\r\n ab+bc+ac = k\r\n\r\n ????\r\nprove a+b+c>=k",
  "Solution_1": "a=1, b=1, c=3\r\nAnd I have proved it wrong.\r\n\r\nFor your information, \r\n$(a+b+c)^2\\ge 3(ab+bc+ca)$\r\nThis can be proved using AM-GM",
  "Solution_2": "k=3\r\n :ninja:",
  "Solution_3": "See my above post, I've solved it.\r\n\r\n[hide=\"Solution\"]$(a+b+c)^2\\ge 3(ab+bc+ca)$\n$\\Rightarrow (a+b+c)^2 \\ge 9$\nThus proved.[/hide]\n\nYou may like to prove the inequality yourself. \n[hide=\"Big Hint\"]\nUse AM-GM[/hide]",
  "Solution_4": "[hide=\"Is this it?\"]Expaninding gives us \\[ a^2+b^2+c^2+2ab+2ac+2bc \\geq 3ab+3ac+3bc \\] So we need to proof that \\[ a^2+b^2+c^2 \\geq ab+ac+bc \\] Apply AM-GM on $\\{a^2,b^2\\}$, $\\{a^2,c^2\\}$ and $\\{b^2,c^2\\}$, getting $\\frac{a^2+b^2}{2} \\geq ab$, $\\frac{a^2+c^2}{2} \\geq ac$ and $\\frac{b^2+c^2}{2} \\geq bc$. Adding those three: \\[ \\frac{a^2+b^2+a^2+c^2+b^2+c^2}{2} = a^2+b^2+c^2 \\geq ab+ac+bc\\ \\ \\square \\] [/hide]",
  "Solution_5": "Yes, it is \"this\" :)",
  "Solution_6": "A simpler way to prove $a^2 + b^2 + c^2 \\geq ab + bc + ac$ is to re-arrange and factor it as\r\n\r\n$1/2 (a - b)^2 + 1/2 (b-c)^2 + 1/2(c -a)^2 \\geq 0$, which is always true."
}
{
  "Tag": [],
  "Problem": "i)x,y mh midenikoi pragmatikoi ari8moi kai  (1-x^2)(1-y^2)=1 Na upologistei to A=(1+|x|)*(1+|y|)!\r\n\r\nii) x^2+y^2+z^2=x+y+z+1 na apodeixth oti dn exei lush stous rhtous\r\n\r\niii)P(x^2)=x^2(x^2+1)P(x) , P(2)=12 na bre8ei to poluwnhmo (P(x) me pragmatikous sidelestes..)",
  "Solution_1": "\u0393\u03b9\u03b1 \u03c4\u03bf \u03c4\u03c1\u03af\u03c4\u03bf \u03b5\u03af\u03bd\u03b1\u03b9 $ p(x)\\equal{}x^4\\minus{}x^2.$ \u03b1\u03bd \u03b4\u03b5\u03bd \u03ad\u03c7\u03c9 \u03ba\u03ac\u03bd\u03b5\u03b9 \u03ba\u03ac\u03c0\u03bf\u03c5 \u03bb\u03ac\u03b8\u03bf\u03c2 \u03c3\u03c4\u03b9\u03c2 \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2....",
  "Solution_2": "swsta tis ekanes..! to prwto einai to pio wraio try it",
  "Solution_3": "\u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bc\u03b1\u03c4\u03b9\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03b7 \u03ba\u03b1\u03b8\u03bf\u03b4\u03bf.....\u0395\u03c7\u03c9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2...\u0391\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c7 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3 y \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03bf \u0391.... :(  :maybe:\u03a4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03b9????\u0391\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5 \u03c4\u03bf \u03bc\u03bf\u03c5 \u03bb\u03b9\u03b3\u03bf \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9....... :)",
  "Solution_4": "\u03a4\u03bf 3\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c0\u03c1\u03c9\u03c4\u03bf \u03b8\u03b5\u03bc\u03b1 \u03b1\u03c0\u03bf \u03c0\u03b1\u03bb\u03b9\u03bf \u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7, \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf\u03c5\u03c4\u03c3\u03b9\u03ba\u03bf\r\n\r\n\u03a3\u03c4\u03bf 1\u03bf \u03bf\u03b9 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03b9 $ |x|^2, |y|^2$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c1\u03b9\u03b6\u03b5\u03c2 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b1\u03c2 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7\u03c2 $ x^2 \\minus{} ax \\plus{} a$ \u03bc\u03b5 $ a \\geq 4$, \u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 \u03c4\u03b5\u03c4\u03bf\u03b9\u03bf $ a$ \u03b7 \u03c0\u03b1\u03c1\u03b1\u03c3\u03c4\u03b1\u03c3\u03b7 \u03b9\u03c3\u03bf\u03c5\u03c4\u03b5 \u03bc\u03b5 $ f(a) \\equal{} \\sqrt {a \\plus{} 2\\sqrt {a}} \\plus{} \\sqrt {a} \\plus{} 1$ (\u03b1\u03bd \u03b4\u03b5\u03bd \u03bc\u03bf\u03c5 \u03be\u03b5\u03c6\u03c5\u03b3\u03b5 \u03ba\u03b1\u03c4\u03b9). \u0397 f \u03c4\u03c9\u03c1\u03b1 \u03c9\u03c2 \u03c3\u03c5\u03bd\u03b5\u03c7\u03b7\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03bd\u03b7\u03c3\u03b9\u03c9\u03c2 \u03b1\u03c5\u03be\u03bf\u03c5\u03c3\u03b1 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7 \u03c4\u03bf\u03c5 $ a$, \u03c3\u03b1\u03c1\u03c9\u03bd\u03b5\u03b9 \u03bf\u03bb\u03bf \u03c4\u03bf \u03b4\u03b9\u03b1\u03c3\u03c4\u03b7\u03bc\u03b1 $ [2\\sqrt {2} \\plus{} 3, \\plus{} \\infty)$, \u03bf\u03c4\u03b1\u03bd \u03bf $ a$ \u03c3\u03b1\u03c1\u03c9\u03bd\u03b5\u03b9 \u03c4\u03bf \u03b4\u03b9\u03b1\u03c3\u03c4\u03b7\u03bc\u03b1 $ [4, \\plus{} \\infty)$. \u0394\u03b5\u03bd \u03ba\u03b1\u03c4\u03b1\u03bb\u03b1\u03b2\u03b1\u03b9\u03bd\u03c9 \u03c4\u03b9 \u03b1\u03bb\u03bb\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03b6\u03b7\u03c4\u03b1\u03b5\u03b9 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1...\r\n\r\nEdit: \u039f\u03c0\u03bf\u03c4\u03b5 \u03b7 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03b7 \u03c4\u03b9\u03bc\u03b7 \u03b5\u03b9\u03bd\u03b1\u03b9 $ 2\\sqrt {2} \\plus{} 3$\r\n\r\n\u03a4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b5 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03bf \u03bc\u03b5 \u03c4\u03bf \u03bd\u03b1 \u03b4\u03b5\u03b9\u03c7\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 \u03c0\u03b1\u03bd\u03c9 \u03c3\u03c4\u03b7 \u03c3\u03c6\u03b1\u03b9\u03c1\u03b1 \u03bc\u03b5 \u03ba\u03b5\u03bd\u03c4\u03c1\u03bf \u03c4\u03b7\u03bd \u03b1\u03c1\u03c7\u03b7 \u03c4\u03c9\u03bd \u03b1\u03be\u03bf\u03bd\u03c9\u03bd \u03ba\u03b1\u03b9 \u03b1\u03ba\u03c4\u03b9\u03bd\u03b1 $ \\sqrt {7}$ \u03c3\u03c4\u03bf\u03bd $ R^3$, \u03b4\u03b5\u03bd \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c3\u03b7\u03bc\u03b5\u03b9\u03bf \u03bc\u03b5 \u03c1\u03b7\u03c4\u03b5\u03c2 \u03c3\u03c5\u03bd\u03c4\u03b5\u03c4\u03b1\u03b3\u03bc\u03b5\u03bd\u03b5\u03c2, \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03c4\u03bf \u03bf\u03c0\u03bf\u03b9\u03bf \u03b5\u03c7\u03c9 \u03c4\u03b7\u03bd \u03b5\u03bd\u03c4\u03c5\u03c0\u03c9\u03c3\u03b7 \u03c0\u03c9\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03b1 \u03b3\u03bd\u03c9\u03c3\u03c4\u03bf \u03b1\u03bb\u03bb\u03b1 \u03c0\u03bf\u03c4\u03b5 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b5\u03c7\u03c9 \u03bc\u03b5\u03bb\u03b5\u03c4\u03b7\u03c3\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c9\u03c1\u03b1 \u03b4\u03b5\u03bd \u03bc\u03c0\u03bf\u03c1\u03c9 \u03bd\u03b1 \u03c6\u03b1\u03bd\u03c4\u03b1\u03c3\u03c4\u03c9 \u03c0\u03c9\u03c2 \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03b1\u03bd\u03c4\u03b9\u03bc\u03b5\u03c4\u03bf\u03c0\u03b9\u03c3\u03c4\u03b5\u03b9, \u03bc\u03b5 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b1\u03c3\u03b5\u03b9\u03c2 \u03c0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03c4\u03c9\u03bd \u03c1\u03b7\u03c4\u03c9\u03bd \u03bc\u03b5 \u03c0\u03b7\u03bb\u03b9\u03ba\u03b1 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03c9\u03bd \u03b4\u03b5\u03bd \u03bd\u03bf\u03bc\u03b9\u03b6\u03c9 \u03bd\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03b4\u03b9\u03bf\u03c4\u03b9 \u03b3\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9 \u03c7\u03b1\u03bc\u03bf\u03c2 \u03b1\u03c0\u03bf \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03b5\u03c2 \u03b5\u03bd\u03c9 \u03bf\u03b9 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b5\u03c2 \u03b4\u03b5\u03bd \u03bf\u03b4\u03b7\u03b3\u03bf\u03c5\u03bd\u03b5 \u03c0\u03bf\u03c5\u03b8\u03b5\u03bd\u03b1, \u03b9\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03b8\u03b5\u03bb\u03b5\u03b9 infinite descent, \u03c4\u03bf \u03bc\u03bf\u03bd\u03bf \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf \u03c0\u03b1\u03bd\u03c4\u03bf\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c4\u03b5 \u03c0\u03b5\u03c1\u03b9 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b7\u03c2 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2....",
  "Solution_5": "\u039a\u03b1\u03bb\u03b1 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03bf \u03b4\u03b5\u03bd \u03c4\u03bf \u03bb\u03b5\u03c2 \u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03b4\u03b9\u03b1\u03b9\u03c4\u03b5\u03c1\u03b1 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03bf \u03c4\u03bf 2\u03bf \u03b1\u03bd \u03ba\u03b1\u03b9 \u03b8\u03b5\u03bb\u03b5\u03b9 \u03c0\u03c1\u03bf\u03c3\u03bf\u03c7\u03b7.\r\n\u039f\u03c5\u03c3\u03b9\u03b1\u03c3\u03c4\u03b9\u03ba\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03bf \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 $ (2x \\minus{} 1)^2 \\plus{} (2y \\minus{} 1)^2 \\plus{} (2z \\minus{} 1)^2 \\equal{} 7$ $ x,y,z$ \u03c3\u03c4\u03bf Q.\r\n\u0395\u03c4\u03c3\u03b9 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c3\u03b1\u03bd \u03bd\u03b1 \u03b8\u03b5\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03b4\u03b5\u03b9\u03be\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 7s^2$ \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 $ a,b,c,s$ (\u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9  \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03bf \u03b1\u03bb\u03bb\u03b1 \u03b5\u03c0\u03b1\u03c1\u03ba\u03b5\u03c2 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03b1)\r\n\u0395\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b2\u03bb\u03b5\u03c0\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c4\u03b5\u03c4\u03c1\u03b1\u03b4\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03bf\u03c5\u03c2($ \\mod 4$) \u03ba\u03b1\u03b9 \u03b1\u03c1\u03b1 \u03bc\u03b5\u03c4\u03b1 $ \\mod 8$ \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c4\u03bf\u03c0\u03bf.\r\n\u0393\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c3\u03c6\u03b1\u03b9\u03c1\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 nick \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03bd\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03b7\u03c3\u03b5\u03b9 \u03bf\u03c4\u03b9 \u03b7 \u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 2 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b5\u03c5\u03b5\u03c4\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03bf\u03c0\u03bf\u03b9\u03bf\u03b4\u03b7\u03c0\u03bf\u03c4\u03b5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf $ 8k \\plus{} 1$ \u03c3\u03c4\u03bf RHS.\r\n\r\nbtw leftmast \u03ba\u03b1\u03bb\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b3\u03b5\u03bd\u03b9\u03ba\u03b1 \u03bf\u03c4\u03b1\u03bd \u03b5\u03c7\u03b5\u03b9\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9 \u03ba\u03b1\u03c4\u03b9 \u03bd\u03b1 \u03b2\u03b1\u03b6\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03c0\u03bf\u03c5 \u03b5\u03c3\u03c4\u03c9, \u03bc\u03b9\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c0\u03bf\u03c5 \u03bf\u03bb\u03bf\u03ba\u03bb\u03b7\u03c1\u03c9\u03bc\u03b5\u03bd\u03b7 \u03bb\u03c5\u03c3\u03b7. \r\n\u0397 \u03b9\u03b4\u03b5\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c3\u03b7\u03bc\u03b1\u03bd\u03c4\u03b9\u03ba\u03b7 \u03b1\u03bb\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03b7 \u03b4\u03b9\u03b1\u03c4\u03c5\u03c0\u03c9\u03c3\u03b7 \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bf\u03b9 \u03bb\u03b5\u03c0\u03c4\u03bf\u03bc\u03b5\u03c1\u03b5\u03b9\u03b5\u03c2 \u03b5\u03c7\u03bf\u03c5\u03bd \u03c3\u03b7\u03bc\u03b1\u03c3\u03b9\u03b1  \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03b5\u03bd\u03bf\u03c2 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03bf\u03c2.",
  "Solution_6": "[quote=\"Athinaios\"]\n\u0395\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b2\u03bb\u03b5\u03c0\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03bc\u03c0\u03bf\u03c1\u03bf\u03c5\u03bc\u03b5 \u03bd\u03b1 \u03ba\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b7\u03bd \u03c4\u03b5\u03c4\u03c1\u03b1\u03b4\u03b1 \u03c0\u03b5\u03c1\u03b9\u03c4\u03c4\u03bf\u03c5\u03c2($ \\mod 4$) \u03ba\u03b1\u03b9 \u03b1\u03c1\u03b1 \u03bc\u03b5\u03c4\u03b1 $ \\mod 8$ \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b1\u03c4\u03bf\u03c0\u03bf.\n[/quote]\r\n\r\n\u03a0\u03bb\u03b1\u03ba\u03b1 \u03ba\u03b1\u03bd\u03b5\u03b9\u03c2 \u03c4\u03c9\u03c1\u03b1.... \u03ba\u03b1\u03b9 \u03c0\u03b1\u03bb\u03b5\u03c5\u03b1 5 \u03c9\u03c1\u03b5\u03c2 \u03c4\u03c9\u03c1\u03b1 \u03bc\u03b5 6 \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b7\u03c4\u03b1\u03bd \u03c4\u03bf\u03c3\u03bf \u03b1\u03c0\u03bb\u03bf... \u03b5\u03c7\u03c9 \u03b1\u03c1\u03c7\u03b9\u03c3\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03be\u03b5\u03c7\u03bd\u03b1\u03c9 \u03b2\u03b1\u03c3\u03b9\u03ba\u03b1 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1 \u03ba\u03b1\u03c4\u03b9 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03b9\u03bd\u03b5\u03c4\u03b1\u03b9....",
  "Solution_7": "Z\u03b7\u03c4\u03c9 \u03c3\u03c5\u03b3\u03b3\u03bd\u03c9\u03bc\u03b7 .....! :blush: \u03a4\u03bf \u03b5\u03ba\u03b1\u03bd\u03b1 \u03b1\u03c3\u03c5\u03bd\u03b1\u03b9\u03c3\u03b8\u03b7\u03c4\u03b1!!!\u03a0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03b7 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7, \u03b1\u03c6\u03bf\u03c5 \u03ba\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5 \u03c4\u03b9\u03c3 \u03b1\u03bd\u03c4\u03b9\u03ba\u03b1\u03c4\u03b1\u03c3\u03c4\u03b1\u03c3\u03b5\u03b9\u03c2 \u03b1\u03bd \u03c4\u03bf \u03c0\u03b1\u03bc\u03b5 mod3 \u03ba\u03b1\u03b9 \u03bc\u03b5\u03c4\u03b1 mod 9 \u03b8\u03b1 \u03b2\u03b3\u03b5\u03b9,\u03b1\u03bd \u03ba\u03b1\u03b9 \u03bc\u03bf\u03c5 \u03b5\u03c0\u03b9\u03b1\u03c3\u03b5 2 \u03c3\u03b5\u03bb\u03b9\u03b4\u03b5\u03c2 \u03c0\u03b5\u03c1\u03b9\u03c0\u03bf\u03c5...\u039c\u03bf\u03bb\u03b9\u03c2 \u03b2\u03c1\u03c9 \u03c7\u03c1\u03bf\u03bd\u03bf \u03b8\u03b1 \u03b2\u03b1\u03bb\u03c9 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03bc\u03bf\u03c5!!!!\u03a0\u03bf\u03bb\u03c5 \u03ba\u03b1\u03bb\u03b7 \u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c3\u03bf\u03c5 Athinaios \u03c0\u03b1\u03bd\u03c4\u03c9\u03c2!!! :)",
  "Solution_8": "[quote=\"lefmast\"]\u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bc\u03b1\u03c4\u03b9\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03b7 \u03ba\u03b1\u03b8\u03bf\u03b4\u03bf.....\u0395\u03c7\u03c9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2...\u0391\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c7 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3 y \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03bf \u0391.... :(  :maybe:\u03a4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03b9????\u0391\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5 \u03c4\u03bf \u03bc\u03bf\u03c5 \u03bb\u03b9\u03b3\u03bf \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9....... :)[/quote]\r\n\r\nlipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P",
  "Solution_9": "[quote=\"KapioPulsar\"][quote=\"lefmast\"]\u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bc\u03b1\u03c4\u03b9\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03b7 \u03ba\u03b1\u03b8\u03bf\u03b4\u03bf.....\u0395\u03c7\u03c9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2...\u0391\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c7 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3 y \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03bf \u0391.... :(  :maybe:\u03a4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03b9????\u0391\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5 \u03c4\u03bf \u03bc\u03bf\u03c5 \u03bb\u03b9\u03b3\u03bf \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9....... :)[/quote]\n\nlipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]\r\n\r\n\u03a4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03c2 \u03c4\u03c9\u03c1\u03b1... \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bd\u03c4\u03c9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b1, \u03b5\u03bd\u03c9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b1\u03bd \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf \u03bc\u03b9\u03ba\u03c1\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf\u03c5 (\u03bf\u03c0\u03c9\u03c2 \u03b5\u03b9\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9, \u03b5\u03b9\u03c7\u03b5 \u03bc\u03c0\u03b5\u03b9 \u03c3\u03b1\u03bd \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c3\u03b5 \u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 \u03c0\u03b1\u03bb\u03b9\u03b1). \u03a4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bf\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03c3\u03c7\u03b5\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b7 \u03c5\u03bb\u03b7, \u03b5\u03b9\u03b4\u03b9\u03ba\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7 \u03c4\u03c9\u03bd \u03c0\u03b1\u03bd\u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03c9\u03bd, \u03b1\u03c6\u03bf\u03c5 \u03bc\u03b5\u03c4\u03b1 \u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03bf \u03b1\u03bd\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03bf\u03bb\u03c5\u03c0\u03bb\u03bf\u03ba\u03b7 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b1 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b1 6 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03c9\u03bd \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03c9\u03bd, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b3\u03b9\u03bd\u03b5\u03b9 \u03c7\u03c1\u03b5\u03b9\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b5\u03c7\u03bd\u03b1\u03c3\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b7 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd, \u03b5\u03b3\u03c9 \u03bf\u03c4\u03b1\u03bd \u03b5\u03b9\u03bc\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf (\u03c3\u03c4\u03bf \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf \u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd) \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03b7\u03ba\u03b5 \u03c0\u03bf\u03c4\u03b5 \u03bf\u03c5\u03c4\u03b5 \u03ba\u03b1\u03bd \u03bf \u03bf\u03c1\u03bf\u03c2 \"\u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1\" \u03c3\u03c4\u03b7\u03bd \u03c4\u03b1\u03be\u03b7...",
  "Solution_10": "[quote=\"NickNafplio\"][quote=\"KapioPulsar\"][quote=\"lefmast\"]\u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bc\u03b1\u03c4\u03b9\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03b7 \u03ba\u03b1\u03b8\u03bf\u03b4\u03bf.....\u0395\u03c7\u03c9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2...\u0391\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c7 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3 y \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03bf \u0391.... :(  :maybe:\u03a4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03b9????\u0391\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5 \u03c4\u03bf \u03bc\u03bf\u03c5 \u03bb\u03b9\u03b3\u03bf \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9....... :)[/quote]\n\nlipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]\n\n\u03a4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03c2 \u03c4\u03c9\u03c1\u03b1... \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bd\u03c4\u03c9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b1, \u03b5\u03bd\u03c9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b1\u03bd \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf \u03bc\u03b9\u03ba\u03c1\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf\u03c5 (\u03bf\u03c0\u03c9\u03c2 \u03b5\u03b9\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9, \u03b5\u03b9\u03c7\u03b5 \u03bc\u03c0\u03b5\u03b9 \u03c3\u03b1\u03bd \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c3\u03b5 \u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 \u03c0\u03b1\u03bb\u03b9\u03b1). \u03a4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bf\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03c3\u03c7\u03b5\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b7 \u03c5\u03bb\u03b7, \u03b5\u03b9\u03b4\u03b9\u03ba\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7 \u03c4\u03c9\u03bd \u03c0\u03b1\u03bd\u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03c9\u03bd, \u03b1\u03c6\u03bf\u03c5 \u03bc\u03b5\u03c4\u03b1 \u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03bf \u03b1\u03bd\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03bf\u03bb\u03c5\u03c0\u03bb\u03bf\u03ba\u03b7 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b1 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b1 6 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03c9\u03bd \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03c9\u03bd, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b3\u03b9\u03bd\u03b5\u03b9 \u03c7\u03c1\u03b5\u03b9\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b5\u03c7\u03bd\u03b1\u03c3\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b7 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd, \u03b5\u03b3\u03c9 \u03bf\u03c4\u03b1\u03bd \u03b5\u03b9\u03bc\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf (\u03c3\u03c4\u03bf \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf \u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd) \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03b7\u03ba\u03b5 \u03c0\u03bf\u03c4\u03b5 \u03bf\u03c5\u03c4\u03b5 \u03ba\u03b1\u03bd \u03bf \u03bf\u03c1\u03bf\u03c2 \"\u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1\" \u03c3\u03c4\u03b7\u03bd \u03c4\u03b1\u03be\u03b7...[/quote] emeis ton anaferame persi sthn 2 lykeiou kai malista kaname kai eisagwgi sto mod alla mexri kai tis diofadikes kaname akoma kai duskoles",
  "Solution_11": "gia thn prwth askhsh exw sto mualo mou mia polu wraia trigwnometrikh lush!  :D",
  "Solution_12": "[quote=\"KapioPulsar\"][/quote][quote=\"NickNafplio\"][quote=\"KapioPulsar\"][quote=\"lefmast\"]\u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bc\u03b1\u03c4\u03b9\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03b7 \u03ba\u03b1\u03b8\u03bf\u03b4\u03bf.....\u0395\u03c7\u03c9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2...\u0391\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c7 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3 y \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03bf \u0391.... :(  :maybe:\u03a4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03b9????\u0391\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5 \u03c4\u03bf \u03bc\u03bf\u03c5 \u03bb\u03b9\u03b3\u03bf \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9....... :)[/quote]\n\nlipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]\n\n\u03a4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03c2 \u03c4\u03c9\u03c1\u03b1... \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bd\u03c4\u03c9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b1, \u03b5\u03bd\u03c9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b1\u03bd \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf \u03bc\u03b9\u03ba\u03c1\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf\u03c5 (\u03bf\u03c0\u03c9\u03c2 \u03b5\u03b9\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9, \u03b5\u03b9\u03c7\u03b5 \u03bc\u03c0\u03b5\u03b9 \u03c3\u03b1\u03bd \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c3\u03b5 \u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 \u03c0\u03b1\u03bb\u03b9\u03b1). \u03a4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bf\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03c3\u03c7\u03b5\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b7 \u03c5\u03bb\u03b7, \u03b5\u03b9\u03b4\u03b9\u03ba\u03b1 \u03bc\u03b5 \u03b1\u03c5\u03c4\u03b7 \u03c4\u03c9\u03bd \u03c0\u03b1\u03bd\u03b5\u03bb\u03bb\u03b7\u03bd\u03b9\u03c9\u03bd, \u03b1\u03c6\u03bf\u03c5 \u03bc\u03b5\u03c4\u03b1 \u03c4\u03bf \u03bc\u03b5\u03c4\u03b1\u03c3\u03c7\u03b7\u03bc\u03b1\u03c4\u03b9\u03c3\u03bc\u03bf \u03b1\u03bd\u03b1\u03b3\u03b5\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03bf\u03bb\u03c5\u03c0\u03bb\u03bf\u03ba\u03b7 \u03b4\u03b9\u03bf\u03c6\u03b1\u03bd\u03c4\u03b9\u03ba\u03b7 \u03b5\u03be\u03b9\u03c3\u03c9\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b1 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b1 6 \u03b1\u03ba\u03b5\u03c1\u03b1\u03b9\u03c9\u03bd \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03c9\u03bd, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03bd\u03b1 \u03b3\u03b9\u03bd\u03b5\u03b9 \u03c7\u03c1\u03b5\u03b9\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03b3\u03bd\u03c9\u03c3\u03b5\u03b9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b5\u03c7\u03bd\u03b1\u03c3\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b7 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd, \u03b5\u03b3\u03c9 \u03bf\u03c4\u03b1\u03bd \u03b5\u03b9\u03bc\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf (\u03c3\u03c4\u03bf \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf \u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd) \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03b7\u03ba\u03b5 \u03c0\u03bf\u03c4\u03b5 \u03bf\u03c5\u03c4\u03b5 \u03ba\u03b1\u03bd \u03bf \u03bf\u03c1\u03bf\u03c2 \"\u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1\" \u03c3\u03c4\u03b7\u03bd \u03c4\u03b1\u03be\u03b7...[/quote][quote=\"KapioPulsar\"] emeis ton anaferame persi sthn 2 lykeiou kai malista kaname kai eisagwgi sto mod alla mexri kai tis diofadikes kaname akoma kai duskoles[/quote]\r\n\r\n\u0397 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03c0\u03bf\u03c5 \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c3\u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03c4\u03b7\u03c2 \u0392 \u039b\u03c5\u03ba\u03b5\u03b9\u03bf\u03c5 \u03b9\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03ba\u03b1\u03bb\u03c5\u03c0\u03c4\u03b5\u03b9 2-3 \u03c3\u03b5\u03bb\u03b9\u03b4\u03b5\u03c2 \u03b1\u03c0\u03bf \u03b5\u03bd\u03b1 \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03b3\u03b9\u03b1 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03c5\u03c2, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b1\u03bd \u03ba\u03b1\u03bd\u03b1\u03c4\u03b5 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03b8\u03b5\u03c9\u03c1\u03b9\u03b1\u03c2 \u03bf\u03c0\u03c9\u03c2 \u03bb\u03b5\u03c2 \u03c4\u03bf\u03c4\u03b5 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b2\u03b3\u03b7\u03ba\u03b1\u03c4\u03b5 \u03c0\u03bf\u03bb\u03c5 \u03b5\u03be\u03c9 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03c5\u03bb\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c7\u03bf\u03bb\u03b5\u03b9\u03bf\u03c5, \u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03c4\u03b1 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03b1\u03be\u03b9\u03b6\u03b5\u03b9 \u03c3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03b7\u03c1\u03b9\u03b1 \u03bf \u03ba\u03b1\u03b8\u03b7\u03b3\u03b7\u03c4\u03b7\u03c2 \u03c3\u03b1\u03c2 \u03b4\u03b9\u03bf\u03c4\u03b9 \u03c3\u03c0\u03b1\u03bd\u03b9\u03b1 \u03ba\u03b1\u03b8\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03ba\u03b1\u03b8\u03b7\u03b3\u03b7\u03c4\u03b5\u03c2 \u03bd\u03b1 \u03b4\u03b9\u03b4\u03b1\u03be\u03bf\u03c5\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c7\u03c1\u03b5\u03b9\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03b5\u03be\u03b5\u03c4\u03b1\u03c3\u03b5\u03b9\u03c2... \u03c0\u03b1\u03bd\u03c4\u03bf\u03c2 \u03b8\u03b1 \u03b7\u03b8\u03b5\u03bb\u03b1 \u03c0\u03bf\u03bb\u03c5 \u03bd\u03b1 \u03b4\u03c9 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b9\u03ba\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b9\u03c0\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03bf\u03c4\u03b9 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03b2\u03c1\u03b5\u03b9.",
  "Solution_13": "[quote=\"NickNafplio\"][/quote][quote=\"KapioPulsar\"][/quote][quote=\"NickNafplio\"][quote=\"KapioPulsar\"][quote=\"lefmast\"]\u03c4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bc\u03b5 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03bf\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bc\u03b1\u03c4\u03b9\u03b1 \u03b2\u03b3\u03b1\u03b9\u03bd\u03b5\u03b9 \u03bc\u03b5 \u03b1\u03c0\u03b5\u03b9\u03c1\u03b7 \u03ba\u03b1\u03b8\u03bf\u03b4\u03bf.....\u0395\u03c7\u03c9 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03b9\u03b5\u03c2 \u03b3\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03bf\u03bc\u03c9\u03c2...\u0391\u03bd \u03b8\u03b5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03ba\u03b1\u03c0\u03bf\u03b9\u03b5\u03c2 \u03c4\u03b9\u03bc\u03b5\u03c2 \u03c3\u03c4\u03bf \u03c7 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03c3\u03bf\u03c5\u03bc\u03b5 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c3 y \u03b4\u03b5\u03bd \u03b8\u03b1 \u03b5\u03c7\u03b5\u03b9 \u03c4\u03b7\u03bd \u03b9\u03b4\u03b9\u03b1 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03bf \u0391.... :(  :maybe:\u03a4\u03b9 \u03b5\u03bd\u03bd\u03bf\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03b9????\u0391\u03bd \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9\u03c2 \u03b5\u03be\u03b7\u03b3\u03b7\u03c3\u03b5 \u03c4\u03bf \u03bc\u03bf\u03c5 \u03bb\u03b9\u03b3\u03bf \u03c3\u03b5 \u03c0\u03b1\u03c1\u03b1\u03ba\u03b1\u03bb\u03c9....... :)[/quote]\n\nlipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]\n\n\u03a4\u03b1 \u03c0\u03b1\u03c1\u03b1\u03bb\u03b5\u03c2 \u03c4\u03c9\u03c1\u03b1... \u03c4\u03bf \u03c0\u03c1\u03c9\u03c4\u03bf \u03ba\u03b1\u03b9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf\u03bd\u03c4\u03c9\u03c2 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b1, \u03b5\u03bd\u03c9 \u03c4\u03bf \u03c4\u03c1\u03b9\u03c4\u03bf \u03bc\u03c0\u03bf\u03c1\u03b5\u03b9 \u03bd\u03b1 \u03bc\u03c0\u03b5\u03b9 \u03ba\u03b1\u03b9 \u03c3\u03b5 \u03b5\u03bd\u03b1\u03bd \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf \u03bc\u03b9\u03ba\u03c1\u03bf\u03c5 \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf\u03c5 (\u03bf\u03c0\u03c9\u03c2 \u03b5\u03b9\u03c0\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03b1\u03c1\u03b1\u03c0\u03b1\u03bd\u03c9, \u03b5\u03b9\u03c7\u03b5 \u03bc\u03c0\u03b5\u03b9 \u03c3\u03b1\u03bd \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c3\u03b5 \u0391\u03c1\u03c7\u03b9\u03bc\u03b7\u03b4\u03b7 \u03c0\u03b1\u03bb\u03b9\u03b1). \u03a4\u03bf \u03b4\u03b5\u03c5\u03c4\u03b5\u03c1\u03bf \u03bf\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03b5\u03c7\u03b5\u03b9 \u03ba\u03b1\u03bc\u03b9\u03b1 \u03c3\u03c7\u03b5\u03c3\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b7 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\u03ba\u03b1\u03b9 \u03c4\u03b5\u03c7\u03bd\u03b1\u03c3\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b7 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03c4\u03c9\u03bd \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd, \u03b5\u03b3\u03c9 \u03bf\u03c4\u03b1\u03bd \u03b5\u03b9\u03bc\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf (\u03c3\u03c4\u03bf \u03b4\u03b9\u03ba\u03bf \u03bc\u03bf\u03c5 \u03bb\u03c5\u03ba\u03b5\u03b9\u03bf \u03c4\u03bf\u03c5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf\u03bd) \u03b4\u03b5\u03bd \u03b8\u03c5\u03bc\u03b1\u03bc\u03b5 \u03bd\u03b1 \u03b1\u03bd\u03b1\u03c6\u03b5\u03c1\u03b8\u03b7\u03ba\u03b5 \u03c0\u03bf\u03c4\u03b5 \u03bf\u03c5\u03c4\u03b5 \u03ba\u03b1\u03bd \u03bf \u03bf\u03c1\u03bf\u03c2 \"\u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03c4\u03bf\u03c4\u03b7\u03c4\u03b1\" \u03c3\u03c4\u03b7\u03bd \u03c4\u03b1\u03be\u03b7...[/quote][quote=\"KapioPulsar\"] emeis ton anaferame persi sthn 2 lykeiou kai malista kaname kai eisagwgi sto mod alla mexri kai tis diofadikes kaname akoma kai duskoles[/quote][quote=\"NickNafplio\"]\n\n\u0397 \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03c0\u03bf\u03c5 \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 \u03c3\u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03c4\u03b7\u03c2 \u0392 \u039b\u03c5\u03ba\u03b5\u03b9\u03bf\u03c5 \u03b9\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03ba\u03b1\u03bb\u03c5\u03c0\u03c4\u03b5\u03b9 2-3 \u03c3\u03b5\u03bb\u03b9\u03b4\u03b5\u03c2 \u03b1\u03c0\u03bf \u03b5\u03bd\u03b1 \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03b8\u03b5\u03c9\u03c1\u03b9\u03b1\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03c9\u03bd \u03b3\u03b9\u03b1 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03c5\u03c2, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b1\u03bd \u03ba\u03b1\u03bd\u03b1\u03c4\u03b5 \u03b4\u03c5\u03c3\u03ba\u03bf\u03bb\u03b1 \u03c0\u03c1\u03bf\u03b2\u03bb\u03b7\u03bc\u03b1\u03c4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03b8\u03b5\u03c9\u03c1\u03b9\u03b1\u03c2 \u03bf\u03c0\u03c9\u03c2 \u03bb\u03b5\u03c2 \u03c4\u03bf\u03c4\u03b5 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b2\u03b3\u03b7\u03ba\u03b1\u03c4\u03b5 \u03c0\u03bf\u03bb\u03c5 \u03b5\u03be\u03c9 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03c5\u03bb\u03b7 \u03c4\u03bf\u03c5 \u03c3\u03c7\u03bf\u03bb\u03b5\u03b9\u03bf\u03c5, \u03b1\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b5\u03c4\u03c3\u03b9 \u03c4\u03b1 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c4\u03bf\u03c4\u03b5 \u03b1\u03be\u03b9\u03b6\u03b5\u03b9 \u03c3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03b7\u03c1\u03b9\u03b1 \u03bf \u03ba\u03b1\u03b8\u03b7\u03b3\u03b7\u03c4\u03b7\u03c2 \u03c3\u03b1\u03c2 \u03b4\u03b9\u03bf\u03c4\u03b9 \u03c3\u03c0\u03b1\u03bd\u03b9\u03b1 \u03ba\u03b1\u03b8\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03ba\u03b1\u03b8\u03b7\u03b3\u03b7\u03c4\u03b5\u03c2 \u03bd\u03b1 \u03b4\u03b9\u03b4\u03b1\u03be\u03bf\u03c5\u03bd \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c7\u03c1\u03b5\u03b9\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b9\u03c2 \u03b5\u03be\u03b5\u03c4\u03b1\u03c3\u03b5\u03b9\u03c2... \u03c0\u03b1\u03bd\u03c4\u03bf\u03c2 \u03b8\u03b1 \u03b7\u03b8\u03b5\u03bb\u03b1 \u03c0\u03bf\u03bb\u03c5 \u03bd\u03b1 \u03b4\u03c9 \u03c4\u03b7 \u03bb\u03c5\u03c3\u03b7 \u03c7\u03c9\u03c1\u03b9\u03c2 \u03c4\u03b5\u03c4\u03c1\u03b1\u03b3\u03c9\u03bd\u03b9\u03ba\u03b1 \u03c5\u03c0\u03bf\u03bb\u03bf\u03b9\u03c0\u03b1 \u03c0\u03bf\u03c5 \u03bb\u03b5\u03c2 \u03bf\u03c4\u03b9 \u03b5\u03c7\u03b5\u03b9\u03c2 \u03b2\u03c1\u03b5\u03b9.[/quote]\r\n\r\nloipon gia to deftero kai egw me mod to elusa alla mou eipe o ka8igiths mou (pou mou thn eixe pei o idos ) oti lunete kai me aples gnwseis giafto kai thn anarthsa bas kai thn brei kapios afthn thn polupo8eth lush :S\r\n\r\ngia to prwto twra h lush mou einai trigwnometrikh :to arxiko ginete $ \\frac {1}{x^2} + \\frac {1}{y^2} = 1$ara yparxei gwnia b anoikoi (0,\u03c0/2) = >1/|x|=sin(b) 1/|y|=cos(b) ara A=(1+1/sin(b))(1+1/cos(b))= 1+(sin(b)+cos(b)+1)/(sin(b)cos(b))=1+2(2^(1/2)*sin(\u03c0/4+b)+1)/(sin(2b))=1+2^(3/2)*(sin(\u03c0/4+b))/sin(2b))+2/sin(2b)\r\npou pernei thn elaxisth timh otan to sin2b einai megisto dld gia b=\u03c0/4 opou $ A=3+2\\sqrt{2}$",
  "Solution_14": "\u03a4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7 \u039c\u03b5\u03c3\u03bf\u03b3\u03b5\u03b9\u03ac\u03b4\u03b1 \u03c4\u03bf\u03c5 2003 (\u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1). \u03a4\u03bf \u03c4\u03c1\u03af\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bd\u03c4\u03c9\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd \u03b1\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7 \u03c4\u03bf\u03c5 2005 \u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9. \u0391\u03bb\u03bb\u03ac \u03bc\u03b7\u03bd \u03c4\u03c1\u03b5\u03bb\u03b1\u03b8\u03bf\u03cd\u03bc\u03b5, \u03c4\u03bf \u03cc\u03c4\u03b9 \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2 \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2 \u03b4\u03b5\u03bd \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ac ! \u03a0\u03c7 \u03c4\u03bf 6\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03c6\u03b5\u03c4\u03b9\u03bd\u03ae\u03c2 \u0399\u039c\u039f \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae \u03ae\u03b8\u03b5\u03bb\u03b5, \u03c0\u03bf\u03c5 \u03b4\u03b9\u03b4\u03ac\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03bb\u03cd\u03ba\u03b5\u03b9\u03bf ... \u03ac\u03c1\u03b1 \u03ae\u03c4\u03b1\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc ? \r\n\u039a\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7 \u03b1\u03ba\u03cc\u03bc\u03b1... \u03a0\u03bf\u03c3\u03c4 \u03c4\u03bf\u03c5 style \"\u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 cauchy-swartz \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1\", \u03c4\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ce \u03b1\u03bd\u03bf\u03cd\u03c3\u03b9\u03b1 \u03ba\u03b1\u03b9 spam",
  "Solution_15": "[quote=\"silouan\"]\u03a4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7 \u039c\u03b5\u03c3\u03bf\u03b3\u03b5\u03b9\u03ac\u03b4\u03b1 \u03c4\u03bf\u03c5 2003 (\u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1). \u03a4\u03bf \u03c4\u03c1\u03af\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03cc\u03bd\u03c4\u03c9\u03c2 \u03b1\u03c0\u03cc \u03c4\u03bf\u03bd \u03b1\u03c1\u03c7\u03b9\u03bc\u03ae\u03b4\u03b7 \u03c4\u03bf\u03c5 2005 \u03c0\u03c1\u03ce\u03c4\u03bf \u03b8\u03ad\u03bc\u03b1 \u03ba\u03b1\u03b9 \u03c0\u03ac\u03bb\u03b9. \u0391\u03bb\u03bb\u03ac \u03bc\u03b7\u03bd \u03c4\u03c1\u03b5\u03bb\u03b1\u03b8\u03bf\u03cd\u03bc\u03b5, \u03c4\u03bf \u03cc\u03c4\u03b9 \u03bb\u03cd\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ad\u03c2 \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2 \u03b4\u03b5\u03bd \u03c3\u03b7\u03bc\u03b1\u03af\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03ac ! \u03a0\u03c7 \u03c4\u03bf 6\u03bf \u03b8\u03ad\u03bc\u03b1 \u03c4\u03b7\u03c2 \u03c6\u03b5\u03c4\u03b9\u03bd\u03ae\u03c2 \u0399\u039c\u039f \u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae \u03ae\u03b8\u03b5\u03bb\u03b5, \u03c0\u03bf\u03c5 \u03b4\u03b9\u03b4\u03ac\u03c3\u03ba\u03b5\u03c4\u03b1\u03b9 \u03c3\u03c4\u03bf \u03bb\u03cd\u03ba\u03b5\u03b9\u03bf ... \u03ac\u03c1\u03b1 \u03ae\u03c4\u03b1\u03bd \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc ? \n\u039a\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b7 \u03b1\u03ba\u03cc\u03bc\u03b1... \u03a0\u03bf\u03c3\u03c4 \u03c4\u03bf\u03c5 style \"\u03c0\u03b9\u03c3\u03c4\u03b5\u03cd\u03c9 \u03cc\u03c4\u03b9 \u03bb\u03cd\u03bd\u03b5\u03c4\u03b1\u03b9 \u03bc\u03b5 cauchy-swartz \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1\", \u03c4\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ce \u03b1\u03bd\u03bf\u03cd\u03c3\u03b9\u03b1 \u03ba\u03b1\u03b9 spam[/quote]  oxi da3i , egw eipa apla oti 8elei mexri gnwseis sxoleiou ade kai tetragwnika upolipa afto apla",
  "Solution_16": "[quote=\"KapioPulsar\"] oxi da3i , egw eipa apla oti 8elei mexri gnwseis sxoleiou ade kai tetragwnika upolipa afto apla[/quote]\r\n\r\nO\u03c0\u03c9\u03c2 \u03c0\u03bf\u03bb\u03cd \u03b5\u03cd\u03c3\u03c4\u03bf\u03c7\u03b1 \u03c0\u03b1\u03c1\u03b1\u03c4\u03ae\u03c1\u03b7\u03c3\u03b5 \u03bf \u03a3\u03b9\u03bb\u03bf\u03c5\u03b1\u03bd\u03cc\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c4\u03bf \u03b3\u03b5\u03b3\u03bf\u03bd\u03ce\u03c2 \u03cc\u03c4\u03b9 \u03c3\u03c4\u03b7 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03b9\u03b1\u03c2 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7\u03c2 ''\u03c7\u03c1\u03b5\u03b9\u03ac\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9'' \u03b3\u03bd\u03ce\u03c3\u03b5\u03b9\u03c2 \u03bb\u03c5\u03ba\u03b5\u03af\u03bf\u03c5 \u03b4\u03b5\u03bd \u03c3\u03b7\u03bc\u03ad\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03ba\u03b1\u03b9 \u03bb\u03c5\u03ba\u03b5\u03b9\u03b1\u03ba\u03bf\u03cd \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5.....\u03a0.\u03c7. \u03c3\u03c4\u03b7 \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03af\u03b1 \u03b1\u03bd \u03ad\u03c7\u03b5\u03b9\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c4\u03b7\u03c1\u03ae\u03c3\u03b5\u03b9 \u03bf\u03b9 \u03c0\u03b9\u03bf \u03c0\u03bf\u03bb\u03bb\u03ad\u03c2 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 \u03c3\u03c4\u03c5\u03c1\u03af\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b1 ''\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03bf\u03cd \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5''...(\u03c0\u03cc\u03c3\u03bf \u03bc\u03ac\u03bb\u03bf\u03bd \u03b1\u03bd \u03b1\u03bd\u03b1\u03bb\u03bf\u03b3\u03b9\u03c3\u03c4\u03b5\u03af\u03c2 \u03cc\u03c4\u03b9 \u03bc\u03ad\u03c3\u03b1 \u03c3\u03c4\u03bf \u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03cc \u03b2\u03b9\u03b2\u03bb\u03af\u03bf \u03b1\u03bd\u03b1\u03c6\u03ad\u03c1\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c4\u03b1 \u03b8\u03b5\u03c9\u03c1\u03ae\u03bc\u03b1\u03c4\u03b1 stewart,ceva \u03ba\u03bb\u03c0).\u0391\u03c5\u03c4\u03cc \u03cc\u03bc\u03c9\u03c2 \u03b4\u03b5\u03bd \u03c3\u03b7\u03bc\u03ad\u03bd\u03b5\u03b9 \u03cc\u03c4\u03b9 \u03bf\u03b9 \u03b1\u03c3\u03ba\u03ae\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03bc\u03c0\u03ad\u03bd\u03bf\u03c5\u03bd \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd\u03c2 \u03b5\u03af\u03bd\u03b1\u03b9 ''\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03bf\u03cd'' \u03b5\u03c0\u03b9\u03c0\u03ad\u03b4\u03bf\u03c5....\r\n\r\n\u0397 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03ae \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b1\u03c0\u03cc \u03b1\u03c5\u03c4\u03ad\u03c2 \u03c0\u03bf\u03c5 \u03b1\u03bd\u03ad\u03b2\u03b1\u03c3\u03b5\u03c2 \u03c0\u03bf\u03c5 \u03b8\u03b1 \u03bc\u03c0\u03bf\u03c1\u03bf\u03cd\u03c3\u03b5 \u03bd\u03b1 \u03c7\u03b1\u03c1\u03b1\u03ba\u03c4\u03b9\u03c1\u03b9\u03c3\u03c4\u03b5\u03af \u03ad\u03c4\u03c3\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03b7 \u03c4\u03c1\u03af\u03c4\u03b7 :) \r\n\r\n\u03a6\u03b9\u03bb\u03b9\u03ba\u03ac\r\n\u0394\u03b7\u03bc\u03ae\u03c4\u03c1\u03b7\u03c2"
}
{
  "Tag": [
    "trigonometry",
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "All I know is arcsin(sin x)=x. How to calculate arcsin(sin x/2), arctan(tan(15*t), etc.?\r\nThe result should not contain trigonometrical functions.\r\n\r\nAnd I've learned calculus.",
  "Solution_1": "If you know that $ \\arcsin\\left(\\sin\\left(x\\right)\\right)\\equal{}x$, then how is $ \\arcsin\\left(\\sin\\left(\\frac{x}{2}\\right)\\right)$ different? Think of $ \\frac{x}{2}$ as a single unit. So...$ \\arcsin\\left(\\sin\\left(\\frac{x}{2}\\right)\\right)\\equal{}\\frac{x}{2}$.",
  "Solution_2": "But be careful: these things are only true within certain limited domains. $ \\arcsin\\left(\\sin\\left(\\frac{x}{2}\\right)\\right)\\equal{}\\frac{x}{2}$ is only true for $ \\minus{}\\pi\\le x\\le\\pi,$ for instance. That function is defined for all real $ x,$ and gives a periodic ziz-zag line graph.",
  "Solution_3": "sorry I meant to say arcsin(cos x/2), things like that\r\nhelp",
  "Solution_4": "use sinx=cos(pi-x/2)"
}
{
  "Tag": [
    "calculus",
    "integration",
    "algebra",
    "polynomial",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "$f(x)$ is a polynomial with the degree of $n$,such that $\\int_0^1 x^kf(x)dx=0(k=1,2,\\cdot,n)$\r\nprove:$\\int_0^1f^2(x)dx=(n+1)^2(\\int_0^1f(x)dx)^2$",
  "Solution_1": "hmm. Let $f(x)=a_nx^n + a_{n-1}x^{n-1} + \\cdots + a_1 x + a_0$\r\n\r\nThus\r\n$\\int_0^1 x^k f(x)dx = \\frac{a_n}{n+k+1} +\\cdots+ \\frac{a_1}{k+2}+\\frac{a_0}{k+1} = \\frac{Q(x)}{(k+1)(k+2)\\cdots(n+k+1)}$ (1)\r\n, where $Q(k)$ is a polynomial of degree $n$.\r\n\r\nAnd we have $Q(k)=0$ for $k=1,2,\\dots,n$. Hence\r\n\r\n$Q(k)=c(k-1)(k-2)\\cdots(k-n)$ (2)\r\n\r\nSubstitute (2) into (1), and multiply $k+1$ in both sides. Let $k=-1$. We obtain that\r\n$a_0 = (-1)^n(n+1)c$ (3)\r\nSubstitute (2) into (1) and ket $k=0$ then we have $\\int_0^1 f(x)dx = \\frac{(-1)^n}{n+1}c$ (4)\r\n\r\nCombine (3) and (4), eliminating $c$, we get $a_0=(n+1)^2\\int_0^1 f(x)dx$.\r\n\r\nNow we are going to prove that $\\int_0^1 f^2(x)dx = a_0\\int_0^1 f(x)dx$. This is quite straight-forward.\r\n$\\int_0^1 f^2(x)dx = \\int_0^1 (a_nx^n +\\cdots +a_1x + a_0)f(x)dx = a_0\\int_0^1f(x)dx$.",
  "Solution_2": "very nice,thank you.liyi"
}
{
  "Tag": [
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Compute the limit of x_n, where \r\nx_n=([sqrt(1)]+[sqrt(3)]+[sqrt(5)]+..+[sqrt(4n^2+4n-1)])/n^3.\r\n\r\nwhere [x]=x-{x}.\r\n\r\ncheers! :D",
  "Solution_1": "Take a_n=  \\sum (k=1, 2n^2+2n) [ \\sqrt 2k-1]. We would like to apply Cesaro-Stolz. But a_(n+1)-a_n=  \\sum (k=2n^2+2n+1, 2n^2+6n+4) [ \\sqrt 2k-1]. Observe that each such integer part is 2n+1 or 2n+2. Thus, a_(n+1)-a_n is between (4n+4)(2n+1) and (4n+4)(2n+2). Thus, a_(n+1)-a_n divided by n^2 is about 8. Thus, a_n/n^3 is 8/3 when n is large.",
  "Solution_2": "suppa duppa! :D:D"
}
{
  "Tag": [
    "geometry",
    "analytic geometry",
    "circumcircle",
    "symmetry",
    "parameterization",
    "2009 USAMO",
    "USAMO"
  ],
  "Problem": "Given circles $ \\omega_1$ and $ \\omega_2$ intersecting at points $ X$ and $ Y$, let $ \\ell_1$ be a line through the center of $ \\omega_1$ intersecting $ \\omega_2$ at points $ P$ and $ Q$ and let $ \\ell_2$ be a line through the center of $ \\omega_2$ intersecting $ \\omega_1$ at points $ R$ and $ S$.  Prove that if $ P, Q, R$ and $ S$ lie on a circle then the center of this circle lies on line $ XY$.",
  "Solution_1": "[hide=\"Solution\"]Let $ PQ$ and $ RS$ meet at $ Z$, let the centers of $ \\omega_{1}$ and $ \\omega_{2}$ be $ O_{1}$ and $ O_{2}$, respectively, and let the center of the circle containing $ P,Q,R,S$ be $ O$.\n\nFirst, observe that by power of a point, $ (PZ)(ZQ)\\equal{}(RZ)(ZS)$. But the LHS is the power of $ Z$ with respect to $ \\omega_{2}$, while the RHS is the power of $ Z$ with respect to $ \\omega_{1}$. Since these are equal, $ Z$ is on the radical axis of $ \\omega_{1}$ and $ \\omega_{2}$, that is, $ X,Y,Z$ are collinear.\n\nIt is well known that $ XY\\perp O_{1}O_{2}$ (this is easy to prove with congruent triangles). Note that $ O$ and $ O_{1}$ are both on the perpendicular bisector of $ RS$, so $ OO_{1}\\perp RS$ and $ OO_{1}\\perp O_{2}S$. Similarly, $ OO_{2}\\perp O_{1}Q$. Thus, $ O_{1}Q\\cap O_{2}S$ is the orthocenter of $ \\Delta OO_{1}O_{2}$, but this is just $ Z$. Thus, $ OZ\\perp O_{1}O_{2}$. But we have $ O_{1}O_{2}\\perp XY$ and $ Z$ is on $ XY$, so thus $ O$ is also on $ XY$, as desired.[/hide]",
  "Solution_2": "Let w be circle passing through PQRS. Let radius of $ w_1,w_2,w$ be $ r_1,r_2,r$ respectively, center be $ O_1,O_2,O$ respectively. Now power of $ O_1$ according to circle w is $ O_1P$*$ O_1Q$=$ O_1O^2\\minus{}r^2$. But $ O_1P$*$ O_1Q$=$ O_1O_2^2\\minus{}{r_2}^2$, so $ O_1O^2\\equal{}r^2\\minus{}{r_2}^2\\plus{}O_1O_2^2$. So the power of O according to $ w_1$ is $ O_1O^2\\minus{}{r_1}^2\\equal{}r^2\\minus{}{r_2}^2\\plus{}O_1O_2^2\\minus{}{r_1}^2$. Similarly, O has same power to $ w_2$. So O lies on radical axis XY.",
  "Solution_3": "let the centres of the circles be O1,O2,O3. (O3 being the circle around PQRS)\r\n\r\nlet O1 be (x1,y1), O2 be (x2,y2), and O3 (x3,y3)\r\nlet the radiuses of the three circles be r1,r2,r3.\r\n\r\nwe construct 3 equations, for each circle.\r\n\r\nby subtracting the equation of O3 from O1, we get the equation of line RS.  -(1)\r\nby subtracting the equation of O3 from O2, we get the equation of line PQ.  -(2)\r\nby subtracting the equation of O2 from O1, we get the equation of line XY.  -(3)\r\n\r\nbut we have another condition: O1 is on PQ and O2 is on RS.\r\n\r\nwe put in the coordinates of O1 into (1), and the coordinated of O2 into (2). then we add them (or subtract them)\r\n\r\nwe get an equation that says O3 must be on line (3), which is XY.\r\n\r\nCAN SOMEBODY VERIFY MY SOLUTION?",
  "Solution_4": "[quote= 'CatalystOfNostalgia']But the LHS is the power of Z with respect to w_2, while the RHS is the power of Z with respect to w_1. Since these are equal, Z is on the radical axis of w_1 and w_2. \r\n[/quote]\r\n\r\n\r\n[quote= 'nine34']we construct 3 equations, for each circle. \r\n\r\nby subtracting the equation of O3 from O1, we get the equation of line RS[/quote]\r\n\r\nHow would you prove either of these assertions? Thanks for the help.",
  "Solution_5": "hmm I thought this was an odd choice for a #1 .. my writeup was done within 15 minutes of the start, but I'm pretty sure I would not have been able to solve this last year.",
  "Solution_6": "My general method:\r\nSuppose $ Z$ was the intersection of $ PQ$ and $ RS$, and $ D$ was the intersection of $ XY$ and $ O_1O_2$, where $ O_1$ and $ O_2$ are the centers of $ \\omega_1$ and $ \\omega_2$ respectively. Also suppose $ C$ is the center of the circle through $ PQRS$.\r\n\r\nStep 1: Prove $ C,Z,D$ are collinear. Check.\r\nStep 2: Prove $ Z$ is on $ XY$. Uhhhhhh..... Nope, I had no idea what a radical axis was during the test :( Too bad, it would've been trivial otherwise.\r\n\r\nAnd plus I missed the parallel case like so many other people T-T. So at most a 2 for me on problem 1.",
  "Solution_7": "yeah, a easy one, just a radical axis and an orthocentre, solved within 40 minutes.",
  "Solution_8": "[quote=\"bzauzmer\"]\n\n[quote= 'nine34']we construct 3 equations, for each circle. \n\nby subtracting the equation of O3 from O1, we get the equation of line RS[/quote]\n\nHow would you prove either of these assertions? Thanks for the help.[/quote]\r\n\r\nWell, since R and S are both on O3 and O1, by subtracting the equation of O3 from O1, we get a equation with both R and S on them.\r\n\r\nbut this equation is linear, so it is a line, and there is only one line that can go thru R and S. (which is line RS)",
  "Solution_9": "[quote=\"Yongyi781\"]My general method:\nAnd plus I missed the parallel case like so many other people T-T. So at most a 2 for me on problem 1.[/quote]\r\n\r\nWhat was the parallel case?  :o",
  "Solution_10": "Let the center of $\\omega_1$ be $ A$, and the center of $\\omega_2$ be $ B$. Let the circumcenter of $ PQRS$ be $ C$. Let $ r_1, r_2, r_3$ be the radii of $ A, B, C$ respectively. $ PQ$ is the radical axis of circles $ A$ and $ C$, $ RS$ is the radical axis of circles $ C$ and $ B$, and $ XY$ is the radical axis of circles $ A$ and $ B$, because the circles intersect and it is a well known fact that the radical axis of intersecting circles goes through the intersection points.\nBy definition of radical axis, the power of $ A$ with respect to circles $ B$ and $ C$, and similarly for the points $ B$ and $ C$ with respect to the other 2 circles, we get:\n$ AC^2 \\minus{} r_3^2 \\equal{} AB^2 \\minus{} r_1^2$\n$ AB^2 \\minus{} r_2^2 \\equal{} AC^2 \\minus{} r_3^2$\nAdding the 2 equations gives us:\n$ AC^2 \\minus{} r_3^2 \\plus{} AB^2 \\minus{} r_2^2 \\equal{} AB^2 \\minus{} r_1^2 \\plus{} AC^2 \\minus{} r_3^2$\n$ BC^2 \\minus{} r_2^2 \\equal{} AC^2 \\minus{} r_1^2$\nThen by defintion,$ C$ has equal power with respect to circles $ A$ and $ B$, so $ C$ lies on the radical axis of circles $ A$ and $ B$, which is $ XY$, so we are done.\n\nBasically what greentreeroad did, use the powers being equal to find the third circle center having the same power with respect to the other 2 circles.\n\nThat takes care of the parallel case since the lengths aren't affected if triangle ABC is degenerate, and the powers being equal still holds.",
  "Solution_11": "Letting the circle containing $ P,Q,R,S$ have center $ C$, I proved that $ C$ had a power of a point equal with respect to $ \\omega_1$ and $ \\omega_2$. But instead of citing radical axes (I only have a vague idea about them, mostly that it is the line $ XY$, so I was careful), I said to consider the extension of line $ CX$ and its two intersections with $ \\omega_1$ and $ \\omega_2$ be $ X_1, X_2$, respectively. Proving the power of a point is equal for both circles proves that $ CX_1\\equal{}CX_2$, implying that they are the same point, i.e. $ Y$. Then this implies that $ C,X,Y$ are collinear.\r\nDoes this logic work?",
  "Solution_12": "Yeah that looks like it works (and it's pretty much how you prove the existence of radical axes [in the case of intersecting circles]).",
  "Solution_13": "a problem very easily, I do not think this is the problem USAMO.",
  "Solution_14": "[quote=\"math10\"]a problem very easily, I do not think this is the problem USAMO.[/quote]\r\n\r\nWhat do you mean?  This was problem #1 on the 2009 USAMO.",
  "Solution_15": "Here I noticed that the intersection of $C_2S$ and $C_1Q$ is the radical axis of $\\omega_{1,2,3}$, but here I don't think the Radical Axis Theorem says that the center of $\\omega_3$ lies on $XY$  :wacko: ",
  "Solution_16": "Let $T = PQ \\cap RS$, $(PRQS) = \\omega_3$, the center of $\\omega_3$ be $O_3$, and $K = XY \\cap O_1O_2$. We first assume $O_1$, $O_2$, and $O_3$ are non-collinear. Notice $$Pow_{\\omega_1}(T) = TR \\cdot TS = Pow_{\\omega_3}(T) = TP \\cdot TQ = Pow_{\\omega_2}(T) \\implies T \\in XY$$ (the Radical Axis of $\\omega_1$ and $\\omega_2$). Since $RS$ is the Radical Axis of $\\omega_1$ and $\\omega_3$, and $PQ$ is the Radical Axis of $\\omega_2$ and $\\omega_3$, we know the intersection of the perpendicular from $O_1$ to $RS$ and the perpendicular from $O_2$ to $PQ$ is $O_3$.\n\nLooking at $\\triangle O_1O_2O_3$, it's obvious that $T$ is its orthocenter. Now, let the foot of the altitude from $O_3$ to $O_1O_2$ be $K'$. Then, $\\angle TK'O_1 = 90^{\\circ}$. Using Radical Axis properties, however, we know $\\angle TKO_1 = 90^{\\circ}$. Thus, $K = K'$ because they both lie on $O_1O_2$. Hence, $O_3, T, K$ are collinear or $O_3 \\in XY$ as desired. \n\nIf $O_1$, $O_2$, and $O_3$ are collinear, then we know $PQ \\parallel RS \\parallel XY$. \n\n[b]Claim:[/b] If $O_1$, $O_2$, and $O_3$ are collinear, then $O_3 = K = XY \\cap O_1O_2$.\n\n[i]Proof.[/i] The claim is equivalent to showing $KP = KQ = KR = KS$ (in this configuration). This is trivial via the Pythagorean Theorem, however, so we're done. $\\square$\n$\\blacksquare$\n\n\n[b]Remark:[/b] In order for the problem to hold when the centers are collinear, $O_3 \\in O_1O_2$ and $O_3 \\in XY$ has to hold, implying $O_3 = XY \\cap O_1O_2 = K$ must be true, motivating the last claim.",
  "Solution_17": "[hide=Solution]Clearly $XY$ is the radical axis of $\\omega_1, \\omega_2$. Define the circumcircle of $PQRS$ to be $\\omega$. Similarly, $SR$ is the radical axis of $\\omega, \\omega_1$. Let $O$ be the center of this circle. Since $O_2$ lies on $RS$, we have $\\text{Pow}_{\\omega} O_{2} = \\text{Pow}_{\\omega_1} O_2$, or $OO_2^2-OP^2=O_1O_2^2-O_1P^2$. Also, we get $OO_1^2-OP^2=O_1O_2^2-O_2Q^2$ from similar methods. Subtracting the first equation from the second gives $OO_1^2-OO_2^2=O_1P^2-O_2Q^2$, or $OO_1^2-O_1P^2=OO_2-O_2Q^2$. This is all that's necessary to finish and we are done.[/hide]",
  "Solution_18": "Let the center of the circle $\\omega$ passing through $P,Q,R,S$ be $O$, and let the length of its radius be $r$. Let the center of $\\omega_1$ be $O_1$ and let the center of $\\omega_2$ be $O_2$, and let the lengths of their radii be $r_1,r_2$ respectively. It suffices to show that $O_1O^2-r_1^2=O_2O^2-r_2^2$. But\n\\[O_1O^2-r^2=O_1P\\cdot O_1Q=O_1O_2^2-r_2^2\\]\n\\[O_2O^2-r^2=O_2R\\cdot O_2S=O_1O_2^2-r_1^2\\]\nso we are done.",
  "Solution_19": "By Radical Center theorem we know O1Q and O2S and XY meet at single point Z.\nLet O3 be center of SPRQ. we have O1O3 \u22a5 SR and O2O3 \u22a5 PQ and O1O2 \u22a5 XY so Z is orthocenter of O1O2O3 and as it lies on XY we have O3 lies on XY as well.\nwe're Done.",
  "Solution_20": "Let $\\omega_3$ and $O_3$ be $(PQRS)$ and its center, respectively. Notice $$O_1O_2^2-r_2^2=\\textit{pow}_{\\omega_2}(O_1)=\\textit{pow}_{\\omega_3}(O_1)=O_1O_3^2-r_3^2$$ and $$O_2O_3^2-r_3^2=\\textit{pow}_{\\omega_3}(O_2)=\\textit{pow}_{\\omega_1}(O_2)=O_2O_1^2-r_1^2.$$ Adding yields $$\\textit{pow}_{\\omega_2}(O_3)=O_2O_3^2-r_2^2=O_1O_3^2-r_1^2=\\textit{pow}_{\\omega_1}(O_3)$$ so $O_3$ lies on $\\overline{XY}.$ $\\square$",
  "Solution_21": "[hide=for fun.]\nLet $O$ be the center of $(PQRS)$, let $O_1,O_2$ are centers of $\\omega_1,\\omega_2$, respectively. By radical axis theorem, $Z=\\overline{PQ}\\cap\\overline{SR}$ lies on $\\overline{XY}\\perp \\overline{O_1O_2}$. Observe that by perpendicular bisectors, $Z$ is the orthocenter of $\\triangle OO_1O_2$. Hence, $\\overline{OZ}\\perp\\overline{O_1O_2}\\perp\\overline{AX}\\implies Z\\in \\overline{XY}$, as desired.\n[/hide]",
  "Solution_22": "imagine doing amo problems \"for fun\" orz",
  "Solution_23": "Let $O_1, r_1$ and $O_2, r_2$ be the centers and radii of $w_1$ and $w_2$, respectively. Also, let $w_3$ be the circle containing $P, Q, R, S$ with a center at $O_3$ and a radius of $r_3$. Since $O_1$ is on the radical axis of $w_2$ and $w_3$, we have:\n$$Pow_{w_2}(O_1)=Pow_{w_3}(O_1)$$\n$$O_1O_2^2-r_2^2=O_1O_3^2-r_3^2$$\nSimilarly, since $O_2$ lies on the radical axis of $w_1$ and $w_3$:\n$$Pow_{w_1}(O_2)=Pow_{w_3}(O_2)$$\n$$O_1O_2^2-r_1^2=O_2O_3^2-r_3^2$$\nSubtracting these two equations gives the following equation, which we continue to simplify.\n$$(O_1O_2^2-r_2^2)-(O_1O_2^2-r_1^2)=(O_1O_3^2-r_3^2)-(O_2O_3^2-r_3^2)$$\n$$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2$$\n$$O_2O_3^2-r_2^2=O_1O_3^2-r_1^2$$\n$$Pow_{w_2}(O_3)=Pow_{w_1}(O_3)$$\nTherefore, $O_3$ lies on the radical axis (line $XY$) of $w_1$ and $w_2$. $\\square$",
  "Solution_24": "Let $O_1, O_2$ be the centers of the circles $w_1,$ and $w_2$ with radius $r_1$ and $r_2$ respectively. Furthermore, let $O_3$ be the center of the circle $w_3$ passing through $P, Q, R,$ and $S$ with radius $r_3$. Then, the problem is equivalent to proving that $P(O_3, w_1) = P(O_3, w_2) \\Leftrightarrow O_3O_1^2 - r_1^2 = O_3O_2^2 - r_2^2$.\n\nNote that $P(O_1, w_2) = P(O_1, w_3) \\implies O_1O_3^2 -r_3^2 = O_1O_2^2 - r_2^2$ and $P(O_2, w_1) = P(O_2, w_3) \\implies O_2O_3^2 - r_3^2 = O_2O_2^2 - r_1^2$. Subtracting the two equations gives $O_1O_3^2 - O_2O_3^2 = r_1^2 - r_2^2 \\implies O_3O_1^2 - r_1^2 = O_3O_2^2 - r_2^2$ which is what we wanted to prove.",
  "Solution_25": "Let $(PQRS)=\\omega_3$ and let $r_i$ and $O_i$ denote the radius and center of circle $\\omega_i$ for $i\\in[1,2,3]$. Since $O_1$ is on the radical axis of $O_2$ and $O_3$ we have $$\\text{Pow}_{\\omega_2}(O_1)=\\text{Pow}_{\\omega_3}(O_1) \\iff O_1O_2^2-r_2^2=O_1O_3^2-r_3^2.$$ Since $O_2$ is on the radical axis of $O_1$ and $O_3$ we have $$\\text{Pow}_{\\omega_1}(O_2)=\\text{Pow}_{\\omega_3}(O_2) \\iff O_2O_1^2-r_1^2=O_2O_3^2-r_3^2.$$ Subtracting yields $$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2 \\iff \\text{Pow}_{\\omega_1}(O_3)=\\text{Pow}_{\\omega_2}(O_3)$$ and we are done.",
  "Solution_26": "Let $O_1,O_2,O_3$ be the centers of $\\omega_1,\\omega_2$, and $(PQRS)$. Let $C$ be the intersection of $PQ$ and $RS$. Note that $C$ also lies on $XY$ since the pairwise radical axes of the three circles are concurrent. Now, note that $O_3$ lies on the perpendicular bisector of both $PQ$ and $RS$. Thus, we have that $O_1O_3$ is perpendicular to $RS$, and $O_2O_3$ is perpendicular  to $PQ$. Therefore, $C$ is the orthocenter of $\\triangle O_1O_2O_3$. This means that $O_3C$ is perpendicular to $O_1O_2$. Since $C$ lies on $XY$, we are done.",
  "Solution_27": ":unamused:\n\nDenote $\\omega_3:=(PQRS)$. Because $O_1\\in PQ$,\n\\[\\text{Pow}_{\\omega_2}(O_1)=\\text{Pow}_{\\omega_3}(O_1)\\implies O_1O_2^2-r_2^2=O_1O_3^2-r_3^2;\\]\nand since $O_2\\in RS$,\n\\[\\text{Pow}_{\\omega_1}(O_2)=\\text{Pow}_{\\omega_3}(O_2)\\implies O_1O_2^2-r_1^2=O_2O_3^2-r_3^2.\\]\nSubtracting,\n\\[r_1^2-r_2^2=O_1O_3^2-O_2O_3^2\\implies O_1O_3^2-r_1^2=O_2O_3^2-r_2^2,\\]\ni.e., $\\text{Pow}_{\\omega_1}(O_3)=\\text{Pow}_{\\omega_2}(O_3)$ or $O_3\\in XY$, as desired. $\\square$",
  "Solution_28": " Let $M=\\overline{SR} \\cap \\overline{PQ}$. Let $O_1,O_2,O_3$ be the centers of $\\omega_1,\\omega_3$ and $(SPRQ)$. Now, it is well known that the radical axis of two circles is perpendicular to the line joining their centers. Then, $SR \\perp O_1,O_2$ and $PQ \\perp O_2O_3$. This means that $M$ must be the orthocenter of $\\triangle O_1O_2O_3$. This means, $O_3M\\perp O_1O_2$ as well.  Clearly $M$ lies on $XY$ ($MP\\cdot MQ = MR\\cdot MS$ so it lies on the radical axis of $\\omega_1$ and $\\omega_2$). But then, $XY$ is the line passing through $M$ and perpendicular to $O_1O_2$. This means that indeed $O_3$ lies on the line $XY$ and we are done.",
  "Solution_29": "Let $O_1$ be the center of $\\omega_1$ and let $O_2$ be the center of $\\omega_3$. Denote the center of $(PRQS)$ to be $O_3$ and let the intersection of $PQ$ and $SR$ be $C$. Now note that $SR$, $PQ$, and $XY$ are the three pairwise radical axes of $\\omega_1$, $\\omega_2$, and $(PRQS)$. \n\nDenote the length of $O_1O_2$ to be $a$ and let the radii of $\\omega_1$, $\\omega_2$, and $(PRQS)$ be $r_1$, $r_2$, and $r_3$, respectively. Since $O_2$ is on the line $PQ$, by the Power of a Point definition of the radical axis, we have that \n\\[a^2-r_1^2=O_2O_1^2-r_1^2=Pow_{\\omega_1}(O_2)=Pow_{\\omega_3}(O_1)=O_2O_3^2-r_3^2,\\]\nwhich gives that $O_2O_3=\\sqrt{a^2-r_1^2+r_3^2}$. Similarly, since $O_1$ is on $RS$, using Power of a Point on $O_1$ with respect to $O_2$ and $O_3$, we get that $O_1O_3=\\sqrt{a^2-r_2^2+r_3^2}$. Finally, using these lengths and Power of a Point one last time, \n\\[Pow_{\\omega_1}(O_3)=O_1O_3^2-r_1^2=a^2+r_3^2-(r_1^2+r_2^2)=O_1O_2^2-r_2^2=Pow_{\\omega_2}(O_3),\\]\nwhich implies that $O_3$ is on the radical axis of $\\omega_1$ and $\\omega_2$, which is $XY$, finishing the problem."
}
{
  "Tag": [
    "\\/closed"
  ],
  "Problem": "Do we have a new administrator nsato, or has he always been here? I understand he's an instructor for some classes.",
  "Solution_1": "Naoki Sato joined the staff of Art of Problem Solving as of yesterday.  He will be a regular instructor.  He was an IMO medalist and worked with the Canadian IMO team for several years, so he is an experienced teacher of top students."
}
{
  "Tag": [
    "function",
    "limit",
    "algebra",
    "domain",
    "calculus",
    "derivative",
    "analytic geometry"
  ],
  "Problem": "Prove that the function $ \\sqrt{x}$ is uniformly continuous for $ x\\in[1,\\infty)$.\r\n\r\nThe hint given was to use the facts that the function continuous on a segment is also uniformly continuous on it, and the fact if the function is continuous for $ x\\in[a,\\infty)$, and the $ \\lim_{x\\to\\infty}$ exists, then it's uniformly continuous. But still I do believe that the word \"finite\" (limit) was dropped out of this last theorem, and that it cannot be applied here.\r\n\r\nIt's an unbounded function on an unbounded domain-it's quite surprising for me that it's uniformly convergent... (but again I have to say that I lack knowledge in more than few topics in analysis...)",
  "Solution_1": "That's the wrong hint at $ \\infty$; what's going on here is that the derivative is bounded for large $ x$. On the other side, you only need a second argument if you extend all the way to zero.",
  "Solution_2": "Oh, I didn't think of the fact that if the derivative is bounded on the interval (and the function is continuous), then the continuity is uniform (that's \"sufficient Lipschitz\", right?)-the theorem is not in the curriculum I have to work by as a UTA (and it would surely be fun to prove it to my students using MVT they've never heard of-oh well, they don't know derivatives yet, so the argument will be a bit misty for them).\r\n\r\nSo I'll try an intuitive-slope-approach, and hope they'll understand  :)\r\n\r\nI just wonder is there a nice $ \\epsilon \\minus{} \\delta$ proof? It seems to me that $ \\epsilon\\equal{}\\delta$ should do the trick, as $ x_1\\minus{}x_2>\\sqrt{x_1}\\minus{}\\sqrt{x_2}$ for $ x_1>x_2$.",
  "Solution_3": "If $ |f'(x)|\\le k$ for all $ x$, $ f$ satisfies the Lipschitz condition $ |f(x)-f(y)|\\le k|x-y|$ by the Mean Value Theorem. That implies uniform continuity; if $ |x-y|<\\frac{\\epsilon}{k}$, $ |f(x)-f(y)|<\\epsilon$.\r\n\r\nIn this case, the derivative is no more than $ \\frac12$, so we use $ \\delta=2\\epsilon$ or smaller.\r\n\r\nIntroducing technical notions like uniform continuity before we know about derivatives and the MVT sounds very weird to me.",
  "Solution_4": "Similarly, the function $ \\sqrt{|x|}$ is uniformly continuous.\r\n$ \\left|\\sqrt{|x|}-\\sqrt{|y|}\\right|\\leq \\sqrt\\left\\parallel{}x|-|y|\\right|}\\leq\\sqrt{|x-y|}$ $ \\forall x,y\\in\\mathbb{R}$.\r\nGiven $ \\varepsilon>0$, we can be sure that $ |f(x)-f(y)|<\\varepsilon$ as long as $ |x-y|<\\varepsilon^{2}$.",
  "Solution_5": "[quote=\"jmerry\"]\nIntroducing technical notions like uniform continuity before we know about derivatives and the MVT sounds very weird to me.[/quote]\r\n\r\nThis Mathematics course is extremely wide, so some things are put upside down /but de facto they're there just pro formae/. First semester has to cover single-variable Calculus and the Analysis basics while in the other semester: multivariable Calculus, DE, Laplace & Fourier, Field theory... Third semester: Complex Analysis, Fourier, Laplace (again)... Quite a blend for young engineers  :)",
  "Solution_6": "[quote=\"hsiljak\"]\nI just wonder is there a nice $ \\epsilon \\minus{} \\delta$ proof? It seems to me that $ \\epsilon \\equal{} \\delta$ should do the trick, as $ x_1 \\minus{} x_2 > \\sqrt {x_1} \\minus{} \\sqrt {x_2}$ for $ x_1 > x_2$.[/quote]\r\n\r\nThis proof is correct on the given interval. Of course, it won't work for $ x \\leq 1$ because then the above inequality won't hold."
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "I have always wondered whether there is a geometric interpretation for a function which is uniform continuity.   For instance, the geometric interpretation for a continuous function from $ R$ to $ R$ is the classic \"a function is continuous if its graph can be drawn without lifting the pencil from the paper\".  Another geometric interpretation of a definition is the one for uniform convergence (I like that because it is very helpful) that says that a sequence of functions converges uniformly to a function $ f$ if I can draw a \"tube\" ,of size epsilon, around $ f$ such that in some moment the tail of the sequence get inside this tube (I hope I explained this well :P).\r\n\r\nI appreciate you can help me with my question and if you want you can add another nice geometric interpretation of some definition.\r\n\r\nThank you!",
  "Solution_1": "My friend and I are in an advanced analysis course and when we first got to uniform continuity and tried to do some proofs, we concluded that a uniformly continuous function can't go \"Blaaah!\"  Now that phrase probably doesn't mean anything to you, but when my friend attempted to draw a function that was not uniformly continuous, he basically drew a line shooting upwards.  In other words, the function can't blow up without control.  For example, $ f(x)=x^{2}$ can't be uniformly continuous because it's getting steeper and steeper so we can't pick a $ \\delta$ that always works.  On the other hand, $ f(x)=\\sqrt{x}$ (and even $ \\sqrt{|x|}$) are both uniformly continuous because with the exception of the cusp at the origin, it's very easy to control the function.\r\n\r\nDid that help at all?",
  "Solution_2": "I guess an o.k. geometric analogy is that a function is uniformly continuous on an interval $ I$ if when you pick a horizontal strip of size $ \\epsilon$, there exists a $ \\delta$ such that no matter where you slide your $ \\epsilon$ strip across the function (as long as it stays in $ I$) any points that are $ \\delta$ units away from each other or less will have images that fit into that epsilon strip. \r\n\r\n\r\nIt's easier to visualize if you draw it out  :P.",
  "Solution_3": "I agree with [b]max_tm[/b]...and with a little searching look what I found...[youtube]oGkgIzSppik[/youtube]",
  "Solution_4": "That's basically the idea behind what I wrote in the first post: the function never blows up without control."
}
{
  "Tag": [
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Need some help.  Consider the lower limit topology defined on $ \\mathbb{R}$. That is, the topology generated by the basis, $ \\{[a,b): a,b\\in\\mathbb{R}\\}$.\r\n\r\nGive a covering of $ [0.1]$ to show that $ [0,1]$ is not compact.",
  "Solution_1": "The obvious covering to try is something about upper limits; how about $ V_n \\equal{} \\left[ 0, 1 \\minus{} \\frac {1}{n} \\right)$?  A finite subcover of this cover is of the form $ \\left[ 0, 1 \\minus{} \\frac {1}{N} \\right)$ and hence cannot cover $ [0, 1]$.",
  "Solution_2": "[quote=\"t0rajir0u\"]$ V_n \\equal{} \\left[ 0, 1 \\minus{} \\frac {1}{n} \\right)$.[/quote]\r\n\r\nBut it is not an open cover of $ [0, 1]$, neither. You may add $ [1, 2)$ to get around this problem.",
  "Solution_3": "Ah...  :lol:  Thanks sos440.",
  "Solution_4": "Is the space locally compact?",
  "Solution_5": "Nope. All compact subsets are at most countable, but all open sets are uncountable (as they are intervals), so no compact set can contain an open set."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "The lengths of the tangents from vertices A, B, C to the incircle are x, y and z respectively. If $a\\geq b\\geq c$, prove that $az+by+cx \\geq \\frac{a^2+b^2+c^2}{2} \\geq ax+by+cz$",
  "Solution_1": "WE have obviously $x+y=c$\r\n                            $x+z=b$ and\r\n                            $y+z=a$\r\nSolving this little system we determine $x,y,z$ and the ineq is then easily proved!\r\nDo not forget the triangle ineq!!!! :)  :D  :D"
}
{
  "Tag": [
    "search",
    "geometry",
    "cyclic quadrilateral",
    "geometry open"
  ],
  "Problem": "[b]\nLet $ ABCD$ be a cyclic quadrilateral and $ I$ denote the point of intersection of the diagonals \n$ AC$ and $ BD$, $ M$ is the midpoint of side $ BC$ and $ N$ is the midpoint of side $ AD$. \nIf $ P \\in AB$ such that $ IP \\bot AB$ and $ R \\in CD$ such that $ IR \\bot CD$ prove that $ MN \\bot PR$.\n[/b]\r\n------------------------------------------------------------------------------------------------------------------\r\n[img]http://img.photobucket.com/albums/v218/hudrea/metrutest.jpg[/img]",
  "Solution_1": "I have seen this problem here ---> http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1549164493&t=55836\r\nyetti's very nice answer  :)",
  "Solution_2": "Thank you very much for your valuable opinion.",
  "Solution_3": "this problem will be solve very easy by the down lemma and congruity triangles. :wink: \r\n\r\n\r\nlemma: if $ ABC$ is a triangle and  $ x ,y$  are the points that $ ABX , ACY$ are rectangular triangles  . ($ BXA \\equal{} CYA \\equal{} 90$)  then if $ M$ is the midpoint of $ BC$  and   $ XBA \\equal{}YCA$   then we have \r\n\r\n[size=150]$ XMY \\equal{} 90$ and $ mx \\equal{} my$[/size]",
  "Solution_4": "It seems this problem is very well known and many people tried it in advance and there are lots of skilled people in mathlinks.\r\nDo you know what is the source of the problem?\r\n\r\nI like finding interesting math problems and I post most many of them here.",
  "Solution_5": "no. i don't know  . i solve this 8month  ago when i was 16. :o  :roll:",
  "Solution_6": ":) it means you have good teacher(s).",
  "Solution_7": "I cant understand u. please speak more cleraly :wink:  :arrow:",
  "Solution_8": "I want to say - if people solve such problems when they are 16 years old - they have good teachers that propose such problems and prepare them to solve these problems.",
  "Solution_9": "YOU  ARE  SO  $ benignity$  :)",
  "Solution_10": "Thank you very much. I said that because when I was 16 old a had no good teacher. But she said me that the Bulgarian Math Olympiad exist and there was good problems in the competitions in their earlier stages. It is the reason for me to invent math problems and to try to find interesting problems and to propose them in forums, magazines and comeptitions."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "For $n>1$ , prove that  $\\sum_{j=1}^{n}\\frac{1}{2j-1}$ can not be integer.",
  "Solution_1": "Analogously with a similar proof for the harmonic series, we can show that the greatest power of $3$ that divides the common denominator is greater than the greatest power of $3$ that divides the common numerator.\r\n\r\nThere is a formulation with $p$-adic numbers, but so far I have been unable to tease it out.  Could somebody who knows about such things sketch it out for me?",
  "Solution_2": "t0rajir0u, Thanks. I saw.",
  "Solution_3": "I wonder that are there other solution except this classical solution? Can anybody help me?",
  "Solution_4": "From Bertrand's Postulate, there exists $k\\le n$ such that $n\\le 2k-1=p\\le 2n-1$ for some prime $p$. Then the fraction (the sum) in lowest terms has denominator divisible by $p$. on the other hand the numerator is equal to $\\text{(some terms divisible by }p)+\\text{(a term not divisible by }p)$. thus it cannot be an integer.",
  "Solution_5": "We can prove the following more general result. \r\n\r\nLet $a_{1},a_{2},...,a_{n}(n>1)\\in\\mathbb{Z}-\\{0\\}$ . Suppose there is a prime $p$ and positive integer $h$ such that $p^{h}|a_{i}$ for some $i$ and $p^{h}\\not |a_{j}\\forall j\\not =i$. Then show that $\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+...+\\frac{1}{a_{n}}\\not\\in\\mathbb{Z}$."
}
{
  "Tag": [],
  "Problem": "What is the value of $ 501^2 \\minus{} 499^2$?",
  "Solution_1": "In general, $ x^2-y^2=(x+y)(x-y)$.  In this case, $ x=501$ and $ y=499$.  Thus\r\n\\begin{align*}501^2-499^2&=(501+499)(501-499)\\\\\r\n&=(1000)(2)\\\\\r\n&=\\boxed{2000}\\end{align*}"
}
{
  "Tag": [
    "number theory",
    "prime numbers",
    "arithmetic sequence",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "proof that $\\xi=0,p_{1}p_{2}p_{3}....$is irrational \r\n$p_{n}$is the $n$ th prime number.",
  "Solution_1": "Same reasoning than in http://www.mathlinks.ro/Forum/viewtopic.php?t=104311 , noting that there are for example abitrary long sequences of zero's in the (decimal) representation because there are primes $\\equiv 1 \\mod 10^{k}$ for all $k$.",
  "Solution_2": "i want a proof and i don't that it's so simple.\r\nmay be the first is simple but this no :D",
  "Solution_3": "Well, if you don't ask for a proof of Dirichlet's theorem (even in the special case mentioned by ZetaX), I don't see what you are wanting...\r\n\r\nPierre.",
  "Solution_4": "I didn't just say it is simple but gave a sketch of a proof...  :maybe: \r\n\r\nA bit more explicit:\r\nAssume it is rational. Then it's (decimal) representation is periodic. By that, there can't be arbitrary long sequences of zeros in it (this could only happen for a periodic representation only consisting of zeros from some point, but this is not the case).\r\nSo if we proof that there are arbitrary long sequences of zeros in it, we are done. So we look for prime numbers of type $a\\cdot 10^{k}+1$ since they have (at least) $k-1$ consecutive zeros in them (they look like $...x000...0001$).\r\nSo it suffices to prove that we can find such a prime for all given $k$, meaing to prove there is a prime $\\equiv 1 \\mod 10^{k}$. But such prime exists by Dirichlets theorem on primes in arithmetic progression, so we are done (note that this case of Dirichlet can be done on an elementary way; for example all prime divisors $\\neq 2,5$ of $x^{4 \\cdot 10^{k-1}}-x^{3 \\cdot 10^{k-1}}+x^{2 \\cdot 10^{k-1}}-x^{10^{k-1}}+1$  work).\r\n\r\nwell, I can post (or link since it surely is on MathLinks) the proof of Dirichlet's theorem for that case if you want..."
}
{
  "Tag": [
    "LaTeX",
    "combinatorics open",
    "combinatorics"
  ],
  "Problem": "$n$ is integer $>0$.let $A={1,2,...n}$.find $f(n)_{max}$ such that when $A$ is divised to $f(n)$ any subset  $A_1,A_2,...A_{f(n)}$st : \r\n1)$A_{i}\\bigcap\\(A_{j}=\\emptyset$ for all $i,j$\r\n2)$A_{1}\\cup\\(A_{2}\\cup\\...\\cup\\(A_{n}=A$\r\n  there exist $A_{k}$ include three elements $x<y<z$ st $x+y\\geq z$",
  "Solution_1": "noone .i think it is very easy :roll:",
  "Solution_2": "[quote=\"dieuhuynh\"]noone .i think it is very easy :roll:[/quote]\r\n\r\nMaybe the problem is easy. But without $\\LaTeX$ it is hard to read and nobody bothers reading it. Think about that.  :)",
  "Solution_3": "yes ,I edited.someone post solution!"
}
{
  "Tag": [
    "inequalities",
    "induction",
    "inequalities unsolved"
  ],
  "Problem": "Let $ x_i \\in \\mathbb{R}^\\plus{}$ for $ i\\equal{}1,2,\\ldots ,n$ such that $ x_1 \\le x_2 \\le \\ldots \\le x_n$. Prove the inequality:\r\n\\[ \\frac{x_1}{x_2}\\plus{}\\frac{x_2}{x_3}\\plus{}\\ldots\\plus{}\\frac{x_{n\\minus{}1}}{x_n}\\plus{}\\frac{x_n}{x_1} \\geq \\frac{x_2}{x_1}\\plus{}\\frac{x_3}{x_2}\\plus{}\\ldots\\plus{}\\frac{x_n}{x_{n\\minus{}1}}\\plus{}\\frac{x_1}{x_n} \\]",
  "Solution_1": "Denote    $ {a_k} \\equal{} \\frac{{{x_k}}}{{{x_{k \\plus{} 1}}}}$ for    $ 1 \\le k \\le n \\minus{} 1$    then      $ 0 < {a_k} \\le 1$\r\nand the inequality became \r\n$ \\sum\\limits_{k \\equal{} 1}^{n \\minus{} 1} {{a_k}}  \\plus{} \\frac{1}{{\\prod\\limits_{k \\equal{} 1}^{n \\minus{} 1} {{a_k}} }}\\ge \\sum\\limits_{k \\equal{} 1}^{n \\minus{} 1} {\\frac{1}{{{a_k}}}}  \\plus{} \\prod\\limits_{k \\equal{} 1}^{n \\minus{} 1} {{a_k}}$\r\nIf $ n\\equal{}2$  the inequality is trivial\r\nLet us assume that the inequality is true for order less than (n-1) \r\nBy induction  we have  for numbers $ {a_1},{a_2},...,{a_{n \\minus{} 2}},{a_{n \\minus{} 1}}{a_n}$\r\n $ \\sum\\limits_{k \\equal{} 1}^{n \\minus{} 2} {{a_k}}  \\plus{} {a_{n \\minus{} 1}}{a_n} \\plus{} \\frac{1}{{\\prod\\limits_{k \\equal{} 1}^n {{a_k}} }} \\ge \\sum\\limits_{k \\equal{} 1}^{n \\minus{} 2} {\\frac{1}{{{a_k}}}}  \\plus{} \\frac{1}{{{a_{n \\minus{} 1}}{a_n}}} \\plus{} \\prod\\limits_{k \\equal{} 1}^n {{a_k}}$\r\nThen $ \\sum\\limits_{k \\equal{} 1}^n {{a_k}}  \\plus{} \\frac{1}{{\\prod\\limits_{k \\equal{} 1}^n {{a_k}} }} \\ge \\sum\\limits_{k \\equal{} 1}^n {\\frac{1}{{{a_k}}}}  \\plus{} \\prod\\limits_{k \\equal{} 1}^n {{a_k}}  \\plus{} {a_{n \\minus{} 1}} \\plus{} {a_n} \\plus{} \\frac{1}{{{a_{n \\minus{} 1}}{a_n}}} \\minus{} {a_{n \\minus{} 1}}{a_n} \\minus{} \\frac{1}{{{a_{n \\minus{} 1}}}} \\minus{} \\frac{1}{{{a_n}}}$\r\n i.e  $ \\sum\\limits_{k \\equal{} 1}^n {{a_k}}  \\plus{} \\frac{1}{{\\prod\\limits_{k \\equal{} 1}^n {{a_k}} }} \\ge \\sum\\limits_{k \\equal{} 1}^n {\\frac{1}{{{a_k}}}}  \\plus{} \\prod\\limits_{k \\equal{} 1}^n {{a_k}}  \\plus{} \\frac{{\\left( {1 \\minus{} {a_{n \\minus{} 1}}{a_n}} \\right)\\left( {1 \\minus{} {a_{n \\minus{} 1}}} \\right)\\left( {1 \\minus{} {a_n}} \\right)}}{{{a_{n \\minus{} 1}}{a_n}}}$\r\nthus   $ \\sum\\limits_{k \\equal{} 1}^n {{a_k}}  \\plus{} \\frac{1}{{\\prod\\limits_{k \\equal{} 1}^n {{a_k}} }} \\ge \\sum\\limits_{k \\equal{} 1}^n {\\frac{1}{{{a_k}}}}  \\plus{} \\prod\\limits_{k \\equal{} 1}^n {{a_k}}$"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "inradius",
    "geometry proposed"
  ],
  "Problem": "Let $ P$ be an arbitrary point inside a given triangle $ ABC$. Let $ A',B',C'$ be the orthogonal projection of $ P$ on $ BC,CA,AB$ respectively. Let $ I$ be the incenter and $ r$ be the inradius of the triangle $ ABC$. Find the least value of the expression\"\r\n\r\n$ PA'\\plus{}PB'\\plus{}PC'\\plus{}\\frac{PI^2}{2r}$",
  "Solution_1": "nobody done??",
  "Solution_2": "Dear Mathlinker,\r\nWhere does this problem come from?\r\nIn Vietnam, it was in [b]Mathematics and Youth[/b] magazine two months ago and still not out of date. So to mod: please lock this topic!"
}
{
  "Tag": [
    "quadratics",
    "function"
  ],
  "Problem": "A merchant has determined that if the selling price of a peach is 15 cents, he will sell 400 each day. He has also determined that for each cent the price is increased, he will sell 16 fewer peaches each day. At what price in cents, should he sell each peach in order to maximize the daily income from peaches?\r\n\r\nI got the answer from guessing and checking, but when I read the formula in the solutions: p(640-16p) = 640p - 16p^2, I didn't completely understand.\r\n\r\nHelp, please and thanks :]",
  "Solution_1": "If he increases the price by $ x$ cents, then the amount of income the merchant will get is $ (15\\plus{}x)(400\\minus{}16x)$.  We wish to maximize this quadratic, which occurs at $ x\\equal{}5$, so $ 15\\plus{}5\\equal{}\\boxed{20}$.\r\n\r\nSomeone wish to explain?",
  "Solution_2": "the reason the maximum occurs at x=5 is when you expand the quadratic you get\r\n\r\n$ \\minus{}16x^2\\plus{}160x\\plus{}6000$\r\n\r\nthe maximum of a negative parabolic function (or the minimum of a positive one) occurs at $ x\\equal{}\\minus{}\\frac{b}{2a}$ so $ x\\equal{}\\minus{}\\frac{160}{(\\minus{}16)}\\equal{}\\boxed{5}$"
}
{
  "Tag": [
    "function",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "In a comutative ring with n^2 elements, n>=2 there exists at most n-2 divisors of 0.Show that the ring is a field.",
  "Solution_1": "Unfortunately, I published this problem in Recreatii Matematice when I was in the 12-th form and I didn't know allmost anything about superiour algebra. So, I didn't know it is well-known and I didn't know it is easy. I just had an idea I thought is nice. But I was wrong. So, the source should be: well-known.",
  "Solution_2": "Does this work?\r\n\r\nWe know that in a finite ring an element is not invertible iff it's a divisor of 0 (even if the ring isn't commutative, but that's just a little bit harder to prove), so in our ring there are at least n<sup>2</sup>-n+1 invertible elements. Let's look at the set x*U(A), where x is a divisor of 0. All the elements of the set are divisors of 0, so |x*U(A)|<=n-2. This means that there is a set M included in U(A) s.t. |M|>=(n<sup>2</sup>-n+1)/(n-2)>n+1 (so |M|>=n+2) and x*M has only one element. Let the elements of M be a<sub>i</sub> with i from 1 to |M|. The elements a<sub>i</sub>-a<sub>1</sub> with i from 2 to |M| are distinct divisors of 0, but there are at least n+1 such elements, so we have a contradiction (because there are at most n-2<n+1 divisors of 0).\r\n\r\nThis shows that there are no divisors of 0 (x, in our case), so the ring is a field.\r\n\r\nI think that with some effort, this could have been proved for non-commutative rings as well, because in all finite rings we have the equivalence a non-invertible iff a divisor of 0 and in all such rings if a is a divisor of 0 and b is invertible then ab is a divisor of 0.",
  "Solution_3": "What is the problem for the non-commutative rings? :)",
  "Solution_4": "Huh? What do you mean?",
  "Solution_5": "What part of your proof is not valid if the ring is not commutative?",
  "Solution_6": "Oh, they're all valid, but the equivalence a non-invertible <=> a divisor of 0 is a bit trickier. \r\n\r\nWe consider the function f:A->A, f(x)=ax. Let's assume a is not a divisor of 0. This means that f is injective, but f is injective iff it?s surjective, so there is an x\\in A s.t. ax=1. Let M be the set of all x\\in A s.t. ax=1. If x\\in M then x(a+1)-1\\in M. Since f is injective, x=x(a+1)-1, so xa=1=ax, so a is invertible. Conversely, if a is invertible, then it's clearly not a divisor of 0.\r\n\r\nThis shows that the proof works fine for any ring with n<sup>2</sup> elements."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "For every n positive reals $a_i (i=1,..,n)$ such that $a_1^2+a_2^2+..+a_n^2=1$ prove that\r\n\r\n$a_1+a_2+...+a_n +\\frac{1}{a_1a_2....a_n} \\geq \\sqrt{n} +(\\sqrt{n})^n$",
  "Solution_1": "First of all I will prove a beautiful lemma (which I cheated from Romanian magazine):\r\n\r\nWe have  e \u2264 a \u2264 b \u2264 f  and we want to prove e+f \u2265 a+b.\r\n\r\n(e-a)(f-a) \u2264 0  => a<sup>2</sup>+ef \u2264 a(e+f)\r\n(e-b)(f-b) \u2264 0   => b<sup>2</sup>+ef \u2264 b(e+f)\r\n\r\nSumming : a<sup>2</sup> + b<sup>2</sup> + 2ef  \u2264 (e+f)(a+b)\r\n\r\nHence if we prove ef  \u2265 ab then (e+f) \u2265 (a+b) \r\n and if we prove e+f \u2264 a+b then ef \u2264 ab.\r\n\r\nSo we have by AM-GM inequality we have:\r\n(a1*a2 an) \u2264 (\u2211ai)^n / n^n \u2264 1/ (\u221an)^n\r\n\r\n(\u2211ai) \u2264  (\u221an) \u2264 (\u221an)^n \u2264 1/(a1*a2 an)\r\n\r\nBy our lemma it is enough to prove:\r\n(\u2211ai)/(a1*a2 an) \u2265 (\u221an)^(n+1)\r\n\r\nBut (\u2211ai)/(a1*a2 an) \u2265 n/(a1*a2 an)^[(n-1)/n] \u2265 (\u221an)^(n+1)\r\n\r\nOr again (\u221an) \u2264 1/(a1*a2 an)^n q.e.d",
  "Solution_2": "Very nice proof.",
  "Solution_3": "I have another proof\r\nI will use this Theorem \r\n${p_1}{a_1}+{p_2}{a_2}+..+{p_n}{a_n} \\ge (\\sum p_i)({a_1}^{p_1}+{a_2}^{p_2}+..+{a_n}^{p_n})^\\frac{1}{\\sum p_i}$  for positive reals $a_i,p_i$\r\nSo rewrite the inequality \r\n$a_1+a_2+..+a_n+\\sqrt{n}^{n+1}\\frac{1}{a_1a_2..a_n\\sqrt{n}^{n+1}}  \\ge   \\sqrt{n}+\\sqrt{n}^n$\r\nUsing the Theorem in the left hand side of the inequality we have\r\n$a_1+a_2+..+a_n+\\sqrt{n}^{n+1}\\frac{1}{a_1a_2..a_n\\sqrt{n}^{n+1}} \\ge (n+\\sqrt{n}^{n+1})\\left(\\frac{\\prod a_i}{(\\prod a_i)^{\\sqrt{n}^{n+1}}\\sqrt{n}^{(n+1)\\sqrt{n}^{(n+1)}}}\\right)^\\frac{1}{n+\\sqrt{n}^{(n+1)}}$\r\nNow we have to prove that \r\n$\\left(\\frac{\\prod a_i}{(\\prod a_i)^{\\sqrt{n}^{n+1}}\\sqrt{n}^{(n+1)\\sqrt{n}^{(n+1)}}}\\right)^\\frac{1}{n+\\sqrt{n}^{(n+1)}}$  $\\ge\\left(\\frac{1}{\\sqrt{n}}\\right)$\r\nUsing the fact that $\\left(\\frac{1}{\\sqrt{n}}\\right)^n \\ge \\prod a_i $ which holds right away from the condition of the problem  one can easily prove the remaining part of the inequality. Of course Prowler's solution is much more easier but i really hope anybody reads mine     ;)",
  "Solution_4": "I hope first inequality is \r\n p1*x1 + p2*x2 +...+pn*xn \u2265 (\u2211pi)*(x1^p1 +x2^p2+...+xn^pn)\r\nSolution is very beautiful, my congratulations.",
  "Solution_5": "Andrei proof is a very nice application of AM-GM generalized inequality.\r\n\r\nSorry, Prowler but I don't understand your post.\r\n\r\nI think that Andrei initial theorem is correct.",
  "Solution_6": "I am sleeping :blush: \r\nI meant p1*x1 + p2*x2 +...+pn*xn \u2265 (\u2211pi)*(x1^p1*x2^p2*...*xn^pn)^1/(\u2211pi)\r\nbecause \r\n[quote=\"Andrei\"]I have another proof\n\n${p_1}{a_1}+{p_2}{a_2}+..+{p_n}{a_n} \\ge (\\sum p_i)({a_1}^{p_1}+{a_2}^{p_2}+..+{a_n}^{p_n})^\\frac{1}{\\sum p_i}$  for positive reals $a_i,p_i$\n\nand uses it in a such way:\n\n$a_1+a_2+..+a_n+\\sqrt{n}^{n+1}\\frac{1}{a_1a_2..a_n\\sqrt{n}^{n+1}} \\ge (n+\\sqrt{n}^{n+1})\\left(\\frac{\\prod a_i}{(\\prod a_i)^{\\sqrt{n}^{n+1}}\\sqrt{n}^{(n+1)\\sqrt{n}^{(n+1)}}}\\right)^\\frac{1}{n+\\sqrt{n}^{(n+1)}}$\n[/quote]\r\n   Or I misunderstand solution.",
  "Solution_7": "[quote=\"prowler\"]First of all I will prove a beautiful lemma (which I cheated from Romanian magazine):\n\nWe have  e \u2264 a \u2264 b \u2264 f  and we want to prove e+f \u2265 a+b.[/quote]\r\nIf $ef \\geq ab$, of course."
}
{
  "Tag": [
    "calculus",
    "integration",
    "inequalities",
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "I am looking for a simple proof of the following inequality:\r\n\r\n${ \\left\\lgroup\\int_{x=a}^{b}\\left\\lgroup \\int_{y=c}^{d}f(x,y)\\,dy\\right\\rgroup^{p}\\,dx\\right\\rgroup^{1/p}\\leq \\int_{y=c}^{d}\\left\\lgroup \\int_{x=a}^{b}f(x,y)^{p}\\,dx\\right\\rgroup^{1/p}\\,dy}$\r\n\r\nwhere $p>1$. If summation sign is used instead of integral sign, this is Minkowski Inequality. Where can I find a proof?  :oops:",
  "Solution_1": "I had this on a problem set in analysis -- the proof there went via Holder, although of course then you would have to prove the Holder integral inequality first.  (All together, than took about 2 pages of official solution from the professor.)",
  "Solution_2": "It goes by the name \"Minkowski integral inequality.\"\r\n\r\nOne way to prove it: \"linearize\" it by making usue of the following fact:\r\n\r\n$\\|f\\|_{p}=\\sup_{\\|g\\|_{p'}=1}\\int fg$\r\n\r\nwhere $p'$ is the dual exponent, $\\frac1p+\\frac1{p'}=1.$",
  "Solution_3": "What is the equality condition for the inequality?\r\nThe following link gives the prove without the equality condition:\r\n\r\nhttp://calclab.math.tamu.edu/~sivan/math663_06a/appendix2.pdf\r\n :roll:",
  "Solution_4": "Here's the complete result:\r\n\r\n$(a)~~{ \\left\\lgroup\\int_{x=a}^{b}\\left\\lgroup \\int_{y=c}^{d}f(x,y)\\,dy\\right\\rgroup^{p}\\, dx\\right\\rgroup^{1/p}\\leq \\int_{y=c}^{d}\\left\\lgroup \\int_{x=a}^{b}f(x,y)^{p}\\,dx\\right\\rgroup^{1/p}\\,dy }$\r\n\r\n$(b)~~{ \\left\\lgroup\\int_{x=a}^{b}\\left\\lgroup \\int_{y=c}^{d}f(x,y)\\,dy\\right\\rgroup^{p}\\, dx\\right\\rgroup^{1/p}= \\int_{y=c}^{d}\\left\\lgroup \\int_{x=a}^{b}f(x,y)^{p}\\,dx\\right\\rgroup^{1/p}\\,dy ~~\\Leftrightarrow~~\\frac{f(x,y)}{f(x,z)}=\\frac{f(u,y)}{f(u,y)}}$\r\n\r\n :P"
}
{
  "Tag": [
    "search",
    "algebra solved",
    "algebra"
  ],
  "Problem": "Find $ f:R--->R $ satisfying :\r\n $ f(f(x-y)) = f(x)-f(y)+f(x)f(y)-xy $",
  "Solution_1": "f(x)=x or sqrt(x^2+2)?",
  "Solution_2": "Can you post your solution ?",
  "Solution_3": "Put x=y, \r\nf(x)^2 - x^2 = f(f(0)) = C (a constant)\r\nThen f(x) = sqrt(x^2+C) or -sqrt(x^2+C)\r\nThen substitute some values of x,y and it can be found that C=0 or 2",
  "Solution_4": "Who are you , Romano ?\r\n This problem is rather easy : \r\n  Take $ x=y --> f(f(0)) = f^2(x)-x^2 $\r\n  $ ----> f(f(0))=f^2(0) $\r\n  Take $ x=f(0) ---> f(0)=0 $ or $ f(0)=\\sqrt2 $\r\n After that , this problem will be easy !!!!",
  "Solution_5": "But if for all $x$, we have $f(x) = \\sqrt {x^2 + 2}$ or $f(x) = x$, it does not mean that we have \r\n$f(x) = x$ for all $x$\r\nor\r\n$f(x) = \\sqrt {x^2 + 2}$ for all $x$.\r\n\r\nPierre.",
  "Solution_6": "So is the problem completely solved? ;)  :D",
  "Solution_7": "No.\r\n\r\nPierre.",
  "Solution_8": "[quote=\"pbornsztein\"]No.\n\nPierre.[/quote]\r\n\r\ndo u mind posting ur solution..?",
  "Solution_9": "I didn't search this problem. I was only reading the proposed answers.\r\n\r\nPierre.",
  "Solution_10": "I hope this is ok.\r\nAssume f(0)=0, then\r\nf(x)^2=x^2\r\nPutting y=0 we have\r\nf(f(x))=f(x)\r\nso f(x-y)=f(x)-f(y)+f(x)f(y)-xy\r\nPut x=0 we have f(-y)=-f(y)\r\nSo f(x+y)=f(x)+f(y)-f(x)f(y)+xy\r\nSum them up and squaring\r\nf(x-y)^2+f(x+y)^2+2f(x-y)f(x+y)=4x^2\r\nf(x-y)/(x-y) * f(x+y)/(x+y)=1\r\nIf we take f(x-y)=x-y, then f(x+y)=x+y\r\nSo f(x)=x for all reals, when we take y tends  to 0.\r\nSimilarly we can have f(x)=-x for all reals too.\r\n\r\nfor this case is it ok..?",
  "Solution_11": "Case 1: Suppose $f(x) = 0$.  It's easy to check that $f(x) = -x \\forall x \\in \\mathbb{R}$ is not a solution, and that $f(x) = x \\forall x \\in \\mathbb{R}$ is.  Select $a \\neq 0$ such that $f(a) = a$, and suppose there is $b \\neq 0$ such that $f(b) = -b$.  Let $x = a, y = b$ in the original equation, then we find that $a=1$.  This means that $f(x) = -x \\forall x \\neq 0, 1$.  But now, setting $x \\neq 0, 1, y = 1$, it's easy to check that this does not work.\r\n\r\nThe other case looks...ugly :( .",
  "Solution_12": "Case 2: $f^2(x) = x^2+2$.\r\n\r\nSuppose that there exists an $a > 0$ such that $f(a) = f(-a)$.  Substituting $x = -y = a$ into the original equation, we obtain $\\sqrt{4a^2 + 4} = 2a^2+2$ (observe that the LHS cannot be $- \\sqrt{4a^2 + 4}$.  But this implies that $a = 0$.  Hence $f(-x) = -f(x) \\forall x \\neq 0$.\r\n\r\nCase 2.1: Suppose that $f(a) = \\sqrt{a^2+2}$ for some $a \\neq 0$.  Let $x = -y = a$, then we get $\\pm \\sqrt{a^2+1} = \\sqrt{a^2+2} - 1$.  It's easy to verify that there are no solutions for both $+$ and $-$.\r\n\r\nCase 2.2: Suppose that $f(a) = -\\sqrt{a^2+2}$ for some $a \\neq 0$.  Using the same method, we get $\\sqrt{a^2+1} = \\sqrt{a^2+2} + 1$, which is even more impossible.",
  "Solution_13": "this case is really ugly..... anyway u solved it :)"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection"
  ],
  "Problem": "particle man is located at $ (2,1)$ and wants to get to the lines $ x\\equal{}3, x\\equal{}0,$ and $ y\\equal{}0$. what is the length of the shortest path he can take?",
  "Solution_1": "Graph and it will seem much, much easeier.",
  "Solution_2": "ive tried reflections",
  "Solution_3": "[hide]Since the lines $ x \\equal{} 0$ and $ x \\equal{} 3$ are $ 3$ units apart and particle man is located $ 1$ unit away from the nearest of these lines, he must move a total of at least $ 4$ units in the $ x$-direction. \nSince he is located $ 1$ unit away from $ y \\equal{} 0$, he must move a total of at least $ 1$ unit in the $ y$-direction. \nThus, the path must be at least $ \\sqrt {4^2 \\plus{} 1^2} \\equal{} \\sqrt {17}$ units long. \n\nReflect $ (2,1)$ over $ x \\equal{} 3$ to $ (4,1)$. Connecting $ (0,0)$ to $ (4,1)$ and reflecting the part such that $ x > 3$ over the line $ x \\equal{} 3$. \nThis gives $ (0,0)$ to $ (3,\\frac {3}{4})$ to $ (2,1)$ which has a total length of $ \\sqrt {17}$.[/hide]"
}
{
  "Tag": [
    "probability",
    "counting",
    "derangement",
    "search",
    "function",
    "probability and stats"
  ],
  "Problem": "A person writes $ n$ mails and he also writes addresses on $ n$ envelopes. What is the probability that neither one mail will reach its destination? What becomes this probability when $ n \\to \\infty$.",
  "Solution_1": "Read [url=http://en.wikipedia.org/wiki/Derangement]here[/url]. You can find also useful links at the end.",
  "Solution_2": "Or put the word \"derangement\" into the search function of this site to find many past discussions."
}
{
  "Tag": [
    "Ramsey Theory",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There are $ n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $ n<13$.",
  "Solution_1": "[quote=\"KenHungKK\"]There are $ n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $ n < 13$.[/quote]\r\n\r\nI can certainly show that $ n < 17$.  Suppose, to the contrary, that there exists a set of $ n \\equal{} 17$ points such that any triangle is colored with exactly $ 2$ colors.  Consider a given point $ A$, and assume without loss of generality that the color most commonly used in lines connected to $ A$ is red.  Then at least $ 6$ points must be joined with a red line to $ A$.  Any two of these points must be joined with yellow or green (lest we have a triangle with all red lines).  Consider a given point in this subset, $ B$.  Assume without loss of generality that the color most commonly used in lines, within this subset, connected to $ B$ is yellow. Then at least $ 3$ points must be joined with a yellow segment to $ B$.  Any two of these points must be joined with green (lest we have a triangle with all yellow lines).  This gives us a triangle with all green lines, which provides a contradiction.",
  "Solution_2": "NoYoo-hooForMe, what you have proved is essentially the Ramsey number $ N(3,3,3,2) \\le 17$. (But here is it is equal to 17 due to the additional condition that any triangle will have two same colored edges.\r\n\r\nI think this problem is quite hard(atleast for me). Can someone give a hint?"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "ratio"
  ],
  "Problem": "A right circular cone whose base has a radius of 4 cm and whose height is greater than 2 cm is inscribed in a sphere with a radius of 5 cm, as shown. What is the ratio of the volume of the cone to the volume of the sphere? Express your answer as a common fraction",
  "Solution_1": "Draw the radius from the center of the sphere to a point on the circumference of the base of the cone. Then, the distance from the center of the sphere to the center of the base of the cone is $ \\sqrt {5^2 \\minus{} 4^2} \\equal{} 3$, so the height of the cone is 3+5=8. The volume of the sphere is $ \\frac {4}{3}5^3\\pi \\equal{} \\frac {500\\pi}{3}$, and the volume of the cone is $ \\frac {1}{3}4^2\\times 8\\pi \\equal{} \\frac {128\\pi}{3}$, so the desired ratio is $ \\frac {\\frac {128\\pi}{3}}{\\frac {500\\pi}{3}} \\equal{} \\frac {32}{125}$",
  "Solution_2": "Isnt this question from the handbook?"
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all natural numbers $n$ less than 200 such that $n^2+(n+1)^2$ is a perfect square.",
  "Solution_1": "$n=k^2-m^2$ and $n+1=2km$\r\nor\r\n$n=2km$ and $n+1=k^2-m^2$ \r\n\r\nand in each of these cases solving it like the Pell's equation."
}
{
  "Tag": [
    "induction",
    "LaTeX",
    "strong induction",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There are two strips. On the first is written letter $ A$, and on the second letter $ B$. In each minute, John writes all the letters presented on one strip on another (in the same sequence), either to the right or to the left of the existing letters. (For example, if at a moment the two strips are $ BAB$ and $ ABB$, John can write to the right of the first strip to make the two strips $ BABABB$ and $ ABB$.) Prove that at any moment, the letters on each strip form two palindromes, one of which could be empty. (For example, $ BBA$ is the palindrome $ BB$ and the palindrome $ A$.)",
  "Solution_1": "[quote=\"hollandman\"]There are two strips. On the first is written letter $ A$, and on the second letter $ B$. In each minute, John writes all the letters presented on one strip on another (in the same sequence), either to the right or to the left of the existing letters. (For example, if at a moment the two strips are $ BAB$ and $ ABB$, John can write to the right of the first strip to make the two strips $ BABABB$ and $ ABB$.) Prove that at any moment, the letters on each strip form two palindromes, one of which could be empty. (For example, $ BBA$ is the palindrome $ BB$ and the palindrome $ A$.)[/quote]\r\n\r\nPosted before, but nobody wrote a solution.\r\n\r\nTo make the problem statement clearer, John doesn't need to copy the two strips one after another. That is, he can copy the second one to right or left of the first strips multiple times successively.\r\n\r\nThis problem is VERY VERY tricky and hard. I had it solved but can only supply some basic ideas. Several suggestions.\r\n\r\n1.Use strong induction on number of steps\r\n2.At the beginning, Let's say John copy B to the first strip p time left and q time right. If p=q, then the thing on the first strip is still a palindrome and we can look at it as a whole. (i.e. BBABB can be simply denoted as C) If p>q, we have on the first strip like BBBAB. Make it into a palindrome and several B's either to left or right. So BBBAB becomes BBC, or BBABBBB becomes CBB. After that, John should start to copy the thing on the first strip to the other one at least once. At this point, things are getting complex. However, the main goal is to show that several steps later, you can always reach a situation that can be achieved using less steps if you change A and B at the beginning to some other palindromes. This is when the strong induction applies.",
  "Solution_2": "Hmm.. I would appreciate it if you could write a complete solution. I'm trying to understand your basic ideas, but they seem complicated.",
  "Solution_3": "Sorry, I will try to explain some detail when I have time. However, since I have few knowledge regarding Latex, it will be very hard for me to draw a diagram.\r\n\r\nHave you at least understand all the thing before the \"After that\" in my second point? If so, just start copy things on the first strip to the second strip and you might see some magic happening."
}
{
  "Tag": [
    "modular arithmetic",
    "algebra",
    "function",
    "domain",
    "quadratics",
    "Ring Theory",
    "algorithm"
  ],
  "Problem": "Let $p$ be a prime number, and suppose that $x$ is an integer such that $x^2 \\equiv -2 \\ mod p$.By cnsidering $u+xv$ for various pairs of $(u,v)$.show that at least one of the equations $a^2+2b^2=p, a^2+2b^2=2p$has a solution.",
  "Solution_1": "First, note that $x^2 + 2\\cdot 1^2 = mp$ for some positive integer $m$. \r\n\r\nChoose $u,v$ from the interval $[-m/2,m/2]$ so that $u\\equiv x \\pmod m$ and $v\\equiv 1\\pmod m$. Then $mr=u^2+2v^2$ where \\[r=\\frac{1}{m}(u^2+2v^2)\\leq \\frac{1}{m}(\\frac{m^2}{4}+\\frac{m^2}{2})<m.\\] Combine $mp=x^2 + 2\\cdot 1^2$  and $mr=u^2+2v^2$ to get $m^2rp=(xu+2v)^2 + 2(xv-u)^2$, where both summands are divisible by $m$. Hence,\r\n\\[rp=\\left(\\frac{xu+2v}{m}\\right)^2 + 2\\left(\\frac{xv-u}{m}\\right)^2.\\] We decreased $m$ to $r$. Similarly we can decrease $r$ farther until $r=1$ producing desired representation of $p$.",
  "Solution_2": "Actually, $a^2+2b^2=p$ always has a solution. \r\n\r\nThis is because $A=\\mathbb Z[\\sqrt{-2}]$ is a unique factorization domain, and since $p|x^2+2=(x+\\sqrt{-2})(x-\\sqrt{-2})$, but it doesn't divide either one of $x+\\sqrt{-2},x-\\sqrt{-2}$, it means that $p$ is not a prime in $A$, which, in turn, means that it's decomposable: we can write it as $(a+b\\sqrt{-2})(c+d\\sqrt{-2})$. \r\n\r\nNow, since $a+b\\sqrt{-2}|p$, we also have $a-b\\sqrt{-2}|p$, and the only possibility left is $a-b\\sqrt{-2}=c+d\\sqrt{-2}$, because both $a+b\\sqrt{-2}, c+d\\sqrt{-2}$ must be primes in $A$ (their norm is $p$, which is a prime in $\\mathbb Z$).",
  "Solution_3": "Nice, these quadratic problems are given all the time repeatedly and the solutions are always the same :).",
  "Solution_4": "[quote=\"maxal\"]We decreased $m$ to $r$. Similarly we can decrease $r$ farther until $r=1$ producing desired representation of $p$.[/quote]\r\nI think there is wrong wth you because you say that always $a^2+ 2b^2=p$ has solution but the problem say another thing. :?",
  "Solution_5": "Maxal and grobber are right! Even if you only know the statement of the problem, on can prove that there are already $a,b$ with $a^2 +2 b^2=p$:\r\n\r\nLet be $c^2 + 2a^2 =2p$ for some $a,c$, then $c(=2b)$ is even. But this gives immediatly $4b^2+2a^2=2p$ or equivalently $a^2+2b^2=p$.",
  "Solution_6": "oh yes you are right ;)",
  "Solution_7": "[quote=\"grobber\"]Actually, $a^2+2b^2=p$ always has a solution. \n\nThis is because $A=\\mathbb Z[\\sqrt{-2}]$ is a unique factorization domain, and since $p|x^2+2=(x+\\sqrt{-2})(x-\\sqrt{-2})$, but it doesn't divide either one of $x+\\sqrt{-2},x-\\sqrt{-2}$, it means that $p$ is not a prime in $A$, which, in turn, means that it's decomposable: we can write it as $(a+b\\sqrt{-2})(c+d\\sqrt{-2})$. \n\nNow, since $a+b\\sqrt{-2}|p$, we also have $a-b\\sqrt{-2}|p$, and the only possibility left is $a-b\\sqrt{-2}=c+d\\sqrt{-2}$, because both $a+b\\sqrt{-2}, c+d\\sqrt{-2}$ must be primes in $A$ (their norm is $p$, which is a prime in $\\mathbb Z$).[/quote]\r\n\r\nhow do you know if $\\mathbb{Z}[\\sqrt{D}], D \\in \\mathbb{Z}$ is a UFD? Do you have to try to prove a Euclidean algorithm for each particular case?",
  "Solution_8": "[quote]Do you have to try to prove a Euclidean algorithm for each particular case?[/quote]\r\n\r\nNo, because most numbers fields with class number 1, that is, with ring of integers which is a UFD, are not Euclidean.  For example, Barnes and Swinnerton Dyer in two papers in [i]Acta Math.[/i] (1952) showed that $Q(\\sqrt{97})$ is not Euclidean.  Davenport, Hua and Inkeri proved that there are only 22 quadratic Euclidean fields.\r\n\r\nA way of showing that a quadratic field has class number 1 is by the theory of binary forms.  An ideal $A$ in a number ring $R$ in a quadratic field is a rank 2 $Z$-module, generated by $a_1,a_2$ say, and $N(a_1X + a_2Y)$ is a quadratic form.  The $SL_2(Z)$-classes of forms correspond to the ideal classes, and the theory of reduction gives an effectively computable way of listing the reduced forms, which are canonical representatives of the classes.  If there is only one reduced form, then the ring is a PID."
}
{
  "Tag": [],
  "Problem": "In how many ways can the letters of the word AFFECTION be arranged, keeping the vowels\r\nin their natural order and not letting the two F\u2019s come together?",
  "Solution_1": "[quote=\"mdk\"]In how many ways can the letters of the word AFFECTION be arranged, keeping the vowels\nin their natural order and not letting the two F\u2019s come together?[/quote]\r\n\r\n\r\nwhen you mean vowels in natural order do you mean A has to be 1st or a vowel has to be first?",
  "Solution_2": "[hide]\nIs it $ \\frac{5!}{2!}$ or $ 60$?[/hide]",
  "Solution_3": "I have the solution :D\r\n\r\n[hide=\"Solution\"]\n\nOk, the first thing I will consider is the two F's:\n\nThere are 9 different places for these F's to be placed. First consider the placing of the first F:\n\nThere 2 different types of places, the ends, and all the places in between. For the 2 ends, there are 7 valid places left over to place the second F, for the 7 places in between, there are 6 places left over to place the second F. This gives the following amount of combinations:\n\n$ 2*7+7*6 = 56$\n\n(This is the edited bit:)\nHOWEVER, here the 2 F's have been considered as different letters! This means that this calculation has counted each combination twice. So I will simply divide by 2:\n\n$ (2*7+7*6)/2 = 28$\n\nAfter the 2 F's are placed, there are 7 spaces left to arrange the remaining letters, and for each arrangement in the seven places, there will be 28 corresponding combinations for application to all the different F placements.\n\nNext I'll consider the vowels in the 7 remaining positions:\n\nIf I first consider every combination that the 4 vowels can be placed, there will be:\n\n$ \\frac{7!}{(7-4)!}$\n\ncombinations. However for each different selection of 4 places to place each vowel, there will be $ 4!$ different ways of placing the vowels in those 4 places, but only 1 of them is right. It follows that only $ \\frac{1}{4!}$ of the first combinations is valid. This means there remains the following amount of combinations:\n\n$ \\frac{7!}{4!(7-4)!}= 35$\n\nThis means there are:\n\n$ 28*35 = 980$\n\nDifferent ways of arranging the F's and the vowels, and there are 3 places left for placing the 3 remaining letters. Now the number of ways of placing the 3 remaining places in the 3 remaining positions is simple:\n\n$ 3! = 6$\n\nThis means the total number of combinations is:\n\n$ \\frac{(2*7+7*6)}{2}* \\frac{7!}{4!(7-4)!}* 6$\n$ = 28*35*6$\n$ = 5880$\n\n[/hide]",
  "Solution_4": "I made a mistake!\r\n\r\nIn the above post, I forgot to divide by 2 in the F's section, as the method I described considers the 2 F's as different letters, counting each combination twice.\r\n\r\nPS: I've now edited the post above."
}
{
  "Tag": [
    "function",
    "parameterization",
    "algebra unsolved",
    "algebra",
    "Fixed point"
  ],
  "Problem": "Suppose that $ u$ is a real parameter with $ 0<u<1$. Define $ f(x)\\equal{}0$ if $ 0 \\le x \\le u$, and $ f(x)\\equal{}1\\minus{}(\\sqrt{ux}\\plus{}\\sqrt{(1\\minus{}u)(1\\minus{}x)})^2$ if $ u \\le x \\le 1$, and define the sequence $ u_n$ recursively by $ u_1\\equal{}f(1)$ and $ u_n\\equal{}f(u_{n\\minus{}1})$ for all $ n >1$. Show that there exists a positive integer $ k$ for which $ u_k\\equal{}0$.",
  "Solution_1": "[quote=\"moldovan\"]Suppose that $ u$ is a real parameter with $ 0 < u < 1$. Define $ f(x) \\equal{} 0$ if $ 0 \\le x \\le u$, and $ f(x) \\equal{} 1 \\minus{} (\\sqrt {ux} \\plus{} \\sqrt {(1 \\minus{} u)(1 \\minus{} x)})^2$ if $ u \\le x \\le 1$, and define the sequence $ u_n$ recursively by $ u_1 \\equal{} f(1)$ and $ u_n \\equal{} f(u_{n \\minus{} 1})$ for all $ n > 1$. Show that there exists a positive integer $ k$ for which $ u_k \\equal{} 0$.[/quote]\r\n\r\nThe key here is to see that $ f(x)<x$ $ \\forall x>0$. So $ u_n$ is a positive non increasing sequence, so converging towards the only fixed point of $ f(x)$, so $ 0$. \r\nSo, for $ n$ great enough, $ u_n\\leq u$ and $ u_{n\\plus{}1}\\equal{}0$"
}
{
  "Tag": [],
  "Problem": "[b]In a class 54 % of the student are girls and 16 % are boys.If the number of girls is 27 , find the total number of student ?[/b]",
  "Solution_1": "where did the other 30% go?\r\nAs if there is none, this is a FAILED question  :(",
  "Solution_2": "The (fixed) problem (I'm guessing) is here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302062",
  "Solution_3": "Didn't you make this thread twice ra ny?\r\n\r\nUh 30% are neither girls nor boys??"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove :  equation $ x^2\\minus{}2y^2\\plus{}8z^3\\equal{}3$ don't have solution $ x,y,z\\in N$",
  "Solution_1": "Check both sides $ mod \\;\\; 4$ we have $ x\\equal{}2a\\plus{}1, y\\equal{}2b\\plus{}1$ are odd.\r\nSo the equation becomes $ 4a^2\\plus{}4a\\minus{}8y^2\\minus{}8y\\equal{}4 \\Leftrightarrow a(a\\plus{}1)\\minus{}2y^2\\minus{}2y\\equal{}1$\r\n$ L.H.S.$ is even but $ R.H.S.$ is odd. So there is no solution.",
  "Solution_2": "please post this in the pre olympiad section"
}
{
  "Tag": [
    "linear algebra",
    "matrix",
    "linear algebra unsolved"
  ],
  "Problem": "Solve the following matrix system: $\\{\\begin{array}{cc}AX-BY=C \\\\ BX-AY=C \\end{array}$ where $A,B,C,X,Y\\in\\mathcal l M_{n}(R)$ and $A+B,A-B$ are not singular matrix.",
  "Solution_1": "No one? C`mon guys :D !",
  "Solution_2": "So  let`s see .... by adding the two equations we get $(A+B)(X-Y)=2C \\rightarrow X-Y=2C(A+B)^{-1} \\rightarrow X=Y+2C(A+B)^{-1}$ and by substracting one from another we get:\r\n$(A-B)(X+Y)=O_n\\rightarrow (A-B)[2Y+2C(A+B)^{-1}]=O_n$ \r\n$\\rightarrow (A-B)Y=-(A-B)C(A+B)^{-1}$\r\n$\\rightarrow Y=-C(A+B)^{-1} \\rightarrow X=C(A+B)^{-1}$",
  "Solution_3": "Well done ;)!"
}
{
  "Tag": [],
  "Problem": "How many factors of 1800 are multiples of 10?",
  "Solution_1": "I'm not so smart, so I did it in probably the longest way possible. :D \r\n\r\n[hide]Anyway, I listed all of the factors and came out with these:\n\n1800: 1, [color=red]1800[/color], 2, [color=red]900[/color], 3, [color=red]600[/color], 5, [color=red]360[/color], 6, [color=red]300[/color], 9, [color=red]200[/color], [color=red]10[/color], [color=red]180[/color], 18, [color=red]100[/color], [color=red]20, 90, 30,  60, 50[/color], 36, [color=red]60, 30, 90, 20, 100,[/color] 18, [color=red]200[/color], 9, [color=red]300[/color], 6, [color=red]600[/color], 3, [color=red]900[/color], 2. That was all I came up with. :( \n\nThen I counted up the factors:\n1800, 900, 600, 360, etc.\n\nFinal Answer: 23 factors are multiples of ten.\n\nPlease tell me if this is wrong, thank you...also tell me any shortcut way[/hide]",
  "Solution_2": "YOu counted numbers like 600 and 900 twice, and there's definetly a better solution (should just take 35 seconds max) with prime factorization.",
  "Solution_3": "Oops...sorry.  :blush: \r\n\r\n[hide]Here's my answer now:\n\nAfter a close examination of my solution, I found out I had repeated a LOT of numbers.\n\nNow my answer is 14 factors.\n\nAnd also, I seriously need to know the 35 seconds way.[/hide]",
  "Solution_4": "No, it is incorrect. This is one of two ways to solve this that I know (without listing). Contribute other ways. \r\n\r\n1800 = 180*10. \r\nSo any factor of 180 times 10 is a factor of 1800 (and a multiple of 10)!. \r\nWe need to find the number of factors in 180. \r\n\r\n180 = $2^2\\cdot3^2\\cdot5$. That means 3*3*2 = 18 factors.",
  "Solution_5": "[hide=\"woot!!!\"]\ni found all the factors of 180 and then multiplied them by 10.  There are 18 factors. ( i don't know how to do this by prime factorization).  So the final answer is [b]28[/b][/hide]",
  "Solution_6": "[quote=\"236factorial\"]How many factors of 1800 are multiples of 10?[/quote]\r\n\r\nUsing prime factorizaton, i get 18",
  "Solution_7": "[hide]Well $1800=2^3\\cdot3^2\\cdot5^2$  We have to have at least one $2$ and one $5$, so from the exponents we have $3\\cdot3\\cdot2=18$[/hide]",
  "Solution_8": "[quote=\"math92\"][hide=\"woot!!!\"]\ni found all the factors of 180 and then multiplied them by 10.  There are 18 factors. ( i don't know how to do this by prime factorization).  So the final answer is [b]28[/b][/hide][/quote]\r\n\r\nDid you make a typo? I can't see where 28 came from.",
  "Solution_9": "[quote=\"236factorial\"][quote=\"math92\"][hide=\"woot!!!\"]\ni found all the factors of 180 and then multiplied them by 10.  There are 18 factors. ( i don't know how to do this by prime factorization).  So the final answer is [b]28[/b][/hide][/quote]\n\nDid you make a typo? I can't see where 28 came from.[/quote]\noops, :blush: \nthat was a typo\n[hide]18 is the real answer[/hide]",
  "Solution_10": "[quote=\"math92\"][quote=\"236factorial\"][quote=\"math92\"][hide=\"woot!!!\"]\ni found all the factors of 180 and then multiplied them by 10.  There are 18 factors. ( i don't know how to do this by prime factorization).  So the final answer is [b]28[/b][/hide][/quote]\n\nDid you make a typo? I can't see where 28 came from.[/quote]\noops, :blush: \nthat was a typo\n[hide]18 is the real answer[/hide][/quote]\r\n\r\ni dont think so, 1800 has 18 different factors , but the factors thats a multiple of 10 is not 18 ( 6 for example isnt a multiple of 10 :s)",
  "Solution_11": "[quote=\"Mo\u00b2\"]i dont think so, 1800 has 18 different factors , but the factors thats a multiple of 10 is not 18 ( 6 for example isnt a multiple of 10 :s)[/quote]\r\nNo, $1800=2^33^25^2$ $4\\cdot3\\cdot3=36$  There has to be a $2$ and a $5$ to make a $10$, so you can have $1,2,3,$ 2's, $0,1,2$ 3's, and $1,2$ 5's.  $3\\cdot3\\cdot2=18$",
  "Solution_12": "I got it!:)",
  "Solution_13": "[quote=\"loohcsnuf\"]I got it!:)[/quote]\r\nthen what did you get?",
  "Solution_14": "[quote=\"loohcsnuf\"]I got it!:)[/quote]\r\nwhat did you get?"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "trigonometry",
    "algebra solved"
  ],
  "Problem": "$ f(x)=a_1cos x + b_1sinx + a_2 cos (2x) + b_2 sin (2x) $\r\n Prove that : If $ f(x) \\leq \\sqrt{a_2^2+b_2^2} $ then $ a_1=b_1=0 $",
  "Solution_1": "Let $\\theta$ satisfies $a_{2}\\cos 2\\theta+b_{2}\\sin 2\\theta=\\sqrt{a_{2} ^2+b_{2} ^2}$.\r\nSubstitute $x=\\theta$ ,and $x=\\theta+\\pi$ and you will get it.\r\n\r\nNo war\r\n\r\n[i]Edited by Myth[/i]",
  "Solution_2": ":D Sorry, the proof is not enough  :D\r\n :cool:     Try again !  :cool:",
  "Solution_3": "Exist $a,b$ st:\r\n$f(x)=usin(x+a)+vcos(2x+b)$\r\nwith $u=\\sqrt{a_1^2+b_1^2}$\r\n       $v=\\sqrt{a_2^2+b_2^2}$\r\n$f(x)\\le v<->v(1-cos(2x+b))-usin(x+a) \\ge 0$ for all $x$\r\n$<->2vsin^2(x+\\frac{b}{2})+usin(x+a+pi) \\ge 0$ for all $x$\r\nthen exist $c$:\r\n$2vsin^2(x-c)+usin(x) \\ge 0$ for all $x$.\r\nwe can assume that $-pi \\le c \\le pi$\r\nFor $x=c->sinc \\ge 0 ->0\\le c \\le pi$\r\nFor $x=c-pi->sin(c-pi)\\ge0->0\\le c-pi\\le pi$\r\nThen $c=0,pi,-pi$\r\n$->2vsin^2(x)+usinx \\ge 0 for all x$\r\n$->u=0$"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "prove that  for prime $ p > 2$ , $ p|[(2 \\plus{} \\sqrt {5})^p] \\minus{} 2^p *2^1$",
  "Solution_1": "We have the question \r\n$ \\Leftrightarrow p|(2\\plus{}\\sqrt{5})^p\\plus{}(2\\minus{}\\sqrt{5})^p\\minus{}2^p \\times 2$\r\n$ \\equal{}2(_p C_2 2^{p\\minus{}2}(5)\\plus{}_p C_4 2^{p\\minus{}4}(5^2)\\plus{}...\\plus{}_p C_{p\\minus{}1} 2^{1}(5^{\\frac{p\\minus{}1}{2}})$\r\n\r\nObviously, $ _p C_k\\equal{}\\frac{p!}{(p\\minus{}k)!(k)!}$, \r\nso for $ 1 \\le k \\le p\\minus{}1$, $ p|nominator$ but $ p \\nmid denominator$ $ \\Rightarrow p|_p C_k$, \r\nSo $ p|[(2 \\plus{} \\sqrt {5})^p] \\minus{} 2^p *2^1$, done.",
  "Solution_2": "stephencheng i think this is what you meant\r\n$ (2 \\plus{} \\sqrt{5})^p \\equal{}$ integer say $ I$ + some number $ k$ where $ k$ is a number $ >0,<1$\r\nso gif of this $ (2 \\plus{} \\sqrt{5})^p$  is $ I$ and  \r\n$ (2 \\minus{} \\sqrt{5})^p$ is a negative fraction say $ h$\r\nadding both we get  \r\n$ (2 \\plus{} \\sqrt{5})^p \\minus{} (2 \\minus{} \\sqrt{5})^p \\equal{}I \\plus{} k \\plus{} h$ but\r\n$ (2 \\plus{} \\sqrt{5})^p \\minus{} (2 \\minus{} \\sqrt{5})^p$ is an integer (by the binomial expansion we get so)\r\nso $ I\\plus{}k \\plus{}h$ is an integer which implies $ k \\plus{} h$ is an integer \r\nwe have $ 1>k>0 ,\\minus{}1<h<0$ so the only integer satisfying this condition is $ k\\plus{}h\\equal{}0$\r\nwhich implies  $ [(2 \\plus{} \\sqrt{5})^p] \\equal{} [(2 \\plus{} \\sqrt{5})^p \\minus{} (2 \\minus{} \\sqrt{5})^p ]$ ,so if we prove that $ [(2 \\plus{} \\sqrt{5})^p \\minus{} (2 \\minus{} \\sqrt{5})^p ]$ is congruent to $ 2^p *2$\r\nfor which you have given the proof"
}
{
  "Tag": [
    "function",
    "Pascal\\u0027s Triangle",
    "algebra",
    "binomial theorem"
  ],
  "Problem": "Find a combinatorial argument to explain the recurrence\r\n\\[\\binom{n+1}{k}= \\binom{n}{k-1}+k\\binom{n}{k}\\]",
  "Solution_1": "uuuh... $\\binom{3}{2}$ does not equal $\\binom{2}{1}+2\\binom{2}{2}$   ...typo?",
  "Solution_2": "Presumably, they mean $\\binom{n+1}{k}=\\binom{n}{k}+\\binom{n}{k-1}$.\r\n\r\nThis can be explained as follows. Suppose you have to pick k beads from a box, of which 'n' are black and 1 is white (but all are distinguishable). Therefore, the number of ways of doing this is the LHS. However, you can split it into two cases: EITHER you pick the white bead in which case there are $\\binom{n}{k-1}$ ways to choose the remaining black beads, OR you don't pick the white bead, in which case there are $\\binom{n}{k}$ ways to choose your k beads from the black beads. Therefore the LHS = the sum of these two numbers!",
  "Solution_3": "[quote=\"Fraxia\"]Find a combinatorial argument to explain the recurrence\n\\[\\binom{n+1}{k}= \\binom{n}{k-1}+\\binom{n}{k}\\]\n[/quote]\r\n\r\nOr you can use [hide=\"explosives\"] Pascal's Triangle. [/hide]",
  "Solution_4": "Hmm, I copied as it is from ACoPS, it's 6.4.17. No wonder I couldn't find a argument... Not sure if Pascal's identity was meant here, but w/e  :P",
  "Solution_5": "There are many ways to show that $\\binom{n+1}{k}=\\binom{n}{k}+\\binom{n}{k-1}$.\r\n\r\n[hide=\"algebraic method\"]\n$\\binom{n}k+\\binom{n}{k-1}=\\frac{n!}{k!(n-k)!}+\\frac{n!}{(n-k+1)!(k-1)!}$\n$\\frac{n!(n-k+1)+n!\\cdot k}{(n-k+1)!k!}$\n$\\frac{(n+1)!}{(n-k+1)!k!}=\\binom{n+1}{k}$[/hide]",
  "Solution_6": "[quote=\"cincodemayo5590\"]Or you can use [hide=\"explosives\"] Pascal's Triangle. [/hide][/quote]\r\n\r\nBut then you need to prove that the entries in Pascal's triangle are given by ${n \\choose k}$, which would be equivalent to proving this lemma in another way.",
  "Solution_7": "find the $x^{k}$ coefficient of $(x+1)^{n+1}=x*(x+1)^{n}+(x+1)^{n}$\r\n\r\nobviously this relies on the same proof for the binomial theorem, but the function proofs are always cool...this is a fast way to remeber it at least..."
}
{
  "Tag": [],
  "Problem": "A simple pendulum is suspended from a peg on a vertical wall. THe pendulum is pulled away from the wall to the horizontal position and released. Rhe ball hit the wall  and the coeffecient of restitution is 2/sqrt(5) . WHat is the minimum number of collissions after which the amplitude of oscillation becomes less than 60 degrees?",
  "Solution_1": "Is it N=4?",
  "Solution_2": "By the definition of coefficient of restitution, the velocity after each impact is $ \\frac{2}{\\sqrt{5}}$ of the velocity before it. Hence, the Kinetic energy after impact is $ \\frac{4}{5}$ of that before. \r\n\r\nTo make amplitude $ \\leq 60^0$, the maximum potential energy (and hence, the total energy) should be halved. \r\n\r\nObserve that $ \\left( \\frac{4}{5}\\right)^4<\\frac{1}{2}$, and so, $ N\\equal{}4$",
  "Solution_3": "Yeah right....\r\n\r\nNihat i would expect u to post the detailed solution when u post ur answers...\r\n\r\n@shreyas-- nice proof",
  "Solution_4": "One request to all of u visiting this forum just hide the answers u post so that everyone gets equal chance :)",
  "Solution_5": "[quote=\"valeriummaximum\"]Yeah right....\n\nNihat i would expect u to post the detailed solution when u post ur answers...\n\n@shreyas-- nice proof[/quote]\r\n\r\nI don't like to risk.It could also be a wrong answer.  :)",
  "Solution_6": "But Nihat, the reason you should post your solution too, is that if its wrong then we can point out the mistake in it.\r\n\r\nIt's not at all embarassing to post a wrong solution. :) \r\n\r\nThis is a forum for discussion after all :D"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "trigonometry",
    "projective geometry",
    "geometry theorems"
  ],
  "Problem": "At the vertices of a triangle $ \\triangle{ABC}$, we draw three tangent lines to its circumcircle. If $ AA$ meets $ BC$ at $ D$, if $ BB$ meets $ AC$ at $ E$, and if $ CC$ meets $ AB$ at $ F$, show that $ D$, $ E$, $ F$ are collinear.\r\n\r\n[i]Remark.[/i] Here, if $ P$ is some point on some circle $ k$, then $ PP$ is an abbreviation for the tangent to the circle $ k$ at the point $ P$.\r\n\r\n[i]Alternative formulation.[/i] Prove that the tangents to the circumscribed circle of a triangle, passing through its vertices, intercept the opposite sides in three collinear points.",
  "Solution_1": "i  dont  know  if  its a  theorem  or not  but this is easy  proved  with  pascal  on  AABBCC ,  so even  if it has a name or not  it doesnt take much  time to  write the proof.  i  think  i saw this problem in a  moldova contest  or something like that though  :lol:",
  "Solution_2": "Well, I got this \"theorem\" from a Chinese website, the name seems Morlain line theorem...... :? \r\n\r\nActually, I didn't think of using Pascual...... I used Menelause on $\\triangle{ABC}$.  :lol:",
  "Solution_3": "[color=blue] [b]Lemma.[/b] Given are a triangle $ABC$ and a point $P$ of the sideline $BC$.\nThen there is the following relation:$\\boxed {\\frac{PB}{PC}=\\frac{AB}{AC}\\cdot \\frac{\\sin \\widehat {PAB}}{\\sin \\widehat {PAC}}}\\ .$[/color]\r\n\r\nHere is a ([color=red]very easily[/color]) solution of the proposed problem ([u]Shobber[/u] is right : without ... [color=brown]atomic bombs[/color]).\r\n\r\n$D\\in BC\\cap AA\\Longrightarrow \\frac{DB}{DC}=\\frac{AB}{AC}\\cdot \\frac{\\sin \\widehat {DAB}}{\\sin \\widehat {DAC}}=\\frac cb\\cdot \\frac{\\sin C}{\\sin (C+A)}=$$\\frac cb\\cdot \\frac{\\sin C}{\\sin B}=\\frac cb\\cdot \\frac cb=\\left(\\frac cb\\right)^2\\ .$\r\n$\\frac{DB}{DC}=\\left(\\frac cb\\right)^2\\ ;\\ \\frac{EC}{EA}=\\left(\\frac ac\\right)^2\\ ;\\ \\frac{FA}{FB}=\\left(\\frac ba\\right)^2\\Longrightarrow$ $\\frac{DB}{DC}\\cdot \\frac{EC}{EA}\\cdot \\frac{FA}{FB}=1\\Longrightarrow F\\in DE\\ .$\r\n\r\n[b]Exercise.[/b] Prove the [color=green]Steiner's theorem[/color]:\r\n\r\nGiven are a triangle $ABC$ and two isogonal rays $[AM$, $[AN$, where the points $M$ and $N$ belong to the sideline $BC$, i.e. the angles ${\\widehat BAC}$ and $\\widehat {MAN}$ have the same angle-bisector.\r\nThen exists the [color=green]Steiner's relation[/color]: $\\boxed {\\frac{MB}{MC}\\cdot \\frac{NB}{NC}=\\left(\\frac{AB}{AC}\\right)^2}\\ \\ (1).$\r\n[hide=\"Here is a solution.\"] $\\frac{MB}{MC}=\\frac{AB}{AC}\\cdot \\frac{\\sin \\widehat {MAB}}{\\sin \\widehat {MAC}}\\ ;\\ \\frac{NB}{NC}=\\frac{AB}{AC}\\cdot \\frac{\\sin \\widehat {NAB}}{\\sin \\widehat {NAC}}\\ .$ The product of these two relations $\\Longrightarrow (1)\\ .$[/hide]",
  "Solution_4": "Finally I found the name of it. It is called the Lemoine Theorem, the line is called the Lemoine line of triangle ABC.",
  "Solution_5": "This line is a polar line of a Lemoine Point. You can easily proof that these tree points are collineral using Pascal Theorem (degenerate case)",
  "Solution_6": "Use ''http://en.wikipedia.org/wiki/Menelaus'_theorem''\r\nIf you can't then I will post my solution  :wink:",
  "Solution_7": "Look at the points $A,A,B,B,C,C$ and apply Pascal.",
  "Solution_8": "[quote=\"N.T.TUAN\"]Use ''http://en.wikipedia.org/wiki/Menelaus'_theorem''\nIf you can't then I will post my solution  :wink:[/quote]\r\nCan you please post your solution?  :blush:",
  "Solution_9": "[b]Problem from different viewpoint, with different notations.[/b] Let triangle $ABC$ and $D,E,F$ be points where the incircle touches with sides($D\\in BC,E\\in CA,F\\in AB$). Suppose $A_{1}=EF\\cap BC, B_{1}=FD\\cap CA, C_{1}=DE\\cap AB$. Prove that $A_{1},B_{1},C_{1}$ lies a line.\r\n\r\n[b]Solution.[/b]\r\n\r\nBy Menelaus's theorem we have \r\n$\\frac{A_{1}B}{A_{1}C}.\\frac{EC}{EA}.\\frac{FA}{FB}=1$( $A_{1},F,E$ lies a line).\r\n\r\n$\\frac{C_{1}A}{C_{1}B}.\\frac{DB}{DC}.\\frac{EC}{EA}=1$($C_{1},D,E$ lies a line).\r\n\r\n$\\frac{B_{1}C}{B_{1}A}.\\frac{FA}{FB}.\\frac{DB}{DC}=1$($B_{1},F,D$ lies a line).\r\n\r\nFrom these we obtain $\\frac{A_{1}B}{A_{1}C}.\\frac{C_{1}A}{C_{1}B}.\\frac{B_{1}C}{B_{1}A}=1$ and have done also by Menelaus's theorem . :)",
  "Solution_10": "thx tuan!  :lol:"
}
{
  "Tag": [
    "inequalities",
    "floor function",
    "inequalities unsolved"
  ],
  "Problem": "There are $ n \\geq 2$ permutations $ P_1, P_2, \\ldots, P_n$ each being an arbitrary permutation of $ \\{1,\\ldots,n\\}.$ Prove that\r\n\r\n\\[ \\sum^{n\\minus{}1}_{i\\equal{}1} \\frac{1}{P_i \\plus{} P_{i\\plus{}1}} > \\frac{n\\minus{}1}{n\\plus{}2}.\\]",
  "Solution_1": "By Cauchy-Schwarz in Engel Form, we have\r\n\\begin{align*}\r\n\\sum^{n-1}_{i=1} \\frac{1}{P_i + P_{i+1}}&>\\frac{(n-1)^2}{2(P_1+P_2+\\ldots+P_n)-P_1-P_n}\r\n\\\\&=\\frac{(n-1)^2}{2(1+2+\\ldots+n)-P_1-P_n}\r\n\\\\&=\\frac{(n-1)^2}{n(n+1)-P_1-P_n}\r\n\\\\&\\ge\\frac{(n-1)^2}{n(n+1)-(n-1)-n}\r\n\\\\&=\\frac{(n-1)^2}{n^2-n+1}\r\n\\\\&>\\frac{(n-1)^2}{n^2+n-2}\r\n\\\\&=\\frac{n-1}{n+2}.\\end{align*}",
  "Solution_2": "So cute :) First, the statement is altogether wrong; there are [b]not[/b] $n$[b] permutations [/b]$P_1,P_2,\\ldots,P_n$, each being an arbitrary permutation of $\\{1,2,\\ldots,n\\}$, since $\\dfrac {1} {P_i + P_{i+1}}$ makes no sense as a number, but rather $n$ [b]numbers[/b] $P_1,P_2,\\ldots,P_n$, being an arbitrary[b] permutation [/b]of $\\{1,2,\\ldots,n\\}$. Second, the proof is quite nice, up to a patently reversed inequality; the right form is\n\\begin{align*}\n\\sum^{n-1}_{i=1} \\frac{1}{P_i + P_{i+1}}&>\\frac{(n-1)^2}{2(P_1+P_2+\\cdots+P_n)-P_1-P_n}\n\\\\&=\\frac{(n-1)^2}{2(1+2+\\cdots+n)-P_1-P_n}\n\\\\&=\\frac{(n-1)^2}{n(n+1)-P_1-P_n}\n\\\\&\\ge\\frac{(n-1)^2}{n(n+1)-1-2}\n\\\\&=\\frac{(n-1)^2}{n^2+n-3}\n\\\\&>\\frac{(n-1)^2}{n^2+n-2}\n\\\\&=\\frac{n-1}{n+2}.\\end{align*}\n\nMost likely the best such permutation is $1,n,2,n-1,\\ldots, \\lfloor n/2\\rfloor +1$.",
  "Solution_3": "Could you use a smoothing argument to turn all the fractions into 1/(n+1)? I turned $p_i < p_j < p_k < p_l $ into $ \\frac{1}{p_i+p_j}+\\frac{1}{p_k+p_l} > \\frac{1}{p_i+p_l} + \\frac{1}{p_j+p_k}$ but I'm not sure if one can treat this as an invariant and solve the problem.\n\n*Also Cauchy-Schwarz in Engel-Form = Titu's Lemma for those who read the solutions above and were a bit confused on what that was. (#me)",
  "Solution_4": "You can see also here for a very similar problem. http://artofproblemsolving.com/community/c6h17331p118699",
  "Solution_5": "$\\noindent \\textit {Sketch}$: use the HM-AM inequality, in which your terms are $P_i+P_{i+1}$, for $i\\in \\{1, 2\\ldots n-1\\}$ (a total of $n-1$ terms)."
}
{
  "Tag": [
    "videos"
  ],
  "Problem": "How will the twin impacts of the Lunar Crater Observation and Sensing Satellite, or LCROSS conducted by NASA affect science?\r\n\r\nOnly five and a half hours remain.Check this out and do express your views on the subject!\r\n\r\n[hide][url]http://www.nasa.gov/mission_pages/LCROSS/main/index.html[/url][/hide]",
  "Solution_1": "The mission seems to have been successfull. Now we just have to wait for the results.",
  "Solution_2": "I totally want to see that being played on the news, and then have someone hack in and edit the video so the moon blows up at the end. \"NASA invented a weapon capable of destroying THE MOON!\""
}
{
  "Tag": [
    "function",
    "trigonometry",
    "logarithms",
    "integration",
    "calculus",
    "calculus computations"
  ],
  "Problem": "(1) Find the general solution of the following differential equation.\r\n\r\n$ \\frac {dy}{dx} \\plus{} P(x)y \\equal{} Q(x)$ where $ P(x),\\ Q(x)$ are functions of $ x$.\r\n\r\n(2) Solve the following differential equation.\r\n\r\n$ x\\frac {dy}{dx} \\plus{} 2y \\equal{} \\sin x$",
  "Solution_1": "For later easier notation, suppose we have:\r\n\r\n$ \\frac {dy}{dx} + y p(x) = q(x)$\r\n\r\nThen\r\n\r\n$ \\frac {dy_h}{dx} + y_h p(x) = 0$\r\n\r\n$ \\frac {dy_h}{ y_h} = - p(x) dx$\r\n\r\n$ \\ln y_h = - \\int p(x) \\,dx + C = - P(x) + C$\r\n\r\n$ y_h = C e^{ - P(x)} \\Rightarrow y = C(x) e ^{ - P(x)}$\r\n\r\n$ \\frac {dy}{dx} = C'(x) e ^{ - P(x)} - C(x) p(x) e ^{ - P(x)}$\r\n\r\nTherefor:\r\n\r\n$ C'(x) e ^{ - P(x)} - C(x) p(x) e ^{ - P(x)} + C(x) e ^{ - P(x)} p(x) = q(x)$\r\n\r\n$ C'(x) e ^{ - P(x)} = q(x)$\r\n\r\n$ C'(x) = q(x) e ^{P(x)}$\r\n\r\n$ C(x) = \\int q(x) e ^{P(x)} \\,dx + K$\r\n\r\nAnd finally:\r\n\r\n$ y = e^{ - P(x)} (\\int q(x) e ^{P(x)} \\,dx +K)$ with $ P(x) = \\int p(x) \\,dx$\r\n\r\nSolving the second problem is easy. We have $ p(x) = \\frac {2}{x}$ and $ q(x) = \\frac {\\sin x}{x}$. This gives us\r\n\r\n$ P(x) = \\int \\frac {2}{x} \\,dx = 2 \\ln x$\r\n\r\nand with this:\r\n\r\n${ y = \\frac {1}{x^2} (\\int \\frac {\\sin x}{x} x^2} \\,dx + C) = \\frac {1}{x^2} (\\sin x - x \\cos x + C) = \\frac {\\sin x}{x^2} - \\frac {\\cos x}{x} + \\frac {C}{x^2}$",
  "Solution_2": "That's correct.\r\n\r\nAlternative Solution for (1):\r\n\r\nMultiply by integerating factor $ e^{\\int P(x)dx}$ both sides of the differential equation, we have \r\n\r\n$ \\frac {d}{dx} e^{\\int P(x)dx}y \\equal{} e^{\\int P(x)dx}Q(x)$\r\n\r\n$ \\therefore y \\equal{} e^{ \\minus{} \\int P(x)dx}\\left(e^{\\int P(x)dx}Q(x)dx \\plus{} C\\right)$.",
  "Solution_3": "I have a feeling these are going to get harder real quick.",
  "Solution_4": "HINT for 2::\r\ndivide the eq. by x.Find Integrating factor which is equal to x^2.\r\n\r\nAnswer of integral ::\r\ny*(x^2)=sin x - x cos (x) + C",
  "Solution_5": "[quote=\"kabi\"]HINT for 2::\ndivide the eq. by x.Find Integrating factor which is equal to x^2.\n\nAnswer of integral ::\ny*(x^2)=sin x - x cos (x) + C[/quote]\r\n\r\nThat's correct. :)",
  "Solution_6": "[quote=\"JRav\"]$ \\boxed{1}$\nSet $ \\mu(x) \\equal{} \\exp\\left(\\int P(x)\\ dx\\right)$.  Then some algebra reveals that\n$ y(x) \\equal{} \\frac {1}{\\mu(x)}\\int\\mu(x)Q(x)\\ dx \\equal{} \\exp\\left( \\minus{} \\int P(x)\\ dx\\right)\\int\\left(\\exp\\left(\\int P(x)\\ dx\\right)Q(x)\\ dx\\right)$\n\nDid I handle this correctly?[/quote]\r\n\r\nYes, you did."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "I like this one:\r\n\r\nFor any finite group G define n(G) as the number of elements of G and s(G) as the number of subgroups of G. Determine whether the following sentences are true or false:\r\n\r\na) For any a>0 there is a finite group G s.t. n(G)/s(G)<a.\r\nb) For any a>0 there is a finite group G s.t. n(G)/s(G)>a.\r\n\r\n[i]Romanian NMO/final round/Alba Iulia 1999/2'nd problem/12'th grade/proposed by Barbu Berceanu[/i]",
  "Solution_1": "For any finite group G define n(G) as the number of elements of G and s(G) as the number of subgroups of G. Determine whether the following sentences are true or false: \r\n\r\na) Take a commutative G where the order of every element is 2 (such groups exist: Z2*Z2*...*Z2 is one example).\r\nlet k=n(G)\r\nFor any x<>y in G-{e} we have that e,x,y,xy is a subgroup of G. There are (k-1)(k-2)/6 such subgroups.\r\nSo s(G)>=(k-1)(k-2)/6 -> n(G)/s(G) <= k/( (k-1)(k-2)/6 ) which goes to 0 as n goes to infinity.\r\n\r\nb) Take (Zp,+) for p prime and get n(G)/s(G) = p/2>a",
  "Solution_2": "Thx! This was the official solution as well. There's a similar problem, only using the number of elements (n) and the number of multiplicative operations which can be defined on Z<sub>n</sub> s.t. it's a ring. Try it (it's also composed by Barbu Berceanu).",
  "Solution_3": "Can there be defined more than one multiplicative operation on Z<sub>n</sub>?",
  "Solution_4": "Of course not :D. Here's the real problem:\r\n\r\n(G,+) is a commutative group with n(G) elements. i(G) is the number of laws (G,*) s.t. (G,+,*) is a ring. Find out if there exists a sequence of groups H<sub>k</sub> (commutative and finite groups) s.t. \r\n\r\n(a) n(H<sub>k</sub>)/i(H<sub>k</sub>)->0 as k->00;\r\n(b) n(H<sub>k</sub>)/i(H<sub>k</sub>)->00 as k->00.",
  "Solution_5": "b) This is the easy part. you only need to take Z<sub>n</sub> and use the fact that only one multiplicative law makes it a ring and get n/1->00  :) \r\n\r\na) I will consider the group G<sub>n</sub>=(Z<sub>2</sub>,+)x(Z<sub>2</sub>,+)x...x(Z<sub>2</sub>,+).\r\nDenote:\r\n  e<sub>1</sub>=(1,0,...,0)\r\n  e<sub>2</sub>=(0,1,...,0)\r\n  ...\r\n  e<sub>n</sub>=(0,0,...,1)\r\nEvery element of G<sub>n</sub> can be written uniquely as a sum of some of this elements. So it is enough to define the multiplicative law for this elements.\r\nIn order to have a unit element, define e<sub>1</sub>e<sub>i</sub>=e<sub>i</sub>e<sub>1</sub>=e<sub>i</sub>\r\nThe other n<sup>2</sup>-2n+1=(n-1)<sup>2</sup> products may take any of the 2<sup>n</sup> values in G<sub>n</sub>.\r\nSo n(G<sub>n</sub>)/i(G<sub>n</sub>)=2<sup>n</sup>/((n-1)<sup>2</sup>*2<sup>n</sup>)=1/(n-1)<sup>2</sup>->0  :)",
  "Solution_6": "I guess it's Ok. This is the 4'th problem from the final round of the RMO in 2003, 12'th grade",
  "Solution_7": "There is something wrong in my proof: for Z<sub>n</sub> we may choose fi(n) multiplicative laws, because the unit element is not necesarily 1. (fi(n) is Euler's function).\r\n\r\nBut the result is still OK as p<sup>n</sup>/fi(p<sup>n</sup>)=p and the set of primes goes to 00. :D",
  "Solution_8": "This is easier than what I tried. I used the fact that \\pi{i=1->n}p<sub>i</sub>/phi(\\pi{i=1->n}p<sub>i</sub>)->00 as n->00, but it's a bit harder to prove (not really hard, but your idea is definitely nicer)."
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "function",
    "calculus computations"
  ],
  "Problem": "Let $ f(x)\\equal{}[x(1\\plus{}(\\ln x)^2)]^{\\minus{}q/p}$.\r\nFix $ p \\in (0,\\infty)$.\r\nFor what  $ q \\in (0,\\infty)$ will the integral of $ f(x)$ over $ (1,\\infty)$ exsists(finite)?\r\nFor what  $ q \\in (0,\\infty)$ will the integral of $ f(x)$ over $ (0,1)$ exsists(finite)?\r\nThanks for any help.",
  "Solution_1": "Hint: $ \\minus{} \\frac {d}{dx}\\frac {1}{\\log x} \\equal{} \\frac {1}{x(\\log x)^2}$, and the presence of the $ 1$ has negligible effect on the function near $ \\infty$ and near $ 0$ (which is where convergence or divergence of this integral is determined).",
  "Solution_2": "I still don't know how to do the second part.\r\nCould someone explain more?",
  "Solution_3": "This is simply a matter of comparison tests for the integrability of a positive function.\r\n\r\nFor large $ x,$ $ f(x)\\approx \\frac1{x^{q/p}(\\ln x)^{2q/p}}.$\r\n\r\nFor $ \\frac{q}{p}\\ne 1,$ it's the power of $ x$ that matters, and we have convergence for $ \\frac{q}p>1$ and divergence for $ \\frac qp<1.$\r\n\r\nThe borderline case is $ \\frac qp\\equal{}1,$ in which case we have essentially $ \\int_?^{\\infty}\\frac1{x(\\ln x)^2}\\,dx.$ That's a convergent integral (which is what JoeBlow was saying).\r\n\r\nFor $ x$ near zero, we also have $ f(x)\\approx \\frac1{x^{q/p}(\\ln x)^{2q/p}}.$\r\n\r\nThis time, we have convergence for $ \\frac qp<1$ and divergence for $ \\frac qp>1.$\r\n\r\nLooking at the borderline case again, and let $ u\\equal{}\\minus{}\\ln x$ :\r\n\r\n$ \\int_0^?\\frac1{x(\\ln x)^2}\\,dx\\equal{}\\int_?^{\\infty}\\frac{du}{u^2},$ which converges.\r\n\r\n(Note: the use of a question mark as a limit of integration simply means some value which doesn't cause us any unnecessary problems.)",
  "Solution_4": "[color=green][Unneeded quote removed by moderator.][/color]\r\n\r\nWhy we can neglect the effect  of $ (\\ln x)^{2q/p}$ near 0?\r\n\r\nOk,I see.\r\nThanks for your help.",
  "Solution_5": "If $ x$ has any exponent other than $ \\minus{}1,$ than we can peel of a sliver of a power of $ x$ to overwhelm any logarithm. That's why we have divergence near zero for $ \\frac qp>1.$ On the other hand, if $ \\frac qp\\equal{}1,$ then we can't neglect the logarithm at all; it really matters. That last line of mine is a change of variables.\r\n\r\nAlso: please don't quote the entire previous post; that's what you're assumed to be replying to anyway. Use the \"reply\" button in most cases; only use \"quote\" to quote a specific portion of a post not the whole thing."
}
{
  "Tag": [],
  "Problem": "$a,b\\in\\mathbb{N}$ and $p=\\frac{b}{4}\\sqrt{\\frac{2a-b}{2a+b}}$ is a prime,\r\nfind the maximum value of $p$(with proof! :) )",
  "Solution_1": "[hide=\" \"]$p=\\frac{b}{4}\\sqrt{\\frac{2a-b}{2a+b}}$ is prime.\n$p=\\frac{b}{8a+4b}\\sqrt{4a^2-b^2}$ is prime. Make c=2a:\n$p=\\frac{b}{4c+4b}\\sqrt{c^2-b^2}$ is prime for $b\\in\\mathbb{N}$ and c is even.\nMultiply by 4 and square:\n$16p^2=\\frac{b^2(c+b)(c-b)}{(b+c)^2}=\\frac{b^2(c-b)}{c+b}$.\nSince $16p^2$ has 15 factors, so does $\\frac{b^2(c-b)}{c+b}$...[/hide]\r\n\r\nNevermind. Just being my ordinary stupid self. I ought to get better at olympiad problems before I use this forum...",
  "Solution_2": "hmm, i don't know where you're going with this, but i'm interested... :)",
  "Solution_3": "it is also on the forum .thats is for iran olympiads.",
  "Solution_4": "yes it is correct, it is from the iran's 16th, (i think 1998) olympiad, second round",
  "Solution_5": "[hide=\"hint\"]\n$p^2=\\frac{b^2}{16}\\frac{2a-b}{2a+b}$\n$b$ must be an even number[/hide]",
  "Solution_6": "I was able to do a little simplification, but not much more.\r\n\r\n[hide]\nb obviously must be even.  Let $b = 2b_0$.  Then it simplifies to \n\\[\np = \\frac{b_0}{2} \\sqrt{\\frac{a-b_0}{a+b_0}}\n\\]\nNow because $\\frac{a-b_0}{a+b_0} = \\frac{a^2 - b_0^2}{(a+b_0)^2}$, we can rewrite it as\n\\[\np = \\frac{b_0}{2} \\cdot \\frac{\\sqrt{a^2 - b_0^2}}{a+b_0}\n\\]\n$(\\sqrt{a^2-b_0^2},b_0, a)$ must be a Pythagorean Triple, so let $a = r^2+s^2$ and $b_0$ can be either $2rs$ or $r^2-s^2$.  No matter which $b_0$ we chose, the equation for $p$ simplifies to the same thing, so for now let $b_0 = 2rs$.  ($r$ and $s$ are integers.)  We have $a^2-b_0^2 = (r^2-s^2)^2$ and $a + b_0 = (r+s)^2$, so it simplifies to\n\\[\np = rs \\cdot \\frac{r^2-s^2}{(r+s)^2} = rs \\cdot \\frac{r-s}{r+s}.\n\\]\n(note how b_0/2 = 2rs/2 = rs) This looks much nicer, but I have no clue what to do with it :P\n[/hide]\r\n\r\nI didn't get any farther than that...",
  "Solution_7": "i dont think that would help you, don't  do the $r$ and $s$  thing... ;)",
  "Solution_8": "you guys, want me to post the proof?",
  "Solution_9": "since no one answered this, i wont hold this anymore, here is my proof( i will do half of the problem and leave the other half to you! :) ):\r\n$ p^2=\\frac{b^2}{16}\\frac{2a-b}{2a+b}$ \r\nlet $b=2c$ so:\r\n$p^2=\\frac{c^2(a-c)}{4(a+c)}\\Rightarrow 4p^2=c^2(\\frac{a+c-2c}{a+c})=c^2(1-\\frac{2c}{a+b})\\Rightarrow 4p^2=c^2-\\frac{2c^3}{a+b} \\Rightarrow c^2-4p^2=\\frac{2c^3}{a+b}\\Rightarrow a+c=\\frac{2c^3}{c^2-4p^2}\\Rightarrow c^2-4p^2|2c^3$\r\n$(c^2-4p^2)(-2c)+(2c^3)=8p^2c-2c^3+2c^3=8p^2c \\Rightarrow c^2-4p^2=(c^2-4p^2, 2c^3)|8p^2c$\r\nnow there is two situations, first: $p\\not |c$ and second: $p|c$\r\nif $p\\not |c$ then $c^2-4p^2$ and $p^2$ are primitive thus:\r\n$c^2-4p^2|8p^2c \\Rightarrow c^2-8p^2|8c$\r\n$(c^2-4p^2)(-8)+(8c)(c)=-8c^2+32p^2+8c^2=32p^2 \\Rightarrow c^2-4p^2=(c^2-4p^2, 8c)|32p^2 \\Rightarrow  c^2-p^2|32$ (because $c^2-4p^2$ and $p^2$ are perimitive)\r\n$c^2-4p^2=(c-2p)(c+2p)|2^5$ because $c-2p$ and $c+2p$ are the same, so both must be even, also $c-2p<c+2p$ so:\r\n$p=3\\Leftarrow\\{\r\n\\begin{array}{cc}\r\nc-2p=2\r\n\\\\\r\nc+2p=4\r\n\\end{array} $\r\n$p=5\\Leftarrow\\{\r\n\\begin{array}{cc}\r\nc-2p=2\r\n\\\\\r\nc+2p=8\r\n\\end{array} $\r\n$p=4.5\\Leftarrow\\{\r\n\\begin{array}{cc}\r\nc-2p=2\r\n\\\\\r\nc+2p=16\r\n\\end{array} (unacceptable)$\r\n$p=3\\Leftarrow\\{\r\n\\begin{array}{cc}\r\nc-2p=4\r\n\\\\\r\nc+2p=8\r\n\\end{array} $\r\nso we can write $p=3,5$ in form of $(a,b\\in\\mathbb{N})p=\\frac{b}{4}\\sqrt{\\frac{2a-b}{2a+b}}$ so now all you have to do is to see is there another prime bigger than $5$ that can be written in this form if not then the answer to the problem is $p=5$\r\nnow you do the second part let $p|c$...\r\nso .... :)",
  "Solution_10": "$p=5$\na=? b=?",
  "Solution_11": "If $p=5$ , $a=39$ and $b=30$."
}
{
  "Tag": [
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "Why would this be?\r\n[img]http://img189.imageshack.us/img189/2840/theoot2.jpg[/img]\r\n\r\nThanks a lot.",
  "Solution_1": "Let $k=-h$ two lines above.",
  "Solution_2": "yes but If the expresion that is red circeled then the numerator must also negatief and why is this so?\r\n\r\nThanks."
}
{
  "Tag": [
    "quadratics",
    "function",
    "conics",
    "parabola"
  ],
  "Problem": "w/o a calculator find \r\na) intercepts \r\nb) vertex\r\nc) max/min\r\n\r\n-4x^2 +16x-15 \r\n\r\n\r\ni completely forget how to do this stuff, dont really care about the answers, just looking for the steps that I need to take in order to get to the answers \r\ntanks",
  "Solution_1": "I dont know if this is correct but as far as i can remember, Quadratic Function in vertex form is\r\n\r\n  f(x) = a(x-h)^2 + k\r\n\r\nand the vertex is ( (-b/2a), ((c - b^2)/4a).\r\n\r\nWhen of a > 0, the parabola opens upward, therefore the vertex is minimum. When a < 0, the parabola opens downward, so the vertex is maximum. I hope this helps.\r\n\r\nTo find the intercepts, if you want to find x, substitute y to 0, the reverse (of course) if otherwise.",
  "Solution_2": "hERES A PROBLEM GIVEN TO US AT mATHpATH...FOR SOME REASON, ITS NO LETTIN ME WRITE IS LOWERCASE\r\n\r\nONE STORMY NIGHT, A PIRATES SHIP GETS CAUGHT IN A STORM.  THE SHIP IS SHIPRECKED BAND EVERYONE DIES...EXCEPT 5.  THEY SWIM TO SHORE SAFELY AND START GATHERING AS MANY COCONUITS AS THEY CAN FIND.  AT THE END IOF THE DAY, THEY PILE ALL THE COCONUTS TOGETHER AND DECIDE TO SPLIT THE PILE TOMMOROW.  THE FIRST SAILER DOESNT TRUST THE  OTHERS SO HE TAKES ONE-FIFTH OF THE PILE AND BURYS IT WILE EVERYONE IS SLEEPING.  THERE IS ONE LEFT OVER SO HE GIVES IT TO THE MONKEY.  SAILER 2 DOES THE SAM...SO DOES 3,4, AND 5\r\n\r\nIN THE MORNING, THEY START TO SLIT THE PILE...NONE OF THEM SAY ANYTHING BECAUSE THEY ALL TOOK SOME..AFTER SPLITTING THE PILE, THERE IS ONE LEFT OVER SO THEY GIVE IT TO THE MONKEY.\r\n\r\nWHAT IS THE SMALLEST POSSIBLE NUMBER OF COCONUTS THEY COULD HAVE FOUND.\r\n\r\np.s.  TRY TO ALSO FIND A NEGATIVE SOLUTION"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "tan^2(1) + tan^2(2) + tan^2(3) + ......... + tan^2(89) = ?\r\n\r\n sin^2(1) + sin^2(2) + sin^2(3) + ..... +sin^2(89) = ?\r\n\r\n sin(20)sin(40)sin(80)  =  ?",
  "Solution_1": "For second use $ sin \\theta \\equal{} \\cos(\\frac {\\pi}{2} \\minus{} \\theta)$\r\nand $ \\sin^{2}\\theta \\plus{} \\cos^{2}\\theta \\equal{} 1$\r\n\r\nFor third use $ 4\\sin A\\sin (60 \\minus{} A)\\sin (60 \\plus{} A) \\equal{} \\sin 3A$\r\nso answer will be $ \\frac {\\sin60^\\circ}{4} \\equal{} \\frac {\\sqrt {3}}{8}$\r\n\r\nand 1^{st }is very trivial",
  "Solution_2": "$ 4\\sin A\\sin (60 \\minus{} A)\\sin (60 \\plus{} A) \\equal{} \\sin 3A$ is a useful identity, but back in the day (hah,  :roll: ) when I was going through high school, people didn't promote that identity much.\r\n\r\nIn general, a useful approach is to try to write the expressing using as few \"weird\" angles and as many \"common\" angles as possible, and then try to simplify the expression. Oftentimes, the \"weird\" angles will cancel out, or sometimes, such as in this case, you'll end up with \"common\" angles.\r\n\r\nLet $ S_\\theta \\equal{} \\sin \\theta$ and $ C _\\theta \\equal{} \\cos \\theta$, for shorthand.\r\n\r\nLet's see how this plays out in this case:\r\n\r\n[hide=\"Weird angle is 10\"]$ S_{20} S_{40} S_{80}$\n\n$ \\equal{} S_{(30 \\minus{} 10)} S_{(30 \\plus{} 10)} C_{10}$\n\n$ \\equal{} (S_{30} C_{10} \\minus{} C_{30} S_{10})(S_{30} C_{10} \\plus{} C_{30} S_{10} ) C_10$\n\n$ \\equal{} \\left( S^2 _{30} C^2 _{10} \\minus{} C^2 _{30} S^2 _{10} \\right) C_{10}$\n\n$ \\equal{} \\left( \\frac {1}{4} c^2 _ {10} \\minus{} \\frac {3}{4} s^2 _{10} \\right) C_{10}$\n\n$ \\equal{} \\frac {1}{4} \\left( C^3 _{10} \\minus{} 3 S^2 _{10} C_{10} \\right)$\n\n$ \\equal{} \\frac {1}{4} C_{(3 \\cdot 10)} \\equal{} \\frac {\\sqrt {3}}{8}$[/hide]"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "One side of a square is 12 + 3x centimeters and another side is 32 + x centimeters. How many centimeters is the length of the diagonal of the square? Express your answer in simplest radical form.",
  "Solution_1": "[hide] We have $ 12\\plus{}3x\\equal{}32\\plus{}x \\implies x\\equal{}10$. Therefore, the side length of the square is $ 32\\plus{}10\\equal{}42$ and the diagonal is $ \\boxed{42\\sqrt{2}}$.[/hide]",
  "Solution_2": "Just set the two expressions equal.  12+3x=32+x\r\n\r\n2x=20, x=10.\r\n\r\nTherefore, the length of the diagonal is $ 42\\sqrt{2}$."
}
{
  "Tag": [
    "function",
    "inequalities",
    "inequalities solved"
  ],
  "Problem": "Show that if $x,y,z>0$ we have:\r\n\r\n$ \\frac{x}{x+2y+2z} + \\frac{y}{2x+y+2z} + \\frac{z}{2x+2y+z} \\geq \\frac{3}{5}$\r\n\r\nI have found this solution:\r\n\r\nWLOG $x+y+z=1$\r\n\r\nwe know that the function $f(x)= \\frac{1}{x}$ is convex (you can show this using derivates) and you apply Jensen's inequality with weights $x,y,z$.\r\n\r\n$ \\frac{x}{x+2y+2z} + \\frac{y}{2x+y+2z} + \\frac{z}{2x+2y+z} \\geq \\frac{1}{1+2( \\sum(xy))} \\geq \\frac{3}{5}$ because $ ( \\sum(x))^2 \\geq 3 \\sum(xy)$\r\n\r\nwhat is your solution?\r\n\r\ncheers",
  "Solution_1": "If you put a=x+2y+2z,b=2x+y+2z,c=2x+2y+2 you obtain\r\n\r\nx=(2b+2c-3a)/5\r\ny=(2c+2a-3b)/5\r\nz=(2b+2c-3a)/5\r\n\r\nso that your inequality is tranformed in\r\n\r\nb/a +a/b +c/a + a/c +b/c + c/b \\geq 6 \r\n\r\nwhich is true as a/b +b/a \\geq 2   by AM-GM."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "algebra",
    "factorization",
    "difference of cubes",
    "special factorizations"
  ],
  "Problem": "Simplify  $ \\frac{\\sqrt[3]{2}}{1\\plus{}\\sqrt[3]{2}\\plus{}\\sqrt[3]{4}}$",
  "Solution_1": "[hide]Multiplying both sides by $ \\sqrt[3]{2}\\minus{}1$ gives $ \\sqrt[3]{4}\\minus{}\\sqrt[3]{2}$[/hide]",
  "Solution_2": "How do I know what to multiply to the denominator to get rid of the radicals?",
  "Solution_3": "[hide=\"sub\"]We let $ \\sqrt[3]{2}\\equal{}x$. Thus we want $ \\dfrac{x}{1\\plus{}x\\plus{}x^2}$. We recognize the denominator from the difference of cubes: $ x^3\\minus{}1\\equal{}(x\\minus{}1)(x^2\\plus{}x\\plus{}1)$. Thus we can multiply the numerator and the denominator by $ \\sqrt[3]{2}\\minus{}1$ to get $ \\dfrac{\\sqrt[3]{4}\\minus{}\\sqrt[3]{2}}{2\\minus{}1}\\equal{}\\sqrt[3]{4}\\minus{}\\sqrt[3]{2}$.[/hide]",
  "Solution_4": "Ok got it thanks",
  "Solution_5": "Mmm okay a new one\r\n\r\nSimplify\r\n\r\n$ \\frac{1}{1\\plus{}\\sqrt{2}\\plus{}\\sqrt{3}}$",
  "Solution_6": "You'll notice the trick used above no longer works.  What to do?\r\n\r\n[hide=\"Idea\"] Get rid of one radical first and then the other.  In other words, let's try multiplying by $ (1 \\plus{} \\sqrt{2} \\minus{} \\sqrt{3})$.  This gives\n\n$ (1 \\plus{} \\sqrt{2})^2 \\minus{} 3 \\equal{} 2 \\sqrt{2}$.\n\nNow we just have to multiply by $ \\sqrt{2}$. [/hide]\n[hide=\"Generalization\"] Rationalize $ a \\plus{} b \\sqrt{c} \\plus{} d \\sqrt{e}$ by multiplying by\n\n$ (a \\plus{} b \\sqrt{c} \\minus{} d \\sqrt{e})(a \\minus{} b \\sqrt{c} \\plus{} d \\sqrt{e})(a \\minus{} b \\sqrt{c} \\minus{} d \\sqrt{e})$. \n\nThese three quantities are the (generalized) [url=http://en.wikipedia.org/wiki/Conjugate_element_%28field_theory%29]conjugates[/url] of $ a \\plus{} b \\sqrt{c} \\plus{} d \\sqrt{e}$. [/hide]"
}
{
  "Tag": [
    "USAMTS"
  ],
  "Problem": "1) If you quote a definition from a book, how should you state the source? (Where would you put it, what would you put--author, title, etc.)\r\n\r\n2) If you use a CAS (computer algebra systems) like Derive, should you explain what it is? What should I write? If it was Mathematica, then you wouldn't need to, but I don't think Derive is as popular as Mathematica, etc.\r\n\r\n3) If you use a calculator (TI-83+) program, should you give a URL for where you can download it? How should you mention it?\r\n\r\nThanks! :)",
  "Solution_1": "The general rule of citation, not just on the USAMTS but in math and science in general, is to give enough information so that the reader can verify your work if he/she wishes.  Meaning if you use a program, you should include the source code and/or the URL where it can be found.  If you use a \"well-known\" software package (and Derive is in that category as far as I'm concerned), you should state it by name.  \r\n\r\nCiting reference books should use a standard citation: author, title, publisher, year, page numbers.",
  "Solution_2": "Thanks. :)"
}
{
  "Tag": [
    "modular arithmetic",
    "summer program",
    "PROMYS",
    "number theory",
    "greatest common divisor",
    "relatively prime",
    "prime numbers"
  ],
  "Problem": "hello i'm new here and don't know where this might go...but anywayshere is a good problem i just solved recently:\r\nif in a game there are only 2 possible scores $a$ and $b$, (u can choose to get either one each time) then:\r\ni)what is the largest score that can not be made by these numbers?(proof needed)\r\nii)what is the number of impossible scores?(proof needed)\r\n\r\nplease don't say for part (i), based on the Chicken McNugget Theorem it has to be $ab-a-b$--it would be unacceptable :) \r\n\r\ngood luck.",
  "Solution_1": "It is important to note that $a$ and $b$ must be relatively prime. Else, their sum would have to be a multiple of some prime number, so there will always be a number not obtainable.\r\n\r\nFor part i (there was a similar proof in AoPS volume 2), write:\\[\\begin{array}[b]{ccccc}\r\n0 & a & 2a & 3a & \\ldots\\\\\r\nb & b+a & b+2a & b+3a & \\ldots\\\\\r\n2b & 2b+a & 2b+2a & 2b+3a & \\ldots\\\\\r\n\\vdots & \\vdots & \\vdots & \\vdots & \\ddots\\\\\\end{array}\\]\r\nYou can get all $n\\equiv0\\pmod{a}$ from the first row.\r\n$n\\equiv b\\pmod{a}$ from the second row for $n\\ge b$.\r\n$n\\equiv 2b\\pmod{a}$ from the third row for $n\\ge 2b$.\r\nEventually, all numbers can be obtainable in $\\bmod{a}$ after row $a$ (that is, we can get all $n\\ge(a-1)b$.\r\nSince it is the last row, that is the smallest number of that remainder in $\\bmod{a}$ that is obtainable. Since all the other remainders are already obtainable, the largest number that cannot be reached is then $ab-a-b$.\r\n\r\nNo idea yet for the second part...",
  "Solution_2": "Just FYI, this is very similar to a question on the PROMYS application. Though PROMYS starts tomorrow, they like to reuse problems. Not accusing or anything, but a heads up to people around here.",
  "Solution_3": "this is the coin exchange problem of Frobenius, and has been discussed before.",
  "Solution_4": "Does a closed form for the second part exist?",
  "Solution_5": "Yes.  One can show that if two nonnegative integers $ k,\\ell$ sum to $ ab \\minus{} a \\minus{} b$, then exactly one of $ k$ or $ \\ell$ is an impossible score (and one is possible).  From this, a closed form for ii) follows immediately.",
  "Solution_6": "So in that case the answer is just $ \\frac{ab\\minus{}a\\minus{}b\\plus{}1}2$, right?\r\n\r\nI think I see the reasoning behind it: it cannot be two numbers that can both be expressible as a non-negative linear combo of $ a,b$ or otherwise $ ab\\minus{}a\\minus{}b$ will be able to also. So it must be at least one that is not expressible, and this number must be in the form $ N\\equal{}mb\\minus{}na>0$ for $ 0<m\\le a\\minus{}1\\in\\mathbb{N}$ and any natural $ n$ such that $ n>0$ (considering the complete residue system mod a $ \\{0,b,2b,...(a\\minus{}1)b\\}$), hence we have $ (a\\minus{}1)b\\minus{}a\\minus{}N\\equal{}(a\\minus{}1\\minus{}m)b\\plus{}(n\\minus{}1)a$, which is a valid non-negative linear combo of $ a,b$.\r\n\r\nIs this proof good enough (of course the formulation for N follows directly from the proof of the chicken mcnugget theorem)?\r\nBTW why is it called chicken mcnugget theorem? Can't they give it a more appetizing name? :P",
  "Solution_7": "People were wondering the largest number of Chicken McNuggets that cannot be attained when you order small, medium, or large orders of Chicken McNuggets, which came in orders of 6, 9, and 20. Hence the name.\r\n\r\nAlso the answer for (ii) seems right because it's odd iff a and b are not both even, and if a and b are even then the GCD is more than 1.\r\n\r\nEDIT: Wait a minute, if $ ab\\minus{}a\\minus{}b$ is even (which it is), then we have it over 2 plus it over 2 equals it. $ \\frac{ab\\minus{}a\\minus{}b}{2}$ can't be both obtainable and unobtainable at the same time...",
  "Solution_8": "[quote=\"1=2\"]EDIT: Wait a minute, if $ ab \\minus{} a \\minus{} b$ is even (which it is)...[/quote]\r\nCorrection: which it never is if $ a$ and $ b$ are relatively prime.",
  "Solution_9": "Just wondering, can this be generalized to more than 2 variables?",
  "Solution_10": "If you look at possible scores of $ a,b,c$, the problem is pretty open. But if you look at possible scores of $ ab,bc,ca$ or $ abc,bcd,cda,dab$, interesting stuff happens.",
  "Solution_11": "[hide=\"idea for three numbers\"] Ask the same question using three relatively prime numbers $ a, b$ and $ c$, where $ a < b < c$. Consider all numbers now $ \\pmod{a}$. Now, using x $ b$s and y $ c$s, where x and y may be as large as needed, construct a complete residue class $ \\pmod{a}$, using the smallest value of $ bx \\plus{} cy$ for each $ d\\pmod{a}$ (note that a immediately gives the $ 0\\pmod{a}$).Then take the largest $ bx \\plus{} cy$, subtract a from it, and use that as your answer.\n\nFor example, consider 4, 9, and 11. We have to find the smallest sum of 4s, 9s, and 11s so we end up with numbers congruent to $ 0, 1, 2, 3 \\pmod{4}$.\n\nSmallest 0 (mod 4): 4\nSmallest 1 (mod 4): 9\nSmallest 2 (mod 4):18\nSmallest 3 (mod 4): 11\n\nOur answer is 18-4=14.\n\nUnfortunately, this may take some time to calculate for larger numbers...[/hide]",
  "Solution_12": "[quote=\"MellowMelon\"]But if you look at possible scores of $ ab,bc,ca$ or $ abc,bcd,cda,dab$, interesting stuff happens.[/quote] Could you please elaborate some interesting stuff about this? Thanks.",
  "Solution_13": "i am very sry to express my ignorance but wat is the chicken mcnuggets theorem?",
  "Solution_14": "[quote=\"The Enigma\"]i am very sry to express my ignorance but wat is the chicken mcnuggets theorem?[/quote]\r\nsome one help me..",
  "Solution_15": "[quote=\"The Enigma\"][quote=\"The Enigma\"]i am very sry to express my ignorance but wat is the chicken mcnuggets theorem?[/quote]\nsome one help me..[/quote]\r\n\r\nhttp://lmgtfy.com/?q=chicken+mcnugget+theorem"
}
{
  "Tag": [
    "geometry",
    "vector",
    "analytic geometry",
    "geometry unsolved"
  ],
  "Problem": "It might be (and I know it actually is) a simple problem but my skill in geometry is not sharp. So, what I have in my program is a line and a segment given with their two points respectively. The question is, how do I find out whether they intersect without computing the intersection point itself? In other words, I need a condition of their intersection.\r\nI'd be very grateful for your help.",
  "Solution_1": "The line intersects the segment iff the two end points of the segment are on opposite sides of the line. How do we know if they are on opposite sides of the line? Denote two arbitrary points on your line A, B and the end points of the segment C, D. Compute (B-A)x(C-A), that is, a vector in the direction of the line crossed with a vector from one of the end points to the line. Then compute (B-A)x(D-A), or the same thing for the other end point. Note that both of these products will have only a z coordinate. If both z coordinates are the same sign (+ or -) then the segment does not intersect the line. Otherwise, they do.\r\n\r\nBy the way, I think this topic would be better suited for the comp sci forums but ah well.",
  "Solution_2": "Yes, it greatly helped, thanks a lot!"
}
{
  "Tag": [
    "conics",
    "projective geometry",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Given three circles in the plane, none of which lies inside another (nonintersecting). Prove that the main diagonals of the convex hexagon determined by three pairs of internal common tangents of the three circles are concurrent. ( where the main diagonal means the one that exactly divides the hexagon into two convex quadrilaterals.)",
  "Solution_1": "My solution:\n\nDenote these three circles as $ (O_1), (O_2), (O_3) $\n\nBy Monge's theorem we get\nthe intersection of two external tangents of $ \\{(O2),(O3) \\}, \\{(O3),(O1) \\}, \\{(O1),(O2)\\} $ are collinear ,\nso by [url=http://mathworld.wolfram.com/FourConicsTheorem.html]Dual of Four Conics theorem[/url] $ \\Longrightarrow $ six internal tangents of $ (O1), (O2), (O3) $ form an envelope of a conic .\nBy Brianchon's theorem we get the main diagonals of the convex hexagon bounded by three pairs of internal common tangents of these three circles are concurrent .\n\nQ.E.D"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Hi all, \r\n\r\nI have the following problem:\r\n\r\nLet $ p \\in \\mathbb{C}[z]$ be a polynomial with complex coefficients, which has no roots in $ S^1$.  Show that the number of zeros of $ p$ in the interior of the unit disc is the degree of the continuous map $ \\hat{p}: S^1 \\rightarrow S^1$ defined by $ \\hat{p}(z)\\equal{}\\frac{p(z)}{|p(z)|}$.\r\n\r\n\r\nOne may factor $ p(z)\\equal{}f(z)g(z)$ where $ f(z)$ has no roots $ z$ with $ |z| \\geq 1$ and $ g(z)$ has no roots $ z$ with $ |z| \\leq 1$.  One may do this because of the Fundamental Theorem of Algebra and by reordering of the linear factors.  If I could show that $ \\hat{p}$ is homotopic to $ f/|f|$, then I would be done since that would imply that $ \\deg(\\hat{p})\\equal{}\\deg(f/|f|)$, which is the number of zeros in the interior of the disc.  But that is where I'm encountering trouble.  How can I do this? \r\n\r\nThanks in advance.",
  "Solution_1": "Your notation is unclear - what does $ S_1$ mean? I thought it meant the unit circle, but clearly that's not right.",
  "Solution_2": "Here, $ S^1: \\equal{} \\{z \\in \\mathbb{C} | |z|\\equal{}1\\}$ is the boundary of the $ 2$-disc.",
  "Solution_3": "Why is that clearly not correct?",
  "Solution_4": "I was getting worried about what would happen to the map p-hat near zeros of p, but the fact that it is continuous immediately implies that all =discontinuities are removable and we fill them in as dictated.",
  "Solution_5": "The fact that $ p$ has no roots on $ S^1$ implies that $ \\hat{p}: S^1 \\rightarrow S^1$ is well-defined.  But how is it homotopic to $ f/|f|$?",
  "Solution_6": "argument principle.\r\nconsider $ log(\\frac{p}{|p|})$, and see how many times will it go around the $ S$ when $ z$ run around $ S$.",
  "Solution_7": "The degree of $ p$ is the integer $ d$ such that $ p_*(x) \\equal{} dx$ for every $ x \\in \\pi_1(S^1)$. \r\n\r\nYou can show that \r\n$ \\mathrm{deg}(pq) \\equal{} \\mathrm{deg}(p) \\plus{} \\mathrm{deg}(q)$: since $ S^1$ is a topological group, multiplication\r\nof loops is homotopic to concatenation, from which this result follows. (This is problem 3 in May Ch. 1; do it \r\nfirst to make things easier.)\r\n\r\nGiven this, you only have to show the result for linear polynomials $ (x \\minus{} z)$. From here this should be much\r\neasier. (There are many ways to proceed - since you are probably getting this problem from May's book,\r\nreference the above argument why $ \\pi_1(S^1) \\equal{} \\mathbb{Z}$.)"
}
{
  "Tag": [
    "probability",
    "geometry",
    "3D geometry",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose that $a_1,...,a_k$ are distinct irrational numbers and $N$ is a positive integer. Is it true that the probability for $n$ to verify that $[na_1]+...+[na_k]$ is congruent to $j$ (mod $N$) is the same for all $j$? Does anyone have any idea about a proof?",
  "Solution_1": "I don't think it's true.  Set $a_1 = \\sqrt{2}$, $a_2 = \\sqrt{2} + 2$, and $N = 2$.",
  "Solution_2": "Sorry, I was idiot, I should have realized that. But what if $a_1,...,a_k$ are linearly independent over $Q$?",
  "Solution_3": "Then the answer is \"Yes\". All you need to do is to use the fact that the points $na_1,\\dots, na_k$ taken modulo $N$ are uniformly distributed in the cube $(0,N)^k$."
}
{
  "Tag": [
    "function",
    "integration",
    "limit",
    "calculus",
    "derivative",
    "real analysis",
    "geometry"
  ],
  "Problem": "Can anyone help prove Riemann Theorem?\r\nThe definition is as following:\r\nAssume the symbols using here are well-defined.\r\nAnd $G(x)$ is a cyclic function with a cycle of $T$.\r\nProve:\r\n$\\displaystyle \\int\\limits_{a}^{b}F(x)G(px)dx=\\frac{1}{T}\\int\\limits_{a}^{b}F(x)dx\\int\\limits_{0}^{T}G(x)dx$",
  "Solution_1": "I forget something,the left hand side should be the limit when $p$ tends to $+\\infty$.\r\nHowever,I don't know how to type that symbol in LaTeX. :blush:",
  "Solution_2": "Is this what you mean?:\r\n\r\n$\\lim_{p\\rightarrow\\infty}\\int_{a}^{b}F\\left(x\\right)G\\left(px\\right)\\,dx=\\frac{1}{T}\\int_{a}^{b}F\\left(x\\right)\\,dx\\int_{0}^{T}G\\left(x\\right)\\,dx$",
  "Solution_3": "Yes,but I didn't know how to type in that limit when I was posting that.",
  "Solution_4": "[Note on language: what you called \"cyclic\" is usually called \"periodic\" in English]\r\n\r\nYou might as well start by subtracting off the average value- write $G(x)=h(x)+c$, where $h$ has integral zero over one period. $\\int_a^b F(x)\\cdot c\\,dx=\\frac1T\\int_a^b F(x)\\,dx\\int_0^T c\\,dx$, and we are left with the taxk of proving that $\\int_a^b F(x)h(px)\\,dx\\to 0$ as $p\\to\\infty$ when $h$ has average value zero.\r\n\r\nIf $F$ is continuously differentiable, we can use integration by parts. Let $H$ be an antiderivative of $h$; since $\\int_0^T h(x)\\,dx=0$, $H(a+T)=H(a)$ and $H$ is bounded. Let $F'=f$.\r\nWe have $\\int_a^b F(x)h(px)\\,dx=\\frac{F(b)H(pb)}{p}-\\frac{F(a)H(pa)}{p}-\\frac1p\\int_a^b f(x)H(x)\\,dx$. Each term is bounded by some constant multiple of $\\frac1p$ by boundedness of $M$, and everything goes to zero.\r\n\r\nFor uglier functions $F$, we use a density result; the \"nice\" functions ($C^1$ in this case) are dense in the space we're working in (such as $L^1$ functions).",
  "Solution_5": "Firstly,thanks for the advice on language.\r\n[quote=\"jmerry\"]\nWe have $\\int_a^b F(x)h(px)\\,dx=\\frac{F(b)H(pb)}{p}-\\frac{F(a)H(pa)}{p}-\\frac1p\\int_a^b f(x)H(x)\\,dx$. Each term is bounded by some constant multiple of $\\frac1p$ by boundedness of $M$, and everything goes to zero.\n[/quote]\nI don't really understand this,can you give a more detailed explanation?\nI can prove the cases that $F(x)$ is a continuous function.But the problem is,$F(x)$ can be a lot uglier.\nThe original expression is that $F(x)$ and $G(x)$ are Riemann integrable...\n[quote=\"jmerry\"]\nFor uglier functions $F$, we use a density result; the \"nice\" functions ($C^1$ in this case) are dense in the space we're working in (such as $L^1$ functions).[/quote]\r\nWhat's the meaning of $C^1$ and $L^1$ :?:\r\nAnd what is the meaning of this density result?",
  "Solution_6": "Firstly,thanks for the advice on language.\r\n[quote=\"jmerry\"]\nWe have $\\int_a^b F(x)h(px)\\,dx=\\frac{F(b)H(pb)}{p}-\\frac{F(a)H(pa)}{p}-\\frac1p\\int_a^b f(x)H(x)\\,dx$. Each term is bounded by some constant multiple of $\\frac1p$ by boundedness of $M$, and everything goes to zero.\n[/quote]\nI don't really understand this,can you give a more detailed explanation?\nI can prove the cases in which $F(x)$ is a continuous function.But the problem is,$F(x)$ can be a lot uglier.\nThe original expression is that $F(x)$ and $G(x)$ are Riemann integrable...\n[quote=\"jmerry\"]\nFor uglier functions $F$, we use a density result; the \"nice\" functions ($C^1$ in this case) are dense in the space we're working in (such as $L^1$ functions).[/quote]\r\nWhat's the meaning of $C^1$ and $L^1$ :?:\r\nAnd what is the meaning of this density result?Can you give an explanation?Thank you.\r\nSorry that I've posted the same thing twice,that's a mistake.",
  "Solution_7": "$C^{1}$ just means that the function's derivative is continuous.  I don't know what $L^{1}$ means...",
  "Solution_8": "$f \\in L^1(\\Omega) \\Rightarrow \\int_{\\Omega} |f(x)|dx <\\infty$\r\n$f$ is any function.\r\n$\\Omega$ is some space (I am used to subspaces of $R^n$ but you can do other spaces if you have a measure).",
  "Solution_9": "And if you want $L^1$ to be a metric space, then its members are not functions but equivalence classes of functions under the relation that two functions are equivalent if they are equal almost everywhere. (Example: the function which is 1 on the rationals and 0 on the irrationals belongs to the equivalence class of the the zero function.)\r\n\r\nFinding a nice dense subset would be saying that for any $\\epsilon>0,$ we can find a \"nice\" function $g$ such that $\\int|f-g|<\\epsilon.$ If you take for granted that you can do that, it shouldn't be that hard to see how to extend jmerry's argument.\r\n\r\nActually proving the density would be part of the Lebesgue theory of integration.",
  "Solution_10": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26568]This[/url] is the best explanation I can find. Look up the Riemann-Lebesgue lemma for more information.",
  "Solution_11": "Thanks a lot.It seems liyi's solution,that classical one, is exactly what I'm looking for.\r\nBy the way,can any of you help give a proof to that dense result?",
  "Solution_12": "You prove the density step by step.\r\n\r\n1. given a positive measurable function, one can construct a sequence of measurable simple functions the increases to the given function. (A simple function is a function with a finite range. It's a linear combination of indicator functions of sets. At this point in the construction, we have no control over the shape of the sets.)\r\n\r\n2. Prove that if this original positive function is integrable ($L^1$), then the sequence of simple functions constructed in #1 converges to the original function not just pointwise but also in the $L^1$ norm. (This proof would use either Fatou's Lemma or the Dominated Convergence Theorem.)\r\n\r\n3. Given a complex-valued integrable function, write it as the linear combination of four functions (positive part of real part, negative part of real part, etc.) Approximate each as in #2. We now have that simple functions are dense in $L^1.$\r\n\r\n4. Now we need this result from measure theory: Given a measurable set $E$ of finite measure and $\\epsilon>0,$ there exists a set $B$ which is a finite union of bounded intervals such that $|E\\Delta B|<\\epsilon.$ Here I'm using $|\\cdot|$ to mean measure and $E\\Delta B=(E\\setminus B)\\cup(B\\setminus E)$ is the symmetric difference of the sets.\r\n\r\n5. Given #4, we can approximate a characteristic function of a set arbitrarily well in $L^1$ by a step function. Now it's just an exercise in epsilon-gymnastics to say that we can approximate any simple function arbitrarily well in $L^1$ by a step function.\r\n\r\n6. Now that we have the step functions dense in $L^1,$ we can proceed along any of several paths. One thought is to \"bevel the edges\" of the step functions to produce continuous functions or even $C^1$ functions. Another is this:\r\n\r\n6'. Suppose we're working on all of $\\mathbb{R}.$ (Cutting down to some interval isn't too hard.) Then translation of functions is continuous in $L^1.$ What I mean by that: Suppose $f\\in L^1(\\mathbb{R}).$ Define $f_t(x)=f(x-t).$ Then $\\lim_{t\\to0}\\|f_t-f\\|_1=0.$ The proof uses #5, the density of step functions.\r\n\r\n7'. Suppose $\\varphi$ is a $C^{\\infty},$ compactly supportd function such that $\\int_{\\mathbb{R}}\\varphi=1.$ Define $\\varphi_{\\epsilon}(x)=\\frac1{\\epsilon}\\varphi\\left(\\frac{x}{\\epsilon}\\right).$ Then $\\varphi_{\\epsilon}*f\\in C^{\\infty}$ (that's convolution) and $\\lim_{\\epsilon\\to0^+}\\|\\varphi_{\\epsilon}*f-f\\|_1=0.$ The proof uses #6'.\r\n\r\nNow we have that $C^{\\infty}$ is dense in $L^1.$"
}
{
  "Tag": [
    "number theory",
    "prime factorization"
  ],
  "Problem": "How many divisors does $ 2\\cdot2\\cdot3\\cdot3\\cdot3$ have?",
  "Solution_1": "Every factor of a number must share its prime factors and have no extra ones. For example, the number 18 has prime factorization $ 2 \\cdot 3^2$. A factor of it, 9, shares those prime factors with no extras: $ 2^0 \\cdot 3^2$.\r\n\r\nThis implies that for the first prime factor of the number with exponent $ n$, there are $ n\\plus{}1$ ways to \"choose\" it. In our 18 example, the 2 could have had an exponent of 0 or 1, and the 3 could have been 0, 1, or 2. Thus, the number of factors is simply the product of the exponents added to 1.\r\n\r\nIn this problem, it's already prime factorized, so we have $ (2 \\plus{} 1)(3 \\plus{} 1) \\equal{} 12$."
}
{
  "Tag": [
    "MATHCOUNTS",
    "AMC",
    "AIME",
    "AMC 10"
  ],
  "Problem": "How do you prepare for State?\r\n\r\nI use the MATHCOUNTS Drills on that one website.\r\n\r\nThe night before the competition, I eat some food with protein.\r\nI also do the past years competition (So I would do the 2005 State Competition for the 2006 State).\r\n\r\nThe morning of the competition, I eat a banana.\r\nI also look over the formula sheet I downloaded and printed from the Formula forum.\r\n\r\nWhat do you do?",
  "Solution_1": "It is a good idea to get a lot of rest several nights before the competition.  You should also eat a good breakfast before the competition so you do not become hungry and have plenty of energy.\r\n\r\nLooking over a formula sheet the morning of the competition won't do you much good.  You should find how the formulas are derived and make sure you understand that, because then you are not very likely to forget the formula, but even if you do, you could derive it.\r\n\r\nIt is a good idea to practice multiple state rounds before the competition, so you have more practice.  I do every sprint round from about 1995 to 2005 (all that I have) and do some twice to make sure I'm ready.  Practicing several different rounds will give you an advantage.  If you have done multiple rounds, you will be likely to be able to pace yourself well so you can finish the test and get the most possible correct.  If you have done only one round as practice, you won't have a good enough idea of the range the of difficulty of the test, so you probably won't do as well.\r\n\r\nAnother good idea is to do all of the warm ups and work outs provided for the current year.  The same people that write the warm ups and work outs write the actual competitions, so if you do them, you will have a better idea of what types of problems might be on the contest.\r\n\r\nBasically, practice will help you do the best.",
  "Solution_2": "Our MATHCOUNTS coach is also our Algebra 2 teacher. 2 weeks before the competiton, we take time from math class and turn it into MATHCOUNTS practice time. We do all the chapter, state, and national competitions.\r\nOne day we'll do all the sprints of one year.\r\nAnother day we'll do all the targets of one year.\r\nWe take home another year of competition for homework.\r\nWe also practice countdown and team after school.\r\nOur coach has finally been able to project the countdown problems on a board, so we can practice it the real way.\r\n\r\nDo you think this will help us at State? (we're in Kentucky and our school has won it six years in a row)(at least I think so)",
  "Solution_3": "I try to win.\r\n\r\nhere are some real tips:\r\nFirst, eat a good dinner the night before.\r\nGet 9+ hours of sleep.\r\nEat a good breakfast.\r\n\r\nTry to stay as relaxed as possible all that day, it really helps.\r\n\r\nSPRINT:\r\nBefore you get the sprint round, make numbered boxes 1-30 on a few sheets of paper.\r\nUse the boxes as scratch paper for specific problems. Circle your answer.\r\nRead each question at least twice, and make sure you get the key information. Trust me, this is important. Many a time I have given radius instead of diameter, speed instead of distance...\r\nDon't spend too long on one problem, if you think it will take more than a minute, skip it and come back later.\r\nEither have a watch or timer to keep time, and with about 5 minutes left, start checking your work. You will probably get more right this way than spending the 5 minutes on the two hardest problems.\r\nMake sure the circled answer that you have on scratch paper is what you have for an answer.\r\nMake sure your answers are in the right form/units.\r\nThen go back and do the problems you skipped.\r\n\r\nTARGET:\r\nYou don't need the numbered paper thing here; you have enough scratch paper on the sheet itself.\r\nDo whatever problem seems easier first.\r\nThen do the harder one.\r\nGo back to the first problem and try it a different way. Do that for the other problem.\r\nKeep confirming your answer in different ways until time is called.\r\n\r\nTEAM:\r\nMost of this is based on what your team's strategy is and how strong you are in comparison of he rest of the group.\r\nHowever, this is something that I think really helps.\r\nMake boxes on paper like in the sprint round, except one sheet of paper per problem.\r\nWrite a decent solution so the other person can follow your steps.\r\nWhen someone or a group is done, give to another person/group to check it over.\r\nWhen there are 2 minutes left, make sure that the captain has all of the answers written on the sheet. Make sure its the captain's sheet though. At nats last year, we almost got 0 on the team round because the captain wrote on the wrong sheet (stupid girl).\r\n\r\nCOUNTDOWN:\r\nIf you make it this far congrats!\r\nBefore you go up, observe who your opponents may be.\r\nNote their speed and whether they are prone to giving a wrong answer too quickly.\r\nWhen you are up there, don't get nervous.\r\nRead the question on the projector (or whatever else) quickly, noting the key words/numbers. If you think you can do it in your head, do so. If you need paper, use it.\r\nMake sure your answer is in the right form, then buzz in. If it isn't, convert it and buzz in immediately before you have the right answer.\r\nIf the other guy buzzes in and gets it wrong, then do the question on paper, and take all of the time you need. Check over it several ways, and buzz in a few seconds before time is called (but don't cut it too close :) )",
  "Solution_4": "Just wondering, what AoPS chapters would be useful for preparing for states?",
  "Solution_5": "I think mainly\r\n$1$ if you forget how to simplify radicals.\r\n$3$\r\n$4$\r\n$\\boxed{5}$ \r\n$7$ but its really easy\r\n$9-18$\r\n$25$\r\n$26$\r\nThats all i think but the rest is useful a bit",
  "Solution_6": "When you start the sprint round, should you scan the first page and look for the easiest question (the one with the least words)?\r\n\r\nI did this at chapter and I got 8th overall.",
  "Solution_7": "[quote=\"Aznwithabrain\"]When you start the sprint round, should you scan the first page and look for the easiest question (the one with the least words)?\n\nI did this at chapter and I got 8th overall.[/quote]\r\n\r\nWell, the shortest problems aren't always the easiest. I would say as a general rule, go through every question in order, and if you need to spend more than 10 seconds visualizing what to do, skip it and come back. What I did at chapter was work on the problems backward (I posted this on a thread somewhere, I just don't remember which one) and I got second on written.",
  "Solution_8": "I wouldn't do that. If you can't answer all of the low numbered problems, then  :oops:",
  "Solution_9": "Well how bout at State?",
  "Solution_10": "at any level.\r\nNot to be mean, but if you can't answer the easy questions, then you aren't gonna win.\r\nFor the last problems, I do that a lot, especially if one looks easy",
  "Solution_11": "Another question\r\n\r\nOther than past competitions, 2006 warm-ups and workouts and etc., and the MATHCOUNTS drill page, and the AoPS books.\r\n\r\nWhat do you use for practice?",
  "Solution_12": "other warmups and workouts",
  "Solution_13": "Is that it?",
  "Solution_14": "mathematical circles a little. also a few amc-10 tests, but the format is so different that >.<\r\nI don't know what else there is :?:",
  "Solution_15": "how often do you meet per week with your team? and what do you do during team practices?  does anyone have suggestions for team round?\r\n\r\nand looking at these schedules, when do you do homework or extracurricular activities?",
  "Solution_16": "state in illinois was in a holiday inn. free jolly ranchers on all the tables. we played poker and used ours as chips. they gave us a boxed lunch, it was okay I guess. Afterwards we had an ice cream social and everyone was coming by congratulating me on the nationals thing...i guess everyone's really nice in illinois!\r\n\r\nWe practiced every monday wednesday friday morning from 7:00 to 7:50. mostly other competitions that our coach has picked up from places. around AMC 10 time she made sure we each understood all the problems, it was really random. sometimes we'd take a session and do countdown (most fun)\r\nfor team round we always set up a group answer sheet the same way. I'm not sure how widespread this is but we take a sheet of paper and make a chart with our initials at the top and rows for 1-10. when we get the answer we put it in the right box. when we get 2 or three answers the same in a row we put it down, when we get 4 different answers we make sure we're covered on all the other ones and work on that...every team round is different (duh) so that's kind of inefficient on some of them. also two people work 10-1 and two work 1-10. That way we at least get to every problem...supposedly.",
  "Solution_17": "We have practice wednesday mornings from 7:15 to 8:00, but it's canceled every other week, and we don't really do anything...",
  "Solution_18": "My team has practice Monday and Friday afternoons and Wednesday mornings from 7:00 to 7:45",
  "Solution_19": "we had 5 practices (literally) before states.\r\n\r\nThey're 90 minutes, but be usually get in only 2 team rounds or countdown,target,team.\r\n\r\nSo we did 5 team rounds total and 3 countdowns, 3 targets (ALL SCHOOL ROUND)",
  "Solution_20": "To prepare for state, I go over everything I've learned the night and day of State",
  "Solution_21": "what place did you get at chapter?",
  "Solution_22": "[quote=\"camathkid\"]To prepare for state, I go over everything I've learned the night and day of State[/quote]\r\n\r\nActually, that's not necessarily a good idea, having to look over all that tires out your brain, and also shows you haven't prepared enough :(.",
  "Solution_23": "How do you prepare?",
  "Solution_24": "i win :)   :D",
  "Solution_25": "very funny",
  "Solution_26": "Yeah, considering that he'll get pwnd at nats.",
  "Solution_27": "Does the handbook with warm-ups and work-outs really help?",
  "Solution_28": "yeah. There are 300 or so problems per year in those, as compared to 144 in chapter, state, national. (sprint target team)",
  "Solution_29": "Warm-ups and Workouts are the building blocks for being good at MATHCOUNTS."
}
{
  "Tag": [
    "LaTeX",
    "\\/closed"
  ],
  "Problem": "The way I save it is click Print when on the transcript, then hit command-s (control-s for windows) and click save as Webpage, complete.\r\n\r\nThe only problem with this is the latex doesnt show up.  It downloads as a folder with all the latex pictures and a html file with the transcript, and i'm wondering if there's an easier way.",
  "Solution_1": "The way I did it was copy-paste onto Word and save as PDF.  OR you can use something like CutePDF and print it to that.",
  "Solution_2": "Use opera, then save it as an MHTML file, it will retain all the images like the original. This should work in IE as well but Opera Pwns IE  :P",
  "Solution_3": "Just copy it onto a Word document...what's so hard about doing that?",
  "Solution_4": "[quote=\"mathemagician1729\"]Use opera, then save it as an MHTML file, it will retain all the images like the original. This should work in IE as well but Opera Pwns IE  :P[/quote]\r\n\r\nI believe AIME15 has Safari.\r\n\r\n@AIME15: Those virtual printers that print into PDF, etc. are probably the best choice.  With me, they retained formatting.  I'm not sure that CutePDF is compatible with Mac, but you can always Google and see."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "function"
  ],
  "Problem": "in real numbers\r\n$x_{1}+x_{2}+....+x_{1997}=1997$\r\n$x_{1}^{4}+x_{2}^{4}+...+x_{1997}^{4}=x_{1}^{3}+x_{2}^{3}+...+x_{1997}^{3}$",
  "Solution_1": "This seems to easy, but if $x_{1},x_{2},x_{3},..x_{1997}=1$   it works",
  "Solution_2": "But there is more than one solution, right?",
  "Solution_3": "I believe it's 1, and -1.",
  "Solution_4": "$-1$ does not work with the either of the statements.",
  "Solution_5": "[quote=\"AstroPhys\"]$-1$ does not work with the either of the statements.[/quote]\r\nHow so?\r\n\r\nI just tried it and it worked..",
  "Solution_6": "[quote=\"ashrafmod\"]in real numbers\n$x_{1}+x_{2}+....+x_{1997}=1997$\n$x_{1}^{4}+x_{2}^{4}+...+x_{1997}^{4}=x_{1}^{3}+x_{2}^{3}+...+x_{1997}^{3}$[/quote]\r\n\r\n[hide=\"An idea?\"]I think from the second equation, $\\sum_{n=1}^{1997}x_{n}^{2}(x_{n}^{2}-x_{n})=0$...so...maybe either $x_{n}=0$...or $x_{n}^{2}-x_{n}=0$...or $x_{n}=1$...? I dunno, that's what I'm thinking...[/hide]",
  "Solution_7": "Wait, asto, I did it wrong.\r\n\r\nNevermind.",
  "Solution_8": "lol its 1. the cube vs. squares thing.",
  "Solution_9": "A less-than-clever way to do it is to use Lagrange multipliers on the function $f = x_{1}+x_{2}+\\cdots+x_{1997}$ subject to the constraint $(x_{1}^{4}-x_{1}^{3})+(x_{2}^{4}-x_{2}^{3})+\\cdots+(x_{1997}^{4}-x_{1997}^{3}) = 0$ which tells us that the extrema must occur when\r\n\r\n$1 = 4x_{i}^{3}-3x_{i}^{2}\\Rightarrow (x_{i}-1)(4x_{i}^{2}+x_{i}+1) = 0$\r\n\r\nfor all $i$, i.e. $x_{i}= 1$ is the only real extremum (we can verify that this is a maximum). Thus we have $f \\le 1997$ with equality iff $x_{i}= 1$ for all $i$.",
  "Solution_10": "If all $x_{i}\\in \\mathbb{R}^{+}$:\r\n\r\n$S_{1}= \\sum_{i=1}^{1997}x_{i}$\r\n$S_{2}= \\sum_{i=1}^{1997}x_{i}^{2}$\r\nsame for $S_{3},S_{4}$\r\n\r\nBy Cauchy-Schwarz we have $S_{1}S_{3}\\ge S_{2}^{2}$.  By AM-GM we have $S_{2}^{2}\\ge 1997S_{4}$\r\n\r\nHence $S_{1}S_{3}= S_{1}S_{4}\\ge 1997S_{4}\\Leftrightarrow S_{1}\\ge 1997$ with equality iff all $x_{i}$ are equal (obviously to 1)\r\n\r\nif one or more are negative then I dunno...maybe try splitting into negative and positive subsets...w/e."
}
{
  "Tag": [],
  "Problem": "Shubrashis Nyogi aka tittu has won the gold medal at Int Bio olympiad!!!!!! :)  :D",
  "Solution_1": "well done tittu!  :clap:  :omighty:",
  "Solution_2": "Congratulations Tittu  :clap2:  :first:  :10:",
  "Solution_3": "congratulations  :thumbup:  :clap:  :first:",
  "Solution_4": "and deserves muccchhhhhh  mooooooreeeee\r\n\r\n:1:",
  "Solution_5": "and you are also going to make us do the same(congratulating) next year :P",
  "Solution_6": "correct skand \r\nbut tittu \r\nvery very very very well done ................................................................"
}
{
  "Tag": [],
  "Problem": "A quarter weighs the same as two pennies. The quarters in one pound of quarters have a total value of $ \\$25$. How many dollars would a pound of pennies be worth?",
  "Solution_1": "the stack of quarters is worth $ \\$25$, which means that there are $ 25 \\times 4$ quarters, or $ 100$ quarters\r\n\r\nsince a quarter weighs $ 2$ pennies, that means there are $ 100 \\times 2$ pennies in that stack, or $ 200$ pennies\r\n\r\n$ 200$ pennies is $ \\$2$, so our answer is $ \\boxed {2}$",
  "Solution_2": "i got this right so i'm gonna explain:the stack is $25 and 2 pennies weigh the same as 1 quarter,2500/25=100,and 100 *2=200,200/100=2 so the answer would be 2"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "function",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "Let\r\n\\[ D \\equal{} \\{z\\parallel{}z| \\le 1\\},\\]\r\n\r\n\\[ A_0 \\equal{} \\{f: D \\to \\mathbb{C}| f\\text{ is analytic in the interior of } D,f\\text{ is continous on D}\\}\\]\r\nDoes the polynomials dense in $ A_0$ with the maximum norm?",
  "Solution_1": "The Cesaro means of the Fourier series of a continuous function converge uniformly. Given $ f(z)\\equal{}\\sum a_nz^n$ (on the open disk), the polynomials $ f_n(z)\\equal{}\\sum_{k\\equal{}0}^{n\\minus{}1} \\frac{n\\minus{}k}{n}a_kx^k$ converge uniformly to $ f$ on the circle, and thus also on the whole disk.",
  "Solution_2": "Another solution is to note that $ f(rz)$ is close to $ f$ if $ r<1$ is close to 1, by uniform continuity. Since $ f(rz)$ is analytic for $ {|z| < 1/r}$, the power series of $ f(rz)$ at 0 converges uniformly to $ f$ on the closed unit disc. This leads to the desired polynomial."
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "$\\frac{\\sin 10^\\circ + \\sin 20^\\circ}{\\cos 10^\\circ + \\cos 20^\\circ}$ equals\r\n\r\nA. $\\tan 10^\\circ + \\tan 20^\\circ$\r\nB. $\\tan 30^\\circ$\r\nC. $\\frac{1}{2} (\\tan 10^\\circ + \\tan 20^\\circ)$\r\nD. $\\tan 15^\\circ$\r\nE. $\\frac{1}{4} \\tan 60^\\circ$",
  "Solution_1": "[hide]we know  $\\sinh +\\ sin b =2sin(a+b)/2*cos(a-b)/2$\nand     \n$\\cosh + \\ cosb = 2cos(a+b)/2*cos(a-b)/2$\nthen the ansewr is D. $\\tan 15^\\circ$[/hide]",
  "Solution_2": "[hide=\"Answer\"]$\\sin 10=\\sin (15-5)=\\sin 15\\cos 5 -\\sin 5\\cos 15$, $\\sin 20=\\sin (15+5)=\\sin 15\\cos 5+\\sin 5\\cos 15$\n$cos 10=\\cos (15-5)=\\cos 15\\cos 5+\\sin 15\\sin 5$, $\\cos 20=\\cos (15+5)=\\cos 15\\cos 5-\\sin 15\\sin 5$\n$\\frac{\\sin 10+\\sin 20}{\\cos 10+\\cos 20}=\\frac{2\\sin 15\\cos 5}{2\\cos 15\\cos 5}=\\tan 15\\Rightarrow \\boxed{D}$[/hide]",
  "Solution_3": "[hide]$\\frac{\\sin 10^\\circ + \\sin 20^\\circ}{\\cos 10^\\circ + \\cos 20^\\circ}$ \n\n$=\\frac{2sin15^\\circ * cos(-5)^\\circ}{2cos15^\\circ * cos(-5)^\\circ}$\n\n$=\\frac{sin15^\\circ}{cos15^\\circ}$\n\n$=tan15^\\circ$\n\n$\\boxed{D}$[/hide]",
  "Solution_4": "wow, we have a calculator for the amc, it takes no thinking to put those into the calculator, and would be faster (maybe)\r\n\r\nbut yes, the other ways are better",
  "Solution_5": "I dont think you could use calculators in 80s AMCs"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Is every real derivable and convex function on an open convex subset A of R\u207f , C\u00b9 ?",
  "Solution_1": "ok no one likes this kind of question : is ..? :oops: \r\njust show that instead in R :\r\n\r\nlet f : R -> R derivable and convex . show that f' is continuous . :roll:",
  "Solution_2": "no one ?  :maybe:  it might be classical one .."
}
{
  "Tag": [
    "function",
    "algebra",
    "domain",
    "inequalities",
    "calculus",
    "derivative",
    "functional equation"
  ],
  "Problem": "How do you figure out if a function one-to-one algebraically instead of graphically (using the horizontal line test).\r\n\r\nThanks!",
  "Solution_1": "Prove the implication \"$ f(a) \\equal{} f(a')\\Rightarrow a \\equal{} a'$\", or its contrapositive \"$ a\\ne a'\\Rightarrow f(a)\\ne f(a')$\"",
  "Solution_2": "[quote=\"10000th User\"]Prove the implication \"$ f(a) \\equal{} f(a')\\Rightarrow a \\equal{} a'$\", or its contrapositive \"$ a\\ne a'\\Rightarrow f(a)\\ne f(a')$\"[/quote]\r\n\r\n\r\ncan u explain it in a better way  please",
  "Solution_3": "[quote=\"b555\"][quote=\"10000th User\"]Prove the implication \"$ f(a) \\equal{} f(a')\\Rightarrow a \\equal{} a'$\", or its contrapositive \"$ a\\ne a'\\Rightarrow f(a)\\ne f(a')$\"[/quote]\n\n\ncan u explain it in a better way  please[/quote]Ex: Show that $ f(x)\\equal{}x^5$ is 1-1.\r\n\r\nIf you graph it, it looks like a snake, a curve similar to $ f(x)\\equal{}x^3$. For any $ a<b$, $ a^5<b^5$ so it passes the horizontal line test.\r\n\r\nTo prove $ f$ is 1-1, suppose that $ f(a)\\equal{}f(a')$. Then, $ a^5\\equal{}(a')^5$. Take fifth roots (we can do that because the domain and range are reals) and you'll have $ a\\equal{}a'$ so $ f$ is 1-1.\r\n\r\nIn general, if you can prove that \"If $ f(a)\\equal{}f(a')$, then $ a\\equal{}a'$\", then $ f$ is 1-1. I wrote the contrapositive of the statement because sometimes it is easier to prove the contrapositive.",
  "Solution_4": "Really?  In my experience it's very rarely easier to prove the contrapositive.  The first way of writing the implication gets used all the time in functional equation problems to show that a function is one-to-one without knowing what it is.",
  "Solution_5": "[quote=\"t0rajir0u\"]Really?  In my experience it's very rarely easier to prove the contrapositive.  The first way of writing the implication gets used all the time in functional equation problems to show that a function is one-to-one without knowing what it is.[/quote]Not trolling or anything. One word: \"sometimes\"",
  "Solution_6": "[quote=\"10000th User\"]To prove $ f$ is 1-1, suppose that $ f(a) \\equal{} f(a')$. Then, $ a^5 \\equal{} (a')^5$. Take fifth roots (we can do that because the domain and range are reals) and you'll have $ a \\equal{} a'$ so $ f$ is 1-1.[/quote]\r\n\r\nI don't get the significance of this.  If you raise something to the fifth power and then take the fifth root, aren't you just left with what you started?",
  "Solution_7": "Exactly.  This means that two real numbers are distinct if and only if their fifth powers are distinct, i.e. taking fifth powers is one-to-one.\r\n\r\nAs for the other point, it is easier to turn an equality into other equalities than to turn an inequality into other inequalities.  In the former case you can apply arbitrary functions; in the latter case you must apply one-to-one functions.",
  "Solution_8": "[quote=\"t0rajir0u\"]Really?  In my experience it's very rarely easier to prove the contrapositive.  The first way of writing the implication gets used all the time in functional equation problems to show that a function is one-to-one without knowing what it is.[/quote]\n\n[quote=\"t0rajir0u\"]As for the other point, it is easier to turn an equality into other equalities than to turn an inequality into other inequalities. In the former case you can apply arbitrary functions; in the latter case you must apply one-to-one functions.[/quote]\r\n\r\nI'm really confused by the above two quotes.  Can someone explain what contrapositive and implication are and how they're applied to this?  Also, I don't get the second quote at all, but may that's because I don't get the first one.\r\n\r\nThanks!",
  "Solution_9": "Suppose you know that $ a \\equal{} b$ and you want to prove that $ c \\equal{} d$.  How would you go about doing this?  \r\n\r\nOne general strategy is to find a function $ f$ such that $ f(a) \\equal{} c$ and $ f(b) \\equal{} d$.  This is the essence of algebraic manipulation.  If you can find any function of this form, you are done.\r\n\r\nHowever, suppose you know that $ a \\neq b$ and you want to prove that $ c \\neq d$.  You have two options:\r\n\r\n1.  Find a one-to-one (that is, injective) function such that $ f(a) \\equal{} c, f(b) \\equal{} d$.  The reason I dislike this strategy is that injective functions are harder to come by than arbitrary ones.  For example, $ x^2$ isn't injective.\r\n\r\n2.  Suppose that $ c \\equal{} d$ and use this to prove that $ a \\equal{} b$; contradiction!  The first strategy applies.  This is my point.\r\n\r\n(Another reason to philosophically dislike using inequalities to prove other inequalities is that inequality isn't [url=http://en.wikipedia.org/wiki/Transitive_relation]transitive[/url]; failure to understand this can occasionally be dangerous.)\r\n\r\nAs for your other two questions:  an implication is a statement of the form \"if P, then Q\" (that is, \"P implies Q.\").  Its contrapositive is the statement \"if not Q, then not P.\"  The two statements are logically equivalent.  For example, \"if it rains, I will not go outside tomorrow\" is logically equivalent to \"if I go outside tomorrow, it didn't rain.\"",
  "Solution_10": "I find the easiest way to prove whether or not a function is one-one is to assume $ x\\neq y$, and then show that either $ f(x)\\equal{}f(y)$ (not one-one), or $ f(x)\\neq f(y)$ (one-one).  This works for more than just functions-it also works in showing uniqueness of certain elements, such as an identity element.",
  "Solution_11": "the method shown by jrav is very right.\r\n\r\nbut there is a shorter method\r\nhere we have 2 prove that the function is 1-1\r\nin other words we have 2 prove that the function is monotic.\r\nso if f(x) is the function\r\nfind f'(x)\r\nthen 1.if f'(x)>o for all x then it is stritcly increasing.\r\n2.if f'(x)<o for all x then it is strictly decreasing.\r\n\r\nso find f'(x).if it satisfies any 1 of the above 2 conditions, it is a 1-1 function.",
  "Solution_12": "[quote=\"VP_91\"]the method shown by jrav is very right.\n\nbut there is a shorter method\nhere we have 2 prove that the function is 1-1\nin other words we have 2 prove that the function is monotic.\nso if f(x) is the function\nfind f'(x)\nthen 1.if f'(x)>o for all x then it is stritcly increasing.\n2.if f'(x)<o for all x then it is strictly decreasing.\n\nso find f'(x).if it satisfies any 1 of the above 2 conditions, it is a 1-1 function.[/quote]You'll have to realize two things:\r\n\r\n1) This is High School Basics Forum, and the material you are discussing is beyond the scope of this subforum and belongs to Calculus & Computational Forum, which is not fair to the intended audience.\r\n\r\n2) A function can be 1-1 without it being monotonic, e.g. $ f(x)\\equal{}\\begin{cases}x&\\text{if x is rational}\\\\\\minus{}x&\\text{if x is irrational}\\end{cases}$, though the other direction is true if it is [i]strictly[/i] monotonic. But, finding derivatives is not the way to [u]prove[/u] whether a function, in general, is 1-1 or not.",
  "Solution_13": "@10,000thuser:i understood my mistake.i believe my logic is applicable mainly for cotinous functions."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "let a,b,c, are base of triagle ABC ,S is the area of  ABC\r\n      prove that:\r\n(2a^2+b^2+2c^2)^2>_128S^2",
  "Solution_1": "you can prove that:\r\n$ (2a^2\\plus{}2b^2\\plus{}2c^2)^2 \\geq 198S^2$",
  "Solution_2": "$ 4(a^2 \\plus{} b^2 \\plus{} c^2)^2\\ge 12abc(a \\plus{} b \\plus{} c)\\ge 192S^2$",
  "Solution_3": "Make it stronger.\r\n$ 16S^2\\equal{}(a\\plus{}b\\plus{}c)(a\\plus{}b\\minus{}c)(a\\minus{}b\\plus{}c)(\\minus{}a\\plus{}b\\plus{}c)\\leq \\frac{(a\\plus{}b\\plus{}c)^4}{27}$ :lol:",
  "Solution_4": "We have:\r\n2a^2 + 2c^2 + b^2 \u2265 [(a+c)^2 \u2013 b^2 ]  + 2b^2\r\n\u22652\u221a[2(a+c-b)(a+b+c)b^2]\r\n\u22652\u221a[2(a+b+c)(b^2-(c-a)^2)(a+c-b)]=8S\u221a2"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Here is a proof of this theorem [url=http://www.its.caltech.edu/~ari/paper-folding.html]Proof[/url].\r\n\r\nIt have an approach of the final length of the paper folded is equal to the initial length.\r\n\r\nI tried to prove this in another way. The area of the paper folded is equal to the area of the initial paper not folded. The area here is a transverse one.\r\n\r\nSo, using the same assumption that the bend in the paper is a semi circle we need to sum all those semi-circles.\r\n\r\nThe radius of each semi circle is $ 2^{i \\minus{} 1}$ where $ i$ is the number of folds. [url=http://www.pomonahistorical.org/12times.htm](See the image of this site if you have doubts about this)[/url]\r\n\r\nSo, by this conservation of area.\r\n\r\n$ \\sum _{i \\equal{} 1}^n \\frac {1}{2} \\pi \\left(2^{i \\minus{} 1} h\\right)^2 \\equal{} L h$\r\n\r\nSolving for $ L$ in the last equation we have:\r\n\r\n$ L \\equal{} \\frac {h \\pi }{6} \\left(2^n \\minus{} 1\\right) \\left(2^n \\plus{} 1\\right)$\r\n\r\nBut the equation founded by Britney Gallivan is:\r\n\r\n$ L \\equal{} \\frac {h \\pi }{6} \\left(2^n \\minus{} 1\\right) \\left(2^n \\plus{} 4\\right)$\r\n\r\nSo, where this difference from 1 to 4 came from?\r\nThe formula above is true, one the length is conserved but the other the area is conserved, in theory the 2 need to be equal and give the same result.\r\n\r\nThe only thing I can think that make this difference is that the length of the part bended is not actually $ j \\pi h$, but less than this or the area is not $ \\frac {1}{2} \\pi \\left(2^{i \\minus{} 1} h\\right)^2$ but a little bit greater (don't know how).\r\n\r\nI would be glad if someone help me to find my mistake and correct the formula to give the correct result.\r\n\r\nNote: If we think about volumes the result would be the same, because in this case the volume is Area times $ w$, and the factor $ w$ would cancell out",
  "Solution_1": "The difference between your and Britney's formula is the difference between the assumption that the area is preserved and the assumption that the length of the outer edge is preserved when making a fold. I do not know which of these two assumptions is more realistic for paper. You both have no mistakes in mathematics but your physical models of the fold are different.",
  "Solution_2": "hm...\r\n\r\nNow I know the difference between. The result needed to be equal, because the paper is conserved itself.\r\n\r\nThe thing is, Britney used the radius of the paper $ j$. So it is the external part of the paper. Physically it would be more realistic to use the middle of the paper insted of the external part, (the mean of the external and internal), so the radius would be $ j \\minus{} \\frac{1}{2}$. After that I found this:\r\n\r\n$ \\sum _{i\\equal{}1}^n \\sum _{j\\equal{}1}^{2^{i\\minus{}1}} \\left(j\\minus{}\\frac{1}{2}\\right) \\pi  h\\equal{}\\sum _{i\\equal{}1}^n \\frac{\\pi  \\left(h 2^{i\\minus{}1}\\right)^2}{2 h}$\r\n\r\nIt give the equality if we change this. The thing confusing is where you mesure the length matter, but the area not.\r\n\r\nThe 2 conservative laws are rigth and actually the same law.\r\n\r\nDo you agree fedja?",
  "Solution_3": "I certainly agree that when you bend metal or plastic, your law makes more sense, so, probably, you are right about paper too. :) I also wonder if the thickness of the paper really changes a bit when folding (at least, I see no reason why it shouldn't happen :?, pehaps I should refresh my very rusty knowledge of \"solid body physics\" or to leave this discussion to someone who knows it better :P).",
  "Solution_4": "Hm... I was also thinking about this.\r\n\r\nIf we fold the paper and unfold it, the thickness in the fold will be less than in others parts of paper. I think this is just sensible for one fold, If you continuosly fold and fold, it will not make a big differente, and if the thickness of the paper is too great this differente in thickness will also be small.\r\n\r\nSo one of the consequences of folding and unfolding a paper is that the length of the paper increass (just a little)\r\n\r\nI just made a very very approximation of the real case (don't know if this hold)\r\n\r\nYou have a paper of length $ L_0$ and thickness $ h$. After you fold it you have basically a retacgle of thickness $ 2 h$ and length $ L_1$ and a semi-circle of radius $ h$. If we use the law of conservation of area (mass) we have.\r\n\r\n$ L_0 h\\equal{} L_1 (2h) \\plus{} \\frac{\\pi }{2} h^2$\r\n\r\nSo, If we unfold the length of the paper would be $ 2 L_1 \\plus{} \\delta L$\r\n\r\nA good approximation of this $ \\delta L$ would be \"Cut the paper when you bend it and the put it thogeher\". So we would joing 2 quartes of circle side by side.\r\n\r\nThe final lenth would be so: $ 2 L_1 \\plus{} 2 h \\equal{} L_0\\plus{}\\frac{1}{2} h (4\\minus{}\\pi )$\r\n\r\nAnd this approximation would've be only valid if $ h$ is small.\r\n\r\nIf the paper has thickness of $ 1mm$ the difference of length would've be only $ 0.43 mm$\r\n\r\nI don't think this is a precise result, but we could've play with it. Find the average thickness of the fold, how may chemical bonds were broken when you fold (this explain why the paper is too sensive in the fold, chemical bonds are broken when you fold a paper)\r\n\r\nThis is not a study in paper, just some interesting models to work out.\r\n\r\nI don't know too much about physics of bending materials, so I can't say that those approximations are good or not."
}
{
  "Tag": [
    "calculus",
    "integration",
    "calculus computations"
  ],
  "Problem": "What is the sum from 1 to infinity of $ \\frac{1}{n \\sqrt{n^2\\plus{}2}}$? Please prove it.",
  "Solution_1": "Numerically, it's about $ 1.1588006.$ Is there some reason why it should be a more recognizable number than that?",
  "Solution_2": "It was on my friend's homework.",
  "Solution_3": "1.  For what class?\r\n\r\n2.  Are you sure it wasn't an integral?\r\n\r\n3.  Are you sure it wasn't an estimate?",
  "Solution_4": "Well, that's what she told me 4 calc bc, no, it's definitely a sum, but the problem was probably just to see whether it converges or not which is trivial, although she asked me 4 the exact value.",
  "Solution_5": "At that level, you're definitely not asking for the exact value on something like this. Most of the time, that exact value can't be written in closed form."
}
{
  "Tag": [
    "induction",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Let $ a_1\\equal{}a_2\\equal{}1$ and \\[ a_{n\\plus{}2}\\equal{}\\frac{n(n\\plus{}1)a_{n\\plus{}1}\\plus{}n^2a_n\\plus{}5}{n\\plus{}2}\\minus{}2\\]for each $ n\\in\\mathbb N$. Find all $ n$ such that $ a_n\\in\\mathbb N$.",
  "Solution_1": "First I get $ na_n\\minus{}(n\\minus{}1)^2a_{n\\minus{}1}\\equal{}\\left\\{\r\n\\begin{array}{ll}\r\n\\minus{}n\\plus{}2\\ (n: odd) &\\quad  \\\\\r\n\\minus{}n\\plus{}3\\ (n: even) &\\quad\r\n\\end{array}\r\n\\right.$",
  "Solution_2": "Prove by induction that \\[ a_n\\equal{}\\frac{(n\\minus{}1)!\\plus{}1}n\\]\r\nThe rest is obvious.",
  "Solution_3": "$ a_{3}=\\frac{2}{3} $. your formula does not work for $ n=1,3 $. your formula works for the sequence $ a_{1}=2,a_{2}=1 $...so it was a typo in the initial problem?",
  "Solution_4": "[quote=\"anonymouslonely\"]$ a_{3}=\\frac{2}{3} $. your formula does not work for $ n=1,3 $. your formula works for the sequence $ a_{1}=2,a_{2}=1 $...so it was a typo in the initial problem?[/quote]\nNo in the initial problem $ a_{1} $ is 2 and $ a_{2} $ is 1.Right!"
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "trigonometry",
    "logarithms",
    "calculus computations"
  ],
  "Problem": "Prove that :$ (sin x)^{cos x}<(cos x)^{sinx}\\  for \\ 0<x<\\frac{\\pi}{4}$",
  "Solution_1": "hello\r\nwell one of much time consuming method:\r\n[hide=\"hint\"]\nconsider a func $ f(x) \\equal{} (cos(x))^{sin(x)} \\minus{} (sin(x))^{cos(x)}$ \n$ g(x) \\equal{} f'(x)$which is coming$ \\minus{}ve$ for the interval (0,\\frac{\\pi}{4})\nthus$ f(\\frac{\\pi}{4})<f(x)$  where $ 0<x<\\frac{\\pi}{4}$\nor$ (cos(x))^{sin(x)}>(sin(x))^{cos(x)}$\n[/hide]\r\nthank u",
  "Solution_2": "[quote=\"pankajsinha\"]Prove that :$ (sin x)^{cos x} < (cos x)^{sinx}\\ for \\ 0 < x < \\frac {\\pi}{4}$[/quote]\r\nYou could do it taking the logarithm.\r\nLet $ f(x) \\equal{} (\\sin x)( \\log \\cos x)$ and $ g(x) \\equal{} (\\cos x)( \\log \\sin x)$\r\nNow divide both by $ \\log \\cos x$.\r\nThat way, you have logarithms with a base and you can compare the values accordingly for the interval $ \\left[0, \\frac {\\pi}{4}\\right]$.",
  "Solution_3": "For $ 0<x<\\frac{\\pi}{4}$, we have $ \\sin x < \\cos x<1$\r\n\r\nHence $ (\\sin x)^{\\cos x} < (\\sin x)^{\\sin x} < (\\cos x)^{\\sin x}$"
}
{
  "Tag": [
    "Ring Theory",
    "number theory",
    "relatively prime",
    "superior algebra",
    "superior algebra solved"
  ],
  "Problem": "suppose R is a commutative ring and P is a prime ideal. then i know that P contains every nil potent element. is the converse true? that is, if an element belongs to every prime ideal then its nilpotent?\r\n\r\n\r\n\r\nok another qn, this is quite unrelated to the previous one. suppose a and b are 2 relatively prime positive integers. show that every integer greater than ab-a-b can b written as a positive linear combination of a and b. by this i mean ax+by=N,N>ab-a-b and x,y>0",
  "Solution_1": "For the first: when I remember correctly: yes (possibly I forgot some other condition that we need)\r\n\r\nFor the second (but please post only one problem per thread in future):\r\nSome people call it (I don't know why) the Chicken McNugget theorem, so searching for it will help.",
  "Solution_2": "I remember posting a proof for the affirmative answer to your first question some time ago, but it's not worth the effort of looking for it :).\r\n\r\nSuppose $x$ is not nilpotent. Our aim is to show that there is a prime ideal which does not contain $x$. Let $\\mathcal I$ be the collection of ideals which do not contain any of the elements $x,x^2,\\ldots$. Since $x$ is not niplotent, the zero ideal is an element of $\\mathcal I$, so, in particular, $\\mathcal I$ is non-empty. It's easy to see that this collection satisfies the hypothesis necessary for applying Zorn's Lemma: every chain (with respect to inclusion) in $\\mathcal I$ has an upper bound, namely the union of all the ideals in the chain. By Zorn's Lemma, $\\mathcal I$ has a maximal element $P$. We're done if we can show that $P$ is prime.\r\n\r\nTake $a,b\\not\\in P$. By the maximality of $P$, we have $x^m\\in P+(a),\\ x^n\\in P+(b)$ for some positive integers $a,b$. We then clearly have $x^{m+n}\\in P+(ab)\\Rightarrow ab\\not\\in P$, so whenever $a,b\\not\\in P$, we get $ab\\not\\in P$, which means precisely that $P$ is prime."
}
{
  "Tag": [
    "inequalities",
    "calculus",
    "derivative",
    "algebra",
    "function",
    "domain"
  ],
  "Problem": "show that for any x natural>=1 we have:\r\n\r\nx=[1/a^2]; where a=arctan(1/sqrt(x)).\r\n\r\nand [t]=t-{x}.",
  "Solution_1": "spoiler:\n\n[hide]\n\nconsider the right triangle below with side lengths 1, :sqrt: x, and :sqrt: x+1.  It is clear that arctan(1/ :sqrt: x)= the angle D.  So what we want to prove is that x = [1/D^2] or equivalently x<=1/D^2<x+1.  The left hand inequality says x<=1/D^2 or D<=1/ :sqrt: x or equivalently D<=tanD, similarly with the rhs inequality, we see D>1/ :sqrt: x+1 or D>sinD, so we need to prove the equivalent identity, for all positive angles D, 0<D<90, sinD<D<=tanD.  For the lhs, the derivative of D-sinD is 1-cosD, which is positve for 0<D<90, so after they intersect at D=0, D>sinD for all D>0.  for the rhs, consider tanD-D with derivative sec^2(D)-1, now this derivative is positve for 0<D<180 since 0<cos^2(D)<1 in that domain, and D=tanD at 0, so tanD > D for 0<D<90, proving our result.[/hide]"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "[hide]I really am kind of lost, and I would like to know, how do you get out of math mode?\n\nEdit: this is what happens.  $ Let n_k\\,be\\,the \\,value \\,of \\,n on the kth turn, let l_k be the value of l on the kth turn, and let r_k be the value of r on the kth turn.$\n\n[code]Let n_k\\,be\\,the \\,value \\,of \\,n on the kth turn,  let l_k be the value of l on the kth turn, and let r_k be the value of r on the kth turn.[/code]\n\nIs the only way to space it to add \"\\,\"?[/hide]\r\n\r\nEDIT 2:  Never mind, I got it.  Can someone please close this thread.   :oops:",
  "Solution_1": ":huh: Don't put text in dollar signs and you won't have this problem.   :|   There are also commands like \\textrm{}, but there's no reason at all to use them on the forum."
}
{
  "Tag": [],
  "Problem": "Tom has a red marble, a green marble, a blue marble, and three identical yellow marbles. How many different groups of two marbles can Tom take?",
  "Solution_1": "the solution says:  There two cases here: either Tom takes two yellow marbles (1 result), or he takes two marbles of different colors ( results). The total number of distinct pairs of marbles Tom can take is 1+6=7.\r\n\r\nbut i can list \r\nrg,by\r\nrg,yy\r\nrb,gy\r\nrb,yy\r\nry,yy\r\nry,gy\r\nry,by\r\ngb,yy\r\ngy,yy\r\nby,yy \r\nso thats eleven different ones already.",
  "Solution_2": "Well, here's what I have.  I haven't checked the official solution, but...\r\n\r\n[i]Case 1: Both marbles are yellow.[/i]\r\n\r\nOne possible way.\r\n\r\n[i]Case 2: Both marbles are not yellow.[/i]\r\n\r\nThis will include only $ 1$ yellow marble in the set, so we have $ \\binom{4}{2}$.\r\n\r\nThus, we have $ 1\\plus{}6$ or $ \\boxed{7}$.",
  "Solution_3": "oh i see, i just misinterpreted the question",
  "Solution_4": "Good thing you see your mistake!",
  "Solution_5": "I don't think that problem should be a level 21 problem(with 111 xp). It seemed pretty easy...",
  "Solution_6": "[font=TIMES NEW ROMAN]@Above I agree: [hide=Easier than it seems]If there were only one yellow marble the answer would be $\\binom{4}{2}=6.$ But since we actually have 3 yellow marbles, we have to add on the $\\text{Yellow, Yellow}$ combination, so our answer is $\\boxed{6+1=7}.$ $$\\definecolor{color}{rgb}{0.5, 0.3, 0.2} \\scriptsize{\\colorbox{color}{\\color{white}{A pifinity solution}}}$$",
  "Solution_7": "[quote=pifinity][font=TIMES NEW ROMAN]@Above I agree: [hide=Easier than it seems]If there were only one yellow marble the answer would be $\\binom{4}{2}=7.$ But we have to add on the $\\text{Yellow, Yellow}$ combination, so our answer is $\\boxed{6+1=7}.$ $$\\definecolor{color}{rgb}{0.5, 0.3, 0.2} \\scriptsize{\\colorbox{color}{\\color{white}{A pifinity solution}}}$$[/quote]\n\n4c2 isnt 7......",
  "Solution_8": "We add the Y,Y combination: [quote]But we have to add on the  $\\text{Yellow, Yellow}$ combination, so our answer is $\\boxed{6+1=7}.$[/quote]",
  "Solution_9": "4c2 should still be $6$...\u2026.. you can add the Y,Y after but u said IF there is only one yellow",
  "Solution_10": "I mean that we take the 6 from 4 choose 2 and add 1: [quote]$\\boxed{6+1=7}.$[/quote]\n\n",
  "Solution_11": "Yes but you dont add $1$ if there is only one yellow ball as u said...\u2026\n",
  "Solution_12": "That's where I ended the condition",
  "Solution_13": "I edited it to make it clearer though. [quote][hide=Easier than it seems]If there were only one yellow marble the answer would be $\\binom{4}{2}=7.$ But since we actually have 3 yellow marbles, we have to add on the $\\text{Yellow, Yellow}$ combination, so our answer is $\\boxed{6+1=7}.$ $$\\definecolor{color}{rgb}{0.5, 0.3, 0.2} \\scriptsize{\\colorbox{color}{\\color{white}{A pifinity solution}}}$$[/quote]",
  "Solution_14": "Ok, im confused now..... just if you are adding one Y,Y to the previous case of only 1 yellow then it sould be 4c2+1=7",
  "Solution_15": "It is. $ $",
  "Solution_16": "If there is only 1 yellow ball, then there are 4c2= 6 possibilities and then u can say, Since there are 3 we need to add the case Y,Y to get 6+1=7",
  "Solution_17": "That's what I edited it to say. Read my post again.",
  "Solution_18": "You cant say there are 4c2=7 possibilities if there is only 1 yellow ball..... You are still saying 4c2 is 7.....",
  "Solution_19": "Oh, that was a typo. Here is a revised version: [quote=pifinity][font=TIMES NEW ROMAN][hide=Easier than it seems]If there were only one yellow marble the answer would be $\\binom{4}{2}=6.$ [b]But since we actually have 3 yellow marbles,[/b] we have to add on the $\\text{Yellow, Yellow}$ combination, so our answer is $\\boxed{6+1=7}.$ $$\\definecolor{color}{rgb}{0.5, 0.3, 0.2} \\scriptsize{\\colorbox{color}{\\color{white}{A pifinity solution}}}$$[/quote]\n\n",
  "Solution_20": "Good thing you see your mistake!\n\n"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "Hi\r\ncananybody tell me from where i can download article.sty files and if i want to type a paper in latex. And the style is atricle.sty then what other files along with article.sty would i be needed to download?\r\n\r\nTanks in advance\r\n\r\njigna",
  "Solution_1": "article.sty is obsolete and part of an old version of LaTeX.\r\nYou can't just download a few files, you need a LaTeX distribution. See [[LaTeX:Downloads]] for details assuming you are using Windows, though you don't say which operating system you are using."
}
{
  "Tag": [
    "function",
    "algebra",
    "domain",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "$U \\in C^2(\\mathbb{R},\\mathbb{R})$ satisfies: $U(x) \\geq 0$, $lim_{|x|\\to \\infty} U(x) = +\\infty$.\r\n\r\nConsider the differential equation:\r\n$(E)$:$X'(t)=-U'(X(t))$, $X(0)=y \\in \\mathbb{R}$ for $t \\geq 0$\r\n1/ show there is a unique solution defined over $\\mathbb{R}^+$\r\n2/ suppose that the set where $U'$ vanishes is discrete. Show that the solution $X(t)$ converges towards a limit $L \\in \\mathbb{R}$ and that $U'(L)=0$\r\n3/ study the same pb with $U \\in C^2(\\mathbb{R}^d,\\mathbb{R})$ ..",
  "Solution_1": "Are we supposed to see what happens to (2) for $u: \\mathbb R^d\\to\\mathbb R$ as well? Because it looks a bit weird, on first sight :).\r\n\r\n(1) works for all $d$ in pretty much the same way. First of all, $u'$ is a $C^1$ funtion, so the equation has the property of [i]local[/i] existence and uniqueness of solutions. This means that the maximal solutions are unique. All we need to do now is prove that maximal solutions are actually global solution, i.e. that every solution can be extended to the entire $\\mathbb R_+$.\r\n\r\nLet $x: [0,k),\\ k<\\infty$ be a solution with $x(0)=y$. We have $\\frac{\\mbox d}{\\mbox dt}u(x(t))=-(x'(t),x'(t))\\le 0$, so $u\\circ x$ is decreasing. Since $u$ goes to infinity as the norm of its argument goes to infinity, this means that the image of $x$ stays within a ball conained in the domain of $u$, and this implies that $x$ can be extended strictly to the right (this is a standard result). \r\n\r\nFor (2) (we're back to the case $d=1$) we could do this:\r\n\r\nSuppose $x(t)=x(t'),\\ t<t'$. Then $u(x(t))=u(x(t'))$, so $u\\circ x$ is constant on $I=(t,t')$. The assumption on the set of zeros of $u'$ now implies that $x$ must be constant on $I$, and from here we conclude that $x$ is monotonic. Its limit $L$ has to be finite, since otherwise $u\\circ x$, which is decreasing, would go to $\\infty$, which is absurd. \r\n\r\nGiven any $\\varepsilon>0$, and any $n\\in\\mathbb N$, we can find $t_n>n$ such that $|x'(t_n)|<\\varepsilon$: simply take $n<t<t',\\ t'-t>1$, with $|x(s)-L|<\\frac\\varepsilon 2,\\ \\forall s\\ge t$, and choose $t_n\\in(t,t')$ such that $(t'-t)x'(t_n)=x(t')-x(t)$. We have $|u'(x(t_n))|=|x'(t_n)|<\\varepsilon$, and the conclusion that $u'(L)=0$ follows.",
  "Solution_2": "nice  :) \r\nand you're right, it's a bit more difficult (i think) for the case $d>1$. But it's still true."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "geometry",
    "circumcircle",
    "3D geometry",
    "algebra",
    "polynomial"
  ],
  "Problem": "[For reference: the contest page: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126603[/url]]\r\n\r\n[size=167]Mock AIME 5 for 2007[/size]\r\n\r\nWritten by Alex Anderson\r\nwith contributions from Jacob Steinhardt\r\n\r\n[hide=\"Answers:\"]1.\t299\n2.\t322\n3.\t972\n4.\t997\n5.\t181\n6.\t204\n7.\t670\n8.\t601\n9.\t275\n10.\t502\n11.\t936\n12.\t003\n13.\t252\n14.\t647\n15.\t219[/hide]\r\n\r\nResults:\r\n\r\n1st place:  :winner_first:  \r\nMysticTerminator, 14/15\r\n2,3,4,5,6,7,8,9,10,11,12,13,14,15\r\n\r\n2nd place:  :winner_second: \r\npaladin8, 12/15\r\n1,2,3,4,5,6,7,8,9,10,12,14\r\n\r\n3rd place [tie]:  :winner_third: \r\nbeta 10/15\r\n2,3,4,5,6,7,8,11,12,13\r\nnat mc 10/15\r\n1,2,3,6,7,9,10,11,12,13\r\n\r\nComments:\r\n\r\nThanks a lot for jsteinhardt for helping me check the test. Also he submitted 4th and 14th problems and thought of the idea for problem 5. \r\n\r\nProblem 12 had a small error...I decided to let the test go on, rather than trying to change anything. It was pretty clear to find the largest lower bound. Regardless, if you even looked at that problem, you should have just asked with a private message; I was quite clear about this.\r\n\r\nI suspect that some people used a calculator for problem 5. I probably should have said that there were 7 prime factors, find the second largest one. \r\n\r\nMany people were confused about problem #3. A geometric sequence of integers with a finite number of terms does not have to have an integer ratio. \r\n\r\nProblem # and percentage of people who answered it correctly:\r\n1.\t75\r\n2.\t87.5\r\n3.\t71.875\r\n4.\t56.25\r\n5.\t50\r\n6.\t68.75\r\n7.\t62.5\r\n8.\t18.75\r\n9.\t46.875\r\n10.\t21.875\r\n11.\t21.875\r\n12.\t12.5\r\n13.\t12.5\r\n14.\t6.25\r\n15.\t3.125\r\n\r\noverall average:\t6.15625 of 15 correct\r\n\r\nDifficulty:\r\n\r\nThe first two problems were significantly easier than the actual AIME. I would say that the next 7 problems [3-9] problems or so were on par with an official AIME. #10 was particularly difficult to find the actual solution. Many people found the pattern and solved that way. Perhaps I could have put this problem later. I feel that many people saw this problem, got discouraged, and stopped working on problems. I feel that more people could have solved #12. Anyway, the later geometry was pretty difficult. Also #14 was tough and interesting problem; one of my favorites on the test.\r\n\r\nI thought that the general content was similar. I did not have any \"annoying casework problems\" that are prevalent on the AIME contests. I also felt that I could have included more counting problems. #6 was pretty interesting, but #8 was a standard application of recursion for expected value.\r\n\r\nSolutions:\r\n\r\nI don't plan to type out official solutions. Every problem has been answered by somebody. Furthermore, I have sent people solutions to some of the harder problems, like #10,14,15. Perhaps I could get some help writing out solutions...I can compile them into a one thread...maybe wiki entries too...\r\n\r\nFinally,\r\nI appreciate the positive feedback. I worked hard on this contest, and I think it turned out well. \r\n\r\n--Alex Anderson",
  "Solution_1": "Cool. I really liked #14 as well :).\r\n\r\nBTW, nice job to Arnav for getting them all except #1 :P.",
  "Solution_2": "Yes, that is rather ironic, isn't it? Problem 1 [b]was[/b] quite an easy problem.\r\nI did 1, 2, 3, and then got too lazy and didn't finish it. I did get all three of them right, though.  :lol:",
  "Solution_3": "I'm willing to bet he submitted 42 or the like for #1.",
  "Solution_4": "Since some people were wondering...\r\n\r\n[hide=\"15\"]\nDenote $DE=x$. Consider just $P$, $A$, $B$, $D$, $E$, and $C$. You can either use similar triangles, or some tricky polar stuff [somebody will have to fill in the pole/polar stuff] to show that $P$, $D$, $E$, and $C$ are in a harmonic progression. In particular this means [without directed lengths] that\n\n\\[\\frac{PD}{DE}*\\frac{EC}{CP}=1\\]\n\nHence $CE=2x$, and using power of a point on $E$, \n\n\\[(x)(2x)=12\\implies x=\\sqrt{6}\\]\n\nIt follows directly that $PD=3\\sqrt{6}$, $PA^{2}=(3\\sqrt{6})(6\\sqrt{6})\\implies PA=PB=6\\sqrt{3}$.\n\nNow we have to deal with $F$. Let $I$ be an inversion centered at $P$, with the circle passing through $A$ and $B$. This maps $w_{1}$ to itself [since the circles are orthogonal, or equivalently using power of a point]. Hence, $I(A)=A$ and $I(D)=C$. Furthermore, we know that the circumcircle of $I(ADFP)$ is a line because any circle through the center of inversion becomes a line, this line being $AC$.  We need to find $I(F)$. We know that this must lie on $PF$ and $AC$. Hence, $I(F)=G$. Then by definition of inversion, \n\n\\[PA^{2}=(PF)(PG)\\implies 108=PF*(6\\sqrt{3}+10)\\implies PF=81\\sqrt{3}-135\\]\n\nIt follows that the answer is $81+3+135=\\boxed{219}$.\n\nComment: Obviously this problem was quite difficult. I wanted to go all out, so to speak, on the geometry at the end.[/hide]",
  "Solution_5": "Woah totally forgot to even look at the questions.\r\n\r\n5 looks way too easy. I don't want to work it out, but can't we be all \"WOAH DIFFERENCE OF SQUARES!!!111lololol\"?",
  "Solution_6": "[quote=\"PenguinIntegral\"]Woah totally forgot to even look at the answers.\n\n5 looks way too easy. I don't want to work it out, but can't we be all \"WOAH DIFFERENCE OF SQUARES!!!111lololol\"?[/quote]\r\n\r\nUnless you can factor 40001 magically, you need some other techniques...I think that these problems are reasonable, and could be on an aime, but since there is nobody preventing the use of a calculator during a mock test, I don't think I will use this type of problem again.",
  "Solution_7": "You can do #15 using only basic angle-chasing/power-of-a-point, but it's still fairly involved.",
  "Solution_8": "[quote=\"probability1.01\"]You can do #15 using only basic angle-chasing/power-of-a-point, but it's still fairly involved.[/quote]\r\n\r\nyea...i realize...but i prefer the concise solution...the harmonic progression directly follows from triangle similarity...the inversion thing is equivalent to showing that $\\triangle APF\\sim\\triangle GPA$...",
  "Solution_9": "One way to factor $40001$ is to write it as $40401-400$.",
  "Solution_10": "[quote=\"Ravi B\"]One way to factor $40001$ is to write it as $40401-400$.[/quote]\r\n\r\n[hide]in general, that would be 4a^4+1=(2a^2-2a+1)(2a^2+2a+1)\n\nthis is the sophie-germain identity[/hide]",
  "Solution_11": "[quote=\"Altheman\"][quote=\"PenguinIntegral\"]Woah totally forgot to even look at the answers.\n\n5 looks way too easy. I don't want to work it out, but can't we be all \"WOAH DIFFERENCE OF SQUARES!!!111lololol\"?[/quote]\n\nUnless you can factor 40001 magically, you need some other techniques...I think that these problems are reasonable, and could be on an aime, but since there is nobody preventing the use of a calculator during a mock test, I don't think I will use this type of problem again.[/quote]\r\nWhy would I need magic? It only took me a few minutes to find 13, 17, and checking everything under the square root of 181 finds that it is prime. What was your intended solution, and could you give a hint for 12?",
  "Solution_12": "[hide=\"hint for 12\"] $x^{3}-x(y^{2}+z^{2}) = y^{3}-y(z^{2}+x^{2})$ tells you a lot more than it looks like it tells you. [/hide]",
  "Solution_13": "[quote=\"paladin8\"][hide=\"hint for 12\"] $x^{3}-x(y^{2}+z^{2}) = y^{3}-y(z^{2}+x^{2})$ tells you a lot more than it looks like it tells you. [/hide][/quote]\r\nSet the difference of cubes equal to the other terms?",
  "Solution_14": "[quote=\"PenguinIntegral\"][quote=\"paladin8\"][hide=\"hint for 12\"] $x^{3}-x(y^{2}+z^{2}) = y^{3}-y(z^{2}+x^{2})$ tells you a lot more than it looks like it tells you. [/hide][/quote]\nSet the difference of cubes equal to the other terms?[/quote]\r\n\r\nYes... and what do you get?",
  "Solution_15": "[quote=\"paladin8\"][quote=\"PenguinIntegral\"][quote=\"paladin8\"][hide=\"hint for 12\"] $x^{3}-x(y^{2}+z^{2}) = y^{3}-y(z^{2}+x^{2})$ tells you a lot more than it looks like it tells you. [/hide][/quote]\nSet the difference of cubes equal to the other terms?[/quote]\n\nYes... and what do you get?[/quote]\r\nSomething like $(x-y)(x^{2}+xy+y^{2})=x(y^{2}+z^{2}-y(z^{2}+x^{2})$\r\n\r\nIf we expand that, $(x-y)(x^{2}+xy+y^{2})=x(y^{2}+z^{2}-y(z^{2}+x^{2})=xy^{2}+xz^{2}-yz^{2}-yx^{2}=xz^{2}-yz^{2}=z^{2}(x-y)$\r\n$(x^{2}+xy+y^{2})=z^{2}$?",
  "Solution_16": "[hide=\"5.\"]1,599,999,999 can be written as $16*a^{8}-1$ if we let $a=10$, then factoring: \n\n\\[16*a^{8}-1=(4a^{4}+1)(4a^{4}-1) =(2a^{2}-2a+1)(2a^{2}+2a+1)(2a^{2}+1)(2a^{2}-1)\\]\n\nHence, we get the factorization $(181)(221)(201)(199)$. It is easy to check these factors for divisibility by $2,3,5,7,11,13$ [we need not go higher because $15^{2}>181,221,201,199$]. We obtain the prime factorization as $3*13*17*67*181*199$. Hence the answer is $\\boxed{181}$[/hide]\n\n[hide=\"12.\"]\n\nSubtracting equations gives:\n\n\\[(x^{3}-y^{3})-z^{2}(x-y)-(xy)(x-y)=(x-y)(x^{2}-2xy+y^{2}-z^{2})\\]\n\nThe second condition gives us that $x\\ne y$, so we have that $(x+y)^{2}=z^{2}$, likewise, we have $(y+z)^{2}=x^{2}$, and $(z+x)^{2}=y^{2}$. Adding these and dividing by 2 gives: \n\n\\[x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=0\\]\n\nSubtracting $(x+y)^{2}=z^{2}$ gives:\n\n\\[2yz+2zx+2z^{2}=0\\implies (z)(x+y+z)=0\\]\n\nWe throw out $z=0$, and have $x+y+z=0$. Plugging this into the last statement gives $x=1$. Then if $y=t$, then $z=-1-t$. We can finish by completing the square:\n\n\\[(1)^{2}+(t)^{2}+(-1-t)^{2}=2t^{2}+2t+2=2\\left(t+\\frac{1}{2}\\right)^{2}+3\\]\n\nThe minimum does not occur...but it can be approached...this was my mistake on the test. Anyway, the answer is $\\boxed{3}$.\n[/hide]",
  "Solution_17": "Is 8 just a recursive mess?",
  "Solution_18": "Oh hah I forgot to submit this...Well lucky for an AIME...no careless mistakes, i got 15!\r\n\r\nI liked it: pretty much all clean problems, no major computation.  But one question I had: I got that there's only onep ossible real part for k in #10...",
  "Solution_19": "[quote=\"K81o7\"]Oh hah I forgot to submit this...Well lucky for an AIME...no careless mistakes, i got 15!\n\nI liked it: pretty much all clean problems, no major computation.  But one question I had: I got that there's only onep ossible real part for k in #10...[/quote]\r\n\r\no rly?\r\n\r\nin retrospect, #10 was not so good for an aime...the point was to prove that the sum was invariant...i could not think of how to phrase the problem so that you would need to find the invariant...\r\n\r\n#8 is pretty simple recursion, if you do it right, it becomes 1.1^n...",
  "Solution_20": "[quote=\"Altheman\"]\n#8 is pretty simple recursion1.1^n...[/quote]\r\nO rly?",
  "Solution_21": "ya rly",
  "Solution_22": "Can someone show me how to do this one: \r\n\r\n\r\n9. If $p(x)=x^{3}-5x^{2}+4x-1$ has roots $a$, $b$, and $c$, then find $\\left(a^{2}-\\frac{1}{a^{2}}\\right)\\left(b^{2}-\\frac{1}{b^{2}}\\right)\\left(c^{2}-\\frac{1}{c^{2}}\\right)$.",
  "Solution_23": "[quote=\"pkothari13\"]Can someone show me how to do this one: \n\n\n9. If $p(x)=x^{3}-5x^{2}+4x-1$ has roots $a$, $b$, and $c$, then find $\\left(a^{2}-\\frac{1}{a^{2}}\\right)\\left(b^{2}-\\frac{1}{b^{2}}\\right)\\left(c^{2}-\\frac{1}{c^{2}}\\right)$.[/quote]\r\n\r\n\r\n[hide=\"9\"]Note that $p(x)=(x-a)(x-b)(x-c)$ by the factor theorem. Then we factor to employ that fact. [The product is taken over a,b, and c].\n\n\\begin{eqnarray*}\\left(a^{2}-\\frac{1}{a^{2}}\\right)\\left(b^{2}-\\frac{1}{b^{2}}\\right)\\left(c^{2}-\\frac{1}{c^{2}}\\right)&=&\\frac{1}{(abc)^{2}}(a^{4}-1)(b^{4}-1)(c^{4}-1)\\\\ &=&\\frac{1}{(abc)^{2}}\\prod (1-a)(-1-a)(i-a)(-i-a)\\\\ &=&\\frac{1}{(abc)}*p(1)p(-1)p(i)p(-i)\\\\ &=&\\frac{1}{(-1)^{2}}*(-1)(-11)(4+3i)(4-3i)\\\\ &=&275\\\\ \\end{eqnarray*}[/hide]\r\n\r\nEDIT: typo",
  "Solution_24": "[quote=\"Altheman\"][hide=\"9\"]Note that $p(x)=(x-a)(x-b)(x-c)$ by the factor theorem. Then we factor to employ that fact. [The product is taken over a,b, and c].\n\n\\begin{eqnarray*}\\left(a^{2}-\\frac{1}{a^{2}}\\right)\\left(b^{2}-\\frac{1}{b^{2}}\\right)\\left(c^{2}-\\frac{1}{c^{2}}\\right)&=&\\frac{1}{(abc)^{2}}(a^{4}-1)(b^{4}-1)(c^{4}-1)\\\\ &=&\\frac{1}{(abc)^{2}}\\prod (1-a)(-1-a)(i-a)(-i-a)\\\\ &=&\\frac{1}{(abc)}*p(1)p(-1)p(i)p(-1)\\\\ &=&\\frac{1}{(-1)^{2}}*(-1)(-11)(4+3i)(4-3i)\\\\ &=&275\\\\ \\end{eqnarray*}[/hide][/quote]\r\n\r\nAh! That's a great solution! Thanks man  :)",
  "Solution_25": "[quote=\"pkothari13\"][quote=\"Altheman\"][hide=\"9\"]Note that $p(x)=(x-a)(x-b)(x-c)$ by the factor theorem. Then we factor to employ that fact. [The product is taken over a,b, and c].\n\n\\begin{eqnarray*}\\left(a^{2}-\\frac{1}{a^{2}}\\right)\\left(b^{2}-\\frac{1}{b^{2}}\\right)\\left(c^{2}-\\frac{1}{c^{2}}\\right)&=&\\frac{1}{(abc)^{2}}(a^{4}-1)(b^{4}-1)(c^{4}-1)\\\\ &=&\\frac{1}{(abc)^{2}}\\prod (1-a)(-1-a)(i-a)(-i-a)\\\\ &=&\\frac{1}{(abc)}*p(1)p(-1)p(i)p(-1)\\\\ &=&\\frac{1}{(-1)^{2}}*(-1)(-11)(4+3i)(4-3i)\\\\ &=&275\\\\ \\end{eqnarray*}[/hide][/quote]\n\nAh! That's a great solution! Thanks man  :)[/quote]\r\nCan't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?",
  "Solution_26": "abc is 1(from vietas), not -1, but it doesnt make a difference when you are squaring it..\r\n\r\nI just kept on playing around with a+b+c, ab+ac+bc, and abc..until you get the answer..",
  "Solution_27": "[quote=\"PenguinIntegral\"]Can't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?[/quote]\r\n\r\nbut that is ugly...and this technique is useful in other contexts...like this year's hmmt algebra #10, for example",
  "Solution_28": "Wow I would have had 11, but I missed #5 because i din't see that $3|201$.  I feel dumb, but not as dumb as arnav missing number one lol.",
  "Solution_29": "[quote=\"Altheman\"][quote=\"PenguinIntegral\"]Can't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?[/quote]\n\nbut that is ugly...and this technique is useful in other contexts...like this year's hmmt algebra #10, for example[/quote]\r\nBut the elegant method takes thought, and it's a nine. I really don't see the AMC putting that kind of textbook question that late in the test.\r\n\r\n@nat mac\r\n\r\nArnav does not \"miss\" math problems. He just typed in the wrong answer.",
  "Solution_30": "Sorry, PenguinIntegral, could you explain your (the AoPS 2) method more fully?\r\n\r\nI solved the problem the Altheman way.",
  "Solution_31": "[quote=\"Ravi B\"]Sorry, PenguinIntegral, could you explain your (the AoPS 2) method more fully?\n\nI solved the problem the Altheman way.[/quote]\r\nI think he means newton's sums",
  "Solution_32": "Heh, I had a pretty ugly solution to 9\r\n\r\n[hide]\nMatching coefficients with the roots, we have\n\\begin{eqnarray*}a+b+c &=& 5\\\\ ab+bc+ac &=&4\\\\ abc &=& 1 \\end{eqnarray*}\n\nExpand out the product to eventually get\n\\[(abc)^{2}-\\left(\\frac{1}{abc}\\right)^{2}+\\frac{a^{4}+b^{4}+c^{4}}{(abc)^{2}}-\\frac{(ab)^{4}+(bc)^{4}+(ac)^{4}}{(abc)^{2}}. \\]\n\nTo find $a^{4}+b^{4}+c^{4}$, apply Newton's sums on the original polynomial. For the last term, the polynomial with roots $ab, bc, ca$ is\n\\begin{eqnarray*}(x-ab)(x-bc)(x-ac) &=& x^{3}-(ab+bc+ac)x^{2}+(a^{2}bc+ab^{2}c+abc^{2})x-(abc)^{2}\\\\ &=& x^{3}-4x^{2}+5x-1 \\end{eqnarray*}\nSo to find $(ab)^{4}+(bc)^{4}+(ac)^{4}$, apply Newton's sum on the above polynomial.\n[/hide]\r\n\r\nOh, and I thought question 1 was kinda hard, relative to its position in the test (I got it wrong,  :blush: )",
  "Solution_33": "I tried Newton's Sums but they don't like me :( . (I had to look them up in AoPS 2 and I still got it wrong :( )",
  "Solution_34": "[quote=\"Ravi B\"]Sorry, PenguinIntegral, could you explain your (the AoPS 2) method more fully?\n\nI solved the problem the Altheman way.[/quote]\r\nLook somewhere in the polynomial section where it shows how to construct polynomials that have, for example, roots that are the reciprocals of the original polynomial.",
  "Solution_35": "Thanks.  I looked at AoPS 2, and see how to construct a polynomial whose roots are reciprocals of the originals, or a multiple of the originals.  But constructing a polynomial with roots $a^{2}-\\frac{1}{a^{2}}$ (and same for $b$ and $c$) seems harder.  At least I don't see a nice method.  Do you have one?",
  "Solution_36": "Hm, I remember having a nice way of doing this, but I don't remember what it was, and I threw away my scratch work... :wallbash:",
  "Solution_37": "[quote=\"Ravi B\"]Thanks.  I looked at AoPS 2, and see how to construct a polynomial whose roots are reciprocals of the originals, or a multiple of the originals.  But constructing a polynomial with roots $a^{2}-\\frac{1}{a^{2}}$ (and same for $b$ and $c$) seems harder.  At least I don't see a nice method.  Do you have one?[/quote]\r\nI'm sure there is a way to do it. I haven't really thought about it much...",
  "Solution_38": "[quote=\"PenguinIntegral\"][quote=\"Altheman\"][quote=\"PenguinIntegral\"]Can't we just use standard AoPS2 methods to find the polynomial with those roots then find the constant term?[/quote]\n\nbut that is ugly...and this technique is useful in other contexts...like this year's hmmt algebra #10, for example[/quote]\nBut the elegant method takes thought, and it's a nine. I really don't see the AMC putting that kind of textbook question that late in the test.[/quote]\r\n\r\nConstructing a polynomial with such roots is complicated, cumbersome, and prone to error. Even for simple problems, it's usually easier to use an alternate approach; a problem with roots as relatively complicated as it was in Altheman's AIME is even more so unwieldy to solve with construction methods. \r\n\r\nOn the other hand, the elegant solution is quick, extremely nice, and FAR more applicable; even simple problems are slaughtered by the factor theorem approach.",
  "Solution_39": "what are the mock AIME for?",
  "Solution_40": "[quote=\"yhh3002\"]what are the mock AIME for?[/quote]\r\npractice and fun"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "I think that the integral $\\displaystyle \\int \\sqrt{ \\tan x} dx$ was discussed last week here. So I was thinking about computing:\r\n\r\n$\\displaystyle \\int \\sqrt[3]{ \\tan x} dx$ and $\\displaystyle \\int \\sqrt[3]{ \r\nctan x} dx$ for $\\displaystyle x  \\in \\left( 0, \\frac{ \\pi}{2} \\right).$\r\n\r\nThen try:\r\n\r\n$\\displaystyle \\int \\sqrt[n]{ \\tan x} dx$ and $\\displaystyle \\int \\sqrt[n]{ ctan x} dx$ for $\\displaystyle x  \\in \\left( 0, \\frac{ \\pi}{2} \\right),$ $n \\geq 2,$ $n \\in N.$\r\n\r\ncheers! :D :D",
  "Solution_1": "Let $\\sqrt[3]{\\tan x} = u$, $x = \\arctan u^3$, thus\r\n$dx = \\frac{3 u^2}{1 + u^6}$\r\n\r\n$\\int \\sqrt[3]{\\tan x} dx = \\int \\frac{3 u^3}{1+u^6} du$\r\n\r\nWe have reduced it to a rational form.\r\n\r\n$\\frac{u^3}{1+u^6} = \\frac{A}{u^2+1} + \\frac{B}{u^2-\\sqrt3 u+1} + \\frac{C}{u^2+\\sqrt3 u+1}$ \r\n\r\nFind $A, B, C$ in this way... and we get the primitive.",
  "Solution_2": "For $A=\\int \\sqrt[n]{\\tan x} dx$, $x=\\arctan u^n$ \r\n$A=n\\int\\frac{u^n}{1+u^{2n}}du$\r\n\r\nPartial fraction in $\\mathbb C[X]$ and coming back on $\\mathbb R[X]$\r\n\r\ngives\r\n\r\nif $m,n\\geq 2$ are positive integer with $m<2n$\r\n\r\n$\\int \\frac{u^{m-1}}{1+u^{2n}}du = \\frac{-1}{2n}\\sum_{k=1}^{n}\r\n\\cos(a_k)\\ln(1-2x\\cos(b_k)+x^2)+\\frac{1}{n}\\sum_{k=1}^{n}\\sin(a_k)\\arctan(\\frac{x-\\cos(b_k)}{sin(b_k)})$\r\n\r\nWith $a_k = \\frac{m\\pi(2k-1)}{2n}$, $b_k=\\frac{(2k-1)\\pi}{2n}$"
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "$ a$, $ b$ are positive non-integers numbers such a+b is an integer:\r\na)is it possible to a^20 + b^20 be an integer?\r\nb)and a^21 + b^21 ?",
  "Solution_1": "If $ a\\plus{}b$ and $ ab$ integers, then $ a^n\\plus{}b^n$ integer too for any natural n. For example $ a\\equal{}3\\minus{}3\\sqrt 2, b\\equal{}3\\plus{}2\\sqrt 2$."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Find all integers n such that\r\n2^(n-1)/n!",
  "Solution_1": "$V_{2}(n!)=\\sum_{k=1}^{+\\infty}[n/2^{k}]$\r\n\r\n$2^{n-1}|n!$  if  $V_{2}(n!)\\geq n-1$\r\n\r\n$V_{2}(n!) < n\\sum_{k=1}^{+\\infty}1/2^{k}=n$\r\n$V_{2}\\ge n-1$   =>  $V_{2}(n!)=n-1$\r\nwe mut have for any k, $E(n/2^{k})=n/2^{k}$ or $E(n/2^{k})=0$\r\n\r\nn must be a power of 2\r\n\r\nS={$2^{m}$ / m de IN}"
}
{
  "Tag": [
    "function",
    "analytic geometry"
  ],
  "Problem": "I'm looking over functions and graphs.  can anyone explain to me why you move the graph $y=x^{2}$, two to the right to graph this:$f(x)=(x-2)^{2}$why is it that you move right when its$x-\\textbf{3}$?  and why does the graph steches horizontally by a factor of 2 when the graph is $f(x)=(2x)^{2}$ ?  seems like everything for x is backwards...\r\n\r\nThank you  :P\r\n\r\n[color=red][size=59]Preview your posts before posting  :wink: \nb-flat[/color][/size]",
  "Solution_1": "Don't EVER think graph. Think COORDINATE SYSTEM.\r\n\r\nIf you put $x-2$ instead of $x$, you effectively move the x-axis for $-2$, that is for $2$ to the left, which appears as the graph moved to the right. Similarly, if you put $x+3$, you move x-axis for $+3$, that is $3$ to the right, which apparently moves the graph $3$ to the left.\r\n\r\nIf you put $2x$ instead of $x$, you multiply every number on x-axis by $2$, which shrinks it twice, and that looks like the graph has been stretched twice.\r\n\r\nAnd so on.\r\n\r\nRemember: Whatever happens to the coordinates, it's applicable to the COORDINATES (not the graph), therefore from the graph point of view, exactly the \"opposite\" happens.",
  "Solution_2": "but that doesnt happen for the y values.  $y=2(x)^{2}$  The numbers on the y axis aren't multiplied by 2.  and why are the numbers on the x axis doubled for $y=(2x)^{2}$ ?  \r\n\r\nthanks for your help",
  "Solution_3": "It DOES happen: Just as you put $2x$ instead of $x$, you put $2y$ instead of $y$, hence the y-axis is shrank twice, meaning the graph is stretched twice along the y-axis.\r\n\r\nTherefore, if you have the graph of $y=f(x)$, then the graph of $y=Cf(ax+b)+D$ is obtained by writing $ax+b$ instead of every $x$ on x-axis and writing $Cy+D$ instead of every $y$ on y-axis. That results in the following: (1) x-axis is shrank $a$ times (\"shrinking\" is to be understood relatively, i.e. it IS shrinking if $a>1$ and becomes stretching if $a<1$) and moved right by $b$ (again relatively: \"right\" by a negative number is \"left\"), hence the graph is stretched $a$ times horizontally and moved left by $b$; (2) y-axis is shrank $C$ times (relatively) and moved up by $D$ (where \"up\" by a negative number is \"down\"), hence the graph is stretched $C$ times vertically and moved down by $D$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "calculus",
    "derivative",
    "geometry",
    "3D geometry",
    "calculus computations"
  ],
  "Problem": "Okay, so I had to estimate the sq. root of 7.9 with a Taylor polynomial of degree three.\r\n\r\nI based the polynomial at 8.\r\n\r\nFirst I calculated the 1st-3rd derivatives of x^(1/2):\r\ng'(x) = 1/2(x)^(-1/2)\r\ng''(x) = -1/4(x)^(-3/2)\r\ng'''(x) = 3/8(x)^(-5/2)\r\nThen there's g(x), which is just the sq. root of 8.\r\n\r\nWhen plugging in 8 for x, you get for the Taylor polynomial of degree three:\r\n\r\nT3(x) = sqrt[8] + 1/2(8)^-1/2 (x-8) + 1/2 * -1/4(8)^(-3/2) (x-8)^2 + 1/6 * 3/8(8)^(-5/2) (x-8)^3\r\n\r\nPlugging in 7.9 for x in the polynomial above, you get 2.810693867, which is not far off from what you get when you enter the square root of 7.9 into a calculcator. When you enter it into the calculator, you get 2.810693865.\r\n\r\nAnd now for the question that is confusing me:\r\nApply Taylor's theorem to find an upper bound on the error in this approximation.\r\n.\r\nIf someone could explain how this is done, I would REALLY appreciate it. Thanks in advance!\r\n\r\n(And one more thing:  I think I might have been wrong in using 8 for approximating the square root of 7.9 since 8 isn't a perfect square.  Would it be more appropriate to use 9 instead?)",
  "Solution_1": "[quote=\"SharkBites77\"]I think I might have been wrong in using 8 for approximating the square root of 7.9 since 8 isn't a perfect square.  Would it be more appropriate to use 9 instead?)[/quote]\r\n\r\nYes.  Notice that in your polynomial you had to evaluate the square root of $8$ implicitly several times, which is not an integer. \r\n\r\nThe choice of $7.9$ is unusual for a square root, though.  Are you sure the problem didn't say [i]cube[/i] root?",
  "Solution_2": "Yeah, it did say sq. root of 7.9.",
  "Solution_3": "Hmm.\r\n\r\nAnyway, once you've found the correct taylor series centered around $9$, consult\r\n\r\nhttp://en.wikipedia.org/wiki/Taylor's_theorem\r\n\r\nTaylor's theorem essentially says that an upper bound for the error when you take the series to $n$ terms is given by the $(n+1)^{th}$ term, except that instead of\r\n\r\n$\\frac{ f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}$\r\n\r\nThe input for $f^{(n+1)}$ is some $\\zeta$ between $a, x$ that maximizes $f^{(n+1)}$.",
  "Solution_4": "Thanks!  I'm a bit confused to how I would find this maximum value, though."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "[color=blue][size=150]a, b and c belongs to the interval [1/3,1].\nprouve that  a/(a+b)+b/(b+c)+c/(c+a)>=7/5 [/size][/color]",
  "Solution_1": "See here http://www.mathlinks.ro/Forum/viewtopic.php?t=134826",
  "Solution_2": "also you can uyse it http://www.mathlinks.ro/Forum/viewtopic.php?t=135954"
}
{
  "Tag": [
    "search"
  ],
  "Problem": "If anybody has links to sites which have practice SATs, it would greatly be appreciated, thanx :)",
  "Solution_1": "Why not buy a book of the \"10 Real SATs\" sort? They're cheap at $20 and contain a lot of tests.",
  "Solution_2": "Kaplan also has good practice SAT's.",
  "Solution_3": "Yes, but I've bought three: Kaplan's and Baron's, and Master the SAT, and finished all of them.",
  "Solution_4": "Ok, [url=http://www.educationplanner.com/education_planner/preparing_article.asp?sponsor=2859&articleName=SAT_Practice_Tests]try this one[/url] :) It has two practice tests (in pdf files). Does it help?\r\n\r\nYou can go to google and search \"SAT practice tests\" etc. ;)",
  "Solution_5": "[quote=\"frt\"]Ok, [url=http://www.educationplanner.com/education_planner/preparing_article.asp?sponsor=2859&articleName=SAT_Practice_Tests]try this one[/url] :) It has two practice tests (in pdf files). Does it help?\n\nYou can go to google and search \"SAT practice tests\" etc. ;)[/quote]\r\n\r\nYup, good link frt  :D",
  "Solution_6": "Really the one and only SAT prep book you need is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0874477182/learninfreed]The Official SAT Study Guide: For the New SAT[/url] by the College Board. It is very easy to waste a lot of time on test prep until test prep becomes your only EC--which makes your college application look kind of lame. Just use the REAL materials from the College Board for test prep, and use the rest of your time for worthwhile activities like reading books you enjoy on subjects you are interested in, participating in your favorite extracurricular activities, keeping up your school grades, and sports. It is DEFINITELY possible to study too much for the SAT. \r\n\r\nGood luck on your tests, and on your college applications.",
  "Solution_7": "Yeah, after a certain amount of work, practicing for the (P)SAT becomes a waste."
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "circumcircle",
    "geometry proposed"
  ],
  "Problem": "Given are triangle $ABC$ and $I$ inside it. Called $R_{a},R_{b},R_{c}$ be circumradius of triangles $BIC,CIA,AIB$ and $R$ is circumradius of $ABC$ prove that\r\n$R\\ge \\min\\{R_{a},R_{b},R_{c}\\}$  :)",
  "Solution_1": "Let the points $O_{A},O_{B},O_{C}$ and $O$ be the circumcentres of the triangles $BIC,AIC,AIB$ and $ABC$.Then what means $R<R_{A}$?That means that $\\angle{BOC}>\\angle{BO_{A}C}$!\r\n=>$2\\angle{A}>360^{0}-2\\angle{BIC}$\r\nSo,if we would have that $R<min(R_{A},R_{B},R_{C})$ then \r\n$180^{0}=A+B+C>3*180^{0}-(\\angle{BIC}+\\angle{AIC}+\\angle{AIB})=180^{0}$ :P  :lol: \r\n\r\nIn the solution above I supossed that all the angles $AIC,BIC,AIB$ are $>90^{0}$.I think the other case is easier.",
  "Solution_2": "Ok that's right, thank Tiks for good solution!   :blush:"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "limit"
  ],
  "Problem": "I am not sure whether I should post this in the Calculus forum or not but I feel that maybe the solution won't require calculus at all.\r\n\r\nI proved the following results.\r\n\r\n1. $\\int^{ 1}_{0}x^{\\alpha}e^{x}. dx < \\frac{3}{\\alpha+1}$ for $\\alpha > 0$\r\n\r\n2. There exist integers $a_{n}$, $b_{n}$ such that $\\int^{1}_{0}x^{n}e^{x}. dx = a_{n}+b_{n}e$ where $n$ is a non-negative integer.\r\n\r\n3. If $r$ is a positive rational number so that $r= \\frac{p}{q}$, $(p,q)=1$ where $p,q$ are positive integers then for all integers $a,b$:\r\n\r\n$|a+br| = 0$ or $|a+br| \\geq \\frac{1}{q}$.\r\n\r\nHence, prove $e$ is irrational. \r\n\r\nI have a feeling that I am missing something completely trivial because it is so late.",
  "Solution_1": "If anyone wants to solve this who doesn't know calculus, here are the relevant facts:\r\n\r\nThere is a function $f$ with the following properties:\r\n\r\n1) $f(\\alpha) < \\frac{3}{\\alpha+1}$ for $\\alpha > 0$\r\n2) For $n \\in \\bf N$, $0 < f(n) = a_{n}+b_{n}e$ for some integers $a_{n}, b_{n}$ dependent on $n$.  (The $0 < f(n)$ was missing in the original question, but it's important.  Maybe this is where the trouble came from.  We also could have put it with given-the-first.)\r\n3) For any rational number $r = \\frac{p}{q}$ and any integers $a, b$, either $|a+br| = 0$ or $|a+br| \\geq \\frac{1}{q}$.\r\n\r\nGiven this knowledge, prove that $e$ is irrational.",
  "Solution_2": "See http://en.wikipedia.org/wiki/Euler's_continued_fraction_formula#The_exponential_function for another interesting approach.",
  "Solution_3": "[hide=\"hint\"]\nUse the well known series for $e$:\n\\[S_{n}=\\sum_{k=0}^{n}\\frac{1}{k!}\\Rightarrow e=\\lim_{n\\rightarrow \\infty}S_{n}\\]\nNow come up with bounds to show $n!(e-S_{n})$ is never an integer for each positive integer $n$, and deduce that $n!\\cdot e$ is never an integer.\n[/hide]"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Let $ n$, $ k$, $ p$ be positive integers with $ 2 \\le k \\le \\frac {n}{p \\plus{} 1}$. Let $ n$ distinct points on a circle be given. These points are colored blue and red so that exactly $ k$ points are blue and, on each arc determined by two consecutive blue points in clockwise direction, there are at least $ p$ red points. How many such colorings are there?",
  "Solution_1": "This is rather easy, no? :maybe: \r\n\r\nWell, if not then my solution is probably wrong.\r\n\r\n[hide]Place, the blue points anywhere.  Now, place in between every $ k$ consecutive pairs of blue points the neccessary $ p$ red points.  Now, we have $ n\\minus{}k\\minus{}kp$ points left, all of which must be colored red.  The distinctness of the coloring will depend on how many red points are between each pair of consecutive blue points.  So, label the blue points around the circle $ b_1,...,b_k$.  Then, let $ p_i$ be the number of red points between $ b_i$ and $ b_{i\\plus{}1}$ excluding the neccessary $ p$.  Thus, we have \n\n$ \\sum_{i\\equal{}1}^k p_i\\equal{}n\\minus{}k\\minus{}kp$\n\nThe number of nonnegative integer solutions for $ \\{p_i\\}$ is $ \\binom{n\\minus{}k\\minus{}kp\\plus{}k\\minus{}1}{k\\minus{}1}\\equal{}\\binom{n\\minus{}kp\\minus{}1}{k\\minus{}1}$.  So, now since we overcounted because of the symmetries of the circle, we retract that by dividing by $ k$.  Thus, there are a total of\n\n$ \\frac{1}{k}\\binom{n\\minus{}kp\\minus{}1}{k\\minus{}1}$\n\ncolorings.[/hide]",
  "Solution_2": "[quote=\"cosinator\"]\n\n[hide]Now, we have $ n \\minus{} k \\minus{} kp$ points left[/hide][/quote]\r\n\r\nisn't it $ n \\minus{} k \\minus{} (k \\minus{} 1)p$ points left?",
  "Solution_3": "[quote=\"cosinator\"]This is rather easy, no? :maybe: \n[/quote]\nI don't think so,though it's not hard.But I still haven't found a proof :blush: .\n[quote=\"cosinator\"]\nWell, if not then my solution is probably wrong.\n[/quote]\r\nIt's definitely incorrect,even the answer is incorrect.Just check some particular cases. :wink:",
  "Solution_4": "woops  :blush:",
  "Solution_5": "Bump ......"
}
{
  "Tag": [],
  "Problem": "Supposed A, B, and C are three numbers for which 1001C-2002A=4004 and 1001B+3003A=5005. The average of the numbers A, B, and C is\r\n(A) 1\r\n(B) 3\r\n(C) 6\r\n(D) 9\r\n(E) not uniquely determined",
  "Solution_1": "[quote=\"Totally Zealous\"]Supposed A, B, and C are three numbers for which 1001C-2002A=4004 and 1001B+3003A=5005. The average of the numbers A, B, and C is\n(A) 1\n(B) 3\n(C) 6\n(D) 9\n(E) not uniquely determined[/quote]\r\n\r\n[hide]Add the two equations and divide by 3003 to get B[/hide]",
  "Solution_2": "Add the two equations to get:\r\n1001A + 1001B + 1001C = 9009\r\nDivide by 3003:\r\n(A+B+C)/3=3\r\nTHE ANSWER IS B.",
  "Solution_3": "[quote=\"vishalarul\"]Add the two equations to get:\n1001A + 1001B + 1001C = 9009\nDivide by 3003:\n(A+B+C)/3=3\nTHE ANSWER IS B.[/quote]\r\n\r\nCant you just divide by 1001 in the first place.",
  "Solution_4": "Yeah, I guess you are right. :|",
  "Solution_5": "Supposed A, B, and C are three numbers for which 1001C-2002A=4004 and 1001B+3003A=5005. The average of the numbers A, B, and C is \r\n(A) 1 \r\n(B) 3 \r\n(C) 6 \r\n(D) 9 \r\n(E) not uniquely determined\r\n\r\n1001C-2002A=4004\r\n= 1001C-4004= 2002A\r\n\r\nAnd \r\n\r\n1001B+3003A-2002A= 5005- (1001C-4004)\r\n\r\n1001B+1001A+1001C= 5005 + 4004\r\n\r\n1001(B+A+C) = 9009\r\nB+A+C=9\r\n9/3= 3\r\nThe answer is B",
  "Solution_6": "[hide]Adding the two equations, dividing by 3003 to get 3.\n\n$3$[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "limit",
    "calculus computations"
  ],
  "Problem": "Find\r\n$ \\lim_{a \\to 0} \\int_{ \\minus{} a}^{a} \\frac {1}{\\sqrt {a^{2} \\plus{} 2ba^{n} \\minus{} x^{2} \\minus{} 2bx^{n}}}dx$,$ n$ is an integer",
  "Solution_1": "case 1: n=1\r\n$ I_a\\equal{}\\int_{ \\minus{} a}^{a} \\frac {1}{\\sqrt {(a\\plus{}b)^2\\minus{}(x\\plus{}b)^2}}dx$\r\n\r\n$ \\equal{}\\frac{1}{a\\plus{}b}\\int_{ \\minus{}a}^{a} \\frac {1}{\\sqrt {1\\minus{}\\left(\\frac{x\\plus{}b}{a\\plus{}b}\\right)^2}}dx$\r\n\r\n$ \\equal{}\\int_{\\frac{b\\minus{} a}{b\\plus{}a}}^{1} \\frac {1}{\\sqrt {1\\minus{}u^2}}dx\\equal{}\\frac{\\pi}{2}\\minus{}\\arcsin \\left( \\frac{b\\minus{} a}{b\\plus{}a}\\right)$\r\n\r\nso we get \r\n\r\n$ \\lim_{a\\to 0} I_a \\equal{}\\frac{\\pi}{2}\\minus{}\\arcsin (1) \\equal{}0$\r\n\r\ncase 2: $ n \\ge 2$\r\n\r\n$ I_a\\equal{}\\int_{ \\minus{} a}^{a} \\frac {1}{\\sqrt {(a^2\\minus{}x^2) \\plus{} 2b(a^n \\minus{} x^n)}}dx$\r\n\r\n$ \\equal{}\\int_{ \\minus{} 1}^{1} \\frac {1}{\\sqrt {(1\\minus{}u^2) \\plus{} 2ba^{n\\minus{}2}(1 \\minus{} u^n)}}du$\r\n\r\nso i would GUESS\r\n\r\n$ \\lim_{a\\to 0} I_a \\equal{} \\int_{ \\minus{} 1}^{1} \\frac {1}{\\sqrt {1\\minus{}u^2}}du \\equal{} \\pi$"
}
{
  "Tag": [
    "LaTeX",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "All these posts have fancy formal notation.  My question is two-fold, where is this formal documentation written up?  In other words, where can I learn it other than deciphering it from the posts and 2), most importantly, how do you type <ABC (angle ABC) in that formal bold lettering, is that in MS word?  What program does that?  \r\n\r\nThanks.  HUGE newbie question.",
  "Solution_1": "I think you're probably alluding to latex and if you look to the very left of your screen and click on Latex , all your questions should be answered.",
  "Solution_2": "okay so \"latex\" is that formal notation program?  I clicked on Latex HElp between the Forum and Classes URL buttons on the mathlinks page, but it said it wasn't listed on the server.",
  "Solution_3": "That was a kind of below average response help; no idea what \"top left of screen\" means, but you directed me to latex.  This is the link that you should give next time some one asks about latex:\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=123690"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "incenter",
    "rectangle",
    "inradius",
    "geometry proposed"
  ],
  "Problem": "let  $ o$ is circumcircle of ABC  triangle and <B=90. $ \\omega$  is tangent to AB,BC and $ o$.\r\nprove that : Radius of $ \\omega$=Diagonal of incircle of ABC.",
  "Solution_1": "Let $ X,Y$ denote the tangency points of $ \\omega \\equiv (V)$ with $ BC,BA.$ It's well-known that the polar of vertex $ B$ WRT the B-mixtilinear incircle $ (V)$ (the line XY), goes through the incenter $ I$ of $ \\triangle ABC.$ If $ \\angle ABC=90^{\\circ}$ $\\Longrightarrow$ $I$ is the center of the rectangle $BXVY.$ Thus, it follows that the radius of $ (V)$ is twice the inradius of $ \\triangle ABC.$"
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find all functions $ f: \\mathbb{R}\\rightarrow\\mathbb{R}$ such that for all $ x\\in \\mathbb R$ and $ y\\in f(\\mathbb{R})$, $ f(x\\minus{}y)\\equal{}f(x)\\plus{}xy\\plus{}f(y)$",
  "Solution_1": "Obviosly $ x\\equal{}y$ give $ f(x)\\equal{}\\frac{\\minus{}x^2}{2}\\plus{}c, c\\equal{}f(0)$, and $ x\\equal{}y\\equal{}0$ give $ c\\equal{}0$.",
  "Solution_2": "$ x\\equal{}y\\in f(\\mathbb{R})$ gives $ f(x)\\equal{}{c\\minus{}x^2\\over 2},c: \\equal{}f(0)$ for all $ x\\in f(\\mathbb{R})$.\r\nIf $ f\\not\\equiv 0$ we find some $ y\\in f(\\mathbb{R}),y\\not\\equal{}0$\r\nand then for all $ z\\in\\mathbb{R}$ we find some $ x\\in\\mathbb{R}$ with\r\n$ z\\equal{}xy\\plus{}f(y)\\equal{}f(x\\minus{}y)\\minus{}f(x)\\equal{}a\\minus{}b$ for some $ a,b\\in f(\\mathbb{R})$.\r\nThis gives\r\n$ f(z)\\equal{}f(a\\minus{}b)\\equal{}f(a)\\plus{}ab\\plus{}f(b)\\equal{}{c\\minus{}a^2\\over 2}\\plus{}ab\\plus{}{c\\minus{}b^2\\over 2}\\equal{}c\\minus{}{(a\\minus{}b)^2\\over 2}\\equal{}c\\minus{}{z^2\\over 2}$\r\nfor all $ z\\in\\mathbb{R}$.\r\nAnd now for $ x\\equal{}f(0)\\in f(\\mathbb{R})$ we have $ f(x)\\equal{}{c\\minus{}x^2\\over 2}\\equal{}c\\minus{}{x^2\\over 2}\\Rightarrow c\\equal{}0$.\r\nSo there are $ 2$ solutions $ f\\equiv 0$ and $ f(x)\\equal{}\\minus{}{x^2\\over 2},x\\in\\mathbb{R}$.",
  "Solution_3": "[quote=\"olorin\"]So there are $ 2$ solutions $ f\\equiv 0$ and $ f(x) \\equal{} \\minus{} {x^2\\over 2},x\\in\\mathbb{R}$.[/quote]\r\n$ f\\equiv 0$ is not solution, consider $ x\\equal{}y\\equal{}1$.",
  "Solution_4": "You forgot the restriction that $ y$ is in the range of $ f$, which is why your solution isn't quite right. $ x \\equal{} y \\equal{} 1$ is not valid because $ 1$ is not a value $ f$ can take."
}
{
  "Tag": [
    "calculus",
    "inequalities",
    "algorithm",
    "symmetry",
    "algebra",
    "polynomial",
    "derivative"
  ],
  "Problem": "a,b,c is real numbers, a,b,c>0 and a+b+c=3\r\nprove that:\r\n                    (a+b^2)(b+c^2)(c+a^2) =<13\r\nhelp me!",
  "Solution_1": "I can prove this if $ a^2\\plus{}b^2\\plus{}c^2 \\le 4$, can anyone prove it given $ a^2\\plus{}b^2\\plus{}c^2 \\ge 4$ ?",
  "Solution_2": "The critical points lie within $ a^2 \\plus{} b^2 \\plus{} c^2 > 4$, so that should prove to be the more difficult part.  One proof strategy might be this:  Prove first that the maximum is achieved when one of the values is $ 0$, then the problem reduces to proving\r\n\\[ a(3\\minus{}a)^2(3\\minus{}a\\plus{}a^2)\\leq 13\\]\r\nHow you would go about accomplishing either of these without calculus, however, is beyond me.  Using calculus, you can find the inequality is a bit tighter actually:\r\n\\[ (a \\plus{} b^2)(b \\plus{} c^2)(c \\plus{} a^2)\\leq 12 \\plus{} \\frac {9308 \\plus{} 8309\\cdot 2^{\\frac {2}{3}} 7^{\\frac {1}{3}} \\minus{} 5936\\cdot 2^{\\frac {1}{3}}7^{\\frac {2}{3}}}{9375}\\approx 12.76496\r\n\\]\r\nthe inequality being strict equality for exactly 3 points.",
  "Solution_3": "Using calculus, how do you prove the max is when one of them is 0?",
  "Solution_4": "yes,why one of them is 0?",
  "Solution_5": "See also here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=58379",
  "Solution_6": "Well if we've given up on a non-calculus solution and taken the gloves off, then the proof sketch is as follows.  Let $ f(a,b) \\equal{} (a \\plus{} b^2)(b \\plus{} (3 \\minus{} a \\minus{} b)^2)(3 \\minus{} a \\minus{} b \\plus{} a^2)$.  $ f$ is of fifth-degree in both $ a$ and $ b$, so $ \\frac {\\partial f}{\\partial a}$ and $ \\frac {\\partial f}{\\partial b}$ are both quartic.  Thus, we can find using any algorithm for quartics exact algebraic expressions for the solutions of\r\n\\[ \\frac {\\partial f}{\\partial a} \\equal{} \\frac {\\partial f}{\\partial b} \\equal{} 0\r\n\\]\r\nAfter some ridiculously tedious algebra, you find the critical points, evaluate $ f$ at those points and find that $ f(a^*,b^*) < 9$ for each of them.  Comparing this to, say, $ f(1,0) \\equal{} 12$, you see that the maximum cannot be at a critical point, and therefore the maximum must be on the boundary of the region $ 0\\leq a \\plus{} b \\leq 3$.  This boundary is composed of three line segments, one of which is $ \\lbrace (a,b): b \\equal{} 0,0\\leq a \\leq 3 \\rbrace$.  You then argue by symmetry that $ f$ takes on the same range of values on each line segment (i.e. interchanging any of a,b,c doesn't change the problem), so the maximum must lie on the line $ \\lbrace (a,b): b \\equal{} 0,0\\leq a \\leq 3 \\rbrace$.  This is simply the one dimensional problem of finding the maximum of\r\n\\[ f(a,0) \\equal{} a(3 \\minus{} a)^2(3 \\minus{} a \\plus{} a^2)\r\n\\]\r\non $ 0\\leq a \\leq 3$.  Again, this is a fifth degree polynomial, so its derivative with respect to $ a$ is quartic, and we can obtain exact solutions.  It has two real solutions, one is $ a \\equal{} 3$, which corresponds to a minimum, the other is\r\n\\[ a \\equal{} \\frac {1}{15}\\left(13 \\plus{} 4\\cdot 2^{\\frac {2}{3}} 7^{\\frac {1}{3}} \\minus{} 2^{\\frac {1}{3}} 7^{\\frac {2}{3}}\\right)\r\n\\]\r\nwhich is where $ f$ obtains its maximum.  Substituting this back into $ f$ yields the upper bound I stated in a previous post.",
  "Solution_7": "[quote=\"Math pro\"]Let $ x,y,z$ are nonnegative real numbers such that $ x \\plus{} y \\plus{} z \\equal{} 3,$ prove that\n\n$ \\left(x \\plus{} y^2\\right)\\left(y \\plus{} z^2\\right)\\left(z \\plus{} x^2\\right) < 13.$[/quote]Using the arithmetic-mean \u2014 geometric-mean inequality,\r\n\r\n$ \\frac {1}{3}\\sum_{cyc}{(3x \\plus{} 2y \\plus{} 4z)^3\\left(x \\plus{} y^2\\right)}\\geq\\prod_{cyc}{(3x \\plus{} 2y \\plus{} 4z)}\\sqrt [3]{\\prod_{cyc}{\\left(x \\plus{} y^2\\right)}}.$\r\n\r\nThis comes down to prove that \r\n\r\n$ 3\\sqrt [3]{13}\\prod_{cyc}{(3x \\plus{} 2y \\plus{} 4z)}>\\sum_{cyc}{(3x \\plus{} 2y \\plus{} 4z)^3\\left(x \\plus{} y^2\\right)}.$\r\n\r\nId est, it remains to the following [color=darkred]quintic[/color] homogeneous inequality:\r\n\r\n$ 47(x \\plus{} y \\plus{} z)^2\\prod_{cyc}{(3x \\plus{} 2y \\plus{} 4z)}>20\\sum_{cyc}{(3x \\plus{} 2y \\plus{} 4z)^3\\left[x(x \\plus{} y \\plus{} z) \\plus{} 3y^2\\right]},$\r\n\r\nsince $ x \\plus{} y \\plus{} z \\equal{} 3$ and $ \\sqrt [3]{13} \\equal{} 2.3513\\cdots > 2.35 \\equal{} \\frac {47}{20}.$ But\r\n\r\n$ 47(x \\plus{} y \\plus{} z)^2\\prod_{cyc}{(3x \\plus{} 2y \\plus{} 4z)} \\minus{} 20\\sum_{cyc}{(3x \\plus{} 2y \\plus{} 4z)^3\\left[x(x \\plus{} y \\plus{} z) \\plus{} 3y^2\\right]}$\r\n\r\n$ \\equal{} \\frac {83}{8}\\sum_{cyc}{x(3x \\minus{} 4y)^2\\left(x^2 \\plus{} 6xy \\plus{} 6y^2\\right)} \\plus{} \\frac {117}{8}\\sum_{sym}{x^5} \\plus{} \\frac {75}{4}\\sum_{cyc}{x^4y} \\plus{} \\frac {95}{4}\\sum_{cyc}{x^3y^2}$\r\n\r\n$ \\plus{} 6441\\sum_{sym}{x^3yz} \\plus{} 8676\\sum_{sum}{xy^2z^2} > 0.$\r\n\r\nsee also here : http://www.mathlinks.ro/viewtopic.php?t=146734"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "Prove that $ \\pi$ is irrational!",
  "Solution_1": "http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational\r\n\r\nThere are more proofs out there, and few can be found searching here on ML (and please try searching before you post these quite well known questions - this one, and the one with Zeta function in College Playground).\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=162010\r\nhttp://www.mathlinks.ro/portal.php?t=162051\r\n\r\n\r\n[hide=\"Something to laugh about\"] \n\nhttp://in.answers.yahoo.com/question/index?qid=20090430032537AA7F1RP\n\n :rotfl: [/hide]"
}
{
  "Tag": [
    "limit",
    "trigonometry",
    "function",
    "calculus",
    "calculus computations"
  ],
  "Problem": "Compute the ff. limits:\r\n\r\n$ 1.)  \\lim_{x\\rightarrow\\infty} \\cos (\\pi \\sqrt{4x^4\\plus{}3x^2\\plus{}1}\\minus{}2\\pi x^2)$\r\n\r\n$ 2.)  \\lim_{x\\rightarrow 0} \\frac{| {\\cos(\\sin(3x)) } | \\minus{}1}{x^2}$",
  "Solution_1": "1. $ L \\equal{} \\lim_{x \\to \\plus{} \\infty} \\cos [\\pi (\\sqrt {4x^{4} \\plus{} 3x^{2} \\plus{} 1} \\minus{} 2 x^{2})]$\r\n\r\n$ \\equal{} \\lim_{x \\to \\plus{} \\infty} \\cos \\left(\\pi \\frac {4x^4 \\plus{} 3x^2 \\plus{} 1 \\minus{} 4x^4}{\\sqrt {4x^{4} \\plus{} 3x^{2} \\plus{} 1} \\plus{} 2 x^{2}}\\right)$\r\n\r\n$ \\equal{} \\lim_{x \\to \\plus{} \\infty} \\cos \\left(\\pi \\frac {3 \\plus{} \\frac {1}{x^2}}{\\sqrt {4 \\plus{} \\frac {3}{x^2} \\plus{} \\frac {1}{x^4}} \\plus{} 2}\\right)$\r\n\r\n$ \\equal{} \\cos \\left(\\frac {3 \\pi}{4}\\right)$\r\n\r\n$ \\equal{} \\minus{} \\frac {1}{\\sqrt {2}}$.\r\n\r\n\r\n2. For x close to zero, the function cos is positive, and the function $ \\frac{\\cos(\\sin(3x)) \\minus{} 1}{x^2}$ is even. Hence,\r\n\r\n$ L \\equal{} \\lim_{x \\to 0} \\frac{\\cos(\\sin(3x)) \\minus{} 1}{x^2} \\equal{} \\minus{} \\lim_{x \\to 0} \\frac{1 \\minus{} \\cos^2(\\sin(3x))}{x^2(1 \\plus{} \\cos(\\sin(3x)))}$\r\n\r\n$ \\equal{} \\minus{} \\frac12 \\lim_{x \\to 0} \\frac{\\sin^2(\\sin(3x))}{x^2} \\equal{} \\minus{} \\frac92  \\lim_{x \\to 0} \\left[\\frac{\\sin(\\sin(3x))}{\\sin(3x)}\\right]^2 \\cdot \\left[\\frac{\\sin(3x)}{3x}\\right]^2 \\equal{} \\minus{} \\frac92$",
  "Solution_2": "yes, that was easy   :)"
}
{
  "Tag": [
    "limit",
    "number theory",
    "prime factorization",
    "number theory unsolved"
  ],
  "Problem": "[i]Is there any elementary proofs to prove that : $ \\lim_{ n \\to \\infty } \\frac{ \\pi(n)}{n} \\equal{} 0$ :maybe: [/i]",
  "Solution_1": "$ \\displaystyle S: =\\lim_{n \\to +\\infty}{\\frac{\\pi(n)}{n}= \\lim_{k \\to +\\infty}{\\prod_{i=1}^k{\\left(1-\\frac{1}{p_i}\\right)} \\ge 0}}$ where $ p_i$ is the $ i$-th prime (it is true by inclusion-esclusion principle)\r\nNow if $ y \\in \\mathbb{R}^+$ then $ 1-y \\le e^{-y}$, with equality iff y=0. It follows that $ S \\le e^{\\displaystyle -\\lim_{k \\to +\\infty}{\\sum_{i=1}^k{\\left(\\frac{1}{p_i}\\right)}}}$.\r\nNow assume that $ \\sum_{i=1}^{\\infty}{\\left(\\frac{1}{p_i}\\right)}$ converges, then there exists $ v \\in \\mathbb{N}$ such that $ \\sum_{i=v+1}^{+\\infty}{\\left(\\frac{1}{p_i}\\right)}<\\frac{1}{2}$. Define $ t: =\\prod_{i=1}^v{p_i}$; because of $ p_j \\nmid nt+1$ forall $ 1 \\le j \\le v$ we must have $ \\displaystyle \\sum_{n=1}^{+\\infty}{\\frac{1}{nt+1}} \\le \\prod_{i=v+1}^{\\infty}{ \\sum_{k=0}^{\\infty}{\\frac{1}{p_i^k}}}$(*), but the last rhs is $ \\le \\prod_{i=v+1}^{\\infty}{\\left(1+\\frac{2}{p_i}\\right)}$ that converges for assumption, contracting that lhs of (*) diverges. So such $ v \\in \\mathbb{N}$ does not exist. It implies that $ \\sum_{i=1}^{\\infty}{\\left(\\frac{1}{p_i}\\right)}$ diverges, and so $ \\displaystyle S: =\\lim_{n \\to +\\infty}{\\frac{\\pi(n)}{n}}$ is exactly 0.",
  "Solution_2": "Your first statement isn't quite rigorous since we don't know that the limit exists; what you're computing is an upper bound on the $ \\limsup$.  \r\n\r\nThe second half of the argument can be made more elementary.  Equivalently, it suffices to show that\r\n\r\n$ \\lim_{k \\to \\infty} \\prod_{i \\equal{} 1}^{k} \\left( \\frac {1}{1 \\minus{} \\frac {1}{p_i} } \\right) \\equal{} \\infty$.\r\n\r\nBut this is classic:  by unique prime factorization we have the bound\r\n\r\n$ \\prod_{i \\equal{} 1}^{k} \\left( \\frac {1}{1 \\minus{} \\frac {1}{p_i} } \\right) \\ge \\sum_{i \\equal{} 1}^{p_k} \\frac {1}{i}$\r\n\r\nand it is elementary to show that the harmonic series diverges, i.e. by Cauchy condensation.  The conclusion follows."
}
{
  "Tag": [
    "function",
    "limit",
    "induction",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "Let the functions f_n:[e,oo)-->R, defined recurently like so f_0(x)=ln(x),\r\nf_(n+1) (x)=ln(x*f_n(x)), n is natural. Then we have:\r\n\r\n \\lim (x-->e) (f_n(x))^(1/(x-e)=e^(n+1)/e, n is natural.\r\n\r\ncheers!!",
  "Solution_1": "I used L'Hospital for this. We are trying to prove that ln(f_(n+1)(x))/x-e tends to (n+2)/e as x->e. It's obvious by induction that f_n(x)->1 as x->e so we have an indetermination so we can apply L'Hospital. \r\nLet's assume that we have proved P_n which states that (g_n)'(x)->(n+1)/e as x->e, where g_n(x)= ln(f_n(x)). Then, by L'Hospital we also have g_n(x)/x-e -> (n+1)/e as x->e. Now let's prove that (g_(n+1))'(x)->(n+2)/e. (g_(n+1))'(x)=(1/ f_(n+1)(x))* f_(n+1)'(x)= (1/ f_(n+1)(x))* (1/x+f_n'(x)/f_n(x)). But we can observe that f_n'(x)/f_n(x) = (g_n)'(x)->(n+1)/e as x->e, and 1/x->1/e as x->e and 1/ f_(n+1)(x)->1 as x->e, so by combining these we get that (g_(n+1))'(x) -> (n+2)/e. So by induction (g_n)'(x)->(n+1)/e as x->e for any n. Now by applying L'Hospital to these we get g_n(x)/(x-e) -> (n+1)/e as x->e. If we put e^(this limit we just got) we obtain the limit we wanted.\r\n\r\nSee ya!\r\n\r\n :D"
}
{
  "Tag": [
    "geometry",
    "power of a point",
    "radical axis",
    "geometry proposed"
  ],
  "Problem": "Let $w=C(O,r)$ be a fixed circle and let $A$ be a fixed point such that $OA>r$.\r\n\r\nFor a variable circle $x,\\ A\\in x,\\ x\\cap w\\ne \\emptyset$, I note the tangent $t$ in the point $A$ to the circle $x$\r\n\r\nand the common chord $c$ of the circles $w,x$. Ascertain the geometrical locus of the point $L\\in t\\cap c$.\r\n\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=54479",
  "Solution_1": "$A$ is a circle of zero radius. The tangent $t$ is the radical axis of the circles $(A, x)$, the chord $c$ is the radical axis of the circles $(w, x)$. $L \\equiv t \\cap c$ is the radical center of the circles $(A, w, x)$ which lies on the radical axis of the circles $(A, w)$ perpendicular to their center line AO. Let the last radical axis intersect the center line AO at a point R.\r\n \r\n$AR^2 = (RO + r)(RO - r) = RO^2 - r^2 = (AO - AR)^2 - r^2$\r\n\r\n$0 = AO^2 - 2AO \\cdot AR - r^2,\\ \\ \\ AR = \\frac{AO^2 - r^2}{2 AO}$"
}
{
  "Tag": [
    "MATHCOUNTS",
    "geometry",
    "3D geometry",
    "prism"
  ],
  "Problem": "Hey who do you think is going to pwn this year?\r\nI practiced and did alot but I think Mark Sellke, Yushi and David are going to pwn. I don't think Vivian Gu does MC this year.\r\n\r\n\r\nBlah got 35th-60th last year, lots of people with 25-28",
  "Solution_1": "duh they're gonna pwn. especially selke, as he should make masters round in '10. sorry for the slight invasion of your forum, guys.",
  "Solution_2": "That's fine as long as we're not not talking about say pythag.\r\n\r\nBy the way is masters top 4 cd or written?",
  "Solution_3": "[quote=\"AwesomeToad\"]That's fine as long as we're not not talking about say pythag.\n\nBy the way is masters top 4 cd or written?[/quote]\r\n\r\nwritten\r\n\r\ncuz I made it...",
  "Solution_4": "dnkwin did u really make the nats?\r\nwhat is ur real name?",
  "Solution_5": "yes he did and he did very well\r\njust pm him instead of posting here\r\nno names to usernames mentioned on the forum\r\n\r\n=)",
  "Solution_6": "bleh.  In 6th grade I got 63rd(23/46. Blah.  Careless mistakes. :( )\r\nNow I'm in 7th grade, trying again......I'm probably gonna get like 25th this time at best.\r\n\r\nI would really love to make nats and that would be so awesome, but I live in Indiana.  CURSE THAT!  IF only I lived in say Oklahoma or North Dakota.......then I'd actually have a chance.  bleh. :(\r\n\r\nAlso, if any of you see Doom Monkey, he's in my school.\r\n\r\nWhat do you think the difficulty of the test will be?  Easyish or hardish?",
  "Solution_7": "[quote=\"AwesomeToad\"]Hey who do you think is going to pwn this year?\nI practiced and did alot but I think Mark Sellke, Yushi and David are going to pwn. I don't think Vivian Gu does MC this year.\n\n\nBlah got 35th-60th last year, lots of people with 25-28[/quote]\\\r\n\r\nI think Vivian Gu is still doing Mathcounts this year. In 2009, she was a 7th grader. :wink:",
  "Solution_8": "[quote=\"maybach\"]Easyish or hardish?[/quote]\r\n\r\nI think it's gonna be middle-ish.",
  "Solution_9": "I think I'm going to win",
  "Solution_10": "[quote=\"ernie\"][quote=\"maybach\"]Easyish or hardish?[/quote]\n\nI think it's gonna be middle-ish.[/quote]\r\n\r\nUm.........similar to, harder than, or easier than the 2008 State?(I hope the latter)",
  "Solution_11": "[quote=\"gauss1181\"][quote=\"AwesomeToad\"]Hey who do you think is going to pwn this year?\nI practiced and did alot but I think Mark Sellke, Yushi and David are going to pwn. I don't think Vivian Gu does MC this year.\n\n\nBlah got 35th-60th last year, lots of people with 25-28[/quote]\\\n\nI think Vivian Gu is still doing Mathcounts this year. In 2009, she was a 7th grader. :wink:[/quote]\r\n\r\nVivian is NOT doing MC this year. I go to her school and she told me. She said something about GMO.",
  "Solution_12": "[quote=\"maybach\"][quote=\"ernie\"][quote=\"maybach\"]Easyish or hardish?[/quote]\n\nI think it's gonna be middle-ish.[/quote]\n\nUm.........similar to, harder than, or easier than the 2008 State?(I hope the latter)[/quote]\r\n\r\nMaybe a bit harder?",
  "Solution_13": "[quote=\"ernie\"][/quote][quote=\"maybach\"][quote=\"ernie\"][quote=\"maybach\"]Easyish or hardish?[/quote]\n\nI think it's gonna be middle-ish.[/quote]\n\nUm.........similar to, harder than, or easier than the 2008 State?(I hope the latter)[/quote][quote=\"ernie\"]\n\nMaybe a bit harder?[/quote]\r\n2008 State was kinda hard... I only got a 35 on it (compared to 46 this year)... so at the rate mathcounts problems are going I would say a LOT easier...",
  "Solution_14": "It's an unwritten rule that they can only repeat every problem five times so they're running out of ideas.\r\n\r\n[quote=\"AwesomeToad\"]Vivian is NOT doing MC this year. I go to her school and she told me. She said something about GMO.[/quote]\r\n\r\n...Do you mean CGMO? I can't think of anything else...\r\n\r\nAlso, that would be a pretty silly reason (bleh maybe not) not to do MC?",
  "Solution_15": "[quote=\"math154\"]It's an unwritten rule that they can only repeat every problem five times so they're running out of ideas.\n\n[quote=\"AwesomeToad\"]Vivian is NOT doing MC this year. I go to her school and she told me. She said something about GMO.[/quote]\n\n...Do you mean CGMO? I can't think of anything else...\n\nAlso, that would be a pretty silly reason (bleh maybe not) not to do MC?[/quote]\r\n\r\nIt is pretty silly... Also wait what unwritten rule? I thought it was two, not five? (exact wording, of course.)",
  "Solution_16": "[quote=\"AwesomeToad\"]\n\n\nBlah got 35th-60th last year, lots of people with 25-28[/quote]\r\n\r\nArgh.  I got 23/46 in state.  But I took 2009 again and got 28/46.  Still total fail though.............",
  "Solution_17": "I think Vivian meant CGMO. There was something else.\r\n\r\nSomeone at my school said the stupidest thing ever:\r\n[quote]Whats the point of doing MAthcounts again if you already got a trip to Orlando?[/quote]",
  "Solution_18": "[quote=\"AwesomeToad\"]I think Vivian meant CGMO. There was something else.\n\nSomeone at my school said the stupidest thing ever:\n[quote]Whats the point of doing MAthcounts again if you already got a trip to Orlando?[/quote][/quote]\r\n\r\ngetting another trip to nationals..............",
  "Solution_19": "Besides, most people at my school don't do math at all (mostly jocks)\r\nI'd like to raise my national ranking if I got another trip",
  "Solution_20": "heh yeah. we all experience these ppl on a daily basis. \r\n\r\nand, other than the whole nats thing, mc is fun! yay",
  "Solution_21": "[quote=\"pinkmuskrat\"]heh yeah. we all experience these ppl on a daily basis. \n\nand, other than the whole nats thing, mc is fun! yay[/quote]\r\n\r\nWait so the nationals is at Disney again this year?\r\n\r\nThe guy who came 3rd at AAPotter (Purdue) chapter goes to my school, came 7th state\r\n\r\nDoes some really weird stuff at my school.\r\n\r\nNot want to mention his name though...",
  "Solution_22": "AwesomeToad, is that the kid who won state countdown?\r\n\r\nI was there.",
  "Solution_23": "[quote=\"maybach\"]AwesomeToad, is that the kid who won state countdown?\n\nI was there.[/quote]\n\nYes.\nExamples: [quote]Oh I hit you. I get 5 points[/quote]\n???\nor\n[quote]I was up all night playing starcraft blah pwns a korean oh i'm braindead go to bed at 5: 60[/quote]",
  "Solution_24": "Yeah.  I was there feeling really sad watching cause' I really wanted to be up there and I actually thought I did a good job(Man I need to practice on sprint, but I've imrpoved my average score 14 points since).  I saw that dude.  He looked like he was praying and he looked really sad when they read \"7th place\".....feel really sorry for him......poor guy :(\r\n\r\nAlso, could I make cd(not nats) for states with 35/46?  Because that's about what I can get on a good day assuming the test was made recently(2006-).",
  "Solution_25": "[quote=\"AwesomeToad\"]Besides, most people at my school don't do math at all (mostly jocks)\nI'd like to raise my national ranking if I got another trip[/quote]\r\n\r\nif you are supposedly in vivian's school, then you are in my school too. i have no clue who you are though.",
  "Solution_26": "People in 8th grade now.\r\nBut the guy who came 7th was sure he got at least a 27 on sprint and perfect target.\r\n\r\nBut he  got 24/14 so yeah",
  "Solution_27": "I totally agree. But I'll win.",
  "Solution_28": "[quote=\"maybach\"]Yeah.  I was there feeling really sad watching cause' I really wanted to be up there and I actually thought I did a good job(Man I need to practice on sprint, but I've imrpoved my average score 14 points since).  I saw that dude.  He looked like he was praying and he looked really sad when they read \"7th place\".....feel really sorry for him......poor guy :(\n\nAlso, could I make cd(not nats) for states with 35/46?  Because that's about what I can get on a good day assuming the test was made recently(2006-).[/quote]\r\n\r\nI hate to say this but don't feel to sorry for him.\r\n\r\nAlso the 35/46 would probably just barely make state cd better not count on it.\r\nOf course, the level of the competition and test also matter",
  "Solution_29": "[quote=\"AwesomeToad\"][quote=\"maybach\"]Yeah.  I was there feeling really sad watching cause' I really wanted to be up there and I actually thought I did a good job(Man I need to practice on sprint, but I've imrpoved my average score 14 points since).  I saw that dude.  He looked like he was praying and he looked really sad when they read \"7th place\".....feel really sorry for him......poor guy :(\n\nAlso, could I make cd(not nats) for states with 35/46?  Because that's about what I can get on a good day assuming the test was made recently(2006-).[/quote]\n\nI hate to say this but don't feel to sorry for him.\n\nAlso the 35/46 would probably just barely make state cd better not count on it.\nOf course, the level of the competition and test also matter[/quote]\r\n\r\nI agree. Only something with a 50/55 would be enough to win state mc.",
  "Solution_30": "Generally on an average test you would need 43/46\r\n\r\nBut this year Sellke got 45 and year before best was 41.\r\nAlso, what do you mean by 50/55?",
  "Solution_31": "Maybe he was referring to the team score?  But team round doesn't matter in state.  Also, it's too bad they cut the scholarship money.  I think with some practice I could get 20th place, but that wouldn't be good enough for a scholarship anymore :(\r\n\r\nAlso, my sister went to MC State in 8th grade at Rose Hulman and got 163rd place(I got 63rd :rofl:).  I thought the Rose Hulman auditorium seemed nicer.  But only 2 kids get scholarships though..........",
  "Solution_32": "Team Round determines the state coach wow my school did better than last year.\r\n\r\n2 people in CD!!/Top 10",
  "Solution_33": "[quote=\"AwesomeToad\"]Team Round determines the state coach wow my school did better than last year.\n\n2 people in CD!!/Top 10[/quote]\r\n\r\nState Coach doesn't really matter. \r\n\r\nAwesome Toad, you're from West Laffayette, right?\r\n\r\nAnyways the state coach will always be Bob Fischer no matter what. \r\n\r\nAlso, I actually got a 4 on target, but I got a 5 because they misread my answer to the rectangular prism question. :)\r\n\r\nStill did horribly though.........",
  "Solution_34": "Are all of you guys 8th graders and below?",
  "Solution_35": "Yeah.  I'm currently in 7th grade!(thank god, with selkie, I know I have know chance this year)",
  "Solution_36": "Sellke, \r\nLiang, and homma are gonna pwn this year.\r\n\r\nVivians not doing mc, so everyone wants the 4th spot",
  "Solution_37": "[quote=\"AwesomeToad\"][quote=\"maybach\"]AwesomeToad, is that the kid who won state countdown?\n\nI was there.[/quote]\n\nYes.\nExamples: [quote]Oh I hit you. I get 5 points[/quote]\n???\nor\n[quote]I was up all night playing starcraft blah pwns a korean oh i'm braindead go to bed at 5: 60[/quote][/quote]\n\nCorrections:\n\nOh I hit you. I get 2 points. (Currently Retired)\n\nI was up 1/3rd of the night playing starcraft, I pwn koreans, I'm braindead go to bed at 1:00. (Retired as of Friday)\n\nAnyways, meh I'm doing more programming than math right now, but I'll still try for 4th  (although there are some really good 8th graders ready to pwn me) :P. Does anyone know what grade 5th/6th place are in?\n\n[quote=\"Maybach\"]Yeah.  I was there feeling really sad watching cause' I really wanted to be up there and I actually thought I did a good job(Man I need to practice on sprint, but I've imrpoved my average score 14 points since).  I saw that dude.  He looked like he was praying and he looked really sad when they read \"7th place\".....feel really sorry for him......poor guy :(\n\nAlso, could I make cd(not nats) for states with 35/46?  Because that's about what I can get on a good day assuming the test was made recently(2006-).[/quote]\r\n\r\nNah, I didn't really expect to make nats (although it would be awesome if it happened). My goal was to pwn people in CD, and not fail too bad in written.",
  "Solution_38": "Alex Barksdale came in 5th if I remember right\r\nCorrect me if I'm wrong...",
  "Solution_39": "Does anyone have a list of the top 16? 2009 State MC?"
}
{
  "Tag": [
    "probability",
    "geometry"
  ],
  "Problem": "A circle of radius $ r$ is concentric with and outside a regular hexagon of side length $ 2$. The probability that $ 3$ entire sides of the hexagon are visible from a random point on the cirlce is $ 1/2$. What is $ r$?",
  "Solution_1": "just to be sure...\r\n\r\nThe area of visibiliy is enclosed by the tangent to the circle at that point and the closest part of the hexagon right??\r\n\r\n\r\n\r\n :)",
  "Solution_2": "Yes thats correct."
}
{
  "Tag": [
    "trigonometry",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Given $ (x_n)$ and $ (y_n)$ defined as\r\n\\[ x_1 \\equal{} y_1 \\equal{} \\sqrt {3}\\ ; \\ x_{n \\plus{} 1} \\equal{} x_n \\plus{} \\sqrt {1 \\plus{} \\left ({x}_n \\right)^2}\\ ; \\ y_{n \\plus{} 1} \\equal{} \\frac {y_n}{1 \\plus{} \\sqrt {1 \\plus{} \\left (y_n \\right)^2}} \\ \\forall n\\ge 2\\]\r\nProve that $ x_n.y_n$ $ \\in$ $ (2,3)$",
  "Solution_1": "[quote=\"mathVNpro\"]Given $ (x_n)$ and $ (y_n)$ defined as\n\\[ x_1 \\equal{} y_1 \\equal{} \\sqrt {3}\\ ; \\ x_{n \\plus{} 1} \\equal{} x_n \\plus{} \\sqrt {1 \\plus{} \\left ({x}_n \\right)^2}\\ ; \\ y_{n \\plus{} 1} \\equal{} \\frac {y_n}{1 \\plus{} \\sqrt {1 \\plus{} \\left (y_n \\right)^2}} \\ \\forall n\\ge 2\\]\nProve that $ x_n.y_n$ $ \\in$ $ (2,3)$[/quote]\r\n\r\nhint :\r\n\r\nfor $ x_n$ : $ \\cot \\frac{x}{2}\\equal{}\\cot x\\plus{} \\sqrt{1\\plus{}\\cot^2 x}$\r\n\r\nfor $ y_n$ : $ \\tan \\frac{x}{2}\\equal{}\\frac{\\tan x}{1\\plus{}\\sqrt{1\\plus{}\\tan^2x}}$",
  "Solution_2": "[quote=\"hophinhan\"][quote=\"mathVNpro\"]Given $ (x_n)$ and $ (y_n)$ defined as\n\\[ x_1 = y_1 = \\sqrt {3}\\ ; \\ x_{n + 1} = x_n + \\sqrt {1 +  ({x}_n )^2}\\ ; \\ y_{n + 1} = \\frac {y_n}{1 + \\sqrt {1 +  (y_n )^2}} \\ \\forall n\\ge 2\\]\nProve that $ x_n.y_n$ $ \\in$ $ (2,3)$[/quote]\n\nhint :\n\nfor $ x_n$ : $ \\cot \\frac {x}{2} = \\cot x + \\sqrt {1 + \\cot^2 x}$\n\nfor $ y_n$ : $ \\tan \\frac {x}{2} = \\frac {\\tan x}{1 + \\sqrt {1 + \\tan^2x}}$[/quote]\r\n\r\nWe have\r\n\r\n$ \\displaystyle 2 < x_{n}\\cdot y_{n} = {\\frac {2 \\cos^2 (\\frac{\\alpha}{2^n})}{2 \\cos^2 (\\frac{\\alpha}{2^n})-1}} < 3 \\qquad n\\ge2, \\alpha=\\frac{\\pi}3$"
}
{
  "Tag": [
    "induction",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "prove that  \r\n                   \r\n           0 < sum g(k)/k from 1 to n - 2n/3 < 2/3\r\n\r\n    where g(k)  is the greatest  odd  divisor of  k",
  "Solution_1": "[quote=\"marko avila\"]prove that  \n                   \n           0 < sum g(k)/k from 1 to n - 2n/3 < 2/3\n\n    where g(k)  is the greatest  odd  divisor of  k[/quote]\r\n\r\nUse induction on n.",
  "Solution_2": "Maybe this is related:\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=73591"
}
{
  "Tag": [],
  "Problem": "Prove that if x is an integer, than x^2+6x+5/x+5 is also an integer.",
  "Solution_1": "note that $\\frac{ x^2+6x+5}{x+5}=x+1 \\in Z$ if $x \\in Z$\r\n\r\nThat's it",
  "Solution_2": "that's correct. nice job!",
  "Solution_3": "what does $\\in Z$ mean?",
  "Solution_4": "It means the number is rational :)\r\nN -> natural\r\nZ -> rational\r\nD -> decimal\r\nQ -> irrational\r\nR -> real\r\n(C -> complex)"
}
{
  "Tag": [
    "calculus",
    "trigonometry",
    "function",
    "search"
  ],
  "Problem": "Okay, so I'm thinking of doing a Family-Feud type of thing.  Math related, of course.  But I need to do some surveys.\r\n\r\n1. Name a word that you will often find in a problem that tells you that it will most likely require calculus to solve.\r\n\r\n2. Besides \u201cQ.E.D.\u201d, name a word or phrase that is often found in proofs.\r\n\r\n3. Name the branch of mathematics that\u2019s the hardest to understand.\r\n\r\n4. Name the most widely known theorem.\r\n\r\n5. Name something that you can do with most graphing calculators, but not with scientific calculators.\r\n\r\n6. Name a value for $x$ between 0 and 90 exclusive that you have $\\sin x^{\\circ}$ memorized.\r\n\r\n7. Name a function that, when graphed, could be identified right away.\r\n\r\n8. Name a mathematical feat that Chuck Norris is capable of.\r\n\r\n9. Name a widely known mathematics competition.\r\n\r\n10. Name a formula that any math student should have memorized.\r\n\r\n11. Name a branch of mathematics that many people hate.\r\n\r\n12. Name a famous mathematician.\r\n\r\n13. Rational or irrational, name one of the most important real numbers.\r\n\r\nPM me your answers, and once we get enough votes, we can play!\r\n\r\nYou may submit multiple answers.\r\nHowever, these answers will be weighted so that everyone who votes gets an equal say.\r\nSo let's say one member votes for answer A, and another nominates answer B and answer C.  Answer A will count for twice as much points as B or C.",
  "Solution_1": "Who is Chuck Norris?",
  "Solution_2": "These should explain it:\r\nhttp://en.wikipedia.org/wiki/Chuck_Norris\r\nhttp://en.wikipedia.org/wiki/Chuck_Norris_Facts\r\n\r\nSo far, 5 people, including me, voted.  I probably will stop once I get a decent size of people, maybe 20 or 25 or something, and then just play around with the numbers.",
  "Solution_3": "[quote=\"bubka\"]Who is Chuck Norris?[/quote]\r\nread my signature. :D search for chuck norris random fact generator on google.",
  "Solution_4": "[quote=\"fireemblem13\"][quote=\"bubka\"]Who is Chuck Norris?[/quote]\nread my signature. :D search for chuck norris random fact generator on google.[/quote]\r\n\r\nOk so your signature has plenty of mathematical feats Chuck Norris is capable of. But that does not help me as regards his identity. I am too lazy for a google search now, so somebody clarify please.",
  "Solution_5": "as far as i know, he's a made-up person, but im not the best person to ask. my friend told me a chuck norris joke(non-mathematical) one day and it was funny... so yea...",
  "Solution_6": "There is a real Chuck Norris and I have seen him on tv ;)   I have signed up, too.",
  "Solution_7": "oh. what did he do?\r\ni sign up too",
  "Solution_8": "He played Walker in \"Walker, Texas Ranger\"",
  "Solution_9": "So far, 11 people voted.  I'm looking for more, still.",
  "Solution_10": "so, once there are a lot of ppl who have done this, what do we do?",
  "Solution_11": "And who is Chuch norris, fireemblem?",
  "Solution_12": "let's see.... the wikipedea link above is pretty useful. basically, he's someone you associate impossible and crazy talsks with. for example, some random facts\r\n1. Chuck Norris doesn't brush his teeth. He points his fist at his mouth and the plaque jumps out. :rotfl:  :rotfl: \r\n2. If you are eating Alphabits and the letters somehow spell Chuck Norris, DO NOT attempt to take another bite if you value your life.\r\n3. Contrary to popular belief, a meteor did not end the dinosaurs' existance. It was Chuck Norris' roundhouse kick; it was so powerful it left a massive crater in the earth. :rotfl:",
  "Solution_13": "Once we get a bunch of votes, we'll start the game.  Most likely in another topic, though.",
  "Solution_14": "How many players are there so far?",
  "Solution_15": "So far, 12 votes.  If you're okay with this, we could get something going.",
  "Solution_16": "Sure. What topic are you going to make?"
}
{
  "Tag": [
    "trigonometry",
    "number theory"
  ],
  "Problem": "resolve this equation \r\n$sin^3$x(1+cotangx)+$cos^3$x(1+tanx) = cos2x",
  "Solution_1": "$\\sin^3x(1+\\cot x)+\\cos^3x(1+\\tan x)=\\cos 2x$\r\n$\\sin^2 x(\\sin x+\\cos x)+\\cos^2 x(\\cos x+\\sin x)=\\cos 2x$\r\n$\\sin x+\\cos x=\\cos 2x$\r\n$\\sin x\\pm\\sqrt{1-\\sin^2 x}=1-2\\sin^2x$\r\n$\\pm\\sqrt{1-\\sin^2 x}=1-\\sin x-2\\sin^2 x$\r\n$1-\\sin^2 x=(1-\\sin x-2\\sin^2 x)^2$\r\nAfter expansion and factorization, we have\r\n$\\sin x(\\sin x+1)(2\\sin^2 x-1)=0$\r\n$\\sin x=0$ or $\\sin x=-1$ or $\\sin x=\\pm\\frac{\\sqrt{2}}{2}$\r\nHowever, from the original equation, we know that $\\sin x$ and $\\cos x$ must be both non-zero or otherwise $\\tan x$ and $\\cot x$ will be undefined. So the case $\\sin x=0$ and $\\sin x=-1$ (which leads to $\\cos x=0$) should be rejected. So the case left is $\\sin x=\\pm\\frac{\\sqrt{2}}{2}$.\r\nWe now get $x=\\frac{\\pi}{4}+\\frac{k\\pi}{2}$ where $k\\in\\mathbb{Z}$\r\nBut when we put x into $\\sin x+\\cos x=\\cos 2x$ adn check, we find that only when $x=\\frac{3}{4}\\pi,\\frac{7}{4}\\pi,etc$ the equation is satified.\r\nTherefore, $x=\\frac{3}{4}\\pi+k\\pi$ where $k\\in\\mathbb{Z}$\r\nPS1 This is my 100th post!  :D \r\nPS2 Is this a number theory problem?",
  "Solution_2": "first of all $x\\neq \\frac{k\\pi}{2},k\\in Z$ ..... [b](*)[/b]\n\nequation is equivalent with:\n\n$\\sin x+\\cos x=\\cos 2x$ \n\n$\\sin x+\\cos x=-(\\sin x+\\cos x)(\\sin x-\\cos x)$ \n\n$(\\sin x+\\cos x)(\\sin x-\\cos x+1)=0$\n\n$(\\sin x+\\cos x)\\left(\\cos\\left(x+\\frac{\\pi}{4}\\right)-\\frac{\\sqrt{2}}{2}\\right)=0$\n\nsecond factor of LHS is $\\neq 0$ (condition [b](*)[/b]), so solution is \n\n$\\boxed{x=m\\pi-\\frac{\\pi}{4}\\ ,\\ m\\in Z}$"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "number theory",
    "relatively prime",
    "AMC"
  ],
  "Problem": "The polynomial \r\n           $ P(x) \\equal{} (1\\plus{}x \\plus{} x^2 \\plus{} . . . \\plus{}x^{17})^2\\minus{}x^{17}$\r\nhas $ 34$ complex zeros of the form $ z_k\\equal{}r_k[cos(2\\pi\\alpha_k) \\plus{}isin(2\\pi\\alpha_k)]$\r\n$ k\\equal{}1,2,3,...,34$ with $ 0<a_1<\\equal{}a_2,a_3 <\\equal{}... a_{34}<1$ and $ r_k>0$. given that  \r\n$ a_1\\plus{}a_2\\plus{}a_3\\plus{}a_4\\plus{}a_5\\equal{}m/n$, where $ m$ and $ n$ are relatively prime positive integers, find $ m\\plus{}n$.",
  "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=378129#378129[/url]"
}
{
  "Tag": [
    "search",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Prove that there are infinitely many positive integers $ n$ such that $ n(n\\plus{}1)$  can be represented as a sum of two positive squares in at least two different ways. (Here $ a^{2}\\plus{}b^{2}$ and $ b^{2}\\plus{}a^{2}$ are considered as the same representation.)",
  "Solution_1": "http://www.mathlinks.ro/viewtopic.php?search_id=820928092&t=122895",
  "Solution_2": "http://www.mathlinks.ro/viewtopic.php?p=695192#695192",
  "Solution_3": "one sol is $n=\\left(\\frac{(2+\\sqrt{5})^{2k}(1+\\sqrt{5})-(2-\\sqrt{5})^{2k}(1-\\sqrt{5})}{2\\sqrt{5}}\\right)^2$ for all $k\\in\\mathbb{N}$.Pretty bad looking,huh? :D\nIn this case,$n(n+1)=p^4+p^2$ and $(p^2-2)^2+(5p^2-4)$.This $5p^2-4$ is a square.",
  "Solution_4": "[quote=sankha012]one sol is $n=\\left(\\frac{(2+\\sqrt{5})^{2k}(1+\\sqrt{5})-(2-\\sqrt{5})^{2k}(1-\\sqrt{5})}{2\\sqrt{5}}\\right)^2$ for all $k\\in\\mathbb{N}$.Pretty bad looking,huh? :D\nIn this case,$n(n+1)=p^4+p^2$ and $(p^2-2)^2+(5p^2-4)$.This $5p^2-4$ is a square.[/quote]\nHey (sorry to revive this thread).. But I seriously wanna ask How the hell did you come up with that? Plz. give the motivation behind it...\n Since there is no reachable solution to this prob on this thread till now here's my idea:-\n[hide]Set $n=m^2$ as then $n(n+1)$ is already a sum of two squares.\nNow if we get $m^2=p^2+q^2,$ then notice that it becomes $(pm+q)^2 - (p-qm)^2$..\nNow it's trivial as it's well know that there are infinite pythagorean triplets...So there are infinite $(m,n,p)$ where m,n,p are integers..\nQED[/hide]\n\n",
  "Solution_5": "This is my solution",
  "Solution_6": "What about this?\n\n$n=4m^4$\n$n^2+n=(4m^4)^2+(2m^2)^2$\n$=(4m^4-2m^2)^2+(4m^3)^2$",
  "Solution_7": "[quote=makar]Prove that there are infinitely many positive integers $ n$ such that $ n(n\\plus{}1)$  can be represented as a sum of two positive squares in at least two different ways. (Here $ a^{2}\\plus{}b^{2}$ and $ b^{2}\\plus{}a^{2}$ are considered as the same representation.)[/quote]\n\nIf n is such a number, then so is 4n(n+1).\n\n"
}
{
  "Tag": [
    "function",
    "geometry"
  ],
  "Problem": "An open box with a square base is required to have a volume of 10 cubic feet.\r\n\r\nNOTE: What exactly is a cubic foot?\r\n\r\n(a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base.\r\n\r\n(b) How much material is required for a base 1 foot by 1 foot?\r\n\r\nNOTE: Is one foot by foot the same as 144 inches?",
  "Solution_1": "The box have a square base.\r\nThe volume of the box is the area of the base times the heigth.\r\nOr\r\n\r\n${x^{2}h = 10}$\r\n\r\nIt is equal 10, because the volume was given.\r\n\r\nThe amout of material is proportional to the sufarce area. so\r\n\r\n${S=x^{2}+4 h x}$\r\n${h=\\frac{10}{x^{2}}}$\r\n${S=\\frac{x^{3}+40}{x}}$\r\n\r\n\r\n${A=\\alpha S}$\r\n\r\nI'm not english. so i dont know how much is one feet, and one inches. but one foot by one foot could be 144 square inches",
  "Solution_2": "dude, omfg.  let me introduce you to my bestest cyber friends, yahoo and his sidekick google.  you could meet them at yahoo.com and google.com.  they would love to meet you because thats how they make their money and im sure they would be extreely helpful to you. \r\n\r\nyou would be saving us a buch of time if you used them\r\n\r\none foot is 12 inches.  that means 1 square foot is 144 square inces and 1 cubic foot is 12^3 cibic inches."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "\\/closed"
  ],
  "Problem": "Hello!\r\n\r\nI have two questions. \r\n\r\n\r\n1. I really don't get the difference between AIME A and AIME B Problem Series, diverting from Aime A is not in the summer. \r\n\r\n2. Which one should I take AMC 12 Problem Series or AIME B? \r\n\r\nI am going to finish Volume 1 and 2 this summer. And I really want to do well on the AIME next year. I also might take WOOT, but I am not positively sure.\r\n\r\nThanks for your help! :lol:  :)",
  "Solution_1": "[quote]Description\nOur two AIME Problem Series classes prepare students for the AIME, which is the second in the series of tests used to determine the United States team at the International Math Olympiad. Many top colleges also request AIME scores as part of the college application process. Each day consists of a discussion of specific problems from past AIME competitions, or from other contests of a similar difficulty level. Important general problem solving strategies and specific math facts are taught through solving these problems.\n\nPlease note, we offer two versions of this class, AIME Problem Series A and AIME Problem Series B. These two courses are roughly the same difficulty, but the problems covered in the two courses are completely different. Many students opt to take both classes. We typically offer one in the summer and one in the fall/winter.[/quote]\r\n\r\nthat should answer question number 1.",
  "Solution_2": "if anyone could clear this up for me at least \r\n\r\n#2 \r\n\r\nthanks! :)",
  "Solution_3": "To answer #2, we'll probably need some more background info on you.",
  "Solution_4": "what would you mean by some background information",
  "Solution_5": "What are your goals? Have you qualified for AIME before? Do you want a higher AIME score or higher AMC score? What grade are you in? etc.\r\n\r\nAlso, I would suggest holding off on WOOT until you are positive that you will qualify for USAMO.",
  "Solution_6": "[quote=\"crazypianist1116\"]What are your goals? Have you qualified for AIME before? Do you want a higher AIME score or higher AMC score? What grade are you in? etc.\n\nAlso, I would suggest holding off on WOOT until you are positive that you will qualify for USAMO.[/quote]\r\n\r\nI agree on more background info needed, but I don't think should avoid WOOT until you are certain you're ready for the USAMO.  We've had plenty of people take WOOT fruitfully before being sure they could pass the AIME.  Before taking on WOOT, you should be sure that you are dedicated enough to put in the time to get the most out of the course.  Mathematically, it would be best if you are confident you can get at least 5 AIME questions.  So, you should have a fighting chance on the AIME before starting WOOT, but you needn't wait until you are sure you can pass the AIME to join."
}
{
  "Tag": [
    "search",
    "probability",
    "expected value"
  ],
  "Problem": "An experiment has n possible outcomes, all equally likely. Each trial is independent. We repeat until we have two trials with the same outcome. For example, we would stop after getting 1, 4, 3, 4. What is the expected number of trials f(n)? Find an asymptotic expression for f(n).",
  "Solution_1": "My apologies for the incorrect link.  Here is the link but a similiar (though different) problem posted eariler.\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1695488873&t=158138",
  "Solution_2": "This is a different (although similar) problem.\r\n\r\n[hide=\"Solution\"]The probability we finish in exactly $ m+1$ trials, $ 0 \\leq m \\leq n$ is $ \\frac{m}{n}\\cdot \\prod_{i = 0}^{m-1}\\frac{n-i}{n}= \\frac{m \\cdot n!}{n^{m+1}(n-m)!}$.  Thus, the answer is $ \\sum_{m = 0}^{n}\\frac{m(m+1) \\cdot n!}{n^{m+1}(n-m)!}$.  I don't, off the top of my head, see a nice form for this expression (although that shouldn't stop others from looking for one).[/hide]"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "ARML",
    "USA(J)MO",
    "USAMO",
    "algebra",
    "functional equation"
  ],
  "Problem": "So I've been working through AoPS vol 1&2 over the summer and have been doing random aime problems (usually around 5~6 tough ones a day). I don't know but I think I'm starting to lose my interest in math since I haven't really been doing much else. I am scared that I will burnout and abandon math altogether. Can someone please tell me how much math I should be doing. Right now I do about 7-8 hours of math per day but most of it is done in the early mornings or at night.\r\n\r\n\r\nThanks in Advance",
  "Solution_1": "There's no amount you \"should\" be doing. Do as much as interests you. I can say that 7-8 hours would be too much for me, at least.",
  "Solution_2": "I would suggest taking a break from math a little bit and start doing something else. I was burned out just after this year's AIME and didn't feel like doing math anymore. Usually, when you take breaks, you start to remember what you enjoyed about math and then you'll do math again. You just need some variety in your life once in a while so it wouldn't hurt to take an entire day or two off of math. And I also do 7-8 hrs per day (during winter, summer, and spring break) so I know how you feel. The chances that you will [b]completely[/b] lose your interest is very low if you really do have a true passion for math.",
  "Solution_3": "You can always diversify and study something like science one day, then math for the next few days.  Take breaks from math, but you don't necessarily have to take breaks from studying.",
  "Solution_4": "Personally, working through a whole problem book is too intimidating for me. I usually do problems from various sources to reduce fatigue (e.g. aops forums).",
  "Solution_5": "the problem is that you're doing 7-8 hours of math a day. you should barely even be awake for that long. add in meals, school, practicing instruments, sports practices, and studying/homework for other subjects and you're up until midnight\r\nyou should probably get some rest",
  "Solution_6": "Yeah, 7-8 is kind of overkill. \r\n\r\nWhen I get bored, I do things I shouldn't be doing (cough group theory/axiomatic set theory/etc...), or I just do olympiad problems instead, which are very different from AIME problems. There's always something out there.",
  "Solution_7": "How quickly you burn out I think depends more on how much you force yourself to do, as opposed to how much you're naturally inclined to do. As an example of the latter, when doing a stupid assignment like writing a paper my idea of a break is finding a functional equation on these forums and busting my brain on it for a bit. So that 7-8 hours is going to read differently for different people. But if you wouldn't normally do something like 4-5 hours per day anyway, unforced, you might want to cut back.\r\n\r\n[OFFTOPIC]\r\n[quote=\"Zhero\"]I do things I shouldn't be doing (cough group theory/axiomatic set theory/etc...)[/quote]\r\n\"Shouldn't be doing\"? That is perfectly legitimate math, maybe more so than contest math, and for the olympiad problems motivated by advanced concepts* you may be well prepared.\r\n\r\n* such as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159989]this IMO problem[/url] whose solution is elementary only in its presentation, though such a problem [i]should[/i] never appear on an olympiad.\r\n[/OFFTOPIC]\r\n\r\nEDIT to below: Of course. I suppose I implicitly meant \"on average\" several times in this post.",
  "Solution_8": "Well the amount you would do \"unforced\" can also vary. For example I might feel inclined to do a lot of math one day and then much less the next.\r\n\r\n@Palmer: Is this not the case for you?",
  "Solution_9": "You should no when you feel tired. After you feel tired you will never remember the concepts you just studied. So if you do not feel tired doing 7-8 hours of math then I guess I should be fine.\r\n\r\n :surf: This is my 50th post(no celebration).",
  "Solution_10": "[quote]\"Shouldn't be doing\"? That is perfectly legitimate math, maybe more so than contest math, and for the olympiad problems motivated by advanced concepts* you may be well prepared. [/quote]\r\nHmm yeah, now that you mention it, I think I'll have to retract my statement. Though I didn't think much of it before, I did benefit from doing axiomatic set theory. For one, it helped me to understand mathematical rigour more (e.g., what axiomatic systems are and how to write formal proofs), and I learned much about functions. \r\n\r\nAnd it certainly provides for a fine break away from the stuff like that the OP's been doing. And it does teach invaluable techniques. \r\n\r\nI suppose this notion arose from my putting competition math above all the other math... in the past few months, in my eyes, all the other math is just a distraction from competition math. \r\n\r\nBut my eyes have opened now. It's summer now, no competitions to worry about. Time to indulge...",
  "Solution_11": "A great way to memorize new formulas/concepts is to do them right before you sleep. Believe it or not, your brain thinks about what you just did while you're asleep, and when you wake up again, what you just learned will stay with you for a long time.",
  "Solution_12": "If I do math right before I sleep, I can't sleep.\r\n\r\nI recommend 1-3 hours of math on weekdays (depending on how much homework you have) and 4-6 hours on weekends. (each day)\r\n\r\nThat's like all I have time for. But the week after ARML (and the month before it) I barely did any math at all, because preparing for USAMO basically made me sick of it. (temporarily, of course)",
  "Solution_13": "Wow, do it like, few hours. 1, 1.5, 2.Something like that. I'd die if I did math stuff 7-8 hours a day. It s like, a part time job for you :D\r\nI'd say forget about maths for like a few weeks, go do some sports, hang out with your buddies more. The reason why you lose interest is because you dedicate too much time to it, you feel a routine in it, it becomes boring. I know what I'm talking about  :roll: \r\n\r\nFor me, I deal with maths in the evening. Open up the unsolved problems section and work my magic. But I can't do it for 7 hours a day :(\r\nYou know if you do maths so long, your head is like, exhausted, usually when I feel that way I go outside jogging or riding a bike or something and when I think about the math problem I had, the answer seems to somewhat more easily reveal itself :P\r\n\r\nSo all in all, you just have to take up something else if you don't wanna lose interest. \r\n\r\nHave fun."
}
{
  "Tag": [
    "linear algebra",
    "matrix"
  ],
  "Problem": "Determine the number of possible values for the determinant of $ A$, given that $ A$ is a $ n*n$ matrix with real entries such that $ A^3\\minus{}A^2\\minus{}3A\\plus{}2I\\equal{}0$, where $ I$ is the identity and $ 0$ is the all-zero matrix.  [/b][/hide][/code]",
  "Solution_1": "All eigenvalues of this matrix are roots of $ x^3 \\minus{} x^2 \\minus{} 3x \\plus{} 2 \\equal{} 0.$ That equation has three real roots, $ 2$ and $ \\frac { \\minus{} 1\\pm\\sqrt {5}}2.$ Let's just call these numbers $ r_1,r_2,$ and $ r_3.$\r\n\r\nThe possible values of $ \\det(A)$ are $ r_1^ir_2^jr_3^k,$ where $ i,j,$ and $ k$ are nonnegative integers such that $ i \\plus{} j \\plus{} k \\equal{} n.$\r\n\r\nWe can form such a product in $ \\binom{n \\plus{} 2}{2}$ ways. The only remaining question is whether different values of $ (i,j,k)$ could result in the same determinant - I don't think so, but I haven't proven that yet.",
  "Solution_2": "Hello,\r\n\r\nI am sure that there is an easier way to finish the last part, but this is one way:\r\n\r\nLet $ i_1\\plus{}j_1\\plus{}k_1\\equal{}n$ and $ i_2\\plus{}j_2\\plus{}k_2\\equal{}n$. Then if we have two values of the determinant that are equal, we get\r\n\r\n$ \\begin{array}{lrl}\r\n &2^{i_{1}}\\left(\\frac{\\minus{}1\\plus{}\\sqrt{5}}{2}\\right)^{j_{1}}\\left(\\frac{\\minus{}1\\minus{}\\sqrt{5}}{2}\\right)^{k_{1}}&\\equal{}2^{i_{2}}\\left(\\frac{\\minus{}1\\plus{}\\sqrt{5}}{2}\\right)^{j_{2}}\\left(\\frac{\\minus{}1\\minus{}\\sqrt{5}}{2}\\right)^{k_{2}}\\\\\r\n\\Leftrightarrow & 2^{i_{i}\\minus{}j_{1}\\minus{}k_{1}\\minus{}(i_{2}\\minus{}j_{2}\\minus{}k_{2})}&\\equal{}\\left(\\minus{}1\\plus{}\\sqrt{5}\\right)^{j_{2}\\minus{}j_{1}}\\left(\\minus{}1\\minus{}\\sqrt{5}\\right)^{k_{2}\\minus{}k_{1}}\\\\\r\n\\Leftrightarrow  &4^{i_{1}\\minus{}i_{2}}&\\equal{}\\left(1\\minus{}5\\right)^{j_{2}\\minus{}j_{1}}\\left(\\minus{}1\\minus{}\\sqrt{5}\\right)^{j_{1}\\minus{}k_{1}\\minus{}(j_{2}\\minus{}k_{2})}\\\\\r\n\\Leftrightarrow  &\\left(\\minus{}1\\right)^{j_{1}\\minus{}j_{2}}4^{i_{1}\\plus{}j_{1}\\minus{}(i_{2}\\plus{}j_{2})}&\\equal{}\\left(\\minus{}1\\minus{}\\sqrt{5}\\right)^{j_{1}\\minus{}k_{1}\\minus{}(j_{2}\\minus{}k_{2})}\\\\\r\n\\Leftrightarrow & \\left(\\minus{}1\\right)^{j_{1}\\minus{}j_{2}}4^{k_{2}\\minus{}k_{1}}&\\equal{}\\left(\\minus{}1\\minus{}\\sqrt{5}\\right)^{j_{1}\\minus{}k_{1}\\minus{}(j_{2}\\minus{}k_{2})}\r\n\\end{array}$\r\n\r\nNow one can argue that the left hand side of the last equation always is rational for integer exponents while the right hand side always is irrational except when the exponent is zero (I am not giving a proof to this statement though) which forces the RHS to be equal to 1. From the LHS we therefore get $ k_1\\equal{}k_2$ which, if put into the condition $ j_1\\minus{}k_1\\equal{}j_2\\minus{}k_2$, shows that $ j_1\\equal{}j_2$, but then $ i_1\\equal{}i_2$ as well and we are done. The answer is thus $ \\binom{n\\plus{}2}{2}$, the one Kent Merryfield anticipated."
}
{
  "Tag": [
    "number theory proposed",
    "number theory"
  ],
  "Problem": "I know this problem is posted before but as I know ,noone posted solution.\r\nProve that for all $ n\\geq 3$ and $ n\\in Z_{\\plus{}}$ we have $ S(9^{n})>9$ .where $ S(n)$ denote sum of all digits of $ n$.",
  "Solution_1": "If n is odd last digit is 9, therefore (because $ 9^{n}>9$) $ S(9^{n})>9$. If n=2k, then $ S(9^{n})\\equal{}S(81^{k})$.\r\n$ 81^{k}\\equal{}1\\plus{}10*k*8\\plus{}100*\\frac{k(k\\minus{}1)}{2}64\\plus{}...$\r\nBy considering many variants of k we can prove, that $ S(81^{k})>9$ for k>1."
}
{
  "Tag": [
    "inequalities open",
    "inequalities"
  ],
  "Problem": "Ask for help!",
  "Solution_1": "[quote=\"teanleen\"]Ask for help![/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=57683",
  "Solution_2": "[quote=\"arqady\"][quote=\"teanleen\"]Ask for help![/quote]\nSee here:\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=57683[/quote]Thank you!\r\nBut ,I don't think it is very nice!"
}
{
  "Tag": [
    "limit",
    "LaTeX",
    "calculus",
    "calculus computations"
  ],
  "Problem": "[color=darkblue]Let $ f(n) \\equal{} (n^2 \\plus{} n \\plus{} 1)^2 \\plus{} 1$, with $ n \\in N$ and $ n\\ge 1$.\n\nLet $ x_n \\equal{} \\frac {f(1).f(3).f(5)...f(2n \\minus{} 1)}{f(2).f(4).f(6)...f(2n)}$, with $ n \\in N$ and $ n\\ge 1$.\n\nLet $ u_n \\equal{} n^2.x_n$\n\nFind $ \\lim_{n \\to \\plus{} \\infty}(u_n)$[/color]",
  "Solution_1": "Hint: $ \\frac{f(2n\\minus{}1)}{f(2n)} \\equal{} \\frac{2n(n\\minus{}1) \\plus{} 1}{2(n\\plus{}1)n \\plus{} 1}$",
  "Solution_2": "$ \\text{\\LaTeX}$ note: use \\cdot for the multiplication dot. $ f(1)\\cdot f(2)$ looks much better than $ f(1).f(2)$. Also, \\dots is a better spaced ellipsis, and \\cdots is a vertically centered one."
}
{
  "Tag": [
    "geometry",
    "area of a triangle",
    "Heron\\u0027s formula"
  ],
  "Problem": "In triangle $ABC$, find $\\frac{\\cot{\\frac{A}{2}}}{s-a}$ where $s$ is the semi-perimeter. (you can answer in side lengths, etc.)",
  "Solution_1": "We have \r\n\r\n\\[\\cot \\frac{A}{2}+\\cot \\frac{B}{2}=\\frac{c}{r}\\]\r\n\r\n\\[\\cot \\frac{B}{2}+\\cot \\frac{C}{2}=\\frac{a}{r}\\]\r\n\r\n\\[\\cot \\frac{C}{2}+\\cot \\frac{A}{2}=\\frac{b}{r}\\]\r\n\r\nso, we get $\\cot \\frac{A}{2}=\\frac{b+c-a}{2r}$,therefore using Heron's formula\r\n\r\n$\\frac{\\cot \\frac{A}{2}}{s-a}=\\frac{1}{r}=\\frac{1}{2}\\sqrt{\\frac{(a+b-c)(b+c-a)(c+a-b)}{a+b+c}}$",
  "Solution_2": "how did you get these realtions??? :(",
  "Solution_3": "I believe you can get it from drawing a triangle, look at $tan {\\frac{A}{2}}$ and then you'll get the relation.",
  "Solution_4": "See the attachment."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra unsolved"
  ],
  "Problem": "Find all $P(x)\\in{R[x]}$ such that \r\n$P(x^{2}+1)=P^{2}(x)-1$",
  "Solution_1": "(*) $P(x^{2}+1)=P(x)^{2}-1$\r\nDeriving (*) we get \r\n(**) $xP'(x^{2}+1)=P(x)P'(x)$.\r\n\r\nTake $g(x)=x^{2}+1$ and for all $n\\geq 0$\r\n$g_{n}(x)=\\underbrace{g(g(\\cdots g(x)))}_{n-{\\text{times}}}$.\r\n\r\nThen $0<g(0)<g_{2}(0)<g_{3}(0)<\\cdots$, as $g(\\lambda)-\\lambda=(\\lambda-{1\\over 2})^{2}+{3\\over 4}>0$ for all $\\lambda\\in\\mathbb{R}$,\r\nand for all $\\lambda\\in\\mathbb{C}$ we have \r\n$P(\\lambda)=0\\ \\ \\Rightarrow P(g(\\lambda))=P(\\lambda)^{2}-1=-1$ and\r\n$P(\\lambda)=-1\\Rightarrow P(g(\\lambda))=P(\\lambda)^{2}-1=0$.\r\n\r\n(1) $\\gcd (P(x),g_{n}(x))=1$ for all $n\\geq 0$:\r\nOtherwise $P(x)$ and $g_{n}(x)$ have a common root $\\lambda\\in\\mathbb{C}$ for some $n\\geq 0$.\r\nBut then $P(\\lambda)=0$ and $P(0)=P(g_{n}(\\lambda))\\in\\{0,-1\\}$.\r\nSo $P(0)=P(g_{2}(0))=P(g_{4}(0))=\\cdots$ and the polynomial $P(x)-P(0)$ has an infinite number of roots.\r\nThis forces $P\\equiv P(0)\\in\\{0,-1\\}$, which contradicts (*).\r\n\r\n(2) $g_{n}(x)\\ |\\ P'(x)$ for all $n\\geq 0$:\r\nWe have $g_{0}(x)=x\\ |\\ xP'(g(x))=P(x)P'(x)$ and so $g_{0}(x)\\ |\\ P'(x)$ by (1)\r\nand from $g_{n}(x)\\ |\\ P'(x)$ follows $g_{n+1}(x)=g_{n}(g(x))\\ |\\ xP'(g(x))=P(x)P'(x)$ \r\nand $g_{n+1}(x)\\ |\\ P'(x)$ for all $n\\geq 0$ again using (1).\r\n\r\nNow $\\deg g_{n}=2^{n}\\mathop{\\longrightarrow}_{n\\rightarrow\\infty}\\infty$, so $P'\\equiv 0$ and $P\\equiv c$ constant. Then (*) gives $c=c^{2}-1$.\r\nSo (*) has $2$ solutions $P\\equiv{1+\\sqrt{5}\\over 2}$ and $P\\equiv{1-\\sqrt{5}\\over 2}$."
}
{
  "Tag": [],
  "Problem": "If $ x$ is positive and $ x^2 \\equal{} 729$, what is the value of $ x$?",
  "Solution_1": "The equation $ x^2\\equal{}729$ has solutions $ x\\equal{}\\pm 27$.  They tell you $ x$ is positive, so $ x\\equal{}\\boxed{27}$",
  "Solution_2": "$ 7+2+9=18=9 \\times 2 \\ \\text{i.e.} \\ 9 \\mid 729$.\r\nSo,\r\n$ \\begin{eqnarray*}729 & = & 9\\times 81 \\\\ & = & 9 \\times 9 \\times 9 \\\\ & = & 9^3 \\\\ & = & \\left(3^2\\right)^2 \\\\ & = & 3^{2 \\times 3} \\\\ & = & 3^{3 \\times 2} \\\\ & = & \\left(3^3\\right)^2 \\\\ & = & 27^2.$",
  "Solution_3": "[quote=\"Kouichi Nakagawa\"]$ 7+2+9=18=9 \\times 2 \\ \\text{i.e.} \\ 9 \\mid 729$.\nSo,\n$ \\begin{eqnarray*}729 & = & 9\\times 81 \\\\ & = & 9 \\times 9 \\times 9 \\\\ & = & 9^3 \\\\ & = & \\left(3^2\\right)^2 \\\\ & = & 3^{2 \\times 3} \\\\ & = & 3^{3 \\times 2} \\\\ & = & \\left(3^3\\right)^2 \\\\ & = & 27^2.$[/quote]\n@Kouichi Nakagawa, for the fourth line in the equation, it should be $=\\left(3^2\\right)^3$, not $=\\left(3^2\\right)^2$.",
  "Solution_4": "[quote=\"JSGandora\"][quote=\"Kouichi Nakagawa\"]$ 7+2+9=18=9 \\times 2 \\ \\text{i.e.} \\ 9 \\mid 729$.\nSo,\n$ \\begin{eqnarray*}729 & = & 9\\times 81 \\\\ & = & 9 \\times 9 \\times 9 \\\\ & = & 9^3 \\\\ & = & \\left(3^2\\right)^2 \\\\ & = & 3^{2 \\times 3} \\\\ & = & 3^{3 \\times 2} \\\\ & = & \\left(3^3\\right)^2 \\\\ & = & 27^2.$[/quote]\n@Kouichi Nakagawa, for the fourth line in the equation, it should be $=\\left(3^2\\right)^3$, not $=\\left(3^2\\right)^2$.[/quote]\n\nYes that's right.\nI've made a typo."
}
{
  "Tag": [
    "calculus",
    "calculus computations"
  ],
  "Problem": "Solve the following, given that y(1) = 1\r\n\r\nx(-5y^2 + 9x^2)y' = 2(y^2)(3x-5y)",
  "Solution_1": "Hello there. This is a first order homogeneous ODE that can be solved using the substitution, y = vx."
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "ratio",
    "geometry unsolved"
  ],
  "Problem": "please post solutions to----\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=238434",
  "Solution_1": "Mmm, It 's easy. You can figure it out for your case. In case, you need the solution. Click on the hidden message.\r\n\r\n[hide=\"Solution\"]\n$ ND$ is the diameter of the circumcircle of triangle $ BFD$ and $ \\triangle up BFD$ is the image of triangle $ ABC$ under the scale of half. (or $ BFD$ is similar to $ ABC$ with ratio $ \\frac{1}{2}$).\n[/hide]"
}
{
  "Tag": [
    "number theory",
    "least common multiple",
    "prime factorization"
  ],
  "Problem": "This should set an example; if you want harder problems look somewhere else.\r\n\r\n1. Two friends are playing tag. The first runs at 20 feet per minute and the other at 25 feet per minute. If the first friend has a minute head start how long will it take the second friend to catch up?\r\n\r\n2. There are two brothers. Right now one is three times as old as the other. In three years he will only be twice as old. What are their ages now?\r\n\r\nI hope these problems show an appropriate level of difficulty for Getting Started problems (they're easy).  ;) \r\n\r\nEdited to change speed (but it doesn't change the answer to the first problem) and to correct spelling.",
  "Solution_1": "These problems look good to me.  Although 20 feet per hour is kind of slow  :D .",
  "Solution_2": "Random question but are American children taught feet and inches in school?!",
  "Solution_3": "Yeah, America mostly uses feet, inches and the like.  I personally believe that we should switch to the metric system(and most other ppl who know science will agree).  We have to spend 1-2 weeks in science class just to learn how to convert from the English system to the metric...all you need to know for the metric system are the prefixes!  We Americans are too stubborn.",
  "Solution_4": "1.\n\n[hide]First guy has a 20 ft head start, since he has a minute's head start.  After one minute, the first guy will have traveled 40 ft, and the 2nd guy will have traveled 25 ft.  So if we keep on going and listing numbers, we find that once 4 minutes are over, they will both have traveled 100 ft.\n\n\n\nEdited, I found a nicer way using the least common multiple of 20 and 25.  I think there's a way to do this problem using the distance formula, but oh well. \n\n20's prime factorization is 5*2 <sup>2</sup> , and 25's is 5 <sup>2</sup> .  So the LCM of these two numbers needs to have a 5 <sup>2</sup>  and a 2 <sup>2</sup> .  Multiplying 25 and 4 out we get 100, our answer from before.  Now, we just need to figure out how long it'll take.  We can use the distance formula I think.  Distance=100, so 100=25x, x=4.  If you wanted to do it for the first guy, his distance is 80 feet, so 80=20x, and both ways we get 4.[hide][/hide]\n\n\n\n2.  \n\n[hide] Let a be the older brother and b be the younger brother.  Thus, a = 3b.  In 3 years a will be twice as old as b.  So a + 3 = 3 (b + 3), which gives a + 3 = 3b + 6.  Using substitution, 3b + 3 = 3b + 6, so b = 3, and a = 9.[/hide][/hide]",
  "Solution_5": "Let's try something. If at time x the faster brother catches up to the slower brother, what's the distance that the faster brother has covered? What's the distance that the slower brother has covered? Make this into an equation and solve for x.",
  "Solution_6": "[quote=\"mikewd\"]Random question but are American children taught feet and inches in school?![/quote]\r\n\r\nYes, as my fellow moderator explained, United States schools still mostly teach English customary units. Interestingly, the metric system is the only system of measurement ever OFFICIALLY acknowledged by Congress (back in the 1860s), but the force of custom is such that most Americans are still much more familiar with inches than with centimeters, with miles than kilometers, and so on.",
  "Solution_7": "[quote=\"joml88\"]Yeah, America mostly uses feet, inches and the like.  I personally believe that we should switch to the metric system(and most other ppl who know science will agree).  We have to spend 1-2 weeks in science class just to learn how to convert from the English system to the metric...all you need to know for the metric system are the prefixes!  We Americans are too stubborn.[/quote]\r\n\r\nEnglish system.. how ironic, we use metric in the UK (well in school at least, my parents don't)",
  "Solution_8": "I tried what you suggested, Osiris, and this is what I got:\n\n[hide]x is the time it takes for 2nd guy to catch up with 1st guy, so 25x = 20 + 20x, since the 1st guy had a 20 ft head start.  Solving for x gives x = 4.  Wow that was a lot easier.[/hide]",
  "Solution_9": "Just cause I'm bored... #1 is sort of ambiguous (is that the right word for this?) because they could make turns, etc..\r\n\r\n#2: Care to clarify?\r\n[quote]In three years he will only be twice as old.[/quote]\r\nTwice as old as he was, or the older brother was, or the older brother will be?",
  "Solution_10": "[quote=\"Duck-Billed Platypus\"]Just cause I'm bored... #1 is sort of ambiguous (is that the right word for this?) because they could make turns, etc..\n\n#2: Care to clarify?\n[quote]In three years he will only be twice as old.[/quote]\nTwice as old as he was, or the older brother was, or the older brother will be?[/quote]\r\n\r\nIn three years the older brother will be twice as old as the younger brother.",
  "Solution_11": "#2: [hide]Let first brother's age be A, second brother's age be B.\n\nOne is 3 times as old as the other, so A=3B.\n\nIn 3 years he will be twice as old as the other, so A+3=2(B+3).\n\nSubstitute.. 3B+3=2B+6\n\nB=3\n\nTheir ages now are 9 and 3.[/hide]",
  "Solution_12": "1.[hide]\n\nThe second friend catches up to the first friend 5 feet / minute, since 25-20 is 5.  Since the first friend has a minute, or 20 feet head start it will take 20/5, or 4 minutes for the second friend to catch up.[/hide]\n\n\n\n2.[hide]\n\na will represent the first brother's age right now, and b will represent the second brother's age right now.\n\n\n\n3a = b\n\n\n\n2(a+3) = b+3\n\n2a+3 = b\n\n\n\n6a+9 = 3b\n\n6a+0= 2b\n\n-------------\n\n0a+9=b\n\n\n\nSo the older brother is 9 years old.  We can substitue that into 3a = b.\n\n\n\n3a = 9\n\na = 3\n\n\n\nSo the younger brother is 3 years old.\n\n[/hide]",
  "Solution_13": "1.[hide]20x + 20 = 25x\n\n20 = 5x\n\n4=x\n\n\n\nthe + 20 is the fact that 20 ft/min* 1 min = 20 ft, thus the y-interecept at 0 min mark of when the faster of the two starts to run.[/hide]\n\n\n\n2.[hide]2 brothers x and \n\n\n\nx = 3y\n\n\n\nx + 3 = 2(3 + y)\n\nx + 3 = 6 + 2y\n\n3y + 3= 6 + 2y\n\ny = 3\n\n\n\nx = 3 (3) = 9\n\n\n\nThe older brother (x) is 9 years of age, while the younger brother (y) is 3 years of age.\n\n[/hide]",
  "Solution_14": "1. [hide] let x = minutes they have been running for. Since the first boy (20ft per minute) has a minute head start, he is 20 feet ahead.  The first boy's distance is represented by 20x, and the second boy's distance is represented by 25x. So I set up an equation:  20 + 20x = 25x. After doing some algebra, x comes out to be 4, so therefore the answer is 4 minutes.[/hide]\n\n\n\n2. [hide] I made the younger one's age x, and the older one's 3x (seeing that he is 3 times the age of the younger one. Three years later, the older one's age will be 3x+3. The problem says at this time, the older one will only be twice the age of the younger one. I set up this equation: 3x+3 = 2(3+x) , or 3x+3=(6+2x). x comes out to be 3, so if you plug that in for the two original ages, you'll get 3 and 9.[/hide]\n\n\n\nThanks for these problems. The other ones in this forum are too hard for me at this point, but these are easy (maybe even a bit too easy)."
}
{
  "Tag": [
    "geometry",
    "parallelogram",
    "geometry unsolved"
  ],
  "Problem": "Could someone give a solid/rigorous proof for the following:\r\n\r\nABCD is a convex quadrilateral. The midpoints of AB, BC, CD, DA are E, F, G, H. Segments AF, BG, CH, DE are drawn. The points at which these segments intersect are W, X, Y, Z. Then the midpoints of WX, XY, YZ, ZW form a parallelogram.\r\n\r\n\r\n\r\n[geogebra]c2f4825bba5facac77e0148a3d467d5d29c0a8f7[/geogebra]",
  "Solution_1": "[quote=\"bbgun34\"]Could someone give a solid/rigorous proof for the following:\n\nABCD is a convex quadrilateral. The midpoints of AB, BC, CD, DA are E, F, G, H. Segments AF, BG, CH, DE are drawn. The points at which these segments intersect are W, X, Y, Z. Then the midpoints of WX, XY, YZ, ZW form a parallelogram.\n\n\n\n[geogebra]c2f4825bba5facac77e0148a3d467d5d29c0a8f7[/geogebra][/quote]\r\nIf $ \\square ABCD$ is a convex quadrialteral then the midpoints of $ AB,BC,CD,DA$ form a parallelogram, so there is probably something wrong in the statement as one is given much more information than needed."
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "Let  $a_k\\geq0$ for all $k=1,2,...,n$.Prove the following inequality:\r\n\r\n               $ e^(a_1+a_2+...+a_n-n)\\geq \\frac^n {a_1+a_2+...+a_n}{n}$ ;)",
  "Solution_1": "Please, learn Latex. I learned quickly,although I know the first notions from you, for which I thank you !",
  "Solution_2": "Oh,levi I'm so lazy.But in the end I must learn it.Anyway,thanks for the advice. ;)"
}
{
  "Tag": [
    "MATHCOUNTS",
    "Pythagorean Theorem",
    "geometry",
    "exterior angle"
  ],
  "Problem": "anyone who has intersting geometric proofs, could ou please post them here?\r\nfor example, the pythagorean theorm.\r\n\r\nIf possible, please post a problem using that proof in your post. only if possible. Thanks.",
  "Solution_1": "For proofs of the Pythagorean Theorem, check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275754]this[/url] thread.",
  "Solution_2": "IDK how intense you are with geometric proofs. Heres a really really basic one:\r\n\r\n\"Show that he shortest distance from a line AB to a point C not on AB is the segment CD, where line CD is perpendicular to line AB and point D lies on line AB.\"\r\n\r\nThats not very interesting, but I came up with it (yeah, obviously someone noticed it waaaaayyyyy before me and its sorta obvious, but I hadnt heard of any theorem like that before, even though im sure there is one out there)",
  "Solution_3": "are you talking about like, MC proofs, if higher move it there mods.",
  "Solution_4": "mathcounts level proofs i mean",
  "Solution_5": "One such example:-\r\n\r\nProve that the sum of the 3 exterior angles of a triangle is equal to 360 degrees.",
  "Solution_6": "To prove arbalas exterior angle prove \r\n\r\nlet say we have a triangle with angle measures a b and c.\r\n\r\nWe know that the exterior angle is equal two the sum of the two interior angles so\r\n\r\nexterior a=b+c\r\nexterior b=a+c\r\nexterior c=a+b \r\n\r\nThe sum is 2a+2b+2c and since a+b+c=180 the sum of the exterior angles ina triangle is 360 degrees",
  "Solution_7": "random proof that if two lines have consecutive interior angles(the ones created by the transversal) that add up to 180, then the lines are parallel: Contradiction- Draw the lines such that they intersect. This creates a triangle with a zero degree angle, which is impossible so we are done.",
  "Solution_8": "you guys can also post formulas also and theorms.",
  "Solution_9": "here's one of the most useful formulas I know:\r\n\r\nThe Pythagorean Theorem states that in a right triangle, where a and b are its legs and c is its hypotenuse, a^2+b^2=c^2\r\n\r\nlol. So basic yet so helpful. :)"
}
{
  "Tag": [
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "Given a circle $ k$ and a point $ P$ outside it, an arbitrary line $ s$ passing through $ P$ intersects $ k$ at the points $ A,B$. Let $ M,N$ be the midpoints of the arcs determined by the points $ A,B$. Let $ C$ be the point on $ AB$ such that $ PC^2\\equal{}PA\\cdot PB$. Prove that $ \\angle MCN$ doesn't depend on the choice of $ s$.",
  "Solution_1": "[img]http://i006.radikal.ru/0804/04/4c243cc021f5.jpg[/img]\r\nLet $ K,L \\in k$, s.t. $ PK$ and $ PL$ are tangent to $ k$. Then we have $ PK^2 \\equal{} PL^2 \\equal{} PA\\cdot PB \\equal{} PC^2$, so $ PK \\equal{} PC \\equal{} PL$, so $ C$ always lie on the circle $ t$ with radius $ PK$. Let $ \\angle{CKL} \\equal{} \\alpha$, then $ \\angle{BPL} \\equal{} 2\\alpha$. But $ 2\\alpha \\equal{} \\angle{BPL} \\equal{} \\dfrac{\\breve{BL} \\minus{} \\breve{AL}}{2} \\equal{} \\dfrac{\\breve{BM} \\plus{} \\breve{ML} \\minus{} \\breve{AL}}{2} \\equal{} \\dfrac{\\breve{LA} \\plus{} 2\\breve{ML} \\minus{} \\breve{AL}}{2} \\equal{} \\breve{ML}$\r\nso $ \\angle{MKL} \\equal{} \\dfrac{\\breve{ML}}{2} \\equal{} \\alpha$ and $ \\angle{CKM} \\equal{} \\alpha \\minus{} \\alpha \\equal{} 0$, so $ K,C,M$ are collinear.\r\n Let $ \\angle{CLK} \\equal{} \\beta$, then $ \\angle{BPK} \\equal{} 2\\beta$. But $ 2\\beta \\equal{} \\angle{BPK} \\equal{} \\dfrac{\\breve{BK} \\minus{} \\breve{AK}}{2} \\equal{} \\dfrac{\\breve{BN} \\plus{} \\breve{KN} \\minus{} \\breve{AK}}{2} \\equal{} \\dfrac{\\breve{AK} \\plus{} 2\\breve{NK} \\minus{} \\breve{AK}}{2} \\equal{} \\breve{NK}$\r\nso $ \\angle{NLK} \\equal{} \\dfrac{\\breve{NK}}{2} \\equal{} \\beta$ and $ \\angle{CLN} \\equal{} \\beta \\minus{} \\beta \\equal{} 0$, so $ N,C,L$ are collinear.\r\n\r\nBecause $ P$ and $ k$ are constant, $ \\angle{KPL} \\equal{} const$. But $ \\angle{MCN} \\equal{} \\angle{KCL} \\equal{} 180^\\circ \\minus{} \\dfrac{\\angle{KPL}}{2} \\equal{} const.$, q.e.d.",
  "Solution_2": "Thank [color=darkblue][b]Ahiles[/b][/color] for nice solution."
}
{
  "Tag": [
    "ARML",
    "algebra",
    "polynomial",
    "geometry",
    "circumcircle",
    "LaTeX",
    "inequalities"
  ],
  "Problem": "Post ARML problems/solutions to previous ARML problems.  To make the difficulty more varried, I suggest we post easier problems (numbers 1-3) first, then moderate difficulty (4,5), then the harder problems (6-8).  The proccess starts over every 10, so just look at the units digit to determine the difficulty of the problem.\r\n\r\n[color=blue][size=200]PROBLEM 1[/size][/color]\r\nSource:  2000 problem 1\r\n\r\nFor $1<x<y$, let $S=\\{1,x,y,x+y\\}$.  Compute the absolute value of the difference between the mean and median of $S$.",
  "Solution_1": "Is this a bit trivial? :D\r\n\r\n[hide=\"Problem 1 Solution\"]\n\nMean = $1+x+y+(x+y)=2(x+y)+1$\nMedian = $\\frac{x+y}{2}$\n\n|Mean-Mode|=$\\frac{3}{2}(x+y)+1$\n\n :maybe: \nIs this a trick question...  :wink: \n[/hide]\r\n\r\n[color=red]You don't need quotations for hide titles.\nb-flat[/color]",
  "Solution_2": "[quote=\"not_trig\"]Is this a bit trivial? :D\n\n[hide=\"Problem 1 Solution\"]\n\nMean = $1+x+y+(x+y)=2(x+y)+1$\nMedian = $\\frac{x+y}{2}$\n\n|Mean-Mode|=$\\frac{3}{2}(x+y)+1$\n\n :maybe: \nIs this a trick question...  :wink: \n[/hide]\n\n[color=red]You don't need quotations for hide titles.\nb-flat[/color][/quote]\r\n\r\nGenerally for mean, you divide by the number of terms  :wink: \r\nNo, I checked my book, and it is exactly the same.",
  "Solution_3": "the answer is $1/4$, it's self explanatory\r\n\r\nuh, I don't really have ARML problems...but I got similar problems I guess or whatever...\r\n\r\nso here's like a #1\r\n[size=150]\nProblem 2:[/size]\r\n\r\nFind the sum of the prime factors of the number which is four more than the square of $625$.",
  "Solution_4": "[hide=\"Problem 2 Solution\"]$625^{2}+4=390,629$, which i think is prime so our answer is $390,629$    \nnot sure if i'm right though[/hide]",
  "Solution_5": "[quote=\"NeverOddOrEven\"][hide=\"Problem 2 Solution\"]$625^{2}+4=390,629$, which i think is prime so our answer is $390,629$    \nnot sure if i'm right though[/hide][/quote]\r\n\r\n390,629 is not prime (I used my calculator here - it is $577\\times677$, making the answer $1254$).  The problem is asking to factor the sum of squares, but I don't know an easy way to do that.",
  "Solution_6": "dude u can't use a calculator",
  "Solution_7": "n^4 + 4",
  "Solution_8": "$=(n^{2}-2n+2)(n^{2}+2n+2)$. When you get simple polynomials like that that don't have real roots, look for difference of squares.\r\n\r\nI don't know who's supposed to post, but I have to leave, so... somebody else do it.",
  "Solution_9": "ohh, that is cool (I can't press a key mutliple times quickly).\r\n\r\n[size=150][color=blue]Problem 3[/color][/size]\r\n$\\triangle ABC$ is inscribed in circle $O$, the radius of $O$ is 12 and $m\\angle ABC=30^\\circ$.  A circle with center $B$ is drawn tangent to the line containing $\\overline{AC}$.  Let $R$ be the region which is within $\\triangle ABC$, but outside circle $B$.  Compute the maximum area of $R$.\r\n\r\nThis was a 1997, problem 2.",
  "Solution_10": "I have solved it after remembering how to divide right :wink:\r\n\r\n[hide=\"solution\"]\n\nLet the radius of circle $B$, which is also the height from $AC$ of $\\triangle ABC$, be $h$.\n\nThe area of the sector of circle $B$ inside $\\triangle ABC$ is a $30^\\circ$ sector, so it has area $(\\frac{1}{12})(\\pi h^{2})$.  \n\n$\\triangle ABC$, on the other hand, has area $(\\frac{1}{2})(12)(h)$, since $AC$ = 12.\n\n$AC$ is 12 because $\\angle AOC$ (let's call O the circumcenter of $\\triangle ABC$) is 60, and $\\triangle AOC$ is isoceles, which is obvious.\n\nSo, optimizing $6h-\\frac{\\pi}{12}h^{2}$, we get $h=\\frac{36}{\\pi}$, and \n\n$R=\\frac{108}{\\pi}$.\n\n[/hide]\r\n\r\nI don't have a problem -- someone else post one. (assuming I'm right...)",
  "Solution_11": "[quote=\"not_trig\"]\nI don't have a problem -- someone else post one. (assuming I'm right...)[/quote]\r\nHey I'm new to NC. I just moved here a week ago to start the tenth grade!\r\n\r\nIf $a^{3}= b^{2}$ and $c^{3}= d^{2}$, where $a, b, c,$ and $d$ are positive integers; and if $c-a = 25$, what are $a, b, c,$ and $d$?",
  "Solution_12": "Congratulations on moving to the ARML national champions' state!\r\n\r\n(thank MysticTerminator)\r\n\r\nYour LaTeX is pretty good.\r\n\r\nBTW, in problem solving marathons, please post the problem number.\r\n\r\n[hide=\"solution to number 4\"]\n\nSince $a^{3}=b^{2}$, $\\sqrt{a^{3}}$ is an integer, $\\sqrt{a}$ must be an integer. Let's call $\\sqrt{a}$ $j$.  Then, $b=j^{3}$.\n\nSimilarly, let $\\sqrt{c}$ be $k$, so $c=k^{2}$ and $d=k^{3}$.\n\nSince $a, c > 0$, $j, k > 0$.\n\nTherefore, $c-a = j^{2}-k^{2}= (j+k)(j-k) = 25 \\cdot 1$.\n\n(Since j, k are positive, 5*5 wouldn't work since k would be 0)\n\nTherefore, $j=13$, $k=12$.\n\nSo, \\[a=169, b=13^{3}, c=144, d=1728\\] (too lazy to calculate $13^{3}$...)  :lol: \n\n[/hide]\r\n\r\nAgain, I have no problems to give  :( \r\n\r\nMaybe I should check my ARML book....",
  "Solution_13": "I feel like such a fool\u2026 The problem above was not an ARML problem (sorry  :blush: ). \r\nAnd I have never taken any national or state math tests. Therefore, I did not make it to the ARML nationals.  :noo: \r\n\r\nOnce again I am sorry for my ignorance and any deceptiveness I may have portrayed on my part. I am just a teenage who likes math.",
  "Solution_14": "[quote=\"Imperial Effect\"]I feel like such a fool\u2026 The problem above was not an ARML problem (sorry  :blush: ). \nAnd I have never taken any national or state math tests. Therefore, I did not make it to the ARML nationals.  :noo: \n\nOnce again I am sorry for my ignorance and any deceptiveness I may have portrayed on my part. I am just a teenage who likes math.[/quote]\r\n\r\ner what it is all ok this is just a forum for like hanging out and stuff",
  "Solution_15": "*ahem* any idea? :oops: \r\n\r\nanyway I have an easy inequality...\r\n\r\n[b]Problem 22[/b]\r\n\r\n$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1 \\Rightarrow x^{2}+y^{2}+z^{2}\\ge 27$",
  "Solution_16": "Let $a=\\frac{1}{x}, b=\\frac{1}{y},$ and $c=\\frac{1}{z}$.\r\n\r\nNow, $f(x)=\\frac{1}{x^{2}}$ is convex in (0,infinity), so\r\n\r\n$f(a)+f(b)+f(c) \\ge 3 \\cdot f(\\frac{a+b+c}{3})=3 \\cdot f(1/3)=27$",
  "Solution_17": "For the people here who aren't familiar with Jensen's (or, for that matter, convexity), QM-HM is probably a more illustrative proof. Yes, I know it is a special case of Jensen's...\r\n\r\n[hide=\":weightlift:\"]$\\sqrt{\\frac{x^{2}+y^{2}+z^{2}}3}\\ge \\frac 3{\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}=3$, from which the result follows immediately.[/hide]\r\nYou got to it first, Jeremy, so I grant you the burden of posting the next problem. ARML marathon, right? :roll:",
  "Solution_18": "Ok, while we are on the subject:\r\n\r\nLet $a,b,c>0$ be such that $a+b+c=1$\r\n\r\nProve that $a^{b}\\cdot b^{c}\\cdot c^{a}\\le ab+bc+ca$\r\n\r\nFor those who want a more ARML-style problem, compute the sum (from k=1 to infinity) of\r\n\r\n$\\frac{F_{k}}{2^{k}}$\r\n\r\nwhere $F_{1}=1$, $F_{2}=1$, and the $F_{k}$ are the Fibonacci numbers",
  "Solution_19": "[quote=\"n^4+4\"]Ok, while we are on the subject:\n\nLet $a,b,c>0$ be such that $a+b+c=1$\n\nProve that $a^{b}\\cdot b^{c}\\cdot c^{a}\\le ab+bc+ca$[/quote]\r\n[hide=\" :rotfl: \"]It's AM-GM on $a,b,c$ with weights $b,c,a$.[/hide]\r\nI think your second problem is harder :P",
  "Solution_20": "[quote=\"not_trig\"]*ahem* any idea? :oops: \n\nanyway I have an easy inequality...\n\n[b]Problem 22[/b]\n\n$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1 \\Rightarrow x^{2}+y^{2}+z^{2}\\ge 27$[/quote]\r\n\r\nwell my solution used AM-GM and GM-HM:\r\n\r\n$\\sqrt[3]{xyz}\\ge\\frac{3}{\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}}=3$; $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\Leftrightarrow xyz=xy+yz+xz \\Rightarrow xy+yz+yz\\ge27$\r\n\r\nnow $x^{2}+y^{2}+z^{2}\\ge xy+yz+xz = 27$\r\n\r\nEDIT: BERMAN YOU FOXY PERSON YOU KEEP USING SMILEYS AS HIDE TITLES",
  "Solution_21": "@boy soprano: out of curiosity, why did you (falsely) change the inequality to be strict?\r\n\r\nActually, I think the second problem is easier... I guess I'm just weird like that. Anyhow, it's pretty cool, so it deserves more than the workout smiley.\r\n\r\n[hide=\":starwars:\"]S=1/2+1/4+2/8+3/16+5/32+...\n[u]2S=1+1/2+2/4+3/8+5/16+8/32+...[/u]\nS+2S=1+2/2+3/4+5/8+8/16+13/32+...\nBut just by quadrupling the original:\n4S=3+2/2+3/4+5/8+8/16+13/32+...\nSubtracting, S=2.[/hide]",
  "Solution_22": "erm i have another problem:\r\n\r\nFind $a-b$ if $a^{2}-b^{2}=45$ adn $a^{3}-b^{3}=945$, $a, b \\in \\mathbb{Z}$.\r\n\r\nSure, this was #2 on our comprehensive regional, but even Vivek skipped it...",
  "Solution_23": "That problem was really easy to cheat on (i.e. just check every one of the 5 answer choices they gave us)\r\n\r\nAs it is I see nothing better than working through about 10 cases",
  "Solution_24": "[quote=\"jb05\"]@boy soprano: out of curiosity, why did you (falsely) change the inequality to be strict?[/quote]\r\nWhat do you mean?\r\n\r\n$\\frac{1}{3}^{1/3}\\frac{1}{3}^{1/3}\\frac{1}{3}^{1/3}= \\frac{1}{3}\\cdot\\frac{1}{3}+\\frac{1}{3}\\cdot\\frac{1}{3}+\\frac{1}{3}\\cdot\\frac{1}{3}= \\frac{1}{3}$?",
  "Solution_25": "When you quoted it, you changed <= to <.\r\n\r\nThis new problem looks disgusting. Who's up for trying all 12 cases? :wink:",
  "Solution_26": "[quote=\"not_trig\"]erm i have another problem:\n\nFind $a-b$ if $a^{2}-b^{2}=45$ adn $a^{3}-b^{3}=945$, $a, b \\in \\mathbb{Z}$.\n\nSure, this was #2 on our comprehensive regional, but even Vivek skipped it...[/quote]\r\n\r\n[hide]The key part that makes it easy is the $a,b\\in\\mathbb{Z}$.\n\n$a^{2}-b^{2}=45$\n\n$(a+b)(a-b)=3^{2}\\cdot5$\n\nThere are only 3 pairs of solutions $(a^{2},b^{2})$ that solve this; $(23^{2},22^{2})$, $(9^{2},6^{2})$, and $(7^{2},2^{2})$.\n\n$(23^{2},22^{2})$ can be thrown out because $23^{3}-22^{3}$ is way greater than 945.\n\n$(9^{2},6^{2})$ produces $(\\pm9,\\pm6)$, which could only possibly be true if $9^{3}+6^{3}=945$, and it does.\n\nThus, $a-b=3$.[/hide]",
  "Solution_27": "[quote=\"jb05\"]When you quoted it, you changed <= to <.\n\nThis new problem looks disgusting. Who's up for trying all 12 cases? :wink:[/quote]\nOh, no, I think it just looks like that because the bottom of the quote box makes it hard to see the equality line.  Compare:\n[quote]$a^{b}\\cdot b^{c}\\cdot c^{a}\\le ab+bc+ca$[/quote]\n[quote]\n$a^{b}\\cdot b^{c}\\cdot c^{a}\\le ab+bc+ca$\n\n[/quote]\r\n\r\n(And in my previous post, I forgot what \"strict\" meant :  I thought strict => sharp => equality included  :roll: )",
  "Solution_28": "The second problem is just plugging 1/2 into the generating function, is it not? So it would be $\\frac{1/2}{1-1/2-1/3}= 2$",
  "Solution_29": "[quote=\"b-flat\"][quote=\"not_trig\"]erm i have another problem:\n\nFind $a-b$ if $a^{2}-b^{2}=45$ adn $a^{3}-b^{3}=945$, $a, b \\in \\mathbb{Z}$.\n\nSure, this was #2 on our comprehensive regional, but even Vivek skipped it...[/quote]\n\n[hide]The key part that makes it easy is the $a,b\\in\\mathbb{Z}$.\n\n$a^{2}-b^{2}=45$\n\n$(a+b)(a-b)=3^{2}\\cdot5$\n\nThere are only 3 pairs of solutions $(a^{2},b^{2})$ that solve this; $(23^{2},22^{2})$, $(9^{2},6^{2})$, and $(7^{2},2^{2})$.\n\n$(23^{2},22^{2})$ can be thrown out because $23^{3}-22^{3}$ is way greater than 945.\n\n$(9^{2},6^{2})$ produces $(\\pm9,\\pm6)$, which could only possibly be true if $9^{3}+6^{3}=945$, and it does.\n\nThus, $a-b=3$.[/hide][/quote]\r\n\r\nlol i think \"use calculator\" would be a slightly easier solution, but this is good too (kinda lucky that $9^{3}+6^{3}=945$)"
}
{
  "Tag": [
    "inequalities",
    "rearrangement inequality"
  ],
  "Problem": "ARGH, I have spent ages trying to prove this.\r\n\r\n\r\nab + bc + ca \u2264  $a^2  + b ^2  + c^2$\r\n\r\nfor all real non-zero numbers a,b,c\r\n\r\nI just cant do it, im missing something simple.\r\n\r\nYour help is appreciated.\r\n\r\nThanks",
  "Solution_1": "Have you heard of the rearrangement inequality?\r\n\r\nWell first assume WLOG that $a \\ge b \\ge c$.  Then by rearrangement, this is trivial.",
  "Solution_2": "Actually, an easier proof for you would be to multiply 2 on both sides, and then subract the $2ab+2bc+2ac$ terms over.  Factor into squares, and then you're done.",
  "Solution_3": "$(a-b)^2 + (b-c)^2 + (c-a)^2 \\geq 0$\r\nBut you should learn \"rearrangement inequality\" too. It's usefull in BW  :)",
  "Solution_4": "WLOG, assume $a \\geq b \\geq c$, then: $a \\cdot a + b \\cdot b + c \\cdot c \\geq a\\cdot b + b \\cdot c + c \\cdot a$.\r\n\r\nBTW, what is BW?",
  "Solution_5": "Probably the Baltic Way competition? Am I right?",
  "Solution_6": "Another solution:$a^2+b^2+c^2-ab-bc-ca=\\left(a-\\frac{b+c}{2}\\right)^2+\\frac{3}{4}(b-c)^2\\geq 0$",
  "Solution_7": "Joining in flogging  ...\r\nif  (x,y,z)  are absolute values  of (a,b,c), and thus non-negative, by AM-GM\r\nwe get  $x^3+y^3+z^3 -  3 xyz  \\ge 0$ \r\nor $(x+y+z)(x^2+y^2+z^2 -xy -yz-zx) \\ge 0$\r\nsince  $x+y+z \\ge 0$ we have  $a^2+b^2+c^2  \\ge  xy+yz+zx$ but $xy \\ge ab $ etc.. so you get what you want..."
}
{
  "Tag": [
    "geometry",
    "inequalities",
    "AMC",
    "AIME",
    "blogs",
    "USA(J)MO",
    "USAMO",
    "\\/closed"
  ],
  "Problem": "I heard that the Olympiad classes (not WOOT, of course):\r\n\r\n- Olympiad Problem Solving\r\n- Olympiad Geometry\r\n- Olympiad Inequalities\r\n\r\nare going to be cancelled. Is this true?  :noo:  :(\r\n\r\nOf course, I may be wrong, and this might just be a whole rumour that's spreading around. If so, I completely apologize and someone can just delete this post. But please say whether the Olympiad classes are going to be continued or not, because I can't afford to pay for WOOT.",
  "Solution_1": "According to the classes information under \"Online School,\" Olympiad Problem Solving will occur during Fall 2006 and Olympiad Geometry and Inequalities will occur during Winter 2006/7.",
  "Solution_2": "We have decided to discontinue the Olympiad classes in order to be able to spend more time on other projects.  We strongly encourage students interested in the Olympiad classes to participate in WOOT.  In order to facilitate this, we have added any early-bird discount on the WOOT price.  Students who sign up by July 15 will receive a 100 dollar discount, as now reflected on the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]Enrollment page[/url].",
  "Solution_3": "WOOT! I think I will probably sign up for WOOT now! I just have to convince my parents, lol :)",
  "Solution_4": "So are the Olympiad classes the only ones being discontinued (of course along with the independent study)?  Will there be any options besides WOOT for those who were planning to take an olympiad class?  I guess those new books are coming along then.",
  "Solution_5": "Does this mean that the classes already planned for this year will be cancelled? (EDIT: Nvm, I see that they've been taken off the classes page.)\r\n\r\nAlso, I won't have the time or money next school year to do WOOT, but I would have enough for olympiad classes, does anyone have any recommendations as to what I should do? I'm considering taking another AIME problem series, doing an intermediate class, though for some of those classes there would be a lot of review (especially seeing as I plan to go through AoPS 2), possibly trying WOOT anyway, or even not taking an AoPS classes.",
  "Solution_6": "While I understand the concerns of those of you who would like to take the old Olympiad classes, I ask that you understand our position as well.  We've been heavily subsidizing the school for a very long time (and will still be subsidizing WOOT).  In order to continue to be able to produce any resources at all, we have to focus more of our time on properly priced resources, instead of spending 50-60% of our time on resources we give away either for free or for a very, very low price.  I've posted some of my thoughts about this in my blog, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=6098]here[/url].",
  "Solution_7": "Thankyou for your response on pricing.\r\n\r\nHowever, assuming that I can afford it, I'm also worried about time, seeing as I'll have a particularly busy sophmore (next) year (I have been planning to do WOOT in junior year). How much time does one have to spend on WOOT? Could I still get as much from WOOT if I didn't spend as much time as I should on it? Or would it be recommended that I do something lighter next year(thinking of AIME Problem Series B as well as A, which I'm doing over the summer)?",
  "Solution_8": "[quote=\"calc rulz\"]Thankyou for your response on pricing.\n\nHowever, assuming that I can afford it, I'm also worried about time, seeing as I'll have a particularly busy sophmore year. How much time does one have to spend on WOOT? Could I still get as much from WOOT if I didn't spend as much time as I should on it? Or would it be recommended that I do something lighter next year(thinking of AIME Problem Series B as well as A, which I'm doing over the summer)?[/quote]\r\n\r\nI think the answer to this depends on how well you can budget your time.  WOOT is somewhat more spread out than our other classes.  As I remember from your earlier postings in the AMC area, you are already pretty proficient with the AIME.  Due to the USAMO expansion, you'll also have some more breathing room.  I'd suggest you start spending more of your time with the more challenging material.  In other words, I'd recommend *not* taking the AIME PS B and focusing more of your efforts on Olympiad material.  If you dive into AIME PS A over the summer, and spend some time with books like Zeitz and such, you should put yourself in a good position with respect to the AIME in 2007.  (I'm assuming you're on your school's math team and will have plenty of opportunity to stay sharp for the AMC.)\r\n\r\nOne of our goals in expanding the USAMO was to get more students to spend some time working on olympiad-style problems.  Far too many would spend far too much time cramming away for the AIME.  I would guess that at least half the students who barely qualified for the AIME (if not more) had never tried an olympiad problem before qualifying for the USAMO.  We'd like to change that, and encourage students to start developing problem solving stamina, writing skills, and the courage to attack problems that appear impossible.",
  "Solution_9": "Very understandable.  Okay. Well, one final question that's been bugging me though:  Why was inversion taught in the olympiad geometry class?  To my understanding [url=http://olympiads.win.tue.nl/imo/]inversion isn't tested at the IMO[/url] but [url=http://en.wikipedia.org/wiki/Mathematical_Olympiad_Program]Wikipedia says[/url] MOP poses questions about it.  Is it just for mathematical enrichment, or can it be useful in some problems although it's not required? Thanks.\r\n\r\nP.S. The \"Naming controvesy\" article on wikipedia is quite humorous",
  "Solution_10": "Inversion was taught for one of the reasons you surmise - it can be extremely useful, but it's not ever required.  (In fact, this is true of a lot of the advanced techniques in all areas of Olympiad problem solving.)\r\n\r\nI also taught it because it's way cool stuff.",
  "Solution_11": "Mr. Rusczyk,\r\n\r\nAlthough I understand the cost issue in cancelling the Olympiad classes, it is a loss nevertheless for students.  I have no doubt WOOT is a great course,  but it will not go as deep as the Olympiad subject classes therefore it is not really able to replace the Olympiad classes.  But again I understand you have no choice.  \r\n\r\nBut I have an idea that may benefit both students and AoPS.  Are you willing or able to sell transcritpts of past Olympiad classes (have you nicely archived those materials?)?   Students purchasing those transcripts can study on their own pace.  It may not benefit as much as actually participating in the classes, but it should be next to the best.  I have no idea how many students will be willing to buy, but I will be willing to pay for it. \r\n\r\nThanks,\r\n\r\nKevin Yuan (I'm using my son Allen's account to post this)",
  "Solution_12": "We will be writing Olympiad books in the coming years, and there will be beginning Olympiad material in the Intermediate texts, the first two of which will be out within a year.  Putting the transcripts in a format (and configuring the ordering system, etc) that would allow selling those is unfortunately too time-consuming to merit what they would produce.  We will be drawing on the highlights of these for WOOT, and I expect the entirety of a few years' WOOT participation will be much broader and deeper than the Olympiad classes were.",
  "Solution_13": "Is there a chance that they will come out in early 2007 or late 2006?  :D",
  "Solution_14": "I am very sorry that this is a little bit off topic and if this has been announced before, although what is the date when the Intro to Number Theory book is going to be released? :?",
  "Solution_15": "[quote=\"Knuth's_check\"]I am very sorry that this is a little bit off topic and if this has been announced before, although what is the date when the Intro to Number Theory book is going to be released? :?[/quote]\r\n\r\nAround the 27th of this month.",
  "Solution_16": "[quote=\"rrusczyk\"]We will be writing Olympiad books in the coming years, and there will be beginning Olympiad material in the Intermediate texts, the first two of which will be out within a year.  Putting the transcripts in a format (and configuring the ordering system, etc) that would allow selling those is unfortunately too time-consuming to merit what they would produce.  We will be drawing on the highlights of these for WOOT, and I expect the entirety of a few years' WOOT participation will be much broader and deeper than the Olympiad classes were.[/quote]\r\n\r\nThen would AoPS mind if we tried to get past transcripts/problems from people who took the courses when they were still offered?",
  "Solution_17": "No, we can't encourage that.  We may bring the olympiad classes back some day, and we'll be drawing on some of that material for WOOT.  Moreover, we cannot encourage any sort of file sharing of our online school material, as it jeopardizes the future of the online school (and has been a contributor to our discontinuing the olympiad classes for now).",
  "Solution_18": "That's perfectly reasonable, of course. Looks like this generation of problem solvers will have to make do without them  :(",
  "Solution_19": "Speaking of books and classes and the like, what are the next books that are going to come out?",
  "Solution_20": "[quote=\"horselvr101\"]Speaking of books and classes and the like, what are the next books that are going to come out?[/quote]Introduction to Algebra and Intermediate Counting & Probability should come out in early 2007.  Intermediate Number Theory should come out in Spring 2007.",
  "Solution_21": "[quote=\"DPatrick\"][quote=\"horselvr101\"]Speaking of books and classes and the like, what are the next books that are going to come out?[/quote]Introduction to Algebra and Intermediate Counting & Probability should come out in early 2007.  Intermediate Number Theory should come out in Spring 2007.[/quote]\r\nWhat exactly will the intro to algebra include?",
  "Solution_22": "Basically all of algebra I, much of the algebraic portion of pre-algebra, and a healthy chunk of algebra II.  \r\n\r\nLinear equations, quadratics, functions, word problems, proportions, special factorizations, arithmetic and geometric sequences and series, radicals, exponentials, plenty of basic analytic geometry.",
  "Solution_23": "What about Intermediate Number Theory? I just took the Intro to Number Theory course and I was wondering where I should go next to continue studying that area...",
  "Solution_24": "[quote=\"horselvr101\"]What about Intermediate Number Theory? I just took the Intro to Number Theory course and I was wondering where I should go next to continue studying that area...[/quote]\r\n\r\nIt'll be a while before we have an Interm. Number Theory book.  The class will probably be offered again next summer.",
  "Solution_25": "Will Interm Algebra come out before or after the class starts?",
  "Solution_26": "Are the olympiad classes ever going to come back or are they gone for good?\r\nWhat is the tentative schedule for books to come out this year?\r\n\r\n\r\n-bpms",
  "Solution_27": "[quote=\"bpms\"]Are the olympiad classes ever going to come back or are they gone for good?\nWhat is the tentative schedule for books to come out this year?\n[/quote]\r\n\r\nThe olympiad classes are closed for the foreseeable future.  Some of the material will be folded into WOOT.\r\n\r\nTentative books schedule: Intro Algebra March, Interm Counting May, Interm Algebra June."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "I've come up with the following conjecture but I'm not sure how to prove it. Take a pre-solved Rubik's cube (normal 3x3x3) and perform any number of moves. Prove that for some number n, you can iterate that sequence n times such that the cube will end in a solved position. I've tried this for several sequences and so far I haven't found one that doesn't solve in less than 200 (note the sequences I've been doing are at most 4-5 moves and not very complicated). If anyone has a program simulating a cube I'd appreciate the source. I don't know too much about orbitals so if they're used to approach this problem, please tell me.",
  "Solution_1": "one i know off top of head; R U' repeated 64 times; 128 moves.",
  "Solution_2": "Let a move be any given sequence of what you normally call a move.\r\n\r\nThe order of the rubik-group is finite, so you will ever reach the solved position (=neutral element). You can even show (by Lagranges theorem) that the number of moves you did is a divisor of the total number of different positions possible.\r\n\r\nTo show it more elementary: you can't get a new constellation everytime after doing the move, since that would mean that the cube has $\\infty$ many positions, which is clearly not to case. So after some time you get a position you already had before, but what is the difference between that position and a full solved cube after some recoloring\u00bf Nothing! So you will ever reach the position you started with (solved one) again.",
  "Solution_3": "Its not too hard to find a maximal period for your re-iterations. The something about it in the strategy section"
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "find square root of:-\r\n\r\n$11+2\\sqrt{7}+\\sqrt{12}+\\sqrt{84}$",
  "Solution_1": "[hide=\"Solution\"]\\begin{eqnarray*}\\sqrt{11+2\\sqrt{7}+\\sqrt{12}+\\sqrt{84}} &=& \\sqrt{11+2\\sqrt{7}+2\\sqrt{3}+2\\sqrt{21}}\\\\ &=& \\sqrt{1^2+\\sqrt{3}^2+\\sqrt{7}^2+2\\cdot 1\\cdot\\sqrt{7}+2\\cdot 1\\cdot\\sqrt{3}+2\\cdot\\sqrt{3}\\cdot\\sqrt{7}}\\\\ &=& \\sqrt{(1+\\sqrt{3}+\\sqrt{7})^2}\\\\ &=& 1+\\sqrt{3}+\\sqrt{7}\\end{eqnarray*}[/hide]",
  "Solution_2": "farenhajt , in that, how did you get to know that it would be consisiting of 3 terms , 1, sqrt3 and sqrt7 ? i mean if it was a much more complex expression then wouldnt it be hard to get to know the terms or to split them like you did?I mean isnt their some sure-shot way like using variables that you can get the answer?",
  "Solution_3": "Well, I think there's no general formula, since the expression must have special properties. In this example it's really obvious that we have $2\\sqrt{3}, 2\\sqrt{7}$ and $2\\sqrt{3}\\cdot\\sqrt{7}$, and that points to a square-of-a-trinomial formula. Once you know where to go, the rest is easy. And of course, you must have working knowledge of polynomial factorization, in order to recognize the patterns."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "not sure where to post this people told me this was one of the good books on glambling and stuff. If anybody has it le me know what you think of it.",
  "Solution_1": "gambling???????",
  "Solution_2": "[quote=\"dreambold22\"]gambling???????[/quote]\r\n\r\nwhy not",
  "Solution_3": "i'm not interested in this topic. good bye.",
  "Solution_4": "[quote=\"Nick31\"][quote=\"dreambold22\"]gambling???????[/quote]\n\nwhy not[/quote]\r\n\r\nErm.....  :what?:  :what?:  :what?:",
  "Solution_5": "Is it a probability book?  :|",
  "Solution_6": "gambling??????? [color=white][size=0]lol copy/paste[/size][/color]",
  "Solution_7": "[quote=\"leoxnlin\"]Is it a probability book?  :|[/quote]\r\n\r\nprobabilities applied to various card games, but I don't have the book so I can't tell exactly"
}
{
  "Tag": [
    "trigonometry",
    "geometry proposed",
    "geometry"
  ],
  "Problem": "Given quadrilateral $ ABCD$, $ AC\\cap BD=O$. Prove that $ ABCD$ is cyclic iff $ OA\\sin A+OC\\sin C=OB\\sin B+OD\\sin D$.",
  "Solution_1": "http://www.mathlinks.ro/Forum/viewtopic.php?t=26474\r\nSee, for the solution,\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=15584 post #2 file IMO_1997_shortlist.pdf or\r\nhttp://www.kalva.demon.co.uk/short/soln/sh9723.html\r\n\r\n  darij"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ x_1,x_2,...,x_n$ be real numbers such that $ x_i > 1$ for $ i \\equal{} 1,2,...,n$. Prove that\r\n\\[ \\frac {x_1x_2}{x_3 \\minus{} 1} \\plus{} \\frac {x_2x_3}{x_4 \\minus{} 1} \\plus{} ... \\plus{} \\frac {x_nx_1}{x_2 \\minus{} 1} \\ge 4n.\r\n\\] Determine when the equality occurs.",
  "Solution_1": "I am so bored today!\r\nNow, we have\r\n$ \\sum_{i \\equal{} 1}^n \\frac {x_ix_{i \\plus{} 1}}{x_{i \\plus{} 2} \\minus{} 1} \\geq 4\\frac {x_ix_{i \\plus{} 1}}{x_{i \\plus{} 2}^2}$\r\nwith the notations are $ n \\plus{} 1 \\equal{} 1$ and $ n \\plus{} 2 \\equal{} 2$\r\nIt is mere to prove that\r\n$ \\sum_{i \\equal{} 1}^n \\frac {x_ix_{i \\plus{} 1}}{x_{i \\plus{} 2}^2} \\geq n$ which is true by the general form of Cauchy in my signature!! :D"
}
{
  "Tag": [
    "inequalities",
    "ratio",
    "geometry",
    "algebra",
    "polynomial",
    "circumcircle",
    "function"
  ],
  "Problem": "Last week, the Nordic Mathematical Contest was held. Here is the problem-set.\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206646#p206646]Problem 1 - link[/url]\r\nFind all positive integers $k$ such that the product of the digits of $k$, in decimal notation, equals \\[\\frac{25}{8}k-211\\]\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206634#p206634]Problem 2 - link[/url]\r\nLet $a,b,c$ be positive real numbers. Prove that \\[\\frac{2a^2}{b+c} + \\frac{2b^2}{c+a} + \\frac{2c^2}{a+b} \\geq a+b+c\\]\r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206644#p206644]Problem 3 - link[/url]\r\nThere are $2005$ young people sitting around a large circular table. Of these, at most $668$ are boys. We say that a girl $G$ has a strong position, if, counting from $G$ in either direction, the number of girls is always strictly larger than the number of boys ($G$ is herself included in the count). Prove that there is always a girl in a strong position. \r\n\r\n[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206638#p206638]Problem 4 - link[/url]\r\nThe circle $\\zeta_{1}$ is inside the circle $\\zeta_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\\zeta_{1}$ also at $B$, and $\\zeta_{2}$ also at $C$. The tangent to $\\zeta_{1}$ at $B$ intersects $\\zeta_{2}$ at $D$ and $E$. The tangents of $\\zeta_{1}$ passing thorugh $C$ touch $\\zeta_{2}$ at $F$ and $G$. Prove that $D$, $E$, $F$ and $G$ are concyclic.",
  "Solution_1": "Can you say please how much time is allowed ?\r\nP.S. the inequality seems to be very easy  :lol:",
  "Solution_2": "4 hours.\r\n\r\nYes, the inequality is very easy!",
  "Solution_3": "Does anyone has recent years' NMO problems?",
  "Solution_4": "Arccosinus, were you at Baltic Way 2004? I might know you from there  ;)",
  "Solution_5": "No portoseb. I wanted to go, but I gave my place in the team to an alternate, because I had to go to a wedding in the U.S.",
  "Solution_6": "I saw the results today, at the Finnish IMO selection in which I take part. I was the first with 18 points, so nobody got full points. As soon as the problems will be put on the internet I will let you know.\r\n\r\nEDIT: here they are:\r\n\r\nhttp://www.math.helsinki.fi/~smy/olympia/PM/NMC19.htm",
  "Solution_7": "19th Nordic Mathematical Contest \t\t\t\t\t\r\nResults\t\t\t\t\t\t\r\n\t\t\t\t\t\t\t\r\n\t\t\tProblem\tSum\r\n\t\t\t1\t2\t3\t4\t\u3000\r\n1.\tDumitrescu, Sebastian\tFIN\t5\t4\t5\t4\t18\r\n2.\tAlanko, Samu\tFIN\t5\t5\t5\t0\t15\r\n\u3000\tPettersson, Ville\tFIN\t5\t5\t5\t0\t15\r\n\u3000\tRennemo, J\u00f8rgen Vold\tNOR\t5\t5\t5\t0\t15\r\n\u3000\tStef\u00e1nsson, \u00d6rn\tICE\t5\t5\t5\t0\t15\r\n6.\tXing, Chen\tSWE\t4\t5\t5\t0\t14\r\n7.\tHalld\u00f3rsson, H\u00f6skuldur P\u00e9tur\tICE\t3\t5\t3\t2\t13\r\n\u3000\tTroeng, Olof\tSWE\t5\t3\t5\t0\t13\r\n9.\tJacobsen, Martin Wedel\tDEN\t5\t5\t0\t2\t12\r\n\u3000\tLiu, Zihan\tSWE\t5\t0\t2\t5\t12\r\n\u3000\tReeh, Sune Nikolaj Precht\tDEN\t5\t5\t0\t2\t12\r\n\u3000\tVass, Marton\tFIN\t5\t0\t5\t2\t12\r\n13.\tEdlund, Erik\tSWE\t5\t5\t1\t0\t11\r\n14.\tKolu, Iiris\tFIN\t5\t4\t1\t0\t10\r\n\u3000\tRudberg, Tore\tSWE\t5\t5\t0\t0\t10\r\n\u3000\tSigstad, Henrik\tNOR\t5\t4\t1\t0\t10\r\n\u3000\tSireni, Jukka\tFIN\t5\t5\t0\t0\t10\r\n\u3000\tTerelius, H\u00e5kan\tSWE\t5\t0\t5\t0\t10\r\n\u3000\tThelin, Sam\tSWE\t3\t5\t2\t0\t10\r\n20.\tHolden, Nina\tNOR\t4\t5\t0\t0\t9\r\n\u3000\tJohnsen, Hilde Galleberg\tNOR\t3\t5\t0\t1\t9\r\n\u3000\tJohnson, Kerstin\tSWE\t4\t5\t0\t0\t9\r\n\u3000\tNors Andersen, Drees\tDEN\t3\t4\t0\t2\t9\r\n24.\tEdlund, Bj\u00f6rn\tSWE\t3\t0\t5\t0\t8\r\n\u3000\tHelgad\u00f3ttir, Inga Steinunn\tICE\t3\t3\t0\t2\t8\r\n\u3000\tMeuller, Simon\tSWE\t0\t3\t5\t0\t8\r\n\u3000\tVesalainen, Esa\tFIN\t5\t3\t0\t0\t8\r\n\u3000\t\u00c5rdal, Atle Rygg\tNOR\t3\t5\t0\t0\t8\r\n29.\tAbrahamsen, Mikkel\tDEN\t4\t0\t2\t1\t7\r\n\u3000\tGrythe, Thomas Berge\tNOR\t2\t5\t0\t0\t7\r\n\u3000\tM\u00fcller, Jeppe Krarup\tDEN\t5\t2\t0\t0\t7\r\n32.\tBl\u00e5sten, Eemeli\tFIN\t0\t3\t3\t0\t6\r\n\u3000\tJacobsen, K\u00e5re\tDEN\t1\t5\t0\t0\t6\r\n\u3000\tKokkala, Janne\tFIN\t1\t5\t0\t0\t6\r\n\u3000\tPeng, Gunnar\tSWE\t1\t4\t0\t1\t6\r\n\u3000\tSiljander, Juhana\tFIN\t5\t0\t1\t0\t6\r\n\u3000\tWilson, Danny\tDEN\t5\t0\t0\t1\t6\r\n38.\tBrandt, Asbj\u00f8rn\tDEN\t5\t0\t0\t0\t5\r\n\u3000\tJensen, Morten Harding\tDEN\t5\t0\t0\t0\t5\r\n\u3000\tL\u00e4hteenm\u00e4ki, Meri\tFIN\t3\t0\t0\t2\t5\r\n\u3000\tM\u00e4ki, Sami\tFIN\t3\t0\t2\t0\t5\r\n\u3000\tPoulsen, Morten Kj\u00e6r\tDEN\t5\t0\t0\t0\t5\r\n\u3000\tSelin, M\u00e5rten\tSWE\t1\t4\t0\t0\t5\r\n\u3000\tSjamsutdin, Aslanbek\tNOR\t3\t0\t1\t1\t5\r\n\u3000\tSpars\u00f8, Eske\tDEN\t5\t0\t0\t0\t5\r\n46.\tEladhari, Matti\tSWE\t4\t0\t0\t0\t4\r\n\u3000\tGu\u00f0mundsson, J\u00f3n Emil\tICE\t3\t0\t0\t1\t4\r\n\u3000\tJohannesen, David\tDEN\t3\t0\t0\t1\t4\r\n\u3000\tLaaksonen, Antti\tFIN\t2\t0\t0\t2\t4\r\n\u3000\tMar\u00edusd\u00f3ttir, \u00deorey Mar\u00eda\tICE\t3\t0\t0\t1\t4\r\n\u3000\tT\u00f8rholm, Sebastian Paaske\tDEN\t3\t0\t0\t1\t4\r\n\u3000\tVester, Steen\tDEN\t4\t0\t0\t0\t4\r\n53.\tAlexanderson, Per\tSWE\t2\t0\t0\t1\t3\r\n\u3000\tEgilsd\u00f3ttir, Salv\u00f6r\tICE\t3\t0\t0\t0\t3\r\n\u3000\tErnvall, Toni\tFIN\t3\t0\t0\t0\t3\r\n\u3000\tGu\u00f0mundsd\u00f3ttir, Mar\u00eda Helga\tICE\t3\t0\t0\t0\t3\r\n\u3000\tGunnarsson, Einar Bjarki\tICE\t2\t0\t0\t1\t3\r\n\u3000\tGyring, Peter\tNOR\t3\t0\t0\t0\t3\r\n\u3000\tHelgason, Einar Axel\tICE\t2\t0\t0\t1\t3\r\n\u3000\tJensen, Ole S.\tDEN\t3\t0\t0\t0\t3\r\n\u3000\tNielsen, Emil Bundgaard\tDEN\t3\t0\t0\t0\t3\r\n\u3000\t\u00d3skarsd\u00f3ttir, Mar\u00eda\tICE\t3\t0\t0\t0\t3\r\n\u3000\tSigur\u00f0sson, Baldur\tICE\t2\t0\t1\t0\t3\r\n\u3000\tSigur\u00f0sson, Dan\u00edel \u00der\u00f6stur\tICE\t3\t0\t0\t0\t3\r\n\u3000\tSun, Lei\tSWE\t3\t0\t0\t0\t3\r\n\u3000\tYilong, Li\tFIN\t0\t3\t0\t0\t3\r\n67.\tJokinen, Kosti\tFIN\t0\t0\t0\t2\t2\r\n\u3000\tKlungre, Vidar\tNOR\t1\t0\t1\t0\t2\r\n\u3000\tLi, Kang\tDEN\t0\t0\t0\t2\t2\r\n\u3000\tMertaniemi, Henrikki\tFIN\t2\t0\t0\t0\t2\r\n\u3000\tNielsen, Kr\u00e6n Vodder\tDEN\t0\t0\t0\t2\t2\r\n\u3000\tNu\u00f1ez, Walter\tSWE\t1\t0\t1\t0\t2\r\n\u3000\tParnfelt, Johan\tSWE\t1\t0\t0\t1\t2\r\n\u3000\tRydberg, Elin\tSWE\t2\t0\t0\t0\t2\r\n\u3000\tStr\u00e5hlman, Christian\tSWE\t2\t0\t0\t0\t2\r\n\u3000\tSveinbj\u00f6rnsson, Benjamin Ragnar\tICE\t2\t0\t0\t0\t2\r\n\u3000\tSvensson, Mats\tSWE\t2\t0\t0\t0\t2\r\n\u3000\t\u00c5strand, Matti\tFIN\t0\t0\t0\t2\t2\r\n79.\tLarsson. Joel\tSWE\t1\t0\t0\t0\t1\r\n\u3000\tLode, P\u00e5l S\u00e6vik\tNOR\t0\t0\t1\t0\t1\r\n\u3000\tMarcher, Jette Louise\tDEN\t0\t0\t0\t1\t1\r\n\u3000\tRantanen, Mikko\tFIN\t1\t0\t0\t0\t1\r\n83.\tBerg, Justus\tSWE\t0\t0\t0\t0\t0\r\n\u3000\tKettunen, Markus\tFIN\t0\t0\t0\t0\t0\r\n\u3000\tLauridsen, Nis Brix\tDEN\t0\t0\t0\t0\t0",
  "Solution_8": "[url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=253&year=2005]Added.[/url]",
  "Solution_9": "[b]Nordic Mathematical Contest 2009[/b]\nApril 2, 2009\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028413#p3028413]Problem 1[/url]\nA point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the given triangle. Prove that $f\\ge\\frac{1}{3}$ and determine those points $P$ for which $f =\\frac{1}{3}$ .\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028409#p3028409]Problem 2[/url]\nOn a faded piece of paper it is possible to read the following:\n\\[(x^2 + x + a)(x^{15}- \\cdots ) = x^{17} + x^{13} + x^5 - 90x^4 + x - 90.\\]\nSome parts have got lost, partly the constant term of the first factor of the left side, partly the majority of the summands of the second factor. It would be possible to restore the polynomial forming the other factor, but we restrict ourselves to asking the following question: What is the value of the constant term $a$? We assume that all polynomials in the statement have only integer coefficients.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028392#p3028392]Problem 3[/url]\nThe integers $1$, $2$, $3$, $4$, and $5$ are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a + b$ and $ab$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are $2009$?\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028405#p3028405]Problem 4[/url]\n$32$ competitors participate in a tournament. No two of them are equal and in a one against one mach the better always wins. Show that the gold, silver, and bronze medal winners can be found in $39$ matches.",
  "Solution_10": "[b]Nordic Mathematical Contest 2004[/b]\nApril 1, 2004\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028451#p3028451]Problem 1[/url]\nTwenty-seven balls labelled from $1$ to $27$ are distributed in three bowls: red, blue, and yellow. What are the possible values of the number of balls in the red bowl if the average labels in the red, blue and yellow bowl are $15$, $3$, and $18$, respectively?\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028439#p3028439]Problem 2[/url]\nThe Fibonacci sequence is defined by $f_1 = 0, f_2 = 1$, and $f_{n+2} = f_{n+1}+f_n$ for $n\\ge 1$. Prove that there is a strictly increasing arithmetic progression whose no term is in the Fibonacci sequence.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028457#p3028457]Problem 3[/url]\nGiven a finite sequence $x_{1,1}, x_{2,1}, \\dots , x_{n,1}$ of integers $(n\\ge 2)$, not all equal, define the sequences $x_{1,k}, \\dots , x_{n,k}$ by\n\\[ x_{i,k+1}=\\frac{1}{2}(x_{i,k}+x_{i+1,k})\\quad\\text{where }x_{n+1,k}=x_{1,k}.\\]\nShow that if $n$ is odd, then not all $x_{j,k}$ are integers. Is this also true for even $n$?\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028454#p3028454]Problem 4[/url]\nLet $a, b, c$ be the sides and $R$ be the circumradius of a triangle. Prove that\n\\[\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{ca}\\ge\\frac{1}{R^2}.\\]",
  "Solution_11": "[b]Nordic Mathematical Contest 2010[/b]\nApril 13, 2010\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029723#p3029723]Problem 1[/url]\nA function $f : \\mathbb{Z}_+ \\to \\mathbb{Z}_+$, where $\\mathbb{Z}_+$ is the set of positive integers, is non-decreasing and satisfies $f(mn) = f(m)f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8)f(13) \\ge (f(10))^2$.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029733#p3029733]Problem 2[/url]\nThree circles $\\Gamma_A$, $\\Gamma_B$ and $\\Gamma_C$ share a common point of intersection $O$. The other common point of $\\Gamma_A$ and $\\Gamma_B$ is $C$, that of $\\Gamma_A$ and $\\Gamma_C$ is $B$, and that of $\\Gamma_C$ and $\\Gamma_B$ is $A$. The line $AO$ intersects the circle $\\Gamma_A$ in the point $X \\ne O$. Similarly, the line $BO$ intersects the circle $\\Gamma_B$ in the point $Y \\ne O$, and the line $CO$ intersects the circle $\\Gamma_C$ in the point $Z \\ne O$. Show that \n\\[\\frac{|AY |\\cdot|BZ|\\cdot|CX|}{|AZ|\\cdot|BX|\\cdot|CY |}= 1.\\]\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029748#p3029748]Problem 3[/url]\nLaura has $2010$ lamps connected with $2010$ buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons simultaneously, so the lamps are lit simultaneously, too.\na) If Richard chooses the buttons to be pressed, what is the maximum number of different combinations of buttons he can press until Laura can assign the buttons to the lamps correctly?\nb) Supposing that Laura will choose the combinations of buttons to be pressed, what is the minimum number of attempts she has to do until she is able to associate the buttons with the lamps in a correct way?\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029757#p3029757]Problem 4[/url]\nA positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n = a_1 \\pm a_2 \\pm a_3 \\pm \\cdots \\pm a_k$ where $a_1, \\dots , a_k$ are simple.",
  "Solution_12": "2004, 2009, 2010 added; thanks for posting! :)",
  "Solution_13": "[b]Nordic Mathematical Contest 2011[/b]\nApril 4, 2011\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029937#p3029937]Problem 1[/url]\nWhen $a_0, a_1, \\dots , a_{1000}$ denote digits, can the sum of the $1001$-digit numbers $a_0a_1\\cdots a_{1000}$ and $a_{1000}a_{999}\\cdots a_0$ have odd digits only?\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029944#p3029944]Problem 2[/url]\nIn a triangle $ABC$ assume $AB = AC$, and let $D$ and $E$ be points on the extension of segment $BA$ beyond $A$ and on the segment $BC$, respectively, such that the lines $CD$ and $AE$ are parallel. Prove $CD \u0015\\ge \\frac{4h}{BC}CE$, where $h$ is the height from $A$ in triangle $ABC$. When does equality hold?\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029951#p3029951]Problem 3[/url]\nFind all functions $f$ such that\n\\[f(f(x) + y) = f(x^2-y) + 4yf(x)\\]\nfor all real numbers $x$ and $y$.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029970#p3029970]Problem 4[/url]\nShow that for any integer $n \\ge 2$ the sum of the fractions $\\frac{1}{ab}$, where $a$ and $b$ are relatively prime positive integers such that $a < b \\le n$ and $a+b > n$, equals $\\frac{1}{2}$.\n(Integers $a$ and $b$ are called relatively prime if the greatest common divisor of $a$ and $b$ is $1$.)",
  "Solution_14": "[b]Nordic Mathematical Contest 2012[/b]\n27 March, 2012\n\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029993#p3029993]Problem 1[/url]\nThe real numbers $a, b, c$ are such that $a^2 + b^2 = 2c^2$, and also such that $a \\ne b, c \\ne -a, c \\ne -b$. Show that\n\\[\\frac{(a+b+2c)(2a^2-b^2-c^2)}{(a-b)(a+c)(b+c)}\\]\nis an integer.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3030023#p3030023]Problem 2[/url]\nGiven a triangle $ABC$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $BC$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $AB$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be the second point of intersection of $k$ and the line $AB$ (if there is no second point of intersection, choose $Q = B$). Prove that $AQ = AC$.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3030035#p3030035]Problem 3[/url]\nFind the smallest positive integer $n$, such that there exist $n$ integers $x_1, x_2, \\dots , x_n$ (not necessarily different), with $1\\le x_k\\le n$, $1\\le k\\le n$, and such that\n\\[x_1 + x_2 + \\cdots + x_n =\\frac{n(n + 1)}{2}\\quad\\text{, and }x_1x_2 \\cdots x_n = n!,\\]\nbut $\\{x_1, x_2, \\dots , x_n\\} \\ne \\{1, 2, \\dots , n\\}$.\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3030045#p3030045]Problem 4[/url]\nThe number $1$ is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a - 1$ and $a + 1$; if the number $0$ occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on the blackboard. Thus the blackboard will show $1$ after $0$ steps; $2$ after $1$ step; $1, 3$ after $2$ steps; $2, 2, 4$ after $3$ steps, and so on. How many numbers will there be on the blackboard after $n$ steps?"
}
{
  "Tag": [
    "calculus",
    "integration",
    "algebra",
    "function",
    "domain",
    "geometry",
    "rectangle"
  ],
  "Problem": "I'm reading about complex analysis and I have some problems translating $\\epsilon$s and $\\delta$s into a geometric interpretation. I have a clear geometric view on Lebesgue/Riemann-Stieltjes integration of real valued functions. I want to accomplish that with complex integration too. I'll state the theorem\r\n\r\n[b]Theorem:[/b] Let $\\gamma: [a,b] \\to \\mathbb{C}$ be of bounded variation and suppose that $f: [a,b] \\to \\mathbb{C}$ is continuous. Then there is a complex number $I$ such that for every $\\epsilon > 0$ there is a $\\delta > 0$ such that when $P=\\{t_{0}< t_{1}< ... < t_{m}\\}$ is a partition of $[a,b]$ with $\\parallel P\\parallel =max\\{(t_{k}-t_{k-1}): 1 \\leq k \\leq m \\}< \\delta$ then\r\n\\[| I-\\sum_{k=1}^{m}f(\\tau_{k})[ \\gamma (t_{k-1})] | < \\epsilon \\]\r\nFirst of all, I hardly believe the author made an awfull typo and he meant to say:\r\n\\[{| I-\\sum_{k=1}^{m}f(\\tau_{k})[ \\gamma (t_{k})-\\gamma(t_{k-1})}] | < \\epsilon \\]\r\nHe calls $I$ the integral of $f$ with respect to $\\gamma$ over $[a,b]$. I don't see the geometric meaning. First of all we are not \"integrating\" over the domain of $f$, which is kinda weird... \r\n\r\nLater on, the author defines the line integral by considering the function $f \\circ \\gamma$. This makes already some more sense to me. But I still don't feel comfortable. If someone can tell me how to make a drawing to clear things up, that would be most helpful :)",
  "Solution_1": "[quote=\"Jacobson\"]\nFirst of all, I believe the author made an awfull typo and he meant to say:\n\\[{| I-\\sum_{k=1}^{m}f(\\tau_{k})[ \\gamma (t_{k})-\\gamma(t_{k-1})}] | < \\epsilon \\]\n[/quote]\nYou are correct :).\n[quote]\nHe calls $I$ [b]the integral of $f$ with respect to $\\gamma$ over $[a,b]$ [/b].\n[/quote]\nThis is the first time I see such a combination of words. Of course, formally it makes perfect sense and allows one to reduce the integration over rectifiable paths to the Stiltjes integration very easily: you just consider your curve $\\gamma$ in the same way as you would consider $g$ in $\\int_{a}^{b}f\\,dg$: you can talk either about Riemann sums or about the (complex-valued) measure generated by $\\gamma$, which is sort of convenient, but the object that arises is of hardly any use outside the particular problem of defining the integral of a function $F: \\mathbb C\\to\\mathbb C$ over $\\gamma$ (for which you just need to apply this construction to $f=F\\circ\\gamma$).\n[quote]\nI don't see the geometric meaning. \n[/quote]\nDo you see the geometric meaning of the Stiltjes integral?\n[quote]\nFirst of all we are not \"integrating\" over the domain of $f$, which is kinda weird... \n[/quote]\nWell, this integral should be written $\\int_{a}^{b}f\\,d\\gamma$, so the integration actually is over the domain of $f$ (which is the interval $[a,b]$). Just don't try to think of it as of integral over $\\gamma$; however strange the definition may seem, it is put together correctly: it is, indeed, \"integral of $f$ over $[a,b]$ with respect to $\\gamma$\" in the same way as the Stiltjes integral $\\int_{a}^{b}f\\,dg$ is \"integral of $f$ over $[a,b]$ with respect to $g$\".\n[quote]\nLater on, the author defines the line integral by considering the function $f \\circ \\gamma$. This makes already some more sense to me. But I still don't feel comfortable. If someone can tell me how to make a drawing to clear things up, that would be most helpful :)[/quote]\r\nNote that when you take $f=F\\circ\\gamma$, you can rephrase the definition of the Riemann sum in the following way. Take the curve $\\Gamma\\subset \\mathbb C$ (the one parameterized by the mapping $\\gamma: [a,b]\\to\\mathbb C$), split it into small pieces by points $z_{0},z_{1},\\dots,z_{n}\\in \\Gamma$ ($z_{k}$ is the same as $\\gamma(t_{k})$); choose a point $\\zeta_{k}$ ($=\\gamma(\\tau_{k})$) on each piece of $\\Gamma$ between $z_{k-1}$ and $z_{k}$ and consider $\\sum_{k}F(\\zeta_{k})(z_{k}-z_{k-1})$. In the case, when $\\Gamma$ is a line segment, you'll get the usual Riemann sum. The problem with the drawing is that the product of two complex numbers doesn't have a clear \"geometric meaning\" unlike the product of two real numbers, which can be represented as a (signed) area of a rectangle, but, I hope, you see that the formula for the Riemann sums didn't change compared to the one for the real-valued functions defined on an interval.\r\n\r\nHope, this helps a bit :).",
  "Solution_2": "thanks a lot :) it helped"
}
{
  "Tag": [
    "function",
    "absolute value",
    "calculus",
    "calculus computations"
  ],
  "Problem": "1. using the definition of cauchy, how do I show that every cauchy sequence of  real numbers is a bounded sequence?\r\n2. if these are true prove them otherwise provide a counterexample:\r\na. every unbounded sequence of positive real numbers diverges to [size=150]\u221e[/size]\r\nb. every unbounded sequence of positive real numbers has a subsequence that diverges to \u221e",
  "Solution_1": "By the Cauchy property, a Cauchy sequence is eventually bounded; there exists $N$ such that $|x_N-x_n|<1$ for $n>N$. (That specialized the property in two ways). Now, the first $N-1$ terms can't make the sequence unbounded; the first $N-1$ terms have a maximum $M$ and a minimum $C$. We have $\\min(x_N-1,C)<x_n<\\max(x_N+1,M)$ for all $n$, and the sequence is bounded.\r\n\r\nThis is an important method; it can be adapted to general metric spaces, and you will probably have reason to use it again.\r\n\r\n2a: The answer is no. Try interlacing an unbounded sequence and a bounded sequence.\r\n\r\n2b: The answer is yes. You want to construct an unbounded subsequence, so start with an element, find something bigger, find something even bigger,...",
  "Solution_2": "1. Show that if a sequence {an} converges to 0, and a sequence {bn} is bounded, then the sequence {an,bn} converges to 0.\r\n\r\n2. give an example of a set S in R that has exactly three limit points.\r\n\r\n3. Suppose {an} from n=1 to [size=200]\u221e[/size] is a cauchy sequence and let bn=3an+7 for each positive integer n. Using the definition of cauchy sequence to show that {bn} from n=1 to [size=150]\u221e[/size].",
  "Solution_3": "3 doesn't make any sense\r\n\r\n$1:$ Say that the terms of $(b_n)_n$ are bounded above in absolute value by $M$.  Given $\\epsilon>0$, pick $N$ large enough that $|a_n|<\\frac{\\epsilon}{M}$ for $n>N$.  Then $|a_nb_n|\\leq \\epsilon$ for $n>N$.\r\n\r\n$2:$  I hope you know how to construct a sequence with a single limit point (  $(n^{-1})_n$ ).  Now mix this with the sequences $(n^{-1}+1)_n$ and $(n^{-1}+2)_n$.  You are going to have to fiddle around with the indices to make it work, but this isn't hard.",
  "Solution_4": "It looks like mikejones left out a few words in #3. I think we can figure out what he meant to say:\r\n\r\nSuppose $(a_n)$ is a Cauchy sequence and that $b_n=3a_n+7.$ Prove that $(b_n)$ is a Cauchy sequence.\r\n\r\nHere's a generalization: suppose $f: \\mathbb{R}\\mapsto\\mathbb{R}$ satisfies a Lipschitz condition. That is, suppose there is a constant $K$ such that for all $x$ and $y,\\,|f(x)-f(y)|\\le K|x-y|.$ Suppose $(a_n)$ is a Cauchy sequence and that $b_n=f(a_n).$ Prove that $(b_n)$ is a Cauchy sequence.\r\n\r\n(The connection between the specific case and the generalization: the function $f(x)=3x+7$ satisfies a Lipschitz condition with constant $K=3.)$\r\n\r\nOr take the generalization one step further out: suppose $f$ is uniformly continuous. Suppose $(a_n)$ is a Cauchy sequence and that $b_n=f(a_n).$ Prove that $(b_n)$ is a Cauchy sequence.",
  "Solution_5": "[quote=\"jmerry\"]By the Cauchy property, a Cauchy sequence is eventually bounded; there exists $N$ such that $|x_N-x_n|<1$ for $n>N$. (That specialized the property in two ways). Now, the first $N-1$ terms can't make the sequence unbounded;...\n[/quote]\nI don't quite see why you can claim the first N-1 can't be unbounded? Can you explain?\n\n[quote=\"jmerry\"]2a: The answer is no. Try interlacing an unbounded sequence and a bounded sequence.\n[/quote]\r\nI have my own question for this one. Do we distiguish between [b]divergent to $\\infty$[/b] and [b]divergent[/b]?",
  "Solution_6": "[quote=\"Hazel\"]I don't quite see why you can claim the first N-1 can't be unbounded? Can you explain?[/quote]\r\n\r\nthey are between $\\min_{i \\in \\overline{1,N-1}} x_i$ and $\\max_{i \\in \\overline{1,N-1}} x_i$.\r\n\r\n\r\nFor the other question, yes, divergent to $\\infty$ means that the sequence has limit $\\infty$, while divergent means that the sequence doesn't have finite limit."
}
{
  "Tag": [
    "function",
    "integration",
    "calculus",
    "derivative",
    "geometry",
    "rectangle",
    "real analysis"
  ],
  "Problem": "Suppose \r\n$u(x,t)$ is continous functions \r\n$\\frac{\\partial u}{\\partial t}$ is continous\r\n$u(x+1,t)=u(x,t), \\; \\frac{\\partial u(x+1,t)}{\\partial x}=\\frac{\\partial u(x,t)}{\\partial x}$ \r\n$\\frac{\\partial^2 u}{\\partial x^2}=\\frac{\\partial^2 u}{\\partial t^2}$\r\n\r\nProve that $E(t)=\\frac{1}{2}\\int_{0}^{1}(\\frac{\\partial u(x,t)}{\\partial t})^2+(\\frac{\\partial u(x,t)}{\\partial x})^2)dx$ is constant independent of $t$",
  "Solution_1": "Here's a formal calculation, assuming we have as many derivatives as we need:\r\n\r\n$\\frac{dE}{dt}=\\int_0^1\\frac{\\partial u}{\\partial t}\\cdot \\frac{\\partial^2u}{\\partial t^2} + \\frac{\\partial u}{\\partial x}\\cdot\\frac{\\partial^2u}{\\partial t\\partial x}\\,dx.$\r\n\r\nNow integrate the second term by parts:\r\n\r\n$\\int_0^1\\frac{\\partial u}{\\partial x}\\cdot\\frac{\\partial^2u}{\\partial t\\partial x}\\,dx = \\frac{\\partial u}{\\partial x}\\cdot\\frac{\\partial u}{\\partial t}|_0^1 - \\int_0^1\\frac{\\partial^2u}{\\partial x^2}\\cdot\\frac{\\partial u}{\\partial t}\\,dx.$\r\n\r\nNow note that the boundary terms in this integration by parts are zero by the periodicity of $u$ and use the fact that $u$ satisfies the wave equation to replace the second derivative with respect to $x$ with the second derivative with respect to $t.$\r\n\r\nHence, $\\frac{dE}{dt}=\\int_0^1\\frac{\\partial u}{\\partial t}\\cdot \\frac{\\partial^2u}{\\partial t^2}-\\frac{\\partial u}{\\partial t}\\cdot \\frac{\\partial^2u}{\\partial t^2}\\,dx = 0.$\r\n\r\nGiven the density of $C^{\\infty}$ functions, this identity would now hold in the appropriate Sobolev space, with the derivatives interpreted as weak derivatives.",
  "Solution_2": "A second calculation, using a similar level of formality. We use the technique of separation of variables to find the general solution of the PDE.\r\n\r\n$u(x,t)=\\sum_{n=-\\infty}^{\\infty} e^{2\\pi inx}\\left(c_ne^{2\\pi i nt}+d_ne^{-2\\pi int}\\right).$\r\n\r\nTake the appropriate partial derivatives:\r\n\r\n$\\frac{\\partial u}{\\partial x}=\\sum_{n=-\\infty}^{\\infty} 2\\pi i ne^{2\\pi inx}\\left(c_ne^{2\\pi i nt}+d_ne^{-2\\pi int}\\right).$\r\n\r\n$\\frac{\\partial u}{\\partial t}=\\sum_{n=-\\infty}^{\\infty} 2\\pi i ne^{2\\pi inx}\\left(c_ne^{2\\pi i nt}-d_ne^{-2\\pi int}\\right).$\r\n\r\nBy Parseval's identity, if $g(x)=\\sum_{n=-\\infty}^{\\infty}a_ne^{2\\pi inx}$ then\r\n\r\n$\\int_0^1|g(x)|^2\\,dx=\\sum_{n=-\\infty}^{\\infty}|a_n|^2.$\r\n\r\nThus, \r\n\r\n$\\int_0^1\\left|\\frac{\\partial u}{\\partial x}\\right|^2\\,dx = \\sum_{n=-\\infty}^{\\infty} 4\\pi^2n^2\\left|c_ne^{2\\pi i nt}+d_ne^{-2\\pi int}\\right|^2$ and\r\n\r\n$\\int_0^1\\left|\\frac{\\partial u}{\\partial t}\\right|^2\\,dx = \\sum_{n=-\\infty}^{\\infty} 4\\pi^2n^2\\left|c_ne^{2\\pi i nt}-d_ne^{-2\\pi int}\\right|^2$\r\n\r\nBy direct calculation, \r\n\r\n$\\left|c_ne^{2\\pi i nt}+d_ne^{-2\\pi int}\\right|^2=|c_n|^2 + c_n\\overline{d_n} e^{4\\pi int}+\\overline{c_n}d_ne^{-4\\pi int}+|d_n|^2$\r\n\r\n$\\left|c_ne^{2\\pi i nt}-d_ne^{-2\\pi int}\\right|^2=|c_n|^2 - c_n\\overline{d_n} e^{4\\pi int}-\\overline{c_n}d_ne^{-4\\pi int}+|d_n|^2$\r\n\r\nAdd these together, and the result is independent of $t$. In the end,\r\n\r\n$E(t) = \\sum_{n=-\\infty}^{\\infty}4\\pi^2n^2\\left(|c_n|^2+|d_n|^2\\right).$",
  "Solution_3": "Now that I look at it, this one has a Green's theorem solution too:\r\nWrite $E(0)-E(T)=\\int_{dR}\\frac12\\left(\\left(\\frac{\\partial u(x,t)}{\\partial t}\\right)^2+\\left(\\frac{\\partial u(x,t)}{\\partial x}\\right)^2\\right)dx+0\\, dt$, where $R$ is the rectangle with vertices $(0,0),(1,0),(1,T),(0,T)$. By Green's theorem, this becomes\r\n$E(0)-E(T)=\\int_R\\frac{\\partial u(x,t)}{\\partial t}\\cdot\\frac{\\partial u(x,t)}{\\partial t^2}+\\frac{\\partial u(x,t)}{\\partial x}\\cdot\\frac{\\partial u(x,t)}{\\partial x\\partial t}\\, dx\\, dt$\r\n$=\\int_0^T\\int_0^1\\frac{\\partial u(x,t)}{\\partial t}\\cdot\\frac{\\partial u(x,t)}{\\partial x^2}+\\frac{\\partial u(x,t)}{\\partial x}\\cdot\\frac{\\partial u(x,t)}{\\partial x\\partial t}\\, dx\\, dt$\r\n$=\\int_0^T\\int_0^1\\frac{\\partial}{\\partial x}\\left(\\frac{\\partial u(x,t)}{\\partial x}\\cdot\\frac{\\partial u(x,t)}{\\partial t}\\right)\\, dx\\, dt=0$\r\nby periodicity of $\\frac{\\partial u(x,t)}{\\partial x}\\cdot\\frac{\\partial u(x,t)}{\\partial t}$ in $x$.\r\n\r\nThis assumes that the mixed partial derivatives are continuous, so we can set them equal to each other."
}
{
  "Tag": [
    "Gauss"
  ],
  "Problem": "Consider the sequence below where each third number is missing.\r\n\r\n$1, 2, 4, 5, 7, 8, 10, 11 ...$\r\n\r\nWhat is the the sum of the first one hundered terms? Show as many solutions as possible.",
  "Solution_1": "[quote=\"guile\"]Consider the sequence below where each third number is missing.\n\n$1, 2, 4, 5, 7, 8, 10, 11 ...$\n\nWhat is the the sum of the first one hundered terms? Show as many solutions as possible.[/quote]\r\n[hide=\"hint\"]\nFirst find the sum when each third number is not missing. And then subtract $3+6+9+...$.\n[/hide]",
  "Solution_2": "[hide=\"Solution\"] I believe it ends in 149 and there are 49 numbers missing, so $\\frac{149(150)}{2}-3\\frac{49(50)}{2}=149\\times75-49\\times25=\\boxed{7500}$. :D I think. [/hide]",
  "Solution_3": "[hide]Well I think that we all know to add the first and last terms together and multiply that by half the number of terms to get 5050 for the sum of 1-100.  We can use the same concept for this.  \n\n$\\displaystyle{(3+99)\\cdot\\116}$  But because there are an odd number of terms we add the middle term, the 17th.\n$\\displaystyle{3\\cdot\\117=41}$  Add these two together to find out much to subtract from 5050.  So we do $\\displaystyle{5050-1673=\\boxed{3377}}$  :D [/hide]",
  "Solution_4": "oops... I answered the sum of 1-100 not including the multiples of 3... sorry :blush:",
  "Solution_5": "You can make this the sum of the first 50 terms of this sequence--3,9,15..., which ends in $3+(49*6)=297$.  Now we find the sum of such terms, $\\frac{(3+297)(50)}{2}=\\fbox{7500}$",
  "Solution_6": "[quote=\"jhollenbeck\"]You can make this the sum of the first 50 terms of this sequence--3,9,15..., which ends in $3+(49*6)=297$.  Now we find the sum of such terms, $\\frac{(3+297)(50)}{2}=\\fbox{7500}$[/quote]\r\n\r\nDarn it, you stole my method!",
  "Solution_7": "[quote=\"DanK\"][quote=\"jhollenbeck\"]You can make this the sum of the first 50 terms of this sequence--3,9,15..., which ends in $3+(49*6)=297$.  Now we find the sum of such terms, $\\frac{(3+297)(50)}{2}=\\fbox{7500}$[/quote]\n\nDarn it, you stole my method![/quote]\r\n\r\nJajajaa!",
  "Solution_8": "well.....\r\n[hide]you add the first and second terms, you get 3. Third and fourth, 9. Fifth and sixth, 15.\nThere are 50 of those pairs.\nYou have $((2*(50*51)/2)-50)*3=50*50*3=[7500]$[/hide]\r\n\r\nsorry, I didn't post all of the solution, because my hands are cold. Oh. I didn't steal anyone's method.",
  "Solution_9": "[quote=\"mbabbitt2003\"]well.....\n[hide]you add the first and second terms, you get 3. Third and fourth, 9. Fifth and sixth, 15.\nThere are 50 of those pairs.\nYou have $((2*(50*51)/2)-50)*3=50*50*3=[7500]$[/hide]\n\nsorry, I didn't post all of the solution, because my hands are cold. Oh. I didn't steal anyone's method.[/quote]\r\n\r\nNice method!!!!",
  "Solution_10": "[hide=\"solution using gauss formula\"]\nyou can make this into another sequence\n$1+2+3+4+5+...$\nby adding $3+6+9...$ to the original sequence\nsince we are adding the first 100 terms of the original sequence\nwe can add all the numbers from 1-150\nand subtract the first 50 multiples of 3\nthe sum of 1-150 is\n$\\frac{(150)(150+1)}{2} = 11325$\nthe 3+6+9 sequence can be expressed as\n$\\frac{3\\times(50)(50+1)}{2} = 3825$\nour answer is then $11325-3825=\\fbox{7500}$[/hide]"
}
{
  "Tag": [
    "calculus",
    "inequalities",
    "calculus computations"
  ],
  "Problem": "We have sets $A$ and $B$ of real numbers.\r\nThe elements of the set $A+B$ are given as $a+b$ where $a\\in{A}$ and $b\\in{B}$.\r\nProve that $supA+supB=sup(A+B)$ where supA=supremum of the set A.",
  "Solution_1": "This is a standard result proved in most good calculus textbooks, e.g. Apostol.",
  "Solution_2": "[quote=\"spider_boy\"]We have sets $A$ and $B$ of real numbers.\nThe elements of the set $A+B$ are given as $a+b$ where $a\\in{A}$ and $b\\in{B}$.\nProve that $supA+supB=sup(A+B)$ where supA=supremum of the set A.[/quote]\r\n\r\nI did not say that it is so hard.I am interested in people's solutions. :)",
  "Solution_3": "[quote=\"spider_boy\"][quote=\"spider_boy\"]We have sets $A$ and $B$ of real numbers.\nThe elements of the set $A+B$ are given as $a+b$ where $a\\in{A}$ and $b\\in{B}$.\nProve that $supA+supB=sup(A+B)$ where supA=supremum of the set A.[/quote]\n\nI did not say that it is so hard.I am interested in people's solutions. :)[/quote]\r\nLet $\\sup A = x$ and $\\sup B = y.$\r\n\r\nWe have $a \\le x \\ \\forall a \\in A$ and $b \\le y\\ \\forall b \\in A,$ which implies $a+b \\le x+y.$ Hence, an upperbound for $A+B$ exists and $x+y$ is one of the upperbounds. \r\n\r\nClaim: $\\sup (A+B) = x+y.$ Suppose, for the sake of contradiction, $\\sup (A+B) = x+y-\\epsilon$  for some $\\epsilon > 0 .$ \r\n\r\nSince $\\sup A = x, \\ \\exists a' \\in A$ such that $a' > x-\\frac{\\epsilon}{2}.$ Again, since $\\sup B = y, \\ \\exists b' \\in B$ such that $b' > y-\\frac{\\epsilon}{2}.$ Thus, we have $a'+b' > x+y-\\epsilon,$ which contradicts the assumption that $x+y-\\epsilon = \\sup (A+B).$ And, we are done.\r\n\r\n[I can't think of a simpler proof. May be there is one.]",
  "Solution_4": "mine was completely this one. :)",
  "Solution_5": "1) max(A)<sup(A) and \r\n2) max(B)<sup(B)\r\n3) max(A)+max(B)<sup(A)+sup(B)\r\n\r\nthen max(A)+max(B)\\in A+B, furthermore it is the max in A+B\r\n\r\nsuppose there is a stronger upper bound (in line 3) ), subtracting one of the inequalities implies 1) or 2) is not the min upper bound, contradiction",
  "Solution_6": "Most of those maximums you refer to don't actually exist.\r\n\r\nFor example, take $X=Y=(0,1)$.  Then $X+Y=(0,2)$, but $\\max(X),\\max(Y),\\max(X+Y)$ all fail to exist."
}
{
  "Tag": [
    "function",
    "inequalities proposed"
  ],
  "Problem": "Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\r\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}.\\]",
  "Solution_1": "My solution:\r\nWe have:\r\n$\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\geq \\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}$\r\nWe need prove that:\r\n$\\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}\\geq a^{2}+b^{2}+c^{2}+d^{2}$\r\n<=>$16 \\geq (a^{2}+b^{2}+c^{2}+d^{2})^{2}$\r\n<=>$(a+b+c+d)^{2}\\geq (a^{2}+b^{2}+c^{2}+d^{2})^{2}$\r\n<=>$a+b+c+d \\geq a^{2}+b^{2}+c^{2}+d^{2}$\r\n<=>$a(1-a)+b(1-b)+c(1-c)+d(1-d) \\geq0$\r\nSuppose that:$a \\geq b \\geq c \\geq d$,then we have:\r\n$1-a \\leq 1-b \\leq 1-c \\leq 1-d$\r\nThen by Chebyshev's inequality we have:\r\n$a(1-a)+b(1-b)+c(1-c)+d(1-d) \\geq0$\r\n(note$a+b+c+d=4$)",
  "Solution_2": "You use Chebyshev inequality in a wrong way. If $a+b+c+d=4$ then certainly\r\n\\[a^{2}+b^{2}+c^{2}+d^{2}\\ge 4.\\]",
  "Solution_3": "Sorry!\r\nI have other solution:\r\nWe have:\r\n$\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}-a^{2}-b^{2}-c^{2}-d^{2}=(\\frac{1}{a}+a)(\\frac{1}{a}-a)+(\\frac{1}{b}+b)(\\frac{1}{b}-b)+(\\frac{1}{c}+c)(\\frac{1}{c}-c)+(\\frac{1}{d}+d)(\\frac{1}{d}-d) \\geq 2(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}-a-b-c-d) \\geq 2(\\frac{16}{a+b+c+d}-a-b-c-d)=0$\r\n(note that $x+\\frac{1}{x}\\geq 2$)\r\nI belive this solution will be true!",
  "Solution_4": "But $(\\frac{1}{a}-a)$ can be $<0$ !!! :wink:",
  "Solution_5": "[quote=\"manlio\"]But $(\\frac{1}{a}-a)$ can be $<0$ !!! :wink:[/quote]\r\nIf $(\\frac{1}{a}-a) <0$,then?",
  "Solution_6": "I believe that this way is hardly true. You should try another one, chienthan.",
  "Solution_7": "For example if $(\\frac{1}{a}-a)<0$ you cannot write $(\\frac{1}{a}+a)(\\frac{1}{a}-a) \\geq 2(\\frac{1}{a}-a)$",
  "Solution_8": "if numbers are in $(0,3)$ i can prove it by convexiti of the function .but i will find an other solution.\r\nhi [b]hungkhtn[/b] can u post ur theorem ( i mean smvt and E?) because i do not now anything about them $thinks$. :D",
  "Solution_9": "About SMV, you can see\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=113248",
  "Solution_10": "[quote=\"hungkhtn\"]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}. \\]\n[/quote]\r\n\r\nI think the following stronger inequality holds for  $a+b+c+d=4$:\r\n\r\n${\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{cd}+\\frac{1}{da}}\\ge a^{2}+b^{2}+c^{2}+d^{2}$.",
  "Solution_11": "Yes, Vasc, it was also on my note few minutes ago. The solution is based on an idea of Vardach.",
  "Solution_12": "I would like to see your solution, Thuan and VASC, if it is not mixing variable.",
  "Solution_13": "i remember mr. Schwarz told us that Vasc's inequality holds even for $5$ variables and for the degree $3,4$ also.",
  "Solution_14": "I proved already that the inequality\r\n\\[\\frac{1}{a_{1}^{2}}+\\frac{1}{a_{2}^{2}}+...+\\frac{1}{a_{n}^{2}}\\ge a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\\]\r\nholds for all $n\\in\\{1,2,...,9\\}$.\r\n\r\nHowever, I would like to see a simple solution for $n=4$, that the reason why I post this problem. I am also very curious of how Cauchy-Schwarz can be applied to prove this case $n=4$ and even $n=5$ as Pohoatza's post. Could you show it? Pohoatza?\r\n\r\nI also wait a solution of VASC and Thuan about their improvement. Hope that I have not to wait long then.\r\n\r\nThe following inequality is true and stronger\r\n\r\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}-4 \\ge \\sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4).\\]",
  "Solution_15": "[quote=\"hungkhtn\"] The following inequality is true and stronger\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}-4 \\ge \\sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4). \\]\n[/quote]\r\nI think the inequality below is true and stronger\r\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}+4 \\ge 2(a^{2}+b^{2}+c^{2}+d^{2}). \\]",
  "Solution_16": "[quote=\"Vasc\"][quote=\"hungkhtn\"] The following inequality is true and stronger\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}-4 \\ge \\sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4). \\]\n[/quote]\nI think the inequality below is true and stronger\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}+4 \\ge 2(a^{2}+b^{2}+c^{2}+d^{2}). \\]\n[/quote]\r\n\r\nNice one, VASC,\r\n\r\nI build this inequality from the general case of $n$ so I don't check the case $n=4$ much. I just use the general result to replace this particular case. For all, we only need to check the inequality when $n-1$ numbers equal, so your problem can be done similarly as mine, VASC, just a bit different in the last step. I prove it by elementary method, not mixing variable, EV or SMV, too.",
  "Solution_17": "Dear VASC and others,\r\n\r\nDo you think that it is time now to show a solution to the initial problem? We did discuss a lot, with strong problem, but a nice solution to the first is still hidden,\r\n\r\n[quote=\"hungkhtn\"]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}. \\]\n[/quote]",
  "Solution_18": "The stronger\r\n\\[\\frac{1}{a\\sqrt{a}}+\\frac{1}{b\\sqrt{b}}+\\frac{1}{c\\sqrt{c}}+\\frac{1}{d\\sqrt{d}}\\ge a^{2}+b^{2}+c^{2}+d^{2}\\]\r\n:)",
  "Solution_19": "[quote=\"Vasc\"][quote=\"hungkhtn\"]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}. \\]\n[/quote]\n\nI think the following stronger inequality holds for  $a+b+c+d=4$:\n\n${\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{cd}+\\frac{1}{da}}\\ge a^{2}+b^{2}+c^{2}+d^{2}$.[/quote]\r\nUsing the known $x^{2}+y^{2}\\le \\frac{(x+y)^{4}}{8xy}\\quad \\forall x,y \\ge 0$ $(*)$, we have\r\n\\[a^{2}+c^{2}\\le \\frac{(a+c)^{4}}{8ac}=\\frac{(a+c)^{4}bd}{8abcd}\\le \\frac{(a+c)^{4}(b+d)^{2}}{32abcd}, \\qquad b^{2}+d^{2}\\le \\frac{(b+d)^{4}(a+c)^{2}}{32abcd}\\]\r\nWe need to prove\r\n\\[{\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{cd}+\\frac{1}{da}}\\ge \\frac{(a+c)^{4}(b+d)^{2}}{32abcd}+\\frac{(b+d)^{4}(a+c)^{2}}{32abcd}\\]\r\nOr\r\n\\[32 \\ge (a+c)(b+d)((a+c)^{2}+(b+d)^{2}) \\]\r\nIt is true by $(*)$. :) Equality holds if and only if $a=b=c=d=1$. :)",
  "Solution_20": "Yes, toanhocmuonmau.  :lol: It is also my solution.",
  "Solution_21": "A general problem:\r\n Let $a_{1},a_{2},...,a_{n}\\geq 0 such that: \\sum a_{i}=n(n\\geq 4)$.\r\nProve that:\r\n  $\\sum\\frac{1}{a_{i}^{n-2}}\\geq \\sum a_{i}^{2}$",
  "Solution_22": "[quote=\"hungkhtn\"][quote=\"Vasc\"][quote=\"hungkhtn\"] The following inequality is true and stronger\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}-4 \\ge \\sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4). \\]\n[/quote]\nI think the inequality below is true and stronger\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}+4 \\ge 2(a^{2}+b^{2}+c^{2}+d^{2}). \\]\n[/quote]\n\nNice one, VASC,\n\nI build this inequality from the general case of $n$ so I don't check the case $n=4$ much. I just use the general result to replace this particular case. [/quote]\r\n\r\nMy general case of $n$ is the following.\r\n\r\n[i]If $a_{1},a_{2},...,a_{n}$ ($n\\ge 4$) are positive numbers such that $a_{1}+a_{2}+...+a_{n}=n$, then\n\\[\\frac{1}{a_{1}^{2}}+\\frac{1}{a_{2}^{2}}+...+\\frac{1}{a_{n}^{2}}-n \\ge \\frac 8{n}(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n). \\]\n[/i]\r\nThe cases $n=4$ and $n=8$ are very interesting.",
  "Solution_23": "In the topic about how to construct an inequality here\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=143631\r\n\r\nI did post the following stronger inequality\r\n\r\n$\\bigstar$ (Hungkhtn's, SI Inequalities A story of Maths, problem 5.3.4)  \r\nLet $a_{1},a_{2},...,a_{n}$ be non-negative real numbers such that $a_{1}+a_{2}+...+a_{n}=n$. Prove that\r\n\\[n^{2}\\left(\\frac{1}{a_{1}^{2}}+\\frac{1}{a_{2}^{2}}+...+\\frac{1}{a_{n}^{2}}-n\\right) \\ge 10(n-1)\\left(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n\\right).\\]\r\n\r\nMy solution is very elementary.",
  "Solution_24": "The following is more sharper than the first\r\n\\[\\sum\\frac{1}{a}\\ge 3+\\frac{3}{4}\\sum a^{2}\\]\r\nand it has 2 equalities :)",
  "Solution_25": "Maybe\r\n\r\n$\\sum \\frac 1{a}\\ge 1+\\frac 3{4}\\sum{a^{2}}$ for $\\sum a=4$.\r\n\r\nThis was posted in 2004 on this forum for $n$ numbers.",
  "Solution_26": "Yes, it is an old inequality\r\n\r\n$\\bigstar$ [5.2.1, SII]\r\nLet $a_{1},a_{2},...,a_{n}\\ (n\\ge 3)$ be non-negative real numbers with sum $n$. Prove that\r\n\\[\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+...+\\frac{1}{a_{n}}-n \\ge \\frac{4(n-1)}{n^{2}}\\left t(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n).\\]",
  "Solution_27": "[quote=\"toanhocmuonmau\"][quote=\"Vasc\"][quote=\"hungkhtn\"]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}. \\]\n[/quote]\n\nI think the following stronger inequality holds for  $a+b+c+d=4$:\n\n${\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{cd}+\\frac{1}{da}}\\ge a^{2}+b^{2}+c^{2}+d^{2}$.[/quote]\nUsing the known $x^{2}+y^{2}\\le \\frac{(x+y)^{4}}{8xy}\\quad \\forall x,y \\ge 0$ $(*)$, we have\n\\[a^{2}+c^{2}\\le \\frac{(a+c)^{4}}{8ac}=\\frac{(a+c)^{4}bd}{8abcd}\\le \\frac{(a+c)^{4}(b+d)^{2}}{32abcd}, \\qquad b^{2}+d^{2}\\le \\frac{(b+d)^{4}(a+c)^{2}}{32abcd}\\]\nWe need to prove\n\\[{\\frac{1}{ab}+\\frac{1}{bc}+\\frac{1}{cd}+\\frac{1}{da}}\\ge \\frac{(a+c)^{4}(b+d)^{2}}{32abcd}+\\frac{(b+d)^{4}(a+c)^{2}}{32abcd}\\]\nOr\n\\[32 \\ge (a+c)(b+d)((a+c)^{2}+(b+d)^{2}) \\]\nIt is true by $(*)$. :) Equality holds if and only if $a=b=c=d=1$. :)[/quote]\r\n\r\nIt is just now I can read your proof. Very nice one, toanhocmuonmau. Congras!\r\n\r\nDo you find a nice proof for $n=5$?",
  "Solution_28": "[quote=\"hungkhtn\"]I proved already that the inequality\n\\[\\frac{1}{a_{1}^{2}}+\\frac{1}{a_{2}^{2}}+...+\\frac{1}{a_{n}^{2}}\\ge a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\\]\nholds for all $n\\in\\{1,2,...,9\\}$.\n[/quote]\nIt's true for $n=10$ and it's wrong for all $n\\geq11$. :wink:",
  "Solution_29": "[b]Ji Chen:[/b]\nLet $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that$$ a^{2}+b^{2}+c^{2}+d^{2}\\leq 2\\left(\\frac{1}{ac}+\\frac{1}{bd}\\right).$$",
  "Solution_30": "[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}.\\][/quote]\nLet $a,b,c,d$ be real numbers such that $abcd=1$. Prove that$$ \\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}+\\frac{1}{d^2}\\geq ab+bc+cd+da$$\n(Proposed by Mircea Becheanu, University of Bucharest, Rom\u00e2nia)\n\nLet $a,b,c$ be real numbers such that $abc=1$. Prove that:$$a^2+b^2+c^2 \\geq a+b+c .$$",
  "Solution_31": "[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}.\\][/quote]\n[url=https://artofproblemsolving.com/community/c6h181229p996138]the stronger one[/url]\n[url=https://artofproblemsolving.com/community/c6h1370494p10534020]here\n[/url]",
  "Solution_32": "Let $a,b,c,d$ be positive real numbers such that $a <b <c <d $ and  $a+b+c+d=4$. Prove that$$4\\leqslant \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{d}\\leqslant \\frac{a}{d}+\\frac{d}{a}+2$$\n(M.S.Vovos)",
  "Solution_33": "ideed strong!",
  "Solution_34": "3-var version:Let $a,b,c$ be positive real numbers such that $a+b+c=3$,then\n\\[\\sum {\\frac{1}{{{a^2}}} \\ge \\sum {{a^2}} }  + 0.548\\sum {{{\\left( {a - b} \\right)}^2}} \\]\nstrong in this form.",
  "Solution_35": "[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}.\\][/quote]\nLet $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. For the integer $n\\geq2$, prove that\n$$\\frac{1}{a^n}+\\frac{1}{b^n}+\\frac{1}{c^n} +\\frac{1}{d^n}\\ge a^n+b^n+c^n+d^n.$$\n[url=https://artofproblemsolving.com/community/c6h1370494p10534020]here[/url]",
  "Solution_36": "[quote=hungkhtn]I proved already that the inequality\n\\[\\frac{1}{a_{1}^{2}}+\\frac{1}{a_{2}^{2}}+...+\\frac{1}{a_{n}^{2}}\\ge a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\\]\nholds for all $n\\in\\{1,2,...,9\\}$.\n\n[hide=*]However, I would like to see a simple solution for $n=4$, that the reason why I post this problem. I am also very curious of how Cauchy-Schwarz can be applied to prove this case $n=4$ and even $n=5$ as Pohoatza's post. Could you show it? Pohoatza?\n\nI also wait a solution of VASC and Thuan about their improvement. Hope that I have not to wait long then.\n\nThe following inequality is true and stronger\n\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}-4 \\ge \\sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4).\\][/hide][/quote]\n[quote=arqady]\nIt's true for $n=10$ and it's wrong for all $n\\geq11$. :wink:[/quote]\n\n...",
  "Solution_37": "See here:\nhttps://artofproblemsolving.com/community/c6h144103p3588112",
  "Solution_38": "[quote=arqady]See here:\nhttps://artofproblemsolving.com/community/c6h144103p3588112[/quote]\nThanks.",
  "Solution_39": "[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that\n\\[\\frac{1}{a^{2}}+\\frac{1}{b^{2}}+\\frac{1}{c^{2}}+\\frac{1}{d^{2}}\\ge a^{2}+b^{2}+c^{2}+d^{2}.\\][/quote]\nLet $ a,b,c,d$ be positive numbers sach that $ a\\plus{}b\\plus{}c\\plus{}d\\equal{}4$. [url=https://artofproblemsolving.com/community/c6h181229p996138]Prove that[/url]\n$$ \\frac 1{ab}\\plus{}\\frac 1{bc}\\plus{}\\frac 1{cd}\\plus{}\\frac 1{da}\\ge a^2\\plus{}b^2\\plus{}c^2\\plus{}d^2$$\n[url=https://artofproblemsolving.com/community/c6h1856322p12541678]h[/url]",
  "Solution_40": "Let $a_1,a_2,\\cdots,a_n  (n\\ge 2)$ be positive numbers such that $a_1a_2\\cdots  a_n=1.$ Prove that $$\\frac{1}{a^n_1}+\\frac{1}{a^n_2}+\\cdots+\\frac{1}{a^n_n}\\geq a_1+a_2+\\cdots+a_n.$$",
  "Solution_41": "[quote=sqing]Let $a_1,a_2,\\cdots,a_n  (n\\ge 2)$ be positive numbers such that $a_1a_2\\cdots  a_n=1.$ Prove that $$\\frac{1}{a^n_1}+\\frac{1}{a^n_2}+\\cdots+\\frac{1}{a^n_n}\\geq a_1+a_2+\\cdots+a_n.$$[/quote]\n\nNice but trivial by Am-Gm inequality multiple times"
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "Prove that if $x,y,z>0$ then we have:\r\n\r\n$ \\frac{3}{2} ( \\sum \\frac{x^{2}}{y^{2}} + 1)  \\geq \\frac{x}{y} +2 \\sqrt{ \\frac{y}{z} }+3 \\sqrt[3]{ \\frac{z}{x} } \\geq 6$.\r\n\r\nmade by Ioan Dancila.\r\n\r\ncheers! :D :D",
  "Solution_1": "3+3sum(x^2/y^2) = sum(x^2/y^2) + 3 + 2sum(x^2/y^2)\r\n>= sum(x^2/y^2) + 3 + 2*3 (xyz/xyz)^2 by AM-GM\r\n= (x^2/y^2 + 1) + (y^2/z^2 + 3) + (z^2/x^2 + 5)\r\n>=2 x/y + 4 (y/z)^1/2 + 6 (z/x)^1/3 by AM-GM on each bracket, proving the left inequality (obviously, equality occurs iff x = y = z = 1).\r\n\r\nTo prove the right inequality, note that \r\n\r\n1/6 x/y + 1/3 (y/z)^1/2 + 1/2 (z/x)^1/3\r\n<=(xyz/xyz)^1/6 = 1, by the weighted AM-GM inequality.  Again, equality occurs iff x = y = z = 1."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "$a+b+c = 1 , a,b,c \\geq 0$\r\n\r\nShow that \r\n\r\n$2(\\frac{b}{a}+\\frac{a}{c}+\\frac{c}{b}) \\geq \\frac{1+a}{1-a}+\\frac{1+b}{1-b}+\\frac{1+c}{1-c}$",
  "Solution_1": "It's \\frac, not frac.\r\nSee http://www.mathlinks.ro/Forum/viewtopic.php?t=25780 .\r\n\r\n  darij"
}
{
  "Tag": [
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "the     following    problem  maybe   a  Olympiad   one:\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1685878#1685878",
  "Solution_1": "Do not double post."
}
{
  "Tag": [],
  "Problem": "Balanced chemical equation:\r\nC3H5(NO3)3(s) \u2192 3CO2(g) + 2.5H2O(g) + 1.5N2(g) + 0.25O2(g)\r\n\r\nI've been thinking for a while, and can't quite answer why it's so explosive.  Any help?",
  "Solution_1": "Hrmm.. your question sparked my interest.. taken from wikipedia:\r\n\r\n\"Nitroglycerin and any or all of the diluents mentioned above can certainly deflagrate, or burn. However, the explosive power of nitroglycerin is derived from detonation: a shock propagates through the fuel-rich medium at a supersonic speed. In other words, the initial decomposition sets up a pressure gradient that induces decomposition in contiguous material, creating a fast-moving transition zone, which (due to the nature of the material) can detonate any unstable or explosive material it encounters. This generates a self-sustained cascade of near instantaneous pressure-induced decomposition into gas of the explosive material which grows upon itself exponentially. This is quite unlike deflagration, which depends solely upon available fuel, regardless of pressure or shock.\"",
  "Solution_2": "A couple of reasons...\r\n\r\nFirst, obviously a very negative $\\Delta H$.\r\n\r\nSecond, a very positive $\\Delta S$\r\n\r\nThird, (very likely) a very low-energy transition state, in turn leading to a high reaction rate.\r\n\r\nFourth, (likely) a low specific heat (Kopp's rule gives $265\\ \\text{J deg}^{-1}\\text{ mol}^{-1}$; the actual value is unknown to me).",
  "Solution_3": "One must be very specific in one's questions if one is to obtain a precise answer.  Unfortunately, in most cases, one must be more familiar to the answer than one is to know enough to ask a percise answer.\r\n\r\nThere are two dimensions to the question about nitroglycerin (the reactive chemical in dynamite that made blasting much safer and fueled the Nobel prizes, and the start of the Nobel chemical company).  The first dimension is the amount of energy released, which is driven by the high bond strength of the formed nitrogen-oxygen bonds formed in the reaction of nitroglycerin with oxygen in burning or exploding.  (This has been addressed in the other responses to your question.)  \r\n\r\nThe second dimension is the speed of the reaction in other words how fast the energy is released.  This is an independent physical property.  One traditional way to measure this is to determine how fast a flame front will progress in an open trench.  This is why nitroglycerin is not used as a fuel for internal combustion engines -- it simply delivers the energy too fast.  It is also not used as a gunpower substitute for the same reason -- it delivers the energy too fast.  The mechanics of converting the released chemical energy is insufficient for mechanical or projectile use.  If one were to put kerosene in a open trench and lit one end of the vapor, one could walk faster than the flame front progressed.  This is a relatively slow release of energy, useful in jet engines and the like.  One would have to run to keep up with an methyl ether fire (a member of my lab did so and got his beaker into the hood before the entire beaker and the flame front coincided, avoiding a more memorable incident.)  If one were to do this with nitroglycerin, the flame front would progress at a speed greater than mach 1- instantaneous for most real world considerations.  In this speed of progression lies the explosive force (and great danger) of nitroglycerin.  The explosion is due to the generation of great quantities of gas that puts great pressure on whatever is about to be impacted, be it a beaker, hand or face.\r\n\r\nThis is too dangerous for casual experiementation.  If one is not working with folks who have experience with these chemicals, one may well end up earning a Darwin Award, perhaps even demonstrating the theory that led to the naming of this award.\r\n\r\nExplosives should not be experiemented with casually.\r\n\r\nExplosives should not be experiemented with casually.\r\n\r\nExplosives should not be experiemented with casually.\r\n\r\nI hope I have addressed your question in a way that will lead to you and the readers of this to not conduct experiements that you and all who love you will regret."
}
{
  "Tag": [
    "geometry",
    "percent",
    "probability",
    "puzzles"
  ],
  "Problem": "A larger square table top divided into 100 squares (on a 10 square by 10 square set-up) of 12 cm. on a side of each smaller square is marked out on the larger sheet of the table top.\r\n\r\nIn order to win a prize, a player has to roll one coin down a chute so that the coin rolls onto the large sheet and lands completely within a smaller square and not touching a line.\r\n\r\nIf a player rolls a 5 cm. diameter coin down the chute, are the chances less than even or greater than even, that the coin will come to rest not touching a line?",
  "Solution_1": "Assuming that the coin stays center on the table:\r\n[hide]\nThe locus of the points where the coin's center can be winning is a lot of 7 by 7 squares in the middle of the grid squares. Area: 49. So, the odds are about $ \\frac{49}{144} \\approx 34.0 percent < \\frac{1}{2}$ for one square. Since this stays the same for all of the interior of the board, and is only slightly different at the edges (that is, if the coin stays completely on the table), my answer would be less than even.\n[/hide]\r\nIs that right? Sorry, can't figure out how to put the percent symbol.",
  "Solution_2": "[quote=\"math_explorer\"]Sorry, can't figure out how to put the percent symbol.[/quote]\r\n\r\nThe code for $ \\%$ is \\%.  :)",
  "Solution_3": "Wait, math_explorer,\r\n\r\nassuming the coin stays on the table,\r\n\r\njust condense this problem down to one 5 cm. diameter coin\r\n\r\nbeing dropped onto one 12 cm. by 12 cm. square.  It must\r\n\r\nstay inside this one square and not touch any of the four\r\n\r\nlines of the square.  Please reconsider with this clarification/hint.",
  "Solution_4": "I thought that was what I did.",
  "Solution_5": "Your post looked right, except you would want \"probability\" instead of \"odds.\"\r\n\r\nGood work.",
  "Solution_6": "Oops... I think my subconscious did that.  :blush:"
}
{
  "Tag": [
    "trigonometry"
  ],
  "Problem": "Prove that\r\n\r\n$ i \\tan(3\\pi/11) \\equal{} (x^3\\minus{}1)/(x^3\\plus{}1) \\equal{} x^{10}\\plus{}x^9\\minus{}x^8\\minus{}x^7\\minus{}x^6\\plus{}x^5\\plus{}x^4\\plus{}x^3\\minus{}x^2\\minus{}x$ \r\n\r\nwhere $ x\\equal{}e^{2\\pi i/11}$.",
  "Solution_1": "Note that $ x^{11}\\equal{}1$, so $ x^{10}\\plus{}x^9\\minus{}x^8\\minus{}x^7\\minus{}x^6\\plus{}x^5\\plus{}x^4\\plus{}x^3\\minus{}x^2\\minus{}x$\r\n$ \\equal{}x^{11}\\plus{}x^{10}\\plus{}x^9\\minus{}x^8\\minus{}x^7\\minus{}x^6\\plus{}x^5\\plus{}x^4\\plus{}x^3\\minus{}x^2\\minus{}x\\minus{}1$\r\n$ \\equal{}(x^9\\minus{}x^6\\plus{}x^3\\minus{}1)(x^2\\plus{}x\\plus{}1)\\equal{}(x^3\\minus{}1)(x^2\\plus{}x\\plus{}1)(x^6\\plus{}1)$\r\n$ \\equal{}\\frac{(x^3\\minus{}1)(x^6\\minus{}1)(x^6\\plus{}1)}{(x\\minus{}1)(x^3\\plus{}1)}\\equal{}\\frac{(x^3\\minus{}1)(x^{12}\\minus{}1)}{(x\\minus{}1)(x^3\\plus{}1)}\\equal{}\\frac{(x^3\\minus{}1)(x\\minus{}1)}{(x\\minus{}1)(x^3\\plus{}1)}\\equal{}\\frac{x^3\\minus{}1}{x^3\\plus{}1}$.\r\n\r\nAlso, $ i\\tan{\\frac{3\\pi}{11}}\\equal{}\\minus{}i\\tan{\\frac{8\\pi}{11}}\\equal{}\\frac{x^7\\minus{}x^4}{x^7\\plus{}x^4}\\equal{}\\frac{x^3\\minus{}1}{x^3\\plus{}1}$, and we are done."
}
{
  "Tag": [],
  "Problem": "A friend of mine spoke to me about someone he knew a few years ago who could track down sports games with mathlab. For instance he could track down where shots were taken from and so on but I have no clue how to do this and neither did my friend, just wondering if any of you have experience doing this.",
  "Solution_1": "anyone has any idea?",
  "Solution_2": "I think I've found it\r\n\r\nhttp://www.cs.ubc.ca/nest/lci/thesis/fhli/videoclips/"
}
{
  "Tag": [
    "factorial",
    "algebra",
    "polynomial",
    "probability and stats"
  ],
  "Problem": "Say $ X$ is distributed Poisson($ \\lambda$). \r\n\r\nWhat is $ E(X^r)$, $ r$ a nonnegative integer?\r\n\r\nMy idea is to express $ E(X^r)$ in terms of $ E(X),E((X)(X \\minus{} 1)),E((X)(X \\minus{} 1)(X \\minus{} 2))$,etc, but I can't find a simple formula for it.",
  "Solution_1": "you're on the right track.\r\n\r\nLet $ (x)_n$ denote the falling factorial, that is\r\n\r\n\\[ (x)_n \\triangleq x(x-1)(x-2)\\ldots(x-n+1)\r\n\\]\r\n\r\n1)  it's straightforward to verify that if $ X\\sim\\text{Poisson}(\\lambda)$, then\r\n\r\n\\[ E((X)_n) = \\lambda^n\r\n\\]\r\n\r\n2)  if you know some combinatorics (i actually had to look this up), then, as polynomials,\r\n\r\n\\[ t^n = \\sum^n_{k=0} S(n,k)(t)_k\r\n\\]\r\n\r\nwhere $ S(n,k)$ are the Stirling numbers of the second kind.\r\n\r\n3)  it follows from linearity of expectations that\r\n\r\n\\begin{eqnarray*}\r\nE(X^n) &=& E\\left(\\sum^n_{k=0} S(n,k)(X)_k\\right)  \\\\\r\n&=&\\sum^n_{k=0} S(n,k)E((X)_k) \\\\\r\n&=&\\sum^n_{k=0} S(n,k)\\lambda^k\r\n\\end{eqnarray*}\r\n\r\nI don't think you're going to get a much more satisfying answer than this."
}
{
  "Tag": [
    "videos"
  ],
  "Problem": "To play this game, watch the video in the previous post and wait until the related videos comes up at the end.  Or just press the menu button.  Pick one of those videos and post it, preferrably [i]not[/i] the same video.  :P \r\n\r\n[youtube]pcYbnxA-qDY[/youtube].",
  "Solution_1": "Okay...\r\n\r\n[youtube]SXx2VVSWDMo[/youtube]",
  "Solution_2": "[youtube]Ooa8nHKPZ5k[/youtube]",
  "Solution_3": "[youtube]Tr1qee-bTZI[/youtube]\r\n\r\nPlease, take 15 minutes out of your day to watch this.",
  "Solution_4": "[quote=\"miyomiyo\"][youtube]Tr1qee-bTZI[/youtube]\n\nPlease, take 15 minutes out of your day to watch this.[/quote]\r\n\r\nActually, for arithmetic, use MCGrawHill, and for Algebra, use AoPS text.\r\n\r\n\r\n[youtube]I-b5D5941AM[/youtube]",
  "Solution_5": "[youtube]VjgidAICoQI[/youtube]\r\n\r\nWhat in the world...?",
  "Solution_6": "[youtube]1bSZslEDUl0[/youtube]",
  "Solution_7": "[youtube]Pfs4Rd5f_IQ[/youtube]."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "If $ a,b,c$ are positive reals such that $ abc \\equal{} a \\plus{} b \\plus{} c$ prove that\r\n\\[ (2ab \\minus{} 1)(2bc \\minus{} 1)(2ca \\minus{} 1)\\ \\geqslant\\ \\frac1{2(a \\plus{} b \\plus{} c)^2}\\]",
  "Solution_1": "Is equivalent with :\r\n\r\n\\[ 2(a\\plus{}b\\plus{}c)(2a\\plus{}2b\\plus{}c)(2a\\plus{}b\\plus{}2c)(a\\plus{}2b\\plus{}2c)\\ge 1\\]",
  "Solution_2": "It seems too weak. \r\n$ 2ab \\minus{} 1 \\equal{} \\frac {2abc}{c} \\minus{} 1 \\equal{} \\frac {2(a \\plus{} b \\plus{} c)}{c} \\minus{} 1 \\equal{} \\frac {2(a \\plus{} b)}{c} \\plus{} 1$\r\nSo the product on the left side must be greater then 1.\r\nSo we have to show $ 2(a \\plus{} b \\plus{} c)^2 \\ge 1$, or $ 2(a \\plus{} b \\plus{} c)^3 \\ge abc$, which is obvious.",
  "Solution_3": "[quote=\"dgreenb801\"]It seems too weak. \n$ 2ab \\minus{} 1 \\equal{} \\frac {2abc}{c} \\minus{} 1 \\equal{} \\frac {2(a \\plus{} b \\plus{} c)}{c} \\minus{} 1 \\equal{} \\frac {2(a \\plus{} b)}{c} \\plus{} 1$\nSo the product on the left side must be greater then 1.\nSo we have to show $ 2(a \\plus{} b \\plus{} c)^2 \\ge 1$, or $ 2(a \\plus{} b \\plus{} c)^3 \\ge abc$, which is obvious.[/quote]\r\n\r\nIndeed , the stronger :\r\n\\[ (2ab \\minus{} 1)(2bc \\minus{} 1)(2ca \\minus{} 1)> \\frac {2916}{(a \\plus{} b \\plus{} c)^2}\\]\r\nis also true for $ a,b,c > 0$ with $ abc \\equal{} a \\plus{} b \\plus{} c$ , but very easy again :( .",
  "Solution_4": "I think\r\n$ (2ab \\minus{} 1)(2bc \\minus{} 1)(2ca \\minus{} 1) \\ge \\frac {3375}{(a \\plus{} b \\plus{} c)^2}$ is true.\r\nWe have\r\n$ 2ab \\minus{} 1 \\equal{} \\frac {2abc}{c} \\minus{} 1 \\equal{} \\frac {2(a \\plus{} b \\plus{} c)}{c} \\minus{} 1 \\equal{} \\frac {2(a \\plus{} b)}{c} \\plus{} 1$\r\nSo we have to show\r\n$ (\\frac {2(a \\plus{} b)}{c} \\plus{} 1)(\\frac {2(b \\plus{} c)}{a} \\plus{} 1)(\\frac {2(c \\plus{} a)}{b} \\plus{} 1) \\ge \\frac {6750}{(a \\plus{} b \\plus{} c)^2}$\r\nBy Holder's inequality,\r\n$ (\\frac {2(a \\plus{} b)}{c} \\plus{} 1)(\\frac {2(b \\plus{} c)}{a} \\plus{} 1)(\\frac {2(c \\plus{} a)}{b} \\plus{} 1) \\ge (2 \\sqrt [3] {\\frac {(a \\plus{} b)(b \\plus{} c)(c \\plus{} a)}{abc}} \\plus{} 1)^3$\r\nBut $ (a \\plus{} b)(b \\plus{} c)(c \\plus{} a) \\ge 8 abc$, so this is $ \\ge 125$. So we have to show\r\n$ 125 \\ge \\frac {3375}{(a \\plus{} b \\plus{} c)^2}$, or $ (a \\plus{} b \\plus{} c)^3 \\ge 27abc$, which is obvious.",
  "Solution_5": "[quote=\"dgreenb801\"]I think\n$ (2ab \\minus{} 1)(2bc \\minus{} 1)(2ca \\minus{} 1) \\ge \\frac {6750}{(a \\plus{} b \\plus{} c)^2}$ is true.[/quote]\r\n\r\nIt fails for $ a\\equal{}b\\equal{}c\\equal{}\\sqrt{3}$ ;)",
  "Solution_6": "Thanks for all your help! :)"
}
{
  "Tag": [
    "geometry",
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Let's start with geometry.\r\n\r\n[size=150]Problem 1[/size]\r\n\r\nLet $ BB', CC'$ be altitudes of triangle $ ABC,$ and assume $ AB\\not\\equal{}AC$. Let $ M$ be the midpoint of $ BC, H$ the orthocenter of $ ABC$, and $ D$ the intersection of $ BC$ and $ B'C'$. Show that $ DH$ is perpendicular to $ AM$.\r\n\r\nPreparation for USAMO  :wink:.",
  "Solution_1": "OK...\r\n\r\n[hide=\"Minor Hint\"]Radical axes either all concur or are parallel.[/hide] \n[hide=\"Hint\"]\nFirst note what happens if we consider the point $ E$ on $ AM$ such that $ EH$ is perpendicular to $ AM$. And then go from there.[/hide]\n[hide=\"Bigger hint\"]Extend $ AM$ to form a parallelogram. Cyclic quads![/hide]",
  "Solution_2": "sweet marathon you have here.\r\n\r\nand no i can't do it.",
  "Solution_3": "I guess I need to start practicing now, since its only 3 weeks away.\r\n\r\nIll try to post as many proof problems as I can.\r\n\r\n^unimpossible, I feel for you and youkow.  You two are probably more deserving than me of making it, seriously.\r\n\r\nBtw, did you guys see David Liang (7th grade) on the list?  That guys my hero, even though Ive never met him."
}
{
  "Tag": [
    "LaTeX",
    "quadratics",
    "arithmetic series",
    "algebra",
    "quadratic formula",
    "arithmetic sequence"
  ],
  "Problem": "Lance Armstrong wants to ride to his friend's house 5000 miles away. He rides his bike 10 miles on September 1. He then rides another 20 miles on September 2. Each subsequent day, he rides 10 miles more than he rode the day before. On what day will he show up at his friend's house?",
  "Solution_1": "I think you can solve this with an arithmetic sequence. I am not good at using latex so I will use regular typing. ar[hide]ith sequence formula. \nan= a1 + (n-1)d\n5000=10+10n-10\nn=50\nI hope I am correct.[/hide]",
  "Solution_2": "[quote]Lance Armstrong wants to ride to his friend's house 5000 miles away.\nHe rides his bike 10 miles on September 1.\nHe then rides another 20 miles on September 2.\nEach subsequent day, he rides 10 miles more than he rode the day before.\nOn what day will he show up at his friend's house?[/quote][hide]\nWe have the [b]sum[/b] of an arithmetic series: $S_n = \\frac{n}{2}[2a + (n-d)d]$\n\nWe have: $a = 10,\\;d = 10,\\;S_n = 5000$\n\nHence: $\\frac{n}{2}[20 + 10(n-1)] = 5000\\;\\;\\;\\Rightarrow\\;\\;\\;n^2 + n - 1000 = 0$\n\nwhich has the positive root: $n \\;= \\;\\frac{-1 + \\sqrt{4001}}{2} \\;= \\;31.126...$\n\n\nTherefore, Lance arrives on the 32nd day: $\\text{October 2.}$[/hide]",
  "Solution_3": "[hide]well first you can divide everything by 10, the numbers will then be a bit easier to work with so you have:\nn(n+1)/2=500\nn^2+n-1000=0\nUsing the quadratic formula, I find x is 31.126...so he will arrive after 32 days. There are 32 days in September, so he will arrive [b]October 2nd[/b][/hide]",
  "Solution_4": "[color=darkred]\nBrute force =D !\n\n[hide=\"yummy solution o_o\"]\n10\n30\n60\n100\n150\n210\n280\n360\n450\n550\n660\n780\n...\ntoo... tired...\n[/hide]\n\n[hide=\"screw that and try a different method o_o\"]\nFormula for triangular numbers:\n\nn(n+1)/2\n\nYou have to find the smallest value of n where that expression > 5000/10.\n\nn(n+1)/2 > 500\nn(n+1) > 1000\nI think n = 32 (please don't eat me if I'm wrong, I don't taste good)\n\nAfter I wonder how the heIl Lance Armstrong managed to bike 310 miles in one day (he doesn't bike a full 320 miles on October 2nd), I got...\nThirty days hath April, September, June and November.\nOctober 2nd.\n[/hide]\n[/color]",
  "Solution_5": "I don't understand what is wrong with my solution though.",
  "Solution_6": "Hello hello!  What you noticed was that the distances he travels are in arithmetic sequence, which is correct.  Thus, you did arrive at the (correct) result that on the 50th day, Lance will travel 5000 miles.  However, you neglected to take into account that at that time he will already have ridden for 49 days, covering a considerable distance.  Thus, the question you needed to solve was not, \"What value of n makes 10n greater than or equal to 5000?\" but rather, \"Which value of n makes 10 + 20 + 30 + ... + 10n greater than or equal to 5000?\"",
  "Solution_7": "[quote]Thus, you did arrive at the (correct) result that on the 50th day, Lance will travel 5000 miles.[/quote]hmm...There is a small typo... missing an extra zero between 500 and 5000?\r\nHope this helps ..",
  "Solution_8": "Haha, I realized that just as I sat down to dinner.  That was another mistake, which made your first seem more reasonable -- you divided 5000 by 10 and got 50.",
  "Solution_9": "Thanks for the corrections. Sorry for showing my mathematical weakness.",
  "Solution_10": "Hey, don't apologise.. making mistakes is the best way to learn!"
}
{
  "Tag": [
    "complex numbers"
  ],
  "Problem": "I'm going to do some self-independent studying and that I'm going to use my textbooks to take notes on important thing and maybe like create a problem and solve by myself. I'm going to upload it on website so other people can see and point out if I'm doing wrong or not.\r\n\r\nI won't post on AoPS. I would post on my own website and I guess it's both good for me and others because it would be like reading notes to other people and getting ideas on in case going wrong during my study.\r\n\r\nI'll start today and if I make some progress, I'll post on here.\r\n\r\nWish me luck :)",
  "Solution_1": "Good luck.\r\n\r\nI'd look at it if you put a link.",
  "Solution_2": "There are two lessons (or notes :) for myself) that I made.\r\n\r\nThe link to there is:\r\n\r\nhttp://www.dhost.info/jwec05/My%20Websites/Number%20one/math%20lesson.htm\r\n\r\nThe first one is polygon and the second one is on complex number. Please read through and find if there is anything wrong.",
  "Solution_3": "What's weird is that some textbooks (like mine) call b the 'imaginary part' of a+bi, and other textbooks (like whatever you're using, apparently) call bi the imaginary part. Maybe you should mention that...",
  "Solution_4": "[quote=\"Silverfalcon\"]I won't post on AoPS. I would post on my own website and I guess it's both good for me and others because it would be like reading notes to other people and getting ideas on in case going wrong during my study.\n\nI'll start today and if I make some progress, I'll post on here.\n\nWish me luck :)[/quote]First of all good luck :) \r\nSecond of all I don't understand how you will not post on AoPS-MathLinks, but still \"I'll post on here\". \r\nFinally, maybe it's hard for people to tell you their opinions when the subject at hand is somewhere else (it would be easier to quote) ;)",
  "Solution_5": "Um.. For the second one, Valentine Vornicu, I posted on my website nothing but in case it will get messed up or it might take a lot of space on this public forum.. You know, this note is my stuff and well, if this takes too much space on the aops forum, it would be like taking spaces off from public forum, ya know..\r\n\r\nThat's all... But thanks! :)",
  "Solution_6": "[quote=\"Silverfalcon\"]Um.. For the second one, Valentine Vornicu, I posted on my website nothing but in case it will get messed up or it might take a lot of space on this public forum.. You know, this note is my stuff and well, if this takes too much space on the aops forum, it would be like taking spaces off from public forum, ya know..\n\nThat's all... But thanks! :)[/quote]I didn't understood what you said. You first said you ain't gonna write it here, then you said you will :P :) \r\n\r\nI think that if you keep everything into a single topic nobody will mind ;)",
  "Solution_7": "[quote=\"LynnelleYe\"]What's weird is that some textbooks (like mine) call b the 'imaginary part' of a+bi, and other textbooks (like whatever you're using, apparently) call bi the imaginary part. Maybe you should mention that...[/quote]\r\n\r\nThe imaginary part of $a+bi$ is $b$. That is quite universal.",
  "Solution_8": "I agree that the imaginary part of $a + bi$ should be $b$, which is how I've seen it in almost all books.  Unfortunately, the AoPS book says that the imaginary part is $bi$.  On the other hand, it says that $\\mathrm{Im}(a + bi) = b$.  That distinction is confusing to me, because I pronounce \"Im\" as \"the imaginary part of\".\r\n\r\nPerhaps the authors felt that, from the point of view of English, it is weird for the imaginary part to be a real number.  I agree it is weird from that point of view, but it is much more convenient mathematically.",
  "Solution_9": "[quote=\"Silverfalcon\"]360 = sum of ANY exterior angles in polygon[/quote]\r\n\r\nI don't suppose it's wrong mathematically, but it sounds funny. Although, if you're using for yourself, it doesn't matter."
}
{
  "Tag": [
    "geometry",
    "calculus"
  ],
  "Problem": "Hi, I was wondering why there are no solid geometry problems in the IMO.",
  "Solution_1": "I was looking thruogh the past years' IMO questions , and there were some really good questions on geometry .... But somehow , the weightage for geometry seems very little .... { I could get only about 1 or 2 geometry questions in one paper. }\r\n :?",
  "Solution_2": "how does that answer my question?",
  "Solution_3": "They chose not to include solid geometry because people didn't like them?",
  "Solution_4": "[quote=\"G-UNIT\"]how does that answer my question?[/quote]\r\n\r\nI laffed.",
  "Solution_5": "my guess is that some of the problems can be greatly simplified with multivariable calculus. and these olympiad stuff promote non-calculus algebra",
  "Solution_6": "[quote]I was wondering why there are no solid geometry problems in the IMO.[/quote]\r\n\r\nSolid geometry is not taught in most countries.  The IMO jury and problem committees are dominated by people from the majority of countries that do not teach this subject, and many of them consider such problems unfair to \"their\" students.",
  "Solution_7": "It's a bit harder to illustrate and put clear proofs for solid geometry.  People would run into complications while grading.",
  "Solution_8": "Solid geometry not being taught in most countries, was the reason stated to me by a couple of IMO team leaders.  I don't know if that is the full story, but I doubt that difficulty of writing up the solutions rigorously has much to do with it (there are several other subjects with the same difficulty, that make it onto the IMO).",
  "Solution_9": "i have noticed that there is at least one solid geometry problem at early IMO.",
  "Solution_10": "Solid geometry was taught in the early-IMO countries at the time of the early IMO.",
  "Solution_11": "[quote=\"fleeting_guest\"]Solid geometry was taught in the early-IMO countries at the time of the early IMO.[/quote]\r\nSolid geometry is an important part in Chinese high school course. What about other countries?"
}
{
  "Tag": [
    "factorial"
  ],
  "Problem": "Evaluate \\[(\\sum^{2002}_{k=1} \\frac{k.k!}{2^k}) - (\\sum^{2002}_{k=1} \\frac{k!}{2^k}) - (\\frac{2003!}{2^{2002}})\\]",
  "Solution_1": "$\\frac{k\\cdot k!}{2^k}-\\frac{k!}{2^k}=\\frac{(k+1-1)k!}{2^k}-\\frac{k!}{2^k}=\\frac{(k+1)!}{2^k}-\\frac{2k!}{2^k}=\\frac{(k+1)!}{2^k}-\\frac{k!}{2^{k-1}}$, Using telescoping method we have \r\n\r\n$\\sum_{k=1}^{2002} \\frac{k\\cdot k!}{2^k}-\\sum_{k=1}^{2002} \\frac{k!}{2^k}-\\frac{2003!}{2^{2002}}$\r\n\r\n$=\\left(\\frac{2!}{2}-\\frac{1!}{2^0}\\right)+\\left(\\frac{3!}{2^2}-\\frac{2!}{2}\\right)+\\cdots+\\left(\\frac{2002!}{2^{2001}}-\\frac{2001!}{2^{2000}}\\right)+\\left(\\frac{2003!}{2^{2002}}-\\frac{2002!}{2^{2001}}\\right)-\\frac{2003!}{2^{2002}}$\r\n\r\n$=-1$",
  "Solution_2": "[quote=\"kunny\"]$\\frac{k\\cdot k!}{2^k}-\\frac{k!}{2^k}=\\frac{(k+1-1)k!}{2^k}-\\frac{k!}{2^k}=\\frac{(k+1)!}{2^k}-\\frac{2k!}{2^k}=\\frac{(k+1)!}{2^k}-\\frac{k!}{2^{k-1}}$, Using telescoping method we have \n\n$\\sum_{k=1}^{2002} \\frac{k\\cdot k!}{2^k}-\\sum_{k=1}^{2002} \\frac{k!}{2^k}-\\frac{2003!}{2^{2002}}$\n\n$=\\left(\\frac{2!}{2}-\\frac{1!}{2^0}\\right)+\\left(\\frac{3!}{2^2}-\\frac{2!}{2}\\right)+\\cdots+\\left(\\frac{2002!}{2^{2001}}-\\frac{2001!}{2^{2000}}\\right)+\\left(\\frac{2003!}{2^{2002}}-\\frac{2002!}{2^{2001}}\\right)-\\frac{2003!}{2^{2002}}$\n\n$=-1-\\frac{2001!}{2^{2000}}$[/quote]\r\n\r\nThe ans. is -1..u can recheck ur calculations...",
  "Solution_3": "Yes, the answer is -1  :D",
  "Solution_4": "Yeh, I was mistaken.I will edit. :lol:"
}
{
  "Tag": [
    "geometry",
    "circumcircle",
    "incenter",
    "angle bisector",
    "geometry proposed"
  ],
  "Problem": "Let $ ABC$ be equilateral triangle inscribed in circumcircle $ (O)$. Let $ D$ be some point on the smaller arc of $ BC$. Let $ I_1,I_2,I_3$ be the incenters of triangle $ DBC,DCA,DAB$, respectively. Prove that the line through $ A,B,C$ perpendicular to $ I_2I_3,I_3I_1,I_1I_2$ concurrent. :D",
  "Solution_1": "[b]hint[/b]: Why don't you use the [b]Carnot theorem[/b] to prove it?? :D",
  "Solution_2": "I shall try solving, before receiving full proof from its author...\r\nFirst, let's re-name I_1, I_2 and I_3 as 1, 2 and 3 respectively.\r\n\r\nIt's known that the perpendiculars from the vertices of a triangle onto the sides of another triangle are concurrent iff the perpendiculars from the vertices of the 2nd triangle onto the sides of the first one are concurrent (is this Carnot's, or just a consequence?).\r\n\r\nI shall show that the perpendiculars from 1 to BC, 2 to AC and 3 to AB concur at the Fermat's point of the triangle 123..\r\n\r\nI name M and N the midpoints of the small arcs BD and CD, M' the point on the circle O such that B is the midpoint of the arc MM', and N' such as C is the midpoint of the arc NN'.\r\nThe points 1 and 3 lie on the circle C(M, BM) and, of course on the angle bisectors of <BAD and <BCD, while 1 and 2 lie on the circle C(N, NC), that is, 13M and 12N are equilateral triangles. The lines M3 and N2 pass through A and the angle <MAN = 30 degs (by construction), this making 13 and 12 perpendicular, or <M12 = <N13 = 150 degs, that is, AM12 and A31N are cyclic and <1M2 = <1A2 ( 1 ) and <1N3 = <1A3 ( 2 ), therefore M, 2 and N' are collinear, and so are N, 3 and M'.\r\nNow it's easy to see that <MAC + <N'MA = 90 degs, that is, 2M and AC are perpendicular, and same are 3N and AB.\r\nBut 2M and 3N pass through the Fermat point of the triangle 123, hence the line joining 1 with Fermat point will be perpendicular to BC.\r\n\r\nBest regards,\r\nsunken rock",
  "Solution_3": "Very nice solution :D [b]sunken rock[/b]. I cannot tell anything, just want to remind you all a very important outcome of [b]Carnot theorem[/b]:\r\n[b]\"The perpendiculars from the vertices of a triangle onto the sides of another triangle are concurrent iff the perpendiculars from the vertices of the 2nd triangle onto the sides of the first one are concurrent\"[/b] :lol:",
  "Solution_4": "If $I$ is incenter of $ \\triangle ABC,$ then according to [url=http://en.wikipedia.org/wiki/Japanese_theorem_for_concyclic_quadrilaterals]Mikami and Kobayashi theorem[/url], $II_1I_2I_3$ is a rectangle. Thus, it remains to prove that $ \\triangle II_2I_3$ and $\\triangle ABC$ are orthologic. Let $ X,Y$ be the projections of $ I_2$ and $ I_3$ onto $ AB$ and $ AC$ and $ Q \\equiv{} I_2X \\cap I_3Y.$ Easy angle chasing gives that complete quadrangle $DI_2QI_3$ has excenter $A$ $\\Longrightarrow$ $AX \\equal{} AY$ $ \\Longrightarrow$ Perpendiculars $ I_2X$ and $ I_3Y$ meet on the angle bisector of $ \\angle BAC.$ Hence, $ \\triangle ABC$ and $ \\triangle II_2I_3$ are orthologic $\\Longrightarrow$ Perpendiculars from $ A,B,C$ to the sidelines of $\\triangle II_2I_3$ concur."
}
{
  "Tag": [
    "function",
    "Putnam",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "Find all functions $ f : Z\\rightarrow Z$ for which we have $ f (0) \\equal{} 1$ and $ f ( f (n)) \\equal{} f ( f (n\\plus{}2)\\plus{}2) \\equal{} n$, for every natural number $ n$.",
  "Solution_1": "[quote=\"tdl\"]Find all functions $ f : Z\\rightarrow Z$ for which we have $ f (0) \\equal{} 1$ and $ f ( f (n)) \\equal{} f ( f (n \\plus{} 2) \\plus{} 2) \\equal{} n$, for every natural number $ n$.[/quote]\r\nSo we don't have any info on negative integers?",
  "Solution_2": "$ f(f(n)) \\equal{} f(f(n \\plus{} 2) \\plus{} 2) \\equal{} n$\r\n$ \\Rightarrow f(f(f(n))) \\equal{} f(f(f(n \\plus{} 2) \\plus{} 2))$\r\n$ \\Rightarrow f(n) \\equal{} f(n \\plus{} 2) \\plus{} 2$\r\nso $ f(n \\plus{} 2) \\equal{} f(n) \\minus{} 2$\r\ngives $ f(2k) \\equal{} f(0) \\minus{} 2k$ and $ f(2k \\plus{} 1) \\equal{} f(1) \\minus{} 2k$\r\nwhere $ f(0) \\equal{} 1\\ and  \\ f(1) \\equal{} f(f(0)) \\equal{} 0$\r\nso $ f(2k) \\equal{} 1 \\minus{} 2k\\ and\\ f(2k \\plus{} 1) \\equal{} \\minus{} 2k$\r\nso $ \\forall n\\in\\mathbb{Z}: \\ f(n) \\equal{} 1 \\minus{} n$",
  "Solution_3": "[quote=\"aviateurpilot\"]$ f(f(n)) \\equal{} f(f(n \\plus{} 2) \\plus{} 2) \\equal{} n$\n$ \\Rightarrow f(f(f(n))) \\equal{} f(f(f(n \\plus{} 2) \\plus{} 2))$\n$ \\Rightarrow f(n) \\equal{} f(n \\plus{} 2) \\plus{} 2$\nso $ f(n \\plus{} 2) \\equal{} f(n) \\minus{} 2$\ngives $ f(2k) \\equal{} f(0) \\minus{} 2k$ and $ f(2k \\plus{} 1) \\equal{} f(1) \\minus{} 2k$\nwhere $ f(0) \\equal{} 1\\ and \\ f(1) \\equal{} f(f(0)) \\equal{} 0$\nso $ f(2k) \\equal{} 1 \\minus{} 2k\\ and\\ f(2k \\plus{} 1) \\equal{} \\minus{} 2k$\nso $ \\forall n\\in\\mathbb{Z}: \\ f(n) \\equal{} 1 \\minus{} n$[/quote]\r\nYou are wrong . \r\nSolution is :\r\n$ f(0)\\equal{}a,f(1)\\equal{}b$ where $ a\\equiv 0 (\\mod 2) ,b\\equiv 1 (\\mod 2)$ and \r\n$ f(2n)\\equal{}a\\minus{}2n$ and $ f(2n\\plus{}1)\\equal{}b\\minus{}(2n\\plus{}1)$ \r\nEasy to check this function satisfy condition.",
  "Solution_4": "[quote=\"TTsphn\"][quote=\"aviateurpilot\"]$ f(f(n)) \\equal{} f(f(n \\plus{} 2) \\plus{} 2) \\equal{} n$\n$ \\Rightarrow f(f(f(n))) \\equal{} f(f(f(n \\plus{} 2) \\plus{} 2))$\n$ \\Rightarrow f(n) \\equal{} f(n \\plus{} 2) \\plus{} 2$\nso $ f(n \\plus{} 2) \\equal{} f(n) \\minus{} 2$\ngives $ f(2k) \\equal{} f(0) \\minus{} 2k$ and $ f(2k \\plus{} 1) \\equal{} f(1) \\minus{} 2k$\nwhere $ f(0) \\equal{} 1\\ and \\ f(1) \\equal{} f(f(0)) \\equal{} 0$\nso $ f(2k) \\equal{} 1 \\minus{} 2k\\ and\\ f(2k \\plus{} 1) \\equal{} \\minus{} 2k$\nso $ \\forall n\\in\\mathbb{Z}: \\ f(n) \\equal{} 1 \\minus{} n$[/quote]\nYou are wrong . \nSolution is :\n$ f(0) \\equal{} a,f(1) \\equal{} b$ where $ a\\equiv 0 (\\mod 2) ,b\\equiv 1 (\\mod 2)$ and \n$ f(2n) \\equal{} a \\minus{} 2n$ and $ f(2n \\plus{} 1) \\equal{} b \\minus{} (2n \\plus{} 1)$ \nEasy to check this function satisfy condition.[/quote]\r\nbut we have $ f(0)\\equal{}1$ in the execice\r\nand we can find $ f(1)\\equal{}fof(0)\\equal{}0$\r\nso your $ a\\equal{}1$ and your $ b\\equal{}0$",
  "Solution_5": "this is putnam 1992 A1\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=185988",
  "Solution_6": "Simultaneous subbing goes brrr...\n\n$n\\to 0 \\implies f(0)=1.$\n$n\\to f(n) \\implies f(n)=f(n+2)+2$\n$ \\implies f(n+2)=f(n)-2.$\nNow induction yields the solution: $\\boxed{f(n)=1-n},$ which works."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "The radius of a circle is 6 meters. A square is inscribed in\nthe circle. What is the number of square meters in the area\nof the square?\n\n[asy]size(70); draw(Circle((0,0),1)); draw((1,0)--(0,1)--(-1,0)--(0,-1)--cycle); draw(\"6 m\",(0,0)--(1,0),N,fontsize(8pt));[/asy]",
  "Solution_1": "The square's diagonal is 12, so its side length is 12/(sqrt2), or 6sqrt2. Square this for the area and you get 72 square meters.",
  "Solution_2": "Alternately, the area of a square is half the square of the length of its diagonal.  Since the length of its diagonal is $ 6\\cdot2 \\equal{} 12$, the area is\r\n\\[ \\frac {12^2}2 \\equal{} \\frac {144}{2} \\equal{} \\boxed{72}\\]"
}
{
  "Tag": [
    "geometry",
    "analytic geometry",
    "AMC"
  ],
  "Problem": "On side AB of square ABCD, right traingle ABF with hypotenuse AB is drawn externally to the square. E is the point of intersection of hte diagonals of the square. If AF=6, BF=8, find EF.",
  "Solution_1": "I used analytic geometry to get umm...7:rt2:.  I'm too tired for explanation.",
  "Solution_2": "Ptolemy's theorem is probably the nicest way to do this . . . but it's certainly doable with analytic geometry, or with sythetic (non-analytic) techniques.",
  "Solution_3": "With Ptolemy's, its like a 1-step thing, right?",
  "Solution_4": "You get the side length is 10, and then it's immediate by Ptolemy."
}
{
  "Tag": [],
  "Problem": "Three people share a car for a period of one year and the mean number of kilometers travelled by each person is 152 per month. How many kilometers will be travelled in one year?",
  "Solution_1": "[quote=\"math92\"]Three people share a car for a period of one year and the mean number of kilometers travelled by each person is 152 per month. How many kilometers will be travelled in one year?[/quote]\r\n\r\n[hide=\"answer\"]\n\n152*3=456\n456*12=$\\framebox{5472}$[/hide]",
  "Solution_2": "[quote=\"math92\"]Three people share a car for a period of one year and the mean number of kilometers travelled by each person is 152 per month. How many kilometers will be travelled in one year?[/quote]\r\n[hide]\nWhat do you mean by sharing the car? Do they each have if for themselves for a part of the year, or do they just use the car the entire year?... im sorta confused..\n\n152x12=1824\n\nor\n\n152x3x12=5472\n\n[/hide]"
}
{
  "Tag": [
    "floor function",
    "number theory",
    "prime numbers"
  ],
  "Problem": "To determine whether a number $ \\textit{N}$ is prime, we must test for divisibility by every prime less than or equal to the square root of $ \\textit{N}$. How many primes must we test to determine whether $ 2003$ is prime?",
  "Solution_1": "$ \\lfloor\\sqrt{2003}\\rfloor\\equal{}44$. The prime numbers $ \\le44$ are $ 2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and there are $ \\boxed{14}$ of them."
}
{
  "Tag": [],
  "Problem": "In the next North American tournament there are 12 more American than Canadian teams (only US and Canada are in). If there are c Canadian teams enrolled, what\u2019s the percentage of Canadian teams?",
  "Solution_1": "Is the answer supposed to be in terms of c?",
  "Solution_2": "[hide=\"Solution?\"]Canadian: $ c$\nAmerican: $ c \\plus{} 12$\nTotal: $ 2c \\plus{} 12$\nCanadian/Total=$ \\boxed{\\frac {c}{200c \\plus{} 1200}}$[/hide]\r\n\r\nI multiplied the denominator by 100 to achieve a percentage...",
  "Solution_3": "You're a bit off (You multiply the numerator by $ 100$ to make it a percentage), so:\r\n\r\n\\[ \\boxed{\\frac{100c}{2c\\plus{}12}}\\]",
  "Solution_4": "Oh, ya....forget my solution, sri340. No idea what I was thinking there. go with LaTeX's."
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "Bob had parked his car under a tree and was heading toward his friend's house when a vicious dog lunged toward him. Luckily for Bob, he was just beyond the dogs reach. Since his friend wasn't home, Bob decided to leave, but he had a slight problem. The dog, straining at the end of the chain, was following Bob's every move. With the chain attached to the tree, the dog had access to both doors. How would Bob be able to get to the door without the dog being able to reach him?",
  "Solution_1": "1) Who is Shadow??\r\n2) [hide]Just have Bob walk in big loops so the dog's chain gets wrapped around the tree, shortening it enough so Bob can walk to his car.\n[/hide]",
  "Solution_2": "How did Bob get from the car to the door in the first place?\r\n\r\nI'll tell ya how....\r\n[hide]The friend's door and the car were on the same side of the dog[/hide]",
  "Solution_3": "[quote=\"besttate\"]1) Who is Shadow??\n2) [hide]Just have Bob walk in big loops so the dog's chain gets wrapped around the tree, shortening it enough so Bob can walk to his car.\n[/hide][/quote]\r\n\r\n1. Shadow? Where does it say shadow? I'm guessing the dog anyways",
  "Solution_4": "[quote=\"moogra\"][quote=\"besttate\"]1) Who is Shadow??\n2) [hide]Just have Bob walk in big loops so the dog's chain gets wrapped around the tree, shortening it enough so Bob can walk to his car.\n[/hide][/quote]\n\n1. Shadow? Where does it say shadow? I'm guessing the dog anyways[/quote]\r\n\r\nihatepie edditted shadow out"
}
{
  "Tag": [
    "pigeonhole principle"
  ],
  "Problem": "The squares of a 9x9 chessboard are fills with numbers $ {1,2,3...,81}$. Prove that there exist two squares with a common side, such that the difference between the numbers in them is at least $ 6$.",
  "Solution_1": "[hide=\"not really pigeonhole but...\"]\nSuppose the contrary: all adjacent squares have difference at most 5.  Notice that if the bottom-left most square is $ x$, then the square above it is at most $ x \\plus{} 5$, the square above that is at most $ x \\plus{} 10$ and so on.  Then the top-left square is at most $ x \\plus{} 35$, and the top-rightmost square is at most $ x \\plus{} 70$.  So any two squares have a difference of at most 70, but there's a square labeled 1 and a square labeled 81 with difference 80, so we have a contradiction.\n\n[/hide]"
}
{
  "Tag": [
    "quadratics",
    "national olympiad"
  ],
  "Problem": "[img]http://i27.photobucket.com/albums/c187/toughtop/Severity.jpg[/img]\r\n\r\n[img]http://i27.photobucket.com/albums/c187/toughtop/Severity1.jpg[/img]\r\n\r\nPlease help guys. Thanks in advance...\r\n\r\n:D Good luck to you all.",
  "Solution_1": "Sketches of the solutions...\r\n[hide=\"19\"]\nThe first one factors to $(x^{2}-x+1)(2x+a)$, which means the roots are $-\\frac{a}{2}, \\frac{1 \\pm i\\sqrt{3}}{2}$, so the difference of the two quadratic roots is nonreal.  Then, the sum of two roots can be equal to the thrid one if and only if $a=-2$\n[/hide]\n[hide=\"20\"]\nI'm assuming it's talking about the $f$ given.\nWe have to solve the equation $x^{3}-3x^{2}+3=x \\iff (x-3)(x-1)(x+1)=0$\nSo the fixed points of that $f$ sum to $-3$.\n[/hide]\n[hide=\"16\"]\nWell, expanding it out, we get the leading coefficient only affected by the $x$ terms, and the constant term only affected by the constant terms.  Then, we get the leading coefficient to be $(-1)^{n+1}3^{(n+1)(2n+1)}$, and the constant term to be $\\frac{(-1)^{n}}{3^{(n+1)(2n+1)}}$.\nThe degree is, of course, $2n+2$, because we have a product of $2n+2$ monomial terms...[/hide]",
  "Solution_2": "I think this is posted in the wrong forum  :wink:"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "3D geometry"
  ],
  "Problem": "A circle is inscribed in a larger regular hexagon, and a smaller regular hexagon is inscribed in the circle. What is the ratio of the area of the smaller hexagon to the area of the larger hexagon? Express your answer as a common fraction.\n\n[asy]draw((5,8.7)--(10,0)--(5,-8.7)--(-5,-8.7)--(-10,0)--(-5,8.7)--cycle);\ndraw((5.8,10)--(11.6,0)--(5.8,-10)--(-5.8,-10)--(-11.6,0)--(-5.8,10)--cycle);\ndraw(Circle((0,0),10));[/asy]",
  "Solution_1": "$ \\frac{\\left(\\sqrt{r^2\\minus{}(r/2)^2}\\right)^2}{r^2} \\equal{} \\left(\\frac{\\sqrt{3}}{2}\\right)^2\\equal{}\\boxed{\\frac{3}{4}}$.",
  "Solution_2": "Sorry, I don't understand what you did at all :blush:",
  "Solution_3": "[quote=\"AIME15\"]Sorry, I don't understand what you did at all :blush:[/quote]\r\n\r\n[asy]size(200);\n\ndraw((5,8.7)--(10,0)--(5,-8.7)--(-5,-8.7)--(-10,0)--(-5,8.7)--cycle);\ndraw((5.8,10)--(11.6,0)--(5.8,-10)--(-5.8,-10)--(-11.6,0)--(-5.8,10)--cycle);\n\ndraw(Circle((0,0),10));\n\ndraw((0,0)--(-8.7,5),red);\nlabel(\"$r$\",(-4.35,2.5),S,red);\n\ndraw((0,0)--(7.5,4.35),blue);\nlabel(\"$\\sqrt{r^2-(r/2)^2}$\",(3.75,2.175),SE,blue);\n\ndraw((0,0)--(5,8.7),blue);\nlabel(\"$r$\",(2.5,4.35),W,blue);\n\nlabel(\"$r/2$\",(6.25,6.525),E,blue);[/asy]\r\n\r\nCf.\r\n[url=http://planetmath.org/encyclopedia/AreaOfAPolygonalRegion.html]area of a polygonal region[/url] - PlanetMath\r\n[url=http://en.wikipedia.org/wiki/Square-cube_law]Square-cube law[/url] - Wikipedia"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Well...I have read the linked post, but I still have a question:\r\n\r\n[b]When are USAMO qualifiers from international papers announced?[/b]\r\n\r\nI mean...how many days before the USAMO.\r\n\r\nLike, today is only 16 days before the test -__-;;;",
  "Solution_1": "This is the same exact question as the one in your previous post.\r\n\r\nI am an American citizen in Taiwan, and I've qualified for the USAMO three years in a row now. I've always gotten it arranged. Have some patience!"
}
{
  "Tag": [
    "vector",
    "linear algebra"
  ],
  "Problem": "Let U be a subspace of R^n and define V to be the set of vectors w such that w belongs to R^n and w is orthogonal to U.\r\n\r\nProve that V is a subspace of R^n and dim(U)+dim(V)=n.",
  "Solution_1": "Proving \"subspace\" requires proving closed under addition and scalar multiple, right?\r\nSo is V closed under addition?  Under scalar multiple?",
  "Solution_2": "Basically it is to prove that V is closed under addition and scalar multiple operations, and in addition to that prove that dim(U)+dim(V)=n.",
  "Solution_3": "So ... it is for YOU to do, not us.",
  "Solution_4": "Basically I could prove that it is closed under scalar multiplication, but not addition (at least it is not so obvious to me)\r\n\r\nAnd there is [b]no need[/b] for you to make such a comment, transseries. I [b]WOULD NOT[/b] be posting this problem in this forum if I had even the slightest idea about solving this problem.[/b]",
  "Solution_5": "[quote=\"simonlin_00\"]And there is [b]no need[/b] for you to make such a comment, transseries. I [b]WOULD NOT[/b] be posting this problem in this forum if I had even the slightest idea about solving this problem.[/quote]\r\nIt was a rude comment, but it had a point. We don't like dealing with people who post routine exercises like this and then act completely helpless. It's much more productive to give a few hints so you can solve it yourself- but we need to know how much you know to do that.\r\n\r\nSaying that a vector is orthogonal to some fixed vector is a linear equation. Saying that it's orthogonal to some fixed collection of vectors is a system of linear equations. Write that down; what does it look like here?"
}
{
  "Tag": [
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Each cell of a 100 x 100 board is painted with one of four colors, so that each row and each column contains exactly 25 cells of each color. \r\n\r\nShow that there are two rows and two columns whose four intersections are all different colors.",
  "Solution_1": "More generally, let us color a $4n x 4n$ square with each row and column having $n$ of each color.  Each column has $(4n)*(3n)/2=6n^2$ pairs of rows in it which are of different color.  Thus there are $24n^3$ pairs of squares which are in the same column, but different colors.  Since there are fewer than $8n^2$ pairs of rows, some pair of rows contains strictly more than $3 n$ columns in which they differ.  \r\n\r\nI claim those two rows can be used to make our rectangle.  To see this, let the colors be labelled 1,2,3,4 and let $f(i,j)$ be the number of columns which are color $i$ in the first row, $j$ in the second.\r\n\r\nIf both $f(1,2)$ and $f(3,4)$ are nonzero, we are done.  Assume WLOG $f(1,2)=0$.  If both $f(1,3)$ and $f(2,4)$ are nonzero, we are done.  Assume WLOG $f(1,3)=0$.  $f(1,4)$ cannot be 0 now, since there are strictly fewer than $n$ columns in our pair of rows which match.  Therefore we are done unless $f(2,3)=0$, which we shall now assume.\r\n\r\nThis gives us the relationships:\r\n\r\n$f(1,1)+f(1,4)=n$ (as there are $n$ squares of color 1 in the first row)\r\n$f(2,2)+f(2,4)=n$ (as there are $n$ squares of color 2 in the first row)\r\n$f(3,3)+f(3,4)=n$ (as there are $n$ squares of color 3 in the first row)\r\n$f(1,4)+f(2,4)+f(3,4) \\leq n$ (there are $n$ squares of color 4 in the second row)\r\n$f(1,1)+f(2,2)+f(3,3)<n$ (the two rows match in fewer than $n$ places.  \r\n\r\nThe sum of the three equations contradicts the sum of the two inequalities.",
  "Solution_2": "Nice solution!  :D"
}
{
  "Tag": [
    "MATHCOUNTS",
    "HMMT"
  ],
  "Problem": "i need to know how to find the date of my mathcounts regional competition so i no it wont interfere with hmmt",
  "Solution_1": "[quote=\"peregrinefalcon88\"]i need to know how to find the date of my mathcounts regional competition so i no it wont interfere with hmmt[/quote]\r\n\r\nI suggest you Google it.  But it's always in February.",
  "Solution_2": "Most places haven't finalized their times for regionals yet.",
  "Solution_3": "You may want to ask a coach or veteran Mathlete, in my chapter, it's always the same weekend in February, this year I have feeling it will be on Valentine's day :P"
}
{
  "Tag": [
    "function",
    "search",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Suppose that $f: \\mathbb{R}\\rightarrow\\mathbb{R}$ fulfils $\\left|\\sum^n_{k=1}3^k\\left(f(x+ky)-f(x-ky)\\right)\\right|\\le1$ for all $n\\in\\mathbb{N},x,y\\in\\mathbb{R}$. Prove that $f$ is a constant function.",
  "Solution_1": "$|f(x+y)-f(x-y)|\\leq \\frac{1}{3},\\forall x,y$ so we are allowed to define $a_n=a_n(x,y)=f(x+ny)-f(x-ny)$.\r\nWe have that $|a_1|\\leq \\frac{1}{3}$(and also $|a_n|\\leq \\frac{1}{3},\\forall n$).Define $c_1=\\frac{1}{3}$.We will search a better upper bound ,named $c_n$,of $|a_1$|(in fact it will be an upper bound for all $|a_k|,\\forall k\\in [1..n]$).We will show that $c_n$ goes to 0. From this it follows that $f$ is constant.\r\n\r\nFrom $\\left|\\sum^{n+1}_{k=1}3^ka_k\\right|\\leq 1$ we get $3^n |a_{n+1}|\\leq c_n(1+3+...+3^{n-1}) +\\frac{1}{3}$.\r\nSo $|a_{n+1}|\\leq \\frac{c_{n}}{2}+\\frac{1}{3^{n+1}}$.So $|a_{1}|\\leq \\frac{c_{n}}{2}+\\frac{1}{3^{n+1}}$.We get a sequence  that  verifies $c_{n+1}\\leq \\frac{c_{n}}{2}+\\frac{1}{3^{n+1}}$.It is easy to check that $c_n$ goes to 0.\r\nI hope it's ok.",
  "Solution_2": "[quote=\"Peter\"]Suppose that $f: \\mathbb{R}\\rightarrow\\mathbb{R}$ fulfils $\\left|\\sum^n_{k=1}3^k\\left(f(x+ky)-f(x-ky)\\right)\\right|\\le1$ for all $n\\in\\mathbb{N},x,y\\in\\mathbb{R}$. Prove that $f$ is a constant function.[/quote]\nLet $x,y \\in \\mathbb{R}$ and\n\\[ T_n := \\sum^n_{k=1}3^k\\left(f(x+ky)-f(x-ky)\\right) \\]\nThen $\\left|T_n-T_{n-1}\\right| \\le \\left|T_n\\right| + \\left|T_{n-1} \\right| \\le 2$, so for all $x,y \\in \\mathbb{R}, n \\in \\mathbb{N}$:\n\\[ \\left |3^nf(x+ny)-f(x-ny)\\right| \\le 2 \\]\nLet $z \\in \\mathbb{R}$ and $x = \\frac{z}{2}, y = \\frac{z}{2n}$ gives for all $n \\in \\mathbb{N}$:\n\\[ \\left | f(z) - f(0) \\right | \\le \\frac{2}{3^n} \\]\nAnd hence $f(z) = f(0)$ for all $z \\in \\mathbb{R}$ and $f$ is constant."
}
{
  "Tag": [
    "recurrence relation",
    "quadratics",
    "algebra",
    "functional equation",
    "IMO Shortlist"
  ],
  "Problem": "Let $ \\mathbb{R}^\\plus{}$ be the set of all non-negative real numbers. Given two positive real numbers $ a$ and $ b,$ suppose that a mapping $ f: \\mathbb{R}^\\plus{} \\mapsto \\mathbb{R}^\\plus{}$ satisfies the functional equation:\r\n\r\n\\[ f(f(x)) \\plus{} af(x) \\equal{} b(a \\plus{} b)x.\\]\r\n\r\nProve that there exists a unique solution of this equation.",
  "Solution_1": "[hide=\"answer\"]Answer \n\n$ f(x) = bx$[/hide]\n[hide=\"solution\"]Solution \n\nPut $ u_0 = k, u_n+1 = f(u_n)$. Then $ u_n+2 - b(a+b)u_n+1 + au_n = 0$. That is a standard linear recurrence relation, so we can immediately solve it. The associated quadratic factorises as $ (x - b)(x + a + b) = 0$, so the general solution is $ u_n = Abn + B(-a-b)n$. Hence $ u_n/(a+b)n = A(b/(a+b))n + (-1)nB$. But $ (b/(a+b))n \u2192 0$ and $ lhs$ is always non-negative. Hence $ B = 0$. So $ A = k$. Hence $ f(k) = u_1 = bk$. It is easy to check that this does indeed satisfy the relation given. [/hide]",
  "Solution_2": "I've got a question what are the neccesary conditions to apply this method since if I use this to solve $f(f(x))=x$ which is equivalent to $a_n+2=a_n$ with $a_0=x$ I get a unique solution $f(x)=x$, but the function $f(x)=1/x$ is also a valid solution. ",
  "Solution_3": "[quote=Ramanujan1057894736842]I've got a question what are the necessary conditions to apply this method since if I use this to solve $f(f(x))=x$ which is equivalent to $a_n+2=a_n$ with $a_0=x$ I get a unique solution $f(x)=x$, but the function $f(x)=1/x$ is also a valid solution.[/quote]\nYou won't get $f(x)=x$. You'll get $f^{2n}(x)=x$ with no information about $f(x)$ (here $f^k(x)$ denotes $k$ times application of $f$)",
  "Solution_4": "Fix $x$, and let $a_n = f^n(x)$ (the $n$-th iterate of $f$). Then, \n\\[a_{n+2} +a.a_{n+1} - b(a+b)a_n = 0,\\]\nwhich has characteristic polynomial \n\\[x^2+ax-b(a+b),\\]\nwhich has roots $-a-b$ and $b$. Hence,\n\\[a_n = u.(-a-b)^n+v.b^n,\\]\nwith $u,v$ constant. Therefore,\n\\[a_n = \\frac{xb-f(x)}{a+2b}(-a-b)^n+\\frac{(a+b)x+f(x)}{2b+a}b^n,\\]\nwhich, unless $f(x) = bx$, is negative for some large $n$. Therefore, $f(x) = bx, \\ \\forall x$, which is a solution.",
  "Solution_5": "Define $f(x_{k-1})=x_k.$ Then the original equation becomes \\begin{align*}x_{k+2}=b(a+b)x_k-ax_{k+1} \\\\ \\implies x_k=c_1b^k+c_2(-a-b)^k,\\end{align*} where $c_1$ and $c_2$ are non-negative constants. We also have $x_0=c_1+c_2,$ and $x_1=c_1b-c_2(a+b).$ Notice that $c_2=0$ for $x_k$ to be non-negative. Thus $$x_0=c_1\\implies f(x_0)=x_1=bc_1=bx_0.$$ Hence the solution is $f(x)=bx ~~\\forall x,$ which fits. "
}
{
  "Tag": [
    "quadratics"
  ],
  "Problem": "Please tell me is this right, if not, where did I made the mistake?\r\n\r\nm^2 + 6m + 9 - n^2\r\n(m^2 + 6m) + (9 - n^2)\r\nm(m + 6) + (9 - n^2)\r\n\r\nThank you all[/code]",
  "Solution_1": "Is this what you're looking for?\r\n$m^2+6m+9-n^2$\r\n$(m^2+6m+9)-n^2$\r\n$(m+3)^2-(n)^2$\r\n$(m+n+3)(m-n+3)$\r\n\r\nEDIT: Fixed. Thanks.",
  "Solution_2": "[quote=\"towersfreak2006\"]Is this what you're looking for?\n$m^2+6n+9-n^2$\n$(m^2+6n+9)-n^2$\n$(m+3)^2-(n)^2$\n$(m+n+3)(m-n+3)$[/quote]\r\n\r\nEverything is right but it should be $6m$ not $6n$",
  "Solution_3": "how about this one? I don't think it's posible, but see if someone can solve this for me\r\n14m^2 + 4mn - 15n",
  "Solution_4": "It can't be factored.\r\nNotice that the expression is a quadratic in $m$, and the discriminant, $\\sqrt{16+4(14)(15)} = 2\\sqrt{214}$, since this is not an integer, the expression can't be factored."
}
{
  "Tag": [
    "trigonometry",
    "geometry",
    "calculus",
    "AMC",
    "AIME",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Many schools finish trig, geometry, and everything else that is tested on the AMC competitions by Grade 10. As printed right on the contest, all problems can be solved using precalculus methods. After that, much of the content we learn cannot be applied to the contests.\r\n\r\nHence, many of us also do not improve significantly in the 11th and 12th grades vs. 10th unless we have practiced since then. I've noticed that at my school, many seniors end up getting approximately the same scores as the 10th and 11th graders.\r\n\r\nSince everyone Grade 10 and above pretty much has an equal chance at the tests (except the fact that you will be more proficient with more practice) why is it that they should get an easier route?\r\n\r\nMost of them do very poorly on the Olympiad anyway. Last year, didn't half of them get 0's?\r\n\r\nAlso, it should be pointed out that many people who qualify for Olympiad-level tests have skipped grades in math, and not every school allows that. My school, for example, has a no-skip policy unless you skipped in middle school and it's beyond their control. When contestants skip grades, they have obvious advantages. I think the cutoffs, should they exist, should be determined by the COURSE contestants are taking, and not their respective ages or grade levels.",
  "Solution_1": "JBL and Complex Zeta have both amazed me with descriptions of the middle school educations they received in geometry, in private schools, but I think the fact remains that most top-level math competitors learn most of their math by self-education, almost more despite school than because of school. The mandatory wastage of time in most school programs is a drag, but everyone has opportunity to use their free time to learn more math, and yet more math, and still more math, so I think it is possible to make it into the top echelon if that is what you really desire. \r\n\r\nOn grade-skipping and math acceleration, I agree with you that current math standards in the United States are too easy (for EVERYONE), but I also point out that you may not want to fall into [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]the calculus trap[/url]. Knowing mathematics well has never been about only knowing the meager topics of the standard curriculum; putting the topics together and finding their unity and understanding their details takes extra time and effort beyond school homework, and can't be rushed. \r\n\r\nSee [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=33140&highlight=#33140]my very first post to this site[/url] for my advice on how a young learner should prepare for deep understanding of mathematics. Some of that advice to parents and to young learners can also be applied by high school students.",
  "Solution_2": "Well, pragmatically, it encourages younger students to study more for AIME / USAMO, investing the the future, and leading to a stronger crop of older USAMO qualifiers and MOPpers two years later.  As far as I'm concerned, anything that gets students interested in proofs at a young age is good. :-)\r\n\r\nIt also rewards students who have natural problem-solving ability but have not had the opportunity to be overly accelerated in math, and students who did not get the head start in problem solving practice provided by MathCounts.\r\n\r\n[quote=\"tAsDoOm\"]\nSince everyone Grade 10 and above pretty much has an equal chance at the tests (except the fact that you will be more proficient with more practice) why is it that they should get an easier route?\n[/quote]",
  "Solution_3": "[quote=\"tAsDoOm\"]\nAlso, it should be pointed out that many people who qualify for Olympiad-level tests have skipped grades in math, and not every school allows that. My school, for example, has a no-skip policy unless you skipped in middle school and it's beyond their control. When contestants skip grades, they have obvious advantages. I think the cutoffs, should they exist, should be determined by the COURSE contestants are taking, and not their respective ages or grade levels.[/quote]\r\n\r\nThat's really rather flawed logic. I (for the most part) agree that cutoffs should be determined, to some extent, by course level; however, claiming that people who have skipped grades in math have obvious advantages. I skipped twice (once in grade 2 and once earlier this year) and it's not helped me much. Second, skipping a full GRADE in school (rather than just a year of maths) is very, very [i]dis[/i]advantageous. Not only does it mean one less year of possibly taking the test for the student in question, but they are already being challenged more in math, so they may not have time to study the extracurricular material necessary to do well on tests like the AIME.\r\n\r\nAnd, contrary to what you seem to be saying, there are advantages to age-based cutoffs. First, all 7th, 8th, 9th, 10th graders get in easier without having to elbow their way through the seniors. Second (and this is much more important) IT GIVES THE YOUNGER STUDENTS AN OPPORTUNITY TO TAKE THE TEST [b]BEFORE[/b] they reach their last year of eligibility. If you think this is a bad thing, OK. Just take the USAMO without practicing at all in a test environment and tell me how you do.",
  "Solution_4": "A student who is exceptional only at mathematics may take math classes beyond their grade level, but probably will never officially \"skip a grade\".\r\n\r\nA student who is very good at all subject areas may officially \"skip a grade\".\r\n\r\nIn MATHCOUNTS, for example, we have 8th grade students who are as young as 12 and other 8th grade students who are as old as 15.\r\n\r\nAMC-10 allows students as old as 17.5 years old to compete.  The AMC-12 cutoff is 19.5 years.\r\n\r\nIs it \"fair\" to the mature 12 year old 8th grader, who is good at math and language arts and science, and is taking 8th grade courses to be competing against a 15-year old 8th grader, who is taking advanced high-school math classes.\r\n\r\nResearch shows a good deal of development occurs in the brain between the ages of 12 and 15, especially in the area of reasoning and critical thinking.  [Not new brain cells, but new connections, and significantly faster nerve impulse activity].\r\n\r\nIn a very real sense, MATHCOUNTS (and AMC-8) penalize the 12 year old because they were also exceptionally good at language arts.\r\n\r\nI can also argue against basing eligibility on what grade of math course the student is taking.  Our school district used to offer Geometry to 8th graders.  Should an 8th grader in our district be disallowed from competing in MATHCOUNTS or AMC-8 because they take Geometry at the high school?  Other districts in our state still offer Geometry to 8th graders.  Should they also be disallowed from competing in MATHCOUNTS or AMC-8?\r\n\r\nNote that I am not meaning to pick on MATHCOUNTS or AMC.  MATHCOUNTS is the premier program for middle-school students.  AMC has the premier competitions for high-school students.\r\n\r\nDifferent programs have slightly different rules.  Most nationwide programs have standardized on using the official \"grade\" as the measure to compare students.  I'm sure they understand the disparities.  I think AMC goes out of their way to recognize the younger students who choose to take higher-level tests.  And their testing and advancement structure encourages students to take the AMC-12 over the AMC-10, once they have the coursework covered by the AMC-12.\r\n\r\nYou just have to know and live by the rules the various competitions have established.  The year MATHCOUNTS was changing from a limit of twice per lifetime to three times per lifetime, I joked with the State MATHCOUNTS champion if he wanted to repeat the 8th grade.  (I don't think he was amused by my suggestion.)",
  "Solution_5": "Since it is the grade that determines eligibility for anything, that is what should be considered.  If being younger bars one from entering a program and being in a certain grade bars the same person from participating in a contest it would be terribly unfair to the child who had little choice in the decisions that were made by his teachers and parents.  Go either by grade or by age.  \r\n\r\nIn sports those who practice with older kids because of the training they received still get to compete with their peers in age or grade depending on the type of sport and level of competition.\r\n\r\nIf the math course determined the eligibility, there would be several 10 and 11 year olds who cannot be moppers.  You have to be a teenager to be a mopper and there are students who will finish all school math offerings by the time they are that age."
}
{
  "Tag": [
    "trigonometry",
    "limit",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Hello peoples,\r\n\r\nI think this is a trick question... well sort of :P\r\n\r\n[img]http://img133.imageshack.us/img133/472/picture8ox1.png[/img]\r\n\r\nfor part (a) i get that the cosine Fourier Series for f(x) = cos(x) to be:\r\n\r\n[img]http://img138.imageshack.us/img138/6114/picture9sq2.png[/img]\r\n\r\ni hope that is ok, but its part (b) that is troubling me...\r\n\r\nis all that happens as [img]http://img138.imageshack.us/img138/7436/picture10gf7.png[/img] is that the cosine Fourier series of cos(x) goes to 0?\r\n\r\ni am guessing i am missing some trick to this question?\r\n\r\nCheers! :D\r\n\r\nSarah",
  "Solution_1": "As $\\alpha\\to\\pi$, most of the terms go to zero by inspection.\r\nThe $n=1$ term in the sum deserves closer inspection: what is $\\lim_{\\alpha\\to \\pi^{-}}\\frac{-2\\alpha\\sin\\alpha}{\\alpha^{2}-\\pi^{2}}$?\r\nThat's a $\\frac{0}{0}$ indeterminate limit, and it should be 1.",
  "Solution_2": "ah yep i see now.  i need to be more careful with the n = 1 term when alpha = pi\r\n\r\nthanks for the help! :D\r\n\r\nSarah"
}
{
  "Tag": [
    "function",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "[color=blue] [b] Problem. [/b]\nLet $a, b\\in\\mathbb{R}$ such that $0< a< b$. Find all functions $f: [a, b]\\to [a, b]$ such that\n\n                                                      \\[ f(x)\\cdot f(f(x))=x^{2}, \\]\n for any $x\\in [a, b]$. \n\n [/color]",
  "Solution_1": "Nice problem, here is my solution:\r\n[hide]\nLet's fix $x$ and we put $a_n=f^{(n)}(x)$ ($f^{(n)}(x)=f(f(f(...(f(x))...)$ n times $f^{(0)}(x)=x$)\nWe get $a_{n+2}a_{n+1}=(a_n)^2$\nPut $b_n=ln(a_n)$ then we get\n$b_{n+2}=-b_{n+1}+2b_n$\nso b is linear recurentional sequence. \n$b_n=c_1*1^n+c_2*(-2)^n$\nbut $a_i \\in[a;b]$, so $b_i \\in[ln(a);ln(b)]$\nbut if $c_2\\neq 0$ we can take such $n$ for which $b_n<ln(a)$, so $c_2=0$ and $b_n=k$, $k=const$, so $a_n=const$, but $a_0=x$, so $f(x)=a_1=a_0=x$. So $f(x)=x$. Check the answer. \n[/hide]"
}
{
  "Tag": [
    "AMC",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "Yay, useless statistics! Number of 2006 USAMO qualifiers by the first letter of first name.\r\n\r\nFirst, see if you can predict which letter is the most common, and which letter is the rarest.\r\n\r\n\r\nHere go:\r\n\r\n[hide][code]j -- 61\na -- 43\nd -- 38\ns -- 35\nm -- 34\nc -- 23\nr -- 20\nt -- 19\nk -- 19\np -- 17\ny -- 17\nb -- 14\ng -- 12\nh -- 12\nw -- 12\ne -- 11\nn -- 11\nz -- 8\nv -- 6\ni -- 6\nf -- 5\nx -- 3\nq -- 3\nl -- 3\nu -- 1\no -- 0[/code]\n[/hide]",
  "Solution_1": "I guessed C. Donno why.",
  "Solution_2": "lol i was close\r\ni guessed A cause it was a vowel :)",
  "Solution_3": "I was surprised by the most common letter too.\r\n\r\nBTW, it seems a lot of Alexes are qualifying. Just something I noticed.",
  "Solution_4": "*runs and changes name to Alex*\r\n\r\nnow im ready to qualify for the usamo next year. :)",
  "Solution_5": "Yay. I'm the top statistic.",
  "Solution_6": "Two other $q$'s qualified?   :P  We should form a club.",
  "Solution_7": "i guessed D with no apparent reason.\r\nyou should do a stat with last names. i think C is most common. and \"chen\" might be the most appearing last name.",
  "Solution_8": "yes! my letter is second!",
  "Solution_9": "I guessed J...but I was biased :P",
  "Solution_10": "The k's will rise...",
  "Solution_11": "yes they will\r\nheh",
  "Solution_12": "score! i guessed J for most common, and U for least common... i was close =D\r\n\r\noh, and btw, my name is alex too =D",
  "Solution_13": "watch the Ks get into jusamo by the dozens\r\nheh ehe",
  "Solution_14": "One of these days the C's are going to beat everyone...\r\nYea.  I guessed C.  :)",
  "Solution_15": "Anyone want to look at statistics over multiple years?",
  "Solution_16": "You asked for it: USAMO Qualifiers by First letter of Last name.\r\n\r\n\r\nSee if you can guess the most and least common letters again! This is a trick question this time: Two letters tied for the most common!\r\n\r\n[hide][code]l -- 47\nc -- 47\ns -- 39\nh -- 30\nb -- 25\nk -- 24\nw -- 23\nm -- 20\nz -- 20\ng -- 18\nf -- 17\np -- 15\nt -- 14\ny -- 13\nj -- 12\nr -- 11\nd -- 11\na -- 8\nn -- 8\nv -- 7\ne -- 5\nx -- 5\no -- 5\ni -- 3\nu -- 3\nq -- 1[/code]\n\n\nThere seem to be a lot of Chens taking the USAMO.[/hide]",
  "Solution_17": "Yay! I got a total of 86 (frequency of first initial+frequency of last initial). If only I had a fun name like Oliver Queen...",
  "Solution_18": "[quote=\"randomdragoon\"]\nThere seem to be a lot of Chens taking the USAMO.[/quote]\r\n\r\nNo surprise. Chen is one of the largest family names in China. Also Li, Zhang, Liu... are the largest. :D",
  "Solution_19": "Nice...s is up there",
  "Solution_20": "First names starting with O:\r\nYear Oleg Others\r\n2005: 1 0\r\n2004: 2 0\r\n2003: 1 1\r\n\r\nClearly, Oleg is the best \"O\" name to have, with three individuals qualifying a total of four times since 2003.",
  "Solution_21": "[quote=\"paladin8\"]Yay. I'm the top statistic.[/quote]\r\n\r\nMe too. Apparently J is for Just Too Good.",
  "Solution_22": "[quote=\"jb05\"]Yay! I got a total of 86 (frequency of first initial+frequency of last initial). If only I had a fun name like Oliver Queen...[/quote]\r\n\r\n108 - beat that! :D\r\n\r\nEh, if only I actually made the USAMO...then it would be 110 :P",
  "Solution_23": "[quote=\"joml88\"][quote=\"jb05\"]Yay! I got a total of 86 (frequency of first initial+frequency of last initial). If only I had a fun name like Oliver Queen...[/quote]\n\n108 - beat that! :D\n\nEh, if only I actually made the USAMO...then it would be 110 :P[/quote]\r\n\r\nWow; that's over twice my total (47.)",
  "Solution_24": "Yay for all Alexes making Usamo!\r\nImagine all the winners being Alexes :rotfl:  :rotfl:",
  "Solution_25": "[quote=\"joml88\"][quote=\"jb05\"]Yay! I got a total of 86 (frequency of first initial+frequency of last initial). If only I had a fun name like Oliver Queen...[/quote]\n\n108 - beat that! :D\n\nEh, if only I actually made the USAMO...then it would be 110 :P[/quote]\r\n\r\nYou didn't make it?\r\n\r\nWow..I deserved to make it...but you definitely did much more...",
  "Solution_26": "[quote=\"rem\"]Yay for all Alexes making Usamo!\nImagine all the winners being Alexes :rotfl:  :rotfl:[/quote]\r\n\r\nYes, that's a good idea. Maybe they should make tiebreakers depend on alphabetical ordering of first name.",
  "Solution_27": "yay! arnav > brian > matt > sherry > stephen > yi > zach",
  "Solution_28": "Who is Matt?\r\n\r\nBy the way, Stephen would actually beat you.  :P",
  "Solution_29": "Perhaps Matt McCutchen...",
  "Solution_30": "hehe\r\nJ definitely wins.\r\ni wonder why y and z are so popular",
  "Solution_31": "Maybe it's because it's 3:45 AM and I've been writing code for the last four hours but I found this topic exceedingly amusing.",
  "Solution_32": "[quote=\"Philip_Leszczynski\"]Who is Matt?\n\nBy the way, Stephen would actually beat you.  :P[/quote]\r\n\r\nturns out he did :lol:\r\n\r\ngood thing I didn't put hesterberg though (could anybody beat him?)",
  "Solution_33": "Sadly, yes--looking through the list, there are two Aarons, an Abraham, and possibly another Adam who would beat me. At least I'd make the team :)",
  "Solution_34": "I, too, was a J who made USAMO  :) .\r\n\r\nMy total is 91.... fairly high but not a 108... JC or JL, i can think of a couple (jeffrey chen)",
  "Solution_35": "[quote=\"Go Beyond\"][quote=\"randomdragoon\"]\nThere seem to be a lot of Chens taking the USAMO.[/quote]\n\nNo surprise. Chen is one of the largest family names in China. Also Li, Zhang, Liu... are the largest. :D[/quote]\r\n\r\nWang is still the most common in my opinion, my mom knows at least  20 Wangs (counting each individual family member though)\r\n\r\nHa, my initials are JC.",
  "Solution_36": "Well my guess J was kinda based on the fact that my name, my brother's name, my other brother's name, my mom's name, one of my uncles' names, one of my cousins' names, and a couple other family names all start with J."
}
{
  "Tag": [
    "modular arithmetic"
  ],
  "Problem": "A=2^29\r\n\r\nA has nine digits\r\nWhich digit doesn't involve in A",
  "Solution_1": "[hide]4[/hide]",
  "Solution_2": "[quote=\"JRav\"][hide]4[/hide][/quote]\r\n\r\nHow did you solve?",
  "Solution_3": "The sum of all $ 10$ digits is $ 45\\equiv 0\\pmod{9}$, and $ 2^{29} \\equal{} 2^2\\cdot(2^3)^9\\equiv 4\\cdot( \\minus{} 1)^9\\equiv 5\\pmod{9}$, hence what's lacking from the lot is $ 4$.",
  "Solution_4": "The problem statement should make clearer that $ A$ has nine digits, all of which are distinct.",
  "Solution_5": "If you look closer into the wording, it's clear - a digiT (not digiTS) missing when there's nine of them can only mean all of the nine digits involved are distinct (otherwise there would be digitS missing). Though I agree it could've been made more explicit.",
  "Solution_6": "Similar problem.\r\n\r\nProve that $ 2^{99}$  involves at least one digit which occurs at least 4 times.\r\nThis is also true for $ 2^{98}$ and $ 2^{97}$."
}
{
  "Tag": [
    "logarithms",
    "ratio",
    "integration",
    "calculus",
    "probability and stats"
  ],
  "Problem": "Let $ X_0$ have a uniform distribution on $ [0,1]$, and for $ n \\ge 1$, $ X_{n \\plus{} 1}$ has a uniform distribution on $ [0,X_n]$. \r\nFind the almost sure limit of $ \\frac {\\ln{X_n}}{n}$.",
  "Solution_1": "try to apply the law of large number ..",
  "Solution_2": "Doesn't the law of large numbers usually involve some kind of sum? :maybe:  Can you give more hints how to solve this?",
  "Solution_3": "If $ Y_n$ are iid random variables that are uniformly distributed on $ [0,1]$, then $ X_n\\equal{}\\prod_0^n Y_j$.",
  "Solution_4": ":blush:  I don't understand how I missed that... That's it from me today, back to learnin PDE  :lol: .",
  "Solution_5": "[quote=\"fedja\"]If $ Y_n$ are iid random variables that are uniformly distributed on $ [0,1]$, then $ X_n \\equal{} \\prod_0^n Y_j$.[/quote]\r\nI m not finding it quite obvious. Can anyone explain how is the above true?",
  "Solution_6": "To say $ X_{n \\plus{} 1}$ is uniformly distributed on $ [0, X_n]$ is the same as saying $ X_{n \\plus{} 1} \\equal{} Y_{n \\plus{} 1} \\cdot X_n$ where $ Y_{n \\plus{} 1}$ is uniformly distributed on $ [0, 1]$ (with appropriate independence).",
  "Solution_7": "JBL, can you explain how is $ X_{n\\plus{}1}\\equal{} Y_{n\\plus{}1}\\cdot X_{n}$?",
  "Solution_8": "Define $ Y_n$ to be the ratio $ \\frac {X_{n \\plus{} 1}}{X_n}$.",
  "Solution_9": "Is it due to the following fact?\r\n\r\n$ P\\left(\\frac{X_{n\\plus{}1}}{X_{n}}\\le x\\right)\\equal{}P\\left(X_{n\\plus{}1}\\le x.X_{n}\\right)$\r\n$ \\equal{}\\int_{0}^{1} \\int_{0}^{x_{0}}.\\, .\\, .\\, \\int_{0}^{x_{n\\minus{}1}}\\int_{0}^{x.x_{n}}1.\\frac{1}{x_{0}}.\\frac{1}{x_{1}}.\\,.\\,.\\,\\frac{1}{x_{n}}\\, dx_{0}\\, dx_{1}\\, .\\, .\\, .\\, dx_{n\\plus{}1}\\equal{}x$\r\n\r\nTherefore $ Y_{n}\\equal{} \\frac{X_{n\\plus{}1}}{X_{n}}$ is uniform$ [0,1]$",
  "Solution_10": "That integral doesn't appear to converge.  We are given that $ X_{n \\plus{} 1}$ is uniform in $ [0, X_n]$, which is equivalent to saying that $ \\frac {X_{n \\plus{} 1}}{X_n}$ is uniform in $ [0, 1]$.\r\n\r\n[Minor edit for grammar.]",
  "Solution_11": "Ya I m convinced. Thank you JBL."
}
{
  "Tag": [
    "Columbia",
    "calculus",
    "geometry",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "counting"
  ],
  "Problem": "This is not a post about the positives and negatives of graduating early.  I am fully aware of what it would take to graduate early and the millions of reasons not to do it.  What I'd like to know is how greatly it would affect my chances of getting into a top school such as Columbia which is where I'd like to go if I can get in after junior year.  I am currently in tenth grade, and am greatly considering the fact of graduating after eleventh.  \r\n\r\nI took AP Calculus BC last year and got a score of 5 (with an AB subscore of 5).  I also took AP Physics C Mechanics and got a score of 5.  This year I am taking AP Physics C Electricity and Magnetism, AP World History, AP Chemistry, AP Economics Micro and Macro, and AP Java AB.  Next year I will be taking AP US Government, AP Statistics, AP Biology, AP English Language, AP Latin Literature, and AP Psychology.\r\n\r\nI am taking the PSAT in October and would take the SAT I in the Spring.  I am going to take the SAT II Math IIc and the SAT II Physics as well within the next few months.  At the end of the year, I'd like to take the SAT II World History and SAT II Chemistry.  My GPA is 97.5/100, unweighted (my school does not weight or use class rank).\r\n\r\nLast summer, I did research in geophysics at MIT.  This coming summer, I plan on continuing research except in more of a theoretical physics area hopefully regarding something like Quantum Mechanics or String Theory.\r\n\r\nMy goal is to make the USAMO this year.  I am assistant captain of my school's math team, captain of my school's chess team, a member of my school's mock trial club, a member of my school's Quizbowl club, and technical manager of my school's newspaper.  I have some math and chess awards, and I expect to have some science research awards by the end of this year.\r\n\r\nHow realistic are my chances?  Thanks.",
  "Solution_1": "You're doing many of the right things, but it's really hard to evaluate chances at a school like Columbia with a low base rate of acceptance. A lot of the applicants who don't make it in are hardly distinguishable from those who do. \r\n\r\nI definitely wish you well in trying to wrap up high school early--I would gladly have done the same at your age if only I had prepared as well as you have. I can't predict the chances for acceptance at the most selective schools, but it does seem you are on the right track. \r\n\r\nDo you have a plan B in the event of plan A not working as you desire?",
  "Solution_2": "I haven't read anything above, so I'm not sure if this is the slightest relevant, nontheless its quite interesting.... :lol:  My local university runs an accelarated, high-school programme called \"transition\".  Its a two-year programme that covers 5-years of highschool. ( In Canada 8th Grade is the beginning of high-school). Meaning transition kids end up graduating at the end of 9th grade, and then proceed onto university.\r\n\r\n  Its a pretty intense programme with very little free-time, my math teacher used to teach there and now considers it child-abuse...",
  "Solution_3": "my plan B would be to stick high school out my senior year and apply again to the highly selective schools where i would have even a greater chance of being accepted (since i would be of the regular age).  of course there are 2nd tier schools such as NYU I could apply to as well.  i'm really hoping that i can get into columbia a year early though, because my senior year in high school would be very pointless.  anything i could do to improve my chances?  unfortunately, i don't do sports or anything like that.  i play cello but not at any great level, so i don't think that would be of much help either.",
  "Solution_4": "FWIW -  There are many  schools (including some Ivy )  who will admit  a promising student  who is in Junior year.  \r\n Apart from Columbia and other Ivy leagues, you may like to check out other schools, say USC  (Resident Honors Program) which will    admit  students  who are in their  Junior year (and may  award  a full tution scholarship - if  they are really impressed with you :))  and work with your high shcool so that you get the highschool diploma  too (( to  work-out any remaining requirtemnts etc)... I knew this was there a few years ago, so check out their site, if the program still exists.\r\n(Of course, other places may have similar programs so check them out too)\r\n\r\nAdded later:   Point to note about  these special programs (eg Resident honors program of USC)  is that there are additional resources (eg special housing, (with less freedom than average freshman :D)  , advisers etc) which  are available for those who are coming in early)",
  "Solution_5": "Stanford sent this to me when I emailed then concerning applying early (I am currently a junior) \r\n\r\n[b]\n\nThank you for contacting the Office of Undergraduate Admission at \nStanford \nUniversity. We have admitted younger applicants who were academically \ncompetitive in the past. We do not discriminate against a student based \non \nage, but we will consider both the academic achievement and the \nstudent's \noverall maturity level and ability to adjust to college life regardless \nof \ntheir age.\n\nHowever, we would strongly recommend taking advantage of the full four \nyears to complete high school while making the most of your time both \nacademically and personally. If you like, you might even consider \ntaking \ncourses at a local community college during your final year.\n\nAs we do not schedule appointments, you are welcomed to drop by the \noffice \nor give us a call. Please bear in mind, however, that we will not be \nable \nto review any materials you may bring to the office.\n\nIf you have any additional questions, please feel free to contact our \noffice at (650) 725-5294.\n\nSincerely,\n\nThe Office of Undergraduate Admission\nStanford University\n\n[/b]\r\n\r\nI decided to just have a really slow and easy senior year instead. [/b]",
  "Solution_6": "It's a bit like the folklore college essay listing activities such as cancer research, international peace negotiations, felling charging elephants with a single shot between the eyes, and so on.   \r\n\r\nThe set of AP exams above is, obviously, overkill for a school at the level of Columbia, especially in tenth grade and with 1 or 2 years left to prolong the list.   Barring a criminal record, or major deficiencies in your grades or bank account, admission is more or less a given.\r\n\r\nBear in mind, though, that  \"research in geophysics\" will be decoded by anyone with a clue as \"interned at a lab\", and that the chances of doing research in string theory at this level (i.e. anything beyond working menial calculations for the actually qualified) are very, very low.  It is generally advisable to discuss such things in realistic, neutral, objective and factual terms rather than exaggerations or speculations about the future, and that is especially true in the context of accelerated college admission, where maturity is a specific consideration.",
  "Solution_7": "You sound way more prepared than I ever was for college.\r\nI'd like to add that in the case that you do not get in after your junior year, you might still be able to take classes at Columbia during your senior year. I took Introduction to Number Theory during the fall term of my senior year at Columbia, with the help of my school's math department assistant principal. Just check with someone in your school.\r\nThe experience was worthwhile, even though I was not officially enrolled in the class and do not plan on claiming credit. You get a feel for the school, know what you might be getting into. Doing something like this can show the colleges that you have the maturity and time management necessary to succeed. In fact, with your credentials, I'd say you might be able to do it next year, and further improve your chances of getting in early."
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection"
  ],
  "Problem": "Pretend that you are camping.  While you are searching for natural berries, you see that your tent is on fire! :) \r\nYou are 200 feet from the river, you're tent is 240 feet downstream from you're current location, and the tent is 120 ft. away from the river bank.  Thankfully, you have a bucket with you.  You must run to the river and to your tent.\r\n\r\n1) What is the shortest distance?\r\n2) How far did you run with the pail full and pail empty?",
  "Solution_1": "[hide]\n\nLet us reflect the image off the river bank. Let A be your current location, B the tent, and B' the reflection. So, we need to find AB'. We have a right triangle with legs 320 and 240 (the distance from the riverbank to A and distance from riverbank to B'; distance tent is downstream). So, the distance is $ \\sqrt{[8(40)]^2\\plus{}[6(40)]^2}\\equal{}\\sqrt{40^2(8^2\\plus{}6^2)}\\equal{}400$ \n\nLet the river be CD, where C is closer to A and D is closer to B. Also, let the intersection of the river and AB' be X. We have $ \\triangle{ACX}\\sim\\triangle{B'DX}$. So, $ \\frac{200}{120}\\equal{}\\frac{5}{3}\\equal{}\\frac{AX}{400\\minus{}AX}$. So, AX=distance without water=250. This distance with water is 400-250=150.\n\n[/hide]"
}
{
  "Tag": [
    "geometry",
    "geometry proposed"
  ],
  "Problem": "Point P lies on side AB of a convex quadrilateral ABCD. Let ! be the incircle of triangle\r\nCPD, and let I be its incenter. Suppose that ! is tangent to the incircles of triangles APD\r\nand BPC at points K and L, respectively. Let lines AC and BD meet at E, and let lines\r\nAK and BL meet at F. Prove that points E, I, and F are collinear.",
  "Solution_1": "see here: http://www.mathlinks.ro/viewtopic.php?p=1186805#1186805"
}
{
  "Tag": [
    "ARML",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "AIME"
  ],
  "Problem": "So like I'm going with the grissom team and so is dynamo. hey I CAN MEET DYNAMO (and ash and evil ed i think). Cool. Hopefully I shall get more than 50% of the questions in individual right this time. hmmm I get the feeling I have a room to myself, since there are 17 people. GO TEAM GRISSOM!\r\n\r\nThis shall be fun. :) I am very excited.\r\n\r\nOh snap if I had made usamo grissom could not be in division B, or I could not go. or we would be in division A, according to the email.",
  "Solution_1": "Haha. I found out I was on ARML this week. I did one practice and got a 3/8  :dry: So if this post contains ARML stupidity it's because idk anything about ARML. \r\nJust wondering, but why are you going with grissom? and do you know, since there are 17 of us, which two are not on the team? \r\nhey I CAN MEET XINKE! The all-famous Xinke...  :o  You won't be impressed by me, I'm not smart, and... I'm white :(\r\nyes, you will meet ash and sid. They are not like each other except for being brown, short, and geniuses so don't expect them to be the same. \r\nhaha yeah go Grissom. weird, 3 of our people dont even go to Grissom. \r\nim excited about the competiton cuz its experience, but i'll weigh us down so i'm nervous.\r\nhow close were you to usamo?",
  "Solution_2": "11 points by amc 10. Horrendous amount of points by amc 12. I need to up my average practice aime score to 12 (I should stop dreaming) since like my real aime score both years have been 2 or 3 less than average practice aime scores....\r\n\r\n\r\n\r\nhmmmm so like i wanted to go to arml and I told Mr. Crawford, and he said Grissom might send a team, and he thought that maybe I could go with grissom, and it worked out well from there.\r\n\r\nHmmmm, you have practiced more for arml than I have this year..... :o \r\n\r\nhmmm you shall be less impressed by me. I like to do problems very slowly, if at all possible, to triple check everything, and the only reason i am ok at math is cuz I kind of work hard at it. I consider you and tony and botong smarter than me.\r\n\r\nby the way they are changing arml rules this year to make it 10 problems instead of 8.",
  "Solution_3": "can i go too? ask your coach dynamo...",
  "Solution_4": "I would have asked you if you wanted to go if the grissom teacher had said that she had more room on the team, but right now, there's already a full team and 2 alternates. Also, registration ended 2 days ago. Sorry."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algorithm",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Given an even degree polynomial in $ \\mathbb{R}[x]$, is there any deterministic way of finding out if it has a real root? ( like we use the discriminant for degree 2 polynomials)",
  "Solution_1": "please see\r\n\r\n[url]http://en.wikipedia.org/wiki/Root-finding_algorithm[/url]",
  "Solution_2": "[url=http://en.wikipedia.org/wiki/Sturm%27s_theorem]Sturm's theorem[/url] is even more relevant here.",
  "Solution_3": "Thanks.\r\n\r\nBut more than an algorithm I was looking for a condition involving the coefficients which would give a conclusion. Rather than evaluating the polynomial at some points.",
  "Solution_4": "I would hazard a guess that Sturm's theorem is the best you're going to get. Be satisfied with that.\r\n\r\nWLOG, assume that the leading coefficient of your $ 2n$ degree polynomial is positive. You're asking whether $ p(x)$ is ever negative. You could phrase that as asking about the minimum value of $ p(x).$ Where is that minimum located? At one of the roots of $ p'(x),$ which is a $ 2n\\minus{}1$ degree polynomial. Can you find the roots of $ p'(x)$ by \"a condition involving the coefficients which would give a conclusion\"? Let's put it this way: you do know about the Abel-Galois impossibility results for the solvability of polynomials of degree 5 and above?\r\n\r\nNow if you want to se if you can find something specifically for the fourth degree case, go look at the Tartaglia-Cardano-Ferrari formulas and see what you can do."
}
{
  "Tag": [
    "search",
    "LaTeX",
    "trigonometry",
    "Olimpiada de matematicas"
  ],
  "Problem": "me disculpo de entrada por que pues no me he puesto a mirar como funciona y por ende no se como usar el latex.el caso es que \r\n si x pertenece al intervalo cerrado cero, pi medios donde x es un angulo dado en radianes demostrar que \r\n\r\n     cos(cosx)>=sex\r\n\r\nnuevamente les pido disculpas por no usar el latex.",
  "Solution_1": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=317043477&t=279103]contestaci\u00f3n[/url]\r\n\r\nEn latex:\r\n\r\nsi $ x$ pertenece al intervalo $ [0,\\frac {\\pi}{2}]$ donde $ x$ es un angulo dado en radianes demostrar que:\r\n \r\n$ \\cos(\\cos(x)) \\geq \\sin(x)$."
}
{
  "Tag": [
    "AMC",
    "AIME",
    "AMC 8",
    "AMC 10"
  ],
  "Problem": "Was the 2003 AMC 10B just really easy? Over winter break, I have averaged about 130 on the AMC 10. This evening, on the AMC 10 test I just did, I got a 145.5, not answering one of them from misreading. Was this to be expected from the difficulty or did I suddenly get smarter?",
  "Solution_1": "Try comparing the AMC 10 statistics.  [url=http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2003-10a/03stats10.html]Here is the 2003 AMC 10/12 statistics.[/url]  The AIME cutoff was 121, which is higher than the usual 120, so I would assume it was slightly easier.",
  "Solution_2": "Same here, I just took it and got a 144, but I usually average a 130-135.",
  "Solution_3": "I just took the 2003 12A and 12B and scored the same as my AMC 10 average, i just think 2003 was a lot easier than all the other years for some odd reason",
  "Solution_4": "I also remember hearing that the 2003 AIME was ridiculously easy compared to other years, so maybe the MAA decided to just make everything easier that year.",
  "Solution_5": "I think that year was slightly easier because my test scores were higher that year.",
  "Solution_6": "2003 amc8 was average though",
  "Solution_7": "Ya, i think it was easier :P\r\nCompare to other people s score",
  "Solution_8": "hmmm... Try the 2004 AMC10 B... It was quite a bit tougher than the 2003 one.",
  "Solution_9": "Um, now, is the AIMe cutoff 120 or higher (100 or higher for AMC 12) or is it 1% (5% for AMC 12)",
  "Solution_10": "[quote=\"AwesomeToad\"]Um, now, is the AIMe cutoff 120 or higher (100 or higher for AMC 12) or is it 1% (5% for AMC 12)[/quote]\r\n\r\nIt's either one, I believe.",
  "Solution_11": "[quote=\"batteredbutnotdefeated\"][quote=\"AwesomeToad\"]Um, now, is the AIMe cutoff 120 or higher (100 or higher for AMC 12) or is it 1% (5% for AMC 12)[/quote]\n\nIt's either one, I believe.[/quote]\r\n\r\nWhichever is more less.",
  "Solution_12": "What do you mean \"more less\"? which one gets most people?",
  "Solution_13": "[quote=\"AwesomeToad\"]What do you mean \"more less\"? which one gets most people?[/quote]\r\n\r\nWhichever has a lesser score.  If the top 1% was people with scores of 140+; then 120+ would be cutoff.  If top 1% was 100+, then I think 100+ would be cutoff."
}
{
  "Tag": [
    "algebra",
    "polynomial"
  ],
  "Problem": "I was looking for roots for this fraction to split it into as many factors as possible.\r\n\r\n$f(x)=\\frac{2x^4+x^3-5x^2+8x-3}{x^3+x^2-2x+3}$\r\n\r\nFirst of i found out that $0.5$ was a root to the nominator.\r\n\r\n$f(x)=\\frac{(x-0.5)(2x^3+2x^2-4x+6)}{x^3+x^2-2x+3}$\r\n\r\n...but from here i couldn't split it further. Maybe some of you can help me one step further.\r\n\r\nThx in advance for your help.",
  "Solution_1": "[hide=\"very big hint\"]try factoring that 2 out of the second polynomial in the numerator[/hide]",
  "Solution_2": "lol, this was really great. Just two seconds after i posted the thread i found the solution to my problem (well, quite simple way when you take a little more time instead of following the book's guidelines). I clicked the delete button but i couldn't deleted my own thread since it is only allowed by the moderators. Then i wrote a reply to answer the result to my own problem and found that in the same time i wrote the reply my internet connection was gone :lol: Didn't help with a restart of my comp. Here is what i wrote in the reply (which i saved):\r\n\r\n------------------------------------------------------------------------------------\r\n\r\nOK, i can't delete it, but now i think i know how to handle this. Plz tell me if this is right, thx.\r\n\r\n$f(x)=\\frac{(x-0.5)(2x^3+2x^2-4x+6)}{x^3+x^2-2x+3}$\r\n\r\n$f(x)=\\frac{2(x-0.5)(2x^3+2x^2-4x+6)}{2(x^3+x^2-2x+3)}$\r\n\r\n$f(x)=2(x-0.5)$\r\n\r\n$f(x)=2x-1$\r\n\r\n------------------------------------------------------------------------------------\r\n\r\nWith the exception that i first timed it up by 2 and then removed the equal fractions instead of just dividing with 2 i seem to have done everything right. Thx for your reply though i didn't need it, but thx anyway (how would you know) ;)"
}
{
  "Tag": [
    "probability",
    "algebra",
    "function",
    "domain"
  ],
  "Problem": "for some who is interested in math, would you suggest acturial science or being an analyst or trader at wall street\r\n\r\nalso which is more high paying?",
  "Solution_1": "If you want to make ungodly sums of money, become a successful investment banker with a big company.  Of course, this requires an MBA.",
  "Solution_2": "[quote=\"JRav\"]If you want to make ungodly sums of money, become a successful investment banker with a big company.  Of course, this requires an MBA.[/quote]\r\nPretty much no Wall Street type jobs [i]require[/i] MBAs anymore. Also, the only way this strategy works is if you're willing to do that for your whole life. Those people at Merrill Lynch or whatever whose ratings of particular stocks can cause huge swings in those stocks once announced have spent decades moving up in those companies. If you want to get rich quick (not super quick but faster than at a big i-bank), work for a hedge fund. \r\n\r\nIf you like programming, being a Wall Street quant type person is probably about the best you can do and stay in the \"math-analytical\" sort of realm. Being an i-banker or a prop trader is not nearly as analytical a profession.",
  "Solution_3": "Wel.....i wanna be an actuary co mathematician is that possible?",
  "Solution_4": "How much do actuaries, quants, etc. get paid?",
  "Solution_5": "[quote=\"123456789\"]How much do actuaries, quants, etc. get paid?[/quote]\r\nIt varies quite a bit, but generally a lot. This should not be your sole motivation for working such a job, though. (Probably, if it is your sole motivation, you won't get one of the best jobs.)",
  "Solution_6": "Just my opinion- having worked for a large (top 10) P&C insurance company.  It was not always the most interesting, but it had it's moments.  What I will say is that the pay was good (10 years ago I started at $ \\$$44K with no exams passed) and the hours were very good - the only extra time I put in was studying for exams. When I decided not to pursue exams, which was fairly accepted in my department, it definitely became a job worth having with a young family. As for making money - they paid$ \\$$2k automatic raise for each exam passed.  If you passed the exam on the first try, there was an added bonus.  There were a few young gung-ho types (under 26) who passed all the exams first time and I know their salary was at least close to $ \\$$100k. I have had a few friends work as investment brokers and they said the money was insane (one claimed over$ \\$$1 mil in a year before bonuses) but the hours could be trying as well.  He basically said there was only one relationship in his life.  But he is in his late 30's and essentially retired.",
  "Solution_7": "The previous poster's discussion of salary is totally consistent with my own personal knowledge of actuary salaries.  Actuaries make a respectable amount of money, but to put it bluntly, it's a pittance compared to what people with comparable math skills are able make in other jobs such as investment banking.  (Career actuaries have to be fairly smart people since they have to pass a series of grueling exams.)  It's also less than what you could make in computer programming or engineering.  On the other hand, actuaries don't have to work particularly hard and have good job security.",
  "Solution_8": "Here is some salary data on actuaries:\r\n\r\n[url]http://www.actuaryjobs.com/salary.html[/url]",
  "Solution_9": "It seems that actuaries stay within probability and statistical application.  Does this field offer much in terms of expanding one's knowledge mathematically?  In other words, is it more of a \"business\" career, or a mathematic career?",
  "Solution_10": "[quote=\"mr.affable\"]It seems that actuaries stay within probability and statistical application.  Does this field offer much in terms of expanding one's knowledge mathematically?  In other words, is it more of a \"business\" career, or a mathematic career?[/quote]\r\n\r\nI think it is more of a business career. Probably you know most of the math but rather have to get used to certain models, assumptions and other domain knowledge.",
  "Solution_11": "I heard actuaries have better job satisfaction? If you're an investment banker or analyst you might only get to sleep from 2am to 7am, and bankers have to work very long hours",
  "Solution_12": "i interned with quants this summer. I mean they know math but not like we do.....(problem solving)",
  "Solution_13": "I think that job satisfaction depends on the person.. everyone likes different things! You cannot conclude that every banker hates his job and has less job satisfaction than the actuaries ;)"
}
{
  "Tag": [
    "LaTeX",
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Determine all integer solutions of the equation $ x^n\\plus{}y^n\\equal{}1994$,where $ n \\geq 2$",
  "Solution_1": "Obviously, $ x,y$ are both odd. It's easy to conclude that $ n\\le3$ because $ 5^5 > 1994$ and $ 2.3^5 < 1994$ and the case $ n \\equal{} 4$ follows easily from Fermat's little theorem.Now we need to check 2cases, which are easy to do.\r\n[hide=\"Some hints\"](the main idea in the case $ n \\equal{} 3$ is to use modulo 7 and conlude that one of the numbers is divisible by $ 7$, say $ x$, and since $ 14^3 > 1994$, so $ x \\equal{} 7$ but then $ y^3 \\equal{} 1651$-impossible)[/hide]",
  "Solution_2": "But x and y need not be positive, right?",
  "Solution_3": "in that case we shall have to solve two equations\r\n$ x^n\\minus{}y^n\\equal{}1994$ and $ x^n\\plus{}y^n\\equal{}1994$\r\n\r\nthe first one is only possible if n is odd cuz else it ll lead to second..\r\nexpand it...ll post the solution as soon as poss..\r\nin a hurry..!",
  "Solution_4": "I don't understand. Why n needs to be smaller than 3 or equal ? \r\n\r\nIf I write : 3^5 < 1994 and 5^5 > 1994 ?",
  "Solution_5": "I missed one case because I thought it's obvious. It's true because we know that x,y are both odd and since $ 5^5>1994$ then they should be 1 or 3. Now, I forgot to write the fact that $ 3^6>1994$ which implies that there aren't any solution for $ n\\ge 5$. But, as I mentioned, the case $ n\\equal{}4$ is impossible because of the Fermat Little Theorem and so $ n\\le3$.",
  "Solution_6": "[size=150]x^2=4a+1=3b+1, thus a=3k, x^2=12h, for some natural h, similarly for b, but 12 does not divide 1994, hence no solutions.....[/size]\r\n\r\nCan someone give me the link for using latex here?[/url]",
  "Solution_7": "for latex, simply put \\$ sign in between text",
  "Solution_8": "if $ n$ even, so simply\r\nif $ n$ odd, then we can assume $ n$ be a prime odd.\r\nSo $ x \\plus{} y|1994$ and $ x \\plus{} y$ be a positive integer even.\r\n$ \\rightarrow x \\plus{} y \\equal{} 2,1994$\r\n1)Case $ x \\plus{} y \\equal{} 1994$\r\nWe have $ x^n \\plus{} y^n \\equal{} (x \\plus{} y)(x^{n \\minus{} 1} \\plus{} x^{n \\minus{} 2}y \\plus{} . . . \\plus{} xy^{n \\minus{} 2} \\plus{} y^{n \\minus{} 1}) > x \\plus{} y \\equal{} 1994$.\r\n2)Case $ x \\plus{} y \\equal{} 2$, assume $ x > 0$.\r\n$ \\rightarrow n|1992$.\r\n$ \\rightarrow n \\equal{} 3,83$.\r\n$ x^n \\minus{} (x \\minus{} 2)^n \\equal{} 1994$.\r\nIf $ x \\minus{} 2\\leq 0$, impossible.\r\nIf $ z \\equal{} x \\minus{} 2 > 0$, we have:\r\n$ 1994 \\equal{} (z \\plus{} 2)^n \\minus{} z^n\\geq 2^n$, so $ n \\equal{} 3$ and we have equation:\r\n$ (z \\plus{} 2)^3 \\minus{} z^3 \\equal{} 1994$ has no root on $ Z$.",
  "Solution_9": "[quote=\"Thjch Ph4 Trjnh\"]We have $ x^n \\plus{} y^n \\equal{} (x \\plus{} y)(x^{n \\minus{} 1} \\plus{} x^{n \\minus{} 2}y \\plus{} . . . \\plus{} xy^{n \\minus{} 2} \\plus{} y^{n \\minus{} 1}) > x \\plus{} y \\equal{} 1994$.[/quote] \r\n\r\nActually, for odd $ n$: $ x^n \\plus{} y^n \\equal{} (x \\plus{} y)(x^{n \\minus{} 1} \\minus{} x^{n \\minus{} 2}y \\plus{} . . . \\minus{} xy^{n \\minus{} 2} \\plus{} y^{n \\minus{} 1})$ (alternating signs). Although, it doesn't really make a difference.\r\n\r\nAnyway, here is a complete solution: \r\n[hide]WLOG, let $ x \\ge y$. If $ x \\equal{} y$, then $ 2x^n \\equal{} 1994 \\Rightarrow x^n \\equal{} 997$, which is impossible. So, $ x > y$. \n\nIf $ x$ and $ y$ are even, then $ 2^n | x^n \\plus{} y^n \\equal{} 1994$, which is a contradiction since $ n \\ge 2$ and $ 4 \\nmid 1994$. \n\nIf one of $ x,y$ is even, and the other is odd, then $ x^n \\plus{} y^n \\equal{} 1994$ is odd, a contradiction. \n\nThus, $ x,y$ are both odd.  \n\nCase I: $ n$ is even. \nThen, $ x^n , y^n > 0$. Also, $ (x,y)$ is a solution iff $ (\\pm x, \\pm y)$ are all solutions. So WLOG, let $ x > y > 0$. \n\nIf $ n \\ge 8$, and $ x \\ge 3$, then $ x^n \\plus{} y^n \\ge 3^8 \\equal{} 6561$, a contradiction. \nThus, $ 1 \\le y < x \\le 1$, a contradiction. \nTherefore, there are no solutions for even integers $ n \\ge 8$\n\nIf $ n \\equal{} 6$ and $ x \\ge 5$, then $ x^6 \\plus{} y^6 \\ge 5^6 \\equal{} 15625$, a contradiction. \nThus, $ 1 \\le y \\le x \\le 3 \\Rightarrow x^6 \\plus{} y^6 \\le 2 \\cdot 3^6 \\equal{} 1458$, a contradiction. \nTherefore, there are no solutions for $ n \\equal{} 6$. \n\nIf $ n \\equal{} 4$ and $ x \\ge 7$, then $ x^4 \\plus{} y^4 \\ge 7^4 \\equal{} 2401$, a contradiction. \nThus, $ 1 \\le y < x \\le 5 \\Rightarrow x^4 \\plus{} y^4 \\le 2 \\cdot 5^4 \\equal{} 1250$, a contradiction. \nTherefore, there are no solutions for $ n \\equal{} 6$.  \n\nIf $ n \\equal{} 2$ and $ x \\ge 45$, then $ x^2 \\plus{} y^2 \\ge 45^2 \\equal{} 2025$, a contradiction. \nThus, $ 1 \\le y < x \\le 43$.\nNote that $ 2y^2 \\le x^2 \\plus{} y^2 \\equal{} 1994 \\le 2x^2$ \nTherefore, $ 33 \\le x \\le 43$ AND $ 1 \\le y \\le 31$. \nTesting all $ 11$ possible values of $ x$ yields $ (x,y) \\equal{} (37,25)$ as the only possible solution with $ x > y > 0$. \nThus, the only solutions for $ n \\equal{} 2$ are $ (x,y) \\equal{} (\\pm 37 , \\pm 25) ; (\\pm 25 , \\pm 37)$.\n\nCase II: $ n$ is odd. \nObviously, either $ x > y > 0$ or $ x > 0 > y$, and $ |x| > |y|$. \n\nSince $ |x|,|y|$ are both odd integers, $ |x| \\ge |y| \\plus{} 2$.\nSince $ x^n \\plus{} y^n \\equal{} 1994$, we must have $ |x|^n \\minus{} |y|^n \\le 1994$. \nSo, we have $ (|y| \\plus{} 2)^n \\minus{} |y|^n \\le |x|^n \\minus{} |y|^n \\le 1994$. \n\nIf $ n \\ge 7$, then we have: $ (|y| \\plus{} 2)^n \\minus{} |y|^n \\ge 3^7 \\minus{} 1^7 \\equal{} 2186$, a contradiction. \nThus, there are no solutions for odd $ n \\ge 7$. \n\nIf $ n \\equal{} 5$, and $ |y| \\ge 3$, then we have: $ (|y| \\plus{} 2)^5 \\minus{} |y|^5 \\ge 5^5 \\minus{} 3^5 \\equal{} 2882$, a contradiction. \nTherefore, we must have $ |y| \\equal{} 1$, and thus, $ x^5 \\pm 1 \\equal{} 1994$ which has no integer solution for $ x$. \nThus, there are no solutions for $ n \\equal{} 5$. \n\nIf $ n \\equal{} 3$, and $ |y| \\ge 19$, then we have: $ (|y| \\plus{} 2)^3 \\minus{} |y|^3 \\ge 21^3 \\minus{} 19^3 \\equal{} 2402$, a contradiction. \nTherefore, we must have $ |y| \\le 17$. \nSince $ y < x$, we have $ 2y^3 < x^3 \\plus{} y^3 \\equal{} 1994 \\Rightarrow y \\le 9$. \nTherefore, $ \\minus{} 17 \\le y \\le 9$. Testing all $ 13$ possible values of $ y$ yield no integer solutions for $ x$. \nThus, there are no solutions for $ n \\equal{} 3$.  \n\nTherefore, the only integer solutions to $ x^n \\plus{} y^n \\equal{} 1994$ with $ n \\ge 2$ are: \n$ (x,y,n) \\equal{} (\\pm 37 , \\pm 25 , 2); (\\pm 25 , \\pm 37 , 2)$. [/hide]",
  "Solution_10": "[quote]\nWe have $ x^n \\plus{} y^n \\equal{} (x \\plus{} y)(x^{n \\minus{} 1} \\plus{} x^{n \\minus{} 2}y \\plus{} . . . \\plus{} xy^{n \\minus{} 2} \\plus{} y^{n \\minus{} 1}) > x \\plus{} y \\equal{} 1994$.\n[/quote]\r\nSorry.:D\r\n$ x^n \\plus{} y^n \\equal{} (x \\plus{} y)(x^{n \\minus{} 1} \\minus{} x^{n \\minus{} 2}y \\plus{} . . . \\minus{} xy^{n \\minus{} 2} \\plus{} y^{n \\minus{} 1})$.\r\n$ 2(x^{n \\minus{} 1} \\minus{} x^{n \\minus{} 2}y \\plus{} . . . \\minus{} xy^{n \\minus{} 2} \\plus{} y^{n \\minus{} 1}) \\equal{} x^{n \\minus{} 1} \\plus{}x^{n \\minus{} 3}(x \\minus{} y)^2 \\plus{} . . . \\plus{}y^{n \\minus{} 3}(x \\minus{} y)^2 \\plus{} y^{n \\minus{} 1}\\geq x^{n \\minus{} 1} \\plus{} y^{n \\minus{} 1} \\geq |x| \\plus{} |y|\\geq 1994$",
  "Solution_11": "[quote=\"soumik\"][size=150]x^2=4a+1=3b+1, thus a=3k, x^2=12h, for some natural h, similarly for b, but 12 does not divide 1994, hence no solutions.....[/size]\n[/quote]\r\n\r\nthis is wrong soumik",
  "Solution_12": ":D I got a bit carried away....."
}
{
  "Tag": [],
  "Problem": "\u039b\u03bf\u03b9\u03c0\u03cc\u03bd , \u03b8\u03b1 \u03c0\u03ad\u03c3\u03b5\u03c4\u03b5 \u03b1\u03c0\u03cc \u03c4\u03b1 \u03c3\u03cd\u03bd\u03bd\u03b5\u03c6\u03b1 !\r\n\r\n  \u039f\u03b9 \u03c4\u03bf\u03cd\u03c1\u03ba\u03bf\u03b9 , \u03bc\u03cc\u03bb\u03b9\u03c2 \u03ad\u03bc\u03b1\u03b8\u03b1\u03bd \u03cc\u03c4\u03b9 \u03c3\u03c4\u03b7\u03bd \u0395\u03bb\u03bb\u03ac\u03b4\u03b1 \u03ba\u03c5\u03ba\u03bb\u03bf\u03c6\u03bf\u03c1\u03b5\u03af \u03bc\u03b5  25 \u03b5\u03c5\u03c1\u03ce \u03c4\u03bf IMO COMPENDIUM, \u03c0\u03b1\u03c1\u03ae\u03b3\u03b3\u03b5\u03b9\u03bb\u03b1\u03bd - \u03c0\u03cc\u03c3\u03b1 \u03bb\u03ad\u03c4\u03b5 ; -  [color=red]20000 [/color]\u03b1\u03bd\u03c4\u03af\u03c4\u03c5\u03c0\u03b1 !!! \u039f\u03b9 \u0388\u03bb\u03bb\u03b7\u03bd\u03b5\u03c2 - \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03af \u03ba\u03b1\u03b9 \u03bc\u03b1\u03b8\u03b7\u03c4\u03ad\u03c2 - \u03b1\u03b3\u03cc\u03c1\u03b1\u03c3\u03b1\u03bd \u03bb\u03b9\u03b3\u03cc\u03c4\u03b5\u03c1\u03b1 \u03b1\u03c0\u03cc 10 \u03b1\u03bd\u03c4\u03af\u03c4\u03c5\u03c0\u03b1.\r\n                \u0395\u03af\u03c0\u03b1\u03c4\u03b5 \u03c4\u03af\u03c0\u03bf\u03c4\u03b1 ;;;\r\n\r\n    \u039c\u03c0\u03ac\u03bc\u03c0\u03b7\u03c2",
  "Solution_1": "\u03a4\u0399 \u039d\u0391 \u03a0\u0395\u0399\u0399\u03a3 ? \u0395\u0394\u03a9 \u0394\u0395\u039d \u03a7\u03a9\u03a1\u039f\u03a5\u039d \u039b\u039f\u0393\u0399\u0391 \u03a0\u0391\u03a1\u0391 \u039c\u039f\u039d\u039f \u03a4\u039f \u039c\u03a5\u039d\u0397\u039c\u0391 \u039d\u0391 \u03a0\u03a1\u039f\u03a4\u03a1\u0395\u03a8\u039f\u03a5\u039c\u0395 \u039f\u039b\u039f\u03a5\u03a3 \u03a4\u039f\u03a5\u03a3 \u039c\u0391\u0398\u0397\u03a4\u0395\u03a3 \u039a\u0391\u0399 \u03a4\u039f\u03a5\u03a3 \u039c\u0391\u0398\u0397\u039c\u0391\u03a4\u0399\u039a\u039f\u03a5\u03a3 \u039d\u0391 \u0391\u0393\u039f\u03a1\u0391\u03a3\u039f\u03a5\u039d \u03a4\u039f \u0392\u0399\u0392\u039b\u0399\u039f  \u0391\u039b\u039b\u0391 \u03a4\u0399 \u039d\u0391 \u0393\u0399\u039d\u0395\u0399 ? \u0394\u0395\u039d \u0391\u03a6\u039f\u03a1\u0391 \u0397 \u03a0\u03a1\u039f\u039f\u0394\u039f\u03a3 \u03a4\u0397\u03a3 \u0395\u039b\u039b\u0391\u0394\u0391\u03a3 \u03a3\u03a4\u0391 \u039c\u0391\u0398\u0397\u039c\u0391\u03a4\u0399\u039a\u0391 \u03a9\u03a3 \u039f\u039c\u0391\u0394\u0391  :mad:  \u03a0\u03a1\u039f\u0397\u0393\u0395\u0399\u03a4\u0391\u0399 \u0392\u039b\u0395\u03a0\u03a4\u0395 \u0397 \u0393\u0399\u039f\u03a5\u03a1\u039f\u0392\u0399\u0396\u0399\u039f\u039d  :mad:  :mad: \r\n\u03a7\u0391\u0399\u03a1\u0395\u03a4\u0391\u0399"
}
{
  "Tag": [
    "trigonometry",
    "inequalities"
  ],
  "Problem": "In $ \\triangle ABC$\r\n\r\n$ AB=t^2, AC=t, BC=1,t>1$\r\n\r\n$ \\angle A=x^o,\\angle C=x^o+60^o,\n\\angle B=120^o-2x^o$\r\n\r\nProve that  $ t <\\sqrt{2} .$",
  "Solution_1": "[hide]Here's my approach which pops up from my head:\nApplying laws of cosines we have: AB^2=1+t^2-2t.cosACB.Therefore, we have:t^4-1-t^2+2t.cosACB=0(3)\nNow, let's calculate cosACB. We see that cos(ACB)=sin(90-ACB)=sin(30-x)(2). In addition we also have:sinx/1=sin(x+60)/t, therefore,t=sin(x+60)/sinx.=cos(30-x)/sinx(1)\nPlugging (1) and (2) into (3) and bashing with algebra, we can find the value of x. From that, we can calculate the value of t, which will be<sqrt(2). \nOr another way to do with it is solving the inequality: cos(30-x)/sinx>1 to get:\nsqrt(3)/2.cosx>1/2.sinx, which means cotg(x)>1/sqrt(3).Using this bound to prove that: [sqrt(2)-1/2].sinx>sqrt(3)/2.cosx, which means: sqrt(3)/(sqrt(2)-1/2)<tg(x) with 0<x<120 is not hard.[/hide]",
  "Solution_2": "Have anyone some other idea?",
  "Solution_3": "hello, the theorem of cosine gives $ \\cos(x)=\\frac{t^2+t^4-1}{2t^3}$ (1), the theorem of sine gives $ \\frac{\\sin(x)}{\\sin(x+60^{\\circ})}=\\frac{1}{t^2}$ (2).\r\nFrom (2) we get $ x=\\arctan(\\frac{\\sqrt{3}}{2t{}^2{}^-1})$, substituting in (1) we have $ \\frac{1}{\\sqrt{1+\\frac{3}{(2t^2-1)^2}}}=\\frac{1}{2}\\frac{t^2+t^4-1}{t^3}$. Solving by $ t$ we have ${ t=\\sqrt{\\frac{1}{4}+\\frac{\\sqrt{13}}{4}+\\frac{\\sqrt{-2+2\\sqrt{13}}}{4}}}<\\sqrt{2}$ as desired.\r\nSonnhard.",
  "Solution_4": "[quote=\"Dr Sonnhard Graubner\"]hello, the theorem of cosine gives $ \\cos(x) = \\frac {t^2 + t^4 - 1}{2t^3}$ (1), the theorem of sine gives $ \\frac {\\sin(x)}{\\sin(x + 60^{\\circ})} = \\frac {1}{t^2}$ (2).\nFrom (2) we get $ x = \\arctan(\\frac {\\sqrt {3}}{2t^2^ - 1})$, substituting in (1) we have $ \\frac {1}{\\sqrt {1 + \\frac {3}{(2t^2 - 1)^2}}} = \\frac {1}{2}\\frac {t^2 + t^4 - 1}{t^3}$. Solving by $ t$ we have ${ t = \\sqrt {\\frac {1}{4} + \\frac {\\sqrt {13}}{4} + \\frac {\\sqrt { - 2 + 2\\sqrt {13}}}{4}}} < \\sqrt {2}$ as desired.\nSonnhard.[/quote]\r\nSorry Dr. Sonnhard Graubner! I think your solution was exactly the same as what I did in my post before. You should give credit to me, Dr. Graubner. :roll:",
  "Solution_5": "[quote=\"Dr Sonnhard Graubner\"]hello, the theorem of cosine gives $ \\cos(x) = \\frac {t^2 + t^4 - 1}{2t^3}$ (1), the theorem of sine gives $ \\frac {\\sin(x)}{\\sin(x + 60^{\\circ})} = \\frac {1}{t^2}$ (2).\nFrom (2) we get $ x = \\arctan(\\frac {\\sqrt {3}}{2t^2^ - 1})$, substituting in (1) we have $ \\frac {1}{\\sqrt {1 + \\frac {3}{(2t^2 - 1)^2}}} = \\frac {1}{2}\\frac {t^2 + t^4 - 1}{t^3}$. Solving by $ t$ we have ${ t = \\sqrt {\\frac {1}{4} + \\frac {\\sqrt {13}}{4} + \\frac {\\sqrt { - 2 + 2\\sqrt {13}}}{4}}} < \\sqrt {2}$ as desired.\nSonnhard.[/quote]\r\n\r\nHave any simple idea to prove it? :D",
  "Solution_6": "[hide]Let $ D$ be a point on $ \\overline{AB}$ such that $ DA \\equal{} DC$ and $ u \\equal{} CD$.\n\nClearly $ 2u\\cos x \\equal{} t$(*),then apply cosine low to $ \\triangle ABC$ and $ \\triangle DBC$,thus we have \n\n$ t^4 \\plus{} t^2 \\minus{} 2t^3\\cos x \\equal{} 1$(**) and $ u^2 \\plus{} 1 \\minus{} u \\equal{} (t^2 \\minus{} u)^2$(***).\n\nby(*) and (**)  $ u \\equal{} \\frac {t^4}{t^4 \\plus{} t^2 \\minus{} 1} < 1$ because $ t > 1$,\n\nthus, by (***) $ t^2 \\equal{} u \\plus{} \\sqrt {u^2 \\minus{} u \\plus{} 1} \\equal{} u \\plus{} \\sqrt {1 \\minus{} u(1 \\minus{} u)} < 1 \\plus{} \\sqrt {1 \\minus{} 0} \\equal{} 2$ as desired.[/hide]",
  "Solution_7": "In Chinese statements version:\r\n\r\nwhere $ r\\equal{}t ,$"
}
{
  "Tag": [
    "search",
    "algebra",
    "binomial theorem"
  ],
  "Problem": "Can anyone help me how to claculate 2137 to the 753th power?",
  "Solution_1": "[quote=\"Zootieroaz\"]Can anyone help me how to claculate 2137 to the 753th power?[/quote]\r\n\r\nExact value? \r\n\r\nThat will be by computer (I don't think calculator can handle that). If you mean last *** digits (*** is number, like two digits, etc), then you can use Fermat's or Binomial Theorem.",
  "Solution_2": "[quote=\"Silverfalcon\"][quote=\"Zootieroaz\"]Can anyone help me how to claculate 2137 to the 753th power?[/quote]\n\nExact value? \n\nThat will be by computer (I don't think calculator can handle that). If you mean last *** digits (*** is number, like two digits, etc), then you can use Fermat's or Binomial Theorem.[/quote]\r\n\r\nhow would u use the binomial theorem to find the last..say 3 digits of that number?",
  "Solution_3": "Well, it doesn't \"always\" work. If you have nice numbers, it's pretty nice.\r\n\r\nIf you want to find out the last three digits of the expansion of $9^{9999}$, we can write this as $(10-1)^{9999}$ and use binomial theorem to consider only few terms that will matter in calculation.",
  "Solution_4": "hmm so can anyone help me?",
  "Solution_5": "the answer is 42 :-)\r\n\r\n(think hitchhiker's guide to galaxy)  \r\n\r\nit would take a super computer to calculate that, then u'd probly spend a year reading the answer, so just be satisfied that it will be really really big.",
  "Solution_6": "what exacltly is this binomial theorem ?  :blush:",
  "Solution_7": "you could always deal with it in terms of logs to make its calculation simpler, or get a quick estimate",
  "Solution_8": "[quote=\"sen\"]what exacltly is this binomial theorem ?  :blush:[/quote]\r\n\r\nhttp://www.google.com/search?client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial_s&hl=en&q=binomial+theorem&btnG=Google+Search\r\n\r\nIt takes TWENTY-EIGHT letters and TWO clicks to find the answer."
}
{
  "Tag": [
    "quadratics",
    "algebra",
    "polynomial"
  ],
  "Problem": "How many solutions does the equation $ \\sqrt{3x}\\equal{}x\\sqrt{3}$ have?",
  "Solution_1": "1) x=0\r\n2) we may divide by sqrt3, so we get x=1\r\n2 solutions",
  "Solution_2": "Square both sides.\r\n\r\n$ 3x \\equal{} 3x^2$\r\n\r\nQuadratics have two solutions.",
  "Solution_3": "Of course, one must be careful to look for double roots.",
  "Solution_4": "Yes, the fundamental theorem of algebra really states: there exists, for a degree n polynomial, at most n roots. The multiplicities of all roots must add up to n.",
  "Solution_5": "For example $ x^2\\minus{}2x\\equal{}\\minus{}1$ has only a solution $ x\\equal{}1$, although it's quadratic. This happens when its discriminant is 0.",
  "Solution_6": "hello, from the range of definition we get $ x\\geq 0$, squaring both sides of the\r\nequation we have $ 3x\\equal{}3x^2$ or $ 0\\equal{}x(x\\minus{}1)$, from here we get the solutions\r\n$ x_1\\equal{}0$ or $ x_2\\equal{}1$, which both fulfill our equation.\r\nSonnhard.",
  "Solution_7": "It's a countdown, you know quadratics have $ 2$ solutions, so you guess $ \\boxed{2}$...",
  "Solution_8": "But that's a critical mistake. Quadratics need not have 2 solutions. A better way is to eye the solutions 0 and 1, then conclude that the answer is 2.",
  "Solution_9": "[quote=\"AIME15\"]It's a countdown, you know quadratics have $ 2$ solutions, so you guess $ \\boxed{2}$...[/quote]\r\n\r\nLook at post #6. You would make a mistake in this case.",
  "Solution_10": "But you know that it's not going to be the square of a binomial, just by looking at it...",
  "Solution_11": "Generally when they ask for how many solutions there are in a quadratic in a countdown, you'd guess $ 2$ and most of the times that'll work.  Of course, if you're wrong, the opponent basically has the answer :P \r\n\r\nOr, hit the buzzer and solve it in the three seconds you have",
  "Solution_12": "Also, the question would have to say \"how many distinct roots\" to be clear, since one could interpret it two ways if there were a double root. This discussion is getting pointless, though...",
  "Solution_13": "Either it's 0 or 1. \nThere are 2 solutions.",
  "Solution_14": " x can equal both 0 or 1.\n"
}
{
  "Tag": [],
  "Problem": "\u03bd\u03b4\u03bf \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c2  99899...9 \u03bc\u03b5 \u03bd+3 \u03c8\u03b7\u03c6\u03af\u03b1 \r\n[u]\u03b4\u03b5\u03bd [/u] \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03bf 37. \r\n\r\n\r\n\r\n :)",
  "Solution_1": "\u039f \u03b1\u03c1\u03b9\u03b8\u03bc\u03cc\u03c2 \u03bc\u03b5\u03c4\u03ac \u03b1\u03c0\u03cc \u03c0\u03c1\u03ac\u03be\u03b5\u03b9\u03c2 \u03b3\u03c1\u03ac\u03c6\u03b5\u03c4\u03b1\u03b9    999 *  10^\u03bd  -1.  \r\n \u03cc\u03bc\u03c9\u03c2 \u03c4\u03bf 37 | 999 \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 \u03c4\u03bf 37 | 999* 10^\u03bd \u03ba\u03b1\u03b9 \u03b4\u03b5\u03bd \u03b4\u03b9\u03b1\u03b9\u03c1\u03b5\u03af \u03c4\u03bf  \r\n 999 *  10^\u03bd  -1  :P"
}
{
  "Tag": [],
  "Problem": "[youtube]cXz6UJL5oDA[/youtube]\r\n\r\nvery inspirational,\r\njorian hoover",
  "Solution_1": "wow\r\njust wow\r\ni'll definitely try to do it",
  "Solution_2": "i know.\r\n\r\ni showed it to my mom today, and she's like \"this makes you feel bad about arguing with me over a restaurant\" and i felt really bad\r\n\r\nso i put it on here\r\n\r\njorian",
  "Solution_3": "where do i send cards to?",
  "Solution_4": "This has also been posted on IBScrewed. \r\n\r\nHere's a link to that page. The address is listed on there. \r\n\r\n[url]http://www.ibscrewed.net/forum/viewtopic.php?t=13019[/url]",
  "Solution_5": "Maybe we could get lots a people from aops to do this if we put this in as many forums as we could, without crossing the line of \"spam\"...."
}
{
  "Tag": [],
  "Problem": "Consider a series starting with digits 1 and 2 and then arriving at the next number by multiplying the previous two digits.\r\n\r\nSo next one is 1*2 that is 2.\r\n\r\nNext is 2*2 that is 4.\r\n\r\nNext is 2*4 that is 8 so on .... \r\n \r\nthus you have\r\n \r\n1 2 2 4 8 32 24 6 4 8 24 24 32 16 8 8 8........\r\n\r\n\r\nThe question is: how many times the digit 9 will appear during the compilation of the first 200 digits??\r\n\r\n :?:  :?:",
  "Solution_1": "$0$\r\nAlso, I believe that your sequence is incorrect.",
  "Solution_2": "[quote=\"LordoftheMorons\"]$0$\nAlso, I believe that your sequence is incorrect.[/quote]\r\n\r\nI am not sure if 0 is the right answer and here is how I have derived the series...\r\n\r\n1 2 2 4 8 32 24 6 4 8 24 24 32 16 8 8 8........\r\n\r\n$1*2 = 2$\r\n$2*2 = 4$\r\n$2*4 = 8$\r\n$4*8 = 32$\r\n\r\nTill now the series is...\r\n1 2 2 4 8 32\r\n\r\nNow multiple $8*3$ and get $24$. Thus, the series updates to...\r\n1 2 2 4 8 32 24\r\n\r\nNow multiple $3*2$ and get $6$. Thus, the series updates to...\r\n1 2 2 4 8 32 24 6\r\n\r\nNow multiple $2*2$ and get $4$. Thus, the series updates to...\r\n1 2 2 4 8 32 24 6 4\r\n\r\nNow multiple $4*2$ and get $8$. Thus, the series updates to...\r\n1 2 2 4 8 32 24 6 4 8\r\n\r\nNow multiple $6*4$ and get $24$. Thus, the series updates to...\r\n1 2 2 4 8 32 24 6 4 24\r\n\r\nand so on....",
  "Solution_3": "Oh, you only multiply the first digit? I see. I still can't see how a $9$ would be introduced into the series, though.",
  "Solution_4": "[quote=\"LordoftheMorons\"]Oh, you only multiply the first digit? I see. I still can't see how a $9$ would be introduced into the series, though.[/quote]\r\n\r\nIts not that I only multiple the first digits... The multiplecation is digit by digit and go on increasing the series for 200 multiplications.\r\n\r\nI also doubt, but still cannot point to it why 9 will not be there?",
  "Solution_5": "[hide=\"solution\"] $0$\nThe largest number you can get is $9*9=81$ So the only 9's attainable are 9 and 49, which come from $3*3$ or $7*7$. 3 is prime so there must be a number ending in 3 and a number beginning in 3 next to each other. There are no numbers ending in 3 that are attainable. Going to $7*7$, since 7 is prime, there must be a number ending in 7 and a number beginning in 7 next to each other. Again, no numbers ending in 7 are attainable. *Note: I am trying to find the first 9, so 27 is not attainable because it requires $9*3$.  :D [/hide]",
  "Solution_6": "[quote=\"lotrgreengrapes7926\"][hide=\"solution\"] $0$\nThe largest number you can get is $9*9=81$ So the only 9's attainable are 9 and 49, which come from $3*3$ or $7*7$. 3 is prime so there must be a number ending in 3 and a number beginning in 3 next to each other. There are no numbers ending in 3 that are attainable. Going to $7*7$, since 7 is prime, there must be a number ending in 7 and a number beginning in 7 next to each other. Again, no numbers ending in 7 are attainable. *Note: I am trying to find the first 9, so 27 is not attainable because it requires $9*3$.  :D [/hide][/quote]\r\n\r\nThanks !"
}
{
  "Tag": [
    "function",
    "geometric sequence",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "find all function $f(x): N \\to  N$satisfy :\r\n$f(m-n)f(m+n)=f(m^2) \\forall m>n$",
  "Solution_1": "This proof works if $n$ can be equal $0$....\r\nBy putting $n=0$ we obtain $f(m)^2=f(m^2)$, and by putting $n=1$ we get $f(m-1)f(m+1)=f(m^2)=f(m)^2$ so \\[ \\frac{f(m+1)}{f(m)}=\\frac{f(m)}{f(m-1}  \\] which means that $f(1), f(2)\\cdots$ form a geometric sequence!!! Then, function $f$ is defined as followes: \\[ f(n)=ab^{n-1}  \\] where $a$ is natural (cos' $f(1)$ is natural) and $b$ is natural, as for each $n$ term $f(n)$ is natural...and as $f(n)^2=f(n^2)$, $a=1$...\r\nOf course if $0$ isn't natural... problem gets complicated ;)  then by putting $n=m-2$ and $n=m-1$ we obtain \\[ \\frac{f(2)}{f(1)}=\\frac{f(2m-1)}{f(2m-2)} \\] and \\[ \\frac{f(3)}{f(2)}=\\frac{f(2m-2)}{f(2m-3)} \\] so if we define $f(1)=a$, $f(2)=f(1)q=aq$ and $f(3)=f(2)w=aqw$ then $f(4)=f(2^2)=f(1)f(3)=a^2qw$ and so on, we can construct next terms step by step... we can easily prove, that function defined like this has a form( for$n\\geq4$) \\[ f(n)=a^2q^{[\\frac{n-1}{2}]}w^{[\\frac{n-2}{2}]} \\] now all wehave to do is to find out which functions of this form are \"good\"... I'll do it ...maybe tomorrow ;)"
}
{
  "Tag": [
    "trigonometry",
    "AMC",
    "USA(J)MO",
    "USAMO",
    "inequalities",
    "calculus",
    "vector"
  ],
  "Problem": "A man with height [b]h[/b] fires a shell with a gun with initial speed [b]v[/b] at an angle [b]A[/b] above the horizontal . The shell hits a target which is a distance of [b]s[/b] from the man . If there is only one  possible value for [b]A[/b] for which this occurs , show that \r\n\r\n$v^2 = g (-h +\\sqrt{h^2 +s^2}) $",
  "Solution_1": "Let $T$ the time required for hitting target by the shell, then we have $s=v(\\cos A)T\\Longleftrightarrow T=\\frac{s}{v\\cos A}$.\r\n\r\nThe hight of the shell at the time $t=T$ can be obtained by $y=h+v(\\sin A)T-\\frac{1}{2}gT^2=h+s(\\tan A)-\\frac{1}{2}g\\frac{s^2}{v^2\\cos ^ 2 A}$. Since $y\\geq 0$, we get\r\n\r\n$v^2\\geq \\frac{1}{2}gs^2\\cdot \\frac{1+\\tan ^ 2 A}{h+s\\tan A}=\\frac{1}{2}g\\cdot \\frac{s^2+(s\\tan A)^2}{h+s\\tan A}$\r\n\r\n$=\\frac{1}{2}g\\left(s\\tan A+h+\\frac{h^2+s^2}{s\\tan A+h}-2h\\right)$\r\n\r\n$\\geq \\frac{1}{2}g\\left(2\\sqrt{(s\\tan A+h)\\cdot \\frac{h^2+s^2}{s\\tan A+h}}-2h\\right)=g(\\sqrt{h^2+s^2}-h)$, we are done.",
  "Solution_2": "wow, that is some nice algebraic insight... and a very nice question!\r\n\r\nThis is the type of question where those USAMO (and up) skills pay off, of which I don't have many.\r\n\r\nNice solution Kunny.",
  "Solution_3": "Thanks, Spoon. :) \r\n\r\nkunny",
  "Solution_4": "[quote=\"kunny\"]Let $T$ the time required for hitting target by the shell, then we have $s=v(\\cos A)T\\Longleftrightarrow T=\\frac{s}{v\\cos A}$.\n\nThe hight of the shell at the time $t=T$ can be obtained by $y=h+v(\\sin A)T-\\frac{1}{2}gT^2=h+s(\\tan A)-\\frac{1}{2}g\\frac{s^2}{v^2\\cos ^ 2 A}$. Since $y\\geq 0$, we get\n\n$v^2\\geq \\frac{1}{2}gs^2\\cdot \\frac{1+\\tan ^ 2 A}{h+s\\tan A}=\\frac{1}{2}g\\cdot \\frac{s^2+(s\\tan A)^2}{h+s\\tan A}$\n\n$=\\frac{1}{2}g\\left(s\\tan A+h+\\frac{h^2+s^2}{s\\tan A+h}-2h\\right)$\n\n$\\geq \\frac{1}{2}g\\left(2\\sqrt{(s\\tan A+h)\\cdot \\frac{h^2+s^2}{s\\tan A+h}}-2h\\right)=g(\\sqrt{h^2+s^2}-h)$, we are done.[/quote]\r\n\r\nWhat happened to the inequality, kunny? I followed your solution but you showed $v^2 \\geq g(\\sqrt{h^2+s^2}-h)$; this isn't precisely what was asked for.",
  "Solution_5": "[quote=\"shyong\"] If there is only one  possible value for [b]A[/b] for which this occurs[/quote]\r\nIf only 1 possible value of [b]A[/b] is allowed, then this must be the A such that the maximum horizontal distance is achieved. Maybe this has something to do with it :\\",
  "Solution_6": "To Dr.No:my solution is exactly same as that of Spoon's.\r\n\r\nkunny",
  "Solution_7": "[quote=\"kunny\"]Let $T$ the time required for hitting target by the shell, then we have $s=v(\\cos A)T\\Longleftrightarrow T=\\frac{s}{v\\cos A}$.\n\nThe hight of the shell at the time $t=T$ can be obtained by $y=h+v(\\sin A)T-\\frac{1}{2}gT^2=h+s(\\tan A)-\\frac{1}{2}g\\frac{s^2}{v^2\\cos ^ 2 A}$. Since $y\\geq 0$, we get\n\n$v^2\\geq \\frac{1}{2}gs^2\\cdot \\frac{1+\\tan ^ 2 A}{h+s\\tan A}=\\frac{1}{2}g\\cdot \\frac{s^2+(s\\tan A)^2}{h+s\\tan A}$\n\n$=\\frac{1}{2}g\\left(s\\tan A+h+\\frac{h^2+s^2}{s\\tan A+h}-2h\\right)$\n\n$\\geq \\frac{1}{2}g\\left(2\\sqrt{(s\\tan A+h)\\cdot \\frac{h^2+s^2}{s\\tan A+h}}-2h\\right)=g(\\sqrt{h^2+s^2}-h)$, we are done.[/quote]\r\n\r\nThats really nice kunny !  :D  :D  :D  I face some problems when trying to group [b]v[/b] out since I use calculus to find the max of [b]A[/b] and then plugging that [b]A[/b]\r\ninside really makes it so ugly .... Your method of solving have solve my problem ! Thanks  :D",
  "Solution_8": "hmm... I don't believe you can solve for this A analytically, someone correct me if I'm wrong!\r\n\r\nAlso, kunny I didn't post a solution  :blush:. Were you refering to my previous post about A being the ange yielding maximum horizontal distance?",
  "Solution_9": "[quote=\"Spoon\"]hmm... I don't believe you can solve for this A analytically, someone correct me if I'm wrong!\n\nAlso, kunny I didn't post a solution  :blush:. Were you refering to my previous post about A being the ange yielding maximum horizontal distance?[/quote]\r\n\r\nActually it can be found that the distance is maximize when \r\n$tanA = \\frac {v}{\\sqrt{v^2 + 2gh}}$",
  "Solution_10": "I'd very much like to see it, if you could please?",
  "Solution_11": "from law of conservation enery , \r\n\r\n1/2 *m*V^2 = 1/2*mv^2 + mgh   \r\n\r\nV^2 = v^2 + 2gh ------(1)    ( here , V is the velocity just before the shell strikes the target or ground  )\r\n\r\nthen suppose the time taken is [b]t[/b] then \r\n\r\nV=gt + v  and  d=vcos[b]A[/b]t----(2)\r\n\r\ndrawing it in vector , and you will have a triangle PQR with QP = v , PR = gt , QR = V .\r\nDrop a perpendicular from Q to PR and let the intersecting point be M . Then angle PQM = [b]A[/b] . Let the area of the triangle PQR be S then \r\n\r\nS = 1/2 * vsin(90-A)gt = 1/2 * vgtcosA=1/2 *dg    (from (2) )\r\n\r\nto maximize [b]d[/b] , you need to maximise S which tell us that triangle PQR must be right angle at Q . Hence \r\n\r\ntan A = PQ/QR = v / V = v / {sqrt (v^2 + 2gh )}     ( from (1) )   ...Q.E.D",
  "Solution_12": "Very simple,shyong.as if Feynman's explanation. :) \r\n\r\nkunny"
}
{
  "Tag": [
    "percent"
  ],
  "Problem": "Express $ 42\\frac{6}{7}\\%$ as a common fraction.",
  "Solution_1": "This can be re-written as $ \\frac{300}{7}$ Since this is a percent, put it over 100, giving you $ \\frac{300}{700}\\equal{}\\frac{3}{7}$"
}
{
  "Tag": [
    "probability",
    "trigonometry",
    "arithmetic sequence",
    "number theory"
  ],
  "Problem": "Well, I took another shot at the 12A recently to see how much I've improved (I became very interested in all types of problem solving shortly after I took the test) and found I am now much better. All of which is thanks to AoPS and every member on these boards ([i]YES[/i], even sharkman). Thanks guys.\r\n\r\n[hide=\"Problem #7\"]Let [i]a[/i], [i]b[/i], [i]c[/i], [i]d[/i], and [i]e[/i] be five consecutive terms in an arithmetic sequence, and suppose that $ a+b+c+d+e=30$. Which of the following can be found?\n(A): $ a$\n(B): $ b$\n(C): $ c$\n(D): $ d$\n(E): $ e$\n[hide=\"Correct answer:\"][i]C[/i]\n\nOw, I thought it was [i]A[/i]. I was under the apparent misconception that $ a+(a+n)+(a+2n)+(a+3n)+(a+4n)=30$. Or is the answer that $ c$ is the mean, and $ c=\\frac{30}{5}$?[/hide][/hide]\n[hide=\"Problem #12\"]Integers [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i], not necessarily distinct, are chosen independently and at random from $ 0$ to $ 2007$, inclusive. What is the probability that $ ad-bc$ is even?\n(A) : $ \\frac{3}{7}$\n(B) : $ \\frac{7}{16}$\n(C) : $ \\frac{1}{2}$\n(D) : $ \\frac{9}{6}$\n(E) :$ \\frac{5}{8}$\n[hide=\"Correct answer:\"][i]E[/i]\n\nI got this. HOWEVER, I brute-forced this answer using a 13 or 14 line program that randomed [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i]. It did the calculation then stored to a constantly updating variable for even or odd respectively 300 times. It got a probability of .634 or something which was really close to [i]E[/i].  Can someone show how to not have to BS this problem? Thanks in advance.[/hide][/hide]\n[hide=\"Problem #16\"]How many three-digit numbers are composed of three distinct digits such that one digit is the sum of the other two?\n\n(A) : $ 96$\n(B) : $ 104$\n(C) : $ 112$\n(D) : $ 120$\n(E) : $ 256$\n[hide=\"Correct answer:\"][i]C[/i]\n\nMy work: .... :read: $ \\Rightarrow$  :stretcher: . The only number I could pit out was $ 96$, and I was pretty sure that was wrong. What am I missing?[/hide][/hide]\n[hide=\"Thoughts:\"]\nOverall, of the 21 problems I completed, 3 were wrong, but I have already taken this test. Most of the problems I seem to be having trouble with include counting, and complex permutations/probability. I am partly finished with [i]Introduction to Number Theory[/i] (AoPS). Will [i]Introduction to Counting and Probability[/i] be highly beneficial?[/hide]\r\nThat's all for now, I still want to try to bust the trigonometry problem ([i]#17[/i]) on my own.",
  "Solution_1": "[quote=\"igiul\"]\nOw, I thought it was [i]A[/i]. I was under the apparent misconception that $ a+(a+n)+(a+2n)+(a+3n)+(a+4n)=30$. .[/quote]\r\nThat's not a misconception.  But, [hide]there you have 2 variables in one equation.  So we can't pinpoint $ a$ or $ n$.\nHowver, simplify that and you get $ 5a+10n=30$  Thus, $ a+2n=6$.  Since $ a+2n=C$, we have solved for C.  \n\nYou can also intuitively solve this based on the fact that the sequence has to be \"balanced\" around 30/2 (EDIT: what!? I meant balanced around the middle).[/hide]",
  "Solution_2": "Oh wow, so I [i]was[/i] right...almost. Thank ya sir.",
  "Solution_3": "[quote=\"igiul\"][hide=\"Problem #12\"]Integers [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i], not necessarily distinct, are chosen independently and at random from $ 0$ to $ 2007$, inclusive. What is the probability that $ ad-bc$ is even?\n(A) : $ \\frac{3}{7}$\n(B) : $ \\frac{7}{16}$\n(C) : $ \\frac{1}{2}$\n(D) : $ \\frac{9}{6}$\n(E) :$ \\frac{5}{8}$\n[hide=\"Correct answer:\"][i]E[/i]\n\nI got this. HOWEVER, I brute-forced this answer using a 13 or 14 line program that randomed [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i]. It did the calculation then stored to a constantly updating variable for even or odd respectively 300 times. It got a probability of .634 or something which was really close to [i]E[/i].  Can someone show how to not have to BS this problem? Thanks in advance.[/hide][/hide][/quote]\n\n[hide=\"Step 1\"] In order for $ ad$ to be odd, both $ a, d$ must be odd, so the probability of $ ad$ being odd is $ \\frac{1}{4}$.  Same with $ bc$.  [/hide]\n[hide=\"Step 2\"] In order for $ ad-bc$ to be even, either both $ ad, bc$ are even or both are odd. [/hide]\n[hide=\"Step 3\"] The probability that $ ad, bc$ are both even is $ \\frac{3}{4}\\cdot \\frac{3}{4}$, and the probability that $ ad, bc$ are both odd is $ \\frac{1}{4}\\cdot \\frac{1}{4}$.  [/hide]\n\n[quote=\"igiul\"]\n[hide=\"Problem #16\"]How many three-digit numbers are composed of three distinct digits such that one digit is the sum of the other two?\n\n(A) : $ 96$\n(B) : $ 104$\n(C) : $ 112$\n(D) : $ 120$\n(E) : $ 256$\n[hide=\"Correct answer:\"][i]C[/i]\n\nMy work: .... :read: $ \\Rightarrow$  :stretcher: . The only number I could pit out was $ 96$, and I was pretty sure that was wrong. What am I missing?[/hide][/hide][/quote]\r\n\r\nWell, your problem statement is not correct.  One digit is supposed to be the [b]average[/b] of the other two.  $ A$ is the correct answer for the sum problem."
}
{
  "Tag": [
    "puzzles"
  ],
  "Problem": "When can you add two to eleven and get one as the correct  \r\nanswer?",
  "Solution_1": "[hide]on a clock[/hide]",
  "Solution_2": "[hide]On a clock, 2 plus eleven is 1.[/hide]",
  "Solution_3": "[quote=\"Cartman\"]When can you add two to eleven and get one as the correct  \nanswer?[/quote]\r\n[hide]Try mod 1, mod 2, mod 3, mod 4, mod 6, and mod 12.\nOr, here's a proof that \n$2+11=1$\n$a=b$\n$a^{2}=b^{2}$\n$a^{2}-b^{2}=0$\n$a+b=0$\n$a=-b$\n$1=-1$\n$2=0$\n$1=0$\n$12=0$\n$11=-1$\n$2=2$\n$2+11=1$[/hide]",
  "Solution_4": "lol @ proof.\r\nBut yeah, either on a clock, or in other bases.",
  "Solution_5": "[quote=\"davidyko\"]lol @ proof.\nBut yeah, either on a clock, or in other bases.[/quote]\r\n\r\nNot other bases, just other moduluses. (Moduli?)"
}
{
  "Tag": [
    "MATHCOUNTS",
    "AMC 10"
  ],
  "Problem": "I think our chapter competition is on the 3rd. How should I prepare? I have limited time (other important things, such as homework and stuff), and I have absolutely no idea as to how to prepare.",
  "Solution_1": "Get good sleep the night before, and practice until saturday, starting now - doesn't matter if it's online here, with a teacher, a friend, or by yourself with a book.",
  "Solution_2": "[quote=\"Zenox\"]Get good sleep the night before, and practice until saturday, starting now - doesn't matter if it's online here, with a teacher, a friend, or by yourself with a book.[/quote]\r\n\r\nGet a good sleep the night before the competition is really important, and do not panic before the competition! You still have several days to get ready. Good luck for the competition! :P",
  "Solution_3": "What you two said is important, but I think he wants to know how to practice for the competition. I would suggest doing a lot of previous chapters and states (found at [url]http://www.mathcounts.org[/url]), and regularly going here and doing Mathcounts level problems.",
  "Solution_4": "Maybe what I'm doing -- go to the AMC 10 website and download those worksheets covering types of problems and doing them without the choices and brushing up on stuff you're not good at.",
  "Solution_5": "Seeing as it's in 5 days, you shouldn't be preparing anymore -- resting would be the best way to prepare.\r\n\r\nBut seriously, I doubt you need to worry about making States or doing well unless you fall into a coma on Friday or something...",
  "Solution_6": "whatever you do, just don't cram the night before.\r\ni have personal experince, and Ubemaya knows exactly what i am talking about.................",
  "Solution_7": "[quote=\"Ubemaya\"]Seeing as it's in 5 days, you shouldn't be preparing anymore -- resting would be the best way to prepare.\n\nBut seriously, I doubt you need to worry about making States or doing well unless you fall into a coma on Friday or something...[/quote]\r\n\r\nYeah I agree, if you want to prepare don't start 5 days before...........",
  "Solution_8": "@vishalarul:  you have done mathcounts before, havent you?  don't you go to miller?  if you do, then you've done more than enough preparation.",
  "Solution_9": "No, he and I go to CMS.",
  "Solution_10": "[quote=\"i_like_pie\"]No, he and I go to CMS.[/quote]\r\n\r\n\r\nLooks like you guys have a strong team.....",
  "Solution_11": "uhh ohh\r\nthat's at least x-1(where x>10 and is an integer)teams better than ours.........we're going to lose\r\n1. Miller\r\n2. Harker\r\n3. Hyde\r\n4. Cupertino\r\n5. Kennedy\r\n6. Challenger Berryesa\r\n7. William Hopkins\r\n8. (reserved)\r\n9. (reserved)\r\n10.(reserved)\r\n......\r\nx. my team",
  "Solution_12": "[quote=\"anirudh\"]Maybe what I'm doing -- go to the AMC 10 website and download those worksheets covering types of problems and doing them without the choices and brushing up on stuff you're not good at.[/quote]\r\nwhere? i couldn't see them anywhere.",
  "Solution_13": "[quote=\"i_like_pie\"]No, he and I go to CMS.[/quote]\r\nthats as clear as me saying i go to a school in central new jersey.\r\nu no, im gonna count how many schools in this country have CMS as its abreviation. I know it stands for Cupertino MS in ilikepie's case, but there are many more. like Community MS in my district, etc\r\n\r\nedit- on the first 10 results from googling CMS middle school got 9 middle schools' websites\r\ni estimate there will be over 30 Middle schools called CMS\r\nthere are actually 2 middle schools with the string \"Community Middle School\" in it.\r\n\r\nif any1's wonderin, i go to a school with abreviation GMS, which is actually short for TGMS",
  "Solution_14": "[quote=\"ProtestanT\"][quote=\"i_like_pie\"]No, he and I go to CMS.[/quote]\nthats as clear as me saying i go to a school in central new jersey.\nu no, im gonna count how many schools in this country have CMS as its abreviation. I know it stands for Cupertino MS in ilikepie's case, but there are many more. like Community MS in my district, etc\n\nedit- on the first 10 results from googling CMS middle school got 9 middle schools' websites\ni estimate there will be over 30 Middle schools called CMS\nthere are actually 2 middle schools with the string \"Community Middle School\" in it.\n\n\nif any1's wonderin, i go to a school with abreviation GMS, which is actually short for TGMS[/quote]yep, i agree. all three middle schools in Carmel are abbreviated with CMS.\r\n\r\nClay MS, Creekside MS, Carmel MS.  :P",
  "Solution_15": "There's a ton of WMSs in my area.",
  "Solution_16": "wow...too much santaclara chapter competition...they should send a lot more teams, considering how big and compeitive our chapter is",
  "Solution_17": "Our chapter competition is extremely competitive!\r\n\r\nThat's why I need to prepare quickly.\r\n\r\nAMC 12's are the way for me.",
  "Solution_18": "[quote=\"anirudh\"]Our chapter competition is extremely competitive!\n\nThat's why I need to prepare quickly.\n\nAMC 12's are the way for me.[/quote]\r\n\r\nWhere is your chapter???",
  "Solution_19": "The way to prepare:\r\n\r\nDo a ton of mathcounts stuff:\r\n\r\n\r\nLike problem of the day, go figure challenge, team round with your team, past sprints and targets, stuff like that...\r\n\r\n\r\nP.S. Squeeze in an AMC 10. But if you get perfects on AMC 10, try doing National sprints and targets.",
  "Solution_20": "No. AMC 10's last 5 are waay harder than MC national sprints/targets. About 10-15 are targets and 1-7 are sprints\r\n\r\nTry amc 12's.",
  "Solution_21": "[quote=\"anirudh\"]No. AMC 10's last 5 are waay harder than MC national sprints/targets. About 10-15 are targets and 1-7 are sprints\n\nTry amc 12's.[/quote]\r\n1. AMC 10s last five are on par with national targets.\r\n2. If you did believe that, why would you be recommending amc 12s?\r\n\r\n[size=75][color=darkred]Edited by nebula42.[/color][/size]",
  "Solution_22": "don't cram too much on the last few days, else you go brain dead................\r\nthree food tips:\r\n1. chew gum. Gum has been known to increase the attemtion span and concentration up to 50%.\r\n2. eat bananas. I have no idea why, but eating bananas on average increases test score. It might be the potassium.\r\n3. no fatty foods. Starting the day before, DO NOT eat fatty foods like pizza, chips, etc. Fat clogs arteries that decrease the amount of blood flow to the brain.\r\none exercise tip:\r\n1. do a workout before the test. Only a small workout is needed, such as Ubemaya's Workout (one pushup and one situp) just to warm up the body.\r\n\r\nI know these tips may sound crazy, but they really do help and have been proven statistically. \r\n\r\nPersonal experience: at Miller Math Marathon, i followed all four steps and ended up x-th place. Judging by the fact that i expected to bomb the test and by how dumb i am, i think they really worked.",
  "Solution_23": "[quote=\"davidlizeng\"]don't cram too much on the last few days, else you go brain dead................\nthree food tips:\n1. chew gum. Gum has been known to increase the attemtion span and concentration up to 50%.\n2. eat bananas. I have no idea why, but eating bananas on average increases test score. It might be the potassium.\n3. no fatty foods. Starting the day before, DO NOT eat fatty foods like pizza, chips, etc. Fat clogs arteries that decrease the amount of blood flow to the brain.\none exercise tip:\n1. do a workout before the test. Only a small workout is needed, such as Ubemaya's Workout (one pushup and one situp) just to warm up the body.\n\nI know these tips may sound crazy, but they really do help and have been proven statistically. \n\nPersonal experience: at Miller Math Marathon, i followed all four steps and ended up x-th place. Judging by the fact that i expected to bomb the test and by how dumb i am, i think they really worked.[/quote]\r\nYAY!!! easy workouts\r\n\r\nwhat do you do if you aren't allowed gum?\r\n\r\ni've heard bannanas and chocolate are good but whatever",
  "Solution_24": "@bpms: I meant AMC12's. Sorry. I accidentally typed 10. :blush:",
  "Solution_25": "[quote=\"#H34N1\"]Maybe what I'm doing -- go to the AMC 10 website and download those worksheets covering types of problems and doing them without the choices and brushing up on stuff you're not good at.[/quote]\r\nwhat did you mean by this.....?",
  "Solution_26": "There's a webpage with a list of worksheets that are downloadable on the amc website.",
  "Solution_27": "[quote=\"funcia\"][quote=\"davidlizeng\"]don't cram too much on the last few days, else you go brain dead................\nthree food tips:\n1. chew gum. Gum has been known to increase the attemtion span and concentration up to 50%.\n2. eat bananas. I have no idea why, but eating bananas on average increases test score. It might be the potassium.\n3. no fatty foods. Starting the day before, DO NOT eat fatty foods like pizza, chips, etc. Fat clogs arteries that decrease the amount of blood flow to the brain.\none exercise tip:\n1. do a workout before the test. Only a small workout is needed, such as Ubemaya's Workout (one pushup and one situp) just to warm up the body.\n\nI know these tips may sound crazy, but they really do help and have been proven statistically. \n\nPersonal experience: at Miller Math Marathon, i followed all four steps and ended up x-th place. Judging by the fact that i expected to bomb the test and by how dumb i am, i think they really worked.[/quote]\nYAY!!! easy workouts\n\nwhat do you do if you aren't allowed gum?\n\ni've heard bannanas and chocolate are good but whatever[/quote]\r\n\r\nchew it anyway...\r\nonce the test starts, there's noway you'll be forced to stop just to spit out gum. personally, i don't care if i get in trouble for chewing gum.........."
}
{
  "Tag": [
    "modular arithmetic",
    "number theory proposed",
    "number theory"
  ],
  "Problem": "Find all pairs of possive integers $ (x;y)$ such that $ y^x\\minus{}1\\equal{}(y\\minus{}1)!$",
  "Solution_1": "A less general problem was posted before:\r\nhttp://www.mathlinks.ro/viewtopic.php?t=187830\r\nbut onfortunately no proof was produced. Could you please let us know what is the source of the problem?",
  "Solution_2": "See http://www.mathlinks.ro/viewtopic.php?t=150674 (containing a small but easy to fix error).",
  "Solution_3": "a similar solution to what ZetaX snet the link is:\r\n\r\nlet $ d\\mid y$ and $ d < y$ then from the equation we have:\r\n\r\n$ d\\mid (y \\minus{} 1)! \\minus{} 1$\r\n\r\nbut on the other hand:\r\n\r\n$ d < y\\Rightarrow d\\leq y \\minus{} 1\\Rightarrow d\\mid (y \\minus{} 1)!$\r\n\r\nthus:\r\n\r\n$ d \\equal{} 1$\r\n\r\nso $ y$ must be a prime number...\r\n\r\nnow according to WILSON theorem we have:\r\n\r\n$ (y \\minus{} 1)!\\equiv \\minus{} 1\\pmod y$\r\n\r\nso we have:\r\n\r\n$ (y \\minus{} 1)! \\minus{} 1 \\equal{} y^x\\Rightarrow \\minus{} 1 \\minus{} 1\\equiv 0\\pmod y\\Rightarrow y\\mid 2\\Rightarrow y \\equal{} 2$\r\n------------------------------\r\nalso another solution which belongs to my friend (ML id:only math) is the following one:\r\n\r\n$ (y \\minus{} 1)! \\minus{} 1 \\equal{} y^x$\r\n\r\n$ \\Rightarrow (y \\minus{} 1)! \\minus{} 2 \\equal{} y^x \\minus{} 1$\r\n\r\n$ \\Rightarrow (y \\minus{} 1)! \\minus{} \\left(y^x \\minus{} 1\\right) \\equal{} 2$\r\n\r\n$ \\Rightarrow (y \\minus{} 1)! \\minus{} (y \\minus{} 1)\\left(y^{x \\minus{} 1} \\plus{} y^{x \\minus{} 2} \\plus{} \\ldots \\plus{} 1\\right) \\equal{} 2$\r\n\r\nbut note that $ y \\minus{} 1\\mid LHS$ so we must have $ y \\minus{} 1\\mid RHS\\Rightarrow y \\minus{} 1\\mid 2$\r\n--------------------------------\r\n\r\nthe problem was also posted here:\r\n[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=183998[/url]",
  "Solution_4": "BaBaK Ghalebi i do not really understand your solution. you wrote that by wilson-theorem we have $ (y\\minus{}1)! \\equiv \\minus{}1 \\bmod{y}$, where $ y \\in \\mathbb{P}$ and using this in the equation we get $ (y\\minus{}1)! \\minus{} 1 \\equiv \\minus{}2 \\not\\equiv 0 \\bmod{y}$. But the equation is $ (y\\minus{}1)! \\plus{} 1 \\equal{} y^x$ and by wlson we get $ (y\\minus{}1)! \\plus{} 1 \\equiv 0 \\bmod{y}$.",
  "Solution_5": "oh,what a stupid mistake,both of my solutions are wrong,those solutions are for the following equation:\r\n\r\n$ (y\\minus{}1)!\\minus{}1\\equal{}y^x$\r\n\r\nsorry :oops:",
  "Solution_6": "I'm sure the following solution is valid:\r\nas I said in my previous post,according to wilson theorem we get that $ y$ must be prime,now let $ y > 5$ and $ x\\not = 0$,note that:\r\n\r\n$ (y - 1)! = (y - 1)\\times (y - 2)\\times\\ldots\\times (\\frac {y - 1}2)\\times\\ldots\\times 2\\times 1$\r\n\r\n$ \\Rightarrow y + 1\\mid (y - 1)!$\r\n\r\nnow according to our equation we have:\r\n\r\n$ 1\\equiv ( - 1)^x\\pmod{y + 1}$\r\n\r\nso $ x$ must be even,so let $ x = 2k$ where $ k\\in\\mathbb{Z}$ now we have:\r\n\r\n$ (y - 1)! = (y^k - 1)(y^k + 1)$\r\n\r\n$ \\Rightarrow (y - 2)! = (y^{k - 1} + y^{k - 2} + \\ldots + y + 1)(y^k + 1)$ (*)\r\n\r\nbut note that $ y - 1$ is a composite number so $ y - 1\\mid (y - 2)!$ thus according to (*) we get that:\r\n\r\n${ 0\\equiv (\\underbrace{1 + 1 + \\ldots + 1}_{k\\textrm{ times }})(1 + 1)}\\pmod {y - 1}$\r\n\r\n$ \\Rightarrow 0\\equiv 2k\\pmod {y - 1}$\r\n\r\n$ \\Rightarrow 0\\equiv x\\pmod {y - 1}$\r\n\r\n$ \\Rightarrow x\\geq y - 1$\r\n\r\nbut it's easy to see that:\r\n\r\n$ y^{x}\\geq y^{y - 1} > (y - 1)! + 1$\r\n\r\nwhich is a contradiction,so we must check the cases $ y = 2,3,5$ and also the case $ x = 0$ which gives us the solution:\r\n\r\n$ (y,x) = (2,1),(3,1),(5,2)$"
}
{
  "Tag": [
    "LaTeX",
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Show that if x,y,z>0 and x+y+z=1 then \r\n64<=(1+1/x )(1+1/y )(1+1/z )\r\n\r\nAlso, can anyone help me with the coding of mathematical symbols?  Is there a place for a tutorial on this page somewhere?  Thanks!",
  "Solution_1": "The first solution that comes to mind is using Holder and obtaining that $ RHS^{1/3}>\\equal{}1\\plus{}1/ \\sqrt[3]{xyz}$ so it suffices to prove that $ 1\\plus{}1/ \\sqrt[3]{xyz}>\\equal{}4\\equal{}>1>\\equal{}3\\sqrt[3]{xyz}\\equal{}>x\\plus{}y\\plus{}z>\\equal{}3\\sqrt[3]{xyz}$ which is true by AM GM inequality.\r\n\r\nAs far as latex is concerned (although I do't know it very well myself) you can start here: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php\r\n\r\nAnd where I used $ \\equal{}>$ i meant it in both directions, but I don't know how to type it!",
  "Solution_2": "Also possible solution to this simple question is:\r\n\r\n$ \\prod (1 \\plus{} \\frac {1}{x}) \\equal{} \\prod (1 \\plus{} \\frac {x \\plus{} y \\plus{} z}{x}) \\equal{} \\prod (1 \\plus{} 1 \\plus{} \\frac {y}{x} \\plus{} \\frac {z}{x})\\geq \\prod 4 \\sqrt [4]{\\frac {yz}{x^{2}}} \\equal{} 4^{3} \\equal{} 64$",
  "Solution_3": "[quote=\"salonikios\"]\n\n\nAnd where I used $ \\equal{} >$ i meant it in both directions, but I don't know how to type it![/quote]\r\n\r\nGo \\iff to get $ \\iff$."
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "combinatorics proposed",
    "combinatorics"
  ],
  "Problem": "Here is a cute one :\r\n\r\nIs it possible to colour all grid points in the plane white and red so that no rectangle with vertices on grid points of one colour and sides parallel to the grid lines has area from the set $\\{1,2,4,8, \\cdots , 2^n,\\cdots, \\}$?\r\n\r\nPierre.",
  "Solution_1": "[hide=Solution]\nYes. Using Cartesian coordinates, color the point $(x,y)$ white iff $x+y\\equiv 0\\pmod{3}$, otherwise color it red.\n\nSuppose to the contrary that such rectangle exists.\nLet $(p,q),(r,q),(p,s),(r,s)$ be the four vertices of the rectangle. We have $|(p-r)(q-s)|\\in \\{ 1,2,4,\\cdots \\}$.\nSince $3\\nmid p-r$, the points $(p,q),(r,q)$ can't be simultaneously white. So the four vertices must be red.\nAlso, we have $p+q\\not\\equiv r+q\\pmod{3}$. This means $(p+q)+(r+q)\\equiv 1+2\\equiv 0 \\pmod{3}$.\nSimilarly, $(p+s)+(r+s)\\equiv 0\\pmod{3}$. But this leads to $q\\equiv s\\pmod{3}\\implies 3\\mid q-s$ which is impossible.\n[/hide]"
}
{
  "Tag": [],
  "Problem": "Eto, nakon IMO-a dolazi i ne\u0161to sitno za nas (ve\u0107im dijelom) umirovljene natjecatelje - IMC se ove godine odr\u017eava u Budimpe\u0161ti od 25. do 30.7.\r\n\r\nNas iz Zagreba dolazi 10, kao \u0161to sam napomenuo ve\u0107 u jednom topicu, a u jednom drugom sam nam spomenuo i imena:\r\n\r\nNikola Ad\u017eaga, druga godina\r\nGoran Dra\u017ei\u0107, tre\u0107a godina\r\nNikola Grubi\u0161i\u0107, \u010detvrta godina\r\n\u017deljko Kereta, tre\u0107a godina \r\nRudi Mrazovi\u0107, \u010detvrta godina\r\nMarin Mi\u0161ur, druga godina\r\nMelkior Ornik, prva godina\r\nPetar Sirkovi\u0107, druga godina \r\nSa\u0161a Stanko, druga godina\r\nLuka \u017duni\u0107, druga godina\r\n\r\n(leader nam je Matija Ba\u0161i\u0107)\r\n\r\nKao da sam negdje pro\u010ditao da dolazi i \u010detvero ljudi iz Ljubljane, a hana je ne\u0161to spominjala da ide i njih \u010detvero.\r\n\r\nEto, toliko. Vi\u0161e nemam \u0161to pametno da ka\u017eem, pa se idem povu\u0107i da gledam Seinfeld.",
  "Solution_1": "Jedan odu\u0161evljeni \u010ditatelj vam \u017eeli sre\u0107u na takmi\u010denju i sretan put  :wink:",
  "Solution_2": "I ovaj \u010ditalac se odu\u0161evljeno pridru\u017euje svim lijepim \u017eeljama",
  "Solution_3": "Cestitke svim osvajacima medalja (i onima koji su to zamalo postali)",
  "Solution_4": "\u010cestitke svima: \u010dini mi se da je svako ne\u0161to donio, od Frst Prajza do Onorabl Men\u0161na, tako da ste opravdali na\u0161a o\u010dekivanja  :lol:",
  "Solution_5": "Hvala na \u010destitkama, bilo je jako lijepo, a i rezultatom sam zadovoljan (zeznuli su mi score u zavr\u0161nim rezultatima  :roll:, ali oprostit \u0107u im kad mi po\u0161alju moju zaslu\u017eenu tre\u0107u nagradu :winner_third:)."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Let a,b be integer integers such that ab-1 divides a^{2}+b^{2}.Show that:\r\n\\frac{a^{2}+b^{2}}{ab-1}=5",
  "Solution_1": "Very very very often posted before.",
  "Solution_2": "Very very very well-known method to solve it...",
  "Solution_3": "Does anyone try it?\r\nPS:ZetaX,do you know Erich Ludendorff?",
  "Solution_4": "very very very very very easy:[url]http://en.wikipedia.org/wiki/Erich_Ludendorff[/url]:wink:",
  "Solution_5": "From history lessons, I'm not that old to be able to have lived with him ;)"
}
{
  "Tag": [
    "inequalities"
  ],
  "Problem": "Let $ x,y > 0$ such that $ x^5 \\plus{} y^5 \\equal{} 2$ . Prove that :\r\n\r\n$ (x^2 \\plus{} y^6)(x^8 \\plus{} y^4) \\plus{} 2\\ge x^{10} \\plus{} y^{10}$",
  "Solution_1": "By Cauchy Schwartz inequality \r\n\r\n$ (x^2 \\plus{}y^6)(x^8 \\plus{}y^4) \\geq (x^5 \\plus{}y^5 )^2$\r\n\r\n$ (x^5 \\plus{}y^5 )^2 \\plus{}2 \\geq x^{10} \\plus{}y^{10}$\r\n\r\n$ (x^5 \\plus{}y^5)^2 \\equal{}x^{10} \\plus{}2x^5 y^5 \\plus{}y^{10}$\r\n\r\nThen \r\n\r\n$ (x^5 \\plus{}y^5 )^2 \\plus{}2 \\equal{}x^{10} \\plus{}2x^5 y^5 \\plus{}y^{10} \\plus{}2 \\geq x^{10} \\plus{}y^{10} \\iff 2x^5 y^5 \\plus{}2 > 0$,no equality case \r\n :wink:",
  "Solution_2": "Is there something wrong with the problem?\r\n\r\n$ (x^2\\plus{}y^6)(x^8\\plus{}y^4)\\equal{}x^{10}\\plus{}x^2y^4\\plus{}x^8y^6\\plus{}y^{10}>x^{10}\\plus{}y^{10}$\r\n\r\nSo apparently the result is true for any x,y...",
  "Solution_3": "[quote=\"Pain rinnegan\"]Let $ x,y > 0$ such that $ x^5 \\plus{} y^5 \\equal{} 2$ . Prove that :\n\n$ (x^2 \\plus{} y^6)(x^8 \\plus{} y^4) \\plus{} 2\\ge x^{10} \\plus{} y^{10}$[/quote]\r\n\r\n\r\n this isn\u00b4t stronger than cauchy, since there is obviously no equality case...",
  "Solution_4": "[quote=\"v235711\"][quote=\"Pain rinnegan\"]Let $ x,y > 0$ such that $ x^5 \\plus{} y^5 \\equal{} 2$ . Prove that :\n\n$ (x^2 \\plus{} y^6)(x^8 \\plus{} y^4) \\plus{} 2\\ge x^{10} \\plus{} y^{10}$[/quote]\n\n\n this isn\u00b4t stronger than cauchy, since there is obviously no equality case...[/quote]\r\n\r\nThe correct version\r\n\r\n$ (x^2 \\plus{} y^6)(x^8 \\plus{} y^4) \\ge x^{10}\\plus{}y^{10}\\plus{}2$ with $ x^5\\plus{}y^5\\equal{}2$"
}
{
  "Tag": [
    "greatest common divisor",
    "number theory",
    "least common multiple",
    "prime factorization",
    "relatively prime"
  ],
  "Problem": "Prove that the product of the greatest common factor and lowest common multiple of two numbers is equal to the product of the numbers.\r\n\r\n[hide=\"Hint\"]Prime factorization[/hide]",
  "Solution_1": "[hide]Let the numbers be $A$ and $B$. Let $g$ denote the GcF of $A$ and $B$. Then let $A$ and $B$  equal $gx$ and $gy$, respectively, where $x$ and $y$ are relatively prime integers. This means that the LcM of $A$ and $B$ is $gxy$. \n\n$gx\\cdot gy=g\\cdot gxy$, so the statement is true.[/hide]",
  "Solution_2": "[hide=\"A formal construction of the proof by prime factorization\"] Let $e_{p}(n)$ denote the exponent of the prime $p$ in the prime factorization of a positive integer $n$.  I give some unproven lemmas involving this function.\n\n[b]Lemma:[/b]  $e_{p}(ab) = e_{p}(a)+e_{p}(b)$\n\n[b]Lemma:[/b]  $e_{p}(a) = e_{p}(b) \\forall p \\in \\mathbb{P}\\Leftrightarrow a = b$\n\n[b]Lemma:[/b]  $e_{p}( gcd(a, b) ) = min( e_{p}(a), e_{p}(b) )$\n\n[b]Lemma:[/b]  $e_{p}( lcm(a, b) ) = max( e_{p}(a), e_{p}(b) )$\n\nLastly, recall that given two numbers $x, y$,\n\n$min(x, y)+max(x, y) = x+y$\n\nThen the proof is rather simple, and takes the following form:\n\n$e_{p}( gcd(a, b) )+e_{p}( lcm(a, b) ) = e_{p}(a)+e_{p}(b) \\forall p \\in \\mathbb{P}$ [/hide]",
  "Solution_3": "For this to work, you must allow for zero exponents.",
  "Solution_4": "$(a,b)\\cdot[a,b]=a\\cdot b$\r\n\r\n[hide]Let $(a,b)=k$. Then we have $a=m\\cdot k$, $b=n \\cdot k$ and $[a,b]=p\\cdot k$.\n$k^{2}\\cdot p = k^{2}\\cdot m \\cdot n$. We divide by $k$ so that $p=m \\cdot n$. Replacing, $k^{2}\\cdot m \\cdot n = k^{2}\\cdot m \\cdot n$ which is true and i'm stuck.[/hide]",
  "Solution_5": "[quote=\"icx\"]$[a,b]=p\\cdot k$.\n$k^{2}\\cdot p = k^{2}\\cdot m \\cdot n$. [/quote]\r\n\r\nEr, you're using the statement you're trying to prove."
}
{
  "Tag": [
    "function",
    "\\/closed"
  ],
  "Problem": "Can one of you please post an announcement that spans all of the forums telling everyone to use the spoiler button? I understand that the spoiler button either wasnt at mathlinks or wasnt used much, but its appreciated tremendously here.\r\nThanx.",
  "Solution_1": "Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.",
  "Solution_2": "[quote=\"jelyman\"]Can one of you please post an announcement that spans all of the forums telling everyone to use the spoiler button? I understand that the spoiler button either wasnt at mathlinks or wasnt used much, but its appreciated tremendously here.\nThanx.[/quote]Basically people do not want in the advanced section the spolier. It is a general belief that at that level you do not need to hide information in that manner, as you can refrain yourself from seeing if you wish so. \r\nSo what I am asking is that people should post without spoilers in the Advanced Section and College Playground. It would be highly appreciated.\r\n\r\nPS you can call me Valentin, I like that better than Vornicu :)",
  "Solution_3": "[quote=\"Valentin Vornicu\"][quote=\"jelyman\"]Can one of you please post an announcement that spans all of the forums telling everyone to use the spoiler button? I understand that the spoiler button either wasnt at mathlinks or wasnt used much, but its appreciated tremendously here.\nThanx.[/quote]Basically people do not want in the advanced section the spolier. It is a general belief that at that level you do not need to hide information in that manner, as you can refrain yourself from seeing if you wish so. \nSo what I am asking is that people should post without spoilers in the Advanced Section and College Playground. It would be highly appreciated.\n\nPS you can call me Valentin, I like that better than Vornicu :)[/quote]\r\n\r\nI'm not sure who are you talking about when you say its a 'general belief', but this is definitely not true of the AoPS people, and as you don't have spoiler at Mathlinks I do not understand what you mean. At AoPS, in every problem forum including the advanced forum, the spoiler function is extremely appreciated and used. \r\nWe would all appreciate it if you continue using the system that has worked so well for us.",
  "Solution_4": "[quote=\"rrusczyk\"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]\r\n\r\nHow do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.",
  "Solution_5": "[quote=\"rrusczyk\"][quote=\"rrusczyk\"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]\n\nHow do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.[/quote]\r\nI like that a lot, if the people who use this option will still put things in the spoilers (it would be easy to forget to use spoilers, if you never saw them).",
  "Solution_6": "[quote=\"rrusczyk\"][quote=\"rrusczyk\"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]\n\nHow do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.[/quote]\r\n\r\nYes, this is a very good idea.",
  "Solution_7": "[quote=\"rrusczyk\"][quote=\"rrusczyk\"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]\n\nHow do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.[/quote]\r\nthat'd be awesome",
  "Solution_8": "This is implemented now.  We will be making a global announcement about this and other items next week.  \r\n\r\nAll old AoPS users have been set to have spoilers still hidden, while old ML users have been set to have them revealed by default.  Go to Options above to change your default."
}
{
  "Tag": [
    "geometry",
    "geometric transformation"
  ],
  "Problem": "This problem was just shown to me by a friend. I don't know if I can restate it as clearly as possible.\r\n\r\nIn a creek, three structures and paths between them. At one side of a creek is a resthouse and at the other side and at the other side a butterfly dome and a hut.  A pond is also to be made  beside the creek.  Suppose that you want to create a path in this order; resthouse, hut, pond and dome build a single bridge, connecting the three structures, and where will you put the bridge and the pond it so that the total distance of your path is minimum?\r\n\r\nSee skeleton drawing below: (Ignore the asterisk, its for spacing)\r\n\r\n*******[b]Resthouse[/b]\r\n                              \r\n______________________________________________\r\n               Creek\r\n______________________________________________\r\n\r\n************************************[b]Dome[/b]\r\n\r\n[b]Hut[/b]",
  "Solution_1": "Well, you should connect the resthouse and the hut with a straight line, and then connect the hut and dome with a straight line.  That gives you the shortest distance for those three.  Then put the bridge over the creek on the line from the resthouse to the hut, and put the pond anywhere on the line from the hut to the dome, and you have your locations on the shortest possible path...",
  "Solution_2": "[quote=\"DanK\"]Well, you should connect the resthouse and the hut with a straight line, and then connect the hut and dome with a straight line.  That gives you the shortest distance for those three.  Then put the bridge over the creek on the line from the resthouse to the hut, and put the pond anywhere on the line from the hut to the dome, and you have your locations on the shortest possible path...[/quote]\r\n\r\nbut you should put the pond beside the river...",
  "Solution_3": "[hide]\nLabel the resthouse $R$, the hut $H$, the creek $c$, and the dome $D$. The shortest path can be found by first finding the shortest distance $RH$ and then finding the shortest path from $H$ to $D$ that touches $c$.\n\nThe first part is easy: the shortest distance is just a straight line, so the bridge should be put at the intersection between $\\overline{RH}$ and $c$.\n\nFor the second part, one will have to touch $c$. Now suppose $c$ is horizontal and $R$ is above $c$. Then one will have to move back down in order to reach $D$. If one moves up instead, the end of the path will be at a point $D^\\prime$ which is found by reflecting $D$ over $c$. So for any path from $H$ to $D$ that touches $c$, there is a path of equal length from $H$ to $D^\\prime$ that touches $c$.\n\nThe minimum distance from $H$ to $D$ touching $c$ is equal to the minimum distance from $H$ to $D^\\prime$ touching $c$. But [i]all[/i] paths from $H$ to $D^\\prime$ touch $c$, so this is just a straight line, like the first part. The pond should be placed at the intersection between $\\overline{HD^\\prime}$ and $c$.\n[/hide]",
  "Solution_4": "[quote=\"scorpius119\"][hide]\nLabel the resthouse $R$, the hut $H$, the creek $c$, and the dome $D$. The shortest path can be found by first finding the shortest distance $RH$ and then finding the shortest path from $H$ to $D$ that touches $c$.\n\nThe first part is easy: the shortest distance is just a straight line, so the bridge should be put at the intersection between $\\overline{RH}$ and $c$.\n\nFor the second part, one will have to touch $c$. Now suppose $c$ is horizontal and $R$ is above $c$. Then one will have to move back down in order to reach $D$. If one moves up instead, the end of the path will be at a point $D^\\prime$ which is found by reflecting $D$ over $c$. So for any path from $H$ to $D$ that touches $c$, there is a path of equal length from $H$ to $D^\\prime$ that touches $c$.\n\nThe minimum distance from $H$ to $D$ touching $c$ is equal to the minimum distance from $H$ to $D^\\prime$ touching $c$. But [i]all[/i] paths from $H$ to $D^\\prime$ touch $c$, so this is just a straight line, like the first part. The pond should be placed at the intersection between $\\overline{HD^\\prime}$ and $c$.\n[/hide][/quote]\r\n\r\nThis is a good explanation, i think im getting the point. But the problem is, i made a mistake with my drawing. The hut should be slightly lower than the dome.. How are we going to compute it then? I have redrawn the sketch, please see the problem again.",
  "Solution_5": "Um, I think his method should still work...",
  "Solution_6": "I got it already, through translation and reflection."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "calculus",
    "calculus computations"
  ],
  "Problem": "What is the solid angle subtended by a cone of semivertical angle $ \\alpha$.In general  ho to find the solid angle",
  "Solution_1": "Draw a sphere of radius $ 1$ centered at the vertex of the cone. Find the area of the part that lies inside the cone. Divide by the total area, which is $ 4\\pi$."
}
{
  "Tag": [
    "function",
    "quadratics",
    "limit",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "Find all functions $f: \\mathbb R \\setminus \\{\\pm 1\\} \\to \\mathbb R$ satisfying\n\\[ f(1-2x) = 2 f\\left(\\frac 1x \\right) + 1, \\quad \\forall x \\in \\mathbb R, x \\neq 0,\\pm 1.\\]",
  "Solution_1": "Ok, I consider that the question is :\r\nFind all functions $ f$ : $ \\mathbb R\\backslash\\{ - 1, + 1\\}\\to\\mathbb R$ such that $ f(1 - 2x) = 2f(\\frac {1}{x}) + 1$ $ \\forall x\\in\\mathbb R\\backslash\\{ - 1,0, + 1\\}$\r\n\r\n\r\n$ f(1 - 2x) = 2f(\\frac {1}{x}) + 1$ $ \\forall x\\in\\mathbb R\\backslash\\{ - 1,0, + 1\\}$ $ \\Leftrightarrow$ $ f(x) = 2f(\\frac {2}{1 - x}) + 1$ $ \\forall x\\in\\mathbb R\\backslash\\{ - 1, + 1, + 3\\}$\r\n\r\nLet $ g(x) = f(x) + 1$ and $ h(x) = \\frac {2}{1 - x}$\r\n\r\nThe problem is now : Find all functions $ g$ : $ \\mathbb R\\backslash\\{ - 1, + 1\\}\\to\\mathbb R$ such that $ g(x) = 2g(h(x))$  $ \\forall x\\in\\mathbb R\\backslash\\{ - 1, + 1, + 3\\}$\r\n\r\nWe'll have (with good conditions over $ x$) : $ g(x) = 2^ng(h(h(h( ...$n times$ ...(x)))) ... )$. And clearly the $ n^{th}$ composit of $ h(x)$ is an homographic function.\r\n\r\nLet then the sequence $ a_n$ defined $ \\forall n\\in\\mathbb Z$ as :\r\n$ a_0 = 0$\r\n$ a_1 = 2$\r\n$ \\forall n > 2$ : $ a_n = a_{n - 1} - 2a_{n - 2}$\r\n\r\n$ \\forall n < 0$ : $ a_n = \\frac {a_{n + 1} - a_{n + 2}}{2}$\r\n\r\nIt's easy to check that $ a_n = \\frac {\\sqrt 2}{2}(b^n - (\\frac { - 1}{b})^n$ (where $ b = 1 + \\sqrt 2$) $ \\forall n\\in\\mathbb Z$ and that $ a_n\\neq 0$ $ \\forall n\\in\\mathbb Z^*$.\r\n\r\nLet then the family of functions defined for any ${ k\\in\\mathbb Z}$ as : $ h_k(x) = 2\\frac {a_{k - 1}x - a_k}{a_kx - a_{k + 1}}$ ($ h_0$ : $ \\mathbb R\\to \\mathbb R$; $ h_k$ : $ \\mathbb R\\backslash\\{\\frac {a_{k + 1}}{a_k}\\}\\to\\mathbb R$)\r\n\r\nIt's easy to check that $ h_0(x) = x$ and $ h_{k + 1} = h\\circ h_k$ $ \\forall k\\in\\mathbb Z$. So, $ g(x) = 2^kg(h_k(x))$\r\nIt's also easy to check that, $ \\forall k\\neq 0$, $ h_k(x)$ has no fixed point (the equation $ h_k(x) = x$ implies a quadratic whose discriminant is $ - \\frac {7a_k^2}{4} < 0$\r\n\r\nLet's then the relation $ \\sim$ defined as $ x\\sim y$ $ \\iff$ $ \\exists$ a unique $ k\\in \\mathbb Z$ such that $ y = h_k(x)$\r\nThis relation is an equivalence one :\r\n$ x\\sim x$ : $ h_0(x) = x$ and no other $ k$ is available since $ \\forall k\\neq 0$, $ h_k(x)$ has no fixed point.\r\n$ x\\sim y$ $ \\implies$ $ y\\sim x$ : $ y = h_k(x)$ $ \\iff$ $ x = h_{ - k}(x)$ and unicity is obtained thru the no_fixed_points property.\r\n$ x\\sim y$ and $ y\\sim z$ $ \\implies$ $ x\\sim z$ :  $ y = h_i(x)$ and $ z = h_j(z)$ imply $ z = h_{i + j}(x)$ and unicity is obtained thru the no_fixed_points property.\r\n\r\nSo we can define a function $ k(x,y)$ defined in each equivalence class of $ \\sim$ as $ y = h_{k(x,y}(x)$\r\nLet $ C(x)$ the equivalence class of $ x$ and $ A = C( - 1)$\r\n\r\nWe have $ - 1 = h(3)$ and $ 1 = h( - 1)$ and so $ \\{ - 1, + 1, + 3\\}\\subset A$\r\n\r\nLet $ r(x)$ any choice function which associates to each real $ x$ a representative of its equivalence class (unique per class).\r\n\r\nThen :\r\n$ \\forall x\\notin A$, we have $ x = h_{k(r(x),x)}(r(x))$ and so $ g(x) = 2^{k(x,r(x))}g(r(x))$ and so we just have to define $ g(r(x))$ as we want.\r\n\r\nThe situation for $ A$ is a little bit different :\r\nWe have $ 0 = h_4(3)$ but we dont have $ g(0) = 2^4g(h_4(3))$ since we dont have $ g(1) = 2g(h(3))$\r\nSo we can choose g(x) freely for $ x = 3$ and for $ x = 0$.\r\n\r\nSo here is the general solution :\r\n================================\r\nLet $ U = \\{r(x)$ $ \\forall x\\notin A = C( - 1)\\}\\cup\\{0,3\\}$\r\nLet $ m(x)$ any function from $ U\\to\\mathbb R$\r\n\r\n$ \\forall x\\notin C( - 1)$ : $ f(x) = 2^{k(x,r(x))}m(r(x)) - 1$\r\n$ \\forall x\\in C( - 1)\\backslash\\{ - 1, + 1\\}$ such that $ k(0,x)\\geq 0$ : $ f(x) = 2^{k(x,0)}m(0) - 1$\r\n$ \\forall x\\in C( - 1)\\backslash\\{ - 1, + 1\\}$ such that $ k(0,x) < 0$ : $ f(x) = 2^{k(x,3)}m(3) - 1$\r\n\r\nAnd obviously this is a general solution.\r\n\r\nIf we want a continuous solution :\r\n================================\r\n\r\nIf we suppose $ g(x)$ continuous in $ \\mathbb R\\backslash\\{ - 1, + 1\\}$, and since $ \\lim_{k\\to - \\infty}h_k(x) = -2b$, we have, using the continuity and $ k\\to - \\infty$, $ g(x) = 0$ $ \\forall x\\in\\mathbb R\\backslash A$. Then, since $ A$ is a countable set, we can always find, in any neighborhood of an element of A, elements of $ \\mathbb R\\backslash A$. And so $ g(x) = 0$ $ \\forall x\\in R\\backslash\\{ - 1, + 1\\}$ (remember $ g(x)$ is not defined in $ \\{ - 1, + 1\\}$ And so the only continuous function solution of the initial problem is $ f(x) = - 1$"
}
{
  "Tag": [
    "inequalities",
    "inequalities proposed"
  ],
  "Problem": "For positive reals $ a,b,c$, prove that\r\n\\[ (a^{2}+ab+b^{2})(b^{2}+bc+c^{2})(c^{2}+ca+a^{2}) \\leq (((b+c)(c+a)(a+b))^{\\frac{2}{3}}-(abc)^{\\frac{2}{3}})^{3}. \\]",
  "Solution_1": "Very nice and good inequality! Congratulations, Kim.",
  "Solution_2": "[quote=\"Sung-yoon Kim\"]For positive reals $ a,b,c$, prove that\n\\[ (a^{2}\\plus{}ab\\plus{}b^{2})(b^{2}\\plus{}bc\\plus{}c^{2})(c^{2}\\plus{}ca\\plus{}a^{2})\\leq (((b\\plus{}c)(c\\plus{}a)(a\\plus{}b))^{\\frac{2}{3}}\\minus{}(abc)^{\\frac{2}{3}})^{3}.\\]\n[/quote]\r\nRewrite the inequality as\r\n$ \\left( 1\\plus{}\\sqrt[3]{\\frac{(a^{2}\\plus{}ab\\plus{}b^{2})(b^{2}\\plus{}bc\\plus{}c^{2})(c^{2}\\plus{}ca\\plus{}a^{2})}{a^{2}b^{2}c^{2}}}\\right)^{3}\\le\\frac{(a\\plus{}b)^{2}(b\\plus{}c)^{2}(c\\plus{}a)^{2}}{a^{2}b^{2}c^{2}}$\r\nBy Holder inequality,\r\n$ \\left( 1\\plus{}\\sqrt[3]{\\prod_{cycl}\\frac{a^{2}\\plus{}ab\\plus{}b^{2}}{ab}}\\right)^{3}\\le\\prod_{cycl}\\left( 1\\plus{}\\frac{a^{2}\\plus{}ab\\plus{}b^{2}}{ab}\\right) \\equal{}\\prod_{cycl}\\frac{(a\\plus{}b)^{2}}{ab}$",
  "Solution_3": "Nice. In fact, my solution uses the similar method.\r\n\r\n$ \\prod (a^{2}\\plus{}ab\\plus{}b^{2}) \\equal{}\\prod ((a\\plus{}b)^{2}\\minus{}ab)\\leq (RHS)$, since $ \\prod (p_{i}\\minus{}q_{i})\\leq ((\\prod p_{i})^{\\frac{1}{3}}\\minus{}(\\prod q_{i})^{\\frac{1}{3}})^{3}$ holds by Holder."
}
{
  "Tag": [
    "number theory",
    "prime factorization",
    "number theory unsolved"
  ],
  "Problem": "For each natural $ n$,we define $ f(n)$ as the exponent of number $ 2$ in the prime factorization of $ n!$.Show that for any natural \r\n$ a$, the equation $ n\\minus{}f(n)\\equal{}a$ has infinitely many solutions.",
  "Solution_1": "We have $ (2n)!\\equal{}2^n \\cdot n!\\cdot1\\cdot3\\cdot5\\cdot...\\cdot(2n\\minus{}1)$ so $ 2n\\minus{}f(2n)\\equal{}2n\\minus{}(n\\plus{}f(n))\\equal{}n\\minus{}f(n)$.\r\nBut $ 1\\minus{}f(1)\\equal{}1 \\Rightarrow 2^a\\minus{}f(2^a)\\equal{}1$ so $ f(2^a\\minus{}1)\\equal{}(2^a\\minus{}1)\\minus{}a$ and $ 2^k(2^a\\minus{}1)\\minus{}f(2^k(2^a\\minus{}1))\\equal{}a$ for every nonnegative integer $ k$.",
  "Solution_2": "[quote=\"mazur89\"]We have $ (2n)! \\equal{} 2^n \\cdot n!\\cdot1\\cdot3\\cdot5\\cdot...\\cdot(2n \\minus{} 1)$ so $ 2n \\minus{} f(2n) \\equal{} 2n \\minus{} (n \\plus{} f(n)) \\equal{} n \\minus{} f(n)$.\nBut $ 1 \\minus{} f(1) \\equal{} 1 \\Rightarrow 2^a \\minus{} f(2^a) \\equal{} 1$ so $ f(2^a \\minus{} 1) \\equal{} (2^a \\minus{} 1) \\minus{} a$ and $ 2^k(2^a \\minus{} 1) \\minus{} f(2^k(2^a \\minus{} 1)) \\equal{} a$ for every nonnegative integer $ k$.[/quote]\r\nNote that $ S_p(n)$ is some  digits  of n in base p.\r\nThen $ v_p(n)\\equal{}\\frac{n\\minus{}S_p(n)}{p\\minus{}1}$\r\nApply this for $ p\\equal{}2$ then \r\n$ f(n)\\equal{}n\\minus{}S_2(n)$ \r\nSo $ n\\minus{}f(n)\\equal{}S_2(n)$\r\nObvious the equation $ S_2(n)\\equal{}a$ have infinite solution on N."
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Prove that the centroid and orthocenter of an equilateral triangle are concurrent.",
  "Solution_1": "[hide]\nThe median of an equilateral triangle is perpendicular to the opposite side, and so is collinear to the altitude. Hence, the intersection of the medians is the same as the intersection of the altitudes, and so the centroid and the orthocenter are concurrent.\n[/hide]"
}
{
  "Tag": [
    "function",
    "abstract algebra",
    "complex numbers",
    "linear algebra",
    "geometry unsolved",
    "geometry"
  ],
  "Problem": "how do we prove that an isometry in the complex plane is always a function of the form az+b or a(conjugate of z)+b, where modulus of a = 1, a and b are complex numbers\r\nAlso, anyone pls explain me what is meant by 'fixed points' of an isometry",
  "Solution_1": "First question: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/isometrycpx.pdf\r\n\r\n(knowing the types of isometric transformations and the way to represent them in complex plane makes your question simple, and the process is given in the paper.)\r\n\r\nSecond: If $ f$ is a transformation, point $ z_0$ for which $ f(z_0)\\equal{}z_0$ is called fixed point of the transformation-it's simply the point which remains intact during the transform.",
  "Solution_2": "thanks for the help hsiljak. Will u pls tell me what is meant by 'kernel'?",
  "Solution_3": "Haven't had much time to study the paper, but I assume that the typical definition of linear operator's kernel and image is what you need.\r\n\r\nAnd this is how it was presented to me in my Linear algebra course:\r\n\r\n[quote]The set of all vectors from vector space $ X$ being mapped in vector space's $ Y$ null-vector is denoted with $ \\text{Ker} A$ and called kernel or zero-subspace of linear operator $ A$.\n\nThe set of all vectors in $ Y$ which are mapping images of elements in $ X$ is denoted with $ \\text{Im} A$ or $ A(X)$ and called image of operator $ A$. \n\nIn other words:\n\n$ \\text{Ker} A : \\equal{} \\{ x \\in X|A(x) \\equal{} 0\\}$\n$ \\text{Im} A : \\equal{} \\{y \\in Y|y \\equal{} A(x) \\text{for some } x \\in X\\}$.[/quote]"
}
{
  "Tag": [
    "geometry",
    "geometry solved"
  ],
  "Problem": "Let $O_1,O_2,O_3$ be the centers of ciclers $(C_1),(C_2),(C_3)$ , respectively.\r\nThe circles $(C_1),(C_2),(C_3)$ intersect at point $O$. Let $A_1,A_2,A_3 $ be point on $(C_1),(C_2),(C_3)$, respectively, such that: $ OA_1||O_2O_3;OA_2||O_3O_1;\r\nOA_3||O_1O_2$.\r\nProve that quadrilateral $OA_1A_2A_3$ ic cyclic. (maybe it was posted before)\r\n\r\n[color=red][Moderator edit: Typo corrected.][/color]",
  "Solution_1": "Grobber, that one is for you  :D\r\n\r\n  darij",
  "Solution_2": "You're too kind.. :D:D\r\n\r\nDo you know how hard I tried not to use inversion? I failed :)\r\n\r\nAfter inversion wrt $O$ (any power you want) you get the following problem (I'll leave out the details of how you get to it: it's straightforward): Given a triangle $ABC$ and a point $O$, assume the perpendicular through $O$ to $OA$ cuts $BC$ in $A_1$. In the same way obtain $B_1,C_1$. Show that $A_1,B_1,C_1$ are collinear.\r\n\r\nThis has been discussed before: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1036&highlight=]here[/url]'s the link.",
  "Solution_3": "Indeed, I don't have a non-inversive solution either...\r\n\r\nJust a remark:\r\n\r\n[quote=\"grobber\"]you get the following problem (I'll leave out the details of how you get to it: it's straightforward): Given a triangle $ABC$ and a point $O$, assume the perpendicular through $O$ to $OA$ cuts $BC$ in $A_1$. In the same way obtain $B_1,C_1$. Show that $A_1,B_1,C_1$ are collinear.[/quote]\r\n\r\nThere is a famous solution of this problem in Jacques Hadamard's geometry book. You can find it in the note \"Problem: The ortho-intercepts line\" on [url=http://de.geocities.com/darij_grinberg]my website[/url].\r\n\r\n  Darij",
  "Solution_4": ":rt5:",
  "Solution_5": "[quote=\"pestich\"]In triangle O1O2O3   we have point  O and lines thru O parallel \n       to the sides.  Parallels intersect the 3rd circle in a point symmetrical to O\n      wrt  perpendicular  thru  center of 3rd circle.   So points are concyclic\n     with the  center of the  resulting circle being orthocenter of O1O2O3.[/quote]\r\n\r\nThis is correct and very beautiful except of the last line, where \"orthocenter\" has to be replaced by \"circumcenter\". Thank you!\r\n\r\n  Darij"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "Let $ x,y,z>0$ . Show that :\r\n\r\n\\[ \\frac{x^2(y\\plus{}z)}{(x\\plus{}y)(z\\plus{}x)}\\plus{}\\frac{y^2(z\\plus{}x)}{(y\\plus{}z)(x\\plus{}y)}\\plus{}\\frac{z^2(x\\plus{}y)}{(z\\plus{}x)(y\\plus{}z)}\\le \\frac{x\\plus{}y\\plus{}z}{2}\\]",
  "Solution_1": "[quote=\"Pain rinnegan\"]Let $ x,y,z > 0$ . Show that :\n\\[ \\frac {x^2(y \\plus{} z)}{(x \\plus{} y)(z \\plus{} x)} \\plus{} \\frac {y^2(z \\plus{} x)}{(y \\plus{} z)(x \\plus{} y)} \\plus{} \\frac {z^2(x \\plus{} y)}{(z \\plus{} x)(y \\plus{} z)}\\le \\frac {x \\plus{} y \\plus{} z}{2}\\]\n[/quote]\n\n[quote=\"Pain rinnegan\"]Let $ x,y,z > 0$ . Show that :\n\\[ \\frac {x^2(y \\plus{} z)}{(x \\plus{} y)(z \\plus{} x)} \\plus{} \\frac {y^2(z \\plus{} x)}{(y \\plus{} z)(x \\plus{} y)} \\plus{} \\frac {z^2(x \\plus{} y)}{(z \\plus{} x)(y \\plus{} z)}\\le \\frac {x \\plus{} y \\plus{} z}{2}\\]\n[/quote]\r\nwithout loss of generality, we may assume that $ x\\ge y\\ge z$.\r\n\\[ \\frac {x^2(y \\plus{} z)}{(x \\plus{} y)(z \\plus{} x)} \\plus{} \\frac {y^2(z \\plus{} x)}{(y \\plus{} z)(x \\plus{} y)} \\plus{} \\frac {z^2(x \\plus{} y)}{(z \\plus{} x)(y \\plus{} z)}\\le \\frac {x \\plus{} y \\plus{} z}{2}\\]\r\n\\[ \\Longleftrightarrow \\sum{\\frac {x (x \\minus{} y) (x \\minus{} z)}{2 (x \\plus{} y) (x \\plus{} z)}} \\ge 0\\]\r\n\\[ \\Longleftrightarrow \\frac{{x\\left( {x \\minus{} y} \\right)^2 }}{{2\\left( {x \\plus{} y} \\right)\\left( {x \\plus{} z} \\right)}} \\plus{} \\frac{{x\\left( {x \\minus{} y} \\right)\\left( {y \\minus{} z} \\right)z}}{{\\left( {x \\plus{} y} \\right)\\left( {y \\plus{} z} \\right)\\left( {z \\plus{} x} \\right)}} \\plus{} \\frac{{z\\left( {y \\minus{} z} \\right)^2 }}{{2\\left( {z \\plus{} x} \\right)\\left( {z \\plus{} y} \\right)}} \\ge 0\\]\r\nClearly."
}
{
  "Tag": [
    "probability",
    "graph theory",
    "combinatorics theorems",
    "combinatorics"
  ],
  "Problem": "Hello everybody. I'm preparing for mathematical olympiads. I want to learn graph theory. Can you suggest me good graph theory books? As far as I know , graph theory is the most important topic in combinatorics. I found some but i decided to ask you for help. thx.",
  "Solution_1": "[b]I have a few recommendations:[/b]\r\n\r\n- Bodendiek, Rainer and Burosch, Gustav: [url=http://www.amazon.de/Streifz%C3%BCge-durch-Kombinatorik-Rainer-Bodendiek/dp/386025393X]Streifz\u00fcge durch die Kombinatorik:[/url]  An excellent book in German using many problems from old shorlists and Russia written for the German IMO preparation. [url=http://www.baltic-college.de/kontakt/42-campus-guestrow/18-prof-dr-gustav-burosch-.html]Gustav Burosch[/url] was team leader of the German Democratic Republic from 1970 to 1990 except for 1976, 1971 and 1981 due to political problems. For other political reasons he could not continue his work after the reunification of the two Germanys after 1990. He specifically discusses how many problems can be phrased in the language of graph theory. Burosch was a world expert on combinatorics and graph theory.\r\n\r\n- L\u00e1szl\u00f3 Lov\u00e1sz, [url=http://www.amazon.com/Combinatorial-Problems-Exercises-Chelsea-Publishing/dp/0821842625/ref=sr_1_1?ie=UTF8&s=books&qid=1247754784&sr=8-1]Combinatorial problems and exercises:[/url] IMO math and more advanced research problem collection\r\n\r\n- Ioan Tomescu, R.A. Melter: [url=http://www.amazon.com/Problems-Combinatorics-Interscience-Discrete-Mathematics/dp/0471801550/ref=sr_1_1?ie=UTF8&s=books&qid=1247754876&sr=1-1]Problems in Combinatorics and Graph Theory, 1985:[/url] Romanian IMO Training from the 80s and 90s\r\n\r\n- Diestel, Reinhard: [url=http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/]Graph Theory:[/url] Good reference book and can can be downloaded from the linked URL. \r\n\r\n- Alon, Noga and Spencer, Joel: [url=http://www.amazon.com/Probabilistic-Wiley-Interscience-Discrete-Mathematics-Optimization/dp/0470170204/ref=sr_1_1?ie=UTF8&s=books&qid=1247754925&sr=1-1]The Probabilistic Method:[/url] This book uses graph theory and also teaches a useful technique which can be seen from the book title. A few sections can be downloaded from [url=http://cs.nyu.edu/cs/faculty/spencer/nogabook/nogabook.html]here.[/url]\r\n\r\n- Engel, Konrad: [url=http://opt.math.uni-rostock.de/engel/sperner.html]Sperner Theory:[/url] This book is also quite specialized but teaches you thinking in combinatorics and graph theory",
  "Solution_2": "[quote=\"orl\"]\n\n- Diestel, Reinhard: [url=http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/]Graph Theory:[/url] Good reference book and can can be downloaded from the linked URL. \n[/quote]\r\n\r\nThank you, and can you tell me how good is the quoted book because I already have it?",
  "Solution_3": "Well, I think it is an excellent book. But it is a book written for university students and not necessarily targeted at olympiad preparing students only.",
  "Solution_4": "[quote=\"fallinlovewithmaths\"]Hello everybody. I'm preparing for mathematical olympiads. I want to learn graph theory. Can you suggest me good graph theory books? As far as I know , graph theory is the most important topic in combinatorics. I found some but i decided to ask you for help. thx.[/quote]\r\nIntroduction to Graph Theory By Gary Chartrand\r\n1.\u3000Introduction . \r\n1.1.\u3000Graphs and Graph Models 1 \r\n1.2.\u3000Connected Graphs 9 \r\n1.3.\u3000Common Classes of Graphs 19 \r\n1.4.\u3000Multigraphs and Digraphs 26 \r\n2.\u3000Degrees \r\n2.1.\u3000The Degree of a Vertex 31 \r\n2.2.\u3000Regular Graphs 38 \r\n2.3.\u3000Degree Sequences 43 \r\n2.4.\u3000Excursion:Graphs and Matrices 48 \r\n2.5.\u3000Exploration:Irregular Graphs 50 \r\n3.\u3000Isomorphic Graphs \r\n3.1.\u3000The Definiition of Isomorphism 55 \r\n3.2.\u3000Isomorphism as a Relation 63 \r\n3.3.\u3000Excursion:Graphs and Groups 66 \r\n3.4.\u3000Excursion:Reconstruction and Solvability 76 \r\n4.\u3000Trees \r\n4.1.\u3000Bridges 85 \r\n.\r\n..\r\n..\r\n.",
  "Solution_5": "What about \r\n\r\nDouglas B. West's\r\n\r\nIntroduction to Graph Theory\r\n\r\nhttp://www.amazon.com/Introduction-Graph-Theory-Douglas-West/dp/0130144002\r\n\r\nBut it is too expensive in English. Nevertheless, we can download by searching in Google Book.",
  "Solution_6": "Also try Andreescu's 102 Combinatorial Problems and A Path to Combinatorics for Undergraduates.. Sure, they are really good books for your training.",
  "Solution_7": "There is a new book on graph theory, specifically aimed towards mathematical olympiads. I don't know how much theory is dicussed in this book, but I browsed the first few pages. The problems discussed are very insightful. I haven't read it, so I can't recommend that you buy it, but it is certainly worth a look.\n\nGraph Theory by Xiong Bin and Zheng Zhongyi, vol.3 of the Mathematical Olympiad Series, published by World Scientific. [url=http://www.amazon.com/Graph-Theory-Mathematical-Olympiad-Xiong/dp/9814271128]Here[/url] is a link.",
  "Solution_8": "[quote=\"fallinlovewithmaths\"]graph theory is the most important topic in combinatorics. [/quote]\nSurely not!\nAnyway, Douglas West's book is good for theory and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=45&t=414427]here[/url] are a few problems  :)",
  "Solution_9": "[url=http://www.amazon.com/Introduction-Graph-Theory-Douglas-West/dp/0130144002]Introduction to Graph Theory by Douglas B. West[/url] or [url=http://www.amazon.com/Graph-Theory-Applications-Adrian-Bondy/dp/0444194517]Graph Theory With Applications by John Bondy and U.S.R Murty[/url] are 2 books that I have and am studying for Iran National Olympiad in Informatics.",
  "Solution_10": "[url]http://www.amazon.com/Walk-Through-Combinatorics-Miklos-Bona/dp/9810249012[/url]\n\nThis book seems to have some chapters in graph theory.\nAre they suitable for Olympiad problems?",
  "Solution_11": "I am searching about graph theory too. You have given really important suggestions about some graph theory books. I just want to know the applications of graph theory. Can you make me get some information about it?",
  "Solution_12": "[quote=\"aestik\"][quote=\"fallinlovewithmaths\"]Hello everybody. I'm preparing for mathematical olympiads. I want to learn graph theory. Can you suggest me good graph theory books? As far as I know , graph theory is the most important topic in combinatorics. I found some but i decided to ask you for help. thx.[/quote]\nIntroduction to Graph Theory By Gary Chartrand\n1.\u3000Introduction . \n1.1.\u3000Graphs and Graph Models 1 \n1.2.\u3000Connected Graphs 9 \n1.3.\u3000Common Classes of Graphs 19 \n1.4.\u3000Multigraphs and Digraphs 26 \n2.\u3000Degrees \n2.1.\u3000The Degree of a Vertex 31 \n2.2.\u3000Regular Graphs 38 \n2.3.\u3000Degree Sequences 43 \n2.4.\u3000Excursion:Graphs and Matrices 48 \n2.5.\u3000Exploration:Irregular Graphs 50 \n3.\u3000Isomorphic Graphs \n3.1.\u3000The Definiition of Isomorphism 55 \n3.2.\u3000Isomorphism as a Relation 63 \n3.3.\u3000Excursion:Graphs and Groups 66 \n3.4.\u3000Excursion:Reconstruction and Solvability 76 \n4.\u3000Trees \n4.1.\u3000Bridges 85 \n.\n..\n..\n.[/quote]\n\nWow! This looks like an excellent resource! However, AoPS Intermediate Counting and Probability only uses one chapter to introduce graph theory to the reader. Is there anything that I should do before I start reading this book or should I go dive into it?"
}
{
  "Tag": [
    "geometry",
    "trapezoid",
    "Gauss"
  ],
  "Problem": "[b][size=100][color=DarkBlue]\u03a0\u03a1\u039f\u03a4\u0391\u03a3\u0397. - \u0394\u03af\u03b4\u03b5\u03c4\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2 \u03c4\u03c1\u03b1\u03c0\u03ad\u03b6\u03b9\u03bf $ ABCD$ \u03bc\u03b5 $ AB\\parallel CD$ \u03ba\u03b1\u03b9 \u03ad\u03c3\u03c4\u03c9 $ P\\equiv AC\\cap BD$ \u03ba\u03b1\u03b9 $ Q\\equiv AD\\cap BC.$ \u039c\u03b5 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c4\u03c1\u03bf\u03c5\u03c2 \u03c4\u03b9\u03c2 \u03b2\u03ac\u03c3\u03b5\u03b9\u03c2 \u03c4\u03bf\u03c5 $ AB,$ $ CD,$ \u03b3\u03c1\u03ac\u03c6\u03bf\u03c5\u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03ba\u03cd\u03ba\u03bb\u03bf\u03c5\u03c2 $ (O_{1}),$ $ (O_{2})$ \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03bf\u03af\u03c7\u03c9\u03c2, \u03bf\u03b9 \u03bf\u03c0\u03bf\u03af\u03bf\u03b9 \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03c4\u03b1 \u03c3\u03b7\u03bc\u03b5\u03af\u03b1 \u03ad\u03c3\u03c4\u03c9 $ E,$ $ F,$ $ ($ $ E,$ $ B,$ $ C,$ \u03c0\u03c1\u03bf\u03c2 \u03c4\u03bf \u03b1\u03c5\u03c4\u03cc \u03bc\u03ad\u03c1\u03bf\u03c2 \u03c4\u03b7\u03c2 $ O_{1}O_{2}$ $ ).$ \u0391\u03c0\u03bf\u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 $ \\angle KEM \\equal{} \\angle NEQ$ \u03ba\u03b1\u03b9 $ PE\\perp EQ,$ \u03cc\u03c0\u03bf\u03c5 $ K\\equiv O_{1}O_{2}\\cap EF$ \u03ba\u03b1\u03b9 $ M\\equiv O_{1}O_{2}\\cap (O_{1})$ \u03ba\u03b1\u03b9 $ N\\equiv O_{1}O_{2}\\cap (O_{2})$ $ ,($ $ M,$ \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03c9\u03bd $ O_{1},$ $ Q$ \u03ba\u03b1\u03b9 $ N,$ \u03bc\u03b5\u03c4\u03b1\u03be\u03cd \u03c4\u03c9\u03bd $ O_{2},$ $ Q$ $ ).$[/color][/size][/b]\r\n\r\n\u039a\u03ce\u03c3\u03c4\u03b1\u03c2 \u0392\u03ae\u03c4\u03c4\u03b1\u03c2.\r\n\r\n[b][size=100][color=DarkBlue]PROBLEM. - An isosceles trapezium $ ABCD$ is given, with $ AB\\parallel CD$ and let be the points $ P\\equiv AC\\cap BD$ and $ Q\\equiv AD\\cap BC.$ We draw the circles $ (O_{1}),$ $ (O_{2}),$ with diameters its bases $ AB,$ $ CD$ respectively, which intersect each other at points $ E,$ $ F,$ $ ($ $ E,$ $ B,$ $ C,$ in to the same place with respect to the line segment $ O_{1}O_{2}$ $ ).$ Prove that $ \\angle KEM \\equal{} \\angle NEQ$ and $ PE\\perp EQ,$ where $ K\\equiv O_{1}O_{2}\\cap EF$ and $ M\\equiv O_{1}O_{2}\\cap (O_{1})$ and $ N\\equiv O_{1}O_{2}\\cap (O_{2}),$ $ ($ $ M,$ between $ O_{1},$ $ Q$ and $ N,$ between $ O_{2},$ $ Q$ $ ).$[/color][/size][/b]\r\n\r\nKostas Vittas.",
  "Solution_1": "\u03a4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03c6\u03b1\u03af\u03bd\u03b5\u03c4\u03b1\u03b9 \u03b4\u03cd\u03c3\u03ba\u03bf\u03bb\u03bf (\u03b2\u03b1\u03c3\u03b9\u03ba\u03ac \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03ac \u03bc\u03c0\u03b5\u03c1\u03b4\u03b5\u03bc\u03ad\u03bd\u03bf \u03c4\u03bf \u03c3\u03c7\u03ae\u03bc\u03b1) \u03b1\u03bb\u03bb\u03ac \u03c4\u03bf \u03b4\u03b5\u03cd\u03c4\u03b5\u03c1\u03bf \u03bd\u03bf\u03bc\u03af\u03b6\u03c9 \u03cc\u03c4\u03b9 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03c7\u03c9\u03c1\u03af\u03c2 \u03ba\u03b1\u03bd \u03c4\u03bf $ ABCD$ \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03c4\u03c1\u03b1\u03c0\u03ad\u03b6\u03b9\u03bf! \r\n\u03a3\u03cd\u03bc\u03c6\u03c9\u03bd\u03b1 \u03bc\u03b5 \u03c4\u03bf \u03b8\u03b5\u03ce\u03c1\u03b7\u03bc\u03b1 Gauss-Bodenmiller \u03c3\u03b5 \u03ad\u03bd\u03b1 \u03c0\u03bb\u03ae\u03c1\u03b5\u03c2 \u03c4\u03b5\u03c4\u03c1\u03ac\u03c0\u03bb\u03b5\u03c5\u03c1\u03bf \u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9 \u03bc\u03b5 \u03b4\u03b9\u03b1\u03bc\u03ad\u03c4\u03c1\u03bf\u03c5\u03c2 \u03c4\u03b9\u03c2 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03af\u03bf\u03c5\u03c2 \u03c4\u03bf\u03c5 \u03ad\u03c7\u03bf\u03c5\u03bd \u03ba\u03bf\u03b9\u03bd\u03cc \u03c1\u03b9\u03b6\u03b9\u03ba\u03cc \u03ac\u03be\u03bf\u03bd\u03b1 \u03bf \u03bf\u03c0\u03bf\u03af\u03bf\u03c2 \u03bc\u03ac\u03bb\u03b9\u03c3\u03c4\u03b1 \u03b4\u03b9\u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 \u03ba\u03b9 \u03b1\u03c0' \u03c4\u03b1 \u03bf\u03c1\u03b8\u03cc\u03ba\u03b5\u03c4\u03c1\u03b1 \u03c4\u03c9\u03bd \u03c4\u03c1\u03b9\u03b3\u03ce\u03bd\u03c9\u03bd \u03c4\u03bf\u03c5 \u03c0\u03bb\u03ae\u03c1\u03bf\u03c5\u03c2 \u03c4\u03b5\u03c4\u03c1\u03b1\u03c0\u03bb\u03b5\u03cd\u03c1\u03bf\u03c5! \u0395\u03b4\u03ce \u03c6\u03b1\u03bd\u03b5\u03c1\u03ac \u03bf\u03b9 \u03b4\u03b9\u03b1\u03b3\u03ce\u03bd\u03b9\u03bf\u03b9 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bf\u03b9 $ AB, CD, PQ$ \u03ba\u03b1\u03b9 \u03bf \u03c1\u03b9\u03b6\u03b9\u03ba\u03cc\u03c2 \u03ac\u03be\u03bf\u03bd\u03b1\u03c2 \u03b7 $ EF$ \u03ac\u03c1\u03b1 \u03bf \u03ba\u03cd\u03ba\u03bb\u03bf\u03c2 \u03bc\u03b5 \u03b4\u03b9\u03ac\u03bc\u03b5\u03c4\u03c1\u03bf $ PQ$ \u03b4\u03b9\u03ad\u03c1\u03c7\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0' \u03c4\u03b1 $ E,F$ \u03ac\u03c1\u03b1 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf :)",
  "Solution_2": "\u0393\u03b9\u03b1 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03b1\u03c1\u03ba\u03b5\u03af \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9:\r\n$ ENM \\equal{}$ \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03c4\u03cc\u03c4\u03b5 $ EF // AB$ \u03ac\u03c1\u03b1 $ EF$ \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c3\u03c4\u03b7\u03bd $ MN$ \u03ba\u03b1\u03b9 \u03ac\u03c1\u03b1 $ EK$ = \u03cd\u03c8\u03bf\u03c2 $ \\Leftrightarrow EK$ = \u03b4\u03b9\u03c7\u03bf\u03c4\u03cc\u03bc\u03bf\u03c2\r\n\r\n\u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9:\r\n[hide]\u03b1\u03bd $ O, P$ \u03c4\u03b1 \u03bc\u03ad\u03c3\u03b1 \u03c4\u03c9\u03bd $ AB, CD$ \u03b5\u03af\u03bd\u03b1\u03b9 $ K \\equal{}$ \u03bc\u03ad\u03c3\u03bf $ OP$ \u03ac\u03c1\u03b1 $ EOP \\equal{}$ \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2,\n\u03b1\u03c6\u03bf\u03cd \u03b5\u03af\u03bd\u03b1\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 $ EK \\equal{}$ \u03ba\u03ac\u03b8\u03b5\u03c4\u03b7 \u03c3\u03c4\u03b7\u03bd $ OP$\n\n\u03ac\u03c1\u03b1 $ \\widehat{KOE} \\equal{} \\widehat{KPE} \\Leftrightarrow$\n(\u03b5\u03c0\u03b5\u03b9\u03b4\u03ae \u03c4\u03b1 \u03c4\u03c1\u03af\u03b3\u03c9\u03bd\u03b1 $ EPN, EOM$ \u03b5\u03af\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ae \u03b1\u03c6\u03bf\u03cd \u03ad\u03c7\u03bf\u03c5\u03bd \u03b4\u03cd\u03bf \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2 \u03c4\u03bf\u03c5\u03c2 \u03af\u03c3\u03b5\u03c2 \u03c9\u03c2 \u03b1\u03ba\u03c4\u03af\u03bd\u03b5\u03c2 \u03c4\u03c9\u03bd \u03ba\u03cd\u03ba\u03bb\u03c9\u03bd $ (O2), (O1)$ \u03b1\u03bd\u03c4\u03af\u03c3\u03c4\u03bf\u03b9\u03c7\u03b1)\n$ \\widehat{OME} \\equal{} \\widehat{PNE}$\n\n\u03ac\u03c1\u03b1 $ ENM \\equal{}$ \u03b9\u03c3\u03bf\u03c3\u03ba\u03b5\u03bb\u03ad\u03c2 :D \n[/hide]\r\n\r\n\r\n\u0388\u03b3\u03c1\u03b1\u03c8\u03b1 \u03c4\u03b9\u03c2 \u03c3\u03ba\u03ad\u03c8\u03b5\u03b9\u03c2 \u03bc\u03bf\u03c5 \u03c0\u03bf\u03bb\u03cd \u03c3\u03c5\u03bc\u03c0\u03b7\u03ba\u03bd\u03c9\u03bc\u03ad\u03bd\u03b5\u03c2 \u03b3\u03b9\u03b1\u03c4\u03af \u03b5\u03af\u03bd\u03b1\u03b9 \u03c3\u03c0\u03ac\u03c3\u03b9\u03bc\u03bf \u03bd\u03b1 \u03c4\u03b9\u03c2 \u03b3\u03c1\u03ac\u03c6\u03c9 \u03c3\u03c9\u03c3\u03c4\u03ac. :blush: \u039a\u03b1\u03b9 \u03c0\u03b1\u03af\u03c1\u03bd\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c5\u03c0\u03cc\u03c8\u03b9\u03bd \u03cc\u03c4\u03b9 \u03b4\u03b5\u03bd \u03ad\u03ba\u03b1\u03bd\u03b1 \u03ba\u03b1\u03bd \u03c3\u03c7\u03ae\u03bc\u03b1 \u03af\u03c3\u03c9\u03c2 \u03bd\u03b1 \u03ad\u03c7\u03c9 \u03bb\u03ac\u03b8\u03b7. \u03a0\u03b5\u03af\u03c4\u03b5 \u03b1\u03bd \u03b2\u03c1\u03b5\u03af\u03c4\u03b5 \u03ba\u03ac\u03c0\u03bf\u03b9\u03bf.",
  "Solution_3": "\u0394\u03c5\u03c3\u03c4\u03c5\u03c7\u03ce\u03c2, \u03be\u03b1\u03bd\u03b1\u03b2\u03bb\u03ad\u03c0\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c4\u03b7\u03bd \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7, \u03c3\u03c5\u03bc\u03c0\u03b5\u03c1\u03b1\u03af\u03bd\u03c9 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03c0\u03c1\u03ce\u03c4\u03bf \u03c5\u03c0\u03bf\u03b5\u03c1\u03ce\u03c4\u03b7\u03bc\u03b1 \u03b4\u03b5\u03bd \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b3\u03b9\u03b1\u03c4\u03af\r\n\u03c3\u03b5 \u03cc\u03bb\u03b5\u03c2 \u03c4\u03b9\u03c2 \u03c3\u03c7\u03ad\u03c3\u03b5\u03b9\u03c2 \u03c0\u03bf\u03c5 \u03c7\u03c1\u03b7\u03c3\u03b9\u03bc\u03bf\u03c0\u03bf\u03af\u03b7\u03c3\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9 \u03b4\u03b9\u03c0\u03bb\u03ae \u03c3\u03c5\u03bd\u03b5\u03c0\u03b1\u03b3\u03c9\u03b3\u03ae \u03ba\u03b1\u03b9\r\n\u03b3\u03b9\u03b1 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 $ K \\equal{}$ \u03bc\u03ad\u03c3\u03bf \u03c4\u03bf\u03c5 $ OP$ \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 $ EO \\equal{} EP$ \u03b1\u03c6\u03bf\u03cd $ EK \\equal{}$ \u03ba\u03ac\u03b8\u03b5\u03c4\u03bf \u03c3\u03c4\u03bf $ OP$\r\n\u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bf\u03b9 \u03b1\u03ba\u03c4\u03af\u03bd\u03b5\u03c2 \u03c4\u03c9\u03bd \u03ba\u03cd\u03ba\u03bb\u03c9\u03bd \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03b5\u03c2, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bf\u03b9 \u03b2\u03ac\u03c3\u03b5\u03b9\u03c2 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03af\u03c3\u03b5\u03c2, \u03b4\u03b7\u03bb\u03b1\u03b4\u03ae \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c1\u03b8\u03bf\u03b3\u03ce\u03bd\u03b9\u03bf \u03c0\u03b1\u03c1\u03b1\u03bb\u03bb\u03b7\u03bb\u03cc\u03b3\u03c1\u03b1\u03bc\u03bc\u03bf, \u03bf\u03c0\u03cc\u03c4\u03b5 \u03b4\u03b5\u03bd \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03c0\u03bb\u03b5\u03c5\u03c1\u03ad\u03c2\r\n\u03b1\u03ba\u03cc\u03bc\u03b1, \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03bf\u03b8\u03b5\u03af \u03b7 \u03b4\u03b9\u03b5\u03c5\u03ba\u03c1\u03af\u03bd\u03b7\u03c3\u03b7 \u03cc\u03c4\u03b9 \u03c4\u03bf \u03cd\u03c8\u03bf\u03c2 \u03c0\u03c1\u03ad\u03c0\u03b5\u03b9 \u03bd\u03b1 \u03b5\u03af\u03bd\u03b1\u03b9 \u03bc\u03b9\u03ba\u03c1\u03cc\u03c4\u03b5\u03c1\u03bf \u03c4\u03bf\u03c5 \u03b7\u03bc\u03b9\u03b1\u03b8\u03c1\u03bf\u03af\u03c3\u03bc\u03b1\u03c4\u03bf\u03c2 \u03c4\u03c9\u03bd \u03b2\u03ac\u03c3\u03b5\u03c9\u03bd \u03b3\u03b9\u03b1 \u03bd\u03b1 \u03c4\u03ad\u03bc\u03bd\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bf\u03b9 \u03ba\u03cd\u03ba\u03bb\u03bf\u03b9,"
}
{
  "Tag": [
    "inequalities",
    "geometry",
    "search",
    "inequalities unsolved"
  ],
  "Problem": "Prove or disprove: $a, b, c > 0 \\rightarrow a^4 + b^4 + c^4 + d^4 + 8abcd \\geq a^2bc + ab^2c + abc^2 + a^2bd + ab^2d + abd^2 + a^2cd + ac^2d + acd^2 + b^2cd + bc^2d + bcd^2$\r\n\r\nAre there any other inequalities with a weak $abcd$ term on the wrong side (which is no schurable)? Edit: I added this line just as harazi replied.",
  "Solution_1": "For positive real numbers, take a look at a more general problem in the open questions area, caller \"Schur for n numbers\", posted by Vasc. I still remember the nice solution I gave to it. Your inequality is the case n=4.",
  "Solution_2": "To answer your second question, the strongest inequality I know with the product in the suffering part of the inequality is Suranyi's inequality.",
  "Solution_3": "Can you write that inequality for me ? Ok.\r\n\r\n\r\n---------------\r\nReyes.",
  "Solution_4": "Just use search button and not only you'll find it, but alsoi its proof.",
  "Solution_5": "Thanks! http://www.mathlinks.ro/Forum/viewtopic.php?highlight=schur&t=14906",
  "Solution_6": "I have read it .It's hard and beautiful.Thanks!"
}
{
  "Tag": [],
  "Problem": "As the subject says, the problem below may seem ugly at first, but there is an excellent solution to it.\r\n\r\nFind the largest integer N such that $N < (\\sqrt{33+ \\sqrt{128}} + \\sqrt{2} - 8)^{-1}$",
  "Solution_1": "Considering that $\\sqrt{128}=8.\\sqrt{2}$:\r\n\r\n$(\\sqrt{33+ \\sqrt{128}} + \\sqrt{2} - 8)^{-1}=({\\sqrt{\\frac{(\\sqrt{ 2}+8)^2}{2}}+\\sqrt{2} - 8})^{-1}=({\\frac{\\sqrt{2}+8}{\\sqrt{2}}+\\sqrt{2} - 8})^{-1}=({5.\\sqrt{2}-7})^{-1}$ \r\nthus:\r\n\r\n${}(5.\\sqrt{2}-7)^{-1}=\\frac{1.(5.\\sqrt{2}+7)}{(5.\\sqrt{2}-7).(5.\\sqrt{2}+7)}=5.\\sqrt{2}+7$\r\n\r\nKnowing that the approached value of $\\sqrt{2}$ is $1,41$ then the approach of $5.\\sqrt{2}+7$ is $14,07$ and the nearest number $N$ of the approach is 14",
  "Solution_2": "FWIW - A \"fun\" way is to notice, as you have shown $5\\sqrt2 - 7$ is less than 1(being reciprocal of $5\\sqrt 2 +7 )$)  and we  see \r\n$(5\\sqrt 2 +7 ) - (5 \\sqrt 2 - 7 )$ is  $14$  (so the requred answer is 14 without sort of  pluging in  the value of $\\sqrt 2$  :)"
}
{
  "Tag": [
    "function",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Prove or supply a counterexample: If the function $f$ from $\\mathbb{R}$ to $\\mathbb{R}$ has both a left limit and a right limit at each point of $\\mathbb{R}$, then the set of discontinuities of $f$ is,at most,countable.",
  "Solution_1": "It's a very well-known problem, and I think it's been discussed before (I remember Myth posting  very similar one, used in a math contest from Russia this year).\r\n\r\nThe set $A_n$ consisting of points $x$ for which $|f_+(x)-f_-(x)|\\ge\\frac 1n$ is discrete and thus countable ($f_+,f_-$ are the right and left limits, respectively), and since the set of discontinuities is the union of the $A_n$'s, that's that.",
  "Solution_2": "[quote=\"grobber\"]The set $A_n$ consisting of points $x$ for which $|f_+(x)-f_-(x)|\\ge\\frac 1n$ is discrete and thus countable ($f_+,f_-$ are the right and left limits, respectively), and since the set of discontinuities is the union of the $A_n$'s, that's that.[/quote]\r\nIt doesn't include all the cases, not $|f_+(x)-f_-(x)|\\ge\\frac 1n$. Use $\\max\\{|f_+(x)-f(x)|,|f_-(x)-f(x)|\\}\\ge\\frac 1n$ instead.",
  "Solution_3": "Right. Sorry, I was thinking about monotonic functions :).",
  "Solution_4": "For monotonic functions, we don't need to do it in that way..."
}
{
  "Tag": [
    "geometry",
    "3D geometry"
  ],
  "Problem": "What is the sum of all the perfect cubes under 100?\r\n\r\n\r\n\r\n\r\nHow do you solve this? Please help. :?:",
  "Solution_1": "This seems too easy so maybe I'm wrong: $ (1)^{3}\\plus{}2^{3}\\plus{}3^{3}\\plus{}4^{3}\\equal{}1\\plus{}8\\plus{}27\\plus{}64\\equal{}100$.",
  "Solution_2": "Well, we know that\r\n\r\n$ 1^3\\plus{}2^3\\plus{}3^3\\plus{}...\\plus{}n^3\\equal{}(1\\plus{}2\\plus{}3\\plus{}...\\plus{}n)^2$. So, since the biggest\r\n\r\nperfect cube less then $ 100$ is $ 64\\equal{}4^3$, $ n\\equal{}4$. \r\n\r\nSo, $ (1\\plus{}2\\plus{}3\\plus{}4)^2\\equal{}10^2\\equal{}100$."
}
{
  "Tag": [
    "geometry",
    "rectangle",
    "induction"
  ],
  "Problem": "Jean is paving a 2 by 10 rectangle with 1 by 2 paving stones. In how many different ways can she do this?",
  "Solution_1": "lol. Did you put this up because you like the question or the pun?",
  "Solution_2": "[hide]Let f(n) be the number of ways to pave a 2xn rectangle. By induction, we found that f(n)=f(n-1)+f(n-2), where f(1)=1, f(2)=2. So f(10)=89.[/hide]",
  "Solution_3": "super008 wrote:[hide]Let f(n) be the number of ways to pave a 2xn rectangle. By induction, we found that f(n)=f(n-1)+f(n-2), where f(1)=1, f(2)=2. So f(10)=89.[/hide]\n\n\n\nsuper, to find your pattern did you calculate some of the numbers and notice the pattern or do you have a reason why it would follow that recurrence? If you have a reason, you should probably explain it.",
  "Solution_4": "I have done this problem before. I first saw the pattern then did the induction.",
  "Solution_5": "[quote=\"zabelman\"]lol. Did you put this up because you like the question or the pun?[/quote]\r\n\r\nJust the pun, sorry.\r\n\r\nThough Super008 should show his induction.",
  "Solution_6": "What I meant is that you should explain why the pattern is true so other people can follow your solution.",
  "Solution_7": "[hide]We observed that f(1)=1, f(2)=2, f(3)=3, f(4)=5, f(5)=8. And they look like consecutive Fibonacci numbers. \n\n\n\nClaim: f(n) = F_(n-1), where F_n is the nth Fibonacci number. \n\nProof: We will prove the claim by induction on n.\n\n\n\nWe have verify that it is true for n=1. \n\nAssume that f(n)=F_(n-1) is true for n=k, we need to prove that it is also true for n=k+1.\n\n\n\nNow in order to pave a 2x(k+1) rectangle, we could either put a 2x1 rectangle down like \"|\", then pave the rest, which is a  2xk rectangle; or we could put down two rectangles across like \"=\", then pave the rest, which is a 2x(k-1) rectangle. Since thses are the only starting position to pave the rectangle, we have f(k+1) = f(k) + f(k-1). \n\n\n\nBy assumption, f(k) = F_(k-1) and f(k-1) = F_(k-2). Substitute them in, we have f(k+1) = F_(k-1) + F_(k-2) = F_k. Thus we have proved that the assumption is true for n=k+1. By induction, we know that f(n)=F_(n-1) is true for positive integer n.[/hide]",
  "Solution_8": "That should actually be F_(n + 1), but good job otherwise."
}
{
  "Tag": [
    "inequalities",
    "inequalities unsolved"
  ],
  "Problem": "suppose a,b,c are side length of a triangle, proof\r\n(1)$ a^3\\plus{}b^3\\plus{}c^3\\plus{}3abc\\minus{}2b^2a\\minus{}2c^2b\\minus{}2a^2c\\ge0$\r\n(2)$ 3a^2b\\plus{}3b^2c\\plus{}3c^2a\\minus{}3abc\\minus{}2b^2a\\minus{}2c^2b\\minus{}2a^2c\\ge0$ :D",
  "Solution_1": "I've also been wondering whether I was really the first one to find them. None of them is hard anyway.\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=35295\r\n\r\n  darij",
  "Solution_2": "Easy :?: .. not at all to me\r\n\r\nwhy $ x^{3} \\plus{} 2xy^{2} \\plus{} y^{3} \\plus{} 2yz^{2} \\plus{} z^{3} \\plus{} 2zx^{2}\\geq 3(x^{2}y \\plus{} y^{2}z \\plus{} z^{2}x)$? (not necessary sides of triangle any more)",
  "Solution_3": "[quote=\"greentreeroad\"]Easy :?: .. not at all to me\n\nwhy $ x^{3} \\plus{} 2xy^{2} \\plus{} y^{3} \\plus{} 2yz^{2} \\plus{} z^{3} \\plus{} 2zx^{2}\\geq 3(x^{2}y \\plus{} y^{2}z \\plus{} z^{2}x)$? (not necessary sides of triangle any more)[/quote]\r\n its easy by EMV theorem :)",
  "Solution_4": "[quote=\"greentreeroad\"]Easy :?: .. not at all to me\n\nwhy $ x^{3} \\plus{} 2xy^{2} \\plus{} y^{3} \\plus{} 2yz^{2} \\plus{} z^{3} \\plus{} 2zx^{2}\\geq 3(x^{2}y \\plus{} y^{2}z \\plus{} z^{2}x)$? (not necessary sides of triangle any more)[/quote]\r\nIf $ x\\equal{}\\min\\{x,y,z\\},$ $ y\\equal{}x\\plus{}u$ and $ z\\equal{}x\\plus{}v$ then $ \\sum_{cyc}(x^3\\minus{}3x^2y\\plus{}2xy^2)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow2(u^2\\minus{}uv\\plus{}v^2)x\\plus{}u^3\\minus{}3u^2v\\plus{}2uv^2\\plus{}v^3\\geq0,$ which obviously true.",
  "Solution_5": "[quote=\"M.A\"]\n its easy by EMV theorem :)[/quote]\r\n\r\nWhat is EMV...  :blush:",
  "Solution_6": "[quote=\"greentreeroad\"][quote=\"M.A\"]\n its easy by EMV theorem :)[/quote]\n\nWhat is EMV...  :blush:[/quote]\r\n its entirely mixing variables method :wink:",
  "Solution_7": "[quote=\"M.A\"][quote=\"greentreeroad\"][quote=\"M.A\"]\n its easy by EMV theorem :)[/quote]\n\nWhat is EMV...  :blush:[/quote]\n its entirely mixing variables method :wink:[/quote]\r\nwhat is entirly mixing variables please post that theory",
  "Solution_8": "[quote=\"Javkhaa\"][/quote][quote=\"M.A\"][quote=\"greentreeroad\"][quote=\"M.A\"]\n its easy by EMV theorem :)[/quote]\n\nWhat is EMV...  :blush:[/quote]\n its entirely mixing variables method :wink:[/quote][quote=\"Javkhaa\"]\nwhat is entirly mixing variables please post that theory[/quote]\r\nDownload file: http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf",
  "Solution_9": "[quote=\"greentreeroad\"]suppose a,b,c are side length of a triangle, proof\n(1)$ a^3 \\plus{} b^3 \\plus{} c^3 \\plus{} 3abc \\minus{} 2b^2a \\minus{} 2c^2b \\minus{} 2a^2c\\ge0$\n[/quote]\r\nSuppose that $ c\\equal{}min\\{a;b;c\\}$\r\nThis ineq can rewrite:\r\n$ (a\\plus{}b\\minus{}c)(a\\minus{}b)^2\\plus{}(a\\plus{}c\\minus{}b)(a\\minus{}c)(b\\minus{}c) \\geq 0$\r\nNote that $ a,b,c$ are side length of a triangle,so we have $ a\\plus{}b\\minus{}c>0;a\\plus{}c\\minus{}b>0$\r\nDone!",
  "Solution_10": "[quote=\"greentreeroad\"]suppose a,b,c are side length of a triangle, proof\n(2)$ 3a^2b \\plus{} 3b^2c \\plus{} 3c^2a \\minus{} 3abc \\minus{} 2b^2a \\minus{} 2c^2b \\minus{} 2a^2c\\ge0$ :D[/quote]\r\nSuppose that $ c\\equal{}min\\{a;b;c\\}$\r\nWe can rewrite this ineq:\r\n$ c(a\\minus{}b)^2\\plus{}(2a\\minus{}b)(a\\minus{}c)(b\\minus{}c) \\geq 0$\r\nWe have $ 2a\\minus{}b \\geq a\\plus{}c\\minus{}b >0$\r\nDone!",
  "Solution_11": "Is this SOS-Schur?"
}
{
  "Tag": [
    "Ramsey Theory",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Find the minimum number of point in plane in which three arbitrary point aren't collinear satisfy for all ways of painting all segment over two point always exit a triangle in which three sides are painted by a same colour!",
  "Solution_1": "Well...it depends on the number of colours... :wink: \r\nFor $ n$ colours, the desired result is the Ramsey number $ R(3,\\cdots,3)$.\r\nIn particular, for two colours, the desired minimum is $ 6$.\r\n\r\nPierre."
}
{
  "Tag": [
    "Columbia",
    "function",
    "integration",
    "algebra",
    "polynomial",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ n$ a positive integer, and $ f: [0,1]\\to\\mathbb{R}$ a continuos function such that \\[ \\int_{0}^{1}x^{k}f(x) dx\\equal{}1\\] for $ k\\equal{}0,1,2,\\ldots,n\\minus{}1$. Prove that \\[ \\int_{0}^{1}f^{2}(x) dx\\geq n^{2}\\]",
  "Solution_1": "See http://www.mathlinks.ro/Forum/viewtopic.php?&t=146373 or http://www.mathlinks.ro/viewtopic.php?p=443980#443980\r\n(The approaches are similar, but I don't understand why $ \\int_{0}^{1}\\left( f(x)\\minus{}q(x)\\right) q(x)\\, dx \\equal{} 0$ in jmerry's solution.)",
  "Solution_2": "It's about orthogonality. Let $ p_{0},p_{1},\\dots,p_{n\\minus{}1}$ be an orthogonal basis for polynomials of degree $ \\le n\\minus{}1$; the hypothesis is equivalent to the system $ \\langle f,p_{i}\\rangle\\equal{}c_{i}$ for some fixed constants $ c_{i}$. When we subtract the solutions, we get $ \\langle f\\minus{}q,p_{i}\\rangle\\equal{}0$ for each $ i$. Since the $ p_{i}$ span the polynomials of degree $ \\le n\\minus{}1$, $ f\\minus{}q$ is orthogonal to all polynomials of degree $ n\\minus{}1$, including $ q$ in particular."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "I had breadsticks and Marinara.(breakfast, not lunch.)",
  "Solution_1": "wtf is up wit tis\r\n\r\nwhatev, ill do it\r\n\r\ni had poha\r\n\r\nguz wat tat is",
  "Solution_2": "I had banana bread, 2 bagels, 1 sports drink, 1 water, and two  small pieces of sandwich...lunch time!",
  "Solution_3": "I haven't had breakfast yet...",
  "Solution_4": "Right now I'm having lunch....chocolate and bananas :wink:",
  "Solution_5": "lunch=grilled cheese and fries",
  "Solution_6": "a bowl of cinnamon toast crunch with 1% milk.",
  "Solution_7": "Nothing, I don't eat breakfast.",
  "Solution_8": "Rice with butter and a fried egg - I leave for college after that.",
  "Solution_9": "i never eat breakfast(seriously)",
  "Solution_10": "poha is good!\r\n\r\ni had puri aloo kudu and hulwa, because its janmashtmi",
  "Solution_11": "I never eat breakfast either.",
  "Solution_12": "[quote=\"star99\"]poha is good!\n\ni had puri aloo kudu and hulwa, because its janmashtmi[/quote]\r\n\r\nGARBA",
  "Solution_13": "I had a chili dog and a pizza pretzel",
  "Solution_14": "I'm eating waffles!!!!! Ego my Lego! Or wtf.\r\nI always thought that meant I have my I say b/c of greek.",
  "Solution_15": "[quote=\"funcia\"]I'm eating waffles!!!!! Ego my Lego! Or wtf.\nI always thought that meant I have my I say b/c of greek.[/quote]\r\nLol! :rotfl: \r\n\r\nSorry, but I find this very funny.",
  "Solution_16": "The first thing I do when I wake up is eat.  I've tried to wait until I'm out of my room to eat since I think it bothers my roommate when I rummage through my cabinet while he's still trying to sleep :P (I mean, I try really hard to be quiet :( ).\r\n\r\nAnyway, today I had an energy bar, a couple pieces of bread with peanut butter, and a banana.\r\n\r\nMy two favorite things for breakfast are muffins and breakfast burritos (from Anna's Taqueria)...mmm.",
  "Solution_17": "I had a chocolate chip cookie dough poptart and strawberries. Now I get to have a bowl of cereal as an almost-1am snack. :jump:  :rotfl:",
  "Solution_18": "I usually have something random: The probability that I'll have oatmeal is definately greater tha 1/2  :(",
  "Solution_19": "Ah, I need to update this with what I ate today:\r\n\r\ntwo bagels (with nutella), 2 pieces of french toast, 1 pancake, a handful of fruits+nuts, 2 glasses of smoothies, 1 slice of ham, 1/2 nutri-grain bar\r\n\r\nSo full....",
  "Solution_20": "Two fried eggs, a pickle, and green tea. :D",
  "Solution_21": "Two boiled eggs, a piece of toast with peanut butter, a bowl of cereal, three glasses of orange juice, a bagel with cream cheese, and half a blueberry muffin.",
  "Solution_22": "NOTHING AT ALL. I have to fast today :(",
  "Solution_23": "Me too, that's why I ate so much yesterday."
}
{
  "Tag": [],
  "Problem": "Go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209860]here[/url] for the rules\r\n\r\nOne difference: there will be no quests.\r\n\r\n14 people to start. (1 king, 1 queen, 2 known knights, 1 unknown knight, 9 citizens)\r\n\r\nSignups:\r\n\r\n1. meewhee009\r\n2. Scamper\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.",
  "Solution_1": "1. meewhee\r\n2.\r\n3.\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.\r\n\r\nOh yeah. /in :)",
  "Solution_2": "Me joins as well.",
  "Solution_3": "1. meewhee009\r\n2. Scamper\r\n3. jjfun1\r\n4.\r\n5.\r\n6.\r\n7.\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.\r\n\r\nSounds fun.\r\n[b]Join.[/b]\r\nW00t, I got my lucky number. :)",
  "Solution_4": "Then why aren't you jjfun3?  This has been a public shaming of jjfun1.  Thank you. [yet another revival]\r\n\r\n1. meewhee009 \r\n2. Scamper \r\n3. jjfun1 \r\n4. ernie\r\n5. \r\n6. \r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12. \r\n13. \r\n14.",
  "Solution_5": "1. meewhee009 \r\n2. Scamper \r\n3. jjfun1 \r\n4. ernie \r\n5. 15!!!\r\n6. \r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12. \r\n13. \r\n14.",
  "Solution_6": "[quote=\"ernie\"]Then why aren't you jjfun3?  This has been a public shaming of jjfun1.[/quote]\r\nMaybe in ernie-land, three and fun rhyme. But not in jjfun1-land. :P",
  "Solution_7": "1. meewhee009 \r\n2. Scamper \r\n3. jjfun1 \r\n4. ernie \r\n5. 15!!! \r\n6. westiepaw\r\n7. \r\n8. \r\n9. \r\n10. \r\n11. \r\n12. \r\n13. \r\n14.",
  "Solution_8": "/in :D  :D  :D",
  "Solution_9": "1. meewhee009\r\n2. Scamper\r\n3. jjfun1\r\n4. ernie\r\n5. 15!!!\r\n6. westiepaw\r\n7. EggyLv.999\r\n8.\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.",
  "Solution_10": "\\me is in.",
  "Solution_11": "1. meewhee009\r\n2. Scamper\r\n3. jjfun1\r\n4. ernie\r\n5. 15!!!\r\n6. westiepaw\r\n7. EggyLv.999\r\n8. spaceguy524\r\n9.\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.",
  "Solution_12": "1. meewhee009\r\n2. Scamper\r\n3. jjfun1\r\n4. ernie\r\n5. 15!!!\r\n6. westiepaw\r\n7. EggyLv.999\r\n8. spaceguy524\r\n9. FlyAgaric\r\n10.\r\n11.\r\n12.\r\n13.\r\n14.",
  "Solution_13": "Sure, I'll play.\r\n\r\n1. meewhee009\r\n2. Scamper\r\n3. jjfun1\r\n4. ernie\r\n5. 15!!!\r\n6. westiepaw\r\n7. EggyLv.999\r\n8. spaceguy524\r\n9. FlyAgaric\r\n10. cf249\r\n11.\r\n12.\r\n13.\r\n14.",
  "Solution_14": "join!\r\n :nhl: \r\nI will beat everyone I want to beat using this hockey stick.\r\nAlso, this reminds me of \"The once and future king\"!\r\n*Read, Read, Journal, Read...*",
  "Solution_15": "Mwahahaha :).\r\n\r\nmeewhee009-20 pts\r\nScamper-21 pts\r\njjfun1-9 pts\r\nwestiepaw-5 pts\r\nEggyLv.999-10 pts\r\nspaceguy524-10 pts\r\nFlyAgaric-10 pts\r\nBramblestar-7 pts\r\nmathling235-18 pts \r\n[b]Scamper for 1[/b]",
  "Solution_16": "When are execution days again?",
  "Solution_17": "[b]Teh king for 1[/b]",
  "Solution_18": "Every third day is execution day.",
  "Solution_19": "When is the next execution day?",
  "Solution_20": "Day 9 is next execution day.",
  "Solution_21": "meewhee009-19 pts \r\nScamper-21 pts \r\njjfun1-9 pts \r\nwestiepaw-5 pts \r\nEggyLv.999-9 pts \r\nspaceguy524-10 pts \r\nFlyAgaric-10 pts \r\nBramblestar-7 pts \r\nmathling235-18 pts \r\n\r\nDay 8 ends.\r\n\r\nDay 9 starts. Everyone gets their salaries:\r\n\r\nmeewhee009-22 pts \r\nScamper-23 pts \r\njjfun1-10 pts \r\nwestiepaw-6 pts \r\nEggyLv.999-10 pts \r\nspaceguy524-11 pts \r\nFlyAgaric-11 pts \r\nBramblestar-8 pts \r\nmathling235-19 pts \r\n\r\nRandom Event: King Execution Day\r\n\r\nThe king may pay 15 points to instantly kill any one person. Under the terms of the new law, EggyLv.999 can be executed for free.",
  "Solution_22": "If we attack the king today, can the king execute us for free or does she have to wait for the next execution day? Or is the law no longer being imposed?\r\n\r\nEggyLv.999 should attack the king.",
  "Solution_23": "meewhee009-22 pts \r\nScamper-23 pts \r\njjfun1-10 pts \r\nwestiepaw-6 pts \r\n[b]EggyLv.999 haz been killedzzz![/b]\r\nspaceguy524-11 pts \r\nFlyAgaric-11 pts \r\nBramblestar-8 pts \r\nmathling235-19 pts \r\n :P \r\nI have a new aim...\r\nhehehehe...",
  "Solution_24": "The king now has to wait until day 12 until another execution can be done.",
  "Solution_25": "meewhee009-22 pts \r\nScamper-23 pts \r\njjfun1-10 pts \r\nwestiepaw-6 pts \r\nspaceguy524-11 pts \r\nFlyAgaric-11 pts \r\nBramblestar-8 pts \r\nmathling235-19 pts \r\n\r\nDay 9 ends.\r\n\r\nDay 10 starts. Everyone gets their salaries\r\n\r\nmeewhee009-25 pts \r\nScamper-25 pts \r\njjfun1-11 pts \r\nwestiepaw-7 pts \r\nspaceguy524-12 pts \r\nFlyAgaric-12 pts \r\nBramblestar-9 pts \r\nmathling235-20 pts\r\n\r\nRandom Event: Small Money Flow\r\n\r\nSo money is becoming scarce in the country, and very little is getting to the king.\r\n\r\n[b]Starting tomorrow, the king only gets 2 points a day in salaries.[/b]",
  "Solution_26": "darnit\r\n\r\ni gots executed",
  "Solution_27": "more activity please",
  "Solution_28": "Does the law passed on Day 9 still apply?",
  "Solution_29": "It does, but meewhee can only execute one of you at a time, and there are 7 of you left..."
}
{
  "Tag": [
    "Pythagorean Theorem",
    "geometry"
  ],
  "Problem": "How do I go about doing this?\r\n\r\nI know how to prove the distance formula by using the pythagorean theorem, however, I  don't know how to prove the mid-point formula by using the distance formula.\r\n\r\nI can prove the mid-point formula by using a graph and relating it to the distance formula, but how do I do this by solely using the distance formula?\r\n\r\nPlease enlighten me.\r\n\r\nRemember:\r\n\r\nD = sqrt((Xsub2 - Xsub1)squared + (Ysub2 - Ysub1)squared) \r\n\r\nand\r\n\r\nM = ((Xsub2 + Xsub1)/2 , (Ysub2 + Ysub1)/2)\r\n\r\nThanks,\r\ndfunk",
  "Solution_1": "Nobody?",
  "Solution_2": "So we have the points $A \\ (x_1,y_1)$ and $B \\ (x_2,y_2)$ and want to prove that $M \\ \\left(\\frac{x_1+x_2}{2},\\frac{y_1+y_2}{2}\\right)$ is the midpoint.  It is sufficient to show that the distance $AM=BM=\\frac 12 AB$.  So we have:\r\n\r\n\\begin{eqnarray*}\r\nAB &=& \\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\\r\nAM &=& \\sqrt{\\left(x_1-\\frac{x_1+x_2}{2}\\right)^2+\\left(y_1-\\frac{y_1+y_2}{2}\\right)^2}\\\\\r\nBM &=& \\sqrt{\\left(x_2-\\frac{x_1+x_2}{2}\\right)^2+\\left(y_2-\\frac{y_1+y_2}{2}\\right)^2}\r\n\\end{eqnarray*}\r\n\r\nSo $AM=\\sqrt{\\left(\\frac{x_1-x_2}{2}\\right)^2+\\left(\\frac{y_1-y_2}{2}\\right)^2}$.  Doing the algebra we easiliy see this is equivalent to $BM$.  Now we only need to prove that this is $\\frac 12 AB$ as follows:\r\n\r\n\\begin{eqnarray*}\r\nAM &=& \\sqrt{\\left(\\frac{x_1-x_2}{2}\\right)^2+\\left(\\frac{y_1-y_2}{2}\\right)^2}\\\\\r\n      &=& \\sqrt{\\frac 14\\left(\\left(\\frac{x_1-x_2}{2}\\right)^2+\\left(\\frac{y_1-y_2}{2}\\right)^2\\right)}\\\\\r\n      &=&\\frac 12 \\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\\r\n      &=& \\frac 12 AB\r\n\\end{eqnarray*}",
  "Solution_3": "Thanks a bunch."
}
{
  "Tag": [
    "geometry",
    "calculus",
    "topology",
    "analytic geometry",
    "graphing lines",
    "slope",
    "geometric transformation"
  ],
  "Problem": "<b>What sorts of topics and questions belong on this forum? </b>\r\n\r\nTo help you decide what the use of this forum is, I have compiled some example questions that might be asked here.  These are not strict guidelines by any means.  Generally, you can post anything here that is math related but does not fall into one of the three other math forums.  If you are unsure, please use the other posts on this forum and the example questions below as a guide.\r\n\r\n(1)  <b>Resources for problems and problem solving.</b>\r\n\ufffd\tWhere can I find a book full of past AMC 12-level geometry problems?\r\n\ufffd\tWhat is a good textbook on Number Theory?\r\n\ufffd\tWould you recommend <u>Mathematical Reasoning: Writing and Proof</u> by Ted Sundstrom for beginners learning how to write proofs?\r\n\r\n(2) <b>Problems from topics that are not generally tested on high school mathematics competitions, like a difficult problem from a textbook. </b>\r\n\ufffd\tI don't understand this calculus question; can you give me a hint?\r\n\ufffd\tI would like to suggest this math-oriented Physics/Chemistry problem as a challenge.\r\n\ufffd\tCan someone help me with this beginning topology question?\r\n\r\n(3) <b>Questions about theory (not a problem).</b>\r\n\ufffd\tI do not understand mathematical induction.  Can you explain the difference between inductive and deductive reasoning?\r\n\ufffd\tWhat is the proof for the Binomial Theorem?\r\n\ufffd\tHow many *Fundamental Theorem of XYZ* are there and what are they?\r\n\r\n(4) <b>Discussion of current events in mathematics. </b>\r\n\r\n(5) <b>Methods of teaching mathematics.</b>\r\n\r\n(6) <b>Studying mathematics.  These should be questions about studying mathematics that do not belong in the \ufffdCollege\ufffd forum.</b>\r\n\ufffd\tWhat is the calculus sequence you learn in college?\r\n\ufffd\tWhat are the other types of mathematics you can learn in college that are not part of the calculus sequence, such as combinatorics, statistics, number theory, and discrete math?\r\n\r\n(7) <b>Discussion of other topics in mathematics which are similar to threads on the \ufffdRound Table\ufffd but are mathematics-oriented.</b>\r\n\ufffd\tShould I take AP Calculus AB, AP Calculus BC, or AP Statistics?\r\n\ufffd\tWhy is the letter <i>m</i> used to represent the slope of a line?\r\n\r\nPlease remember that these are not strict guidelines.\r\n\r\nThanks, and please use the forum responsibly!",
  "Solution_1": "That's a great list of topics, and well reflects how this forum has been used over the last school year. I would only add that if the topic has to do with math, it really belongs [b]here[/b], on Other Problem Solving Topics, and [b]not[/b] on the Round Table, which at its best is more for issue-oriented political such as several of the highest-volume threads posted there recently. \r\n\r\nI'm glad to see that you've come on board as a co-moderator here.",
  "Solution_2": "Yes, I think that that is a wonderful guide.  Welcome to the elite corps of moderators! ;)",
  "Solution_3": "Thanks!  I feel elite already  :D !  I will edit it w/Tokenadult's suggestion."
}
{
  "Tag": [
    "ceiling function",
    "logarithms",
    "number theory",
    "greatest common divisor",
    "relatively prime",
    "number theory proposed"
  ],
  "Problem": "Given $a_0 > 1$, the sequence $a_0, a_1, a_2, ...$ is such that for all $k > 0$, $a_k$ is the smallest integer greater than $a_{k-1}$ which is relatively prime to all the earlier terms in the sequence.\r\nFind all $a_0$ for which all terms of the sequence are primes or prime powers.",
  "Solution_1": "No odd numbers larger than 3 can be included unless the number following is a power of two, and the number following that is a power of three. 2 and 3 work in this set, as 4 is a power of a prime, 2, and all following numbers will be prime (as they are relatively prime to 3 and 4, and every number before that relatively prime to 3 and 4 etc.). 7 also works, because directly above 7 is a power of 2, and above that is a power of 3, and a factor of 5 will be included at 25, also a power of a prime.\r\n\r\nI don't think any other numbers exist, as there it must go in the order (prime, power of 2, power of 3, power of 5, etc) all the way up to a power of the prime below the starting prime, which can be shown (I think) cannot happen.",
  "Solution_2": "A hint: show that if n is sufficiently large then $a_n$ is prime.",
  "Solution_3": "Does anyone have a full solution to the problem?",
  "Solution_4": "All solutins are $a_{0}=2,3,4,7,8.$ It is easy to check that if $a_{0}<10$ only these solutions. For prove, that it had not another solutions sufficiently consider equations $2^{k}\\pm1=3^{l}$, $2^{k}\\pm 1=5^{l}$ (k>3) and $3^{k}\\pm 2=5^{l}$(k>1). This equations had not solutions. It give if $a_{0}>9$ had not another solutions.",
  "Solution_5": "Suppose $a_1,a_2,...,$ is a sequence such that the $a_i$ are all prime powers. (I'm starting at $a_1$ because starting with $a_0$ is annoying)\n\n[b]Lemma 1:[/b] Every prime divides a term in the sequence.\n[b]Proof:[/b] Take any prime $p$. Suppose $k$ is the least integer such that $a_1 \\le p^k$. Now let $z$ be the least such integer such that $a_z \\ge p^k$. If $a_z \\neq p^k$, then we have $p$ divides one of $a_1,a_2,...,a_{z-1}$. But then we're done.\n\n[b]Corollary:[/b] If $p$ is prime, then $p^{\\lceil \\log_p{a_1} \\rceil}$ is one of the $a_i$.\nIt immediately follows for sufficiently large $n$, we have the $a_n$ are just a bunch of consecutive primes. Note that by the corollary we have that if $a_n > a_0^2$ then $a_n$ is prime.\n\nI claim we can pick two primes $p,q$ such that $pq \\in [a_1,a_1^2]$ but $p^{\\lceil \\log_p{a_1} \\rceil},q^{\\lceil \\log_q{a_1} \\rceil} > pq$\n\nBy Bertrand's Postulate we can find a prime $p$ such that $\\frac{\\sqrt{a_1}}{2} - 1 < p < \\sqrt{a_1}$ when $a_1 \\ge 16$. Let $\\frac{\\sqrt{a_1}}{p} = z$. Then we can also find a prime $q$ such that $z\\sqrt{a_1} < q < 2z\\sqrt{a_1}$.\nNow, clearly $pq  > a_1$. However, $p^{\\lceil \\log_p{a_1} \\rceil} \\ge p^3 > pq$. We also have $q^{\\lceil \\log_q{a_1} \\rceil} = q^2 > pq$. Thus it easily follows $pq$ must be one of the $a_i$, contradiction and thus we have $a_1 < 16$. Checking which of these work is just stupid casework.\n\nEDIT : So I just realized $16$ is not big enough as then in strange cases we can have $q \\ge p^2$. This method actually yields something like $a_1 \\le 256$ which is pretty hard to check up to... though working with [b]proglote[/b] a bit on this we deduced $a_0$ must be of the form $2^{2^n}$ except in some exceptional cases so I guess its not too bad.",
  "Solution_6": "Here's another solution with some casework:\n\nIf $a_0$ is odd, the next terms in the sequence are clearly $a_0 + 1$ and $a_0 + 2$. Since $a_0 +1$ is even, it must be a power of two, and either $a_0$ or $a_0 + 2$ is a power of three. So it remains to solve $2^{a} \\pm 1 = 3^{b}$ which is easy, giving $a = 2,3 \\implies a_0 \\in \\{3,7\\}.$\nIf $a_0$ is even, then the next terms in the sequence are clearly $a_0 +1, a_0 +3, a_0 +5$ (though we can't say anything further since $a_0 + 7$ and $a_0 + 1$ may both be multiples of 3). Also note that $a_0 = 2^a$ for some positive integer $a$.\nIf $a$ is 1 mod 3, then $7|2^{a} +5$, i.e. $a_0 + 5$ is a power of $7.$ So $2^a + 5= 7^b$, which implies $a=1$, so $a_0 = 2.$\nIf $a$ is 2 mod 3, then similarly it remains to solve $2^a + 3 = 7^b$, and the only solution is clearly $a =  2$, so $a_0 = 4.$\nIf $3|a$, then $a_0 + 1= 2^{3a'} + 1$ for some positive integer $a'.$ Note that $x^3 + 1 = (x+1)(x^2 - x + 1)$ and $(x+1, x^2 - x + 1) = (x+1, 3).$ Since $2^{3a'} + 1$ is a prime power, it cannot be factored into two coprime integers greater than 1, so the gcd of the two factors is 3 and we get $2^a + 1 = 3^b$ for some $b.$ So the only solution is $a=3$, i.e. $a_0 = 8.$\nTherefore, $a_0 \\in \\{2,3,4,7,8\\}.$",
  "Solution_7": "[quote=proglote]Here's another solution with some casework:\n\nIf $a_0$ is odd, the next terms in the sequence are clearly $a_0 + 1$ and $a_0 + 2$. Since $a_0 +1$ is even, it must be a power of two, and either $a_0$ or $a_0 + 2$ is a power of three. So it remains to solve $2^{a} \\pm 1 = 3^{b}$ which is easy, giving $a = 2,3 \\implies a_0 \\in \\{3,7\\}.$\nIf $a_0$ is even, then the next terms in the sequence are clearly $a_0 +1, a_0 +3, a_0 +5$ (though we can't say anything further since $a_0 + 7$ and $a_0 + 1$ may both be multiples of 3). Also note that $a_0 = 2^a$ for some positive integer $a$.\nIf $a$ is 1 mod 3, then $7|2^{a} +5$, i.e. $a_0 + 5$ is a power of $7.$ So $2^a + 5= 7^b$, which implies $a=1$, so $a_0 = 2.$\nIf $a$ is 2 mod 3, then similarly it remains to solve $2^a + 3 = 7^b$, and the only solution is clearly $a =  2$, so $a_0 = 4.$\nIf $3|a$, then $a_0 + 1= 2^{3a'} + 1$ for some positive integer $a'.$ Note that $x^3 + 1 = (x+1)(x^2 - x + 1)$ and $(x+1, x^2 - x + 1) = (x+1, 3).$ Since $2^{3a'} + 1$ is a prime power, it cannot be factored into two coprime integers greater than 1, so the gcd of the two factors is 3 and we get $2^a + 1 = 3^b$ for some $b.$ So the only solution is $a=3$, i.e. $a_0 = 8.$\nTherefore, $a_0 \\in \\{2,3,4,7,8\\}.$[/quote]\n\nmacetei"
}
{
  "Tag": [
    "geometry",
    "perimeter",
    "geometry solved"
  ],
  "Problem": "Could anybody prove the following theorem\r\n\r\n[color=green][b]Theorem.[/b] Let P and Q be two convex polygons such that P is on the non-externally region of Q, then prove that perimeter(P)=< perimeter(Q).[/color]\r\n\r\nUsing this lema (I proved this lema)\r\n\r\n[color=green][b]Lema. [/b]Let Q be a convex polygons and let P be a triangle such that P is on the non-externally region of Q, then prove that perimeter(P)=< perimeter(Q).[/color]\r\n\r\nThanks a lot..\r\n\r\nJos\u00e9 Carlos",
  "Solution_1": "In [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=197437#p197437]http://www.mathlinks.ro/Forum/viewtopic.php?t=31419 post #3[/url], this was proven for the case when the external figure Q is a circular disc rather than a polygon, but for a polygon the proof is exactly the same.\r\n\r\n  Darij"
}
{
  "Tag": [
    "limit",
    "trigonometry",
    "inequalities",
    "real analysis",
    "ratio",
    "quadratics",
    "continued fraction"
  ],
  "Problem": "Is it true that\r\n\\begin{eqnarray*} \\lim_{n \\rightarrow \\infty} \\frac{1}{n \\sin n} & = & 0 \\end{eqnarray*}\r\nWhat about\r\n\\begin{eqnarray*} \\underset{n \\longrightarrow \\infty}{\\lim \\sup} \\frac{1}{n \\sin n} & = & ? \\end{eqnarray*}",
  "Solution_1": "I should think not. We can find some $m$ (infinitely many in fact) so that $|\\pi - \\frac{n}{m}| < \\frac{1}{m^2}$. Then $\\frac{1}{n\\sin(n)}$ is about $\\frac{1}{n(\\frac{1}{m^2})} \\approx \\frac{m}{\\pi}$, which is arbitrarily large. \r\n\r\n(for rigor you should probably take two terms of the taylor series for sine so that the \"about\" is an inequality in the direction we want, and then bound from each side by a constant times $m$)\r\n\r\nI bet it does converge in some weaker sense, like Cesaro.",
  "Solution_2": "The first fact is a corollary of Hurwitz theorem, as far as I know.\r\nOtherwise, I don't think you really need to mess up with Taylor series. Some basic inequalities should suffice.",
  "Solution_3": "For any $c$, $\\limsup \\left|\\frac1{n\\sin cn}\\ge\\frac1{\\pi}\\right|$. This limsup is $\\infty$ when the continued fraction for $\\frac c{\\pi}$ has arbitrarily large quotients or $\\frac c{\\pi}$ is rational.\r\n\r\nFor any integer $m$, $|\\sin cn|=|\\sin(cn-m\\pi)|< |cn-m\\pi|=n\\pi\\left|\\frac{c}{\\pi}-\\frac mn\\right|$. If $\\frac pq$ is a convergent (in lowest terms) of the continued fraction for $\\frac c{\\pi}$, $\\left| \\frac{c}{\\pi}-\\frac pq\\right|<\\frac1{q^2}$. There are infinitely many convergents $\\frac pq$; choosing these as our $m,n$, we have $n\\pi\\left|\\frac{c}{\\pi}-\\frac mn\\right|<\\frac{\\pi}n$. Taking the reciprocal and dividing by $n$, we have $|\\frac1{n\\sin cn}|> \\frac1{\\pi}$.\r\n\r\nXevarion's argument misses one of these factors of $n$; whether the $\\limsup$ is finite in any particular case is an interesting question. It is always nonzero.\r\n\r\nIf we drop the absolute value signs, we can show that $\\limsup \\frac1{n\\sin cn}\\ge\\frac1{4\\pi}$ and $\\liminf \\frac1{n\\sin cn}\\le-\\frac1{4\\pi}$. The convergents $\\frac pq$ alternate between being too large and too small; if we then let $n=2q$, $\\sin 2cn$ will be positive when $\\frac pq$ is low and negative when $\\frac pq$ is high. In both cases, we lose a factor of 4 by doubling.",
  "Solution_4": ":blush: \r\nThanks for correcting my error jmerry. Can you please give some more detail on the case where the continued fraction for $\\frac{c}{\\pi}$ has arbitrarily large quotients? I'm afraid I don't see immediately why it's so ...  :oops:",
  "Solution_5": "If the quotient after the convergent $\\frac pq$ is $r$, we have $\\left|\\frac c{\\pi}-\\frac pq\\right|\\le\\frac1{rq^2}$.\r\n\r\nI think so, anyway; it's been a while since I worked out the details. The two quantities are comparable in any case.\r\nAs a corollary, large quotients indicate unusually good rational approximations. As a famous example, here's the first few terms of the continued fraction for $\\pi$:\r\n3;7,15,1,292\r\nIf we cut off just before the 292, we get $\\frac{355}{113}$ as a convergent.",
  "Solution_6": "Oh, I get it, that's really cool! Thanks  :D \r\n\r\nAre there general methods for knowing whether the continued fraction of some irrational has arbitrarily large quotients (other than actually knowing the general term, like in case of the golden ratio)?",
  "Solution_7": "[quote=\"Xevarion\"]... like in case of the golden ratio)?[/quote]\r\nTheorem: the quotients of the continued fraction of a number are periodic if and only if the number is a quadratic irrational - that is an irrational number that can be written in the form $\\frac{a+\\sqrt{b}}c$ or $\\frac{a-\\sqrt{b}}c$ for integers $a,b,c.$ (If the number is rational, then the continued fraction representation terminates.) Of course, the golden ratio is a quadratic irrational.\r\n\r\nSo quadratic irrationals cannot have arbitrarily large quotients, because the quotients are periodic. But we can't say anything about the converse."
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "If $ a,b,c>0$ such that $ a\\plus{}b\\plus{}c\\equal{}3$, then: \r\n\r\n$ (a\\plus{}\\frac1{b^2\\plus{}c^2})(b\\plus{}\\frac1{a^2\\plus{}c^2})(c\\plus{}\\frac1{b^2\\plus{}a^2})\\le\\frac{27}8$",
  "Solution_1": "[quote=\"akai\"]If $ a,b,c > 0$ such that $ a \\plus{} b \\plus{} c \\equal{} 3$, then: \n\n$ (a \\plus{} \\frac1{b^2 \\plus{} c^2})(b \\plus{} \\frac1{a^2 \\plus{} c^2})(c \\plus{} \\frac1{b^2 \\plus{} a^2})\\le\\frac {27}8$[/quote]\r\nTry $ a\\equal{}b\\equal{}0.001,c\\equal{}2.998$.\r\nIf you meant that\r\n\\[ (a \\plus{} \\frac1{b^2 \\plus{} c^2})(b \\plus{} \\frac1{a^2 \\plus{} c^2})(c \\plus{} \\frac1{b^2 \\plus{} a^2})\\ge\\frac {27}8\\]\r\nThen try $ a\\equal{}b\\equal{}\\frac{1}{2},c\\equal{}2$."
}
{
  "Tag": [],
  "Problem": "if f(z)=z+1/z-1, then find f to the 1991 power (2+i).",
  "Solution_1": "[hide]\nTry the first two values to find that f^2(2+i) = 2+i, so f^1991(2+i) = f^1(2+i) = (3+i)/(1+i).[/hide]"
}
{
  "Tag": [
    "geometry",
    "incenter",
    "geometry unsolved"
  ],
  "Problem": "$ O$ is incircle of a $ \\triangle ABC$. An another circle $ O'$ tangent to rays $ AB,AC$.If center of $ O'$ is lie circle $ O$, prove that $ \\triangle ABC$ is isosceles. \r\n\r\n                                                                          H.\u0130BRAH\u0130M AYANA",
  "Solution_1": "i don't see why this needs to be true:\r\n[asy]size(200); pair A,B,C,D,X,Y,Z; C=origin; B=(50,20); A=(30,70); draw(A--B--C--cycle); draw(incircle(A,B,C)); D=incenter(A,B,C); dot(D); X=(13.5,32); Y=(43,37); draw(incircle(A,X,Y)); Z=incenter(A,X,Y); dot(Z); label(\"A\",A,N); label(\"B\",B,E); label(\"C\",C,W); label(\"O\",D,S); label(\"O'\",Z,N);[/asy]"
}
{
  "Tag": [
    "MATHCOUNTS",
    "modular arithmetic"
  ],
  "Problem": "1. (AoPS) Factor 141, 1441, and 14441.\r\n\r\n2. (AoPS) Find the base 8, 9, and 16 representations of $47_{10}$.\r\n\r\n3. (AoPS) Find the base 10 equivalents of $BEE_{16}$, $DEF_{16}$, and $A1_{16}$.\r\n\r\n4. (MATHCOUNTS 1986) Find the value of digit $A$ if the five-digit number 12A3B is divisible by both 4 and 9, and $A\\ne B$.\r\n\r\n5. (MATHCOUNTS 1988) In how many ways can a debt of 69 dollars be paid exactly using only 5 dollar bills and 2 dollar bills?\r\n\r\n6. (MATHCOUNTS 1988) Find the sum of all four digit natural numbers of the form $4AB8$ which are divisible by 2, 3, 4, 6, 8, and 9.\r\n\r\n7. (Mu Alpha Theta 1990) If $a\\ne 1$ and $\\sqrt[a]{10000_{a}}= 10_{a}$, find $a$.\r\n\r\nGood luck!",
  "Solution_1": "[quote=\"chess64\"]1. (AoPS) Factor 141, 1441, and 14441.\n\n2. (AoPS) Find the base 8, 9, and 16 representations of $47_{10}$.\n\n3. (AoPS) Find the base 10 equivalents of $BEE_{16}$, $DEF_{16}$, and $A1_{16}$.\n\n4. (MATHCOUNTS 1986) Find the value of digit $A$ if the five-digit number 12A3B is divisible by both 4 and 9, and $A\\ne B$.\n\n5. (MATHCOUNTS 1988) In how many ways can a debt of 69 dollars be paid exactly using only 5 dollar bills and 2 dollar bills?\n\n6. (MATHCOUNTS 1988) Find the sum of all four digit natural numbers of the form $4AB8$ which are divisible by 2, 3, 4, 6, 8, and 9.\n\n7. (Mu Alpha Theta 1990) If $a\\ne 1$ and $\\sqrt[a]{10000_{a}}= 10_{a}$, find $a$.\n\nGood luck![/quote]\r\n\r\n[hide=\"1.\"]\n$141 = 47 \\cdot 3$\n$1441 = 11 \\cdot 131$ \n$14441 = 7 \\cdot 2063$\n[/hide]\n\n[hide=\"2.\"]\n$48_{10}= 57_{8}$\n\n$48_{10}= 52_{9}$\n\n$48_{10}= 2F_{16}$\n\ni want my ranking to be fixed...\n[/hide]",
  "Solution_2": "About time somebody did them...\r\n\r\nBoth correct. :)",
  "Solution_3": "I'm not in middle school but I'll try these anyway.  :) \r\n\r\n[hide=\"#4\"]\nTo be divisible by 4, the last two digits must form a number divisible by 4, so B=2 or B=6.  To be divisible by 9, the sum of the digits must be a multiple of 9.  1+2+3+6=12, so if B=6, A=6, which is not allowed.  1+2+3+2=8, so if B=2, A=1.  This is the only solution, so A=1.\n[/hide]",
  "Solution_4": "uhh.. I hope you don't have to be from Maryland to answer these but here's #7.\r\n[hide=\"7\"]Changinge everything to base 10, we get $\\sqrt[a]{a^{4}}=a\\rightarrow a^{4}=a^{a}\\rightarrow a=4$[/hide]",
  "Solution_5": "well actually you have to be from cockeysville middle school, but since they aren't answering them you can :lol:",
  "Solution_6": "Hmm I understand the divisibility stuff but like number 5 how can you use those rules and apply them to real life based problems?",
  "Solution_7": "[hide=\"5\"]\nWe have $5a+2b=69$ for some positive integers $a,b$. Modulo 5, the equation becomes $2b\\equiv 4\\pmod{5}$ so $b\\equiv 2\\pmod{5}$. Therefore, the smallest possible value of $b$ is 2, and when $b=2$, $a=13$. Notice that decreasing $a$ by 2 and increasing $b$ by 5 will give us another solution. Therefore, the solutions are (13, 2), (11, 7), (9, 12), (7, 17), (5, 22), (3, 27), (1, 32). So the answer is $\\boxed{7}$.\n[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "function",
    "real analysis",
    "calculus computations"
  ],
  "Problem": "How can one prove that\r\n\\[ \\int f(x) \\, \\mbox{sgn} \\left( f(x) \\right)  \\, \\mbox{d}x \\equal{} \\mbox{sgn} \\left( f(x) \\right) \\int f(x) \\, \\mbox{d}x\\]\r\nIt is trival, but I don't have slightest idea how to prove it in a nice way :(\r\n\r\nIs it enough when we consider intervals when f>0 and f<0  :?:",
  "Solution_1": "What is sgn? Does it stand for the sign (+ or - symbol) of a function? And if I'm interpreting that correctly, the proof is indeed trivial, write the LHS as a Riemann sum and factor out the sign of the function.",
  "Solution_2": "sgn stands for sign function.\r\nOk, thanks.",
  "Solution_3": "If you look carefully the equation doesn't make sense (at least without some conditions). How do you 'factor out' the sign function while keeping the variable the same?",
  "Solution_4": "$ f sgn(f) \\equal{} |f|$"
}
{
  "Tag": [
    "AoPSwiki",
    "AMC",
    "AIME",
    "LaTeX",
    "articles",
    "USA(J)MO",
    "USAMO"
  ],
  "Problem": "If you haven't checked out the [url=http://www.artofproblemsolving.com/Wiki/]AoPSWiki[/url] yet, you really ought to!\r\n\r\nOne of the ongoing projects in the AoPSWiki is the development of the [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC_Problems_and_Solutions]AMC Problems/Solutions[/url].  This is not a small task, so we will need a lot of help.  Anyone can help.  You just use your AoPS account for the AoPSWiki.\r\n\r\nI've already started with the [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I]2006 AIME I[/url].  If you look around, you should get a feel for it.\r\n\r\nI haven't put the solutions in for the 2006 AIME I.  Also, if you want to put links to message board posts with the problems, that would be a good idea.  Just make sure to put the links in the ''See also'' section of the article.\r\n\r\nThe biggest advantage of the wiki over the contest directory is that anyone can edit it at any time.\r\n\r\nThanks.",
  "Solution_1": "Another note for entering problems into the wiki:\r\n\r\nLook at the [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems]2006 AIME 1 Problems[/url].  It has all the problems listed as their own section. After the problem statement it has a solution link which brings you to a page dedicated solely for that problem.  When making these solution pages you MUST name them like the following: 2006 AIME I Problems/Problem 1.   What this does is it creates a subarticle for that problem (instead of making a whole new article).\r\n\r\nThanks.",
  "Solution_2": "COOL! :coolspeak:",
  "Solution_3": "An additional note:\r\n\r\nLook at the way I entered the [url=http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam]USC exam[/url] into the wiki.  I think the AMC's should follow this format (it's open to improvement though if you have suggestions).\r\n\r\nAlso, note that on the individual problem pages ( e.g. http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_2 ) I have links to the previous problem, next problem, and back the exam page.  I also added the [[Category:...] tags for all of them.",
  "Solution_4": "Sorry to bump....\r\n\r\nbut the AIME's certainly still need more work!!!\r\n\r\nToday I fixed around a couple things about the AIME's (preliminaries for like 3 or 4 tests, one completely added)\r\nFor many, the Latex is up but it just need to have <math> tags added and split into problems.\r\n\r\nWhen you work on an AIME, still try to post here :)",
  "Solution_5": "for editing the AIME pages, is it ok to copy and paste the \"official AoPS solution\", which is linked to the right of the problem statement in the Contests section? and if so should the original solver's name be posted?",
  "Solution_6": "Seeing as those aren't truly \"official AoPS solutions\" but solutions that members posted and mods edited, yes, its okay, and yes the solver's name should be up there.\r\nAlso if you have a different solution you can post it as a second solution.",
  "Solution_7": "[quote=\"me@home\"]Seeing as those aren't truly \"official AoPS solutions\" but solutions that members posted and mods edited, yes, its okay, and yes the solver's name should be up there.\nAlso if you have a different solution you can post it as a second solution.[/quote]\r\nYou think so?  When I come up with my own solutions I usually don't give myself credit\u2014in fact, I usually don't give anybody credit at all.  This is probably okay because it is quite possible for several people to come up with practically identical solutions to the same problem.  In any case, rather than copying and pasting, you would probably do better to learn the solution and rewrite it yourself, especially since the proofs posted on the forums are sometimes poorly written (and sometimes written in an odd sort of slang).  Plus, you might as well link to the problem's discussion on the forum, where anybody interested can see who posted what solutions.",
  "Solution_8": "Can someone start putting up solutions to this year's AIME I?  I typed up the problems [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems]here[/url] but don't have time to put up all my solutions.",
  "Solution_9": "[quote=\"joml88\"]The biggest advantage of the wiki over the contest directory is that anyone can edit it at any time.\n\nThanks.[/quote]That's also a disadvantage in my view but it doesn't matter :P",
  "Solution_10": "That's a little pessimistic... :(",
  "Solution_11": "A pessimist is just a realist who's not high...\r\n\r\nBut w/e.  Do we have MAA's permission to break copyright? I know that the Kalva archive has to avoid that.\r\n\r\nOn second thought, if the problems are split up it should be legal...at least I think that's what the pamphlet said.",
  "Solution_12": "Why do we have a wiki and contest resources section?\r\nWe should combine those as some problems aren't entered in the one but is in the other and it gets annoying to filp back and forth.\r\nAlso alot of pictures and solutions are missing in the contest resources and typos are present.",
  "Solution_13": "Time and energy...you're more than welcome to do it :)",
  "Solution_14": "Aren't only certain moderators able to edit the resources section?",
  "Solution_15": "[quote=\"ceegers\"]So obviously both the wiki and the resources sections are incomplete. Now I've got some extra time in here for a couple months, so I quite possibly will go ahead and add some stuff to the wiki.[/quote]\n\nThat is great!  :)  \n\n[quote=\"ceegers\"]So... is AoPS still concerned with updating the competition resources section? If so, should I maybe post problems that aren't in the resources section in the appropriate forum so they can be linked? Anything else I should do? If you are still updating both, which has priority to be updated, the wiki or the resources?[/quote]\r\n\r\nThe resources section is more frequently updated and is of greater priority. If you have problems that don't already appear in the resources section, of course do go ahead and post them. That being said, many of the American contests have already been uploaded to the resources section; the wiki by contrast is missing many pages, including both problems and articles, and it is much easier to directly make changes in the wiki. Chances are you'll find that much more work needs to be done with the wiki.\r\n\r\n\r\nAnd post 1000.",
  "Solution_16": "Is this the place to report mistakes in the posted problems, because either I am reading something wrong, or Problem 15 on the 2005 AIME II has an error in it.  I think the question is asking for the smallest possible positive value.\n\nI'd better put why I think it's wrong inside of spoiler space, in case somebody wants to work on the problem without a hint:\n[hide]The ellipse of possible centers for the circle goes through the Y-axis, which means that if you choose a point where X = a very small negative value, a < -1,000,000, so the answer > 1000.  The answer makes it look like they are looking for a positive slope.[/hide]\n\n-- Don",
  "Solution_17": "You post this in the discussion section of that particular thread.\n\nOr you could just edit it yourself.",
  "Solution_18": "what happens if someone edits wrongly?",
  "Solution_19": "If it's an accident, people will probably change it back to what it's supposed to be.  If it's on purpose, then that's a whole other thing.",
  "Solution_20": "What's difference between AoPS wiki and the solutions found on the \"contest section\" with all the different questions?",
  "Solution_21": "The questions in the contest section link you back to the original thread from whence it came (where users have posted solutions).  AoPSWiki contains solutions in which users type in there, not necessarily the same as the solution in the original thread.",
  "Solution_22": "http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems\n\nLike how all the links to the solutions are to Friday?",
  "Solution_23": "Who is this HyperSet guy???",
  "Solution_24": "[quote=\"Herp\"]http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems\n\nLike how all the links to the solutions are to Friday?[/quote]\nWhat?  I don't get it...the links to the solutions are fine.\n[quote=\"1=2\"]Who is this HyperSet guy???[/quote]\nWhere did that come from?",
  "Solution_25": "[quote=\"JSGandora\"][quote=\"Herp\"]http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems\n\nLike how all the links to the solutions are to Friday?[/quote]\nWhat?  I don't get it...the links to the solutions are fine.[/quote]\n\nAs of 12:30 PM Eastern Time, the links to the solutions of all the problems were links to the \"Friday\" video.\n\n[quote=\"JSGandora\"][quote=\"1=2\"]Who is this HyperSet guy???[/quote]\nWhere did that come from?[/quote]\n\nHe last edited the article before the actual solutions came up.",
  "Solution_26": "Uh what\nsomeone changed it again",
  "Solution_27": "[quote=\"AoPS wiki\"]# (cur) (prev)  17:23, 13 May 2011 Ksun48 (Talk | contribs) (6,902 bytes) (Undo revision 38515 by Xantos C. Guin (Talk)) (undo) [/quote]\n\nNice try, ksun48.",
  "Solution_28": "[quote=\"1=2\"]Who is this HyperSet guy???[/quote]\n\nhttp://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=102351",
  "Solution_29": "He/She has a history of putting youtube links in to pages, as well as putting nonsense in pages (http://www.artofproblemsolving.com/Wiki/index.php?title=Demonic_Possession ).  I will block him/her from editing the wiki and will revert the vandalism."
}
{
  "Tag": [
    "inequalities solved",
    "inequalities"
  ],
  "Problem": "Find the least value of the expression \r\n\r\nx<sup>20</sup>y<sup>-11</sup> + y<sup>20</sup>z<sup>-11</sup> + z<sup>20</sup>x<sup>-11</sup>\r\n\r\nwhere x, y, z are positive real numbers satisfying x + y + z = 2001.",
  "Solution_1": "We have x^20/y^11+y^20/z^11+z^20/x^11=( x^2/y)^10/y+... \\geq (  \\sum _ciclyc (x^2/y)^5)^2/(sum x). But  \\sum  x^2/y \\geq  \\sum x. So, the expression is at least (5*(  \\sum x^2/y /5)^5)^2/sum (x^2/y). Since  \\sum x^2/y is at least 2001, the minimum is attained when x=y=z"
}
{
  "Tag": [
    "LaTeX"
  ],
  "Problem": "The first few terms of a sequence are: $3, 12, 4, 16, \\frac{16}{3}, \\ldots$. Each even-numbered term is obtained by multiplying the preceding term by 4, whereas each odd-numbered term, after the first, is obtained by dividing the preceding term by 3.  If the 101st term is represented as $\\frac{4^{a}}{3^{b}}$, determine the ordered pair $(a,b)$.",
  "Solution_1": "This is my first time on the forum, so I don't know how to do Latex so I'll have to post this solution in just normal writing.\r\n\r\nWe want to put the sequence in terms, of 4^x/3^y, which will let us figure out the 101st term.  3=4^0/3^-1, 12= 4^1/3^-1, 4= 4^1/3^0.  This means that the exponents in front of x and y increase alternately.  Every odd numbered term can be expressed as 4^((x-1)/2)/3^((x-1)/2-1)). When you let x=101, you find a=50, b=49 so the a+b is 99",
  "Solution_2": "We're only asked to find (a,b), not a+b"
}
{
  "Tag": [],
  "Problem": "Imi poate spune si mie cineva despre ecuatia claselor de echivalenta a unui grup G?\r\nAcea relatie intre Z(G) si clasele care il dau pe G.",
  "Solution_1": "Uite [url=http://www.mateforum.ro/viewtopic.php?t=377]aici[/url] e un post bun; poti gasi si o demonstratie acolo."
}
{
  "Tag": [],
  "Problem": "Use four distinct digits to form a four-digit number which does not end in the digit \u201c0\u201d.  Reverse those four digits to write a second four-digit number.  What is the maximum possible positive difference between those two numbers?\r\n\r\nA.3087\tB.8532\tC.8622\tD.8712\tE.8802",
  "Solution_1": "[hide=\"Solution\"]Write the number as $ ABCD$. We want to maximize $ ABCD \\minus{} DCBA$, which is the equivalent of maximizing $ 999(A\\minus{}D) \\plus{} 9(B\\minus{}C)$, with $ D \\neq 0$. Let $ A \\equal{} 9$, $ D\\equal{}1$, $ B \\equal{} 8$, and $ C \\equal{} 0$. We therefore get $ 999 \\times 8 \\plus{} 90 \\times 8 \\equal{} \\boxed{8712}$. The answer is $ \\boxed{D}$. \n\n[/hide]"
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "Let a,b,c be real numbers such that $ |a^3|\\leq bc$. Prove that $ b^2 \\plus{} c^2 \\geq {\\frac {1}{3}}$ whenever $ a^6 \\plus{} b^6 \\plus{} c^6 \\geq{\\frac {1}{27}}$.",
  "Solution_1": "Have $ |a^3|\\le bc\\Rightarrow a^6 \\plus{} b^6 \\plus{} c^6 \\equal{} b^6 \\plus{} c^6 \\plus{} (a^3)^2\\le b^6 \\plus{} c^6 \\plus{} b^2c^2$\r\nNow, $ a^6 \\plus{} b^6 \\plus{} c^6 \\geq{\\frac {1}{27}}\\Rightarrow b^6 \\plus{} c^6 \\plus{} b^2c^2\\ge \\frac {1}{27}$\r\nThen, $ b^2c^2\\ge \\frac {1}{27} \\minus{} (b^6 \\plus{} c^6)$   (*)\r\nOn the other hand, $ (b^2 \\plus{} c^2)^3 \\equal{} b^6 \\plus{} c^6 \\plus{} 3b^2c^2(b^2 \\plus{} c^2)$\r\nThen, if $ b^2 \\plus{} c^2 \\equal{} x, b^6 \\plus{} c^6 \\equal{} y$ have $ b^2c^2 \\equal{} \\frac {x^3 \\minus{} y}{3x}$\r\nSubstituting $ b^2c^2 \\equal{} \\frac {x^3 \\minus{} y}{3x}$ into (*) and replacing $ b^6 \\plus{} c^6$ by $ y$ get:\r\n$ \\frac {x^3 \\minus{} y}{3x}\\ge \\frac {1}{27} \\minus{} y$.\r\nAs $ 3x \\equal{} 3(b^2 \\plus{} c^2) > 0$ (we can't have $ 3(b^2 \\plus{} c^2) \\equal{} 0$ as then $ b \\equal{} c \\equal{} 0$ so we do not get $ a^6 \\plus{} b^6 \\plus{} c^6\\ge \\frac {1}{27}$), we can multiply both sides by $ 3x$:\r\n$ x^3 \\minus{} y\\ge \\frac {1}{9}x \\minus{} 3xy$\r\n$ \\Leftrightarrow (x^3 \\minus{} \\frac {1}{9}x) \\plus{} (3xy \\minus{} y)\\ge 0$\r\n$ \\Leftrightarrow x(x \\plus{} \\frac {1}{3}(x \\minus{} \\frac {1}{3})) \\plus{} 3y(x \\minus{} \\frac {1}{3})\\ge 0$\r\n$ \\Leftrightarrow (x \\minus{} \\frac {1}{3})(x(x \\plus{} \\frac {1}{3}) \\plus{} 3y)\\ge 0$ (**)\r\nBecause $ x,y\\ge 0$ then $ x(x \\plus{} \\frac {1}{3}) \\plus{} 3y\\ge 0$ so in order for (**) to be true, must have $ x \\minus{} \\frac {1}{3}\\ge 0\\Rightarrow b^2 \\plus{} c^2\\ge \\frac {1}{3}$ QED."
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "sphere",
    "vector",
    "topology",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "which is the definicion of spherical tangent bundle????? :blush:",
  "Solution_1": "I am not sure. May be it would be easy if you can tell us the context. But, what I am guessing is that it is a tangent bundle of a sphere. I could be wrong!\r\nPeace Up",
  "Solution_2": "google arxiv.org/pdf/math.GT/0207219",
  "Solution_3": "I think this like Ajoye23 thought.\r\nHope it is not wrong.",
  "Solution_4": "Do you mean tangent sphere bundle?\r\n\r\nA sphere bundle is the set of vectors in a bundle of norm 1. You need to use the fact that you have a Riemannian metric on a manifold to get an inner product (or some other inner product here).\r\n\r\nSo,\r\n\r\nST(M)|U = { v in TM_U | ||v|| = 1 } \r\n\r\n\r\nThis is a useful object in showing bundles are orientable and also shows up in characteristic class theory."
}
{
  "Tag": [
    "limit",
    "integration",
    "trigonometry",
    "logarithms"
  ],
  "Problem": "$\\alpha$\u3092$0\\le \\alpha < 1$\u306a\u308b\u5b9f\u6570\u3068\u3059\u308b\u3002\r\n\r\n$n$\u500b\u306e\u5b9f\u6570$\\sqrt{1},\\sqrt{2},\\sqrt{3},\\cdots,\\sqrt{n}$\u306e\u3046\u3061\u3001\u305d\u306e\u5c0f\u6570\u90e8\u5206\u304c$\\alpha$\u4ee5\u4e0b\u3067\u3042\u308b\u3082\u306e\u306e\u500b\u6570\u3092$N(n)$\u3068\u304a\u304f\u3068\u304d\u3001\r\n\u3000\u3000\u3000\u3000$\\displaystyle\\lim_{n\\to\\infty}\\frac{N(n)}{n}=\\alpha$\r\n\u3068\u306a\u308b\u3053\u3068\u3092\u793a\u305b\u3002",
  "Solution_1": "\u9ad8\uff11\u306e\u6642\u306b\u3053\u308c\u3092\u805e\u304d\u3001\u6570\u5b66\u3092\u597d\u304d\u306b\u306a\u3063\u305f\u304d\u3063\u304b\u3051\u306e\u554f\u984c\u3067\u3059\u3002\u305d\u306e\u6642\u306f\u7d50\u679c\u304c\u3068\u3066\u3082\u7dba\u9e97\u3060\u3068\u601d\u3044\u307e\u3057\u305f :P\u4eca\u898b\u308b\u3068\u7279\u306b\u4f55\u3067\u3082\u306a\u3044\u554f\u984c\u306a\u3093\u3067\u3059\u304c\u2026 :noo:",
  "Solution_2": "$n$\u304c\u7121\u9650\u5927\u306e\u6642\u3092\u8003\u308b\u3068\r\n\u5b9f\u6570$\\sqrt k$\u3000(k=1,2...)\u304c\u5b9f\u6570$\\alpha$\u3092\u5c0f\u6570\u90e8\u5206\u306b\u6301\u3064\u78ba\u7387\u306f\u7b49\u3057\u3044\u306e\u3067\r\n$\\alpha$\u4ee5\u4e0b\u306e\u5c0f\u6570\u90e8\u5206\u3092\u3082\u3064\u78ba\u7387\u306f$\\alpha$\u306b\u306a\u308b\u3002\r\n\r\n\u3088\u3063\u3066$\\lim_{n \\to \\infty} \\frac{N(n)}{n}=\\alpha$",
  "Solution_3": "\u305d\u3093\u306a\u304a\u304a\u3056\u3063\u3071\u306b\u89e3\u3044\u3066\u3057\u307e\u3046\u89e3\u6cd5\u3082\u3042\u308b\u306e\u3067\u3059\u306d\uff01\u305d\u306e\u65b9\u304c\u30e9\u30af\u3067\u3059\u306d :P \r\n\u50d5\u306f\u5730\u9053\u306b\u306f\u3055\u307f\u3046\u3061\u306e\u89e3\u6cd5\u3057\u304b\u77e5\u308a\u307e\u305b\u3093\u3067\u3057\u305f :lol:",
  "Solution_4": "\u6570\u5b66\u7684\u306b\u3053\u308c\u3067OK\u306a\u306e\u304b\u3069\u3046\u304b\u306f\u5fae\u5999\u306a\u306e\u3067\u3059\u304c\u3002\r\n\r\nn\u304c\u7121\u9650\u5927\u306e\u6642\r\n\u4f8b\u3048\u3070100000000000000000001\u3068100000000000000000002\u306e\u5e73\u65b9\u6839\u306e\u5c0f\u6570\u90e8\u5206\u306f\u50c5\u304b\u306e\u9055\u3044\u3057\u304b\u3082\u305f\u306a\u3044\u306e\u3067\r\nn\u304c\u7121\u9650\u5927\u306e\u6642\u306e\u78ba\u7387\u306f\u540c\u7b49\u3068\u898b\u306a\u3057\u3066\u3082\u307e\u3041\u3044\u3044\u304b\u306a\uff1f\u3068\u3044\u3046\u611f\u3058\u3067\u30fb\u30fb",
  "Solution_5": "\u3067\u3082\u7406\u5c48\u3068\u3057\u3066\u306f\u30a4\u30b1\u3066\u305d\u3046\uff1f\u3067\u3059\u3088\u306d :P \r\n\r\n\u306f\u3055\u307f\u3046\u3061\u3092\u7528\u3044\u305f\u89e3\u6cd5\u3082\u304b\u3044\u3066\u304a\u304d\u307e\u3059 :) \r\n\r\n\u307e\u305a\u3001$\\sqrt{1},\\sqrt{2},\\sqrt{3},\\cdots$\u306e\u3046\u3061\u3001\u305d\u306e\u6574\u6570\u90e8\u5206\u304c$l$\u3067\u3001\u5c0f\u6570\u90e8\u5206\u304c$\\alpha$\u4ee5\u4e0b\u3067\u3042\u308b\u3082\u306e\u306e\u500b\u6570\u304c$[2\\alpha l +\\alpha^2]+1\\cdots (\\ast)$\u3067\u8868\u3055\u308c\u308b\u3053\u3068\u3092\u793a\u3057\u307e\u3059\u3002\r\n$\\alpha$\u3092\u56fa\u5b9a\u3057\u3066\u3001$\\sqrt{1},\\sqrt{2},\\sqrt{3},\\cdots$\u306e\u4e2d\u306e\u5c0f\u6570\u90e8\u5206\u304c$\\alpha$\u4ee5\u4e0b\u3067\u3042\u308b\u3082\u306e\u306e\u3046\u3061\u3001\u6574\u6570\u90e8\u5206\u304c$l$\u3067\u3042\u308b\u3082\u306e\u306e\u500b\u6570\u3092$a_l$\u3068\u3059\u308b\u3002\u307e\u305f\u3001\u6b63\u6574\u6570$k$\u306b\u5bfe\u3057\u3066\u3001\r\n\u3000\u3000\u3000\u3000$l \\le \\sqrt{k} \\le l + \\alpha \\Longleftrightarrow l^2 \\le k \\le l^{2}+2\\alpha l +\\alpha ^2$\r\n\u3000\u3000\u3000\u3000$\\Longleftrightarrow l^2 \\le k \\le l^{2}+[2\\alpha l +\\alpha ^2]$\r\n\u3000\u3000\u3000\u3000$\\Longleftrightarrow k=l^{2},l^{2}+1,l^{2}+2,\\cdots,l^{2}+[2\\alpha l +\\alpha ^2]$\r\n\u3067\u3042\u308b\u306e\u3067\u3001\r\n\u3000\u3000\u3000\u3000$a_{l}=(l^{2}+[2\\alpha l +\\alpha ^2])-l^{2}+1=[2\\alpha l +\\alpha^2]+1$\r\n\u3088\u3063\u3066$(\\ast)$\u306f\u793a\u3055\u308c\u305f\u3002\r\n\r\n\u3055\u3066\u3001$p=[\\sqrt{n}]$ $(p\\le \\sqrt{n}< p+1)$\u3000\u3067$p$\u3092\u5b9a\u3081\u308b\u3068\u3001$n$\u500b\u306e\u5b9f\u6570$\\sqrt{1},\\sqrt{2},\\sqrt{3},\\cdots,\\sqrt{n}$\u306e\u6574\u6570\u90e8\u5206\u306f$1,2,\\cdots,p$\u3067\u3042\u308b\u304b\u3089\u3001$N(n)$ \u306f\u3001\r\n\u3000\u3000\u3000\u3000$\\displaystyle\\sum_{l=1}^{p-1}a_{l} < N(n) \\le \\displaystyle\\sum_{l=1}^{p}a_{l}$\r\n\u3000\u3000\u3000\u3000$\\Longleftrightarrow \\displaystyle\\sum_{l=1}^{p-1}[2\\alpha l +\\alpha^2]+1 < N(n) \\le \\displaystyle\\sum_{l=1}^{p}[2\\alpha l +\\alpha^2]+1$   $\\cdots (\\star)$\r\n\u307e\u305f\u3001$2\\alpha l +\\alpha^2 - 1 <[2\\alpha l +\\alpha^2] \\le 2\\alpha l +\\alpha^2$\u3000\u3067\u3042\u308b\u304b\u3089\u3001\r\n\u3000\u3000\u3000\u3000$(\\star)\\Longleftrightarrow \\displaystyle\\sum_{l=1}^{p-1}2\\alpha l +\\alpha^2 < N(n) \\le \\displaystyle\\sum_{l=1}^{p}2\\alpha l +\\alpha^2+1$\r\n\u3000\u3000\u3000\u3000$\\Longleftrightarrow \\alpha (p-1)p+\\alpha^2 (p-1) < N(n) \\le \\alpha p(p+1)+(\\alpha^2 +1)p$\r\n\u3000\u3000\u3000\u3000$\\Longleftrightarrow \\alpha\\cdot\\displaystyle\\frac{(p-1)p}{n} + \\alpha^2 \\cdot\\displaystyle\\frac{p-1}{n}<\\displaystyle\\frac{N(n)}{n}\\le \\alpha\\cdot\\displaystyle\\frac{p(p+1)}{n}+(\\alpha^2 +1)\\cdot\\displaystyle\\frac{p}{n}$ $\\cdots(\\sharp)$\r\n\u307e\u305f\u3001\r\n\u3000\u3000\u3000\u3000$p\\le \\sqrt{n}< p+1\\Longleftrightarrow p^2 \\le n< p+1\\Longleftrightarrow 1\\le \\displaystyle\\frac{n}{p^2}<\\displaystyle\\left(1+\\frac{1}{p}\\right)^2$\r\n\u3067\u3042\u308a\u3001$n\\to\\infty$\u306e\u3068\u304d$p\\to\\infty$\u3067\u3042\u308b\u304b\u3089\u3001$\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\frac{n}{p^2}=1$\u3000\u3088\u3063\u3066$\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\frac{p^2}{n}=1$\u3000\u3068\u306a\u308b\u3002\u3088\u3063\u3066$(\\sharp)$\u306b\u304a\u3044\u3066\r\n\u3000\u3000\u3000\u3000$\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\frac{p}{n}=\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\frac{p^2}{n}\\cdot\\displaystyle\\frac{1}{p}=0$\r\n\u3068\u306a\u308b\u304b\u3089\u3001\u5f93\u3063\u3066\u3001$(\\sharp)$\u3067$n\\to\\infty$\u3068\u3059\u308b\u3053\u3068\u3067\u3001\u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u306b\u3088\u308a\r\n\u3000\u3000\u3000\u3000$\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\frac{N(n)}{n}=\\alpha$\r\n\u3092\u5f97\u308b\u3002$(Q.E.D.)$",
  "Solution_6": "\u5927\u962a\u5e02\u7acb\u5927\r\n\r\n$ \\boxed {3a}$ 1996\u5f8c\u671f(\u6539)\r\n$ a_{1} \\equal{} 2,a_{n} \\equal{} 2^{n \\minus{} 1}$$ (n\\ge2)$\u306b\u5bfe\u3057\u3066\u3001\r\n\u3000\u3000\u3000\u3000$ \\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{k \\equal{} 1}^{n}\\displaystyle\\int_{ \\minus{} \\pi}^{\\pi}(\\cos a_{1}x)(\\cos a_{2}x)\\cdots(\\cos a_{k}x)dx$\r\n\u3092\u6c42\u5024\u305b\u3088\u3002\r\n\r\n\r\n$ \\boxed {3b}$ 2005\u524d\u671f(\u6539)\r\n$ M$\u30922\u4ee5\u4e0a\u306e\u81ea\u7136\u6570\u3068\u3059\u308b\u3002$ \\mathbb{N}$\u3092\u81ea\u7136\u6570\u5168\u4f53\u306e\u96c6\u5408\u3068\u3057\u3001$ n\\in\\mathbb{N}$\u306b\u3064\u3044\u3066\u3001\u96c6\u5408$ A_{n} \\equal{} \\displaystyle\\left\\{m|m\\in\\mathbb{N},m\\le\\displaystyle\\frac {M}{n}\\right\\}$\u306e\u8981\u7d20\u306e\u500b\u6570\u3092$ a_{n}$\u3068\u3059\u308b\u3002$ S(M) \\equal{} \\displaystyle\\sum_{i \\equal{} 1}^{M}a_{i}$\u3068\u304a\u304f\u3068\u304d\u3001\r\n\u3000\u3000\u3000\u3000$ \\displaystyle\\lim_{M\\to\\infty}\\frac {S(M)}{M\\log M}$\r\n\u3092\u6c42\u5024\u305b\u3088\u3002\r\n\r\n\r\n\u7b54\u3048\u3060\u3051\u2026 :P \r\n[hide=\"answer 3a\"]$ 2\\pi$[/hide]\u3000\u3000[hide=\"answer 3b\"]1[/hide]\n\n$ \\boxed {4a}$\n$ \\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{k \\equal{} 1}^{n}\\log\\displaystyle \\left(1 \\plus{} \\displaystyle\\frac {k}{n^2}\\right)$\n\n[hide=\"ans\"]$ \\frac {1}{2}$[/hide]\n\n$ \\boxed {4b}$\n$ \\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{k \\equal{} 1}^{n}\\displaystyle\\frac {\\sqrt {k}}{n}\\sin\\displaystyle \\left(\\displaystyle\\frac {\\sqrt {k}}{n}\\right)$\n\n[hide=\"ans\"]$ \\frac {1}{2}$[/hide]\r\n\r\n$ \\boxed {4c}$\r\n$ \\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{k \\equal{} 1}^{n}\\displaystyle\\sin\\displaystyle \\left(\\displaystyle\\frac {1}{n \\plus{} k}\\right)$\r\n\r\n[hide=ans]$ \\log{2}$"
}
{
  "Tag": [
    "function",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "The function f(x,y) satisfies the conditions:\r\n(1) f(0,y)= y+1\r\n(2) f(x+1,0)= f(x,1)\r\n(3) f(x+1,y+1)= f(x, f(x+1,y))\r\nfor all non-negative integers x,y. \r\nFind f(4,1981)",
  "Solution_1": "This is IMO 1981[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=60818[/url]"
}
{
  "Tag": [
    "limit",
    "inequalities",
    "calculus",
    "ratio",
    "real analysis",
    "real analysis solved"
  ],
  "Problem": "if ${\\lim }\\limits_{n \\to \\infty } \\frac{a_n}{a_{n+1}}=l$,prove  ${\\lim }\\limits_{n \\to \\infty } a_n^\\frac {1}{n}=l$.\r\nand if ${\\lim }\\limits_{n \\to \\infty } a_n^\\frac {1}{n}=l$.can we also prove  ${\\lim }\\limits_{n \\to \\infty } \\frac{a_n}{a_{n+1}}=l$\r\n$(a_i>0)$",
  "Solution_1": "Your first excercise is very standart.\r\n\r\nSecond one is false. We can easily construct example satisfying  $\\lim_{n \\to \\infty } a_n^{1/n}=l$ and $\\lim_{n\\to\\infty}a_{n+1}/a_n$ doesn't exist.",
  "Solution_2": "Is nt the 1st one called the Cauchy's 2nd limit theorem.\r\nI might be wrong..",
  "Solution_3": "As [b]Myth[/b] told it can be easily constructed a series s.t. exists $\\lim_{n{\\rightarrow}\\infty}{\\sqrt[n]{a_{n}}}$,but it does not exists $\\lim_{n{\\rightarrow}\\infty}{\\frac{a_{n+1}}{a_n}}$(I'm sure that [b]Myth[/b] had that example).Here is one:$1,\\frac{1}{2},1,\\frac{1}{3},1,\\frac{1}{4},...,...$.\r\nNow more generalised inequality\r\n$\\liminf_{n{\\rightarrow}\\infty}{\\frac{a_{n+1}}{a_n}}{\\leq}\\liminf_{n{\\rightarrow}\\infty}{\\sqrt[n]{a_n}}{\\leq}\\limsup_{n{\\rightarrow}\\infty}{\\sqrt[n]{a_{n}}}{\\leq}\\limsup_{n{\\rightarrow}\\infty}{\\frac{a_{n+1}}{a_n}}$.\r\nIt is not hard to show.Indeed this inequality proves that the Dalamber's test of converging of positive series is weaker than Causchy's one.",
  "Solution_4": "[quote=\"LevonNurbekian\"]Indeed this inequality proves that the Dalamber's test of converging of positive series is weaker than Causchy's one.[/quote]\r\nTo translate this into the terminology usually used in American calculus books, \"The ratio test for the convergence of series is weaker than the root test.\"",
  "Solution_5": "thank you. :)"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Is it possible to tile an equilateral triangle with several equilateral triangles who have all different sizes ?  :)",
  "Solution_1": "I couldn't figure out how to approach the problem myself, but the answer appears to be\r\n\r\n[hide]no[/hide]\n\nas can be seen at\n\n[hide]http://www.math.niu.edu/~rusin/known-math/99/equilateral[/hide]",
  "Solution_2": "Thank you Kevinatcausa  :D \r\n\r\nI've find another approch more geometric .",
  "Solution_3": "Would you please post the geometric approach you found?",
  "Solution_4": "Require that we can cut the equilateral triangle in equilateral triangles which are all in  different sizes . We call \u00ab piece \u00bb a set of two of these equilateral triangles which share a vertex and an edge . The common vextex is call the origin of the piece and the vextex in the  common edge is call the end of the piece . Two pieces $ P$ and $ Q$ are independant  if  the interior of $ P\\cap Q$ is empty . Let $ P$ a piece of end $ E$ , the piece $ D$ is a daughter of $ P$ if $ D$ and $ P$ are independant and $ E$ is the origin of $ D$ . $ M$ is a mother of $ P$ if $ P$ is a daughter of $ M$ . The initial triangle is convex so every piece have exactly one daughter and every piece have at most one mother . Now , let a piece build with the smallest triangle and its daughter , the daughter of its daughter \u2026 The number of pieces is finite so in the list there are two identical pieces . Every peace have only one mother so the last daugther is the initial piece . It\u2019s not possible because the mother of the first piece will be build with a smaller peace of the smallest : contradiction . \r\n\r\nSorry for my bad english  :(\r\n\r\n[img]http://img115.imageshack.us/img115/6320/pieceau8.jpg[/img]",
  "Solution_5": "You say\r\n\r\nIt\u2019s not possible because the mother of the first piece will be build with a smaller peace of the smallest : contradiction .\r\n\r\nbut why?  there's no guarantee that the pieces will be getting smaller.",
  "Solution_6": "Okay , it's not correct ( sorry ) . What I have wanted to say : if the first piece have a mother , we can find a smaller piece of the smallest  :lol:   \r\n\r\n[img]http://img409.imageshack.us/img409/3892/piecesmotherny5.jpg[/img]"
}
{
  "Tag": [
    "trigonometry",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "solve equation $ 3^x\\cos(x\\arccos\\frac {1}{3}) \\equal{} 3^x \\minus{} 16x$    [color=green]its new ))[/color]   :D",
  "Solution_1": "[quote=\"Pirkuliyev Rovsen\"]solve equation $ 3^x\\cos(x\\arccos\\frac {1}{3}) \\equal{} 3^x \\minus{} 16x$    [color=green]its new ))[/color]   :D[/quote]\r\n\r\n$ 0$\r\n$ 2.90904131026685...$\r\n$ 4$\r\n$ 5.61158243898924...$\r\n$ 10.1531915815343...$\r\n$ 10.2610927724147...$\r\n$ 15.3088919940067...$\r\n$ 15.3168873649031...$\r\n$ 20.4169175734575...$\r\n$ 20.4174768414224...$\r\n$ 25.521477621273...$\r\n$ 25.5215154995391...$\r\n$ 30.6257946159122...$\r\n$ 30.6257971294984...$\r\n$ 35.7300951022729...$\r\n$ 35.7300952667275...$\r\n$ 40.8343944609037...$\r\n$ 40.8343945105003...$\r\n$ 45.938693783541...$\r\n$ 45.9386938220228...$\r\n$ 51.0429931077106...$\r\n$ 51.042993134429...$\r\n...\r\n\r\nOnce again certainly an invented problem and not a serious one coming from a contest or a regular training.\r\n\r\nSo no serious problem, no serious answer .",
  "Solution_2": "Here my solution  ;$ 3\\cos(x\\arccos\\frac{1}{3})\\equal{}3^2\\minus{}18x\\plus{}2x$\r\n$ \\cos(x\\arccos\\frac{1}{3})\\equal{}1\\minus{}2x\\frac{9}{3^x}\\plus{}2x\\frac{1}{3^x}$\r\n$ \\cos(x\\arccos\\frac{1}{3})\\equal{}1\\minus{}2x\\frac{1}{3^x\\minus{}2}\\plus{}2x\\frac{1}{3^x}$\r\n$ \\cos\\alpha\\equal{}\\frac{1}{3}\\implies\\alpha\\equal{}3\\arccos\\frac{1}{3}$\r\n$ \\cos(x\\arccos\\frac{1}{3})\\equal{}\\cos2\\alpha$\r\n$ \\cos2\\alpha\\equal{}1\\minus{}2x\\cos^x\\minus{}2 \\alpha\\plus{}2x\\cos^x \\alpha$\r\nIt are known  $ \\cos4\\alpha\\equal{}8\\cos^4\\alpha\\minus{}8\\cos^2\\alpha\\plus{}1$\r\n$ \\Longrightarrow  x\\equal{}4$",
  "Solution_3": "[quote=\"Pirkuliyev Rovsen\"]Here my solution  ; \n[...]\n$ \\Longrightarrow x \\equal{} 4$[/quote]\r\n\r\nYou just forgot infinitely many other solutions. You should better check your solution. Something is surely missing ...  :wink:"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "least common multiple"
  ],
  "Problem": "Let a and b be integers for which:\r\n\r\n1 + 1/2 + 1/3 + ... + 1/2004 = a/b\r\n\r\nSuppose that exactly one of a and b is even. Determine which of a and b is even, and find the largest integer r such that 2<sup>r</sup> divides ab.\r\n\r\n(As always, put your solutions in spoiler mode)",
  "Solution_1": "well so far, i found out that.... \n\n\n\n[hide]if a/b is in lowest terms then b = [1,2,3,...,2004] = 2^10*3^6*5^4....\n\n\n\nso b has 10 factors of 2 and must be even[/hide]\n\n\n\nehhh but im stuck on the rest",
  "Solution_2": "[hide]Sum(i=1,2004)(1/i) = Sum((2004!/i)/2004!).\n\n\n\nSay 2004! = 2^k * m, where a is odd.\n\n\n\nwe see that Sum(2004!/i) | 2^(k-10), since there isn't any i in the denomenator that factors to a power of 2 greater than 10.\n\n\n\nHowever, we see that if we take Sum(2004!/i) mod 2^(k-9), all terms in the series drop out (since they lack a power of 2 greater than 9) EXCPECT when i=1024.  Thus, we have Sum(2004!/i)  ! :equiv: 0 mod 2^(k-9).  \n\n\n\nThus, sum(2004!/i) = 2^(k-10) * n, where n is odd.\n\n\n\nSo, we have Sum((2004!/i)/2004!)= 2^(k-10) * n / (2^k * m) =      n/(2^10 * m).  Where n is odd.  Thus, we have a=n, b=2^10m.\n\nab=nm*2^10.  Since nm is odd, we have that 2^10 is the largest power of two that divides the product.  r=10.[/hide]",
  "Solution_3": "i just thought of an easy way to do this....\n\n[hide]\n\nsince a/b is in lowest terms, b is the least common multiple of 1,2,3,...2004 so the largest factor of 2 in b that is less than or equal to 2004 is 2^10=1024\n\n\n\nthis means b is even and a is odd (or else a/b would reduce even more)\n\n\n\nso a has no factors of 2, b has 10, thus ab has 10 factors of 2 and r=10[/hide]",
  "Solution_4": "Interesting. [hide]b=the least common multiples of set(1,2,3,4,5....2004)=2^10*3^6...... and the top is an odd integer, therefore, r=10. [/hide]",
  "Solution_5": "ThAzN1 wrote:i just thought of an easy way to do this....\n[hide]\nsince a/b is in lowest terms, b is the least common multiple of 1,2,3,...2004[/hide]\n\n\n\nHow do you know this is true?",
  "Solution_6": "errr i didn't mean lowest terms... [hide]i meant that only one of them has any factor of 2\n\n\n\nwell... if they weren't in lowest terms with one even and one odd, that wouldn't affect the amount of factors of 2 in ab anyway[/hide]\n\nwhoops...",
  "Solution_7": "ThAzN1 wrote:errr i didn't mean lowest terms... [hide]i meant that only one of them has any factor of 2[/hide]\n\n\n\nBut how do you know that is true? I think in essense this is what the problem is asking you to prove.",
  "Solution_8": "gauss202 wrote:ThAzN1 wrote:errr i didn't mean lowest terms... [hide]i meant that only one of them has any factor of 2\n\nwell... if they weren't in lowest terms with one even and one odd, that wouldn't affect the amount of factors of 2 in ab anyway[/hide]\n\nSure it would. It would depend on how many factors of 2 you had in a. The more there were, the less you would have in a/b when it was reduced.\n\n\n\n??? but since... one of a or b is odd and the other is even, it would be impossible to reduce a/b by a factor of 2?",
  "Solution_9": "I know. I reread what you wrote and edited my post concerning your assumption that only one of a and b is even for that choice of b.",
  "Solution_10": "I agree with ThAzN1 if you read my solution. b=2^10*3^6....... so each term equals b/n/b, where n=1,2,3,...2004. when n=1024, we will have an odd number for b/n, for other ns, we have all even for b/n, so a=b/1+b/2+b/3....b/2004 which is an odd number.",
  "Solution_11": "I agree with ThAzN1 if you read my solution. [hide]b=2^10*3^6....... so each term equals b/n/b, where n=1,2,3,...2004. when n=1024, we will have an odd number for b/n, for other ns, we have all even for b/n, so a=b/1+b/2+b/3....b/2004 which is an odd number.[/hide]",
  "Solution_12": "[hide]for my choice of b = [1,2,3,...2004]... a = b + b/2 + b/3... + b/2004, thus all of the b/something will have some factor of 2 with the exception of 1024, which is odd since b has 10 factors of 2 and so does 1024... that means b/1024 has no factor of two...\n\n\n\nso 2003 even integers + 1 odd means that a is odd when b = [1,2,3,...2004][/hide]\n\n\n\nwow... i never thought about all this before...",
  "Solution_13": "Nowww you've got it. ;-) It wasn't that what you were saying was untrue - it's just that you hadn't justified it. Your last post does that though. Good job! :)",
  "Solution_14": "Now what about in general:  If you write\n\n1/1 + 1/2 + 1/3 + ... + 1/n in lowest terms as a/b,\n\nIs a or b a multiple of 2.  What is the largest integer value of r such that 2^r divides ab?\n\n\n\n[hide]The same thing will always happen as when n=2004.\n\nSay k is the largest positive integer such that 2^k <= n.  Then we can write the fraction with a common denominator or 1*3*5*...([(n+1)/2]*2-1) * 2^k.\n\n[(n+1)/2]*2-1 is just a way of saying the greatest odd integer no greater than n.\n\n\n\nThen all the terms in the sum will have even numerators except for 1/(2^k).\n\nSo again the numerator of the fraction is odd, and the denominator has a factor of 2^k.\n\n[/hide]\n\n\n\nWhat if instead we ask about whether a or b is a multiple of 3, and what is the largest value of r such that 3^r divides ab?\n\n\n\n--Dan",
  "Solution_15": "well i think that in general...\n\n\n\n[hide]if 1+1/2+1/3...+1/n = a/b and n :ge: 2\n\nuhh then a = b+b/2+b/3...+b/n\n\nand b = [1,2,3,...n]\n\nso for every prime p :le: n... the terms b, b/2, b/3..., b/n will all be divisible by p^something except for the fraction b/p^x where x is the power of p in [1,2,3,...n] because all of the p's cancel out\n\n\n\nthis means that a is not divisible by p so a/b is in lowest terms if the process above is repeated for all primes :le: n\n\n\n\n....sooo b would be a multiple of 3 and the largest r that divides ab would be the power of 3 in [1,2,3,...n]\n\n\n\nwhich is 6 when n=2004[/hide]\n\n\n\nyea... im not really sure if all this is right tho"
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "algebra unsolved"
  ],
  "Problem": "For a polynom p, p(x)^5 - x is divided by (x-1)(x-2)(x-3). Find p(x)",
  "Solution_1": "[quote=\"fallinlovewithmaths\"]For a polynom p, p(x)^5 - x is divided by (x-1)(x-2)(x-3). Find p(x)[/quote]\r\nLet $ Q(x) \\equal{} (p(x))^5 \\minus{} x$, then $ Q(1)\\equal{}Q(2)\\equal{}Q(3)\\equal{}0$ or\r\n$ p(1)^5 \\equal{} 1$, $ p(2)^5 \\equal{} 2$, $ p(3)^5 \\equal{} 3$. Or equivalently:\r\n$ p(1) \\equal{} 1$, $ p(2) \\equal{} \\sqrt[5]{2}$ $ p(3) \\equal{} \\sqrt[5]{3}$. And there are infinitly many polynomials which satisfies that. Should there be another restricition?"
}
{
  "Tag": [
    "calculus",
    "integration",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "I've been figuring this out for the past two days. No luck. :(\r\n\r\nIntegral of: arctan[1/x] dx\r\n\r\nI must use integration by parts.\r\n\r\nAny ideas? Thank you very much for your time.",
  "Solution_1": "$\\arctan(x^{-1})=\\text{arccot}(x)$.  Put $du=dx$, $v=\\text{arccot}(x)$ and see what you get.",
  "Solution_2": "Question, does that mean:\r\n\r\narcsin(1/x) = arccsc(x)\r\narccos(1/x) = arcsec(x) ? Thanks!",
  "Solution_3": "Yes, up to periodicity/uniqueness issues."
}
{
  "Tag": [],
  "Problem": "The positive integer 182634795 has the property that the digits 1 to 5 are in natural ascending order, but 1 to 6 are not, and each digit from 1 to 9 is used precisely once.\r\n\r\nHow many such positive integers exist?",
  "Solution_1": "Put 12345, then you have 6 spaces to put the remaining four numbers in, but you have to subtract the case where 6 is in the last space.\n\n\n\nI think the answer is \n\n\n\n[hide]\n\n\n\n\n\n\n\n[/hide]",
  "Solution_2": "anyone can post a solution and explain please",
  "Solution_3": "First place the digits 1 to 5 in their natural order, as requested by the problem. In the obtained number, their are six \"slots\" where you may place other digits: $\\overline{a1b2c3d4e5f}$ We can see immediately that their are 5 slots, namely $a$ to $e$ that may contain the digit $6$, hence 5 possibilities to place the digit $6$. We then obtain a six-digit number with 7 slots. Now consider three cases:\r\n[list]\n[*] The remaining three digits occupy one single slots. As they may be permutated, there are $7\\cdot3!$ possibilities.\n[*] The remaining three digits occupy different slots. There are $3!\\cdot\\left(\\begin{array}{c}7\\\\3\\end{array}\\right)$, as, for each permutation of $\\{7;8;9\\}$, three slots among seven have to be selected.\n[*] Finally, suppose the three digits occupy two slots. There are $\\left(\\begin{array}{c}7\\\\2\\end{array}\\right)$ choices for the slots. Furthermore, the three digits may be permutated and partitionned in $2\\cdot 3!$ ways.[/list]\r\nFinally, the total number of such integers is $5\\cdot\\left[7\\cdot3!+3!\\cdot\\left(\\begin{array}{c}7\\\\3\\end{array}\\right)+2\\cdot 3!\\cdot\\left(\\begin{array}{c}7\\\\2\\end{array}\\right)\\right]$ ways.\r\n\r\nThe casework may seem exaggerated, by as I am awfully bad at combinatorics, I prefer to be prudent.",
  "Solution_4": "After you find the 5 possibilities to place the digit 6, you can do this.\r\n\r\nLet a,b,..,g be the remaining spots to place the other digits.\r\na+b+c+d+e+f+g = 3\r\nWhich has $\\binom{9}{3}$ solutions. Finally you can permute those 3 last digits.\r\n\r\n$\\boxed{5\\cdot 3! \\binom{9}{3}}$",
  "Solution_5": "yes, that seems to be easier.  :D  \r\n\r\nBy the way, a slightly more interesting problem is the following one:\r\n\r\n[i]The positive integer 5016283497 has the property that the digits 0 to 4 are in natural ascending order, but 0 to 5 are not, and each digit from 0 to 9 is used precisely once. How many such positive integers exist? (We suppose that a number does not begin with 0.)[/i]",
  "Solution_6": "[quote=\"Lepuslapis\"]yes, that seems to be easier.  :D  \n\nBy the way, a slightly more interesting problem is the following one:\n\n[i]The positive integer 5016283497 has the property that the digits 0 to 4 are in natural ascending order, but 0 to 5 are not, and each digit from 0 to 9 is used precisely once. How many such positive integers exist? (We suppose that a number does not begin with 0.)[/i][/quote]\r\nThe integer looks like this:\r\n\r\n[] 0 [] 1 [] 2 [] 3 [] 4 []\r\n\r\nThe \"[]\" represents an optional place for any number of the remaining digits, or for none at all.\r\n\r\nThere are 5 places to assign the digit 5 to (Anywhere except the end of the number, because 1 to 5 don't occur in their natural order).\r\n\r\nThen there are 7 places to assign the digit 6 to, because it can go in any of the places, and at either side of the 5.  Afterwards, there are 8 places to place the 7 into, 9 places for the 8 and 10 places for the 9.  This means that there are 5 x 7 x 8 x 9 x 10 = 25 200 integers of this kind, including those which begin with 0.\r\n\r\nTo exclude those which begin with zero, do the same process again, but now, nothing can go in the first empty position.\r\n\r\nSo there are 4 x 6 x 7 x 8 x 9 = 12 096 of them which begin with zero.\r\n\r\nThis leaves 25 200 - 12 096 = 13 104 valid integers.\r\n\r\nIs this correct?",
  "Solution_7": "Yes, I think so. I had thought of a slightly different approach: (1) The digit 5 is placed in the first place. Then, continue normally. (2) The digit 5 is not placed in the first slot. So we have to place another digit in that slot (there are 4 possibilities remaining). Then, continue normally.",
  "Solution_8": "[hide]_1_2_3_4_5\nThere are 5 places to put 6.\nThen there are 7 places, at the front, back, or in between two numbers, to place 7.\nThen there are 8 places to place 8.\nThen there are 9 places to place 9.\n\n$5\\cdot7\\cdot8\\cdot9=2520$.[/hide]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "logarithms",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "I'm very sorry if it had been posted .\r\nCalculus the integral : \r\n$ \\int {x^x}dx$",
  "Solution_1": "There is no simple solution to the integral. Still, you could be interested in this:\r\n\r\n[url]http://mathworld.wolfram.com/SophomoresDream.html[/url]",
  "Solution_2": "I cannot open the link to Mathworld...but ususally they dont have very in-depth reasoning.\r\n\r\nBut\r\n $ \\int_0^1\\frac{dx}{x^x}$\r\n\r\n$ =\\int_0^1{e^{x\\ln\\left(\\frac{1}{x}\\right)}dx}$\r\n$ =\\int_0^1\\sum_{n=0}^{\\infty}\\frac{x^n\\ln^n\\left(\\frac{1}{x}\\right)}{n!}$\r\n\r\nNow if we consider\r\n\r\n$ \\int_0^1x^n\\ln^n\\left(\\frac{1}{x}\\right)dx$\r\n\r\nThe change of variable $ \\ln\\left(\\frac{1}{x}\\right)=z$\r\n\r\n\r\nGives\r\n\r\n$ \\int_0^{\\infty}e^{-(n+1)z}z^ndz$\r\n\r\nFrom there if we let $ (n+1)z=m$ we get\r\n\r\n$ \\frac{1}{(n+1)^{n+1}}\\int_0^{\\infty}e^{-m}m^ndm$\r\n\r\nWhich we should realize is \r\n\r\n$ \\frac{\\Gamma(n+1)}{(n+1)^{n+1}}=\\frac{n!}{(n+1)^{n+1}}\\quad\\forall{n}\\in\\mathbb{N}$\r\n\r\n$ \\therefore\\sum_{n=0}^{\\infty}\\frac{1}{n!}\\int_0^1(-x\\ln(x))^ndx$\r\n\r\n$ =\\sum_{n=0}^{\\infty}\\frac{n!}{n!(n+1)^{n+1}}$\r\n\r\n$ =\\sum_{n=1}^{\\infty}\\frac{1}{n^n}$"
}
{
  "Tag": [
    "LaTeX",
    "calculus",
    "algebra unsolved",
    "algebra"
  ],
  "Problem": "I am completely baffled on what to do with fractional exponents like in the following problem.\r\n\r\nfactor and simplify \r\n\r\n(x^2 + 5)^ -1/4 - (x^2 + 5)^ -5/4\r\n\r\nCan anyone point me in the right direction?\r\n\r\nthanks",
  "Solution_1": "It might help to let $u = (x^{2}+5)^{-\\frac{1}{4}}= \\frac{1}{ \\sqrt[4]{ x^{2}+5}}$.  Then the original expression is\r\n\r\n$u-u^{5}= u(1-u^{4})$\r\n\r\nWhich is\r\n\r\n$\\frac{1-\\frac{1}{x^{2}+5}}{ \\sqrt[4]{x^{2}+5}}= \\frac{x^{2}+4}{ (x^{2}+5) \\sqrt[4]{x^{2}+5}}$\r\n\r\nThere's... really not much you can do past that point, as far as I can see.",
  "Solution_2": "[quote=\"t0rajir0u\"]It might help to let $u = (x^{2}+5)^{-\\frac{1}{4}}= \\frac{1}{ \\sqrt[4]{ x^{2}+5}}$.  Then the original expression is\n\n$u-u^{5}= u(1-u^{4})$\n\nWhich is\n\n$\\frac{1-\\frac{1}{x^{2}+5}}{ \\sqrt[4]{x^{2}+5}}= \\frac{x^{2}+4}{ (x^{2}+5) \\sqrt[4]{x^{2}+5}}$\n\nThere's... really not much you can do past that point, as far as I can see.[/quote]\r\n\r\nthanks... also did you use special board code to make it pretty print?",
  "Solution_3": "It's called $\\LaTeX$.  Most of the time, you just need to put dollar signs around your math.  There's a whole section of the site devoted to learning how to use it.",
  "Solution_4": "[quote=\"t0rajir0u\"]It's called $\\LaTeX$.  Most of the time, you just need to put dollar signs around your math.  There's a whole section of the site devoted to learning how to use it.[/quote]\r\n\r\nah interesting... I am a going back to school for a Computer Science degree, which requires a lot of math, so I might be posting/reading this forum a lot in the future.  Nice site.  I have not had any math for over a decade and I am jumping into College Algebra.  A lot of it is coming back, but I still have some catching up to do.  I guess it's a matter of re-learning.  I made it through Calculus before, but it's been awhile...  \r\n\r\nNow I don't fully understand what your are telling me.  This is what I did.\r\n\r\n$(x^{2}+5)^{-}1/4$ - $(x^{2}+5)^{-}5/4$\r\n\r\n      $1$\r\n_____________\r\n$(x^{2}+5)^{1}/4$ - $(x^{2}+5)^{5}/4$\r\n\r\n   $1$\r\n_____________\r\n$(x^{2}+5)^{-}4/4$ =\r\n\r\n    $1$\r\n-  _____________\r\n $(x^{2}+5)$\r\n\r\nthis probably isn't right, but that was my logic.  What did I do wrong if it isn't right?\r\n\r\nthanks",
  "Solution_5": "It's that very first step that's problematic.  $a^{-1}+b^{-1}\\neq \\frac{1}{a+b}$. \r\n\r\nRather, $a^{-1}+b^{-1}= \\frac{1}{a}+\\frac{1}{b}$, a very different expression, so that first step should result instead in\r\n\r\n$\\frac{1}{ (x^{2}+5)^{ \\frac{1}{4}}}-\\frac{1}{(x^{2}+5)^{ \\frac{5}{4}}}$"
}
{
  "Tag": [
    "inequalities proposed",
    "inequalities"
  ],
  "Problem": "In an triangle $ ABC$, denote by $ l_a,l_b,l_c$ the internal angle bisectors, $ m_a,m_b,m_c$ the medians, and $ h_a, h_b, h_c$ the altitudes to the sides $ a, b, c$ of the triangle. Prove that\r\n\r\n$ \\frac{m_a}{h_b\\plus{}l_b}\\plus{}\\frac{m_b}{h_c\\plus{}l_c}\\plus{}\\frac{m_c}{h_a\\plus{}l_a}\\geq\\frac{3}{2}$",
  "Solution_1": "[quote=\"filomen\"]In an triangle $ ABC$, denote by $ l_a,l_b,l_c$ the internal angle bisectors, $ m_a,m_b,m_c$ the medians, and $ h_a, h_b, h_c$ the altitudes to the sides $ a, b, c$ of the triangle. Prove that\n\n$ \\frac {m_a}{h_b \\plus{} l_b} \\plus{} \\frac {m_b}{h_c \\plus{} l_c} \\plus{} \\frac {m_c}{h_a \\plus{} l_a}\\geq\\frac {3}{2}$[/quote]\r\nIt's obviously true.",
  "Solution_2": "[quote=\"arqady\"][quote=\"filomen\"]In an triangle $ ABC$, denote by $ l_a,l_b,l_c$ the internal angle bisectors, $ m_a,m_b,m_c$ the medians, and $ h_a, h_b, h_c$ the altitudes to the sides $ a, b, c$ of the triangle. Prove that\n\n$ \\frac {m_a}{h_b \\plus{} l_b} \\plus{} \\frac {m_b}{h_c \\plus{} l_c} \\plus{} \\frac {m_c}{h_a \\plus{} l_a}\\geq\\frac {3}{2}$[/quote]\nIt's obviously true.[/quote]\r\n\r\nArqady, can you detail , please?  :(",
  "Solution_3": "filomen, because, $ h_b\\leq m_b,$ $ l_b\\leq m_b$ and AM-GM. :wink:"
}
{
  "Tag": [
    "induction"
  ],
  "Problem": "Give a combinatorial proof of the assertion that $p|(2^{p}-2)$ for all positive primes $p$.",
  "Solution_1": "This looks like a chinesse hypothesis that I heard about.\r\nAnyway, the proof that you are searching for is widely known, it can be found in Arthur Engel's book, and as I am sure in many many other books.\r\nHere is a link http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem",
  "Solution_2": "[quote=\"drunner2007\"]Give a combinatorial proof of the assertion that $p|(2^{p}-2)$ for all positive primes $p$.[/quote]\r\n[hide]Just let 2=1+1 and plug into the binomial theorem. The two 1s cancel out since your subtracting 2, and everything else has a factor of p.[/hide]\n[hide=\"or you could\"]try induction[/hide]"
}
{
  "Tag": [
    "AMC 10",
    "AMC"
  ],
  "Problem": "What to you think the AMC 10A 2008 passing score will be?",
  "Solution_1": "I'd suggest adding a poll.",
  "Solution_2": "The only realistic options are 114, 115.5, 117, and 118.5. Also, why is there no option for 120?\r\n\r\nAlso, many of the options overlap, if you think it will be 99 do you vote less than 105 or 90-100?",
  "Solution_3": "sorry, i screwed up the poll, please ignore!",
  "Solution_4": "Hmm, im just going to guess:\r\n\r\n96 for AMC 12, 118 for AMC 10\r\n\r\nit seemed like the amc 12 might have been harder than previous years amc 12s, and amc 10 about even, but ill just take down 2 points because there was no calcs.\r\n\r\nbut its most likely not right",
  "Solution_5": "I don't know... that AMC 10A seemed harder than last year's.  I'm going with 115.5",
  "Solution_6": "[quote=\"the future\"]Hmm, im just going to guess:\n\n96 for AMC 12, 118 for AMC 10\n\nit seemed like the amc 12 might have been harder than previous years amc 12s, and amc 10 about even, but ill just take down 2 points because there was no calcs.\n\nbut its most likely not right[/quote]\r\n\r\nwon't the cutoff be a possible score. for example, it is impossible to get a score of 118 as the score 118.5 is possible. but i don't know.",
  "Solution_7": "[quote=\"cognos599\"][quote=\"the future\"]Hmm, im just going to guess:\n\n96 for AMC 12, 118 for AMC 10\n\nit seemed like the amc 12 might have been harder than previous years amc 12s, and amc 10 about even, but ill just take down 2 points because there was no calcs.\n\nbut its most likely not right[/quote]\n\nwon't the cutoff be a possible score. for example, it is impossible to get a score of 118 as the score 118.5 is possible. but i don't know.[/quote]\r\n\r\nYep, you're right  :lol:"
}
{
  "Tag": [],
  "Problem": "Here is my rather holey proof of 1+1=3\r\n\r\nWe know that 2+2=4,\r\nbut 2=II in roman numerals,\r\nand, since i=sqr(-1),\r\n2+2=-1+-1=-2=-(II)=-(-1)=1\r\nSo, 2+2=1.\r\n\r\nSubtract 1 from both sides\r\n3=0\r\n\r\nDivide by 3\r\n1=0\r\n\r\nAdd two to both sides\r\n3=2\r\n\r\nSince 2=1+1,\r\n3=1+1\r\n\r\nAnyone want to improve?",
  "Solution_1": "Relevant to what?",
  "Solution_2": ":shock: \r\n\r\nRelevant for everyday life today in the modern world.\r\n\r\nDuh! <jk>",
  "Solution_3": "Isn't math [i]supposed[/i] to be useless to ``everyday life'' and stuff like that?"
}
{
  "Tag": [
    "function",
    "limit",
    "real analysis",
    "real analysis theorems"
  ],
  "Problem": "In an analysis book i found this proof for the uniqness of a limit of a function.\r\n\r\n\r\n\r\n\r\n\"It is easy to show that when a Limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that\r\n\r\n\r\nlim f(z) =l and lim f(z)=m as z----> a (z goes to a).\r\n\r\n\r\nThen,for any positive number \u03b5,there are positive numbers r,\u03b4 such that \r\n\r\n\r\n\r\n  \r\n           |f(z)-l |< \u03b5   whenever 0< |z-a |<r\r\n\r\n                       and\r\n\r\n             |f(z)-m |<\u03b5 whenever 0< |z-a |<\u03b4\r\n\r\n\r\n     So if   0< |z-a |<\u03b8 ,where \u03b8 denotes the smaller of the two Nos r and \u03b4,we find that\r\n\r\n |l-m |= |(f(z)-m)-(f(z)-l) |=< |f(z)-l | + |f(z)-m | < \u03b5 +\u03b5=2\u03b5                                                                                               \r\n\r\n        \r\n          But   |l-m | is a nonnegative constant, and \u03b5 can be chosen arbitrarily small.\r\nHence\r\n\r\nl-m =0 , or l=m.\" \r\n\r\n\r\nis that proof correct??",
  "Solution_1": "What doubts do you have about it?  Does every step make sense to you?",
  "Solution_2": "[quote=\"dallasboy\"]What doubts do you have about it?  Does every step make sense to you?[/quote]\r\n\r\n\r\nWhat the Author of that proof really proves here is:\r\n\r\n\r\nIf limf(z) = l and limf(z) = m as z goes to a ,then given \u03b5>0 and also GIVEN\r\n\r\n0<lz-al < \u03b8 ,where \u03b8 denotes the smaller of the two Nos r and \u03b4,\r\n\r\nTHEN l=m.\r\n\r\n...............and not.\r\n\r\n\r\nIf limf(z) = l and limf(z) = m as z goes to a, THEN l=m\r\n\r\nNote the Author in the last step of his proof ,he assumes the theorem:\r\n\r\nIF, for all \u03b5>o <\u03b5 ,then l=m\r\n\r\nMany of analysis books give a wrong proof following ,more or less the lines of the above proof.\r\n\r\nDo you agree??",
  "Solution_3": "You can prove by contradiction that if $ |x| < \\epsilon$ for all $ \\epsilon > 0$ then $ x \\equal{} 0$. Since $ \\epsilon$ was arbitrary in the proof this property will hold.",
  "Solution_4": "I know the classical proofs of the uniqueness of limit values, but I was wondering if this would hypothetically \"cut it\". Let f be a real valued function. Let $ \\lim_{x\\to{c}}f(x)$ exist and equal $ p$. Also let $ \\lim_{x\\to{c}}f(x)$ equal $ p'$. So we know that $ \\lim_{x\\to{c}}f(x)\\minus{}\\lim_{x\\to{c}}f(x)\\equal{}p\\minus{}p'$, but because the existence of both limits we would have that $ \\lim_{x\\to{c}}f(x)\\minus{}\\lim_{x\\to{c}}f(x)\\equal{}\\lim_{x\\to{c}}\\left\\{f(x)\\minus{}f(x)\\right\\}\\equal{}0$ which implies that $ p\\minus{}p'\\equal{}0\\implies p\\equal{}p'$. I know that is not as rigorous as usual, but is that not technically correct?",
  "Solution_5": "You're implicitly using the uniqueness of limits to prove the uniqueness of limits.  There's nothing stopping $ \\lim_{x \\to c} f(x) \\minus{} \\lim_{x \\to c} f(x) \\neq 0$ because you can't show additivity of limits until you show limits are unique!",
  "Solution_6": "[quote=\"t0rajir0u\"] There's nothing stopping $ \\lim_{x \\to c} f(x) \\minus{} \\lim_{x \\to c} f(x) \\neq 0$ because you can't show additivity of limits until you show limits are unique![/quote]\r\nDang additivity! Here is another one. This one is shady at best and very specific, proceed to shoot down if need be. Let $ f: \\mathbb{R}\\mapsto\\mathbb{R}$ and $ f\\in\\mathcal{C\\left(\\mathbb{R}\\right)}$. Since $ f$ is a continuous function we may redefine $ f(x) \\equal{} \\lim_{t\\to{x}}f(t)$, now suppose that $ \\lim_{t\\to{x}}f(t)$ was multi-valued, this would contradict our definition of $ f$ as a function. I know, that was pretty bad, but you cannot learn without getting your hands dirty.\r\n\r\nEDIT: Wait, same thing. We cannot establish that $ f\\in\\mathcal{C}\\implies\\lim_{x\\to{c}}f(x)\\equal{}f(c)$ before we establish uniqueness. Sorry to waste your time...long night  :roll:"
}
{
  "Tag": [
    "function",
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "There is a connected network with $ 2008$ computers, in which any of the two cycles don't have any common vertex. A hacker and a administrator are playing a game in this network. On the $ 1st$ move hacker selects one computer and hacks it, on the $ 2nd$ move administrator selects another computer and protects it. Then on every $ 2k\\plus{}1th$ move hacker hacks one more computer(if he can) which wasn't protected by the administrator and is directly connected (with an edge) to a computer which was hacked by the hacker before and on every $ 2k\\plus{}2th$ move administrator protects  one more computer(if he can) which wasn't hacked by the hacker and is directly connected (with an edge) to a computer which was protected by the administrator before for every $ k>0$. If both of them can't make move, the game ends. Determine the maximum number of computers which the hacker can guarantee to hack at the end of the game.",
  "Solution_1": "I have a qustion: Can a the hacker hack a computer twice?\r\nAnother quesiton: Is the network a undirected graph or directed graph?",
  "Solution_2": "A (maybe implicit) hint:[hide]If the network is undirected connnected and both hacker and administrator can hack and protect a page twice(which is a reasonable guess), then the answer of the generalizede question about $ N$ pages , if I am not wrong, is around $ \\frac {N}{3}$, though I haven't calculated the detail.The extreme case is that there are two cycles with $ 2k$ and $ k$ vertice and have excalty a common vertex. \nAnd the key method to solve this problem I used is construct a graph $ T(V,E)$, each vertex $ v$ of whic represents a cycle in the network and each edge $ (v_i,v_j)$ of which represent the cycles represented by $ v_i$ and $ v_j$ have a common computer.It follows that $ T$ should be a tree. And we can estabish a function $ f$ on $ V$ such that $ f(v_i)$ is the number of computers of the cycle represented by $ v_i$. Then we can find an edge of $ T$ such that the two sub-tree created by eliminating it has the least difference of the sum of $ f(v)$, and then give some algebric analysis.  [/hide]",
  "Solution_3": "It is undirected.\r\nActually, there was nothing about hacking or protecting a computer twice in the exam paper. Yet, I think it doesn't make sense. Because hacking a computer twice brings you nothing. \r\n\r\n[hide]@Hawk Tiger: Cycles have no common vertex as it is given in the question. So your extreme case is wrong, though your answer prediction is correct. (In fact, even if this graph is accepable, it is still wrong.) The extreme case is interesting, one of the main parts of the solution. \nBesides, it is true that the solution requires some algebraic work, but the solution I know isn't using the turning the cycles into vertices idea, in spite of the fact that it seems like a good idea to start. [/hide]\r\n\r\nI'm going out of the town for more or less a week. When I come back I'll write the solution if nobody writes it.",
  "Solution_4": "[quote=\"Umut Varolgunes\"]It is undirected.\nActually, there was nothing about hacking or protecting a computer twice in the exam paper. Yet, I think it doesn't make sense. Because hacking a computer twice brings you nothing. \n\n[hide]@Hawk Tiger: Cycles have no common vertex as it is given in the question. So your extreme case is wrong, though your answer prediction is correct. (In fact, even if this graph is accepable, it is still wrong.) The extreme case is interesting, one of the main parts of the solution. \nBesides, it is true that the solution requires some algebraic work, but the solution I know isn't using the turning the cycles into vertices idea, in spite of the fact that it seems like a good idea to start. [/hide]\nI'm going out of the town for more or less a week. When I come back I'll write the solution if nobody writes it.[/quote]\r\nI think I have misread the problem. I thought it is \"Cycles have no common edges\". And I don't know why if my graph is accepable, it is still wrong(I haven't done problems for long time, and may be stupier now...). And I think my solution is still useful, for my premary condition is weaker.",
  "Solution_5": "sorry, it must be \"acceptable\". (Does accepable exist?) In your solution you are finding an edge instead of a vertice. I'm giving the extreme case and you can see that your idea doesn't work, because you must start with a specific vertice of a cycle, instead of considering a cycle only with its number of vertices. \r\nin this graph, hacker should hack one of the six vertices of hexagon and after, admin. protects the opposite vertice. The admin can guarantee to protect two of the three main vertices.",
  "Solution_6": "Could somebody post the full solution?",
  "Solution_7": "the answer is 1004 , isn\u00b4t it?",
  "Solution_8": "We claim that the answer is $\\boxed{671}$. First, let's show that the hacker cannot do any better. Take a cycle of six vertices, say $v_1 v_2 v_3 v_4 v_5 v_6.$ Then, draw a path of $669$ vertices which starts at $v_1$, a path of $668$ vertices which starts at $v_3$, and a path of $668$ vertices which starts at $v_5.$ Draw these paths in a way so that they share no vertices amongst each other, and each share only one vertex with the cycle. In this case, it can be seen that the hacker's optimal result is getting $671$ computers. Notice that every computer must eventually be hacked or protected because of connectivity, and so a winning strategy for the hacker is equivalent to a strategy which restricts the administrator to at most $1337$ computers. \n\nLet's now show that the hacker can always get $671$ computers. Let's consider the obvious graph $G$ which corresponds to the network of computers. Consider now the graph $G'$ which results when we condense all cycles of $G$ into vertices. This is doable by the condition of the problem. For example, if $G$ is the example exhibited above, $G'$ becomes three paths of $668, 667, 667$ edges glued together at one vertex. Observe that $G'$ is acyclic and connected, and so therefore is a tree. \n\nFor a subgraph of $G'$, define its [b] size [/b] to be the number of vertices it has, after \"un-condensing\" all vertices of it which correspond to cycles of $G'$. For instance, in our example, $G'$ has size $2008$ and not $1 + 668 + 667 + 667 = 2003$.\n\nCall a vertex $v$ of $G'$ [b] tasty [/b] if its removal leaves only connected components of $G'$ which have sizes $< 1338.$ We now consider two cases. \n\n[b] Case 1. [/b] There does not exist a tasty vertex of $G'$.\n\nWe will apply the following operation on every vertex $v$ of $G'.$ Since $v$ is not tasty, there exists a connected component of size at least $1338$ resulting from the removal of $v$. Let $w$ be the unique vertex of this connected component which is adjacent to $v.$ Then, draw an arrow from $v$ to $w.$ Because there are only $|G'| - 1$ edges of $G$ and $|G'|$ arrows drawn, by Pigeonhole there must be an edge $ab$ so that there is an edge from $a$ to $b$ and from $b$ to $a.$ \n\nNow, consider the effect of removing the edge $ab.$ From the definitions of these arrows, we know that the two resulting connected components must both have size at least $1338.$ However, it's clear that the sum of their sizes must be at most $2008$, and so this case is actually an absurdity.\n\n[b] Case 2. [/b] There exists a tasty vertex of $G'$.\n\nCall this vertex $v.$ If this vertex corresponds to a vertex of $G$ (and not a cycle), then there is literally no way that the hacker can lose. Indeed, the administrator can only protect the computers in one connected component of $G$ resulting from the removal of $v$, and so this is at most $1337$ computers. Hence, the hacker can ensure at least $2008 - 1337 = 671$ hacked computers because the remainder of the graph is connected. \n\nOtherwise, suppose that $v$ corresponds to a cycle $v_1 v_2 \\cdots v_k$ of $G.$ In the cases when $k = 2, 3, 4, 5,$ or $6$, it is easy to find a winning strategy for the hacker and so these cases are left as exercises to the reader. Henceforth, assume that $k > 6.$ \n\nWe'll first exhibit a strategy for the hacker to hack at least half of the vertices of the cycle. Indeed, this is trivial, as the hacker simply keeps on hacking computers in the cycle arbitrarily as long as he/she can. \n\nConsider a set $S$ of contiguous vertices in the cycle. We will define the [b] quality [/b] of $S$ as follows. Consider the subgraph of $G$ which results when we remove the vertices of the cycle and all edges involving any of these vertices. It is a collection of connected components, some of which were attached to vertices of $S$ in the original graph $G.$ Call these connected components [b] branches [/b] of $S.$ The quality of $S$ is then the sum of the numbers of vertices of all its branches plus the number of vertices in $S.$ Notice that, in a sense, quality is additive. \n\nLet's now consider two subcases.\n\n[b] Subcase 1. [/b] There exists a contiguous set of at least half of the vertices of the cycle with quality at most $670.$\n\nWLOG let this contiguous set of vertices be $\\{v_1, v_2, \\cdots, v_a\\}$ for some $a \\ge \\frac{k}{2}$. Then, it's clear that the quality of $\\{v_{a+1}, v_{a+2}, \\cdots, v_k\\}$ is at least $1338.$ By \"discrete continuity,\" there must exist some $a+1 \\le b \\le k$ so that the quality of $\\{v_{a+1}, \\cdots, v_b\\}$ and the quality of $\\{v_b, v_{b+1}, \\cdots, v_{k}\\}$ are each at least $669.$ The hackers strategy is now simple. Start by hacking $v_b.$ Now, it's clear that the hacker can ensure that either the entirety of the set $\\{v_{a+1}, \\cdots, v_b\\}$ is hacked or the entirety of the set $\\{v_b, v_{b+1}, \\cdots, v_{k}\\}$ is hacked. Indeed, simply start hacking clockwise around the cycle or start hacking counterclockwise around the cycle, depending on which computer the administrator first protectts. In either case, the hacker will have hacked a contiguous set of vertices of quality at least $669$ eventually. After the hacker ensures that he/she has hacked a contiguous set of at least $669$ vertices as described above, then the hacker will continue to hack vertices of the cycle arbitrarily. Afterwards, we consider two cases. If the administrator has protected a computer in the branches of the set of computers that the hacker has hacked, then the hacker wins because the administrator can protect at most one branch with at most $1337$ vertices (by the tastiness of $v$). Otherwise, the administrator proceeds to infect all of the branches of the set of computers he has hacked so far. This ensures that the hacker has hacked at least $669$ computers.\n\nWe will now work on improving this bound. If one of the sets consists of at least half of the vertices of the cycle, then it's clear that the other set consists of only one vertex. The hacker alters his strategy as follows. Start by infecting this vertex, say $\\alpha$. From here, the hacker will set out to infect at least half of the vertices of the cycle. The hacker will hack at least two more computers in the cycle, because $k \\ge 7.$ Again, if the administrator protects any of the branches of $\\{\\alpha \\}$, then he/she can again protect at most $1337$ computers, and the hacker wins. Otherwise, the hacker hacks all of the branches of $\\{\\alpha\\}$. After this, the hacker will have hacked at least $669$ computers (quality of $\\{\\alpha\\}$) plus the additional two computers, for at least $671$ computers.\n\nOtherwise, both sets consist of less than half of the vertices of the cycle. If both sets have at most $\\left \\lceil \\frac{k}{2} \\right \\rceil - 2$ vertices, then the hacker can win with a similar strategy as in the above paragraph (infect an entire set and then get some more computers in the cycle). Otherwise, suppose that one of the sets has at least $\\left \\lceil \\frac{k}{2} \\right \\rceil - 1$ vertices. This means that the other set has at most $2$ vertices. If it has one vertex, then the above argument works. Suppose it has $2$ vertices. There then exists a set of two vertices with quality at least $669$. Call these two computers the [b] legends. [/b] Observe that if the hacker is able to hack both of the legends as his/her first two moves, then he/she wins. Indeed, he/she then simply focuses on infecting additional vertices of the cycle, and then we finish with the same \"if the administrator protected a branch of the legends, he/she loses; otherwise, we just infect the branches of the legends\" logic. Therefore, if the hacker takes one of the legends as his first move, the administrator must respond by protecting the other legend. This leads to the following solution. Split the cycle into two contiguous sets of vertices differing in size by at most $1$ so that the legends are in different set. The hacker then hacks the legend which in the set of quality at least $1004$, call this set $\\Gamma$. After the hacker protects the other legend, it's clear that the hacker can go on to infect all of the other vertices in $\\Gamma$, and from there can infect all of the branches of $\\Gamma.$ As the quality of $\\Gamma$ is at least $1004$, the hacker will have infected at least $1004 \\ge 671$ computers, and so he/she wins. \n\n[b] Subcase 2. [/b] There does not exist such a set.\n\nThen, the hacker simply keeps on picking vertices of the cycle arbitrarily until he/she cannot. At this point, the hacker will have infected a set of at least half of the computers of the cycle with quality at least $671$, by the assumption of this case. If the administrator ever protected any computer among the branches of this set of vertices, then since $v$ is tasty he/she will have protected at most $1337$ computers and the hacker can therefore win by infecting all other computers. Otherwise, the hacker can infect all branches of this set as well, for at least $671$ computers. \n\n\nAs we've exhausted all cases, we've shown that the hacker can always infect $\\ge 671$ computers, and so we're done.\n\n$\\square$"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "tetrahedron",
    "circumcircle",
    "sphere",
    "IMO Shortlist",
    "IMO Longlist"
  ],
  "Problem": "In a tetrahedron, all three pairs of opposite (skew) edges are mutually perpendicular. Prove that the midpoints of the six edges of the tetrahedron lie on one sphere.",
  "Solution_1": "This can basicly be reduced to a plane geometry problem.\r\n\r\nThe vertex $D$ is projected onto the orthocenter $H$ of $ABC$ (this follows from the conditions $AB\\perp CD,\\ AC\\perp BD,\\ AD\\perp BC$). This means that the midpoints $U,V,T$ of $DA,DB,DC$ are projected onto the midpoints $X,Y,Z$ of $HA,HB,HC$. Let $M,N,P$ be the midpoints of $BC,CA,AB$. The points $M,N,P,X,Y,Z$ liue on the nine-point circle of $ABC$, so the points $U,V,T$ lie on a circle of radius equal to $\\frac R2$ ($R$ is the circumradius of $ABC$) which lies on a plane parallel to $ABC$ and which has its center directly above the nine-point center of $ABC$. It's is now clear that $M,N,P,U,V,T$ lie on a sphere which has its center in the midpoint of the segment formed by the centers of $(MNP),(UVT)$."
}
{
  "Tag": [
    "MATHCOUNTS"
  ],
  "Problem": "Express in simplest form:\r\n$\\left(1+\\frac{1}{2}\\right)+\\left(1+\\frac{1}{3}\\right)+\\left(1+\\frac{1}{4}\\right)+...+\\left(1+\\frac{1}{7}\\right)$\r\n\r\nSprint round = no calculators =  :( = I need a fast way to do this!\r\n\r\n[size=150][b]EDIT:[/b][/size]It is actually$\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right)...\\left(1+\\frac{1}{7}\\right)$\r\nSorry :blush:",
  "Solution_1": "[hide]\n$1+1+1+1+1+1+\\frac{(6(5(4(2+3)+3!)+4!)+5!)+6!}{7!}$\n\nIt's really nasty to calculate, but it works...\n\nBtw, the ans is $7 \\frac{22}{35}$\n[/hide]",
  "Solution_2": "i think simplest form means common fraction, not mixed number",
  "Solution_3": "[quote=\"xscapezaer\"][hide]\n$1+1+1+1+1+1+\\frac{(6(5(4(2+3)+3!)+4!)+5!)+6!}{7!}$\n\nIt's really nasty to calculate, but it works...\n\nBtw, the ans is $7 \\frac{22}{35}$\n[/hide][/quote]\r\nThat is not right. And how would you do that calculation without a calculator?\r\n\r\nEdit: Sorry, I accidentally put the wrong equation in... look at first post... :blush:",
  "Solution_4": "[hide=\"Hint\"]Look for a pattern.[/hide]\n\n[hide=\"Solution\"]If you add the first parentheses, you'll get$\\frac{3}{2}$.When you add the next one you get $\\frac{4}{3}$. So the denominator is going to be the previous numerator and the numerator is going to be one more than it. And this happens for 6 times. So its just $\\frac{3}{2}\\cdot\\frac{4}{3}\\cdot\\frac{5}{4}\\cdot\\frac{6}{5}\\cdot\\frac{7}{6}\\cdot\\frac{8}{7}=4$.[/hide]\r\nSorry if it is too hard to follow(if it is).",
  "Solution_5": "[quote=\"Quevvy\"]$\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right)...\\left(1+\\frac{1}{7}\\right)$[/quote]\r\n[hide=\"Solution\"]$\\frac32\\cdot\\frac43\\cdot\\frac54\\cdot\\frac65\\cdot\\frac76\\cdot\\frac87$\n\n$\\frac{\\not3}2\\cdot\\frac{\\not4}{\\not3}\\cdot\\frac{\\not5}{\\not4}\\cdot\\frac{\\not6}{\\not5}\\cdot\\frac{\\not7}{\\not6}\\cdot\\frac8{\\not7}$\n\n$\\frac82=\\boxed{4}$[/hide]",
  "Solution_6": "Oh you were supposed to multiply those? Oh well...",
  "Solution_7": "[quote=\"Quevvy\"]Express in simplest form:\n$\\left(1+\\frac{1}{2}\\right)+\\left(1+\\frac{1}{3}\\right)+\\left(1+\\frac{1}{4}\\right)+...+\\left(1+\\frac{1}{7}\\right)$\n\nSprint round = no calculators =  :( = I need a fast way to do this!\n\n[size=150][b]EDIT:[/b][/size]It is actually$\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right)...\\left(1+\\frac{1}{7}\\right)$\nSorry :blush:[/quote]\r\n\r\nnow its vway easier, with the multiplication.\r\nit is simply 3/2 times 4/3 times 5/4 times (denominator + 1)/(denominator) up to 7, so the 3, 4, 5, 6, and 7 cancel out, and that leaves $\\frac{8}{2}=4$"
}
{
  "Tag": [
    "calculus",
    "geometry"
  ],
  "Problem": "I have a general question about learning mathematics.\r\n\r\nWhat type of learning methodology do you feel is more effective: the \"top-down\" approach, or the \"bottom-up\" approach?\r\n\r\nBy top-down I mean learning a subject by first looking at theorems and their proofs, and then running through many examples and applying the theorems directly, and deductively coming up with new results from the theorems.\r\n\r\nBy bottom-up I mean learning a subject by working on specific problems that involve new concepts, learning the concepts through the problems, and slowly building up, culminating in a theorem or complete generalization.\r\n\r\nWhy do you think one approach is more effective than the other? Are different approaches more or less suited for particular subjects, or at particular times? How does one's mathematical maturity affect the effectiveness of each of these approaches?\r\n\r\nI know this is a long list of questions, but I want to try to get a good understanding of the topic. Thanks",
  "Solution_1": "This is a very interesting topic. I learned pretty much all of high school math using the bottom-up approach, and I think it really helped me gain a solid understanding. However, top-down is much faster. I think the main difference is that bottom-up builds intuition, whereas top-down just gets the ideas across. You should go bottom-up in very new subjects, but for something like analysis after calculus, top-down is probably more efficient.",
  "Solution_2": "[quote=\"probability1.01\"]This is a very interesting topic. I learned pretty much all of high school math using the bottom-up approach, and I think it really helped me gain a solid understanding. However, top-down is much faster. I think the main difference is that bottom-up builds intuition, whereas top-down just gets the ideas across. You should go bottom-up in very new subjects, but for something like analysis after calculus, top-down is probably more efficient.[/quote]\r\n\r\ni agree with most of what you said. but it seems like at a certain point all math classes take the top-down approach (even in subjects that students have never seen before).",
  "Solution_3": "ET Bell, speaking of Carl Gustav Jacob Jacobi, says:\r\n\r\n\"Jacobi seems to have been the first regular mathematical instructor in a university to trin students in research by lecturing on his own latest discoveries and lettign the students see the creation of a new subject taking place before them. He belived in pitching young men into the icy water to learn to swim or drown by themsleves. Many students put off attempting anything on their own account till they have mastered everything relating to their problem that has been done by others. The result is that but a few ever acquire the knack of independent work. Jacobi combated this dilatory erudition. To drive home the point to a gifted but diffident young man who was always putting off doing anything until he had learned soemthing more, Jacobi delivered himself of the following parable: \"Your father would never have married, and you wouldn't be here now, if he had insisted on knowing [i]all[/i] the girls in the world before marrying [i]one[/i].\"\"",
  "Solution_4": "[quote=\"bubka\"]ET Bell, speaking of Carl Gustav Jacob Jacobi, says:\n\n\"Jacobi seems to have been the first regular mathematical instructor in a university to trin students in research by lecturing on his own latest discoveries and lettign the students see the creation of a new subject taking place before them. He belived in pitching young men into the icy water to learn to swim or drown by themsleves. Many students put off attempting anything on their own account till they have mastered everything relating to their problem that has been done by others. The result is that but a few ever acquire the knack of independent work. Jacobi combated this dilatory erudition. To drive home the point to a gifted but diffident young man who was always putting off doing anything until he had learned soemthing more, Jacobi delivered himself of the following parable: \"Your father would never have married, and you wouldn't be here now, if he had insisted on knowing [i]all[/i] the girls in the world before marrying [i]one[/i].\"\"[/quote]\r\n\r\nwow that's pretty interesting. i notice that i to feel like attaining master of a subject before i venture out to discover/research new things. with the new book i'm working in though, it forces you to discover important results on your own. so maybe i can change that tendency.",
  "Solution_5": "Bottom up requires lots of time, but develops deeper understanding imo. Classes seem to want to get large amounts of work through to students as quickly as possible..",
  "Solution_6": "I would say going over and studying a topic at least 3 times during the study time will force you to learn the material thoroughly. So I think the top-down approach is a better strategy for true understanding.",
  "Solution_7": "actually now that I think about it, doing problems is where the real understanding comes from. So a bottom up teaches you problem solving skils (i.e. looking at problems first, seeing if you can do them, and then looking at book for help)",
  "Solution_8": "Here in India , it's mostly a top-down approach that most people take..... I've never really seen anyone go by a bottom-up approach here....  :|",
  "Solution_9": "most of the AOPS books are centered around the bottom up approach, right?",
  "Solution_10": "[quote=\"drunner2007\"]most of the AOPS books are centered around the bottom up approach, right?[/quote]\r\nThat has been my impression (from an exhaustive sampling of [hide=\"...\"] one book  :roll: [AoPS Volume 2][/hide]).  Art of Problem Solving is extremely unusual in their efforts to actually teach students all the reasoning behind the mathematics, as opposed to force-feeding them information.\r\n\r\nI can also attest to the fact that the above-mentioned book caused my AMC score to jump by nearly twenty points.",
  "Solution_11": "When teaching myself from the book I use the \"top-down\" approach, whereas my math teacher uses the \"bottom-up\" appproach when teaching my class.  \r\n\r\nI can work through more material at a faster rate using the \"top-down\" system, but I think that I understand material better the way my teacher does it (or maybe that is just because I have a teacher, I don't know).  I think that it really depends on the person and the material, what works best and what they can handle.",
  "Solution_12": "In the 1960's, my dad did his dissertation on the question of which approach was more effective for the teaching of middle-school mathematics.  His study showed that it's more effective to do bottom-up, but takes longer.\r\n\r\nWhen I study mathematics, I tend to use a combination approach by starting with top-down to narrow the area I want to think about, and then use bottom-up to get a grasp on that area.  This has been effective for me in producing research in subfields outside my specialty.\r\n\r\nWhen I'm teaching, I tend to vary my approach based on the mathematical maturity and interest of the students.  I usually spend class time on bottom-up activities, and have students do any top-down reading outside of class.  But the more mature/interested the students, the less I assign reading and the more I assign investigations."
}
{
  "Tag": [
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ \\Omega \\subset \\mathbb{R}^n$ open set, and let $ f_1,...,f_r \\in \\mathcal{ C}(\\Omega):$ $ \\displaystyle{\\sum_{j\\equal{}1}^r}|f_j(x)|>0, \\forall x \\in \\Omega$. Find $ g_1,...,g_r \\in \\mathcal{C}(\\Omega)$ so that $ \\displaystyle{\\sum_{j\\equal{}1}^rg_j(x)f_j(x)}\\equal{}1, \\forall x \\in \\Omega$",
  "Solution_1": "Can you please find titles that say [i]something[/i] about the content of your threads?"
}
{
  "Tag": [],
  "Problem": "So I was on the plane listening to music on random mode, where it just randomly plays songs without regard to whether the song has already been played or not.  So first I hear Song 5, then I hear Song 17, and then I hear Song 5 again, even though I had practically just heard it.  If I have 80 songs on my CD, then on average, what is the number of songs I need to listen to in order to hear every single song at least once?",
  "Solution_1": "there must have been a similar problem posted before.\r\nthe answer is [tex]n\\left(1+\\frac{1}{2}+\\cdots+\\frac{1}{n}\\right)[/tex]",
  "Solution_2": "see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14034[/url]\r\nthe \"coupon collector\" problem at the near bottom of the page.",
  "Solution_3": "Intutive and short reply:  let us take general case with 'n'  songs (n=80)\r\n\r\nIf you want to hear  one particular song you have to play (on average) n songs.\r\nSuppose there are 'm' songs of  one type (say rock-and-roll), then to hear  one (that is, any one) song  of that type..  you have to  (on average) play    n/m   songs.\r\n\r\nRight?\r\nNow let us say the by the type I mean  \"fresh\" of 'not_heard_before':\r\nSo for first  \"fresh\" song  =>            1    (obviiously  first song is always  new)\r\nFor the second \"fresh\" song ==>    n/(n-1)    \r\nFor the next fresh song        ==>     n/(n-2)\r\n\r\nSo the answer = 1+  n/(n-1)+ n/(n-2) + n/(n-3) + ......   n/1\r\nWhich is same as given above.. (Detail argument is given in the link shown there) ..."
}
{
  "Tag": [
    "probability"
  ],
  "Problem": "In a class of 28 sutdents, the teacher selects 4 people at random to participate in a geography contest. What is the probability that this group of four students includes at least 2 of the top 3 geography students in the class? Express your answer as a common fraction.",
  "Solution_1": "We examine two cases.\r\n\r\n[b]Case 1[/b]: 2 of the top 3 students\r\n\r\n$ 2 \\times 1 \\times 26 \\times 25 \\equal{} 1300$\r\n\r\n$ \\binom{28}{4} \\equal{} 20475$\r\n\r\n$ \\frac{1300}{20475} \\implies \\frac{4}{63}$\r\n\r\n[b]Case 1[/b]: 3 of the top 3 students\r\n\r\n$ 3 \\times 2 \\times 1 \\times 25 \\equal{} 150$\r\n\r\n$ \\binom{28}{4} \\equal{} 20475$\r\n\r\n$ \\frac{150}{20475} \\implies \\frac{2}{273}$\r\n\r\n$ \\frac{4}{63} \\plus{} \\frac{2}{273} \\equal{} \\boxed{\\frac{58}{819}}$\r\n\r\nPlease correct me if I'm wrong.",
  "Solution_2": "[hide]\nIn total, there are $ {28\\choose 4}$ ways to choose them.\nTo include at least 2 of the top 3 geography students,\n1- We only choose 2\n$ 3{25\\choose 2}$\n2- We choose all three.\n$ 25$\n\nSo $ \\dfrac{925}{{28\\choose 4}}$, i don't feel like calculating.[/hide]\r\nEDIT: you can also use complementary counting."
}
{
  "Tag": [
    "group theory",
    "abstract algebra",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let $ (G,*)$ a group and $ (H_1,*)$, $ (H_2,*)$  two subgroups of $ (G,*)$\r\n\r\nIt's obvious that the intersection of $ H_1$ and $ H_2$ give a subgroup of G\r\n\r\nnow find a counterexample wich proves that The union of two subgroups is not always a subgroup.",
  "Solution_1": "the union of two subgroups $ (H_1,*) (H_2,*)$ is a subgroup iff $ H_1 \\subset H_2$ or$ H_2 \\subset H_1$.\r\nproof:\r\nsuppose that the union is subgroup and there exist $ a \\in H_1 \\minus{} H_2$ and $ b \\in H_2 \\minus{} H_1$.since the union is subgroup then $ ab \\in H_1 \\cup H_2$.\r\n if $ ab \\in H_1$ and $ a \\in H_1$ then $ b$ is also in $ H_1$ which it is contardiction and we the same way we obtain  a contradiction in the other case .",
  "Solution_2": "Take $ 2\\in 2\\mathbb{Z}$ and $ 3\\in 3\\mathbb{Z}$, then $ 5\\not\\in 2\\mathbb{Z}\\cup 3\\mathbb{Z}$, so we don't \r\nhave closure...",
  "Solution_3": "thank you chebychev and vl4d :)"
}
{
  "Tag": [
    "function"
  ],
  "Problem": "If a+b=n, where n > 0 and where a > 0 and b > 0, obviously n > a and n > b \r\n\r\nProve that max ab= (n^2)/4",
  "Solution_1": "This might be too easy for pre-olympiad  :D  \r\nIt's just equivalent to $4ab \\leq n^{2}= (a+b)^{2}$, i.e. $0 \\leq (a-b)^{2}$.",
  "Solution_2": "Another way:\r\nLet $a=\\frac{n}{2}+x$; then $b=\\frac{n}{2}-x$ so $ab=(\\frac{n}{2}+x)(\\frac{n}{2}-x)=\\frac{n^{2}}{4}-x^{2}\\le \\frac{n^{2}}{4}$.",
  "Solution_3": "Yeah, I know. I should have looked at it for a while on my own. I also meant to say to find the maximum of ab, but it is essentially the same. I now know that since nb-b^(2) is a function and quadratic. It, therefore, can be written as -(b-n/2)^(2) + N^2/4."
}
{
  "Tag": [
    "geometry proposed",
    "geometry"
  ],
  "Problem": "The vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite arc $BC$ is a semicircle while arc $BC$ and arc $AC$ are supplementary. To each of three arcs, we draw a tangent such that its point of tangency is the mid point of that portion of the tangent intercepted by the extended lines $AB,AC$. More precisely, the point $D$ on arc $BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where tangent at $D$ intersects the extended lines $AB,AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$. Prove that triangle $DEF$ is equilateral.",
  "Solution_1": "We have\r\n\\[ \\left\\{\\begin{array}{c}\\widehat{D''DC}=\\widehat{DAC}=\\widehat{DD''A}\\\\ \\widehat{E''EC}=\\widehat{EAC}=\\widehat{EE''A}\\end{array}\\right\\| \\] \\[ \\Rightarrow\\widehat{DCE}=\\widehat{DCA}+\\widehat{ACE}=2\\cdot\\left(\\widehat{DAC}+\\widehat{CAE}\\right)=2\\cdot\\widehat{DAE}\\]\r\nOn the other hand, the quadrilateral $ AECD$ is cyclic, so $ \\widehat{DCE}+\\widehat{DAE}=180^\\circ\\Rightarrow\\widehat{DAE}=60^\\circ$\r\n\r\nHence $ \\widehat{DFE}=\\widehat{DAE}=60^\\circ.$ Similarly, $ \\widehat{DEF}=60^\\circ.$ \r\n\r\nIt implies that triangle $ DEF$ is equilateral.",
  "Solution_2": "[quote=\"April\"]We have\n\\[ \\left\\{\\begin{array}{c}\\widehat{D''DC}=\\widehat{DAC}=\\widehat{DD''A}\\\\ \\widehat{E''EC}=\\widehat{EAC}=\\widehat{EE''A}\\end{array}\\right\\| \\]\n\n\\[ \\Rightarrow\\widehat{DCE}=\\widehat{DCA}+\\widehat{ACE}=2\\cdot\\left(\\widehat{DAC}+\\widehat{CAE}\\right)=2\\cdot\\widehat{DAE}\\]\nOn the other hand, the quadrilateral $ AECD$ is cyclic, so $ \\widehat{DCE}+\\widehat{DAE}=180^\\circ\\Rightarrow\\widehat{DAE}=60^\\circ$\n\nHence $ \\widehat{DFE}=\\widehat{DAE}=60^\\circ.$ Similarly, $ \\widehat{DEF}=60^\\circ.$ \n\nIt implies that triangle $ DEF$ is equilateral.[/quote]\r\n\r\n[b] Nice problem - Nice solution. I add the figure to it.[/b]\r\n\r\n A! Now I see a question : Why \r\n\r\n   $ \\widehat{ACE}=2\\widehat{CAE}$ ?\r\n\r\n I do not deduce it obviously :(\r\n\r\n[color=red] [b]EDIT :  The figure in the official solution is different. Please do not look at it. I will change it , if it is wrong.[/b][/color]"
}
{
  "Tag": [
    "calculus",
    "integration",
    "number theory",
    "modular arithmetic"
  ],
  "Problem": "i am a math teacher that is working on her master's.  My teacher gave me this problem to work and other than going one by one to find a pattern, I am not sure how to do this.  This is due tomorrow!   :oops:  the problem is:\r\n\r\n Find every positive integer (n) such that n^3+n^2+n^1+1 is square\r\n\r\nthank you!!!",
  "Solution_1": "We can factor the given equation to $(n^2+1)(n+1)$ Now we are given this number is square. We note that the given equation is also equal to $\\frac{n^2+1}{n+1}(n+1)^2$, thus since the result is also square, $\\frac{n^2+1}{n+1}$ is square. Now modulus 4 is fun for squares: We know that every square is 0 or 1 mod 4. We can perform casework upon that fraction to find when it is in the set {0,1} mod 4.\r\n\r\nn==0 mod 4: we have the result equal to 1 mod 4, so a possibility\r\nn==1 mod 4: we have 3/2, which is not a perfect square mod 4.\r\nn==2 mod 4: we have 5/3, which isn't a perfect square mod 4.\r\nn==3 mod 4: we have 10/4, which isn' a perfect square mod 4. \r\n\r\nThus n==0 mod 4. Now let n=4a. We have now $\\frac{16a^2+1}{4a+1}$ square. And now we can evaluate mod 16, which results in $\\frac{1}{4a+1}$. Now we know that the denominator is integral, thus it must be 1 for the fraction to be a square integer thus a=0. Therefore the only possibility is n=0.\r\n\r\nActually I have a minor little hole, but you can use infinite descent to plug it.",
  "Solution_2": "N^3+N^2+N^1+1= (N^4-1)/(n-1) for n>1. \r\nDoes that help?",
  "Solution_3": "N=0,1, and 7 works... but I'm not sure what you did wrong Sodafid farms. Can you explain to me why you put it in mod 16? And why that means that n has to equal 0? I understand the mod 4 part.",
  "Solution_4": "that gives me a formula--so do I simply start plugging a number in for (N).  Is there no true pattern?  And why is 1 not working--When I plug 1 into my problem 1^3+1^2+1^1+1 that =4 which is a square.  But it's not working in your formula.  Thanks!!",
  "Solution_5": "solafidefarms - you are examining the equation in modular arithmetic, so you cannot conclude that $n = 0$ because you have the top and bottom mod 16 (for example, the top of the fraction can be 17).",
  "Solution_6": "[quote=\"georgia03\"]And why is 1 not working--When I plug 1 into my problem 1^3+1^2+1^1+1 that =4 which is a square.  But it's not working in your formula.  Thanks!![/quote]\r\n$n^3+n^2+n+1=\\frac{n^4-1}{n-1}$ is not true for $n=1$ because the fraction is undefined if $n=1$."
}
{
  "Tag": [
    "function",
    "calculus",
    "integration",
    "complex analysis",
    "complex analysis unsolved"
  ],
  "Problem": "I'm looking for help in integrating \r\n\r\nint(dz/z*)\r\n\r\nWhere z* denotes complex conjugate.\r\n\r\nThanks very much.",
  "Solution_1": "Since that's not an analytic function, it has no antiderivative. The integral is strongly path-dependent, and should be handled using real-variable methods. Green's theorem may be useful.",
  "Solution_2": "Thanks..\r\n\r\nCould you explain how I could use real variable methods and Green theorem to go about it??"
}
{
  "Tag": [
    "search",
    "pigeonhole principle",
    "number theory",
    "number theory solved"
  ],
  "Problem": "Looking at the IMOshortlist2002 at number theory I found an interesting problem(well.. all problems are.. but this one caught me :D) because it may be a generalization of a problem given at the first selection test for the OIM 2003 and OBM 2003 12 march 2003 in the Republic Moldavia, if it is true: \r\n\r\n\r\nN4. Does the equation 1/a + 1/b + 1/c + 1/(abc) = m/(a + b + c) have infinitely many solutions in positive integers a, b, c for any positive integer m? \r\n\r\n\r\nthe problem was like so at the test:\r\n\r\nProve that the equation 1/a + 1/b + 1/c + 1/(abc) = 12/(a + b + c) has infinitely many solutions in natural numbers! so it is actually m=12.\r\n\r\nwhat do you think?  cheers! :D",
  "Solution_1": "I think I managed to slve the problem at the test (for m=12). I just observed some solutions and then I guessed how to produce more:\r\n\r\nWe can see that (1,1,1); (1,1,2); (1,2,3) are solutions. Seeing these I thought we can find a sequence of numbers a1,a2,...,an,... such that any 3 consecutive terms of the sequence form a sol fo our eqn. We have a1=a2=a3=1, a4=2, a5=3. I found the next sol: (2,3,7). Then I observed that an is alternatively (a(n-1)+a(n+1))/2 or (a(n-1)+a(n+1))/3. You can just check and you'll see it works :D . We can also observe that an*a(n+1)+1=a(n-1)*a(n+2). I just thought this might help. Here are some terms:1,1,1,2,3,7,11,26,41,...\r\n\r\nAs for problem in the shortlist, in that form, it's easy:the answer is NO. Think about what happens if you put m=1. The left part is obviously > than the right part. Actually we can prove without any difficulty that any  natural m for which we have sol must be at least 7. I think it should ask for which m's the eqn has sols, and for which it doesn't. What do you think?",
  "Solution_2": "hey\r\nnice solution grobber!!! i wud like to some of the solutions that \"grobber\" has already given.we search for particular solutions in artihmetic progression,we have \r\na=x-d,b=x,c=x+d.then after some calculations this reduces to pell's equations x^2-3d^2=1 which ,as is well known,as infinitely many solutions.\r\ncheers!(this is for the case m=12)",
  "Solution_3": "How about the general case?\r\n\r\nDoesn't this way work for other m's? As long as they are large enough, of course :D",
  "Solution_4": "Hey, sorry for asking such a stupid question :D I misunderstood the question from the shortlist. I thought that it was asking if there were an infinity of solutions for ANY m. Boy, that was stupid of me!",
  "Solution_5": "How exactly is the question worded? If it's worded this way the solution is easily no since there's no solution for m=1,2,3",
  "Solution_6": "No solution when n<=9, very possible to have infinite solution when n=10 or n>=12 (involoves some generalisation of Pell's Equation that I don't know the results), unsure about n=11",
  "Solution_7": "There are also infinitely many solutions for m=12 if you set x=a, y=b, z=a+b.  Also we can find infinitely many m for which this gives infinitely many solutions.  For if m = 2k this gives\r\n\r\n1/a + 1/b + 1/(a+b) + 1/(ab(a+b)) =k/(a+b) or\r\n\r\n(a+b)^2 + ab + 1 = kab, or\r\n\r\na^2  + b^2 - (k-3)ab = 1, so we get a Pellian equation of the form a^2+b^2-rab=1.  But  1 = a^2+b^2-rab = (a+b*(r+sqrt(r^2-4))/2)*(a+b*(r-sqrt(r^2-4)/2) so it suffices for (a+b*(r+sqrt(r^2-4))/2) to be a unit in the ring of integers in [bold]Q[/bold](sqrt(r^2-4)).  Let r^2-4 = a^2*d with d squarefree.  Then as long as d>1 it is a theorem than there are infinitely many units in the ring of integers in [bold]Q[/bold](sqrt(d)).  So if d=r^2-4, we're home free since any such unit can be written in the desired form (a+b*(r+sqrt(r^2-4))/2).   If not, we're still done because infinitely many of the units are congruent to 1 mod r+sqrt(r^2-4)/2 (pigeonhole principle) and those ones will give us solutions for a, b.\r\n\r\n--Alison"
}
{
  "Tag": [
    "AMC",
    "AIME",
    "ARML",
    "HMMT",
    "Harvard",
    "college",
    "MIT"
  ],
  "Problem": "which problems are harder, AIME or ARML?\r\nwould it be good practice for AIME if i did old ARML problems, or are they a step-down in difficulty?\r\n\r\nthanks! :)",
  "Solution_1": "It probably depends on which round of ARML. I've never done ARML, just practiced from tests posted online. I haven't worked all of the kinds of rounds, but I personally think the Individual round is slightly easier than AIME, and the Relay round is a bit easier than that.",
  "Solution_2": "Just a suggestion, but it would be good practice for AIME if you did past AIME(!) [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]problems[/url]. Plus many of the problems are very nice. (I don't know much about ARML.)",
  "Solution_3": "[quote=\"carpo\"]Just a suggestion, but it would be good practice for AIME if you did past AIME(!) [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]problems[/url]. Plus many of the problems are very nice. (I don't know much about ARML.)[/quote]\r\n\r\nyeah, i know, but what if i finish all past AIME problems? i mean, im going to have to do a lot of them this year..and what happens when i run out of old aime questions? i just wanted to know of what contests held a similar range of difficulty as the AIME",
  "Solution_4": "Try HMMT (Harvard-MIT Math Tournament) problems",
  "Solution_5": "ARML Team Round problems are probably good practice for the AIME.  Note, though, that ARML is one of the few competitions that makes you pay for its tests (with the exception of the previous year's).  I second the recommendation of HMMT.",
  "Solution_6": "[quote=\"mihail911\"][quote=\"carpo\"]Just a suggestion, but it would be good practice for AIME if you did past AIME(!) [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]problems[/url]. Plus many of the problems are very nice. (I don't know much about ARML.)[/quote]\n\nyeah, i know, but what if i finish all past AIME problems? i mean, im going to have to do a lot of them this year..and what happens when i run out of old aime questions? i just wanted to know of what contests held a similar range of difficulty as the AIME[/quote]\r\nIf you actually finish all past AIME problems, you should be moving onto past Olympiads; this is what I did last year, except that I never even got around to the AIMEs from the mid to late 1990s.",
  "Solution_7": "ARML problems tend to rely more on tricks.  If you have solutions, learning those tricks can be useful.",
  "Solution_8": "Yea, I tend to think of ARML problems as: \"one step then don't make a computation error\" problems.",
  "Solution_9": "The other ARML problems (back to 2004) are up on the website, they just aren't linked to anywhere. Try taking the links and replacing the current year with a previous one.",
  "Solution_10": "[quote=\"mihail911\"]yeah, i know, but what if i finish all past AIME problems? i mean, im going to have to do a lot of them this year..and what happens when i run out of old aime questions? i just wanted to know of what contests held a similar range of difficulty as the AIME[/quote]\r\n\r\nI echo MellowMelon. If you manage to get through all past AIME problems, then either you have gone through them too fast, in which case you should go over the difficult problems again; or you will be extremely well-prepared for the contest. (By the way, it's a good idea anyway, whenever you find a problem you can't solve and read the solution, to go over it a few days later, since otherwise you may forget how to solve it.) If you diligently go through only half of the problems then you are probably ready to move on to olympiad problems. I speak from experience: I printed of all the AIMEs and planned to do them, but only ever got from 1983 to about 1994."
}
{
  "Tag": [
    "limit",
    "algebra proposed",
    "algebra"
  ],
  "Problem": "$ p_{\\minus{}1}\\equal{} 1 ,q_{\\minus{}1}\\equal{} 0$\r\n$ p_{0}\\equal{} q_{0}\\equal{} 1$\r\n$ p_{n}\\equal{} 2p_{n\\minus{}1}\\plus{}(2n\\minus{}1)^{2}p_{n\\minus{}2}$\r\n$ q_{n}\\equal{} 2q_{n\\minus{}1}\\plus{}(2n\\minus{}1)^{2}q_{n\\minus{}2}$\r\nFind $ lim\\frac{p_{n}}{q_{n}}$",
  "Solution_1": "[quote=\"TTsphn\"]$ p_{\\minus{}1}\\equal{} q_{\\minus{}1}\\equal{} 1$\n$ p_{0}\\equal{} q_{0}\\equal{} 1$\n$ p_{n}\\equal{} 2p_{n\\minus{}1}\\plus{}(2n\\minus{}1)^{2}p_{n\\minus{}2}$\n$ q_{n}\\equal{} 2q_{n\\minus{}1}\\plus{}(2n\\minus{}1)^{2}q_{n\\minus{}2}$\nFind $ lim\\frac{p_{n}}{q_{n}}$[/quote]\r\nObviosly $ p_{n}\\equal{}q_{n}$, therefore $ lim\\equal{}1$.",
  "Solution_2": "I edited it.",
  "Solution_3": "Obviosly $ p_{n}\\equal{} (2n\\plus{}1)!!,n\\ge 0$. Let $ q_{n}\\equal{} x_{n}p_{n}$, then\r\n$ x_{0}\\equal{} 1;x_{1}\\equal{}\\frac{2}{3}, x_{n}\\equal{} x_{n\\minus{}2}\\plus{}\\frac{2(x_{n\\minus{}1}\\minus{}x_{n\\minus{}2})}{2n\\plus{}1}$.\r\nLet $ y_{n}\\equal{} x_{n}\\minus{}x_{n\\minus{}1}$, then $ y_{1}\\equal{}\\minus{}\\frac{1}{3},\\ y_{n}\\plus{}y_{n\\minus{}1}\\equal{}\\frac{2}{2n\\plus{}1}y_{n\\minus{}1}$ or\r\n$ y_{n}\\equal{}\\minus{}\\frac{2n\\minus{}1}{2n\\plus{}1}\\equal{} ... \\equal{}\\frac{3}{2n\\plus{}1}(\\minus{}1){n\\minus{}1}y_{1}\\equal{}\\frac{(\\minus{}1)^{n}}{2n\\plus{}1}$, therefore $ x_{n}\\equal{} 1\\plus{}\\sum_{k \\equal{} 1}^{n}\\frac{(\\minus{}1)^{k}}{2k\\plus{}1}$. Let $ a \\equal{}\\sum_{k \\equal{} 0}^{\\infty}\\frac{(\\minus{}1)^{k}}{2k\\plus{}1}\\equal{} arctg(1)$, then $ \\lim_{n\\to\\infty}\\frac{p_{n}}{q_{n}}\\equal{}\\frac{1}{a}\\equal{}\\frac{4}{\\pi}$.",
  "Solution_4": "I think, that these problem is best your problem.",
  "Solution_5": "No I have some other problems and I will post later\r\nHave fun."
}
{
  "Tag": [
    "topology",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Let $ M$ be an open set in $ \\mathbb{R}$ that is unbounded above. Does there exist a point $ x \\in M$ such that $ \\{nx\\}_{1}^{\\infty}$ intersects $ M$ in an uncountable set of points? (I can prove it for the infinite, but countable set of points.)",
  "Solution_1": "For any $ x$, the set of all $ nx$ is countable. There's no chance."
}
{
  "Tag": [
    "function",
    "advanced fields",
    "advanced fields unsolved"
  ],
  "Problem": "Let $X$ and $Y$ be Banach spaces and $\\Omega\\subset X$ be open and $\\Phi\\in C^{1}(\\overline\\Omega,Y)$. The space $C^{1}(\\overline\\Omega,Y)$ is endowed with the norm $\\|\\Phi\\|: =\\sup_{x\\in\\Omega}\\|\\Phi(x)\\|+\\sup_{x\\in\\Omega}\\|\\Phi'(x)\\|_{B(X,Y)}$ with $B(X,Y)$ being the space of all bounded linear operators from $X$ to $Y$.\r\n\r\nNow let $y\\in Y$ be a regular value of $\\Phi$ (i.e. $\\Phi'(x)\\in B(X,Y)$ is isomorphism for each $x\\in\\Phi^{-1}(y)$). Suppose that the equation\r\n\\[\\Phi(x)=y \\]\r\nhas $n$ solutions in $\\Omega$.\r\n\r\nProve that there exists $\\epsilon>0$ such that the equation\r\n\\[\\Phi_\\epsilon(x)=y \\]\r\nalso has $n$ solutions if $\\|\\Phi_\\epsilon-\\Phi\\|\\le\\epsilon$.\r\n\r\n[b]Comment:[/b] Do not use degree theory.",
  "Solution_1": "The Inverse Function Theorem is your friend.",
  "Solution_2": "The Inverse Function Theorem is local result. Anyway I don't know how to solve the problem using that theorem. I solved it using Banach contraction principle instead. Can you solve the problem using the Inverse Function Theorem?",
  "Solution_3": "The problem is a local problem.  Remember that the IFT is a corollary of the contraction mapping principle, so the IFT solution is probably essentially the same as yours.\r\n\r\nThe basic idea is to apply the IFT to a new function $\\Psi: X\\times\\mathbb{R}\\to Y\\times\\mathbb{R}$ given by $\\Psi(x,t)=(\\Phi(x)+{t\\over\\epsilon}(\\Phi_\\epsilon(x)-\\Phi(x)), t)$."
}
{
  "Tag": [
    "algebra",
    "polynomial",
    "Vieta",
    "complex numbers"
  ],
  "Problem": "(****)\r\n\r\n42. Solve the system where $ a,b,c$ are distinct complex numbers:\r\n\\[ a^3 \\plus{} 7b \\plus{} 7c \\plus{} 6 \\equal{} 0\r\n\\]\r\n\r\n\\[ b^3 \\plus{} 7c \\plus{} 7a \\plus{} 6 \\equal{} 0\r\n\\]\r\n\r\n\\[ c^3 \\plus{} 7a \\plus{} 7b \\plus{} 6 \\equal{} 0\r\n\\]",
  "Solution_1": "Subtract the second equation from the first to get $ (a\\minus{}b)(a^2\\plus{}ab\\plus{}b^2\\minus{}7)\\equal{}0$, since $ a\\not \\equal{} b$, $ a^2\\plus{}ab\\plus{}b^2\\equal{}7$. Similarly, $ b^2\\plus{}bc\\plus{}c^2\\equal{}7$. Subtracting, $ (a\\minus{}c)(a\\plus{}b\\plus{}c)\\equal{}0$, since $ a\\not \\equal{} c$, $ a\\plus{}b\\plus{}c\\equal{}0$. The original equation become $ a^3\\minus{}7a\\plus{}6\\equal{}(a\\minus{}1)(a\\minus{}2)(a\\plus{}3)\\equal{}0$, so $ (a,b,c)$ can be any permutation of $ (1,2,\\minus{}3)$",
  "Solution_2": "A question from a lower-level math guy...\r\nHow do you recognize what will and won't factor? Do you just know it, see certain patterns, or use trial and error?",
  "Solution_3": "That is a well-known technique for solving this type of problem. If you notice that you switch a and b in the first equation, then you get the second equation. If you subtract them, you have a polynomial in a that has a root of b, so (a-b) is a factor.\r\n\r\nI was trying to illustrate another technique for this problem, I forgot to check this technique.",
  "Solution_4": "The nicest solution method for a symmetric system, in my experience, is a symmetric manipulation of the equations.  In this case, we have a symmetric manipulation (taking the three pairwise differences) that gives us three two-variable equations instead of three three-variable equations.",
  "Solution_5": "Setting the LHS's of the equations equal in pairs, we find that $ a^{3}\\minus{}7a\\equal{}b^{3}\\minus{}7b\\equal{}c^{3}\\minus{}7c$. Since they are distinct, they must be the roots of $ x^{3}\\minus{}7x\\plus{}C\\equal{}0$, for some constant $ C$. Thus, by Vieta, $ a\\plus{}b\\plus{}c\\equal{}0$, and $ \\minus{}7a\\equal{}7b\\plus{}7c$. Substituting this into the first equation gives $ a^{3}\\minus{}7a\\plus{}6\\equal{}0$, etc."
}
{
  "Tag": [
    "abstract algebra",
    "invariant",
    "inequalities",
    "set theory",
    "superior algebra",
    "superior algebra unsolved"
  ],
  "Problem": "Let F be a free module of infinite rank $ \\alpha$ over a ring R that has invariant dimension property. Prove: for each cardinal $ \\beta$ such that $ 0 \\leq \\beta \\leq \\alpha$, F has infinitely many proper free submodules of rank $ \\beta$.\r\n\r\nAttempted Solution:\r\n\r\nI'm not too good with set theory, so there is probably something wrong with this.\r\n\r\nFirst, I think that inequality $ 0 \\leq \\beta$ should be strict. There is only one empty basis...\r\n\r\nLet F be free on the set X. So X is a basis for F over R and $ |X| \\equal{} \\alpha$.\r\n\r\nIf $ \\beta$ is finite, then this is pretty clear. There are infinitely many ways to take $ \\beta$ elements out of $ X$. Given a subset Y of X of size $ \\beta$, consider the submodule it generates. This submodule is free on Y and has invariant dimension property and has dimension $ |Y| \\equal{} \\beta$ and it must be proper since otherwise X is not linearly independent.\r\n\r\nNow for the infinite case. Given $ \\beta \\geq \\aleph_0$, we find a proper subset $ Y$ of $ X$ with cardinality $ \\beta$. $ Y$ will generate a submodule with invariant dimension property that has have dimension $ \\beta$, that is free on Y, and is proper like in the finite case. Now we can throw out infinitely many elements of $ Y$ to get infinitely many proper subsets of $ Y$ each of which generate submodules of dimension $ \\beta$.",
  "Solution_1": "you have to keep attention that your submodules are different. I don't think that your easy approach works. translate the problem into matrices, then it's pretty clear.",
  "Solution_2": "[quote=\"-oo-\"]you have to keep attention that your submodules are different. I don't think that your easy approach works.[/quote]For $ A\\subset B\\subset X$ it holds that the modules generated by A and B, resp., are equal iff $ A\\equal{}B$, since $ X$ is linearly independent. Assuming that you inductively construct a family of strictly decreasing subsets $ Y_{n\\plus{}1}\\subsetneqq Y_n\\subsetneqq X$ with $ |Y_n|\\equal{}\\beta$, then I cannot see where the approach fails to work, since the induced modules are strictly decreasing."
}
{
  "Tag": [
    "inequalities unsolved",
    "inequalities"
  ],
  "Problem": "let a+b+c+d=0 and a ^3 +b ^3 +c ^3 +d ^3 >0\r\nprove that a^5 + b^5+c^5+d^5>0\r\ni have found the answer for it but it's too long\r\ncan you give me a short solution.",
  "Solution_1": "Let S_1 = a+b+c+d, S_2 = ab+bc+cd+da+ac+ad, S_3 = abc+acd+abd+bcd and S_4 = abcd.\r\nLet P(x) = (x-a)(x-b)(x-c)(x-d)\r\n= x <sup>4</sup> - S_1*x <sup>3</sup>  + S_2*x <sup>2</sup>  - S_3*x + S_4.\r\nSince a,b,c,d are the roots of P(x) and from S_1 = 0, we deduce that :\r\na <sup>4</sup> = -S_2*a <sup>2</sup>  + S_3*a - S_4, \r\nthus\r\na <sup>5</sup>  = -S_2*a <sup>3</sup> + S_3*a <sup>2</sup>  - S_4,\r\nand we have similar relations with b,c,d. Summing, it leads to :\r\n \\sum a <sup>5</sup>  = -S_2* \\sum a <sup>3</sup>  + S_3* \\sum a <sup>2</sup>  , (1).\r\n\r\nIn another hand :\r\n0 = (a+b+c+d) <sup>3</sup> =  \\sum a <sup>3</sup> + 3 \\sum a <sup>2</sup> (b+c+d) + 6S_3\r\n= \\sum  a <sup>3</sup>  + 3 \\sum a <sup>3</sup> (-a) + 6S_3\r\n= - 2 \\sum a <sup>3</sup> + 6S_3\r\nthus 3*S_3 =  \\sum a <sup>3</sup> .\r\nMoreover :\r\n0 = (a+b+c+d) <sup>2</sup>  then 2*S_2 = - \\sum a <sup>2</sup> .\r\n\r\nFrom (1), we deduce that :\r\n \\sum  a <sup>5</sup> /5 =  (\\sum a <sup>3</sup> /3)*( \\sum a <sup>2</sup>  /2), from which the result follows easily.\r\n\r\nPierre.",
  "Solution_2": "40 th British  mo , 2004 , round  2.  No  i dont  have  an  easier solution  Pierre. :)"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "homothety",
    "ratio",
    "inequalities",
    "limit",
    "Putnam"
  ],
  "Problem": "Let $I_k =[a_k,b_k]$ such that $0<a_1<b_1<a_2<...<a_k<b_k<...$ and $a_k\\to\\infty$.\r\nProve that there exists $x>0$ such that the sequence $\\{nx\\}$ intersects infinitely many $I_k$.\r\n\r\n[i]Edited by Myth[/i]",
  "Solution_1": "Let $I'_{k_1}=I_1$. Consider the image of $I'_{k_1}$ through a homothety of ratio $n$ (and center $O$, the origin of the real line). For some $n$, there is a $k$ s.t. $I_k\\cap nI'_{k_1}\\ne \\emptyset$, and let $I'_{k_2}=\\frac 1n\\cdot I_k\\cap nI'_{k_1}$. What we did for $I'_{k_1}$ we do now for $I'_{k_2}$ (we can find an $n$ larger than the previous one, of course), and we find an $I'_{k_3}$, and so on. We can thus find a decreasing sequence of compact intervals ($I'_{k_1}\\supseteq I'_{k_2}\\supseteq I'_{k_3}\\supseteq\\ldots$), with non-void intersection (Cantor's Theorem). Take $x$ to be a point belonging to this intersection.\r\n\r\nI bet you think it's not clear enough. That's my opinion as well :).",
  "Solution_2": "We can WLOG suppose $b_k-a_k\\to 0$\r\n\r\nLemma. For each $[u,v]\\subset[0,1]$ there exists $n$ and $k$ s.t. $[a_k,b_k]\\in [nu,nv]$.\r\n$\\square$ Proof. Indeed, for all $n>N$ the following inequality holds: $nu+v<nv$. Let's find $I_k$ s.t. $b_k-a_k<v-u$ and $nu<a_k<(n+1)u$, $n>N$. Then $b_k<a_k+(v-u)<(n+1)u+(v-u)<nu+v<nv$.$\\blacksquare$\r\n\r\nLet $[u_0,v_0]=[0,1]$. Due to lemma we can find $[a_{k_1},b_{k_1}]\\subset [n_0u_0,n_0v_0]$. Define $[u_1,v_1]=[a_{k_0}/n_0,b_{k_0}/n_0]$. And so on. We have obtained infinite sequence of embedded segments. Therefore we conclude there is $x$, which lies in all this segments. It means $\\{nx\\}$ intersects infinitely many $I_k$.",
  "Solution_3": "There is \"somehow\" related problem :) \r\n\r\n Is is possible that :\r\n- $f$ is continuous\r\n- $\\lim_{n\\rightarrow \\infty}{f(nx)} = \\infty$  for all $x$\r\n- We don't have $\\lim_{x \\rightarrow \\infty}{f(x)} = \\infty$ for all $x$ in $\\mathbb{R}$\r\n\r\n This one from Putnam",
  "Solution_4": "And I think it's been discussed before, receiving similar solutions. I don't know if it was precisely this one though. I remember something very similar, but with limits $0$ instead of $\\infty$.",
  "Solution_5": "Actually, it is direct consequence of initial problem.\r\nJust suppose contrary, then we will able to find sequence of segments $I_k$, s.t. $f$ is bounded on $I_k$ with the same constant $C$.\r\nThat's all.\r\nAnd yes, Levon posted problem with 0 instead of $+\\infty$ some time ago.",
  "Solution_6": "[quote=\"grobber\"]And I think it's been discussed before, receiving similar solutions. I don't know if it was precisely this one though. I remember something very similar, but with limits $0$ instead of $\\infty$.[/quote]\r\nI'm just saying that they are two \" somehow related\" problems!!  :?  \r\n No intention to ask anyone to solve  :D",
  "Solution_7": "And I remember giving basicly the same solution as the one Myth briefly presented here :). The problem has been posted more than once, I think. I seem to remember alekk posting it as well."
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Let $ G$ be a graph with $ n$ vertices, and let $ d_1$, $ d_2$, ..., $ d_n$ be the degrees of these $ n$ vertices of $ G$. Let $ e$ be the number of edges of the graph $ G$. Prove that the number of triangles in $ G$ and $ \\overline G$ (that is, the sum of the number of triangles in $ G$ with the number of triangles in $ \\overline G$) is:\r\n$ {n \\choose 3} \\minus{} \\left(n \\minus{} 2\\right)e \\plus{} \\sum_{i \\equal{} 1}^n {d_i\\choose 2}$.",
  "Solution_1": "We calculate the number of triples $(a,b,c)$ of vertices of graph $G$, which do not form a triangle neither in $G$, nor in $\\overline{G}$. We must prove that the number of such triples equals $(n-2)e-\\sum{d_i\\choose 2}$. Note that for each such triple, there exists two vertices in it, which are joined by the edge with exactly 1 other vertice in this triple (say, if $a$ is joined with $b$ and $c$, and $b$ and $c$ are not joined, such vertices are $b$ and $c$). Call such vertices nice vertices of the triple. On the other hand, every vertex $a$ of degree $d$ is nice in exactly $d(n-1-d)$ triples (we may add to the triple arbitrary vertex $b$, joined with $a$, and arbitrary vertex $c$, not joined with $a$). So, our number equals $\\frac12(\\sum d_i(n-1-d_i))=(n-2)e-\\sum{d_i\\choose 2}$ ."
}
{
  "Tag": [
    "pigeonhole principle"
  ],
  "Problem": "Prove that in any convex 2n-gon  there is a diagonal not parallel to any side.",
  "Solution_1": "[hide]There are $ n$ pairs of parallel sides. Label the vertices, in clockwise (or counterclockwise) order, $ A_{1},A_{2},\\ldots,A_{2n}$. Consider the side $ A_{1}A_{2}$, and its corresponding parallel side, $ A_{n+2}A_{n+1}$. There are $ n-2$ diagonals parallel to these, namely $ A_{2n}A_{3},A_{2n-1}A_{4},\\ldots,A_{n+3}A_{n}$. Hence, there are $ n(n-2)$ diagonals parallel to a side. There are $ \\frac{2n(2n-3)}{2}=n(2n-3)$ diagonals in total, and $ n(n-2)<n(2n-3)$ for all $ n>1$, so by the Pigeonhole Principle there must exist some diagonal not parallel to any side.[/hide]",
  "Solution_2": "You seem to have assumed the polygon is equiangular.",
  "Solution_3": "[quote=\"t0rajir0u\"]You seem to have assumed the polygon is equiangular.[/quote]\r\n\r\noooh, whoops, I can't read. But wouldn't that be the extreme case?",
  "Solution_4": "[quote=\"13375P34K43V312\"][hide]There are $ n$ pairs of parallel sides. Label the vertices, in clockwise (or counterclockwise) order, $ A_{1},A_{2},\\ldots,A_{2n}$. Consider the side $ A_{1}A_{2}$, and its corresponding parallel side, $ A_{n+2}A_{n+1}$. There are $ n-2$ diagonals parallel to these, namely $ A_{2n}A_{3},A_{2n-1}A_{4},\\ldots,A_{n+3}A_{n}$. Hence, there are $ n(n-2)$ diagonals parallel to a side. There are $ \\frac{2n(2n-3)}{2}=n(2n-3)$ diagonals in total, and $ n(n-2)<n(2n-3)$ for all $ n>1$, so by the Pigeonhole Principle there must exist some diagonal not parallel to any side.[/hide][/quote]\r\n\r\nyeah i think i came to the same result. the equilateral case should be the extreme case with the largest number of pairs of parallel diagonals."
}
{
  "Tag": [],
  "Problem": "My brother is 4 x as old as iam right now.  in 6 years he will be 2x as old as i am. how old am i?",
  "Solution_1": "Set ur age as X\r\nUr brother is 4X\r\n4X+6=2(X+6)\r\nX=3",
  "Solution_2": "[b] Just Subtitute  and you will get your age as 3[/b]",
  "Solution_3": "Let your age be x.\r\nYour brother's age=4x.\r\n\r\n4x+6=2(x+6)\r\n4x+6=2x+12\r\n4x-2x=12-6\r\n2x=6\r\nx=6/2=3\r\n\r\nx=3",
  "Solution_4": "Guys, please don't post the same solution over and over again if you use the exact same method.\r\nAnd don't revive a thread that's a month old.",
  "Solution_5": "1st set teh variables\r\n                  age now      | age in 6 years                   \r\nYou           |     x           |x+6\r\nyour brother|     4x         |2x\r\n2nd make teh equation\r\n4x+6=2(x+6)\r\nsolve\r\n2x+3=x+6\r\nx=3\r\nso you r 3 years old",
  "Solution_6": "Hi lalagundam! I see you're new to AoPS.\r\n\r\nAs future reference, try not to post solutions to problems that already have solutions unless you have an alternate method OR you find problems with the previous solutions. Also, \"reviving\", as AoPSers commonly call it, is frowned upon. \r\n\r\nWelcome to AoPS, and....have fun! :)"
}
{
  "Tag": [
    "calculus",
    "derivative",
    "analytic geometry",
    "graphing lines",
    "slope",
    "function",
    "algebra"
  ],
  "Problem": "Hi, How do you do the following question?\r\n\r\nFind the equation of all lines that are tangent to both y=x^2+2 and y=-2-x^2. Indicate the points where the tangent line intersect the curve. \r\n\r\nIs y^2=4x^2 the solution? \r\n\r\nThanks",
  "Solution_1": "Let $f(x) = x^2 + 2$ and $g(x) = -x^2 - 2$.  At any point $(c,f(c))$, the tangent to $f$ is given by:\r\n\r\n\\[ y = f'(c)(x-c) + f(c) = 2c(x-c) + c^2 + 2 = (2c)x + (2-c^2) \\]\r\n\r\nLikewise, at any point $(d,g(d))$, the tangent to $g$ is given by:\r\n\r\n\\[ y = g'(d)(x-d) + g(d) = -2d(x-d) + -d^2 + -2 = (-2d)x + (d^2-2) \\]\r\n\r\nWhat the question is asking for is to find the values of $c$ and $d$ for which the tangent lines described above are the same.  So we have a system of equations where both the slopes and the y-intercepts must be the same:\r\n\r\n$2c = -2d$\r\n\r\n$2-c^2 = d^2 - 2$\r\n\r\nSolve this to get $c = \\pm\\sqrt{2}$\r\n\r\nThus, using tangent equation $y = f'(c)(x-c) +f(c)$, the two desired tangent lines are:\r\n\r\n\\[ y = 2\\sqrt{2}(x-\\sqrt{2}) + 4 \\]\r\n\\[ y = -2\\sqrt{2}(x+\\sqrt{2}) + 4 \\]\r\n\r\nThink about the graph of the two functions, and it makes sense that there are only two such tangents.  Hope this helps.\r\n\r\nmadpenguin"
}
{
  "Tag": [],
  "Problem": "The sum of a certain number of consecutive positive integers is 1000. Find these integers.\r\n\r\nThis problem looks easy.. but it's a little bit tricky.",
  "Solution_1": "[hide]There are actually 4 different sets.  I'll give one of them:  198+199+200+201+202... The rest shouldn't be [i]too[/i] hard to find :P [/hide]",
  "Solution_2": "It's not hard...just a little bit of number theory.\r\n\r\n[hide]\n$n(2a+n-1)=2000$ and $2a+n-1>n$ :)\n[/hide]",
  "Solution_3": "Any other more elegant solution? :P",
  "Solution_4": "[hide=\"Somewhat Elegant Solution\"]Case 1: Odd number of integers.\n\nLet $n$ be the average of the integers.  The integers are\n\n$n-k,n-k+1,n-k+2\\ldots n-1,n,n+1\\ldots n+k-2,n+k-1,n+k$\n\nThere are $2k+1$ terms.  The sum is $n(2k+1)$.  Therefore, $n(2k+1)=1000$, and $2k+1$ is an odd factor of $1000$.  We have\n\n$(k,n)=(0,1000),(2,200),(12,40),(62,8)$\n\nThe sets of numbers for this case are\n\n$1000$\n$198$ to $202$\n$28$ to $52$\n\nThe fourth ordered pair produces some negative integers, so it doesn't work.\n\nCase 2: Even number of integers.\n\nThe integers are\n\n$n-k-\\frac{1}{2},n-k-\\frac{1}{2}+1\\ldots n-\\frac{1}{2},n+\\frac{1}{2}\\ldots n+k+\\frac{1}{2}-1,n+k+\\frac{1}{2}$\n\nSince these are integers, $n=\\frac{2m+1}{2}$ for some integer $m$.  The average of the integers is $n$, and there are $2k+2$ integers.  The sum is $n(2k+2)$.  Therefore,\n\n$\\frac{2m+1}{2}(2k+2)=1000$\n$(2m+1)(k+1)=1000$\n\n$2m+1$ is an odd factor of $1000$.\n\n$(m,k)=(0,999),(2,199),(12,39),(62,7)$\n$(n,k)=(\\frac{1}{2},999),(\\frac{5}{2},199),(\\frac{25}{2},39),(\\frac{125}{2},7)$\n\nOnly the fourth ordered pair produces only positive integers.  The set of numbers for this case are\n\n$55$ to $70$.\n\nThere are four solutions.[/hide]",
  "Solution_5": "[hide]\nlet $1000=n+(n+1)+...+(n+m)$\nwe then get\n$1000=(m+1)n+\\frac{m(m+1)}{2}$\nso that goes to:\n$1000=(m+1)(n+\\frac{m}{2})$\nnow we factor 1000, $1000=2^3*5^3$\nso now we look for $(m+1)$ to be equal to one of the divisors of 1000, however, by looking at the $\\frac{m}{2}$ in the equation we see that m+1 is some multiple of 5; 5, 25, 125.  Looking at 25 we find $m=24$.  Then by plugging in $n=28$, so, $1000=28+29+...+52$\nto find other values you change the m; 4, 24 and 124 are possible values so there are 3 possible nontrivial series.  To find them just pug into the main equation\n[/hide]",
  "Solution_6": "Nice solution, demji!\r\n\r\n[quote=\"demji\"]\n$1000=(m+1)(n+\\frac{m}{2})$\n[/quote]\r\n\r\nAt this point, I would multiply by $2$ to get $(m+1)(2n+m)=2000$, so that you can forget about the possibility of fractions.",
  "Solution_7": "however, by keeping the 2 as a divisor we limit ourselves to powers of 5, i didn't count $5^0=1$ since that is a trivial answer.  Its much faster to keep the 2 since you can automatically eliminate cases where m is odd, meaning that m+1 is an even divisor.  The only odd divisors would be powers of 5, giving u 4 total possibilities, 3 non-trivial."
}
{
  "Tag": [
    "function",
    "probability and stats"
  ],
  "Problem": "Suppose I have a uniformly distributed r.v. on [0,1] and the cumulative distribution function is strictly increasing on [0,1]. What steps would I take to come up with a new r.v. on [a,b] with a known cdf F that is also strictly increasing on [a,b]?\r\n\r\nI was thinking that if the original r.v were X: S -> R on [0,1]\r\n\r\nThe new r.v. is, say Y: S -> R on [a,b], and defined over the same S, so Y(s): F^{-1}(X(s))\r\n\r\nhowever, I need this to map to [a,b]. So, say we generate X(s_i) = r_i. Then I need to map each of these r to some number in [a,b]. Does a linear transformation work? How would I define this transformation?",
  "Solution_1": "X is uniformly distributed on [0,1].\r\nF-1 (X) will be a random variable with distribution function F in [a,b]. :)"
}
{
  "Tag": [
    "geometry",
    "AMC"
  ],
  "Problem": "So... How can I apply for it? I would appreciate an answer from an admin. or another student... Thank you in advance",
  "Solution_1": "Here's what the MAA site says:\r\n[quote]What if my school doesn\u2019t offer the AMC tests?\nA. Urge your principal, math teacher, gifted education coordinator or anyone else you can think of to help you out and register for the contest.  If your school doesn\u2019t offer the AMC tests to you, then you can offer the tests to it.  You could make your local school an offer it would be silly to refuse by signing up for the AMC test of your choice at your own expense, and then letting the local school know you have an opportunity for other students at the school that is already paid for.  Please make arrangements with your school prior to registering as we must send the contest materials to the school directly.  Many colleges and universities also host the contests, particularly the \u201cB\u201d date of the AMC 10/12.  Check our web pages for a list of participating Institutions of Higher Learning. [/quote]\r\nIMO, If your school doesn't offer it, your best bet is probably to contact other schools in your area and see if they'll let you take it there. If not, try the above.",
  "Solution_2": "I had just read that, but thanks anyways, mate... Have a good New Years"
}
{
  "Tag": [],
  "Problem": "One cookie and two brownies cost the same as two cookies and one brownie. One cookie and one brownie together cost $ \\$$1. What is the cost, in cents, of one cookie?",
  "Solution_1": "If one cookie and two brownies cost the same as two cookies and one brownie, then one cookie costs the same as one brownie (Why? Think logic.). Thus, if two of them cost $ \\$1$, one of them would cost $ \\fbox{50}$ cents.",
  "Solution_2": "Here's the \"why\":\r\n\r\n$ c\\plus{}2b\\equal{}2c\\plus{}b$\r\n$ c\\equal{}b$."
}
{
  "Tag": [
    "number theory unsolved",
    "number theory"
  ],
  "Problem": "Solve the equation with x,y are integer  :\r\n$x(x+1)(x+2)(x+3)(x+4)(x+5)+1=y^2$",
  "Solution_1": "No solutions ?  :(  :(",
  "Solution_2": "The obvious one is $x=0,y=1$",
  "Solution_3": "[quote=\"pontios\"]The obvious one is $x=0,y=1$[/quote]\r\n\r\nwell .... when y = 1 or -1 , x =0,-1,-2,-3,-4,-5\r\n\r\nanother one is y=71 , x=2",
  "Solution_4": "I think we can use the fact that 6 consecutive integer can be divided by 6!=720 hence \r\n\r\ny^2 = 720k + 1    where k is any non-negative integer \r\n\r\nbut then I have no idea how to continue .... :blush:  :blush:"
}
{
  "Tag": [
    "combinatorics unsolved",
    "combinatorics"
  ],
  "Problem": "Let $ S$ be a finite set of natural numbers. Among arbitrary 3 numbers from $ S$ we can choose 2 of them who's sum belongs to $ S$. What is the maximum possible number of elements for $ S$.?",
  "Solution_1": "Check that if we have $ a_1 > a_2 > a_3 > a_4 > 0$ as the four largest elements of $ S$ then we can't have the given condition for both the triples $ \\{a_1, a_2, a_3\\}$ and $ \\{a_1, a_2, a_4\\}$.  Then conclude.\r\n\r\nWhat if we replace \"3\" by \"4?\""
}
{
  "Tag": [
    "probability",
    "probability and stats"
  ],
  "Problem": "what formula do i need to solve the following question?:\r\n\r\n\"The grades for a biology pop quiz are normally distributed with a standard deviation of 12.5% and a mean of 58.6%. Give the probability that a student received a mark of 71% or more.\"\r\n\r\n\r\nthanx",
  "Solution_1": "You first find the Z-Value, which is just the \"(sample percentage - mean percentage)/ (Standard Deviation)\". After you get the Z- Value, use Table A to find the probability. \r\n\r\nNote: If you have no clue what I'm talking about then this isn't the right problem for you.",
  "Solution_2": "isn't table A for z-scores that are negative? i got a z-score of .992",
  "Solution_3": "No it doesnt have to be negative. There is a positive and negative side."
}
{
  "Tag": [],
  "Problem": "Anyone know how to tell the number of ligands that a transition metal ion will accept?  Thanks.",
  "Solution_1": "It's been a while and I may be wrong on some of these-\r\n\r\nFirst case- the ion has a filled D subshell - the ion will have 4 ligands in a tetrahedral shape (the 1 s- and 3 p- orbitals are used)\r\nSecond case- the ion has a D subshell with 8 electrons- the ion will have 4 ligands in a square planar shape- (one of the P orbitals is left empty)\r\nThird case- the ion has a D subshelll that has 6 electrons or fewer such that 2 D orbitals can be opened up and a total of 2 d, 1 s, and 3 p orbitals may be used for the complex to have a octahedral shape.",
  "Solution_2": "Great, thanks.",
  "Solution_3": "my chemistry teacher always used to tell me the number of ligands in a coordinated complex is usually twice the charge. so like Fe(OH)[size=75]4[/size] 2-\r\n\r\nbut there are always exceptions. so you might wanna look that up"
}
{
  "Tag": [
    "ratio",
    "geometry",
    "similar triangles"
  ],
  "Problem": "In the triangle $ ABC$, $ AC \\equal{} 2BC$, $ \\angle C \\equal{} 90^{\\circ}$ and $ D$ is the foot of the altitude from $ C$ onto $ AB$. A circle with diameter $ AD$ intersects the segment $ AC$ at $ E$. \r\nFind $ AE: EC$.",
  "Solution_1": "[hide=\"Solution\"]\n[asy]unitsize(3cm);\n\npair C=(0,0);\npair B=(0,1);\npair A=(2,0);\npair D=waypoint(B--A,0.2);\npair O=midpoint(D--A);\npair F=foot(O,A,C);\npair E=reflect(O,F)*A;\n\nlabel(\"$B$\",B,N);\nlabel(\"$C$\",C,SW);\nlabel(\"$A$\",A,SE);\nlabel(\"$D$\",D,NE);\nlabel(\"$O$\",O,NE);\nlabel(\"$E$\",E,SW);\nlabel(\"$F$\",F,S);\n\ndraw(C--B--A--cycle);\ndraw(C--D);\ndraw(O--E);\ndraw(O--F);\ndraw(Circle(O,arclength(D--O)));[/asy]\nLet $ BC \\equal{} x$ and $ AC \\equal{} 2x$. By similar triangles, $ DA \\equal{} \\frac {4\\sqrt5}{5}x$, so $ OA \\equal{} \\frac {2\\sqrt5}{5}x$ and thus $ FA \\equal{} \\frac {4}{5}x$. By HL we know that $ \\triangle OFA$ and $ \\triangle OFE$ are congruent, so $ FE \\equal{} FA \\equal{} \\frac {4}{5}x$. So $ \\frac {AE}{EC} \\equal{} \\frac {\\frac {8}{5}x}{2x\\minus{}\\frac{8}{5}x} \\equal{} \\boxed{4}$.[/hide]",
  "Solution_2": "[b]worthawholebean[/b] \u2014\r\n[hide=\"Correction\"]$ EC \\neq 2x$, rather $ EC \\equal{} 2x \\minus{} \\frac{8}{5}x \\equal{} \\frac{2}{5}x$. Since they ask you for the ratio $ AE : EC$,  the actual answer would be $ \\frac{8}{5}x : \\frac{2}{5}x \\Rightarrow\\boxed{4: 1}$, or simply $ 4$.[/hide]",
  "Solution_3": "Following relations hold: $AC^2=AD\\cdot AB; BC^2=BD\\cdot AB$; dividing side by side: $\\left(\\frac{AC}{BC}\\right)^2=\\frac{BD}{AD}\\ (\\ 1\\ )$.\n$DE\\bot AC\\iff DE\\parallel BC\\implies \\frac{AE}{CE}=\\frac{AD}{BD}=\\left( \\frac{AC}{BC}\\right)^2=\\frac{4BC^2}{BC^2}=4$.\n\nBest regards,\nsunken rock",
  "Solution_4": ":) [quote=\"sunken rock\"] $DE\\bot AC\\iff DE\\parallel BC\\implies \\frac{EA}{EC}=\\frac{DA}{DB}=\\left( \\frac{CA}{CB}\\right)^2\\implies$ $\\boxed{\\frac {EA}{EC}=k^2}$ , where $CA=k\\cdot CB$ .[/quote]\nYes, it is the shortest proof !"
}
{
  "Tag": [
    "geometry",
    "3D geometry",
    "probability"
  ],
  "Problem": "Three faces of a cube are randomly selected. What is the probability that they have a common vertex? Express your answer as a common fraction.",
  "Solution_1": "[quote=\"ccy\"]Three faces of a cube are randomly selected. What is the probability that they have a common vertex? Express your answer as a common fraction.[/quote]\r\n[hide]this problem is familliar... but I still forgot how... 4/5?[/hide]",
  "Solution_2": "[hide=\"Hint\"]How many faces of a cube meet at a single vertex?[/hide]",
  "Solution_3": "[hide]Ummmm\nwell by choosing a random face\nyou have four other faces that can potentially share the same vertex with another face. $\\frac{4}{5}$. \nThen you have two other faces that share the same vertex with the other two. $\\frac{2}{4}$\n$\\frac{4}{5}\\cdot\\frac{2}{4}=\\boxed{\\frac{2}{5}}$[/hide]",
  "Solution_4": "[hide=\"this is probably wrong\"]$6*5*4$ choices. 6 vertices, each with one possible combination.\n$\\frac{6}{6*5*4}=\\boxed{\\frac{1}{20}}$.\nI somehow know that's wrong.[/hide]",
  "Solution_5": "[quote=\"ccy\"]Three faces of a cube are randomly selected. What is the probability that they have a common vertex? Express your answer as a common fraction.[/quote]\r\n\r\nso what was the answer? post it in hide",
  "Solution_6": "[hide]There are $\\binom63=20$ ways of choosing 3 faces.\n\nThere are only 8 vertices so there are only 8 combinations.\n\n$\\frac{8}{20}=\\frac25$[/hide]",
  "Solution_7": "[hide]\n8 vertices, 8 ways for the faces to come together.\n\n$\\binom 63=20$ ways to choose the faces, so the answer is\n$\\frac{8}{20}=\\frac{2}{5}$\n$\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\boxed{\\frac 25}}}}}}}}}}}}}}}$[/hide]"
}
{
  "Tag": [
    "geometry",
    "geometric transformation",
    "reflection",
    "ratio",
    "trapezoid",
    "perpendicular bisector",
    "geometry unsolved"
  ],
  "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14718[/img]\r\n\r\n\r\n :(",
  "Solution_1": "Let the perpendicular bisector $ m$ of the segment $ AB$ cut the line $ r$ at $ M.$ Let $ C \\in r$ on the same side of $ m$ as $ A$ be arbitrary, so that $ CA < CB,$ and let $ D$ be reflection of $ C$ in $ m.$ Internal and external bisectors of the angle $ \\angle BCA$ cut the line $ AB$ at $ Z, Z'.$ Ratio $ \\frac {PA}{PB} \\equal{} \\frac {CA}{CB}$ is constant for $ P$ on the C-Apollonius circle $ (K)$ of the $ \\triangle ABC$ with diameter $ ZZ',$ centered at the midpoint $ K$ of the segment $ ZZ'.$ Since $ C \\in r,$ $ \\frac {CA}{CB}$ is minimum, when the C-Apollonius circle $ (K)$ is tangent to $ r$ at $ C.$ $ A$ is a reflection (inversion) of $ B$ in $ (K)$ and the other way around $ \\Longrightarrow$ $ KA \\cdot KB \\equal{} KC^2,$ the right $ \\triangle AKC \\sim \\triangle CKB$ are oppositely similar and the angles $ \\angle CAK \\plus{} \\angle CBK \\equal{} 90^\\circ$ are complementary. $ ABCD$ is isosceles trapezoid,\r\n\r\n$ \\angle CAD \\equal{} 180^\\circ \\minus{} (\\angle CAK \\plus{} \\angle BAD) \\equal{} 180^\\circ \\minus{} (\\angle CAK \\plus{} \\angle CBK) \\equal{} 90^\\circ.$\r\n\r\nSince $ \\frac {CA}{CB}$ is minimum, $ \\frac {DA}{DB} \\equal{} \\frac {CB}{CA}$ is maximum. To construct $ C, D,$ draw a circle with center $ M$ and radius $ MA \\equal{} MB,$ intersecting $ r$ at $ C, D.$",
  "Solution_2": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14779[/img]\r\n\r\n :thumbup:"
}
{
  "Tag": [
    "geometry"
  ],
  "Problem": "Given three circles, construct a fourth one which bisects the circumference of the other three, ie the 4th circle cuts each of the other circles in diametrically opposite points.",
  "Solution_1": "Does something like this work... I think that even if the circles were randomly placed, you would first take the opposing endpoints of the diameter of each circle and then find a point equidistant from all 6 points and you should have your circle"
}
{
  "Tag": [
    "calculus",
    "integration",
    "trigonometry",
    "logarithms",
    "derivative",
    "real analysis",
    "real analysis unsolved"
  ],
  "Problem": "Find general expressions for the two following integrals:\r\n\r\n1) $ \\int_0^{\\pi/2} \\frac{(\\tan x)^p \\minus{} 1}{\\ln(\\tan x)}\\,dx$, 0 < p < 1.\r\n\r\n2) $ \\int_0^{\\infty} \\sin(x^n)\\,dx$, n natural, n > 1.",
  "Solution_1": "carcul,  you must be psychic !  :o i asked   (1) just 3 weeks back on  grade 8's midterm quiz!\r\n\r\n(1) here is the solution i gave later. i gave 0 to anyone who did not provide the answer like i wanted.\r\n lots of them did as a matter of fact ...i was   disappointed....  :noo: \r\n\r\n$ \\textbf I\\equal{}\\int_0^{\\dfrac{\\pi}{2}} \\;\\dfrac {\\tan^p\\,(x)\\minus{}1 }{ \\ln\\,(\\tan\\, x)} \\;\\textbf dx\\equal{}\\int_0^p\\int_0^{\\dfrac{\\pi}{2}}\\tan^p\\,(x)\\;\\textbf dx\\;\\textbf dp$\r\n$ \\left[ we\\quad know \\quad \\int_0^{\\dfrac{\\pi}{2}}\\;\\tan^p\\,(x)\\;\\textbf dx\\equal{}\\dfrac{\\pi}{2\\;\\sin\\,\\left[\\frac{\\pi}{2}\\,(p\\plus{}1)\\right]}\\qquad (for\\quad 0<p<1)\\right]$\r\n$ so,\\quad \\textbf I\\equal{}\\int_0^p\\dfrac{\\pi}{2}\\text{cosec}\\,\\left[ \\dfrac{\\pi}{2}(p\\plus{}1)\\right]\\;\\textbf dp$\r\n$ \\equal{}\\int_0^{\\dfrac{\\pi}{2}p} \\sec\\,(z)\\;\\textbf dz\\qquad (where\\quad z\\equal{}\\dfrac{\\pi}{2}p)$\r\n$ \\equal{}\\boxed{\\ln\\,\\left(\\tan\\,\\left[\\dfrac{\\pi}{4}(p\\plus{}1)\\right]\\right)}$\r\n\r\n(2) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158083",
  "Solution_2": "Yes, that's right. For the first one you can also differentiate in order to p.",
  "Solution_3": "Carcul can u please post ur solution using the differentiation please :)",
  "Solution_4": "I think this is what Carcul might have meant.\r\n\r\n$ I(P) \\equal{} \\int_0^{\\pi/2} \\frac{(\\tan x)^p \\minus{} 1}{\\ln(\\tan x)}\\,dx$\r\n\r\n$ I'(P) \\equal{} \\int_0^{\\pi/2} (\\tan x)^p\\,dx$ (By Leibnitz method)\r\nand then continue with misan's steps.",
  "Solution_5": "Yes, that's right. Next you are to show that $ \\int_0^{\\pi/2}(\\tan x)^p\\,dx \\equal{} \\frac{\\pi}{2} \\sec\\left(\\frac{p \\pi}{2}\\right)$, 0 < p < 1.",
  "Solution_6": "2) Let $ I_n=\\int_0^{\\infty} \\sin(x^n)\\,dx$\r\n\r\nSubstitute $ u=x^n$ to get $ I_n=\\frac{1}{n}\\int_0^{\\infty} u^{\\frac{1}{n}-1}\\sin u\\,du$.\r\n\r\nNow put $ J_n = \\frac{1}{n}\\int_0^{\\infty} u^{\\frac{1}{n}-1}e^{-iu}\\,du$ so that $ I_n = -\\mbox{Im}\\left[J_n\\right]$. Put $ w=iu$ in $ J_n$ to get\r\n\r\n$ J_n = -\\frac{1}{n}(-i)^{\\frac{1}{n}}\\int_0^{\\infty} w^{\\frac{1}{n}-1}\\e^{-w}\\,dw$\r\n$ = -\\frac{1}{n}(-i)^{\\frac{1}{n}}\\Gamma\\left(\\frac{1}{n}\\right)$\r\n$ = -\\frac{1}{n}\\left[\\cos \\left(\\frac{\\pi}{2n}\\right)-i\\sin\\left(\\frac{\\pi}{2n}\\right)\\right] \\Gamma\\left(\\frac{1}{n}\\right)$\r\n$ \\Rightarrow I_n= -\\mbox{Im}\\left[J_n\\right] = \\frac{1}{n}\\sin\\left(\\frac{\\pi}{2n}\\right)\\Gamma\\left(\\frac{1}{n}\\right) = \\boxed{\\Gamma\\left(1+\\frac{1}{n}\\right)\\sin\\left(\\frac{\\pi}{2n}\\right)}$"
}
{
  "Tag": [],
  "Problem": "No teams got this problem correct at the tournament :).\r\n\r\n$a,b,c,d,x,y,z$ are real numbers such that\r\n\r\n\\begin{eqnarray*}ax+by+cz &=& d,\\\\ bx+cy+dz &=& a,\\\\ cx+dy+az &=& b,\\\\ dx+ay+bz &=& c. \\end{eqnarray*}\r\n\r\nIf $a\\neq c$, find all possible values of $y$.",
  "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119313[/url]\r\n\r\nquestion 10 someone posted it...",
  "Solution_2": "[hide]Subtracting the third equation from the first, we get\n\\[(a-c)x+(b-d)y+(c-a)z = (d-b) \\ \\Rightarrow \\ (a-c)(x-z) = (d-b)(y+1).\\]\nSubtracting the fourth equation from the second, we get\n\\[(b-d)x+(c-a)y+(d-b)z = (a-c) \\ \\Rightarrow \\ (b-d)(x-z) = (a-c)(y+1).\\]\n\nThe product of these two equations is\n\\[(a-c)(b-d)(x-z)^{2}= (a-c)(d-b)(y+1)^{2}.\\]\nAssuming $a \\neq c$, we can divide both sides by $a-c$.  Re-arranging, we get\n\\[(b-d)[(x-z)^{2}+(y+1)^{2}] = 0.\\]\n\nIf the first factor is zero, then $b = d$.  Then $(b-d)(x-z) = (a-c)(y+1) = 0$ and $a \\neq c$, so $y =-1$.  If the second factor is zero, then again $y$ must be $-1$.  So the only possible value of $y$ is $-1$.\n\nThere are many possible solutions when $y =-1$.  For example, take $x = y = z =-1$, and $a$, $b$, $c$, $d$ any real numbers such that $a+b+c+d = 0$ (and $a \\neq c$).[/hide]\r\n\r\nNice problem!",
  "Solution_3": "That is correct  :) ."
}