{ "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "AMC 10" ], "Problem": "This year's USAMO has 20 qualifiers from the 5% club.\r\n\r\nSomething about this strikes me as very odd. Looking at the Summary of High School Results and Awards 2006, I found that only one member of the 5% club scored over 120 on the AMC 10 or the AMC 12 last year and 12 members scored over 100 on the AMC 10 or 12.\r\n\r\nHowever, none of these members are the ones listed in the 2007 USAMO qualifier list. \r\n\r\nSecondly, no student from the 5% club last year scored over 5 points on the AIME. \r\n\r\nI hold nothing against the 5% Club, but is there some explanation for this? (Perhaps massive recruiting or rigorous problem solving courses?) Otherwise, I find the fact that 20 students qualified to be very unlikely.", "Solution_1": "It's not a school, and it appears that the students in question probably do not reside in the U.S. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291&start=120]this post[/url] by kohjhsd explains at least some of it.", "Solution_2": "Does this mean that the 5% Club is using the fact that they're based in Los Angeles (even though they took the test in Korea) to qualify for the USAMO?\r\n\r\nUnless the qualifiers are all US Citizens, shouldn't these spaces be given to US qualifiers?", "Solution_3": "[quote=\"Kent Merryfield\"]It's not a school, and it appears that the students in question probably do not reside in the U.S. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291&start=120]this post[/url] by kohjhsd explains at least some of it.[/quote]\n[quote=\"AIME Teacher's Manual Page 5\"]US and Canadian Citizens and students residing in the United States (with qualifying scores) are eligible to take the USAMO.[/quote]\n[quote=\"fourierseries\"]Unless the qualifiers are all US Citizens, shouldn't these spaces be given to US qualifiers?[/quote]\r\nOn the face, your conclusion would appear to be valid. But we don't know all the facts. Maybe the CAMC didn't ask where the students reside. Maybe there is a special agreement between CAMC and this group? Maybe these students intend to move to the USA before April 24?", "Solution_4": "The 5% Club is an academic preparation and enrichment program in Los Angeles CA. The 5% Club registered for the 2006 AMC 10/AMC 12, with about 150 participants in each level contest. The 5% Club registered again for the 2007 AMC 10/AMC 12 contest with about 10 times as many participants for each contest. \r\n\r\nFrom conversations with the Contest Manager, I know that the 5% Club was providing 2007 AMC Contest administration both in the United States and in Korea. It appears that in 2007 the 5% Club adminstered the contests in Los Angeles, Irvine, Victorville, and San Diego, and possibly Chicago as well as Korea.\r\n\r\nRegistration from within the US for a school outside the US is a acceptable practice in registration for the AMC contests and has been done for years. In past years, registration for schools outside the US was usually with organizations representing embassy schools, \"American Schools\" and some schools with APO/FPO addresses. The reason is that mailing the contests is faster, cheaper and more reliable than mailing directly. The registration form the 5% Club was much larger than we have seen in the past from these \"embassy and American school\" organizations, but it is not different in principle or custom.\r\n\r\nAll the registration information about the students indicates that the registration was from California.\r\nOur records indicate that the students from the 5% Club did not provide addresses on their answer forms. The address on the answer form is the address we use for direct correspondence with the students. It also provides a cross-check on residence within the US for puposes of USAMO invitation. Therefore, we do not know where these students came from. I am attempting to determine that now.\r\n\r\nPlease note that not providing the address on the answer form is common. Many of the correspondents on this forum have contacted me personally to inform me that they forgot to include some piece of information. So not providing an address is not an automatic reason for disqualification . Until we have information on the residence of all students, we will use the best available information.\r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions", "Solution_5": "If they reside in Korea, wouldn't it be possible that they took the AIME after the answers were already posted on AoPS? I know AMCDirector mentioned that certain schools in Asia were taking the AIME later...", "Solution_6": "If, hypothetically, something did disqualify one or more of them from taking the USAMO officially, would another person take their place or would their be fewer qualifiers?", "Solution_7": "Unless they received the AIME packets later, Korea should have taken AIME before the US because they are across the international date line, putting them over ten hours ahead of New York.", "Solution_8": "I doubt that they can change the index so that only 1 or 2 more people qualify to take the USAMO to replace those disqualified because i bet there are quite a few people who missed the index by a little bit.", "Solution_9": "Apparently there are only 12 of those people. See Mr. Dunbar's post, \"Revised USAMO list is being prepared\". Well, 12 who barely missed the floor tiebreaker, anyway." } { "Tag": [], "Problem": "Two hunters A and B got 10 birds for shooting.Sum of the square of their shots fired is 2880 and the product of the number of shots fired by each was 48 times the product of the number of birds killed by each.If A had fired as often as B and B would have fired as often as A,then B would have shot five more birds than A.Find the number of birds killed by each.", "Solution_1": "each got 10 birds or total they got that much???", "Solution_2": "I haven't tried the question yet but that's the way the question is....I thought it was wrong at first sight :D", "Solution_3": "shadysaysurspammed wrote:Two hunters A and B got 10 birds for shooting.Sum of the square of their shots fired is 2880 and the product of the number of shots fired by each was 48 times the product of the number of birds killed by each.If A had fired as often as B and B would have fired as often as A,then B would have shot five more birds than A.Find the number of birds killed by each.\n\n\n\na = # of birds from hunter A\n\nb = # of birds from hunter B\n\nx = # of shots from hunter A\n\ny = # of shots from hunter B\n\n\n\n[hide]\n\na = 10 - b\n\nx^2 + y^2 = 2880\n\nxy = 48ab\n\nay/x = bx/y + 5\n\n\n\nx and y have to be whole numbers\n\nso either (24, 48) or (48, 24), but they're interchangeable (you just have to adjust a and b accordingly), so we'll say x = 24 and y = 48\n\n\n\nx = 24\n\ny = 48\n\n\n\n2a = b/2 + 5\n\n2a = 20 - 2b\n\n15 = 5b/2\n\n\n\nb = 6\n\na = 4\n\n\n\nHunter A killed four birds and fired 24 times\n\nHunter B killed six birds and fired 48 times\n\n[/hide]", "Solution_4": "sorry if this is dumb but if [quote=\"shady\"]B would have shot five more birds than A\"[/quote]\r\nthen number of birds shot by A=$a$ so $2a+5=10$ this is $a=2.5$????", "Solution_5": "[quote=\"riddler\"]sorry if this is dumb but if [quote=\"shady\"]B would have shot five more birds than A\"[/quote]\nthen number of birds shot by A=$a$ so $2a+5=10$ this is $a=2.5$????[/quote]\r\n\r\nExactly same doubt here too :oops:", "Solution_6": "[quote=\"riddler\"]sorry if this is dumb but if [quote=\"shady\"]B would have shot five more birds than A\"[/quote]\nthen number of birds shot by A=$a$ so $2a+5=10$ this is $a=2.5$????[/quote]\r\n\r\nYEahh....I don't really get the use of as often as..its bugging me.", "Solution_7": "so is treething right,shady??? tell the answer" } { "Tag": [ "quadratics", "Vieta", "algebra", "algebra open" ], "Problem": "Given the quadratic equation $ ax^2\\plus{}bx\\plus{}c \\equal{} 0$ which has $ r$ and $ s$ as its roots. I know that we can find the quadratic equation with $ (r\\plus{}1)$ and $ (s\\plus{}1)$ as its roots and even $ r^3$ and $ s^3$ as its roots. But, is there a way to find a quadratic equation with $ 2^r$ ans $ 2^s$ as its roots?", "Solution_1": "A quadratic always has 2 roots (whether or not they are real, rational, equal,... is another thing). Therefore, we can always write\r\n$ (x-2^{r})(x-2\\&{s})$. Carry out that multiplication and you have your equation.", "Solution_2": "I understand that but is it possible to find it in terms of a, b and c?\r\n$ r\\plus{}s\\equal{}\\minus{}\\frac{b}{a}$ and $ rs\\equal{}\\frac{c}{a}$.\r\nWhat about for $ 2^r\\plus{}2^s$ and $ (2^r)(2^s)$?", "Solution_3": "If you want to find your coefficients $ a$,$ b$ and $ c$, expand JRav's factorization-and voila, you have your form $ ax^2 \\plus{} bx \\plus{} c$.\r\n\r\nOn the other hand, do you know Vieta's formulae? It looks like you know it (from your last post, which is a bit confusing). I assumed you wanted to show that you can find the coefficients if solutions are $ r$ and $ s$, and you're asking if the principle is same for this particular problem ($ 2^r$,$ 2^s$). Answer to that is yes. \r\n\r\n$ \\frac \\minus{}{b}{a} \\equal{} x_1 \\plus{} x_2$\r\n\r\n$ \\frac {c}{a} \\equal{} x_1x_2$\r\n\r\nwhatever the form of $ x_1$ and $ x_2$ is." } { "Tag": [ "geometry", "inradius" ], "Problem": "In $ \\triangle ABC$, $ \\angle C\\equal{} 90 \\deg$ and $ CD$ is the altitude to $ AB$.\r\n\r\n$ r_1$, $ r_2$ and $ r$ are the in-radii of $ \\triangle CAD$, $ \\triangle CBD$ and $ \\triangle ABC$. \r\n\r\nProve that $ r_1 \\plus{} r_2 \\le r \\times \\sqrt(2)$. When does the equality occur?", "Solution_1": "[hide=\"Hint\"] \nIn right-angled triangle with sides of lenghth $ a$, $ b$ and $ c$ and inradius $ r$ holds:\n\n$ r\\equal{}\\frac {a\\plus{}b\\minus{}c}{2}$\n\n[/hide]" } { "Tag": [ "geometry", "perimeter", "SFFT", "special factorizations" ], "Problem": "What is \"Simon's favorite factoring trick\"? How is it used?", "Solution_1": "I believe its $a+ab+b+1=a(b+1)+(b+1)=(a+1)(b+1)$, or something like that.", "Solution_2": "nevermind, i found it in another post. in case you're wondering:\r\n\r\n\\[(x-k)(y-k)=xy-kx-ky+k^2\\], where $k$ is a constant and $x$ and $y$ are variables. That's cool!", "Solution_3": "\\begin{eqnarray*}\r\n3m+4n&=&mn-5\\\\\r\n5&=&mn - 3m - 4n\r\n\\end{eqnarray*}\r\n\r\nNow what? How do you do it if the coefficients of $m$ and $n$ are different?", "Solution_4": "[hide] mn - 3m - 4n +12 = 5+12\n\n(m-4) (n-3) = 17. Is this right?[/hide]", "Solution_5": "[hide]\\begin{eqnarray*}\n3m+4n&=&mn-5\\\\\n5&=&mn - 3m - 3n - n\\\\\n5+n&=&mn - 3m - 3n\\\\\n5+n&=&(m-3)(n-3) - 9\\\\\n14+n&=&(m-3)(n-3)\n\\end{eqnarray*}\n\nWhat do I do with the extra $n$ on the LHS?[/hide]", "Solution_6": "[quote=\"Phelpedo\"][hide] mn - 3m - 4n +12 = 5+12\n\n(m-4) (n-3) = 17. Is this right?[/hide][/quote]\r\n\r\nI think it is. Since 17 is a prime number, $m-4=1$ and $n-3=17$ or $m-4=17$ and $n-3=1$. In the first case, $m=5$ and $n=20$. Then $3m+4n=mn-5$, so $15+80=95$, which is true! In the second case, $m=21$ and $n=4$. Then $3m+4n=mn-5$, so $63+16=79$, which is also true! Good job Phelpedo.", "Solution_7": "Here's another one:\r\n\r\nWhat is the minimum value of $a+b+c$ where a,b, and c, are distinct positive integers, if $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{9}{10}$.\r\n\r\n[hide=\"Solution\"]\nBecause the final value of the fraction is $\\frac{9}{10}$, the common denomenator has a factor of 2 and a factor of 5. One case is $a=5$. Then $\\frac{1}{b}+\\frac{1}{c}=\\frac{7}{10}$. \n\n\\begin{eqnarray*}\n\\frac{1}{b}+\\frac{1}{c}=\\frac{7}{10}\\\\\n\\frac{b+c}{bc}=\\frac{7}{10}\\\\\n10b+10c=7bc\\\\\n0=7bc-10b-10c\\\\\n0=bc-\\frac{10}{7}b-\\frac{10}{7}c\\\\\n\\frac{100}{49}=bc-\\frac{10}{7}b-\\frac{10}{7}c+\\frac{100}{49}\\\\\n0=(b-\\frac{10}{7})(c-\\frac{10}{7})-\\frac{100}{49}\\\\\n\\frac{100}{49}=(\\frac{7b-10}{7})(\\frac{7c-10}{7})\\\\\n100=7(7b-10)(7c-10)\\\\\n\\end{eqnarray*}\n\nNone of these are integers. Let's try the second case. If $a=2$:\n\\begin{eqnarray*}\n\\frac{1}{b}+\\frac{1}{c}=\\frac{2}{5}\\\\\n\\frac{b+c}{bc}=\\frac{2}{5}\\\\\n5b+5c=2bc\\\\\n0=2bc-5b-5c\\\\\n0=bc-2.5b-2.5c\\\\\n6.25=bc-2.5b-2.5c+6.25\\\\\n0=(b-2.5)(c-2.5)-6.25\\\\\n6.25=(b-2.5)(c-2.5)\\\\\n625=100(b-2.5)(c-2.5)\\\\\n25=(4b-10)(4c-10)\n\\end{eqnarray*}\n\nNow we need to find pairs of numbers that multiply to give 25: (1,25) and (5,5). Only (1,25) will work because (5,5) will not give us distinct numbers. So $b=\\frac{11}{4}$ and $c=\\frac{35}{4}$, but none are integers.\n\nHeh well, looks like this problem I just made up has no distinct integer solutions! (I think)\n[/hide]", "Solution_8": "[quote=\"Rep123max\"]Here is a less obvious use of Simon's Favorite Factoring Trick: \n\nThree non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length 1. The polygons meet at a point A in such a way that the sum of the three interior angles at A is $360^\\circ$. Thus the three polygons form a new polygon with A as an interior point. What is the largest possible perimeter that this polygon can have?[/quote]\r\n\r\nErr I didn't really use Simon's Favorite Factoring Trick but I have a solution.\r\n[hide]The sum of one of the interior angles of the three polygons have to be equal to 360, and two of them must be congruent.\n\n$\\frac {360n - 720}{n} + \\frac {180p - 360}{p} = 360$\n\n$\\frac {2n - 4}{n} + \\frac {p - 2}{p} = 2 \\Rightarrow 3np - 4p - 2n = 2np$\n\nSolving for $p$ in terms of $n$ we get \n\n$p = \\frac {2n}{n-4} \\Rightarrow p = 2 + \\frac 8{n-4}$\n\nThe perimeter of the new polygon would be equal to $2n + p - 6$ because 6 of the sides are in the interior of the polygon, and the rest will count. The largest value of $n$ is 12 which gives us the maximum perimeter of $21$.[/hide]", "Solution_9": "chess64, you forgot that $a$ could equal 10, but i tried it out and there aren't any solutions to it anyway :P nice problem though!\r\n\r\nwhat is the factoring way to figure out Rep123Max's problem?", "Solution_10": "Check this out: [url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=1069.com/Forum/weblog_entry.php?e=10[/url]\r\n\r\nIt uses the factoring trick to determine how you can tesselate certain polygons. kinda cool, eh?", "Solution_11": "This is how you do Rep123max's problem with the factoring trick:\r\n\r\n[hide] Allow n to be the # of sides of the figure which is duplicated and x to be # of sides of the figure which is not. Three angles, two coming from the duplicated figures and one from the non-duplicated one, meet at one point. This gives us:\n\n(360(n-2))/n+(180(x-2))/x=360\n\nMultiplying out and simplifying gives us nx-4x-2n=0. This is where the factoring trick comes in play. \n\nFactoring gives: \n\n(n-4)(x-2)=8\n\nFor the perimeter to be as large as possible, the values n and x have to be the farthest apart, because each time we increase n by 1, we add 2 to the perimeter, but then subract 1 because this requires us to decrease x by 1. Thus, we add 1 to the perimeter each time we increase n and decrease x by 1 each. This proves that n and x have to be as far apart as possible. Allowing n-4=8 and x-2=1, we get n=12 and x=3, thus giving us a perimeter of:\n\n[b]21[/b][/hide]" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "If ACED and CBFG(F lies on EC) are squares, H is the midpoint ot AC and HE = HG, prove that F is the midpoint of EC.", "Solution_1": "I call CH=x,CB=y\r\n $ \\rightarrow x^2\\plus{}(2x)^2\\equal{}(x\\plus{}y)^2\\plus{}y^2$\r\n $ \\rightarrow 2x^2\\minus{}y^2\\minus{}xy\\equal{}0$\r\n $ \\rightarrow (x\\minus{}y)(2x\\plus{}y)\\equal{}0$\r\n $ \\rightarrow x\\equal{}y$" } { "Tag": [ "search", "function", "algebra", "polynomial", "quadratics", "geometry", "geometric transformation" ], "Problem": "for a sequence like $ 1,3,5,7,9\\cdots$ the odd numbers \r\n$ U_n\\equal{}2n\\minus{}1$ which comes from the arithmetic sequence formula $ U_n\\equal{}a\\plus{}(n\\minus{}1)d$\r\n\r\nfor this sequence $ 2,3,6,11,18,27\\cdots$ where does $ U_n\\equal{}n^2\\minus{}2n\\plus{}3$ comes from ? can we get this from the formula $ U_n\\equal{}a\\plus{}(n\\minus{}1)d$", "Solution_1": "[quote=\"binomial_4eva\"]for this sequence $ 2,3,6,11,18,27\\cdots$ where does $ U_n \\equal{} n^2 \\minus{} 2n \\plus{} 3$ comes from ? can we get this from the formula $ U_n \\equal{} a \\plus{} (n \\minus{} 1)d$[/quote]\r\n\r\nThis is neither an arithmetic nor a geometric sequence. However, its [b]first difference[/b] $ 3 \\minus{} 2 \\equal{} 1, 6 \\minus{} 3 \\equal{} 3, 11 \\minus{} 6 \\equal{} 5, 18 \\minus{} 11 \\equal{} 7, 27 \\minus{} 18 \\equal{} 9$ is an arithmetic (that is, linear) sequence, so we can write an equation for $ U_n$ that is quadratic.\r\n\r\nThis is a standard lemma. Look up the [url=http://www.google.com/search?hl=en&q=method+of+finite+differences&btnG=Google+Search]method of finite differences[/url]. \r\n\r\nAlternately, because\r\n\r\n$ 3 \\equal{} 2 \\plus{} 1$\r\n$ 6 \\equal{} 2 \\plus{} 1 \\plus{} 3$\r\n$ 11 \\equal{} 2 \\plus{} 1 \\plus{} 3 \\plus{} 5$\r\n$ 18 \\equal{} 2 \\plus{} 1 \\plus{} 3 \\plus{} 5 \\plus{} 7$\r\n$ 27 \\equal{} 2 \\plus{} 1 \\plus{} 3 \\plus{} 5 \\plus{} 7 \\plus{} 9$\r\n\r\nwe can consider this to be a sequence of [b]partial sums[/b] of the arithmetic sequence $ 1, 3, 5, 7, 9, ...$ (ignoring the initial $ 2$), and work from there using what we know about how to evaluate arithmetic [b]series[/b]. In particular, we know that \r\n\r\n$ 1^2 \\equal{} 1$\r\n$ 2^2 \\equal{} 1 \\plus{} 3$\r\n$ 3^2 \\equal{} 1 \\plus{} 3 \\plus{} 5$\r\n...\r\n$ n^2 \\equal{} 1 \\plus{} 3 \\plus{} 5 \\plus{} ... \\plus{} (2n \\minus{} 1)$\r\n\r\nso we can easily find an equation for $ U_n$ from there. I say \"an\" equation because, as we have discussed, such equations are not unique to a given sequence.", "Solution_2": "how many kinds of sequences are there ? like these 2, arithmetic and geometric.", "Solution_3": "There are a lot, but arithmetic and geometric are the two most important ones. Another important one is the harmonic series, but that's mostly in the context of calculus.\r\n\r\nThere really isn't a specific definition for what constitutes a special sequence, and here's an example why. I could make up one that starts with $ 1$ and the $ n$th term ($ n > 1$) is the $ n\\minus{}1$th term with the digital representation of $ n!$ on it, so the sequence would go like $ 1, 21, 621, 24621$, etc. Then I can be all smug and be like \"Check out this new sequence I discovered that will become cutting edge mathematics\" and preemptively call it the Melon Sequence. But most people including me would agree that the above is silly and insignificant.\r\n\r\nTo start learning about more advanced sequences:\r\n[url=http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]Harmonic Series[/url]\r\n[url=http://en.wikipedia.org/wiki/Riemann_zeta]Riemann Zeta Function[/url]\r\n[url=http://en.wikipedia.org/wiki/Fourier_series]Fourier Series[/url]\r\nThis is by no means a comprehensive list.", "Solution_4": "[quote=\"binomial_4eva\"]how many kinds of sequences are there ?[/quote]\r\n\r\nThere are more kinds of sequences than you can possibly imagine names for.", "Solution_5": "$ (1)$\r\ndoes this sequence $ 2,3,6,11,18,27\\cdots$ have a name ?like this sequence , what are those sequences called that the second last difference is always an arithmetical sequence,or geometrical sequence.\r\n\r\n$ (2)$\r\nhow do you get the last differences (the constant) of a sequence without writing all the differences of that sequence down?\r\n\r\n$ (3)$\r\nif a sequence has a polynomial formula, what does the root of that polynomial tell us about that sequence? example $ 1,4,9,16,25,36\\cdots$ $ U_n\\equal{}n^2$ what does the root $ (x\\equal{}0)$ tell us about the sequence ?.\r\n\r\n$ (4)$\r\ncan you get the coefficient of the polynomial formula for a sequence if you dont know the elements in the sequence but you only know that the fourth difference is constant", "Solution_6": "(1) You might as well call it a \"quadratic sequence\" or (less specifically) a \"polynomial sequence,\" since it's generated by a quadratic polynomial. (Polynomial sequences are the only ones whose difference tables eventually reach a constant row.)\r\n\r\n(2) You can't unless you already know the formula used to generated the sequence, in which case writing it in terms of the basis of falling powers* will help.\r\n\r\n(3) Not very much. If we're only talking about polynomials over the integers and $ n \\in \\bf Z$ is a root then $ x \\minus{} n$ is a factor so you know that the $ m$th term of the sequence is divisible by $ m \\minus{} n$. More generally, I don't think you really learn much of anything. (Divisibility means nothing over $ \\bf R$ or $ \\bf C$.)\r\n\r\n(4) Of course not: every fourth-degree polynomial has constant fourth difference and not all fourth-degree polynomials have the same coefficients.\r\n\r\n\r\n* Every polynomial (say in the variable $ x$) can be written as a unique linear combination (that is, a sum of multiples) of the elements of $ \\{1, x, x^2, x^3, \\ldots\\}$ -- this is what we're normally used to seeing when we think of a polynomial in expanded form. It's also possible to write every polynomial (in the variable $ x$) as a unique linear combination of the elements of $ \\{1, x, x(x \\minus{} 1), x(x \\minus{} 1)(x \\minus{} 2), \\ldots\\}$. These are called the [i]falling powers[/i] of $ x$. For example, we can write $ x^3 \\plus{} 3x \\plus{} 2 \\equal{} x(x \\minus{} 1)(x \\minus{} 2) \\plus{} 3x(x \\minus{} 1) \\plus{} x \\plus{} 2$. Look at the difference table for the sequence generated by $ x^3 \\plus{} 3x \\plus{} 2$ to see what my comment (2) means.", "Solution_7": "apart from question (2) i get everything else you said, thank u very very much.\r\n\r\nthe sequence of $ x^3\\plus{}3x\\plus{}2$ is \r\n\r\nthe seq $ 6,16,38,78,142$\r\n1th diff $ 10,22,40,64$\r\n2th diff $ 12,18,24$\r\n3th diff $ 6,6$\r\n\r\ni couldnt figure out how to get the constant (6) from the formula.", "Solution_8": "Okay, I actually made two mistakes there:\r\n\r\nMistake 1: $ x^3 \\plus{} 3x \\plus{} 2 \\equal{} x(x\\minus{}1)(x\\minus{}2) \\plus{} 3x(x\\minus{}1) \\plus{} 4x \\plus{} 2$.\r\n\r\nMistake 2: I shouldn't have used falling powers; instead, I should have use binomial coefficients. There's an easy translation between falling powers and binomial coefficients: $ \\binom{n}{k}$ is the $ k$th falling power of $ n$ divided by $ k!$. Then we get $ x^3 \\plus{} 3x \\plus{}2 \\equal{} 6\\binom{x}{3} \\plus{} 6\\binom{x}{2} \\plus{} 4\\binom{x}{1} \\plus{} 2\\binom{x}{0}$.\r\n\r\nNow, even with that information, you probably still don't see what this has to do with your sequence, and reasonably so. That's because I start my sequences at $ 0$ instead of at $ 1$, so one or the other of us has to change. I'd rather make you change ;) -- in that case, your difference table gets an extra first column, making it look like this:\r\nthe seq $ 2, 6,16,38,78,142$\r\n1th diff $ 4, 10,22,40,64$\r\n2th diff $ 6, 12,18,24$\r\n3th diff $ 6,6,6$\r\nOkay, now you should be able to see the relation, yes? (Obviously, this isn't a [i]proof[/i] of the relation -- it's a demonstration of a particular case. The proof isn't hard, exactly, but it's a decent amount of work to write it all down properly.)\r\n\r\nIf you can figure out how to correctly formulate the relation my example has suggested, you should be able to state for yourself the relationship between the polynomial that generates a sequence and the eventual row of constants in its difference table. Once you think you've formulated the general result, test your claim with a few different polynomials. For example, I recommend:\r\n$ f_1(x) \\equal{} ax \\plus{} b$\r\n$ f_2(x) \\equal{} x^3 \\plus{} x$\r\n$ f_3(x) \\equal{} 7x^3$\r\n$ f_4(x) \\equal{} \\minus{}x^3 \\minus{} x$\r\n$ f_5(x) \\equal{} 3x^4 \\plus{} 2x^2 \\plus{} 1$\r\nbut I'm sure you can come up with a bunch of polynomials yourself ;)", "Solution_9": "[quote=\"JBL\"]in that case, your difference table gets an extra first column, making it look like this:\nthe seq $ 2, 6,16,38,78,142$\n1th diff $ 4, 10,22,40,64$\n2th diff $ 6, 12,18,24$\n3th diff $ 6,6,6$[/quote]\r\n\r\nThis reminds me of differentiation :wink:", "Solution_10": "[quote=\"not_trig\"][quote=\"JBL\"]in that case, your difference table gets an extra first column, making it look like this:\nthe seq $ 2, 6,16,38,78,142$\n1th diff $ 4, 10,22,40,64$\n2th diff $ 6, 12,18,24$\n3th diff $ 6,6,6$[/quote]\n\nThis reminds me of differentiation :wink:[/quote]\r\n\r\nIndeed. This is the [url=http://en.wikipedia.org/wiki/Difference_operator]forward difference operator[/url], which shares properties in common with differentiation.", "Solution_11": "In addition, with the forward difference operator, the falling power takes the place of normal powers. That is, if $ x^^{\\underline{k}}$ is the falling power $ x$ with $ k$ terms, that is $ x(x-1)(x-2)(\\cdots)(x-k+1)$, then $ \\Delta x^{^{\\underline{k}}} = kx^{^{\\underline{k-1}}}$. Also, $ 2$ take the place of $ e$. $ \\Delta 2^x = 2^x$.", "Solution_12": "i am stuck in this part \r\n[quote]Then we get $ x^3 \\plus{} 3x \\plus{} 2 \\equal{} 6\\binom{x}{3} \\plus{} 6\\binom{x}{2} \\plus{} 4\\binom{x}{1} \\plus{} 2\\binom{x}{0}$.[/quote]\r\n\r\nhow did you get that ? can we write any polynomial in that form (i think you said we can) ? how ?\r\nsorry .i dont get it. can you write a few more poly....s in that form.\r\n\r\nnow i undrestand where the constant comes from, the $ 6,6,4,2$ in the equation are the first elements of the sequence and differences $ (2,4,6,6)$", "Solution_13": "So, first, I hope you've understood how to write a polynomial in terms of the falling powers. It's really a pretty straightforward process: the $ k$th falling power (what Hamster wrote $ x^{\\underline k}$, which is the product $ x(x-1)\\cdots(x -k + 1)$) is a polynomial of degree $ k$ and leading coefficient 1. If you have any polynomial (Let's take $ 4x^3 - x^2$ as an example), first you choose the highest-degree term (in this case $ x^3$), and you need to match its coefficient with some falling power. So, in our example, we need to start with $ 4x^{\\underline{3}} = 4x(x - 1)(x - 2) = 4x^3 -12x^2 + 8x$. Now, move on to the next highest power: we're supposed to have $ x^2$ with a coefficient of -1, but instead we have it with a coefficient of -12, so we need to add $ 11 x^{\\underline{2}}$. This gives us \r\n\\begin{align*}\r\n4x^{\\underline{3}} + 11 x^{\\underline{2}} & = (4x^3 -12x^2 + 8x) + 11(x(x - 1)) \\\\\r\n& = (4x^3 -12x^2 + 8x) + 11x^2 - 11x \\\\\r\n& = 4x^3 - x^2 - 3x.\r\n\\end{align*}\r\nNow we move on to the next highest power: we're supposed to have $ x$ with a coefficient of 0, but instead we have it with a coefficient of -3. So, we need to add a $ 3x^{\\underline{1}}$ term. But luckily $ x^{\\underline{1}} = x$ so we don't need to do any arithmetic for this one. Then we have to move on to the next highest power, which in this case is the constant term -- we already have the right constant (0), so we don't have to add anything here, either. This gives us the final expression\r\n\\[ 4x^3 - x^2 = 4x^{\\underline{3}} + 11 x^{\\underline{2}} + 3x^{\\underline{1}}.\r\n\\]\r\n\r\nNow, what does this have to do with binomial coefficients? A lot, as it turns out: when $ x$ is an integer and $ k \\leq x$, \r\n\\begin{align*}\r\nx^{\\underline k} & = x(x-1)(x-2)\\cdots(x-k+2)(x-k+1) \\\\\r\n& = \\frac{x\\cdots(x-k+1)(x-k)(x-k-1)\\cdots(2)(1)}{(x-k)(x-k-1)\\cdots(2)(1)} \\\\\r\n& = \\frac{x!}{(x-k)!}\r\n\\end{align*}\r\nand so $ \\binom{x}{k} = \\frac{x!}{(x-k)!k!} = \\frac{x^{\\underline{k}}}{k!}$, or equivalently $ x^{\\underline{k}} = \\binom{x}{k}\\cdot k!$. Thus, if you can write something in terms of falling powers, you can translate it immediately into binomial coefficients by using this substitution. Thus is our example we have immediately that\r\n\\begin{align*}\r\n4x^3 - x^2 & = 4x^{\\underline{3}} + 11 x^{\\underline{2}} + 3x^{\\underline{1}} \\\\\r\n& = 4\\cdot 3!\\binom{x}{3} + 11\\cdot 2! \\binom{x}{2} + 3\\cdot 1!\\binom{x}{1} \\\\\r\n& = 24\\binom{x}{3} + 22\\binom{x}{2} + 3\\binom{x}{1}.\r\n\\end{align*}\r\n\r\nNow, here's an interesting thing to notice: when we go from a \"regular\" polynomial to an expression in terms of falling powers, we never change the leading coefficient. And when we go from the falling powers to the binomial coefficients, we change it in a predictable way. In other words, if we have a polynomial that looks like\r\n$ P(x) = ax^n + \\ldots$ for some $ a \\neq 0$, where $ n$ is the highest-degree term, then if we write $ P$ in terms of falling powers we get $ P(x) = ax^{\\underline{n}} + \\ldots$ (where the thing in the $ \\ldots$ has changed from one step to the next). If we then write this in terms of binomial coefficients, we get $ P(x) = a\\cdot n! \\binom{x}{n} + \\ldots$ (where again something has happened to the stuff in the $ \\ldots$). Now, as you've already noted (although, again, this is something no one has proven on this thread), the final row of constants is just equal to the leading coefficient in this expression, which is $ an!$. So if [i]all[/i] you're interested in is what that bottom constant row in the difference table is, you can get it just by looking at the largest-degree term in your polynomial.", "Solution_14": "dunno how to thank you ,really nice explanation, thanks JBL. i learned loads of things", "Solution_15": "Glad to be of service! While I'm here, I thought I'd add a little more of interest (like, say, some actual proofs).\r\n\r\nConsider the difference table for the polynomial $ f(x) \\equal{} \\binom{x}{n}$. (Here $ n$ is just some constant.) We know that $ f(0) \\equal{} f(1) \\equal{} f(2) \\equal{} \\ldots \\equal{} f(n \\minus{} 1) \\equal{} 0$ and that $ f(n) \\equal{} 1$. So, if we start the difference table, it looks like this:\r\nThe seq: $ 0, 0, 0, \\ldots, 0, 0, 1, \\ldots$ (and there are $ n$ 0s there before the 1)\r\n1st dif: $ 0, 0, 0, \\ldots, 0, 1, \\ldots$ (now there are only $ n \\minus{}1$ 0s)\r\n2nd dif: $ 0, 0, \\ldots, 1, \\ldots$ (now $ n \\minus{} 2$)\r\n$ \\vdots$\r\n$ n\\minus{}1$ dif: $ 0, 1, \\ldots$\r\n$ n$th dif: $ 1, \\ldots$\r\n\r\nBut here's the thing: our sequence in generated by a degree-$ n$ polynomial, so the $ n$th differences are just a constant! So, $ f(x) \\equal{} \\binom{x}{n}$ gives us a difference table so that the first diagonal ($ f(0), \\Delta f(0), \\ldots$) is all-0, except that the $ n$th difference at 0 is equal to 1. So what? Well, here's the nice part: if you have two sequences and you add them, their difference tables add. If you have a sequence and you multiply it by a constant, the entire difference table gets multiplied by the same constant. And, of course, when our sequence is generated by a polynomial, all you need is that first diagonal to be able to reconstruct the entire sequence. So, suppose you have a polynomial sequence with a difference table that looks like this:\r\n\r\nThe seq: $ a_0(0), \\ldots$\r\n1st dif: $ a_1(0), \\ldots$\r\n2nd dif: $ a_2(0), \\ldots$\r\n$ \\vdots$\r\n$ k$th dif: $ a_k(0), \\ldots$ (and this last row is known to be constant, i.e. we're dealing with a degree-$ k$ polynomial to generate our sequence),\r\n\r\nI can say to you, \"Hey, I know a polynomial that generates exactly this difference table: to get the $ a_0(0)$ at the beginning of the first row, I need to take $ a_0\\binom{x}{0}$; to get the $ a_1(0)$ at the beginning of the second row, I need to take $ a_1\\binom{x}{1}$; and so on.\" The key idea here is that the difference tables of the $ \\binom{x}n$ are so nice that they don't interact with each other -- when I add in a multiple of $ \\binom{x}{m}$ for some $ m$, I know it will [i]only[/i] change the value in the $ m$th row along that first diagonal. Then, since two sequences with the same difference table are the same, we have exactly that my polynomial (written in terms of the $ \\binom{x}{n}$) and whatever polynomial we started with must be the same. Going the other direction, writing a polynomial in terms of these binomial coefficients gives me exactly the difference table I've claimed in my previous post.\r\n\r\n\r\nNow, a quick aside on something Hamster noted, using some calculus: the function $ \\exp(x) \\equal{} e^x$ has the nice property that $ \\frac{d}{dx}\\exp(x) \\equal{} \\exp(x)$. Why exactly does it have this property? Well, $ e^x \\equal{} 1 \\plus{} x \\plus{} \\frac{x^2}{2!} \\plus{} \\frac{x^3}{3!} \\plus{} \\ldots$ is a sum of polynomials of the form $ p_n(x) \\equal{} x^n/n!$. What's special about these polynomials? They are the unique polynomials such that if we take the $ k$th derivative $ p_n^{(k)}(x)$ and look at the value at $ 0$, we get that $ p_n^{(k)}(0) \\equal{} \\delta_{k, n}$. This is fancy-talk for saying that the value is 0 unless $ k \\equal{} n$, in which case the value is 1. Now, I just described some polynomials that $ f_n(x)$ that have the property that the $ k$th difference $ \\Delta^kf_n(x)$ taken at $ x \\equal{} 0$ gives $ \\Delta^kf_n(x) \\equal{} \\delta_{k, n}$, i.e. the $ k$th difference at 0 is 0 unless $ k \\equal{} n$, in which case it's 1. So, this suggests that the natural analogue of $ \\exp(x)$ should be the sum $ E(x) \\equal{} \\sum_{n \\geq 0} f_n(x) \\equal{} \\sum_{n \\geq 0}\\binom{x}{n}$. What is this sum? When $ x$ is an integer (the only case we really deal with for finite difference tables), $ \\binom{x}{n} \\equal{} 0$ whenever $ n > x$. So, we can actually rewrite $ E(x) \\equal{} \\sum_{n\\equal{} 0}^x \\binom{x}{n}$. But a well-known combinatorial identity tells us that this is just $ E(x) \\equal{} 2^x$. So, this (I hope) gives some intuition on why it should be that $ 2^x$ is the analogue of $ e^x$, i.e. it's more than just a freak accident but actually has a lot to do with the underlying frameworks.", "Solution_16": "how do we find the constant term of this $ U_n\\equal{}2^{n\\minus{}1}$ or more generally $ U_n\\equal{}a(r)^{n\\minus{}1}$", "Solution_17": "Polynomial sequences are the only sequences that have an eventual constant row. (Neither direction of the equivalence is very hard to prove.) You could easily verify for yourself that the first differences of the sequence $ V_n \\equal{} 2^n$ are $ 2^{n \\plus{} 1} \\minus{} 2^n \\equal{} 2^n$, i.e. its first differences (and so all further differences) just spit the same sequence back to you. Your $ U_n$ is just my $ V_n$ shifted, so to get one from the other you shift the entire difference table.\r\n\r\nFor $ r^n$, you can do the same calculation to see that the first differences are $ r^{n \\plus{} 1} \\minus{} r^n \\equal{} (r \\minus{} 1)r^n$, i.e. the difference operator is the \"multiply by $ r \\minus{} 1$\" operator for this sequence. (Multiplying by a constant changes nothing, of course.) Extending the analogy from the end of my previous post, this says that in some sense $ r \\minus{} 1$ is the difference table equivalent of $ \\ln r$ (since $ \\frac {d}{dx}(r^x) \\equal{} (\\ln r)r^x$).", "Solution_18": "sorry i couldnt explain the problem completly in my previous post, i was trying to ask these questions ,\r\n\r\n$ (1)$how can we find the constant term of a sequence that ends in a geometric sequence. a sequence like this :\r\n\r\nThe seq:$ 3,3,6,24,192,3072,\\dots$\r\n1st diff: $ 1,2,4,8,16,\\ldots$\r\n2nd diff:$ 2,2,\\ldots$\r\n\r\n$ (2)$ is there a general formula for the $ n_{th}$ term of all sequences that ends in geometric sequence like this one \r\nThe seq:$ 3,3,6,24,192,3072,\\dots$\r\n1st diff: $ 1,2,4,8,16,\\ldots$\r\n2nd diff:$ 2,2,\\ldots$\r\nthat ends in a geometric sequence.\r\n\r\nlike this one that i found for the sequences that ends in an arithmetics sequence\r\nif we expand the general arithmetic formula $ U_n \\equal{} a \\plus{} (n \\minus{} 1)d$ it becomes \r\n$ U_n \\equal{} a \\plus{} \\frac {1}{1!}(n \\minus{} 1)d_1 \\plus{} \\frac {1}{2!}(n \\minus{} 1)(n \\minus{} 2)d_2 \\plus{} \\cdots \\plus{} \\frac {1}{t!}(n \\minus{} 1)\\cdots(n \\minus{} t)d_t \\equal{} a \\plus{} {{n \\minus{} 1}\\choose1}d_1 \\plus{} {{n \\minus{} 1}\\choose2}d_2 \\plus{} \\cdots \\plus{} {{n \\minus{} 1}\\choose t}d_t \\equal{} a \\plus{} \\sum^{t}_{i \\equal{} 1}\\binom{n \\minus{} 1}{i}d_{i}$ so $ U_n \\equal{} a \\plus{} \\sum^{t}_{i \\equal{} 1}\\binom{n \\minus{} 1}{i}d_{i}$ \r\ni love this kind of general formulas\r\n\r\nso for question $ (2)$ what i am asking is if we can expand the geometric formula $ G_n \\equal{} a(r)^{n \\minus{} 1}$ like the one above.\r\n\r\n$ (3)$ As you said the constant term of sequences that ends in arithmetic sequence is $ C \\equal{} an!$\r\nwhat would the formula for the constant term of a sequence that ends in geometric sequence be ?", "Solution_19": "According to the posts above, there will be no constant difference row in geometric sequence:\r\n$ f(x)\\equal{}a\\cdot r^x$\r\n$ \\Delta[f](x)\\equal{}a\\cdot r^x(r\\minus{}1)$\r\n$ \\Delta^n[f](x)\\equal{}a\\cdot r^x(r\\minus{}1)^n$\r\nWhen $ r\\equal{}2$, for all natural number $ n$ there is $ \\Delta^n[f](x)\\equal{}a\\cdot 2^x$, as @JBL said.", "Solution_20": "[quote=\"timwu\"]According to the posts above, there will be no constant difference row in geometric sequence:\n$ f(x) \\equal{} a\\cdot r^x$\n$ \\Delta[f](x) \\equal{} a\\cdot r^x(r \\minus{} 1)$\n$ \\Delta^n[f](x) \\equal{} a\\cdot r^x(r \\minus{} 1)^n$\nWhen $ r \\equal{} 2$, for all natural number $ n$ there is $ \\Delta^n[f](x) \\equal{} a\\cdot 2^x$, as @JBL said.[/quote]\r\ni am confused, of course not constant [size=150]difference[/size], but has a constant [size=150]term[/size], its $ 2$ isnt it ?\r\nThe seq:$ 3,3,6,24,192,3072,\\dots$\r\n1st diff: $ 1,2,4,8,16,\\ldots$\r\n2nd diff:$ 2,2,\\ldots$", "Solution_21": "What do you mean by \"constant term\"? You're being very unclear.\r\n\r\nAnd your example doesn't help at all. You're not taking the differences between consecutive terms in the sequence; you're taking the [i]quotients[/i], which isn't what we're talking about. (If you take the base-$ 2$ logarithm of the sequence, you'll get JBL's result for polynomial sequences, which is also not what we're talking about.) \r\n\r\nIf a sequence is polynomial, then its \"constant term\" is the coefficient of the $ x^0$ term. But if a sequence is not polynomial there is not necessarily an unambiguous choice of \"constant term.\" Consider, for example, the sequence\r\n\r\n$ a_n \\equal{} 2 \\cos^2 \\frac{\\pi n}{6} \\equal{} \\cos \\frac{\\pi n}{3} \\plus{} 1$" } { "Tag": [], "Problem": "According to my Algebra 2/Trig teacher, many lines are tried and each time the calculator calculates the distance from each point to the line and squares this distance. It then adds up all these squares and records the values. After it tries enough lines it finds the line with the least sum of all the squares of the distances and gives you this line. Does anybody know how the coefficient of variation, r is calculated?", "Solution_1": "From my statistics book:\r\n\r\n\\[r=\\frac{1}{n-1}\\sum\\left(\\frac{x_i-\\overline{x}}{s_x}\\right)\\left(\\frac{y_i-\\overline{y}}{s_y}\\right)\\]\r\n\r\nn is the number of values in the list\r\n\r\n$\\overline{x}$ is the mean of the x's(same with the y's)\r\n\r\n$s_x$ is the standard deviation of the x's(ditto with y)" } { "Tag": [], "Problem": "if i hav a math i dea, how does it become an \"official\" conjecture?", "Solution_1": "you write it up and ask some scientists or mathematician ppls and see if ur anywhere near possibly right.. and then.. u try to prove it, but if you can't then it becomes one of those famous unprovable things that take centuries for someone to prove... =) Good luck if ur planning on doing something like that! sounds cool.. hehe", "Solution_2": "dont you get a prize or anything?", "Solution_3": "For making up a question you can't solve?", "Solution_4": "theres alot of mathematicians that made up unsolvable problems like fermat and goldbach and there famous now. and what about those unsolvable millenium problems?" } { "Tag": [ "function", "algebra", "polynomial", "linear algebra", "matrix", "probability" ], "Problem": "I have a question on statistics, and I don't know which forum to post it on.\r\n\r\nI have two large sets of numbers, each with the same size, say 1000 elements.\r\n\r\nBut to simplify let's say A{1,5,7,4} and B{1.5,6,14,3.5}\r\n\r\nI want to find some kind of formula for \"how close\" the two sets are to each other.\r\n\r\nThese are the characteristics I would prefer:\r\n(1) the differences |A_i - B_i| are important, not the ratios. (In other words, if A_i=4 and B_i=5, that would give the same result as A_i=0.01, B_i=1.01)\r\n(2) the value returned is large when the sets are close and small when the sets are very different\r\n(3) if two points are far apart, even if all the others are really close, this has a very dramatic effect on the function - it pushes it down a lot. Therefore averaging the differences will not be a good idea.\r\n(4) it is not a computational nightmare. For instance, if I make a Lagrange polynomial for one, then it will have a degree of about 1000 and that would not be good.\r\n\r\nThank you for anyone who can help me find such a function. (There may be some other restrictions that I haven't thought of yet, but I think I found all of them.)", "Solution_1": "My first reaction would be to do a $\\chi^2$ test of homogeneity with the two sets as two rows of a matrix (you'd first order the sets in increasing or decreasing value). Then either the resulting $\\chi^2$ test statistic or the p-value associated with it could be used as a measure of the \"difference\" between the two sets (with the p-value more precisely indicating the probability that if the two sets were samples from propulations with identical relative distributions, that the resultings samples would be at least as different as the sets you come up with). The p-value would satisfy condition 3, since it will be large if the sets are similar to each other and small if they are very different.", "Solution_2": "Do you mean for us to consider the sets as ordered? One thing you might consider is taking a large power mean of the differences, rather than the arithmetic mean. This will be more sensitive to a single large difference. How large a power to choose depends how sensitive you want it to be.\r\n\r\n\r\nPower mean: for positive reals $x_1, x_2, \\ldots, x_n$, the $q$th power mean\r\n$f_q = (\\frac{x_1^q + x_2^q + \\ldots + x_n^q}{n})^{\\frac 1q}$ is a continuous, increasing function for $q$ on the positive or negative reals. It has the properties that as $q$ goes to infinity, $f_q \\to \\max{x_i}$, as $q$ goes to negative infinity, $f_q \\to \\min{x_i}$ and as $q\\to 0$, $f_q$ approaches the geometric mean of the $x_i$." } { "Tag": [], "Problem": "How many numbers between $ 1$ and $ 200$ inclusive are divisible by the square of a prime number?", "Solution_1": "[hide=\"Solution\"]Just so I can be lazy, I'm not going to add floor symbols for any division, so pretend that any divisions I take are done by a computer that does integer division. :lol: \n\nWe use inclusion-exclusion:\n\nThe prime squares less than $ 200$ are $ 4$, $ 9$, $ 25$, $ 49$, $ 121$, and $ 169$.\n\nCase 1: Squares comprising one prime. $ 200/4 \\equal{} 50$, $ 200/9 \\equal{} 22$, $ 200/25 \\equal{} 8$, $ 200/49 \\equal{} 4$, $ 200/121 \\equal{} 1$, $ 200/169 \\equal{} 1$.\n\nCase 2: Squares comprising two primes.$ 200/36 \\equal{} 5$, $ 200/100 \\equal{} 2$, $ 200/196 \\equal{} 1$\n\n$ 4\\times 9 \\times 25 > 200$, so no squares comprising three primes exist.\n\n$ 50\\plus{}22\\plus{}8\\plus{}4\\plus{}1\\plus{}1\\minus{}5\\minus{}2\\minus{}1 \\equal{} 78$\n\nTherefore, there are exactly $ \\boxed{78}$ numbers between $ 1$ and $ 200$ divisible by a prime square.\n\n[/hide]", "Solution_2": "Thanks darkdieuguerre." } { "Tag": [ "logarithms", "integration", "Euler", "calculus", "calculus computations" ], "Problem": "this is the last question of the day from me, i promise. :D :P \r\n\r\n(since i asked this in class, thought i would also ask here, LOL!)...... have fun.... :) \r\n\r\n\r\nshow\r\n\r\n$ \\dfrac{1}{2}\\cdot \\sum_{k\\equal{}1}^\\infty\\; \\dfrac{(\\minus{}1)^{k\\plus{}1}}{ \\dbinom{2k}{k}\\cdot k^2} \\;\\;\\equal{}\\;\\; \\boxed{ \\left(\\, \\sinh^{\\minus{}1}\\,\\left(\\dfrac{1}{2}\\right)\\, \\right)^2}$", "Solution_1": "I think you love gamma and beta function's. :lol: \r\nanother way: \r\nmore generally $ your sum ... \\equal{} \\frac {1}{2}\\sum ( \\minus{} 1)^{k \\plus{} 1}\\frac {(k!)^{2}}{(2k)!k^{2}}$\r\nwe already know that :\r\n$ f(x)\\equal{} \\sum_{n \\equal{} 0}^{\\infty}\\frac {(n!)^{2}}{(2n)!}x^{n} \\equal{} \\frac {4}{4 \\minus{} x} \\minus{} \\frac {4\\sqrt {|x|}}{(4 \\minus{} x)^{3/2}}\\ln \\frac {\\sqrt {|x|} \\minus{} \\sqrt {4 \\minus{} x}}{2}$ for $ x < 0$\r\nanswer will be :\r\n$ \\minus{} \\int\\frac {\\int f(x)/x}{x}|_{x \\equal{} \\minus{} 1}$ or something like this :)\r\nfew calcualtion and we are done", "Solution_2": "[quote=\"Extremal\"]I think you love [b][size=150][color=blue]gamma[/color] [/size][/b]and beta function's. :lol: [/quote]\r\n\r\nyes sir !! :D that i most certainly do. :thumbup: \r\n\r\ni have never seen anything more elegant and beautiful ! (after all i worship euler and ramanujan... (not literally of-course) :) ) \r\n\r\ni have this huge poster of ramanujan in my studyroom (a colleague sent it to me from cambridge), it says\r\n\r\n\r\n$ 2\\,\\sqrt [4]{\\; \\dfrac{3^2}{3^2\\minus{}1} \\cdot \\dfrac{5^2\\minus{}1}{5^2} \\cdot \\dfrac{7^2}{7^2\\minus{}1} \\cdot \\dfrac{9^2\\minus{}1}{9^2} \\cdot \\dfrac{11^2}{11^2\\minus{}1}\\cdot \\quad \\ldots\\;} \\;\\;\\equal{}\\;\\; \\boxed { \\dfrac{ \\Gamma\\left( \\dfrac{1}{4}\\right) }{\\sqrt\\pi} }$" } { "Tag": [ "function", "topology", "advanced fields", "advanced fields unsolved" ], "Problem": "Let $(X,d)$ be a compact metric space and $f: X\\rightarrow X$ be a function such that $\\forall x, y \\in X$ we have:\r\n$d(f(x),f(y)) \\leq d(x,y)$ ,where equation holds only when $x=y$.\r\nProve that the function $f$ has a unique fixed point.", "Solution_1": "already posted many times .. $x\\to d(x,f(x))$ is continuous and achieves its minimum somewhere in the compact. From here it is easy.", "Solution_2": "it is Banach fixed point theorem.\r\nf is a contraction here." } { "Tag": [ "probability", "function", "real analysis", "absolute value", "probability and stats" ], "Problem": "If $ P$ and $ Q$ are two probability measures on $ \\mathbb{N}_0\\equal{}\\{0,1,2, \\, . \\, . \\, .\\}$ then $ \\parallel{} P\\minus{}Q\\parallel{} : \\equal{}\\sup_{B\\subset \\mathbb{N}}|P(B)\\minus{}Q(B)|$\r\nCan someone show that how is the above equal to $ \\frac{1}{2} \\sum_{k\\ge 0}|P(k)\\minus{}Q(k)|$?\r\nThanks.", "Solution_1": "hint: introduce the set $ A\\equal{}\\{i: P(i) > Q(i) \\}$", "Solution_2": "Hi all, \r\n\r\nI must have missed something because I can't find the 1/2 factor. \r\nIntuitively we have that the \"sup\" is reached for $ B \\equal{} \\mathbb{N}$ otherwise you could add some i not in B and have a bigger \"sup\" if P(i) is not equal to Q(i). \r\n\r\nSo I get $ \\parallel{} P \\minus{} Q\\parallel{}: \\equal{} \\sup_{B\\subset\\mathbb{N}}|P(B) \\minus{} Q(B)| \\equal{} \\sum_{i\\ge 0}|P(i) \\minus{} Q(i)|$\r\n\r\nCould someone show me where I am wrong \r\n\r\nthx", "Solution_3": "[quote=\"TheBridgeTwo\"]Intuitively we have that the \"sup\" is reached for $ B \\equal{} \\mathbb{N}$ otherwise you could add some i not in B and have a bigger \"sup\" if P(i) is not equal to Q(i).[/quote]\r\nI think, this is where you may be wrong!\r\n\r\nFirst, we observe that $ \\sum_{i: P(i) > Q(i)}\\left(P(i) \\minus{} Q(i)\\right)\\equal{}\\sum_{i: P(i) < Q(i)}\\left(Q(i) \\minus{} P(i)\\right)$\r\n\r\nTherefore, $ \\sum_{k\\ge 0}|P(k)\\minus{}Q(k)|\\equal{}\\sum_{i: P(i) > Q(i)}\\left(P(i) \\minus{} Q(i)\\right)\\plus{}\\sum_{i: P(i) < Q(i)}\\left(Q(i) \\minus{} P(i)\\right)$\r\n\r\n$ \\equal{}2. \\sum_{i: P(i) > Q(i)}\\left(P(i) \\minus{} Q(i)\\right)$\r\n\r\n$ \\equal{}2.\\sup_{B\\subset\\mathbb{N}_{0}}|P(B) \\minus{} Q(B)|$\r\n\r\n$ \\equal{}2.\\parallel{} P \\minus{} Q\\parallel{}$\r\n\r\nAm I right alekk?", "Solution_4": "yes, this is the idea. But you might want to justify a little bit more carefully this step:\r\n\\[ \\ldots \\equal{} \\sum_{i: P(i) > Q(i)}\\left(P(i) \\minus{} Q(i)\\right) \\equal{} \\sup_{B\\subset\\mathbb{N}_{0}}|P(B) \\minus{} Q(B)| \\]\r\nwhich is essentially the most important one :)", "Solution_5": "[quote=\"TheBridgeTwo\"]Could someone show me where I am wrong[/quote] More specifically, $ |P({\\bf N}) \\minus{} Q({\\bf N})| \\equal{} |1 \\minus{} 1| \\equal{} 0$ is certainly not the supremum of a sometimes-positive function :). (You are treating $ |P({\\bf N}) \\minus{} Q({\\bf N})|$ as if it were a sum of absolute values, but in fact it is the absolute value of a sum.)", "Solution_6": "thx JBL,Alekk, and limsup\r\n\r\nFor the point underlined by alekk \r\n\r\nlet B an event and A the set defined in alekk's post \r\n\r\nthen we have \r\n\r\n$ |P(B) \\minus{} Q(B)| \\equal{} |\\sum_{i\\in B}P(i) \\minus{} Q(i)|$\r\n\r\n$ \\equal{} Max(\\sum_{i\\in B\\cap A}P(i) \\minus{} Q(i) \\minus{} \\sum_{i\\in B\\cap A^c}P(i) \\minus{} Q(i), \\minus{} [\\sum_{i\\in B\\cap A}P(i) \\minus{} Q(i) \\minus{} \\sum_{i\\in B\\cap A^c}P(i) \\minus{} Q(i)])$\r\n\r\n$ \\le Max(\\sum_{i\\in A}P(i) \\minus{} Q(i),\\sum_{i\\in A^c}Q(i) \\minus{} P(i)) \\equal{} \\sum_{i \\in A}P(i) \\minus{} Q(i) \\equal{} |P(A) \\minus{} Q(A)|$ \r\n\r\nbecause we have $ \\sum_{i\\in B\\cap A}P(i) \\minus{} Q(i) \\equal{} \\sum_{i\\in B\\cap A^c}Q(i) \\minus{} P(i)$ which can be seen from the fact that \r\n\r\n$ 1 \\equal{} \\sum_{i\\ge 0}P(i) \\equal{} \\sum_{i\\in B\\cap A}P(i) \\plus{} \\sum_{i\\in B\\cap A^c}P(i) \\equal{} \\sum_{i\\in B\\cap A}Q(i) \\plus{} \\sum_{i\\in B\\cap A^c}Q(i) \\equal{} \\sum_{i\\ge 0}Q(i) \\equal{} 1$\r\n\r\nSo the sup is indeed reached with $ B \\equal{} A$.", "Solution_7": "I think your arguments are true only when $ B \\equal{} \\mathbb{N}_{0}$.", "Solution_8": "Would you please elaborate limsup I don't see what's wrong with my proof \r\n\r\nthx", "Solution_9": "[quote=\"TheBridgeTwo\"]$ 1 \\equal{} \\sum_{i\\ge 0}P(i) \\equal{} \\sum_{i\\in B\\cap A}P(i) \\plus{} \\sum_{i\\in B\\cap A^c}P(i) \\equal{} \\sum_{i\\in B\\cap A}Q(i) \\plus{} \\sum_{i\\in B\\cap A^c}Q(i) \\equal{} \\sum_{i\\ge 0}Q(i) \\equal{} 1$[/quote]\r\nHow is this true unless $ B\\equal{}\\mathbb{N}_{0}$?", "Solution_10": "I cannot edit the post so here is the corection :\r\n\r\n[quote]\nbecause we have \n\n$ \\sum_{i\\in A}P(i) \\minus{} Q(i) \\equal{} \\sum_{i\\in A^{c}}Q(i) \\minus{} P(i)$ which can be seen from the fact that \n\n\n$ 1 \\equal{} \\sum_{i\\ge 0}P(i) \\equal{} \\sum_{i\\in A}P(i) \\plus{} \\sum_{i\\in A^{c}}P(i) \\equal{} \\sum_{i\\in A}Q(i) \\plus{} \\sum_{i\\in A^{c}}Q(i) \\equal{} \\sum_{i\\ge 0}Q(i) \\equal{} 1$\n\n[/quote]\r\n\r\nThis was used in the equality :\r\n\r\n$ Max(\\sum_{i\\in A}P(i) \\minus{} Q(i),\\sum_{i\\in A^{c}}Q(i) \\minus{} P(i)) \\equal{} \\sum_{i\\in A}P(i) \\minus{} Q(i)$" } { "Tag": [ "analytic geometry", "topology", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Decide whether the following subset of $ \\ell^2$ is compact:\r\n$ A: \\equal{}\\{a\\in\\ell^2|\\ \\ |a_k|\\leq\\frac{1}{k}\\mbox{ for all }k\\}$\r\n\r\nDirect from the definition (every open covering has a finite subcovering), or using completeness+precompactness seems not convenient, so I expect I'll need to use \"every sequence in A has a converging subsequence\". After some playing with it, I suspect it is compact (i.e. I didn't manage to come up with a sequence without converging subsequence). \r\n\r\nTake a sequence $ (a_n)_n$ in A, i.e. the n-th term is a sequence $ (a_n^k)_k$ with $ |a_n^k|\\leq\\frac{1}{k}$. I need to construct a converging subsequence... This is so general, that I don't know how to begin. \r\nAny help? Thanks in advance!", "Solution_1": "Use the metric properties to your advantage and construct a Cauchy sequence. Since the product of finitely many intervals $ [\\minus{}\\frac1k,\\frac1k]$ is compact, we can find a subsequence for which the first $ n$ coordinates converge. Then the remaining coordinates contribute at most $ \\frac1n$ to the squared distance between points, and we can go far enough out so that the squared distances are at most $ \\frac2n$. Pick some elements, then apply this same idea to that bounded subsequence with a larger $ n$. Repeating infinitely, this constructs a Cauchy subsequence, which must converge.\r\n\r\nAlternately, this product is compact in the product topology, where convergent sequences are those that converge pointwise. Show that a sequence that converges pointwise which is dominated by an element of $ \\ell^2$ converges in $ \\ell^2$.", "Solution_2": "Thanks! That's very helpful :)\r\nI guess $ A': \\equal{} \\{a\\in\\ell^{2}|\\ \\ |a_{k}| < \\frac {1}{k}\\mbox{ for all }k\\}$ (with strict inequality) is not compact then?", "Solution_3": "It's not. A compact set does need to be closed." } { "Tag": [ "abstract algebra", "topology", "group theory", "limit", "geometry", "geometric transformation", "superior algebra" ], "Problem": "Let $ A$ and $ B$ be two Abelian groups, and define the sum of two homomorphisms $ \\eta$ and $ \\chi$ from $ A$ to $ B$ by \\[ a( \\eta\\plus{}\\chi)\\equal{}a\\eta\\plus{}a\\chi \\;\\textrm{for all}\\ \\;a \\in A\\ .\\] With this addition, the set of homomorphisms from $ A$ to $ B$ forms an Abelian group $ H$. Suppose now that $ A$ is a $ p$-group ( $ p$ a prime number). Prove that in this case $ H$ becomes a topological group under the topology defined by taking the subgroups $ p^kH \\;(k\\equal{}1,2,...)$ as a neighborhood base of $ 0$. Prove that $ H$ is complete in this topology and that every connected component of $ H$ consists of a single element. When is $ H$ compact in this topology? [L. Fuchs]", "Solution_1": "every descending chain $ U_k$ of subgroups of a group $ G$ defines the structure of a topological group with these subgroups as a neighborhood basis of $ 0$. there is only one way to define the topology and this works: a subset $ S$ is open in $ G$ if for every $ s \\in S$ there is some $ k$ such that $ s \\plus{} U_k \\subseteq S$. in particular, $ H$ is a topological group with the $ p^k H$ as a neighborhood basis of $ 0$.\r\n\r\nthe completeness states that the canonical homomorphism $ H \\to \\lim_n H / p^k H$ is an isomorphism. if $ h$ is in the kernel, and $ a \\in A$, we may choose $ k$ such that $ p^k a \\equal{} 0$. since $ h \\in p^k H$, this shows $ h(a) \\equal{} 0$. now let $ (\\overline{h_k})_k$ be an element in the projective limit, i.e. $ h_k \\in H$ such that $ h_i \\equiv h_k mod p^k H$ for all $ i \\geq k$. if $ a \\in A$, choose again $ k$ such that $ p^k a \\equal{} 0$ and define $ h(a) : \\equal{} h_k(a)$. this is well-defined since we are free to make $ k$ larger without changing $ h_k(a)$. this yields our desired preimage $ h \\in H$.", "Solution_2": "[b]- Every connected component of $ H$ is a singleton: [/b]\r\n\r\nThe connected component $ C$ of the zero element is a closed subgroup of $ H$. If we can show that $ C$ is contained in every $ p^nH$, then we are done, since then $ C\\subseteq \\cap_n p^nH\\equal{}\\{0\\}$ and the connected component of $ h\\in H$ is $ h\\plus{}C$.\r\n\r\nProof of $ C\\subseteq p^nH$: Write $ C$ as the disjoint union of cosets of the subgroup $ C_n: \\equal{}C\\cap p^nH$ of $ C$. $ p^nH$ is an open subgroup of $ H$ such that, with $ C_n$, every coset $ h\\plus{}C_n$ ($ h\\in C$) is an open subset of $ C$ (equipped with the topology inherited from $ H$). By connectedness of $ C$, just one coset is sufficient to cover $ C$. But $ C_n\\equal{}C$ is equivalent to $ C\\subseteq p^nH$.\r\n\r\n\r\n[b]- $ H$ is compact iff $ H/p^nH$ is finite for every $ n$:[/b]\r\n\r\nLet $ H$ be compact. The disjoint union of cosets of the open subgroup $ p^nH$ is an open cover of $ H$. Since this covering is disjoint, it cannot be refined such that there are only finitely many cosets, hence the index $ [H: p^nH]$ is finite. \r\nIf all quotients $ H/p^nH$ are finite, they are compact when equipped with the discrete topology. By Tychonov/Axiom of Choice, the product space $ \\prod_nH/p^nH$ is compact. $ H$ is compact as a closed subspace of the product space.", "Solution_3": "[quote=\"Third Edition\"]Since this covering is disjoint, it cannot be refined such that there are only finitely many cosets, hence the index $ [H: p^nH]$ is finite. [/quote]\r\ncould you explain this argument?", "Solution_4": "You could use the following lemma:\r\n\r\nLet $ H$ be a compact topological group and $ H'$ an open subgroup of H. Then $ [H: H']$ is finite.\r\n[hide=\"Proof:\"] Let $ (h_i)_{i\\in I}$ be a set of representatives of $ H/H'$. Then $ h_iH'$ is an open subset of $ H$ (since, for fixed $ h_0\\in H$, the translation $ h\\mapsto h_0h$ is a homeomorphism of $ H$) and we have the disjoint union \\[ H\\equal{}\\stackrel{\\cdot}{\\bigcup}_{i\\in I}h_iH'\\]Because this is an open covering of the compact space $ H$, there is a finite subcovering, i.e. there is a finite subset $ I'\\subset I$ such that \\[ H\\equal{}\\stackrel{\\cdot}{\\bigcup}_{i\\in I'}h_iH'\\]The initial covering is disjoint, hence we must have $ I\\equal{}I'$, thus $ [H: H']\\equal{}|I|\\equal{}|I'|$ is finite.[/hide]", "Solution_5": "ok thx (I knew the lemma but forgot the compactness...)." } { "Tag": [ "calculus", "integration", "trigonometry", "function", "vector", "Support", "real analysis" ], "Problem": "Let $ |\\ |\\equal{}|\\ |_{C^0}$ be the supremum norm on $ C^0$.Define an integral transformation $ T: C^0\\longrightarrow C^0$ by $ T(f)(X)\\equal{}\\int_0^xf(t)dt$\r\n$ i)$ show that $ T$ is continuous and find its norm.\r\n$ ii)$ find $ T(\\cos (nt)),\\ n\\equal{}1,2,...$\r\n$ iii)$is the set of functions $ K\\equal{}\\{T(\\cos (nt)\\equal{}n\\in \\mathbb{N}$ closed?bounded?compact?\r\n$ iiii)$ is $ T(K)$ compact?", "Solution_1": "[hide=\"Hint for (i)\"]Let $ (V,\\left\\|\\cdot\\right\\|_{V})$ and $ (W,\\left\\|\\cdot\\right\\|_{W})$ be normed vector spaces, and let $ T: V\\rightarrow W$ be a linear map. Then the following are equivalent:\n1) T is Lipschitz\n2) T is uniformly continuous\n3) T is continuous\n4) T is continuous at $ 0\\in V$\n5) there is a constant $ C<\\infty$ such that $ \\left\\|T(x)\\right\\|_{W}\\leq C\\left\\|x\\right\\|_{V}$ for all $ x\\in V$.\n\nShow one of these is true and you're done.[/hide]", "Solution_2": "Normally $ C^0$ means either the space of continuous functions with compact support on the entire real line or the space of the functions that tend to $ 0$ at $ \\infty$. In both cases $ T$ doesn't even map $ C^0$-functions to $ C^0$-functions, which means that your definition of $ C^0$ can be neither of those. So, what is $ C^0$ in your problem?", "Solution_3": "[quote=\"fedja\"]Normally $ C^0$ means either the space of continuous functions with compact support on the entire real line or the space of the functions that tend to $ 0$ at $ \\infty$. In both cases $ T$ doesn't even map $ C^0$-functions to $ C^0$-functions, which means that your definition of $ C^0$ can be neither of those. So, what is $ C^0$ in your problem?[/quote]\r\n$ |f|_{c^0}\\equal{}\\max\\{|f(t)|: t\\in [0,1]\\}$defined on the infinite-dimensional vector space $ C^0$ of continuous functions $ f: [0,1]\\rightarrow \\mathbb{R}$" } { "Tag": [ "symmetry", "combinatorics unsolved", "combinatorics" ], "Problem": "24 points are equally spaced on a circle so that the distance between any two adjacent points is\r\n1. How many different ways can we choose 8 points so that the arc length between any two chosen\r\npoints is not 3 or 8.", "Solution_1": "I get 258.\r\n\r\n$\\begin{array}{llllllll}1&4&7&10&13&16&19&22\\\\9&12&15&18&21&24&3&6\\\\17&20&23&2&5&8&11&14\\end{array}$\r\n\r\nIf we write the numbers 1 to 24 in an array where adjacent terms differ by three or eight then the problem is equivalent to finding a 'path' through the array, containing one number from each column and no two adjacent numbers from the same row (where the last and first columns are considered adjacent).\r\n\r\nIf we start with $1$, we notice that the number of partial paths ending at any given element is the sum of the number of partial paths ending at each of the two elements in the previous column but not in the same row. This allows us to create an array containing the number of paths ending at each point:\r\n\r\n$\\begin{array}{llllllll}1&0&2&2&6&10&22&42\\\\0&1&1&3&5&11&21&43\\\\0&1&1&3&5&11&21&43\\end{array}$\r\n\r\nThe $86$ paths which do not end on $22$ are all valid, and by symmetry it is easy to see that there are the same number of paths starting with $9$ and $17$, for a total of $256$.\r\n\r\nIncidentally, it is possible to derive general formulae for the entries in the second array, which would hold for larger values, and could be extended to similar problems. In particular, the entries in the top row are $\\frac{2^{n-1}-2}{3}$ for $n$ even, and $\\frac{2^{n-1}+2}{3}$ for $n$ odd, whilst the corresponding entries in the other two rows are $\\frac{2^{n-1}+1}{3}$ and $\\frac{2^{n-1}-1}{3}$ respectively. However, with such a small array in this particular problem, it is easier to compute by hand.", "Solution_2": "I have an idea a little different.\r\nFirst, draw the first table of Diarmuid. The number of paths are equal to number of that:\r\nwe have a circle which is cut into 8 sectors. We have 3 colours for them. Now the adjacent sectors must be coloured in different colours.\r\nAnd this is a classical problem of number sequence." } { "Tag": [ "group theory", "abstract algebra", "inequalities", "superior algebra", "superior algebra unsolved" ], "Problem": "Let c(G) be the number of conjugate classes for finite group G and H be a subgroup of G. Whether or not c(H)<2c(G)?", "Solution_1": "Not necessary. Take $ H \\equal{} G$.\r\n\r\n(I read the inequality too fast. I'm blind.)", "Solution_2": "Well, if H=G we do have c(H) < 2c(H), right? \r\n\r\nBut in fact, the answer to the original question is \"no\". There are finite groups G with proper subgroups H such that H has more than twice as many conjugacy classes as G. I believe the smallest counterexample is a group of size 57. In GAP notation it is SmallGroup(57, 1). It has 9 conjugacy classes but it has a subgroup with 19 conjugacy classes (EDIT: of course, this subgroup is a cyclic group of order 19).\r\n\r\nOther examples are SmallGroup(68,3), SmallGroup(75, 2) and there are others. SmallGroup(93, 1) has 13 conj classes but has an element of order 31 (and therefore, a cyclic subgroup with 31 conjugacy classes).\r\n\r\n- R" } { "Tag": [], "Problem": "Hey! i found this weird pattern but i dont know what good it is! so if someone else has ever seen this pattern EXPLAIN to me why it works! or i will [b]hunt you down and annoy you until you do.[/b]\r\n\r\nwell look at this:\r\n\r\n0^2 + 0^3=0\r\n0^2+1^3=1\r\n1^2+2^3=9=3^2\r\n3^2+3^3=6^3\r\n.......\r\n.......\r\n........\r\n\r\nhey notice how the the first and last numbers are triangle numbers? well yea the pattern continues but why does this work?", "Solution_1": "1) You've posted this in a wrong section of the forum. (I've now moved it.)\r\n\r\n2) This appears to be a consequence of the fact that $ 1 \\plus{} 2 \\plus{} \\ldots \\plus{} n \\equal{} \\frac {n(n \\plus{} 1)}{2}$ and $ 1^3 \\plus{} 2^3 \\plus{} \\ldots \\plus{} n^3 \\equal{} \\left(\\frac {n(n \\plus{} 1)}{2}\\right)^2$, but you haven't given enough information about what this pattern is for me to be sure. The two results I've quoted can be proven inductively or with a variety of cute tricks (try searching the forum for more info) -- I don't think there's any \"deep\" reason why the two formulas are so closely related.", "Solution_2": "That looks like the right result. The last line should be $ 6^2$ (clearly $ 6^3$ is far too large). See [url=http://mathworld.wolfram.com/NicomachussTheorem.html]Nichomachus' theorem[/url]. As far as I know, this result is unique among results about power sums.", "Solution_3": "look if we put letters in this we have this pattern:\r\n\r\n1^2+2^2=3^2 than\r\n3^2+3^3=6^2\r\n\r\nif you see:\r\n\r\n1 2 3\r\n3 3 6\r\n6 4 10\r\n10 5 15\r\n15 6 21\r\n\r\ntriangle numbers", "Solution_4": "That pattern is explained by exactly the identity given by JBL. Plug in a few values of $ n$ and see what happens." } { "Tag": [ "AMC", "AIME", "AIME I", "\\/closed" ], "Problem": "When I go to look at a transcript of a class, sometimes the background is dark blue instead of the creamy white color. This happens to about 1 out of 3 transcripts. Then after a while, it changes to light then to dark etc.\r\nI am using Mac OS 10.3.9 , Firefox 2.0.0.11 , I use ML skin on AoPS, but that shouldn't affect this.\r\nIs this a CSS problem?\r\nI attached a PDF that has the menu and copyright as an example.", "Solution_1": "If you refresh, do the colors correct themselves? By about once in every three times, do you mean on the same transcript or that 1 in every 3 unique transcripts have this problem?", "Solution_2": "Possibly same problem as http://www.mathlinks.ro/viewtopic.php?t=195102\r\nYour browser might be doing this to you, in which case we can't help you (we can recommend you upgrade your browser to the latest version though, or use another one instead like Firefox or Opera).", "Solution_3": "This is probably related:\r\n\r\nPart of the 2008 AIME I Math Jam transcript's background is dark. Half of the transcript has a white background (note: this is directly in the middle in terms of height), while the rest (above and below the white part) is dark.\r\n\r\nThis might involve some kind of limit on the length of the white part. It gets displayed as large as possible, and centered in terms of height.\r\n\r\nBrowser/OS: Firefox 2.0.0.13 on Ubuntu, also confirmed on Windows XP." } { "Tag": [], "Problem": "if \r\n$a+b+c=0$\r\nshow that\r\n$\\frac{1}{4}(a^{4}+b^{4}+c^{4})=\\frac{1}{8}(a^{2}+b^{2}+c^{2})^{2}$", "Solution_1": "I don't now if I'm am correct..\r\nif a+b+c=0=>\r\n(a+b+c)^2=0=> a^2+b^2+c^2=-2(ab+ac+bc)\r\nthen (a^2+b^2+c^2)^2=(-2(ab+ac+bc))^2 =>\r\na^4+b^4+c^4+2(ab)^2+2(bc)^2+2(ca)^2=4((ab)^2+(bc)^2+(ca)^2)\r\n+8abc(a+b+c)=4((ab)^2+(bc)^2+(ca)^2)\r\nthen a^4+b^4+c^4=2((ab)^2+(bc)^2+(ca)^2)\r\n=>(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(ab)^2+2(bc)^2+2(ca)^2=\r\n2(a^4+b^4+c^4) What we want!!!" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "If $f(\\dfrac{x}{x^2+x+1})=\\dfrac{x^2}{x^4+x^2+1}$ then\r\nfind $f(x)$.\r\n\r\nSaman Gharib", "Solution_1": "Take $a = \\dfrac{x}{x^2 + x + 1}$, then $\\dfrac{1}{a} - 2 = \\dfrac{x^2 + x + 1}{x} - 2 = \\dfrac{x^2 - x + 1}{x}$. Hence $\\dfrac{x}{x^2 - x + 1} = \\dfrac{a}{1 - 2a}$. Also, $\\dfrac{x^2}{x^4 + x^2 + 1} = \\dfrac{x}{x^2 + x + 1}\\cdot\\dfrac{x}{x^2 - x + 1} = a \\cdot \\dfrac{a}{1 - 2a} = \\dfrac{a^2}{1 - 2a}.$ So, the solution is $f(x) = \\dfrac{x^2}{1 - 2x}$.", "Solution_2": "thank you arne for your nice solution :)" } { "Tag": [ "symmetry" ], "Problem": "i dont know which forum this belongs on, it might be too easy for this one but how many symmetry planes does a football have if u ignore the laces and writing. also, how many symmetry planes does an oblique cylinder have? thanks.", "Solution_1": "[quote=\"led14pa2\"]i dont know which forum this belongs on, it might be too easy for this one but how many symmetry planes does a football have if u ignore the laces and writing. also, how many symmetry planes does an oblique cylinder have? thanks.[/quote]\r\n\r\nI don't know if I am understanding your question, but I would say that the answer is infinity in both cases. \r\n\r\nIn the case of a football (assuming that you are talking about american football, and not soccer), then you can take any plane that contains the line from extreme to extreme (infinity of them). Also you can consider the plane containing the equator of the ball.\r\n\r\nIn the case of an oblique cylinder, (assuming that the perpendicular projection of the upper circle in the plane containing the lower circle gives you two concentric circles) then, you can consider any plane that passes for the segment containing those two circles.", "Solution_2": "[img]http://www.gototem.com/Webgifs/ClipArt/Sports/Football.GIF[/img]", "Solution_3": "I kinda doubt you made that yourself.", "Solution_4": "[quote=\"RC-7th\"]I kinda doubt you made that yourself.[/quote]\r\n\r\nBecause of the \"Totem Graphic International\" written there?", "Solution_5": "Actually, because of the \"Totem Graphics Incorporated\" written there. :D", "Solution_6": "Psst. It's called Google Images. I was usin' this as an example.", "Solution_7": "maybe he meant a soccer ball, the one made of 12 pentagons and 20 exagons (am i right with the number of faces?), the same one as fullerene :)\r\nthis could be more interesting, maybe!", "Solution_8": "sorry for the confusion, i meant an american football, but u guys think it is infinitely many for both the american football and the oblique cylinder", "Solution_9": "led14 said ignore the writing and the laces, but I don't think there's anything there about ignoring the seams. . .", "Solution_10": "oops sorry, i meant to say to ignore the seams as well" } { "Tag": [], "Problem": "Suppose today is Tuesday. What day of the week will it be 100 days from now?", "Solution_1": "[hide=\"da answer\"]\n$\\frac{100}{7}=14 R2$\n\nThat means 14 weeks and 2 days will have passed.\n14 weeks from Tuesday is Tuesday.\n2 days after tuesday is [b]Thursday[/b][/hide]", "Solution_2": "this is problem 1 in elementary school matholympiads.", "Solution_3": "[hide]The answer is thursday because 100/7 is 14 r2, so tuesday+2 is thursday.[/hide]", "Solution_4": "[hide]Thursday because 100/7 = 14 Remainder 2 14 weeks from tuesday would be tuesday and 2 days after tuesday is thursday.[/hide]", "Solution_5": "[quote=\"MCrawford\"]Suppose today is Tuesday. What day of the week will it be 100 days from now?[/quote]\r\n\r\n[hide]$\\frac{100}{7}$ has a remainder of 2\n\nso it is Thursday[/hide]", "Solution_6": "thursday :)", "Solution_7": "I'm sorry if this sounds really stupid, but what's the point of posting if so many people before you have posted the answer with a relatively same explanation? Unless you have a very different solution, or if you have a different answer...?" } { "Tag": [ "articles", "analytic geometry", "graphing lines", "slope", "LaTeX" ], "Problem": "Hi,\r\nWhen I use the minipage environment (for side-by-side figures or side-by-side text and figure), the whole matter gets shifted into the left margin. Is this because I am specifying too much width in the minipage option (I did try reducing this but got the same result) or is there something else I need to fix? Any help is appreciated.\r\nSSN.", "Solution_1": "Sorry for bumping this post, but I thought that if I post some code it might help you understand/diagnose the problem better.\r\n\r\n[code]\n\\documentclass[12 pt]{article}\n\n\\usepackage{graphicx}\n\\usepackage{enumerate}\n\\usepackage{indentfirst}\n\\usepackage[fleqn,intlimits]{amsmath}\n\\usepackage{setspace}\n\\usepackage{hhline}\n\\usepackage[headings]{fullpage}\n\n\\begin{document}\n\n\\begin{enumerate}\n\\item For the three fracture surfaces in Figures \\ref{2}, \\ref{3} and\n\\ref{4}\n\\begin {enumerate}\n\\item 1, 2, 3 are all ductile fractures \\item 1, 2, 3 are all brittle\nfractures \\item 2 \\& 3 are intergranular fractures \\item 2 \\& 3 are\ntransgranular fractures \\item can't tell\n\\end {enumerate}\n\n\\begin{figure}[!h]\n\\begin{minipage}{3.5 in}\n\\item Which is false\n\\begin {enumerate}\n\\item if material is loaded up to a point on the curve between points 2 and\n3, necking will be observed. \\item material is not deformed\npermanently upon loading up to point 1. \\item plastic deformation\nstarts at point 2. \\item if the slope of the linear region is\ndecreased, the resistance to elastic deformation is decreased.\n\\end{enumerate}\n\\end{minipage}\n\\hfill\n\\begin{minipage}{2.5 in}\n\\begin{center}\n\\includegraphics[angle=270,width=2.5 in]{Figure5.eps}\n\\caption{Question 6} \\label{5}\n\\end{center}\n\\end{minipage}\n\\end{figure}\n\n\\end{enumerate}\n\\end{document}\n\n [/code]\r\n\r\nThe problem I have is that the minipage environment is shifted to the left by one tab space (with respect to the regular document matter), no matter how much width I specify in the environment argument. If one of you could try the code (I have posted the figure as an attachment) and see if you experience the same problem or, even better, have a solution it'd be a great help. Thanks.\r\nSSN.", "Solution_2": "The source of the problem appears to be the figure environment. If you remove it or start it just after \\begin{document} you don't get the same effect. OK, so you get \"\\caption outside float\" error but that [i]might[/i] be solved using the ccaption package." } { "Tag": [ "quadratics", "algebra unsolved", "algebra" ], "Problem": "Given any four distinct positive real numbers, show that one can choose three numbers $A,B,C$ from among them, such that all three quadratic equations \\begin{eqnarray*} Bx^2 + x + C &=& 0\\\\ Cx^2 + x + A &=& 0 \\\\ Ax^2 + x +B &=& 0 \\end{eqnarray*} have only real roots, or all three equations have only imaginary roots.", "Solution_1": "Let the four numbers be $a, b, c, d$ with $0\\dfrac{1}{4}$ and so $bd$ and $cd$ are also greater than $\\dfrac{1}{4}$. The former implies that $1-4ab$, $1-4ac$, $1-4bc$ are all nonnegative and so setting $A=a$, $B=b$, $C=c$ gives all equations real roots. The latter implies that $1-4bc$, $1-4bd$, $1-4cd$ are all negative and so setting $A=b$, $B=c$, $C=d$ gives all equations negative roots.", "Solution_2": "I may be wrong....but from my sources the INMO 1999 Question 3 reads the same as Rushil's except that the fact that the reals are positive is not given.....can this be right??", "Solution_3": "@Rijul saini:In the official question paper,it does say that the reals are positive!", "Solution_4": "INMO 1999 Question 3 was wrong since it did not mention that A,B,C were positive. it is possible only when A,B,C are positive. a counter example is -2,-1,1,2", "Solution_5": "Indeed positivity is needed. A simple approach runs like that, for the numbers being $0 1/4$ then $ad>1/4$ and we can choose $a,c,d$. If $bc > 1/4$ then $bd>1/4$ and we can choose $b,c,d$. Otherwise $ac\\leq 1/4$ and $bc \\leq 1/4$, and we can choose $a,b,c$.", "Solution_6": "The problem is a simple application of PHP\nNow, let the four numbers be $a,b,c,d$\nNow, look at the numbers of the form $1-4xy$that can be formed by the numbers a, b,c,d.\nNow, the to tall possible such numbers is $4C2=6$\nnow, these numbers can be either positive or negative.\nSo we need to divide these $6$ numbers in two intervals.Say one of theses numbers is already settled now a apply PHP on these $5$numbers.Hence, by PHP $3$ numbers fall in the same interval.Now,select any three of the numbers that fall in the same interval. \nNow, note that if three numbers of the form. $1-4ab,1-4ac,1-4ad$ cannot be selected.Now, if three of the numbers fall in the same interval then select three numbers from the other three interval,while if this interval has any other number apart from these three numbers then select that number and two of these numbers.And, we are done.:)" } { "Tag": [ "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be a group,\r\nif $ \\forall a,b \\in G (ab)^{x\\plus{}1}\\equal{}a^{x\\plus{}1}b^{x\\plus{}1}$ then $ \\forall a,b \\in G a^xb \\equal{} ba^x$", "Solution_1": "[quote=\"coma\"]Let $ G$ be a group,\nif $ \\forall a,b \\in G (ab)^{x \\plus{} 1} \\equal{} a^{x \\plus{} 1}b^{x \\plus{} 1}$ then $ \\forall a,b \\in G a^xb \\equal{} ba^x$[/quote]\r\nfor which $ x$? if for all then by using fact $ (ab)^{ \\minus{} 1} \\equal{} b^{ \\minus{} 1}a^{ \\minus{} 1}$\r\nand let $ x \\equal{} \\minus{} 2$ then we have \r\n$ (ab)^{ \\minus{} 1} \\equal{} a^{ \\minus{} 1}b^{ \\minus{} 1}$\r\nand in finnaly $ b^{ \\minus{} 1}a^{ \\minus{} 1} \\equal{} a^{ \\minus{} 1}b^{ \\minus{} 1}$ and $ ab \\equal{} ba$ which means that your group is comutative. then I can say that \r\n$ a^{x}b \\equal{} ba^{x}$", "Solution_2": "$ x$ is a fixed integer ...", "Solution_3": "What if $ G$ is the group of octonions?\r\nThen $ G$ has order 8 and its center is $ {\\minus{}1,\\plus{}1}$. If we choose $ x\\equal{}7$ it follows that $ a^7$ is in the center for all $ a\\in G$, which is not true.", "Solution_4": "That's the quaternion group- octonions aren't associative.\r\n\r\nThe example works for any non-abelian finite group; if $ n$ is the order, $ (ab)^n\\equal{}a^nb^n\\equal{}e$ but $ a^{n\\minus{}1}b\\equal{}a^{\\minus{}1}b$ and $ ba^{n\\minus{}1}\\equal{}ba^{\\minus{}1}$ are not always equal.", "Solution_5": "yes, of course jmerry. \r\nThanks for the correction.", "Solution_6": "[hide=\"hint\"][url]http://www.mathlinks.ro/Forum/viewtopic.php?p=926932#926932[/url][/hide]", "Solution_7": "That's a completely different problem. Unlike this one, it's correct." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Find all triples $ (p,q,n)$ such that\r\n$ q^{n\\plus{}2} \\equiv 3^{n\\plus{}2} \\pmod{p^n}$\r\n$ p^{n\\plus{}2} \\equiv 3^{n\\plus{}2} \\pmod {q^n}$\r\nwhere $ p,q>0$ are odd primes and $ n>1$ is an integer.", "Solution_1": "It had been already submitted in Own & Proposed Problems.", "Solution_2": "[quote=\"praneeth\"]It had been already submitted in Own & Proposed Problems.[/quote]\r\nOh :blush: ... I am really sorry, I assume problems from national olympiad will post in unsolved problems section, so I think nobody had post it before.Please delete this post together with [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=184715]this[/url] if necessary." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "Can anyone kindly go through once of the hypersurface section in the lecture notes ?\r\n\r\nhttp://www.math.ust.hk/%7Emamyan/ma204/lecture.pdf\r\np.260-262\r\n\r\nThank you.", "Solution_1": "Can you be more specific in your request?\r\n\r\nVery good lecture notes by the way! The teacher sounds very good.", "Solution_2": "I'm very confused about the definition of tangent space here. In particular, I don't understand why for the case the preimage represents the hypersurface, the tangent space is like that.\r\n\r\nBy the way, his lecture notes is worse than his lectures. :lol:" } { "Tag": [ "geometry", "logarithms", "function", "limit", "perimeter", "pigeonhole principle", "calculus" ], "Problem": "Hi. \r\nI tried to do some problems on an introduction to proofs packet given at my high school math team. When I asked the co-captain when and where I could ask about the questions if I was stumped, he said \"you can always just post aops\"... so here:\r\nCould you point me in the right direction or give me suggestions with these problems? All of my work on these problems are trivial... I'm stumped on all of them. \r\n\r\n[b]1. Prove that an irrational raised to an irrational power can be rational.[/b]\r\n\r\nI'm clueless. What does it actually mean to raise something to an irrational power? You can't express such a number with exponents and nth roots... The statement isn't even intuitive... or is it?\r\n\r\n[b]2. Prove that $ \\sqrt {2}$ is irrational. [/b]\r\n\r\nI think the question wants a more rigorous proof. All my life I've thought of an irrational number as a number that can't be expressed as the quotient of two integers. \r\n$ \\frac {a}{b} \\equal{} \\sqrt {2}$ where a and b are integers. \r\nI hoped I could find some contradiction with that statement.\r\n\r\n[b]3. Show that you cannot cover a circular disk with two circular disks of smaller diameter.[/b]\r\n\r\nSince this has to do with area, I thought of pigeon hole like that box problem in aops volume 1. I don't know how to break up a circular disk in a clever way, though.\r\n\r\n[b]4. Show that any set of n integers has some subset whose sum is divisible by n.[/b]\r\n\r\nIt works for one in the set. It works for two...", "Solution_1": "let a/b=$ \\sqrt{2}$ by definition, a and b are relatively prime integers if we assume that sqrt 2 is rational.\r\n\r\nthis means that $ a^2\\equal{}2b^2$. so a must be a multipl of 2. let a=2m.\r\n\r\nthen $ 4m^2\\equal{}2b^2\\rightarrow2m^2\\equal{}b^2$ However, this means that b must also be a multiple of two. but this is impossible because we assumed a and b were relatively prime.", "Solution_2": "1. $ \\sqrt {2}^{\\log_{\\sqrt {2}} 10} \\equal{} 10$. I'll leave it to you to prove that $ \\log_{\\sqrt {2}} 10$ is irrational.\r\n\r\nAnd these problems probably would belong in Intermediate, if not HSB.", "Solution_3": "The definition of raising something to an irrational power comes from the fact that we want exponential functions to be continuous. So, what you do to evaluate $ a^b$ is to take a sequence of rational numbers $ b_1, b_2, b_3, \\ldots$ which converge to $ b$. Then, $ a^b \\equal{} \\lim_{n\\to\\infty}a^{b_n}$. So, for example, $ \\sqrt {2}^{\\sqrt {2}}$ is what the sequence $ \\sqrt {2}^{1}, \\sqrt {2}^{1.4}, \\sqrt {2}^{1.41}, \\sqrt {2}^{1.414}, \\ldots$ converges to.\r\n\r\nFor number 3, my favorite way is to use perimeter. A smaller disk cannot cover half or more of the perimeter of the larger disk, so therefore some of the perimeter must be left uncovered by two disks. You can also consider lines perpendicular to the line connecting the centers of the smaller disks.\r\n\r\nNumber four is pigeonhole. Choose a nice collection of $ n$ sets, then try to create a set from them whose sum is divisible by $ n$.", "Solution_4": "Thank you all for responding. I'm still very confused, though.\r\n\r\nFirst, aside from the way GoColts0626 showed, are there methods to prove a number irrational or rational?\r\n\r\nWhat does it mean for $ b_1,b_2,b_3...$ to converge to b?\r\nI mean, $ b_n$ doesn't equal b las n approaches the limit if the the terms from $ b_1$ to $ b_n$ add to b. How do you know that $ \\sqrt {2}^{1}, \\sqrt {2}^{1.4}, \\sqrt {2}^{1.41}, \\sqrt {2}^{1.414}, \\ldots$ converges to $ \\sqrt {2}^{\\sqrt {2}}$?\r\n\r\nIf the perimeter of the larger circle cannot be covered, does it imply that neither can the area? Why or why not?\r\n\r\nFinally, I'm confused about number four. Why do we want to choose a collection of n sets? Aren't we looking for numbers within a single set with n integers which add to a multiple of n? How do we apply pigeonhole?", "Solution_5": "$ b_1,b_2,b_3,\\ldots$ converging to $ b$ means that for every $ \\epsilon > 0$, there exists an $ N$ such that for every $ m>N$, $ |b\\minus{}b_m|<\\epsilon$. Basically the terms of the sequence approach $ b$ and would equal $ b$ if we continued it infinitely far.\r\n$ \\sqrt{2}^1, \\sqrt{2}^{1.4}, \\sqrt{2}^{1.41},\\sqrt{2}^{1.414},\\ldots$ converges to $ \\sqrt{2}^{\\sqrt{2}}$ because $ 1,1.4,1.41,1.414,\\ldots$ is a sequence of rational numbers converging to $ \\sqrt{2}$, so our definition kicks in.\r\n\r\nIf the perimeter can't be covered, the thing can't be covered because the perimeter is part of the area.\r\n\r\nThere are $ n$ different remainders upon division by $ n$. If all of the remainders of the $ n$ sets are different, then one is $ 0$. Otherwise two are the same by pigeonhole. See if you can finish it from there.", "Solution_6": "An alternative way for 3:\r\n\r\nIt suffices to show that for any two points in a circle, there is a third point in the circle closer (or equally close to) to the center than to either of the two original points. This holds because you can always draw a diameter such that both points are on the same side (or one or both are on the diameter itself).\r\n\r\nFor #4...well this is a well-known trick, but it's nice when you see it. I'll leave you to look at Hamster's hint, because it's gratifying when you figure it out.", "Solution_7": "[quote=\"Hamster1800\"]The definition of raising something to an irrational power comes from the fact that we want exponential functions to be continuous. So, what you do to evaluate $ a^b$ is to take a sequence of rational numbers $ b_1, b_2, b_3, \\ldots$ which converge to $ b$. Then, $ a^b \\equal{} \\lim_{n\\to\\infty}a^{b_n}$. So, for example, $ \\sqrt {2}^{\\sqrt {2}}$ is what the sequence $ \\sqrt {2}^{1}, \\sqrt {2}^{1.4}, \\sqrt {2}^{1.41}, \\sqrt {2}^{1.414}, \\ldots$ converges to.[/quote]\r\n\r\nI don't like this way of defining general exponentiation (in particular, it's not at all clear how to show that a sequence of algebraic numbers converges to a number that may or may not be transcendental). It's much more useful to define\r\n\r\n$ a^b \\equal{} e^{b \\ln a}$\r\n\r\nwhere $ e^x, \\ln x$ can be defined in other ways that require calculus and are totally independent of the intuitive concept of exponentiation. In fact, this is how a calculator evaluates square roots.", "Solution_8": "For the irrationals, we can just use $ e^{\\ln N} \\equal{} N$.\r\n\r\n[hide=\"Circles\"]Construct a rectangle such that the longer sides are both tangent to both circles (which may or may not overlap), and the shorter sides are each tangent to one circle. The circles are fully enclosed in this rectangle, but the shorter dimension of the rectangle is shorter than the diameter of the original circle, so this rectangle, and thus the circles inside it, can never fully cover the square.[/hide]\n\n[hide=\"Subsets\"]Let $ b_i \\equal{} \\sum _{k\\equal{}1}^i a_k$. There are $ n$ different such sums. If, for $ j > i$, $ b_i \\equiv b_j \\pmod n$, then $ \\sum _{k\\equal{}i\\plus{}1} ^ \\equal{} b_j \\minus{} b_i \\equiv 0 \\pmod{n}$. If this is not the case, all of the sums have different remainders in $ \\pmod{n}$, and each of $ 0, 1, 2, \\ldots n\\minus{}1$ must be represented, so one of these sums is itself divisible by $ n$.[/hide]" } { "Tag": [], "Problem": "1. How do you name an alkene with more than one double bonds present?\r\n\r\ne.g. $ (CH_3)_2C\\equal{}CHCH_2CH_2C(CH_3)\\equal{}CHCH_2CH\\equal{}C(CH_3)CH\\equal{}CH_2$\r\n\r\n2. It is known that the stability of $ \\text{trans}$ isomers are greater than that of their $ \\text{cis}$ isomers counterpart due to steric strain. Is there an analogous rule with $ E,Z$ configuration?\r\n\r\n3. What's the differentce between $ \\text{steric strain}$ and $ \\text{steric crowding}$?\r\n\r\n4. What is $ \\text{hyperconjugation}$ and how does it explain the fact that more highly substituted alkenes are stabler (and also for carbocations).", "Solution_1": "[hide=\"Some tips\"]1. The purpose is to number the longest carbon chain (the one that includes the biggest number of double bonds) in order to give the lowest numbers to the double bonds. In your example the name would be\n\n[hide=\"Name of 1\"]3,7,11-trimethyldodeca-1,3,6,10-tetraene.[/hide]\n\n2. E,Z-configuration and cis-trans configurations are the same thing. However, the \"E,Z\" designation is more recent and the one in common use nowadays.\n\n3. They are the same concept.\n\n4. Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161878]here[/url] (post number 4).[/hide]", "Solution_2": "Hyperconjugation: The stabilizing effect due to orbital overlap between the carbon-carbon $ \\pi$ bond and a correctly oriented carbon-hydrogen $ \\sigma$ bond\r\n\r\nIs that a good enough definition?\r\n\r\nAlso, how does this stabilizing effect actually arise? I mean most of the sources I read simply states that it exists.\r\n\r\nI don't think $ \\text{E,Z}$ and $ \\text{trans \\minus{} cis}$ are the same convention because how would you name:\r\n\r\n$ \\text{(Z) \\minus{} 1 \\minus{} bromo \\minus{} 1 \\minus{} chloro \\minus{} 2 \\minus{} iodo \\minus{} 1 \\minus{} propene}$ using $ \\text{cis \\minus{} trans}$?", "Solution_3": "[quote=\"BanishedTraitor\"]Is that a good enough definition?[/quote]\n\nIt doesn't have to be with a carbon-hydrogen sigma bond, it can also be with a carbon-carbon sigma bond. And the stabilizing effect refers to empty or half-filled p orbitals, not pi orbitals.\n\n[quote=\"BanishedTraitor\"]Also, how does this stabilizing effect actually arise?[/quote]\n\nMost probably, it is simply an inductive effect: alkyl groups are electron donating.\n\n[quote=\"BanishedTraitor\"]I don't think E,Z and trans-cis are the same convention[/quote]\r\n\r\nZ = cis and E = trans: what is the meaning of each? They mean that the groups with highest priority (according to Cahn-Ingold-Prelog rules) are in opposite sides of the double bond (trans or E) or on the same side (Z or cis)." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "For $ a>0,\\ b>0,\\ c>0,$ prove :\r\n\r\n$ (a\\plus{}b\\plus{}c)(b\\plus{}c\\minus{}a)(c\\plus{}a\\minus{}b)\\plus{}(a\\plus{}b\\plus{}c)(c\\plus{}a\\minus{}b)(a\\plus{}b\\minus{}c)$\r\n$ \\plus{}(a\\plus{}b\\plus{}c)(a\\plus{}b\\minus{}c)(b\\plus{}c\\minus{}a)\\minus{}(a\\plus{}b\\minus{}c)(b\\plus{}c\\minus{}a)(c\\plus{}a\\minus{}b)\\leqq (a\\plus{}b)(b\\plus{}c)(c\\plus{}a)$", "Solution_1": "[quote=\"kunny\"]For $ a > 0,\\ b > 0,\\ c > 0,$ prove :\n\n$ (a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\plus{} (a \\plus{} b \\plus{} c)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)$\n$ \\plus{} (a \\plus{} b \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a) \\minus{} (a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b)\\leqq (a \\plus{} b)(b \\plus{} c)(c \\plus{} a)$[/quote]\r\n\r\n$ (a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\plus{} (a \\plus{} b \\plus{} c)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)$\r\n$ \\plus{} (a \\plus{} b \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a) \\minus{} (a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b)\\equal{}8abc\\equal{}2\\sqrt{ab}2\\sqrt{ac}2\\sqrt{bc}\\leq(a\\plus{}b)(a\\plus{}c)(b\\plus{}c)$\r\n\r\n$ a\\equal{}b\\equal{}c$", "Solution_2": "[quote=\"zaya_yc\"]\n$ (a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\plus{} (a \\plus{} b \\plus{} c)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)$\n$ \\plus{} (a \\plus{} b \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a) \\minus{} (a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\equal{} 8abc$[/quote]\r\n\r\nHow did you calulate that?", "Solution_3": "I think it will work although I haven't tried yet!\r\nUsing homogenous. we can assume that a+b+c=1\r\nThen we have:(1-2a)(1-2b)+(1-2c)(1-2a)+(1-2b)(1-2c)<=(1-c)(1-b)(1-a)+(1-2a)(1-2b)(1-2c)\r\nAfter expansion, we seem to get the solution(maybe using AM-GM 1 or 2 times more). Another way is let c=min(a,b,c) and a=c+x,b=c+y and proving the coefficient>0!In general, it's ugly inequality :lol:", "Solution_4": "[quote=\"kunny\"][quote=\"zaya_yc\"]\n$ (a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\plus{} (a \\plus{} b \\plus{} c)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)$\n$ \\plus{} (a \\plus{} b \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a) \\minus{} (a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\equal{} 8abc$[/quote]\n\nHow did you calulate that?[/quote]\r\n\r\n$ P\\equal{}\\frac{a\\plus{}b\\plus{}c}{2}$ --->\r\n\r\n $ (a \\plus{} b \\plus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\plus{} (a \\plus{} b \\plus{} c)(c \\plus{} a \\minus{} b)(a \\plus{} b \\minus{} c)\\plus{}(a \\plus{} b \\plus{} c)(a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a) \\minus{} (a \\plus{} b \\minus{} c)(b \\plus{} c \\minus{} a)(c \\plus{} a \\minus{} b) \\equal{}8p(p\\minus{}a)(p\\minus{}b)\\plus{}8p(p\\minus{}b)(p\\minus{}c)\\plus{}8p(p\\minus{}c)(p\\minus{}a)\\minus{}8(p\\minus{}a)(p\\minus{}b)(p\\minus{}c)\\equal{}$\r\n$ \\equal{}8p(p^{2}\\minus{}(a\\plus{}b)p\\plus{}ab)\\plus{}8p(p^{2}\\minus{}(b\\plus{}c)p\\plus{}bc)\\plus{}8p(p^{2}\\minus{}(a\\plus{}c)p\\plus{}ac)\\minus{}8(p^{3}\\minus{}p^{2}(a\\plus{}b\\plus{}c)\\plus{}p(ab\\plus{}ac\\plus{}bc)\\minus{}abc)\\equal{}8p(3p^{2}\\minus{}2(a\\plus{}b\\plus{}c)p\\plus{}ab\\plus{}ac\\plus{}bc)\\minus{}8(p^{3}\\minus{}2p^{3}\\plus{}p(ac\\plus{}ab\\plus{}bc)\\minus{}abc)\\equal{}8p(3p^{2}\\minus{}2p^{2}\\plus{}ab\\plus{}ac\\plus{}bc)\\minus{}8p^{3}\\minus{}8p(ab\\plus{}ac\\plus{}bc)\\plus{}8abc\\equal{}8p^{3}\\plus{}8p(ab\\plus{}ac\\plus{}bc)\\minus{}8p^{3}\\minus{}8p(ab\\plus{}ac\\plus{}bc)\\plus{}8abc\\equal{}8abc$", "Solution_5": "Thanks. Here is my solution.\r\n\r\nLet $ a\\plus{}b\\minus{}c\\equal{}x,\\ b\\plus{}c\\minus{}a\\equal{}y,\\ c\\plus{}a\\minus{}b\\equal{}z$, we have $ a\\plus{}b\\plus{}c\\equal{}x\\plus{}y\\plus{}z.$ Thus $ L.H.S\\equal{}(x\\plus{}y)(y\\plus{}z)(z\\plus{}x)\\equal{}8abc$, which follows your solution, zaya_yc. :lol:" } { "Tag": [ "function", "LaTeX", "algebra", "functional equation", "algebra solved" ], "Problem": "Find all funtions $ f: R_{+} \\rightarrow R_{+} $ such that\r\n\r\n$f(x+yf(x))=f(x)f(y) \\quad \\forall x,y \\in R_{+}$\r\n\r\nand there exist only finitely many $x \\in R_{+}$ such that $f(x)=1$", "Solution_1": "Does $R_+$ contain $0$? (It does in Romanian notation; the one without $0$ is $R_+^*$ here :))", "Solution_2": "[quote=\"grobber\"]Does $R_+$ contain $0$? (It does in Romanian notation; the one without $0$ is $R_+^*$ here :))[/quote]\r\n\r\nno, it unfortunately doesn't :) \r\n\r\nthe one including $0$ is often denoted by $R_{0}^{+}$ in italy", "Solution_3": "Assume $x_0$ is a solution to the equation $f(x)=1$. Then make $x=x_0$ in the functional equation, and get $f(x+y)=f(y),\\ \\forall y\\in\\mathbb R_+^*$ (I'll use the Romanian notation :)). This means that $f$ is a periodic function of period $x$, so $f(nx)=1,\\ \\forall n\\in \\mathbb N^*$, so there are infinitely many $x$ for which $f(x)=1$, a contradiction, so there is no solution to the eqn $f(x)=1\\ (*)$.\r\n\r\nNow assume there are $a,x>0$ s.t. $f(x+a)=f(x)\\Rightarrow f(\\frac a{f(x)})=1$, which contradicts $(*)$. This means that $f$ is injective.\r\n\r\n$f(x)f(y)=f(x+yf(x))=f(y+xf(y))$ because $f(x)f(y)$ is symmetric in $x,y$. This means that $x+yf(x)=y+xf(y)\\Rightarrow f(x)=kx+1$, and the conditions show that $k>0$. Conversely, each function of the form $f(x)=kx+1,\\ k>0$ satisfies the conditions, so this is the answer.", "Solution_4": "Bonjour (:-)) 14 07 0409h21 Saint Jean de Trzy France\r\n\r\nVous trouverez votre quation fonctionnelle dans l'Enseignement mathmatique\r\n(Revue Suisse internationale Case postale 240 CH-1211 Genve 24 Suisse)\r\nJanvier 1964 page 147", "Solution_5": "Moubi said to me that the langage of this forum is english :\r\nThe roumanian do not understand french ?\r\nWhy I have not french flag ? \r\n\r\nGrobber wrote : javascript:emoticon(':(')\r\n\"\r\nNow assume there are s.t. f(x+a)=f(x) ====> f(a/f(x))=1\"\r\nI do not understand if f(x)=0 ?", "Solution_6": "Moubi said to me that the langage of this forum is english : \r\nThe roumanian do not understand french ? \r\nWhy I have not french flag ? How copy in text the smileys in mathlinks\r\n\r\nYou will find your functional equation in the mathematical Education (Seen again(Revised) international Switzerland POB 240 CH-1211 Geneva 24 Switzerland(Swiss)) in January, 1964 p 147", "Solution_7": "[quote=\"VIDIANI\"]\nNow assume there are s.t. f(x+a)=f(x) ====> f(a/f(x))=1\"\nI do not understand if f(x)=0 ?[/quote]\r\n\r\nAs talpuz said, $R_+$ doesn't contain $0$, and $f$ takes values in $R_+$, so..", "Solution_8": "wow!\r\n\r\nthanks grobber for your answer\r\n\r\ni noticed that if the function was injective the solution would be trivial, but i couldn't prove that f was injective :)", "Solution_9": "???? 'What's the sound of one hand clapping ?'\r\nI do not understand ?\r\n\r\nIn french $R+=[0,+_infty[ $ and $R+^*=]0,+\\infty[$\r\nT+ roumanian is R+^* french\r\n\r\nThat is the international notation for $[0,+\\infty[$ and $]0,+\\infty[$ ????", "Solution_10": "Come on Vidiani! Go to the topic 'training in Latex' and the 'Training in English' one too :D :D \r\nYou may preview you post before posting by using the appropriate button (near the one to send the post). Thus, you may avoid some latex problems.\r\n\r\nPierre.", "Solution_11": "i think that $R_+$ is the international symbol for the set of POSITIVE reals, and $R_0^+$ denotes the set of NON-NEGATIVE reals (including $0$)\r\n\r\n(at least, Arthur Engel agrees with me ;) )" } { "Tag": [ "floor function", "induction", "number theory unsolved", "number theory" ], "Problem": "I hope you can help me with this:\r\n\r\nProve that for every positive integer $m$ there exists a positive integer $a$ such that $\\lfloor (1+\\sqrt{3})^{2m+1}\\rfloor= 2^{m+1}a$, where $\\lfloor . \\rfloor$ is the integer part of a number.\r\n\r\n\r\nAdriana", "Solution_1": "In other words, we want to prove that $2^{m+1}|[(1+\\sqrt{3})^{2m+1}]$. Consider the sequence $(a_{n})$ given by:\r\n$a_{0}=2, a_{1}=2$ and $a_{n+2}= 2a_{n+1}+2a_{n}$. Standard method for solving recurrence relations gives us $a_{n}=(1+\\sqrt{3})^{n}+(1-\\sqrt{3})^{n}$. If $n$ is odd then $-1 < (1-\\sqrt{3})^{n}<0$ so we must have $[(1+\\sqrt{3})^{2m+1}]=(1+\\sqrt{3})^{2m+1}+(1-\\sqrt{3})^{2m+1}$ (since $a_{n}$ is an integer, for every $n$). So, we are left with proving that $2^{m+1}|a_{2m+1}$. But that's easy. Using induction and the recurrence relation we can prove that $2^{[\\frac{n}{2}]+1}|a_{n}$ for every $n \\geq 0$.", "Solution_2": "[hide]\nConsider the sequence\n\\[\\frac{\\left( 1+\\sqrt{3}\\right)^{2m+1}+\\left( 1-\\sqrt{3}\\right)^{2m+1}}{2^{m+1}}\\]\n\\[=\\left( \\frac{1+\\sqrt{3}}{2}\\right) \\left( 2+\\sqrt{3}\\right)^{m}+\\left( \\frac{1-\\sqrt{3}}{2}\\right) \\left( 2-\\sqrt{3}\\right)^{m}\\]\n(This is so because $\\left( 1\\pm \\sqrt{3}\\right)^{2}=2\\left( 2\\pm \\sqrt{3}\\right)$.) Let this sequence be $a_{m}$. Then we have\n\\[a_{0}=1,\\ a_{1}=5,\\ a_{m+2}=4a_{m+1}-a_{m}\\]\n(the recurrence is straightforward induction and follows from the fact that $x=2+\\sqrt{3},\\ y=2-\\sqrt{3}$ implies $x+y=4,\\ xy=1$). In particular, $a_{m}$ is integer for all nonnegative $m$, so\n\\[\\left( 1+\\sqrt{3}\\right)^{2m+1}+\\left( 1-\\sqrt{3}\\right)^{2m+1}\\]\nis an integer multiple of $2^{m+1}$. But also $\\left( 1-\\sqrt{3}\\right)^{2m+1}$ lies strictly between -1 and 0, so we have\n\\[\\left\\lfloor \\left( 1+\\sqrt{3}\\right)^{2m+1}\\right\\rfloor =\\left( 1+\\sqrt{3}\\right)^{2m+1}+\\left( 1-\\sqrt{3}\\right)^{2m+1}\\]\nwhich completes the proof. (We can also show that the integer is [i]not[/i] divisible by $2^{m+2}$ since $a_{m}$ is always odd.)\n[/hide]" } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all the function $ f: \\mathbb{R}\\rightarrow \\mathbb{R}$ such that $ f(xf(y)\\plus{}f(x))\\equal{}2f(x)\\plus{}xy$", "Solution_1": "hi!\r\nfor $ x\\equal{}1$ we have $ f(f(y)\\plus{}f(1))\\equal{}2f(1)\\plus{}y$ ==> f is bijective.\r\nnow let $ f(0)\\equal{}a$ and $ f(\\minus{}1)\\equal{}b$\r\nfor $ x\\equal{}y$ we have: $ f(f(x)(x\\plus{}1))\\equal{}2f(x)\\plus{}x^{2}$ $ (1)$\r\nand for $ x\\equal{}\\minus{}1$ ==> $ a\\equal{}2b\\plus{}1$ $ (2)$\r\n\r\nsince f is surjective,so there exist $ c$ such that: $ f(c)\\equal{}0$\r\nso for $ x\\equal{}c$ and $ y\\equal{}0$ we have: $ f(ac)\\equal{}0\\equal{}f(c)$\r\nf is injective so $ ac\\equal{}c$ ==> $ a\\equal{}1$ and from $ (2)$ $ b\\equal{}0$\r\nso $ f(\\minus{}1)\\equal{}0$ and $ f(0)\\equal{}1$.\r\nfor $ y\\equal{}\\minus{}1$ we have $ f(f(x))\\equal{}2f(x)\\minus{}x$\r\nnow if we consider a sequence $ (x_{n})$ defined by: $ x_{0}\\equal{}x$ and $ f(x_{n})\\equal{}x_{n\\plus{}1}$\r\nwe have : $ x_{n\\plus{}2}\\equal{}2x_{n\\plus{}1}\\minus{}x_{n}$,then we solve the equation $ r^{2}\\minus{}2r\\plus{}1\\equal{}0$.\r\nthis equation has one solution $ r\\equal{}1$ so : $ x_{n}\\equal{}\\alpha\\plus{}\\beta*n$ and for $ n\\equal{}1$ we have :\r\n$ x_{1}\\equal{}f(x)\\equal{}x\\plus{}\\beta$ (because $ x_{0}\\equal{}x\\equal{}\\alpha$).\r\nthen we put our function in the initial f.e we conclude that $ \\beta\\equal{}1$\r\nso $ f(x)\\equal{}x\\plus{}1$.", "Solution_2": "[i]How you can get $ c \\ \\neq \\ 0$ :maybe: [/i]", "Solution_3": "if $ f(0) \\equal{} 0$ we get: $ f(f(x)) \\equal{} 2f(x)$ ==> $ f(x) \\equal{} 2x$ and that is wrong, so $ c\\neq0$ ;)", "Solution_4": "How do we know f is surjective?", "Solution_5": "f is bijective===> f is injective and surjective :maybe:", "Solution_6": "Okay, I get how $ f(f(y)\\plus{}f(1))\\equal{}2f(1)\\plus{}y$ proves its injective but not how it proves it's surjective", "Solution_7": "[quote=\"dgreenb801\"]Okay, I get how $ f(f(y) \\plus{} f(1)) \\equal{} 2f(1) \\plus{} y$ proves its injective but not how it proves it's surjective[/quote]\r\nPlugging in $ y\\minus{}2f(1)$ for $ y$ gives $ f(f(y\\minus{}2f(1))\\plus{}f(1))\\equal{}y$ for all $ y$; thus, the function can take on any value.", "Solution_8": "an other proof :\r\nwe have $ f(f(y)\\plus{}f(1))\\equal{}2f(1)\\plus{}y$ \r\nsince $ h(y)\\equal{}2f(1)\\plus{}y$ is bijective so $ f(f(y)\\plus{}f(1))$ is bijective too==> $ f$ is surjective and $ f(y)\\plus{}f(1)$ is injective.\r\nthen,since $ g(y)\\equal{}f(y)\\plus{}f(1)$ is injective so $ f(y)$ is injective too.\r\n\r\nconclusion: f is bijective.", "Solution_9": "[i]Ok , correct solution :lol: [/i]" } { "Tag": [ "induction", "calculus" ], "Problem": "Well.This problem is [i]very[/i] well known and its more like an exercise but its good to have a rigorous proof for it.\r\n\r\nProblem.Let $A$ and $B$ be two points. Let $P$ be a continous path between them with lenght $\\ell$.\r\nShow that \r\n\\[\\ell \\geq AB \\]\r\nWhere AB is the lenght of the segment.\r\n\r\nPS.Problem is easy but its not obvious.", "Solution_1": "[quote=\"lomos_lupin\"]Well.This problem is [i]very[/i] well known and its more like an exercise but its good to have a rigorous proof for it.\n\nProblem.Let $A$ and $B$ be two points. Let $P$ be a continous path between them with lenght $\\ell$.\nShow that\n\\[\\ell \\geq AB \\]\nWhere AB is the lenght of the segment.\n\nPS.Problem is easy but its not obvious.[/quote]\r\n\r\nWow...i didn't even know that this could be a [i]problem[/i]! Hmmm....rigourous.....well....\r\n\r\nAssume wlog, that $P$ is [i]\"piecewise linear\". [/i]Prove induction on the number of pieces $n$ :lol: \r\n\r\nInduction between $n$ and $n+1$: use triangle inequality...and base case=1=when $P=AB$", "Solution_2": "Yeah this problem is obvious. You just have to take into account that $P$ can be a curve. Then you can look at a curve as an infinite number of line segments and then we proceed to Chen's solution. :)", "Solution_3": "Suppose we have determined $M$ to be the shortest continuous curve with endpoints $A$ and $B$, for some unequal points $A$ and $B$. Consider the curve $M'$ with endpoints $A$ and $C$ where $C$ is any point on $M$. By hypothesis, $M'$ is homothetic to $M$. So $A, B, C$ are collinear. It follows $M$ must be a line segment.", "Solution_4": "[quote] Assume wlog, that is \"piecewise linea\", \n[/quote]\n[quote=\"rem\"]that $P$ can be a curve. Then you can look at a curve as an infinite number of line segments [/quote]\r\nThis is not rigorous by any means.\r\n\r\n\r\nBut yet There exists a rigorous calculus base solution\r\nBut its brute force. \r\nThis result was known [i]long [/i] before calculus was invented.\r\n\r\nAnother approach, Similar to kileel's idea, is that\r\n you can [i] transform [/i] your curve into a part of a cricle." } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "a, b, c and d are in N. let:\r\n\r\n(P) : a+b < c+d\r\n(Q) : (a+b) (c+d) < ab + cd\r\n(R) : (a+b) cd < ab (c+d)\r\n\r\nprove that at least one of (P) or (Q) or (R) is false.", "Solution_1": "Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!" } { "Tag": [ "inequalities", "algebra", "polynomial", "function" ], "Problem": "Ok, here I go ... :) \r\n\r\nProve that if a, b, c, d > 0 then\r\n\r\na/(b+2c+3d) + b/(c+2d+3a) + c/(d+2a+3b) + d/(a+2b+3c) >= 2/3.", "Solution_1": "(a+b+c+d)^2/4(ab+bc+cd+da+ac+bd) :ge: 2/3\r\n3 :Sigma: a^2+6 :Sigma: ab :ge: 8 :Sigma: ab\r\n(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+(a-c)^+(b-d)^2 :ge: 0", "Solution_2": "Maverick's first step follows from Cauchy !\r\nThe second step follows from maclaurin's inequalities.", "Solution_3": "What are maclaurin's inequalities?", "Solution_4": "Do you know symmetric polynomials ? If not, let me know.\r\n\r\nFor a certain positive integer n, the symmetric polynomials of n positive numbers a(1), a(2), ..., a(n) are denoted s(1), s(2), ..., s(n).\r\nFor example, if n = 3 then we have\r\na(1) = (a + b + c)/3\r\na(2) = (ab + bc + ca)/3\r\na(3) = abc\r\nNow for any n we have :\r\na(1) >= a(2)^(1/2) >= a(3)^(1/3) >= ... >= a(n)^(1/n).\r\nThis is Maclaurins inequality.\r\n\r\nTry to apply this to the above posts !\r\nIf you don't know how to do this, let me know ...", "Solution_5": "[color=cyan]Maverick, it really isn't helpful at all if you don't write out full solutions. Putting in the important steps does nothing for many of us who have less familiarity with these types of problems.[/color]", "Solution_6": "Ah, yes, I have indeed seen what you call maclaurin's inequalities before, just under a different name and have not yet had a chance to apply them.\r\n\r\nThanks!", "Solution_7": "I simply had to make this post!!\r\n\r\nYou know.. this is a very clear example of Jensen.\r\nConsider WLOG that a+b+c+d=1 and apply Jensen to the convex function f: (0,+oo)-->R, f(x)=1/x with a,b,c and d as weights.\r\n\r\ncheers! :D :D", "Solution_8": "I'll work out Lagrangia's idea ...\r\n\r\nSince the inequality is homogenous in a, b, c and d we can assume WLOG that a + b + c + d = 1. Now notice that the function f(x) = 1/x is convex and increasing on the interval ]0,+:inf:[ so we have by Jensen's inequality :\r\n\r\na f(b + 2c + 3d) + b f(c + 2d + 3a) + ... + d f(a + 2b + 3c)\r\n:ge: f(4(ab + bc + cd + da + ac + bd)) \r\n:ge: f(24(a + b + c + d)/16)\r\n= 2/3\r\n\r\nwhere 4(ab + bc + ... + bd) :ge: 24(a + b + c + d)/16 follows immediately from Maclaurin's inequalities. :)" } { "Tag": [ "calculus", "integration" ], "Problem": "Find all possible integral solutions $ (x,y,z)$ such that \r\n\r\n$ x^2\\plus{}y^2\\plus{}z^2\\plus{}2xy\\plus{}2x(z\\minus{}1) \\plus{}2y(z\\plus{}1)$ is a perfect square.", "Solution_1": "Note that $ x^2\\plus{}y^2\\plus{}z^2\\plus{}2xy\\plus{}2x(z\\minus{}1)\\plus{}2y(z\\plus{}1)\\equal{}(x\\plus{}y\\plus{}z)^2\\plus{}2(y\\minus{}x)$. If $ y>x$, $ x^2\\plus{}y^2\\plus{}z^2\\plus{}2xy\\plus{}2x(z\\minus{}1)\\plus{}2y(z\\plus{}1)\\ge(x\\plus{}y\\plus{}z\\plus{}1)^2$, we easily get contradiction. Similarly if $ y(2(a+b+c)-abc)^2\\leq 100[/tex]\r\nBy Cauchy schwart we have:\r\n[tex](2(a+b)+(2-ab)c)^2\\leq (8+(ab)^2-4ab)(9+2ab)=100+(2ab-7)(ab+2)^2[/tex](check it)\r\nBut[tex] 2ab\\leq a^2+b^2=9-c^2\\leq 6<7[/tex]\r\nAnd we have done", "Solution_10": "yeah,you are right.\r\nbut your inequality can be real numbers.\r\nBTW,nice solution.", "Solution_11": "[quote=\"zhaobin\"]yeah,you are right.\nbut your inequality can be real numbers.\nBTW,nice solution.[/quote]\r\nI don't think so...a,b,c have to be positive", "Solution_12": "[quote=\"keira_khtn\"][quote=\"zhaobin\"]yeah,you are right.\nbut your inequality can be real numbers.\nBTW,nice solution.[/quote]\nI don't think so...a,b,c have to be positive[/quote]\r\nwhy?\r\nyour solution is okay for reals", "Solution_13": "Here is my solution for the first one:The given inequal is equivalent to:\r\n[tex](a^2+b^2+c^2)(10(a^2+b^2+c^2)^{\\frac12}-6(b+c-a))\\geq 27abc[/tex].Notice:the equal occurs iff [tex]b=c=2a[/tex] then we wrote:[tex]\\frac{10}{3}\\sqrt{(a^2+b^2+c^2)(1+2^2+2^2)}-6(b+c-a)\\geq \\frac{10}{3}(a+2b+2c)-6(b+c-a)=\\frac{28}{3}a+\\frac{2}{3}b+\\frac23 c[/tex]\r\nApplying AM-Gm we obtain:[tex]a^2+4.\\frac{b^2}{4}+4.\\frac{c^2}{4}\\geq \\frac{9}{4^{\\frac89}}a^{\\frac29}b^{\\frac89}c^{\\frac89}(1)[/tex]\r\n[tex]\\frac{28}{3}a+\\frac{2}{3}b+\\frac23c=7.\\frac43a+\\frac{2}{3}b+\\frac23c\\geq 3.4^{\\frac89}a^{\\frac79}b^{\\frac19}c^{\\frac19}(2)[/tex]\r\nMultiping (1) and (2) it follows!\r\nAnd we have to have a,b,c>0 to use Am-Gm" } { "Tag": [ "calculus", "integration", "search", "number theory unsolved", "number theory" ], "Problem": "$a_{n},(n=1,2,\\dots)$ is a integeral sequence such that $a_{k+1}$ is the smallest postive integer which are different from $a_{1},a_{2},\\dots,a_{k}$ and has the prperty $(k+1)|S_{k}+a_{k+1}$.\r\nIs every positive integer a term of this sequence?\r\n[hide]I think they are.[/hide]\r\nEdit:I don't know if there is a solution.\r\nEdit:$S_{k}=a_{1}+a_{2}+\\dots+a_{k}$\r\nEdit:more,for which $a_{1}$ every positive integer is a term of this sequence.", "Solution_1": "$S_{k}= a_{1}+a_{2}+...+a_{k}$ \u00bf", "Solution_2": "Yes,it is.", "Solution_3": "I investigated the $a_{1}= 4$ case and I don't think $1$ appears anywhere in the sequence. I'm too lazy to prove it, though.", "Solution_4": "[quote=\"t0rajir0u\"]I investigated the $a_{1}= 4$ case and I don't think $1$ appears anywhere in the sequence. I'm too lazy to prove it, though.[/quote]\r\nthank you.I have investigated the case $a_{1}=1$ and I found every postive integer $k\\leq 30$ appear in the sequence.", "Solution_5": "If $a_{1}= 1$, then $a_{a_{n}}= n$ for all positive integer $n$. There are chances that this special case was already solved on the forum. See also http://mathforum.org/wagon/fall96/p818.html and http://www.research.att.com/~njas/sequences/?q=1%2C3%2C2%2C6%2C8%2C4&sort=0&fmt=0&language=english&go=Search\r\n\r\nP.S: I will be unable to answer to any replies for the next two week." } { "Tag": [ "geometry", "rectangle", "probability", "MATHCOUNTS" ], "Problem": "[hide=\"#14\"]The letters A, B, C and D represent four distinct integers from 0 through 9. When A is added to B the result is C. When B is subtracted from A the result is D. How many possible ways are there to assign values to A, B, C and D?[/hide]\n[hide=\"#24\"]There is a 3 by 4 rectangle filled by 1 by 1 squares. The top row and rightmost column of the big rectangle has stars(only one star in each square) and the 3 squares on the rightmost column are shaded.(6 stars total w/3 shaded). \n Henri stands in the square on the bottom left corner marked \"start.\" Tresa flips a fair coin. If the coin lands Heads, Henri moves one square upward: if it lands tails, henri moves one square to the right? This process is continued until henri moves into a square with a star, at which time the game ends. If this final square witht he star is shaded, Teresa wins; if it is not shaded, Henri wins. What's the probablility that Teresa wins? Express as a common fraction. (i dont know how to draw a picture, so i described it best i can. sorry if the discription sucks-it'll help if you draw the thing first)[/hide]\n[hide=\"#28\"]There is a number line of integers from -1 to 12. Jeff will pick a card at random from ten cards numbered 1 through 10. The number on this card will indicate his starting point on the number line. He will then spin a fair spinner with 3 equal sectors, one of which says \"move 1 space left\", and two which says \"move 1 space right\", and follow the instruction indicated by his spin. From this new point he will spin the spinner again and follow the resulting instruction. What is the probability that he ends up at a multiple of 3 on the number line? express as a common fraction\n<------------------------------------>\n-1 0 1 2 3 4 5 6 7 8 9 10 11 12\n\n[/hide]", "Solution_1": "#14 is easily brute forcable.. but\r\n[hide]if a-b is greater than 0 then a is greater than or equal to b, so when a=0, there is 1 choice, a=1, there are 2 choices, when a=2, 3 choices.. all the way to a=4, 5 choices, but when a=5, a+b is less than 9 so there are again 5 choices, so when a=5, 5 choices, a=6, 4 choices..a=9, 1 choice so 1+2+3+4+5+5+4+3+2+1=30, however, if a b c and d are distint, then a cannont equal b or b cannot equal 0, there are 14 sceniros when this happens, s 30-14=16 [/hide]", "Solution_2": "remember that for #28 there's two types of numbers that can possibly result ina multiple of 3\r\n\r\n-jorian", "Solution_3": "art of owna for #14, [hide]16[/hide] is correct", "Solution_4": "16 isn't correct, it is 13 as written in teh Mathcounts Book.", "Solution_5": "[quote=\"mathgeniuse^ln(x)\"]16 isn't correct, it is 13 as written in teh Mathcounts Book.[/quote]\r\n\r\nwhy dont YOU explain it then?-16 is what all my peers were telling me and what i got", "Solution_6": "ii brute forced number 28. i made it sorta like a casework problem.\r\n\r\ni supposed that you started with each of them for example, if you started on -12, there are x ways. if you start on -11, there are y ways. it sounds like it takes a long time, but it took me about 2 minutes and i got it right :D .", "Solution_7": "[quote=\"espark52\"][quote=\"mathgeniuse^ln(x)\"]16 isn't correct, it is 13 as written in teh Mathcounts Book.[/quote]\n\nwhy dont YOU explain it then?-16 is what all my peers were telling me and what i got[/quote]\r\n\r\nOkay\r\n\r\n[hide]D is at least 1, since the integers have to be distinct. Also, none of the integers can be 0, since then this would imply that two of the integers are the same.\n\nSo if D=1, then (A,B,C)=(3,2,5), (4,3,7), (5,4,9)\nD=2, then (A,B,C)=(3,1,4),(5,3,8)\nD=3, then (A,B,C)=(4,1,5), (5,2,7),\nD=4, then (A,B,C)=(5,1,6), (6,2,8),\nD=5, then (A,B,C)=(6,1,7), (7,2,9)\nD=6, then (A,B,C)=(7,1,8)\nD=7, then (A,B,C)=(8,1,9)\nAnd that is all there is.\n\nCount 'em 13.[/hide]", "Solution_8": "Mathgenius, you are right.. Not only can B not equal A or 0, B cannont be half of A because then B would equal D, so there are 17 cases-(for A and B)-(0,0 1,0 1,1 2,0 2,1 2,2 3,0 3,3 4,1 4,2 4,4 5,0 6,0 6,3 7,0 8,0 9,0) So 30-17=13.. \r\n\r\nMaybe I shouldve brute forced it..", "Solution_9": "[quote=\"espark52\"][hide=\"#24\"]There is a 3 by 4 rectangle filled by 1 by 1 squares. The top row and rightmost column of the big rectangle has stars(only one star in each square) and the 3 squares on the rightmost column are shaded.(6 stars total w/3 shaded). \n Henri stands in the square on the bottom left corner marked \"start.\" Tresa flips a fair coin. If the coin lands Heads, Henri moves one square upward: if it lands tails, henri moves one square to the right? This process is continued until henri moves into a square with a star, at which time the game ends. If this final square witht he star is shaded, Teresa wins; if it is not shaded, Henri wins. What's the probablility that Teresa wins? Express as a common fraction. (i dont know how to draw a picture, so i described it best i can. sorry if the discription sucks-it'll help if you draw the thing first)[/hide][/quote]\r\n\r\nIs this what it looks like?", "Solution_10": "[hide=\"24\"]If you assign the number of ways to get to each square in the rectangle, you get\n\n1 3 6 10\n1 2 3 4\n1 1 1 1\n\nEach way is equally likely, so there are $1+3+6+10+4+1=25$ ways to play and finish the game. Of these ways, $10+4+1=15$ ways result in Teresa's winning. So, our probability is $\\frac{15}{25}=\\frac{3}{5}$.[/hide]", "Solution_11": "[quote=\"vishalarul\"][quote=\"espark52\"][hide=\"#24\"]There is a 3 by 4 rectangle filled by 1 by 1 squares. The top row and rightmost column of the big rectangle has stars(only one star in each square) and the 3 squares on the rightmost column are shaded.(6 stars total w/3 shaded). \n Henri stands in the square on the bottom left corner marked \"start.\" Tresa flips a fair coin. If the coin lands Heads, Henri moves one square upward: if it lands tails, henri moves one square to the right? This process is continued until henri moves into a square with a star, at which time the game ends. If this final square witht he star is shaded, Teresa wins; if it is not shaded, Henri wins. What's the probablility that Teresa wins? Express as a common fraction. (i dont know how to draw a picture, so i described it best i can. sorry if the discription sucks-it'll help if you draw the thing first)[/hide][/quote]\n\nIs this what it looks like?[/quote]\r\n\r\n\r\n\r\nsorry i took so long. yea that picture is correct.\r\n(i was grounded from the computer for a couple weeks)", "Solution_12": "[quote=\"archimedes1\"][hide=\"24\"]If you assign the number of ways to get to each square in the rectangle, you get\n\n1 3 6 10\n1 2 3 4\n1 1 1 1\n\nEach way is equally likely, so there are $1+3+6+10+4+1=25$ ways to play and finish the game. Of these ways, $10+4+1=15$ ways result in Teresa's winning. So, our probability is $\\frac{15}{25}=\\frac{3}{5}$.[/hide][/quote]\nThat doesn't work because...\n[hide]Once you get to a star, you finish. So it looks like this\n\n1 2 3 0\n1 2 3 3\n1 1 1 1\n\n6 possible winning scenarios for Henri and 4 possible winning scenarios for Teresa scenarios, so it's 2/5\n [/hide]", "Solution_13": "[quote=\"Zenox\"][quote=\"archimedes1\"][hide=\"24\"]If you assign the number of ways to get to each square in the rectangle, you get\n\n1 3 6 10\n1 2 3 4\n1 1 1 1\n\nEach way is equally likely, so there are $1+3+6+10+4+1=25$ ways to play and finish the game. Of these ways, $10+4+1=15$ ways result in Teresa's winning. So, our probability is $\\frac{15}{25}=\\frac{3}{5}$.[/hide][/quote]\nThat doesn't work because...\n[hide]Once you get to a star, you finish. So it looks like this\n\n1 2 3 0\n1 2 3 3\n1 1 1 1\n\n6 possible winning scenarios for Henri and 4 possible winning scenarios for Teresa scenarios, so it's 2/5\n [/hide][/quote]\r\nOops, you're right, I missed that part.", "Solution_14": "14\r\nA>B\r\nA+B<10\r\nUsing only these rules we get \r\nb=1, a=2~8. 7 numbers\r\nb=2, a=3~7, 5 numbers and etc so that the sum is 16\r\nHowever, if A is twice B, B=C so we get rid of a=2,b=1\r\na=4,b=2 and a=6,b=4 (note that there is no more because a=8. b=4 is not possible)\r\n\r\n24\r\n1 2 3 0\r\n1 2 3 3\r\n1 1 1 1\r\n\r\nways to get to certain squares are denoted by the number on that respective square\r\nGuys remember the formula for probability? number of stuff we want/total outcomes\r\n(3+1)/(1+2+3+3+1)\r\n\r\n4/10=2/5\r\n\r\n28\r\nfirst we add up the probabilities of landing on a multiple of 3 with the cards 1-10 and divide it over 10 since each card is equally likely. ALso note that each card has exactly 1 way of getting onto a multiple of 3 --- either by going left 2 times (1/9) going right 2 times(4/9) and staying in the same place (going right then left or left then right 4/9) so its (4+1+4+4+1+4+4+1+4+4)/9*(1/10)\r\n31/90\r\n\r\n\r\n\r\nEZZZZZZ\r\n\r\nbtw how do u do that hidden content thing (help out the new member)", "Solution_15": "On solution, they say answer is 5/16. Can someone pls explain????", "Solution_16": "The probability for each of the scenarios is different, so you have to calculate the probability of ending at each square, not just the number of paths. ", "Solution_17": "@machack you see the eye crossed out? you click that and when it says the hide things put the hide=solution into the first one.", "Solution_18": "hey can someone send me the answer key? how do you find the answer key to these sprints\n" } { "Tag": [ "inequalities", "logarithms" ], "Problem": "Prove that $e^x \\geq x^e$, $x\\geq 0$", "Solution_1": "The inequality is trivial. It's equivalent with the well known lnx<=x/e, x>=0.", "Solution_2": "I love proofs!!!", "Solution_3": "A proof of that inequality is as follows.\r\n\r\n$f(x)=\\ln x$ is convex, and hence it lies below a given tangent to it at all points. Thus at $x=e$, we get\r\n\r\n$\\ln x\\leq \\frac {1}{e}(x-e)+1=\\frac {x}{e}$", "Solution_4": "I think lnx is concave.", "Solution_5": "Yes. Don't worry, that was just a typo - the rest of my post reads correctly if you replace convex with concave." } { "Tag": [ "function", "search", "floor function", "number theory", "totient function", "relatively prime" ], "Problem": "Evaluate:\n\\[ \\sum_{n=1}^{64}\\sigma (n) \\]", "Solution_1": "Whats the symbol represent?", "Solution_2": "$\\sigma(n)$ is the number of [proper] divisors of $n$. I'm not sure about the \"proper\"; check mathworld.", "Solution_3": "[quote=\"eryaman\"]Whats the symbol represent?[/quote]\n\n$\\sigma (n)$ is the sum of the divisors of number n.\n\nEDIT: I learnt this from the link in your signature, chess64. :lol:", "Solution_4": "Ah, sorry I got them mixed up.\r\n\r\nI also find that link very useful, so I put it in my sig. so people would be able to benefit from it. Unfortunately, as soon as I did, the mandelbrot people removed it from their site, and now I have to use google's cache :(", "Solution_5": "NVM, doesn't work.", "Solution_6": "[quote=\"chess64\"]$\\sigma(n)$ is the number of [proper] divisors of $n$. I'm not sure about the \"proper\"; check mathworld.[/quote]That's phi, or the totient function I believe.\r\n\r\nI think I figured out the problem. Not the actual answer, but how to do it.\r\n[hide=\"hint\"]How many numbers from 1-64 does 1 divide into? All of them, so add 64 1's=64\nHow many numbers from 1-64 does 2 divide into? 32 of them, so add 32 2's=64\nHow many numbers from 1-64 does 3 divide into? 21 of them, so ad 21 3's=63.\n\nHmmmm....So it seems that if $n|64$ then its sum will be 64, and if it doesn't....[/hide]", "Solution_7": "No, $\\phi$ is the relatively prime one.", "Solution_8": "[quote=\"eryaman\"][quote=\"chess64\"]$\\sigma(n)$ is the number of [proper] divisors of $n$. I'm not sure about the \"proper\"; check mathworld.[/quote]That's phi, or the totient function I believe.\n\nI think I figured out the problem. Not the actual answer, but how to do it.\n[hide=\"hint\"]How many numbers from 1-64 does 1 divide into? All of them, so add 64 1's=64\nHow many numbers from 1-64 does 2 divide into? 32 of them, so add 32 2's=64\nHow many numbers from 1-64 does 3 divide into? 21 of them, so ad 21 3's=63.\n\nHmmmm....So it seems that if $n|64$ then its sum will be 64, and if it doesn't....[/hide][/quote]\r\nThat's not totient function. You can use Search feature of ML/AoPS and look up what totient function means before posting a false claim.", "Solution_9": "Just to make things clear...\n$d(n)$ represents the number of divisors of a number n.\n$\\sigma (n)$ represents the sum of the divisors of a number n.\n$\\phi (n)$ represents the number of positive intergers less than n relatively prime n.", "Solution_10": "[quote=\"chess64\"]Ah, sorry I got them mixed up.\n\nI also find that link very useful, so I put it in my sig. so people would be able to benefit from it. Unfortunately, as soon as I did, the mandelbrot people removed it from their site, and now I have to use google's cache :([/quote]\n\nThanks. I printed it out for portable usage of the titles to excel. :lol:", "Solution_11": "well, i think that a proper divisor of n is every divisor except 1 and n, so we find all the numbers that are divided by 2,3,4,5,etc then subtract 63 from that because we overcount each divisor once because a proper divisor only includes divisors that are not n or 1\r\n\r\n$-63+\\sum_{k=2}^{64}\\left\\lfloor \\dfrac{64}{k}\\right\\rfloor$\r\n\r\nthis can be simplified because we know every number above 32 will just give 1 so\r\n\r\n$-63+32+\\sum_{k=2}^{32}\\left\\lfloor \\dfrac{64}{k}\\right\\rfloor$\r\n\r\ni suppose a calc from here would work since we only have to do 31 numbers" } { "Tag": [ "function", "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "h(x) = x^3 - x^2 + 2x - 4\r\n\r\nHow would I find out what \"x\" is when h(x) = 0 ?\r\nI tried factoring x^3 - x^2 + 2x - 4, but I couldn't get it simple enough that I could understand how to find what h(x) would be, if I knew what \"x\" was.\r\nI sketched the graph, but I wanted to know how to solve for h(x) exactly when knowing what \"x\" is.\r\nPlease explain your work. Please be detailed. Thank you very much.\r\n\r\nAnother function, is g(x) = x^4 - 2x^3 + x - 3. I have the same problem with this one too. Any help would be much appreicated. If there is a certain method that is applied to equations like these, please tell me. :)", "Solution_1": "[hide]for h(x) you can use Cubic Formula\nhttp://mathworld.wolfram.com/CubicFormula.html\n[/hide]", "Solution_2": "Wow, I must of spent like 10 mins trying to figure out what all that means and I still don't have a clue. Thank you felipesa for providing that information. Any help is appreciated, but I forgot to mention am not really a math wiz... if it can explain it in simpler terms (if possible) it would help a lot. As far as the Cubic Formula, I'll look more into that.", "Solution_3": "Avani?\r\n\r\nAnyway, first you set the polynomial expression equal to zero by using the transitive property. Then, it's best to try and factor it. Use rational root theorem (look it up) to look for roots of the equation. Since the equation is to the third degree, you should find 3 roots." } { "Tag": [ "logarithms" ], "Problem": "Find a positive solution to $x^{x^{2006}}=2006$\r\n\r\nPlease more experienced guys leave it for the less experienced.\r\n\r\n(I define more experienced by: you solved this in less than 10 minutes!)", "Solution_1": "[hide]\n$x^{x^2006}=2006\\Rightarrow x^{2006}*\\log x=log 2006\\Rightarrow \\log x=\\log_{x^2006} 2006$.\n$\\log x=2006\\log_x 2006\\Rightarrow \\frac{(\\log x)^2}{\\log 2006}=2006$.\n$\\Rightarrow \\log x=\\sqrt{2006*\\log{2006}}\\Rightarrow x=\\boxed{10^\\sqrt{2006\\log 2006}}$\n[/hide]", "Solution_2": "so i sorta simplified it to (not sure if it helps at all though)\r\n[hide]\n$x^{2006}ln(x)=ln(2006)$\n[/hide]\r\nbut am not able to do anymore. could u provide a hint or something.", "Solution_3": "Hmmm...I got $\\sqrt[2006]{2006}$...it checks out too...", "Solution_4": "It's x^x^2006, not x^2006 :P", "Solution_5": "[quote=\"Treething\"]It's x^x^2006, not x^2006 :P[/quote]\r\n\r\nSo, it still works out because we simplify the exponents first and so $\\sqrt[2006]{2006}^{\\sqrt[2006]{2006}^{2006}}=\\sqrt[2006]{2006}^{2006}=2006$ ;)\r\n\r\nEDIT: Oops..never mind..simplified the exponents wrong :blush: :fool:", "Solution_6": "[quote=\"Iversonfan2005\"][quote=\"Treething\"]It's x^x^2006, not x^2006 :P[/quote]\n\nSo, it still works out because we simplify the exponents first and so $\\sqrt[2006]{2006}^{\\sqrt[2006]{2006}^{2006}}=\\sqrt[2006]{2006}^{2006}=2006$ ;)\n\nEDIT: Oops..never mind..simplified the exponents wrong :blush: :fool:[/quote]\r\n\r\nNo, you did it right. If seamus meant (x^x)^2006, he would have written $(x^x)^{2006}$.\r\n$x^{x^{2006}}$ means x^(x^2006), which is what you did.", "Solution_7": "Oh okay...for a second, I thought I needed to review exponents again :!: :rotfl:", "Solution_8": "haha that seems like an obvious answer. :rotfl: \r\n\r\ni am wondering if there is a proof or some method you used to arrive at this solutoin.", "Solution_9": "$x=y^{\\frac{1}{2006}}$\r\n$y^{\\frac{y}{2006}}=2006$\r\n$y^y=2006^{2006} \\implies y=2006 \\implies x=\\sqrt[2006]{2006}$", "Solution_10": "Good one guys.\r\n\r\nMy first though was to sub the 2006 back into the tower of x's to obtain:\r\n\r\n$x^{x^{x^{x^{2006}}}}=2006$\r\n\r\nand continue doing this to get an infinate number of x's in the exponent.\r\n\r\nThis looked as if the solution to $x^{2006}=2006$ is the same as the solution to the one i posted. Subbing it in, it worked!" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $x,y,z$ be positive real numbers with $x^{2}+y^{2}+z^{2}\\geq 3$.\r\nShow\r\n\\[\\frac{x^{n}}{(1+y)(1+z)}+\\frac{y^{n}}{(1+z)(1+x)}+\\frac{z^{n}}{(1+x)(1+y)}\\geq \\frac{3}{4}\\]\r\nfor a positive integer $n \\geq 3$.", "Solution_1": "[quote=\"giancarlo\"]Let $x,y,z$ be real numbers with $x^{2}+y^{2}+z^{2}\\geq 3$.\nShow\n\\[\\frac{x^{n}}{(1+y)(1+z)}+\\frac{y^{n}}{(1+z)(1+x)}+\\frac{z^{n}}{(1+x)(1+y)}\\geq \\frac{3}{4}\\]\nfor a positive integer $n \\geq 3$.[/quote]\r\n\r\nFor $x,y,z$ real?\r\n\r\nCounter example: $x=y=z=-2, n=3$ :D", "Solution_2": "indeed, in the rush i forgot to write 'positive'. :-)", "Solution_3": "wts : $x^{n}(x+1)+y^{n}(y+1)+z^{n}(z+1) \\ge \\frac{3}{4}(x+1)(y+1)(z+1)$\r\n$\\iff 4x^{n}+4y^{n}+4z^{n}+4x^{n+1}+4y^{n+1}+4z^{n+1}\\ge 6+3(x+y+z)+3(xy+yz+zx)$\r\nIt is easy to show that $x^{n}+y^{n}+z^{n}\\ge x+y+z$, $x^{n+1}+y^{n+1}+z^{n+1}\\ge xy+yz+zx$\r\nand AM-GM, we can solve this inequality." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Please help !!", "Solution_1": "It can be proved using Bertrand's Postulate", "Solution_2": "Using Bertrand is an overkill. :P Here's how I would proceed. The proof of the hint is standard. Now for a sufficiently large number $ n$ (eliminate the first couple values) take the largest prime $ p\\leq n$. Use the hint, and proceed from here." } { "Tag": [], "Problem": "what is the value of $(1+\\frac{1}{2})\\cdot(1+\\frac{1}{3})\\cdot(1+\\frac{1}{4})\\cdot...\\cdot(1+\\frac{1}{2005})$?", "Solution_1": "The answer is $1003$", "Solution_2": "Yay! correct (no surprise). Can you show your solution now?", "Solution_3": "$P(n)=(1+\\frac{1}{2})\\cdot(1+\\frac{1}{3})\\cdot(1+\\frac{1}{4})\\cdot...\\cdot(1+\\frac{1}{n})=\\frac{3}{2}\\cdot\\frac{4}{3}\\cdot...\\cdot\\frac{n}{n-1}\\cdot\\frac{n+1}{n}=\\frac{n+1}{2}$\r\n\r\nThe answer is $P(2005)=1003$.", "Solution_4": "wow, nice job kunny, but i don't get your logic. :?", "Solution_5": "I get it now! :idea:", "Solution_6": "[quote=\"math92\"] \nwow, nice job kunny, but i don't get your logic. :? [/quote]\r\n\r\nThe trick to this problem is converting mixed numbers to improper fractions, so we cancel out execessive numbers.\r\n\r\n Doing so yields:\r\n\r\n\r\n$\\frac{3}{2} \\cdot \\frac{4}{3} \\cdot \\frac{5}{4} \\cdot \\frac{6}{5} \\cdot \\frac{7}{6} \\cdots \\frac{2005}{2004} \\cdot \\frac{2006}{2005}$\r\n\r\n Now we notice that every number cancel outs except for \"2\" and \"2006\" which gives us $\\frac{2006}{2}=\\boxed{1003}$\r\n\r\n Algebraicly this looks like:\r\n\r\n$\\frac{n+1}{n} \\cdot \\frac{n+2}{n+1} \\cdot \\frac{n+3}{n+2} \\cdot \\frac{n+4}{n+3} \\cdots \\frac{n+r-1}{n+r-2} \\cdot \\frac{n+r}{n+r-1}=\\boxed{\\frac{n+r}{n}}$" } { "Tag": [], "Problem": "A basketball team has won 50 games of 75 played. The team still has 45 games to play. How many of the remaining games must the team win in order to win 60% of all games played during the season?", "Solution_1": "$\\frac{50+x}{75+45}=\\frac{6}{10}\\iff x=22$", "Solution_2": "Essientially the same:\r\n[hide] There are going to be 75+45=120 total games played. They have to win .6*120=72 games. Therefore, they have to win 72-50=$22$ more games. [/hide]", "Solution_3": "Great replies and easy methods for solving this kind of question.\r\n\r\nThanks!" } { "Tag": [], "Problem": "Would someone who has the Art of Problem Solving: volume 1 basics book be willing to post the problems and contents of page 46?\r\n\r\nThank you very much.", "Solution_1": "You know the book is copyrighted by AoPS so that wouldn't be right.\r\n\r\nIsn't it cheap looking at the free excerpts and asking for the following pages after the excerpt ends?" } { "Tag": [], "Problem": "Jane is 25 years old. Dick is older than Jane. In $ n$ years, where $ n$ is a positive integer, Dick's age and Jane's age will both be two-digit number and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $ d$ be Dick's present age. How many ordered pairs of positive integers $ (d,n)$ are possible?", "Solution_1": "[hide=\"Answer\"]We are looking for the number of 2 digit numbers greater than 25 such that the ones digit is greater than the tens digit.\n\nFor tens digits 3 through 8, when the tens digit is k, there are 9-k digits higher than k that can be the ones digits. When the tens digit is 2, the ones digit can be 6, 7, 8, or 9.\n\nThis makes a total of 4+6+5+4+3+2+1 = 25 ordered pairs.[/hide]" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Given $ a0$, ineq becomes\r\n\r\n$ 3 a^2 t\\plus{}3 a t^2\\plus{}t^3\\minus{}3 t\\plus{}4 >0,$\r\n\r\nand $ \\Delta \\equal{} (3t^2)^2\\minus{}4(3t)(t^3\\minus{}3 t\\plus{}4)\\equal{}\\minus{}3 (t\\minus{}2)^2 t (t\\plus{}4) \\le 0,$ so ineq holds.", "Solution_2": "Why can we directly use Discriminant in this case :wink: ?", "Solution_3": "[quote=\"kunny\"]Why can we directly use Discriminant in this case :wink: ?[/quote]\r\n\r\nit is a correct solution?", "Solution_4": "The inequality is equvialent to $ (b\\minus{}a)(b^2\\plus{}ab\\plus{}a^2\\minus{}3) \\ge \\minus{}4$. If $ b^2\\plus{}ab\\plus{}a^2 \\ge 3$ we are done. Let $ b^2\\plus{}ab\\plus{}a^2 <3$. Then we have by AmGm $ (b\\minus{}a)(3\\minus{}a^2\\minus{}ab\\minus{}b^2) \\le \\frac{(3\\minus{}a^2\\minus{}ab\\minus{}b^2\\minus{}a\\plus{}b)^2}{4}$ Hence it's left to prove that $ a^2\\plus{}ab\\plus{}b^2\\plus{}a\\plus{}1 \\ge b$, which is equivalent to $ \\frac{1}{2} \\left( (a\\plus{}b)^2\\plus{}(a\\plus{}1)^2\\plus{}(b\\minus{}1)^2 \\right) \\ge 0$, which is clearly true. Btw, equality does never occur." } { "Tag": [ "ratio", "geometry", "rectangle" ], "Problem": "Find the ratio of the area of triangle $ XZV$ to the area of the rectangle $ XYZW$.\n\n[asy]draw((0,0)--(3,0)--(3,1.5)--(0,1.5)--cycle);\nlabel(\"X\",(0,0),SW);\nlabel(\"W\",(3,0),SE);\nlabel(\"Y\",(0,1.5),NW);\nlabel(\"Z\",(3,1.5),NE);\nlabel(\"2c\",(0,0)--(2,0),N);\nlabel(\"c\",(2,0)--(3,0),N);\ndraw((3,1.5)--(0,0));\ndraw((3,1.5)--(2,0));[/asy]", "Solution_1": "The area of the rectangle is $ 3c \\times x$. \r\nThe area of the triangle is $ \\frac{2c \\times x}{2}$.\r\n$ \\frac{\\frac{2c \\times x}{2}}{3c \\times x} \\equal{} \\boxed{\\frac{1}{3}}$.", "Solution_2": "hello, let $ x$ the side length of the given rectangle, then we have this area as\r\n$ 3cx$ and the area of the triangle is $ \\frac{1}{2}2cx$ and the ratio of both is $ \\frac{1}{3}$.\r\nSonnhard." } { "Tag": [ "algebra", "polynomial", "function", "conics", "parabola", "geometry", "geometric transformation" ], "Problem": "Solve the equation\r\n\r\n[tex]\\displaystyle\\sqrt{5-x}=5-x^2[/tex]\r\n\r\nwithout having to manipulate/factor an ugly quartic polynomial. (BTW, no ti89's or computer algebra systems either!)\r\n\r\nEDIT --- solve the equation for REAL x.", "Solution_1": "[hide]x^2= 5- :sqrt: (5-x)\n\nx= :pm: :sqrt:(5-:sqrt:(5-x))\n\n\n\nResubstitute it back into itself...\n\n\n\nx= :pm: :sqrt: ((5- :sqrt: (5- :sqrt: 5-.....)\n\n\n\nSquare it to get x^2=5-x\n\n\n\nSolve.... x^2+x-5=0.... (1 :pm: :sqrt: (21))/2[/hide]", "Solution_2": "Very interesting method, but you're not quite done. There could be other solutions than the ones you found (think about the quartic that forms when you square the equation... there are 4 possible solutions, not 2) AND you have to check all 4 solutions to cast out the extraneous ones. That IS a very nice method, however both of mine are different.", "Solution_3": "This problem has one of the nicest, cleverest solutions of anything I have ever seen! That is a nice method, by the way ... I wonder what happened to lose two roots.", "Solution_4": "I did it the not-so-ugly quartic polynomial way, using an interesting fact.\n\n[hide]\n\nThe first thing I noted is that +/-LHS is the inverse function of RHS. I decided to find the solutions of +/-LHS=RHS and then toss out the extraneous ones. Roughly graphing +/-LHS and RHS, you see that you have 2 nice parabolas intersecting at 4 points. Interestingly, all 4 intersection points are in separate quadrants. Knowing that inverse functions are reflections across the line y=x, I realized that the intersection points in quadrant I and quadrant III must be on the line y=x. Solving for the x-values:\n\n\n\n5-x^2=x\n\nx^2+x-5=0\n\n\n\nI won't use the quadratic formula yet, but I note that I want the positive root.\n\n\n\nNow I find the ugly quartic polynomial to be x^4-10x^2+x+20. Then I use long division to divide it by x^2+x-5 to get x^2-x-4. This time I want the negative root.\n\n\n\nUsing the quadratic formula to find the positive root of x^2+x-5=0 and the negative root of x^2-x-4, I find that:\n\n\n\nx=(-1+sqrt(21))/2 or x=(1-sqrt(17))/2\n\n[/hide]\n\n\n\nAnd then, because of my submission to the machine, I check my answer with a TI-89 .", "Solution_5": "Ragingg wrote:[hide]\nx= :pm: :sqrt: ((5- :sqrt: (5- :sqrt: 5-.....)\n\nSquare it to get x^2=5-x\n[/hide]\n\n\n\nYou lost the two solutions in this step.", "Solution_6": "The solution Yasha gave --- noticing the inverse functions --- is one of my 2 cool solutions. Good job Yasha! But my other one is far cooler in my opinion. Can anyone come up with it?", "Solution_7": "[quote=\"Yasha\"]I did it the not-so-ugly quartic polynomial way, using an interesting fact....[/quote]\r\n\r\nI don't see how that makes the problem easier, you can just square both side of the equation and factor it out, that might have been less complicated. \r\n\r\nI am working on figuring the cool solution now.", "Solution_8": "[quote=\"beta\"][quote=\"Yasha\"]I did it the not-so-ugly quartic polynomial way, using an interesting fact....[/quote]\n\nI don't see how that makes the problem easier, you can just square both side of the equation and factor it out, that might have been less complicated. \n\nI am working on figuring the cool solution now.[/quote]\r\n\r\nI don't know how you would go about factoring the quartic polynomial without knowing any more information or using a TI-89. None of the roots are rational and you wouldn't be able to factor by grouping either.", "Solution_9": "For this problem, an ugly way would be to factor the quartic polynomial with guess and check. BLEH, who would ever want to do that? Basically, a 'elegant' way would be to sidestep the necessity to factor by guessing. Yasha's way of noticing the inverse functions gave him a couple of solutions to the quartic WITHOUT having to guess at them, which is why I like that method. The other method is even cooler...", "Solution_10": "Square both side we have:\r\n\r\n[tex]5-x=(5-x^2)^2[/tex]\r\n[tex]5-x=25-10x^2+x^4[/tex]\r\n[tex]x^4-10x^2+x+20=0[/tex]\r\nWhich factors into\r\n[tex](x^2-x-4)(x^2+x-5)=0[/tex]\r\n\r\nSolve each quadratic will give the solution, however that's not elegant and is brute-force factoring.\r\n\r\n{EDIT}- Zabelman beats me by 2 minutes.", "Solution_11": "[quote=\"beta\"]Which factors into... [/quote]\r\n\r\nWhile this solution works, the reason I don't like it is... How'd you come up with that factorization? (I don't mind quadratic formula. At least the quadratic formula isn't pulled out of... er... you know where)\r\n\r\nBeta, don't take this the wrong way. That solutions is perfectly fine, and there is nothing in the world wrong with it. However, there is an extremely elegant solution that I'm waiting for.", "Solution_12": "I am not sure if this is what you are looking for:\r\n\r\nAdd 20 to both side of the equation we have\r\n[tex]20+\\sqrt{5-x}=(5-x)(5+x)[/tex]\r\nLet [tex]k=\\sqrt{5-x}[/tex]\r\n\r\n[tex]20+k=k^2(10-k^2)[/tex]\r\nSimplify it we have\r\n[tex]k^4-10k^2+k+20=0[/tex]\r\nIf we square the original equation and rearrange it\r\n[tex]x^4-10x^2+x+20=0[/tex]\r\n\r\nk=x is clearly a solution, which means [tex]x^2+x-5=0[/tex].\r\nSynthetic division give us the other quadratic.", "Solution_13": "This surprisingly seems very similar to my solution. Here you solved x=LHS to get x^2+x-5=0, whereas I equated RHS=x to get the same quadratic.", "Solution_14": "My solution is not really \"cool\" since it's pure algebraic manipulation, yours is a bit cooler. I think the coolest solution is more geometric than algebraic.", "Solution_15": "For my solution... I noticed that when I resubstituted it back into itself, I only used the +... If I used the -, would the other two solutions come out of the strange new pattern?", "Solution_16": "Since no one has posted here in a few days, I think it's time to show ya'll the awesome solution...\r\n\r\n\r\n\r\nSquare the original equation to get\r\n\r\n[tex]5^2+5(-2x^2-1)+(x^4+x)=0[/tex].\r\n\r\nThen we use quadratic formula. You may think, \"But it isn't quadratic in x!\" That is correct. But it IS quadratic in 5. So, we can use the quadratic formula to solve for 5.\r\n\r\n[tex]\\displaystyle 5=\\frac{2x^2+1\\pm\\sqrt{(-2x^2-1)^2-4(1)(x^4+x)}}{2}[/tex]\r\n\r\nwhich simplifies to\r\n\r\n[tex]\\displaystyle 5=x^2+x[/tex] or [tex]5=x^2-x+1[/tex].\r\n\r\nAll that's left now is to solve these 2 quadratic equations for x and then to check the solutions.", "Solution_17": "Wow... that is really cool. It makes the problem seem so much easier. Where did you first see this problem?", "Solution_18": "[quote=\"zabelman\"]\n[tex]5^2+5(-2x^2-1)+(x^4+x)=0[/tex].\n\nThen we use quadratic formula. You may think, \"But it isn't quadratic in x!\" That is correct. But it IS quadratic in 5. So, we can use the quadratic formula to solve for 5.\n[/quote]\r\n\r\nThis is pretty cool! \r\n\r\nIf you can get a \"clean\" answer by using the quadratic formula, it is usually factorable in a clean way. So here goes,\r\n\r\nNotice that [tex]x^4+x=x(x^3+1)=x(x+1)(x^2-x+1)[/tex]. If we let [tex]Y=5[/tex] in the above equation, then \r\n\r\n[tex]Y^2+(-2x^2-1)Y+x(x+1)(x^2-x+1)=0[/tex]. Cross multipyling, the LHS factors as:\r\n\r\n[tex](Y-x(x+1))(Y-(x^2-x+1))[/tex].", "Solution_19": "[quote]If you can get a \"clean\" answer by using the quadratic formula, it is usually factorable in a clean way.[/quote]\r\n\r\nExactly! Quadform is just a simple way to find the factors (if you are fortunate enough to see it).", "Solution_20": "Nice solution!\r\n\r\nBut what other further implications does that have exactly as a factoring tool? Does this mean that for any given expression which is nicely factorable into two factors, [b]there exists a constant[/b] for which if we define the constant to be the \"variable,\" the discriminant (of the quadratic with the constant as the variable) will simplify nicely? Seems that the answer is a definite yes... but then how can we guarantee that this constant appears somewhere in the question? Is the constant the square root of any numerical (as opposed to algebraic) perfect square, like it was in this question (5, from 25)? How do we [b]know[/b]?" } { "Tag": [ "trigonometry" ], "Problem": "Try this one, found it in an old highschool textbook! :D\r\n\r\nUse cos(3x) = cos(2x + x) to prove cos(3x) = 4cos^3x - 3cosx\r\n\r\nHint: Never forget the basics!", "Solution_1": "[hide=\"Proof\"]\\[ \\begin{gathered} \\cos 3x = \\cos (2x + x) \\hfill \\\\ = \\cos 2x\\cos x - sin2xsinx \\hfill \\\\ = (\\cos ^2 x - \\sin ^2 x)\\cos x - 2\\sin x\\cos x\\sin x \\hfill \\\\ = \\cos ^3 x - \\sin ^2 x\\cos x - 2\\sin ^2 x\\cos x \\hfill \\\\ = \\cos ^3 x - (1 - \\cos ^2 x)\\cos x - 2(1 - \\cos ^2 x)\\cos x \\hfill \\\\ = \\cos ^3 x - (\\cos x - \\cos ^3 x) - 2(\\cos x - \\cos ^3 x) \\hfill \\\\ = \\cos ^3 x + (\\cos x - \\cos ^3 x)( - 1 - 2) \\hfill \\\\ = \\cos ^3 x - 3\\cos x + 3\\cos ^3 x \\hfill \\\\ = 4\\cos ^3 x - 3\\cos x{\\kern 1pt} \\quad \\hfill \\\\ \\end{gathered} \\][/hide]", "Solution_2": "Congratulations, very nice work! Nice and neat too, I like that! :lol:", "Solution_3": "Wrong forum - try intermediate topics.", "Solution_4": "Well, let's give it a Pre-Olympiad solution :D \r\n\r\n[hide=\"Solution\"]\n$\\cos 3x = \\Re \\left( (\\cos x + i \\sin x)^3 \\right) = \\Re \\left( \\cos^3 x + 3i \\cos^2 x \\sin x - 3\\cos x \\sin^2 x - i \\sin^3 x \\right) = \\cos^3 x - 3\\cos x (1 - \\cos^2 x) = 4 \\cos^3 x - 3 \\cos x$\n\nI'd give a general solution for $\\cos nx$, but it requires transforming the real terms with an even $\\sin^k x$ power and it's rather ugly :| \n[/hide]", "Solution_5": "moved" } { "Tag": [], "Problem": "I just noticed that the Math Jam isn't happening until next week. :o Does this mean that we have to wait until then to discuss the problems?", "Solution_1": "[quote=\"calc rulz\"]I just noticed that the Math Jam isn't happening until next week. :o Does this mean that we have to wait until then to discuss the problems?[/quote] \r\n\r\nyes, see other topics", "Solution_2": "That's not necessarily true. In the past they've allowed discussion before the Math Jam if it's later than normal.\r\n\r\nEdit: nevermind, in this case it was confirmed for after the Math Jam; however, keep in mind that sometimes they do change this policy." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "The sequence has $ a_1 \\equal{} 0$ and $ a_{n \\plus{} 1} \\equal{}\\pm(a_n \\plus{} 1)$ for all $ n$. Show that the arithmetic mean of the first $ n$ terms is always at least $ \\minus{} \\frac12$.", "Solution_1": "$ a_{n\\plus{}1}\\equal{}\\sqrt{(a_n \\plus{} 1)^2}$\r\n\r\n$ a_n \\equal{} (a_{n\\plus{}1}^2 \\minus{} a_n^2 \\minus{}1)/2$\r\n\r\n$ a_1 \\plus{} a_2 \\plus{} ... \\plus{} a_n \\equal{} (a_{n\\plus{}1}^2 \\minus{} n)/2$\r\n\r\n$ M \\equal{} a_{n\\plus{}1}^2/2n \\minus{} 1/2 >\\equal{} \\minus{}1/2$" } { "Tag": [ "conics", "hyperbola" ], "Problem": "A hyperbola whose foci are (0, 3 -sqrt{3}) and (0, 3 + sqrt{3}) has a conjugate axis of length 2. Write an equation of the hyperbola in standard form.\r\n\r\nThe textbook tells me that the center (h, k) = (0, 3).\r\n\r\nWhere did that come from?\r\n\r\nI need someone to please explain STEP BY STEP how to do questions like this one.\r\n\r\nWhat about if the vertices were given and NOT the foci?\r\n\r\nHow would I solve something like that?\r\n\r\nWhat is the transverse axis?", "Solution_1": "The textbook tells me that the center (h, k) = (0, 3). \r\nWhere did that come from? \r\nThe centre of the hyperbola is the midpt of the focii and the vertices.\r\nWhat about if the vertices were given and NOT the foci? \r\nDiscussed above.\r\nIn genl focii are (+or-ae,0) or (0,+or-ae) depending on the major axis.\r\nThe hyperbola has 2 axes the major and the transverse axis .\r\n :)" } { "Tag": [ "IMO", "IMO 2007" ], "Problem": "Here are some more partial results\r\n\r\n[code]SER 77 71 77 71 -1 61 (problems 4&5)\nRUS 7 7 7 7 7 7 (problem 4)\nSWE 7 - 7 - 7 7 (problem 4)\nJAP - - 3 - 1 - (problem 3)\nTPE 7 7 1 7 1 1 (problem 5)\nNZL 7 4 7 7 7 3 (problem 4)[/code]\r\n\r\nThe minus sign represents either a 0 or an yet not coordinated paper.", "Solution_1": "[quote=\"Valentin Vornicu\"]Some current partial scores that I can see on the tables here:\n\n[code]BUL 7 7 3 3 7 7 on problem 1\nMEX 7 7 7 6 7 7 on problem 4\nPER 5 7 - - 1 6 on problem 5\nHKG 7 7 7 7 7 7 on problem 2\nHUN 6 7 - 7 7 1 on problem 2\nEST 4 - 1 4 1 1 on problem 2\nCOL 1 2 1 1 1 - on problem 2[/code][/quote]", "Solution_2": "what's 'problem x' means?\r\nand SER's score seems strange. :o", "Solution_3": "So based on these partial results was the marking scheme of problem 1 3+4 ??\r\nto mathpk: The Serbians have two problems corrected :wink:", "Solution_4": "Valentin , any more partial results ?", "Solution_5": "If you get results of Ukraine, tell please.", "Solution_6": "Can you tell me the results of [color=darkred]Kazakhstan[/color] team please?", "Solution_7": "What about Israel resulrs, please?", "Solution_8": "Valentin,\r\n\r\n are the partial results you posted in order , i.e. the first score corresponds to the first contestant?", "Solution_9": "Could you post results of the Macedonian team, please?", "Solution_10": "Marko: yes the order is the one given by the IMO site to the contestants. I've uploaded the latest table with results, check the main page for that.", "Solution_11": "Check main page? What main page?", "Solution_12": "I mean http://www.mathlinks.ro/portal.php?t=160319 \r\n\r\nOr you can do \r\n\r\nhttp://www.artofproblemsolving.com/Forum/files/result_2.htm\r\nand \r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=9914", "Solution_13": "Wait, did no one from China get full marks!?", "Solution_14": "Only 3 contestants resolved problem 6 up to now... :| \r\n\r\nI think nobody will make a perfect score this year :o", "Solution_15": "[quote=\"Nico0O\"]Only 3 contestants resolved problem 6 up to now... :| \n\nI think nobody will make a perfect score this year :o[/quote]\r\n\r\nI think nobody will get more than 37 points", "Solution_16": "ITA 6 and RUS 4 can have 37 [or more]... (also PER 1, SRB 1, UNK 4, NOR 6 and AUS 2; but we haven't got their score on problem 3 and 6 yet...)\r\n\r\nWe'll see :P", "Solution_17": "Guys did any one calculate the cutoffs approximately?", "Solution_18": "Any new results from belgian team? Can someone check the scores for P2 and P3 for BEL4-6 please?", "Solution_19": "I think the cutoff for gold will be around 27 based on the partial results given... :maybe:", "Solution_20": "[quote=\"Albanian Eagle\"]Guys did any one calculate the cutoffs approximately?[/quote]\r\n\r\nI did the following: I took the list of Joseph Myers and added for each pending result the average points for this problem. After sorting I got following estimated cutoffs:\r\n\r\nRank 43 (552/12) has 24 points.\r\nRank 131 (552/4) has 18 points.\r\nRank 261 (552/2) has 12,55 points.\r\n\r\nI don't say this will be the cutoffs, just telling you the numbers. Especially the gold/silver cutoff may be higher as in my list from 44 to 48 there are 5 contestants (RUS1, RUS5, RUS6, POL6, PRK6) with 22 points and only problem 5 pending (estimated 23,86 points).", "Solution_21": "Gold : 28\r\nSilver : 21-22 (In my statistics, 26.23% have >= 21 And 24.71% have >= 22)\r\nBronze : 13-14 (In my statistics, 50.95% have >= 13 And 48.48% have >= 14)\r\n\r\nBut it's just statistics :D !\r\n\r\nAnd the winner will certainly be RUS 4 or SRB 1 (if he answered correctly problem 6...)\r\n\r\nBut I haven't got results of China team... (except problem 6)\r\n\r\n\r\nXerus, there are 526 contestants.. not 552 (I think)", "Solution_22": "Xerus, your method will probably work to give an estimate for the bronze cutoff, however the people scoring around the Silver and Gold cutoffs are better than average and so will probably higher than what you predict. So not only those 5 but people you predict will be on 20 can easily score higher than 25 (and from the Chinese team only one result was known so you can easily assume you underestimate their scores). Moreover I count only 522 contestants as not all teams have 6 members.", "Solution_23": "[quote=\"Nico0O\"]\n\nAnd the winner will certainly be RUS 4 or SRB 1 (if he answered correctly problem 6...)\n\n[/quote]\r\n\r\nI think the winner will be Peter from Germany with 36 points.", "Solution_24": "Perhaps...\r\nBut if Konstantin Matveev (RUS 4) have 7 on problem 5, he will have 37 :| .", "Solution_25": "Can anyone send me the complete results for the macedonian team", "Solution_26": "[quote=\"Nico0O\"]\nAnd the winner will certainly be RUS 4 or SRB 1 [/quote]\r\nSRB1 has 30 points,final score", "Solution_27": "[quote=\"Valentin Vornicu\"]and \nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=9914[/quote]\r\n\r\nWhat does P mean at problem 3 mostly?", "Solution_28": "[quote=\"KMN\"]\nWhat does P mean at problem 3 mostly?\n[/quote]\r\n\r\nI think it means [b]P[/b]ending", "Solution_29": "There is a rumour, The chinese team has 2 silver.", "Solution_30": "Does that mean cutoffs are settled?", "Solution_31": "Can someone please post the final results of India? Eagerly awaiting some news here.", "Solution_32": "Results of the russian team is out:\r\n \t\r\n\r\n Russia RUS4\r\n\t7 \t7 \t2 \t7 \t7 \t7 \t37 \tGold Medal \r\n\r\n \tRussia \tRUS3 \t\r\n\t7 \t7 \t5 \t7 \t7 \t1 \t34 \tGold Medal \r\n\r\n\tRussia \tRUS2 \t\r\n\t7 \t7 \t2 \t7 \t7 \t1 \t31 \tGold Medal \r\n\r\n\tRussia \tRUS1 \t\r\n\t7 \t7 \t1 \t7 \t7 \t0 \t29 \tGold Medal \r\n\r\n \tRussia \tRUS5 \t\r\n\t7 \t7 \t1 \t7 \t7 \t0 \t29 \tGold Medal \r\n\r\n\tRussia \tRUS6 \t\r\n\t7 \t7 \t1 \t7 \t2 \t0 \t24 \tSilver Medal", "Solution_33": "[quote=\"color\"]Results of the russian team is out:\n \t\n\n Russia RUS4\n\t7 \t7 \t2 \t7 \t7 \t7 \t37 \tGold Medal \n\n \tRussia \tRUS3 \t\n\t7 \t7 \t5 \t7 \t7 \t1 \t34 \tGold Medal \n\n\tRussia \tRUS2 \t\n\t7 \t7 \t2 \t7 \t7 \t1 \t31 \tGold Medal \n\n\tRussia \tRUS1 \t\n\t7 \t7 \t1 \t7 \t7 \t0 \t29 \tGold Medal \n\n \tRussia \tRUS5 \t\n\t7 \t7 \t1 \t7 \t7 \t0 \t29 \tGold Medal \n\n\tRussia \tRUS6 \t\n\t7 \t7 \t1 \t7 \t2 \t0 \t24 \tSilver Medal[/quote]\r\n\r\nWell, I see that [b]1 point from problem 3[/b] is really the gold :) A very good experience is that ... writing everything you can, for even problems you can not solve :D That sometimes give you the Gold medal, not the Silver :lol:", "Solution_34": "Yes those cut-offs can be really cruel if you end up just below it :(", "Solution_35": "are the scores available anywhere?", "Solution_36": "TPE = 149 (total team scores) :) \r\n\r\nCan someone please post the final results of TPE (each contestant's score)? \r\nThank you very much in advance.\r\n :)", "Solution_37": "Are the complete scores out as yet? If not could someone do us all a favour and put them up. Also, if that too is not possible - could you tell me the results of the India team?\r\nReally awaiting results here - results that don't seem likely to come soon :(\r\nAshwath", "Solution_38": "final result sheet is available as below: :) \r\nhttp://www.imo2007.edu.vn/statics/result/result%20imo2007.htm", "Solution_39": "ohhhhhh...", "Solution_40": "Matching [url=http://www.imo2007.edu.vn/statics/result/result%20imo2007.htm]names[/url] with [url=http://www.mathlinks.ro/viewtopic.php?mode=attach&id=9914]results[/url] I am quite surprised, not only by the defeat of the Chinese team but especially the individual scores are partially unexpected. The previously mentioned fact that problem 3 and 6 are quite different in difficulty level than the rest of the problems played an important factor whether or not to get a gold medal, i.e. carefulness on problems 1/2/5/6 avoiding easy gaps/mistakes and hoping for partial credit on the hard problems. People successfully implementing this strategy got a gold medal but others who could have solved harder problems than 1/2/5/6 but not yet problem 3 or 6 might have missed out the gold medal due to levity." } { "Tag": [ "MATHCOUNTS" ], "Problem": "in secter OAB, where O is the center of the circle and A and B are on the ends of the circle making it a quarter circle. the radius of the circle is 3 and there is a circle inscribed in the sector tangent to all three sides. find the radius of the smaller circle.", "Solution_1": "i remeber this from a few years ago\n\ni was stuck for days then BINGO i found out\n\nhere is the solution\n\n[hide]you see a 1/4 circle draw the circle inside mark the center x lets have perpendicular lines drawn at x to the two straight sides. lets define r as the radius of circle. we the diangoal line through x that cuts the 1/4 circle in half is 3 (radius of big circle). and we know that trainlge formed from the lines of a base and the diagonal through x is a 45-45-90 triangle so 3=r+r*sqrt(2) so 3/(1+sqrt(2)) and you get -3+3sqrt(2) i believe[/hide]", "Solution_2": "yep, did that today in science" } { "Tag": [], "Problem": "Does anybody know the term for the moment when you say something that your brain is thinking, but you don't want to say out loud? I forgot it... I think it started with an F? and ended with Lapse?\r\nlike F------ Lapse?", "Solution_1": "Perhaps what you're looking for is \"Freudian slip\" - named for Sigmund Freud, of course. And pronounce that \"eu\" as it's pronounced in German, like the English \"oy\". But the second word of that phrase is always \"slip,\" not \"lapse.\"" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "[color=darkblue]Let $ x>1, y>1, z > 1$ and $ \\frac {1}{x} \\plus{} \\frac {1}{y} \\plus{} \\frac {1}{z} \\equal{} 2$. Prove that $ \\sqrt {x \\plus{} y \\plus{} z} \\ge \\sqrt {x \\minus{} 1} \\plus{} \\sqrt {y \\minus{} 1} \\plus{} \\sqrt {z \\minus{} 1}$[/color]", "Solution_1": "$ \\left(\\sqrt{\\frac{x\\minus{}1}{x}\\plus{}{\\frac{y\\minus{}1}{y}}\\plus{}{\\frac{z\\minus{}1}{z}}}\\right) (\\sqrt{x\\plus{}y\\plus{}z} ) \\ge (\\sqrt{x\\minus{}1}\\plus{}\\sqrt{y\\minus{}1}\\plus{}\\sqrt{z\\minus{}1})$\r\n but \r\n$ \\left({\\frac{x\\minus{}1}{x}\\plus{}{\\frac{y\\minus{}1}{y}}\\plus{}{\\frac{z\\minus{}1}{z}}}\\right)\\equal{}3\\minus{}2\\equal{}1$", "Solution_2": "[color=darkblue]It's nice, thank!:lol:[/color]", "Solution_3": "I think this problem is in Iran Olimpiad.\r\nIt is killed by B.C.S", "Solution_4": "Yes, this is an inequality from Iran Mathematical National Olympiad 1998. It's has been solved similarly in my colection of inequality \"Inequaltiy: elementary and advenced techniques\"" } { "Tag": [], "Problem": "From what I have heard, you can only find Quant jobs on Wall Street and in London. Is this true? Are there any jobs for quant available in other cities like Chicago(which has North America's second largest \"Bank Market\") and Toronto(which has North America's third largest \"Bank Market\")? \r\n\r\nAlso, if one cannot find a job as a quant, can he/she work as something else like a financial manger in a bank? \r\nThanks for your help.", "Solution_1": "i believe jane street has branches in chicago and tokyo.", "Solution_2": "There is at least one rather large investment bank here in Minnesota, too.", "Solution_3": "by the way, houston has a decent amount of quants working at energy companies that run var on energy trades and price out the optionality of power plants... although it is not as prestigious as wall st, etc they still pay pretty well and require a good head on your shoulders.", "Solution_4": "[quote=\"satyam\"]From what I have heard, you can only find Quant jobs on Wall Street and in London. Is this true? Are there any jobs for quant available in other cities like Chicago(which has North America's second largest \"Bank Market\") and Toronto(which has North America's third largest \"Bank Market\")? \n\nAlso, if one cannot find a job as a quant, can he/she work as something else like a financial manger in a bank? \nThanks for your help.[/quote]\r\n\r\nThere are quant jobs all over, but most in the US are in NYC. \r\n\r\nFinancial manager in a bank is miles away from being a quant. If you are trained for quantitative work in general, there are lots of fields for you to consider if being a quant in a financial firm doesn't work out for you.", "Solution_5": "[quote]Financial manager in a bank is miles away from being a quant. If you are trained for quantitative work in general, there are lots of fields for you to consider if being a quant in a financial firm doesn't work out for you.[/quote]\r\n\r\nrrusczyk, can you please give me some examples of what kind of fields I will be able to work in financial industry if I am not hired as a quant.\r\n\r\nThanks.", "Solution_6": "The largest is consulting. Others include government work (NASA, NSA, etc., all hire lots of quants), startups of various kinds (particularly if you know how to program), or starting your own company. Many biotech companies now require serious modeling as well, though that may require more education. Speaking of education, academia is also an option.", "Solution_7": "NYC & London - by far the most jobs, and funds\r\nGreenwich - 2nd...but basically NYC\r\nChicago - the big C and a spin off, BofA took some of their quants back to NYC\r\nDallas - about 4 funds, one of them grew 200% this past year\r\nHouston - as mentioned, tons of energy work there (NO State tax for Dallas and Houston...)\r\nSan Fran - decent amount of work, one of the super funds built a branch there, 5 other funds I know of\r\nMinnesota - there is even a couple funds there!", "Solution_8": "[quote]Speaking of education, academia is also an option.[/quote]\r\n\r\nPositions in academia are more selective and require more \r\ntime to pursue, than Wall St type quant. I don't know what could possibly be meant by \"academia is also an option\" for a question like this. \r\nIf academia is an option then so is quant but not the other way around.", "Solution_9": "I work as a hedge fund quant in NYC, and despite the fact that I love my job, I am a bit frustrated by the lack of quant jobs outside the New York area. They say NYC is the capital of the United States for finance jobs in general. This holds even more true for quant jobs, IMO." } { "Tag": [ "function", "trigonometry", "complex numbers" ], "Problem": "An complex # takes form of z= x+iy where x is Re and iy is Im.\r\n\r\nThis can be written in the form (cos theta + i sin theta) = z = 1 as polar coordinates.\r\n\r\n\r\nmy problem is how would you write z = 4+5i in polar coordinates. [/img]", "Solution_1": "[quote=\"kdog3682\"]This can be written in the form (cos theta + i sin theta) = z[/quote]\r\n\r\nSomewhat incomplete. The form you actually want is\r\n\r\n$z = r (\\cos \\theta+i \\sin \\theta)$\r\n\r\nWhere $r = |z|$, the \"length\" of $z$. Plotting a complex number $z = x+yi$ as the point $(x, y)$, the complex numbers of the form $\\cos \\theta+i \\sin \\theta$ correspond only to the points on the unit circle $x^{2}+y^{2}= 1$. \r\n\r\nHaving said that,\r\n\r\n[b]To convert from rectangular to polar:[/b]\r\n\r\n$r = \\sqrt{x^{2}+y^{2}}$\r\n$\\theta = \\arctan \\frac{y}{x}$ (be careful about quadrants!)\r\n\r\n[b]To convert from polar to rectangular:[/b]\r\n\r\n$x = r \\cos \\theta$\r\n$y = r \\sin \\theta$", "Solution_2": "For $z = 4+5i$ we have that $x = 4$ and $y = 5$, and so\r\n\r\n$r = \\sqrt{4^{2}+5^{2}}= \\sqrt{41}$ and $\\theta = \\arctan(\\frac{5}{4}) = 51.3^{o}$.\r\n\r\nSo: $z = \\sqrt{41}[\\cos(51.3^{o})+i\\sin(51.3^{o})]$.\r\n\r\nNote that the angle $\\theta$ must belong to the first quadrant because x > 0 and y > 0." } { "Tag": [ "topology", "vector", "group theory", "abstract algebra", "invariant", "advanced fields", "advanced fields unsolved" ], "Problem": "Let G be a smooth manifold which is also a group.\r\nMoreover suppose the group structure and the differentiable structure are compatible, that is, the multiplication and inversion maps are smooth.\r\nThen G is called a smooth group.\r\nSuppose p: G' -> G is a covering map.\r\nThen the question is Under which conditions is G' also a smooth group?\r\nIf I am not wrong when G' is the universal covering of G then G' can be equipped with a smooth group structure.\r\nHowever, what if G' is not the universal covering of G?\r\nThank you in advance for your help buddy...", "Solution_1": "The usual name for that structure is a Lie group. We should also assume that $ G$ is connected.\r\n\r\nLet's use those tools, particularly the Lie algebra. Given any vector $ x$ at the identity of $ G$, we can move it to $ a$ by multiplying by $ a$; this gives a vector field $ X$ on $ G$, which we take to be an element of the Lie algebra $ H$. This vector field can be naturally pulled back to $ G'$.\r\nThis also gives us the exponential map $ \\exp$ from $ H$ to $ G$. We can pull that back to a map $ \\exp'$ from $ H$ to $ G'$: the path $ \\exp(tX)$ in $ G$ can be pulled back to a path $ \\exp'(tX)$ in $ G'$, and we just pick out the individual points on that path.\r\nWhat can we do with that? We can follow the flow of a (complete) vector field, and use that to construct (left) multiplication by $ \\exp(X)$ or $ \\exp'(X)$. That isn't necessarily everything, but it does contain a neighborhood of the identity, and thus generates everything. This is automatically associative and includes inverses $ \\exp(X)^{\\minus{}1}\\equal{}\\exp(\\minus{}X)$. \r\n\r\nSo, the answer is yes. $ G'$ is also a Lie group, and we can construct its multiplication from the multiplication on $ G$, together with the covering map.", "Solution_2": "[quote=\"jmerry\"]The usual name for that structure is a Lie group. We should also assume that $ G$ is connected.\n\nLet's use those tools, particularly the Lie algebra. Given any vector $ x$ at the identity of $ G$, we can move it to $ a$ by multiplying by $ a$; this gives a vector field $ X$ on $ G$, which we take to be an element of the Lie algebra $ H$. This vector field can be naturally pulled back to $ G'$.\nThis also gives us the exponential map $ \\exp$ from $ H$ to $ G$. We can pull that back to a map $ \\exp'$ from $ H$ to $ G'$: the path $ \\exp(tX)$ in $ G$ can be pulled back to a path $ \\exp'(tX)$ in $ G'$, and we just pick out the individual points on that path.\nWhat can we do with that? We can follow the flow of a (complete) vector field, and use that to construct (left) multiplication by $ \\exp(X)$ or $ \\exp'(X)$. That isn't necessarily everything, but it does contain a neighborhood of the identity, and thus generates everything. This is automatically associative and includes inverses $ \\exp(X)^{ \\minus{} 1} \\equal{} \\exp( \\minus{} X)$. \n\nSo, the answer is yes. $ G'$ is also a Lie group, and we can construct its multiplication from the multiplication on $ G$, together with the covering map.[/quote]\r\n\r\nThank you jmerry. \r\nI have one more question Why do you need G to be connected?\r\nWhere does your argument fail if G is not connected?\r\nThank you in advance", "Solution_3": "The subgroup generated by the image of the exponential is the connected component of the identity. If that isn't everything, there's no canonical extension to other components by this method. Now, that's an open, closed, normal subgroup; if we quotient by it, we get a discrete group, which just gets duplicated some number of times. We can extend the group structure there, but how we do it isn't canonical.\r\n\r\nI also note that your musings about universal coverings assume connectedness, as do all invocations of covering spaces I've ever seen.", "Solution_4": "Here is another approach using the universal covering and some more natural identifications around covering mappings in the category of connected pathwise connected topological groups which is larger than the category of Lie groups. If $ G$ is additionally assumed to carry a structure of a Lie group, then the mappings which will be derived in the larger category automatically descend to smooth mappings.\r\n\r\nThis approach has the advantage to globally define the topological group structure on $ G'$, whereas the method using the image of the exponential map uses the differential structure and the gap of being well-defined needs to be filled. The latter problem needs to be tackled if the image of exp is not surjective.\r\n\r\n\r\nSo let $ G$ be a connected topological group and $ p: G'\\to G$ a covering map. Then let $ \\pi': \\widehat{G}\\to G'$ be a universal covering of $ G'$. Note that $ \\pi: \\equal{} p\\circ \\pi': \\widehat{G}\\to G$ is a universal covering of $ G$. Picking an element $ e_{\\widehat{G}}\\in \\pi^{ \\minus{} 1}(\\{e_G\\})$ we can lift the group structure of $ G$ to a unique structure on $ \\widehat{G}$ as a topological groups with identity element $ e_{\\widehat{G}}$ such that $ \\pi$ is a group homomorphism (monodromy theorem).\r\n\r\nNow we want to push down the group structure of $ \\widehat{G}$ to $ G'$ via $ \\pi'$: For this we use the [b]natural[/b] group isomorphism $ \\Phi_G: \\pi_0(G)\\cong ker(\\pi)$ between the fundamental group of $ G$ and the kernel of $ \\pi$: where $ \\Phi_G([\\alpha]) \\equal{} \\widehat{\\alpha}(1)$ with the lift $ \\widehat{\\alpha}$ w.r.t. $ \\pi$ of a representing element of $ [\\alpha]\\in\\pi_0(G)$. This gives the idea of how to construct the group structure of $ G'$ by stepping through the following road map:\r\n[list]1 - $ p$ induces an injection $ \\pi_0(G')\\hookrightarrow\\pi_0(G)$ providing a subgroup $ H$ of $ ker(\\pi)$. \n\n2 - $ H$ is a discrete subgroup of $ \\widehat{G}$, hence it is a normal subgroup such that the abstract quotient group $ \\widehat{G}/H$ carries a unique structure of a topological space such that the quotient map $ q_H: \\widehat{G}\\to \\widehat{G}/H$ is a covering map.\n\n3 - The group structure of $ \\widehat{G}/H$ is automatically continuous such that $ \\widehat{G}/H$ is a topological group and $ q_H$ is a group homomorphism of topological groups. \n\n4 - $ \\pi'$ is $ H$-invariant, i.e. $ \\pi'\\circ L_h \\equal{} \\pi'$ for all $ h\\in H$, thus $ \\pi'$ factors through $ \\pi'': \\widehat{G}/H\\to G'$.\n\n5 - $ \\pi''$ is a covering mapping such that $ \\pi_0(\\pi''): \\pi_0(\\widehat{G}/H)\\to \\pi_0(G')$ is an isomorphism (use $ \\pi_0(\\widehat{G}/H)\\cong H$). This implies that every fibre of $ \\pi''$ has exactly one element, thus $ \\pi''$ is a homeomorphism.\n\n6 - Define the group structure on $ G'$ such that $ \\pi''$ is an isomorphism of topological groups.\n\n7 - Finally show that the group structures are compatible with the covering mappings, i.e. the covering mappings are group homomorphisms.[/list]" } { "Tag": [ "ratio", "calculus", "integration", "number theory", "prime factorization" ], "Problem": "(1)For the number 420, what is the ratio of the number of positive odd integral factors to the number of positive even integral integral.\r\n\r\nI listed out all the factors, but it takes up a lot of time. Is there a way to do this using the prime factorization?", "Solution_1": "That's basically what he did. \r\n\r\nHint: Odd times odd is even; even times even is even; odd times even is even.", "Solution_2": "Taking the prime factorization as nick42 did, $ 420 \\equal{} 2^2\\cdot{3}\\cdot{5}\\cdot{7}$. The even factors of 420 must have a two as one of there factors. So they either have to use $ 2^1$ or $ 2^2$(Thats two possibilities). The odd factors all use $ 2^0$ (1 possibilty). So the answer is $ \\frac {1}{2}$.\r\n\r\nDoes that make sense?", "Solution_3": "positive even factors need to have at least one 2.\r\n\r\ntherefore, the number of positive even factors of 420 is number of factors of $ 2*3*5*7$ which is 16.\r\n\r\nand the total number of factors of 420 is $ 2^2*3*5*7\\equal{}24$\r\n\r\nhence, the answer is $ \\frac{8}{16}$\r\n\r\nanswer: $ \\frac{1}{2}$", "Solution_4": "[quote=\"isabella2296\"]That's basically what he did. \n\nHint: [b]Odd times odd is even[/b]; even times even is even; odd times even is even.[/quote]\r\n\r\nEnough said.", "Solution_5": "Fail.\r\n\r\nMeant odd times odd is odd.", "Solution_6": "[quote=\"FantasyLover\"]positive even factors need to have at least one 2.\n\ntherefore, the number of positive even factors of 420 is number of factors of $ 2*3*5*7$ which is 16.\n\nand the total number of factors of 420 is $ 2^2*3*5*7 \\equal{} 24$\n\nhence, the answer is $ \\frac {8}{16}$\n\nanswer: $ \\frac {1}{2}$[/quote]\r\n\r\nWhen you find the even factors of 210, you still have to list them out, right? What I was thinking is using the prime factorization $ 2^2\\times3\\times5\\times7$ and multiply all the odds with odds and evens with evens and odds. So I did $ 3$, $ 5$, $ 7$ amongst themselves and found out there are 7 odds ($ 3, 5, 7, 15, 35, 21, 105$). 420 has $ 24$ factors, so the number of even factors must be $ 24 \\minus{} 7 \\equal{} 17$, and $ \\frac {7}{17}$ doesn't equal $ \\frac {1}{2}$.\r\n\r\nEDIT!!!::: Wait....for the odd factors I forgot $ 1$!!!, so that must mean: $ \\frac {7 \\plus{} 1}{17 \\minus{} 1} \\equal{} \\frac {8}{16} \\equal{} \\boxed{\\frac12}$!!!\r\n\r\nI don't know id this is what you did FantasyLover....I kinda didn't get what you were saying. Yay....\r\n\r\n\r\nYay is for me since I solved the problem...not because I didn't get your post. yay...", "Solution_7": "420 has 8 odd factors: 1, 3, 5, 7, 15, 21, 35, 105\r\n\r\nEDIT: sry im not that good a teacher...", "Solution_8": "[quote=\"FantasyLover\"]420 has 8 odd factors: 1, 3, 5, 7, 15, 21, 35, 105\n\nEDIT: sry im not that good a teacher...[/quote]\r\n\r\nyeah....when I was doing it from the prime factorization, i forgot to include $ 1$, that was my mistake. But thanks anyway!!", "Solution_9": "You don't even need to find the number of odd and even factors. The answer is always $ \\frac{1}{n}$ where $ n$ is the power of the 2 in the number's prime factorization", "Solution_10": "[quote=\"tinytim\"]You don't even need to find the number of odd and even factors. The answer is always $ \\frac {1}{n}$ where $ n$ is the power of the 2 in the number's prime factorization[/quote]\r\n\r\nVery helpful. BTW: You copied Yongi's avatar, right?", "Solution_11": "Sorry to revive this, but tinytim why is that trick of yours true?", "Solution_12": "Because if it is 2^n, the factors are odd for 2^0, and even for 2^1,2^2...2^n. Thus for every one odd factor, there are n even ones." } { "Tag": [ "function" ], "Problem": "$ P(x)\\plus{}P(2x)\\equal{}5x^2\\minus{}18$\r\n$ P(4)\\equal{}?$", "Solution_1": "[hide=\"Hint\"]\nSubstitute values of $ x$ until you get $ P(4)$ on a side of the equation and then solve for it. Usually, $ 0$ is a good choice to plug in first.\n[/hide]", "Solution_2": "i know, solve it then.", "Solution_3": "Note that we must have $ \\deg P\\ge2$. Assume that $ \\deg P\\equal{}n>2$ such that $ P(x)\\equal{}a_nx^n\\plus{}a_{n\\minus{}1}x^{n\\minus{}1}\\plus{}\\cdots$ so that in order for\r\n\r\n$ \\deg P(x)\\plus{}\\deg P(2x)\\equal{}2$, we must have $ a_n\\plus{}a_n\\cdot2^n\\equal{}0$, and since $ a_n\\neq0$, $ 1\\plus{}2^n\\equal{}0$, which is clearly impossible.\r\n\r\nThus, we can only have $ \\deg P\\equal{}2\\implies P(x) \\equal{} ax^2 \\plus{} bx \\plus{} c$ for constants $ a,b,c$, so that\r\n\r\n$ P(x) \\plus{} P(2x) \\equal{} 5ax^2 \\plus{} 3bx \\plus{} 2c \\equal{} 5x^2 \\minus{} 18$\r\n\r\n$ \\implies(a,b,c) \\equal{} (1,0, \\minus{} 9)\\implies P(x) \\equal{} x^2 \\minus{} 9\\implies P(4) \\equal{} \\boxed7$." } { "Tag": [ "linear algebra", "matrix", "vector", "linear algebra unsolved" ], "Problem": "Let $A \\in R^{k \\times n}$ be a matrix of full rank $k7308$\r\n$ p^2 \\geq 7225$\r\n$ p \\geq 85$\r\n\r\nSince\r\n$ 85 \\geq p \\geq 85$\r\n$ p\\equal{}85$\r\n\r\nRepeat with m.\r\n$ p\\equal{}85,m\\equal{}83,m\\plus{}p\\equal{}\\boxed{168}$", "Solution_8": "a + u * u = 7308 (1)\na * a + u = 6974 (2)\n\n(1) - (2) we got:\n\na + u * u - a * a - u = 334\na - u + (u - a) ( u + a) = 334\n(u-a) ( u + a - 1) == 334 = 167 * 2\n\nso u + a - 1 = 167, thus the answer is 168", "Solution_9": "LATEX:\nLet Great Aunt Minnie's age be $ m$.\nLet Great Uncle Paul's age be $ p$.\n$$m + p^2 = 7308$$\n$$m^2 + p = 6974$$\nSubtracting the 2 equations gets:\n$$m + p^2 - m^2 -p=334$$\n$$m - p+(p-m)(p+m) = 334$$\n$$(m-p)(p+m-1) =334 =167*2$$\nso $m+p-1 = 167$, thus the answer is $\\boxed {168}$" } { "Tag": [ "induction" ], "Problem": "$ a_1\\equal{}1$ and for every integer $ n\\geq 2$, $ |a_n| \\equal{} |a_{n\\minus{}1}\\plus{}2|$ is given. Find the minimum value of $ \\sum_{i\\equal{}1}^{2000}a_i$.\r\n\r\nA) -4000 B) -3000 C) -2000 D) -1000 E) None", "Solution_1": "[hide=\"Turkey Sandwich\"]$ |a_2| = |1 + 2| = 3$\n\n$ a_2 = - 3$\n\n$ |a_3| = | - 3 + 2| = | - 1| = 1$\n\n$ a_3 = - 1$\n\n$ |a_4| = | - 1 + 2| = 1$\n\n$ a_4 = - 1$\n\n.\n.\n.\n\nSince we want the minimum, we wanted to make as many of them negative. We see that $ a_i = - 1$, for $ i \\geq 3$. Therefore,\n\n$ \\displaystyle\\sum_{i = 1}^{2000} a_i = 1 - 3 + ( - 1) \\cdot (1998) = \\boxed{C ) \\& - 2000}$[/hide]", "Solution_2": "The way $ 1,3,\\minus{}5,\\minus{}3,\\minus{}1,\\minus{}1,...$ also gives the same result. So taking the negative is not a winning strategy. I think you need some proof.", "Solution_3": "I got \r\n$ \\sum_{i\\equal{}1}^{n}a_{i} \\ge\\frac{\\minus{}n}{2}$\r\nso the correct answer is $ D$.\r\n[hide=\"Hint\"]Square and sum it till $ n$ terms[/hide]", "Solution_4": "[hide]\nLet $ b_n \\equal{} a_n \\plus{} 1$. Then the sequence becomes $ b_1 \\equal{} 2$ and for $ n\\geq 2$, $ |b_n \\minus{} 1| \\equal{} |b_{n \\minus{} 1} \\plus{} 1|$. In particular, given $ b_k$, the possible values for $ b_{k \\plus{} 1}$ are $ \\{ \\minus{} b_k,b_k \\plus{} 2\\}$.\n\n[b]Lemma:[/b] $ \\sum_{k \\equal{} 1}^n b_k\\geq 0$ for all integers $ n\\geq 1$.\n[b]Proof.[/b] The base case, $ n \\equal{} 1$, is clear. Also, for $ n \\equal{} 2$, the sum is either $ 2 \\plus{} 4$ or $ 2 \\plus{} ( \\minus{} 2)$, both of which are nonnegative.\n\nNow, for $ n\\geq 3$, if all elements are nonnegative, we are done, so consider the first negative element $ b_k$. Then $ b_{k \\minus{} 1}$ is positive, so $ b_k \\equal{} \\minus{} b_{k \\minus{} 1}$ and $ b_{k \\minus{} 1} \\plus{} b_k \\equal{} 0$. Therefore we can form a smaller sequence by excluding $ b_k$ and $ b_{k \\minus{} 1}$ and the sum is nonnegative by the induction hypothesis. Also, $ b_{k \\minus{} 2}\\in\\{ \\minus{} b_k \\minus{} 2,b_k\\}$ and $ b_{k \\plus{} 1}\\in\\{b_k \\plus{} 2, \\minus{} b_k\\}$, so this new sequence still follows the rules defined by the sequence.\n\nNow, $ \\sum_{k \\equal{} 1}^n a_k \\equal{} \\left(\\sum_{k \\equal{} 1}^n b_k\\right) \\minus{} n\\geq \\minus{} n$, and so for 2000 elements the lower bound is $ \\minus{} 2000$. We can easily show that this can be achieved by either modularmarc101's or xeroxia's examples, hence the answer is $ \\textbf {(C)}$.[/hide]", "Solution_5": "[quote=\"xeroxia\"]$ a_1 \\equal{} 1$ and for every integer $ n\\geq 2$, $ |a_n| \\equal{} |a_{n \\minus{} 1} \\plus{} 2|$ is given. Find the minimum value of $ \\sum_{i \\equal{} 1}^{2000}a_i$.\n\nA) -4000 B) -3000 C) -2000 D) -1000 E) None[/quote]\r\n\r\nPlz tell which is the correct answer,mine or Yongi's?", "Solution_6": "Two other users already showed how to get a sum of -2000, so the minimum value of the sum can't possibly be -1000 :roll:" } { "Tag": [ "calculus" ], "Problem": "Anyone know of an AoPS-like calculus book that starts at the introductory level? I'm starting calculus soon and it'd be really nice to have a good book, not just joe textbook.", "Solution_1": "I really liked Schaum's outline of Beginning Calculus for its clarity. I didn't always get the notation part of it (like $\\frac{d}{dx}$) at first, but besides that it's one of the best books i've used (probably next to AoPS and AaCoPS). Preferably you should have someone you can at least ask for help but you'll be fine either way.", "Solution_2": "I personally like the book \"A First Course in Calculus\" by Serge Lang, Springer publications.", "Solution_3": "CALCULUS, by Michael Spivak.", "Solution_4": "I like the books written by Anton or Stewart.\r\nThey are not like AoPS style, but Anton does a great job of providing proof on why stuff works and Stewart had great practice problems." } { "Tag": [], "Problem": "hi i'm new to this forum so i'm not quite sure if set problems belong to the intermediate section\r\n\r\nit's not excatly an problem but more or less a concept question\r\n\r\nis it possible to prove the principle of inclusion/exclusion for four intersecting sets?\r\n\r\ni understand how it is proven with three sets and my txtbook says knowing the principle for three sets means you automatically should know the proof for the principle of 4 sets, without much explanation, therefore i am lost... :( \r\n\r\nany help is appreciated, thx :)", "Solution_1": "yes it is, but i forgot and i'm too lazy to derive it :|", "Solution_2": "You can just prove it for $n$ sets with a counting argument.", "Solution_3": "You can prove it for the general case if you can prove the identity \\[ n-\\binom{n}{2}+\\binom{n}{3}-\\binom{n}{4}+\\cdots+(-1)^{r-1}\\binom{n}{n}= 1\\]", "Solution_4": "[quote=\"Fraxia\"]You can prove it for the general case if you can prove the identity \\[n-\\binom{n}{2}+\\binom{n}{3}-\\binom{n}{4}+\\cdots+(-1)^{r-1}\\binom{n}{n}= 1\\] [/quote]\r\nUse Newton's binomial formula: $0=(1+(-1))^{n}=\\sum_{i=0}^{n}\\binom{n}{i}(-1)^{i}$.", "Solution_5": "Treething, stop spamming and be quiet.", "Solution_6": "[quote=\"Fraxia\"]You can prove it for the general case if you can prove the identity \\[n-\\binom{n}{2}+\\binom{n}{3}-\\binom{n}{4}+\\cdots+(-1)^{r-1}\\binom{n}{n}= 1\\] [/quote]\r\n\r\ni'm sorry but what exactly does that mean? :huh: \r\n\r\nwhat does n stand for?\r\n\r\nand what is happening to the n and the number inside the brackets? is that a fraction...", "Solution_7": "[quote=\"princevegetam\"][quote=\"Fraxia\"]You can prove it for the general case if you can prove the identity \\[n-\\binom{n}{2}+\\binom{n}{3}-\\binom{n}{4}+\\cdots+(-1)^{r-1}\\binom{n}{n}= 1\\] [/quote]\n\ni'm sorry but what exactly does that mean? :huh: \n\nwhat does n stand for?\n\nand what is happening to the n and the number inside the brackets? is that a fraction...[/quote]\r\n\r\nIn this case, $n$ is any positive integer. The thing in the parentheses, for example $\\binom{n}{4}$ means the number of ways to choose 4 things out of a set of $n$. For example, if I have $n$ colored balls, each a different color, and I want to pick out 4 of the balls, $\\binom{n}{4}$ is the number of different ways I can do this. There are mathematical ways to compute this.\r\nAs to your original question, you can prove the principle of inclusion-exclusion for 4 sets by using the one for 3 sets on the first 3 sets you have, then the one for 2 sets. Like this:\r\n$a \\cup b \\cup c \\cup d = (a \\cup b \\cup c) \\cup d$\r\nHope that helps.", "Solution_8": "oh, thanx alot :)", "Solution_9": "To find the total number of distinct elements among any number of sets, you alternate between adding and subtracting the intersections.\r\n\r\nThe total number of elements equals:\r\n\r\n$(1)$ the sum of the number of elements in each individual set\r\n\r\n$(2)$ minus the sum of the number of elements in each pair of sets\r\n\r\n$(3)$ plus the sum of the number of elements in each set of three sets\r\n\r\netc.\r\n\r\nTo prove that this works, you want to show that each element, no matter how many sets its part of, is counted exactly once in the above expression.\r\n\r\nFor an element in $n$ sets, in $(1)$, it's counted $n$ times. In $(2)$, it's subtracted $ \\binom{n}{2}$ times. In $(3)$, it's added back in $ \\binom{n}{3}$ times etc... You can prove that this expression equals one by plugging in $x=1, y=-1$ in the expansion of $(x+y)^{n}$ with the Binomial Theorem." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "[b]Problem[/b]. Let $ a,b,c$ be non-negative real numbers and $ 0 \\le k \\le \\frac {1}{2}$.Prove that\r\n\r\n$ \\prod (a \\plus{} k\\sqrt {bc}) \\ge \\prod ((k \\minus{} 1)a \\plus{} b \\plus{} c)$\r\n\r\n :)\r\n\r\nI think this problem is true and It's old problem but I donn't know author :)", "Solution_1": "[quote=\"vuthanhtu_hd\"][b]Problem[/b]. Let $ a,b,c$ be non-negative real numbers and $ 0 \\le k \\le \\frac {1}{2}$.Prove that\n\n$ \\prod (a \\plus{} k\\sqrt {bc}) \\ge \\prod ((k \\minus{} 1)a \\plus{} b \\plus{} c)$\n\n :)\n\nI think this problem is true and It's old problem but I donn't know author :)[/quote]\r\n\r\n\r\n$ a \\equal{} x^2,b \\equal{} y^2,c \\equal{} z^2$\r\n\r\n$ \\prod (a \\plus{} k\\sqrt {bc}) \\ge \\prod ((k \\minus{} 1)a \\plus{} b \\plus{} c)$\r\n\r\n$ \\Longleftrightarrow f(k) \\equal{} ( \\minus{} x^2y^4 \\plus{} xzy^4 \\plus{} 3y^2z^2x^2 \\plus{} yzx^4 \\minus{} z^4x^2 \\minus{} x^4z^2 \\minus{} y^4z^2$ $ \\minus{} z^4y^2 \\minus{} x^4y^2 \\plus{} xyz^4)k^2 \\plus{}$\r\n\r\n$ (z^4y^2 \\plus{} y^3x^3 \\plus{} y^4z^2 \\plus{} z^4x^2 \\plus{} x^3z^3 \\minus{} y^6 \\plus{} x^4y^2 \\plus{} x^4z^2 \\minus{} 6y^2z^2x^2 \\plus{} y^3z^3 \\plus{}$ $ x^2y^4 \\minus{} z^6 \\minus{} x^6)k$ \r\n\r\n$ \\minus{} z^4x^2 \\plus{} x^6 \\plus{} y^6 \\plus{} z^6 \\minus{} y^4z^2 \\minus{} x^2y^4 \\minus{} x^4y^2 \\minus{} z^4y^2 \\plus{} 3y^2z^2x^2 \\minus{} x^4z^2\\geq 0$\r\n\r\nNote\r\n\r\n$ \\minus{} x^2y^4 \\plus{} xzy^4 \\plus{} 3y^2z^2x^2 \\plus{} yzx^4 \\minus{} z^4x^2 \\minus{} x^4z^2 \\minus{} y^4z^2$ $ \\minus{} z^4y^2 \\minus{} x^4y^2 \\plus{} xyz^4$ \r\n\r\n$ \\equal{} \\minus{} (x \\minus{} y)^2(y \\minus{} z)^2(z \\minus{} x)^2 \\minus{} 2\\sum(y^2z^2(x \\minus{} y)(x \\minus{} z))$ $ \\minus{} xyz\\sum(x(x \\minus{} y)(x \\minus{} z)) \\minus{} xyz\\sum((y \\plus{} z)(x \\minus{} y)(x \\minus{} z))\\le 0$\r\n\r\nwe only prove $ f(0)\\geq 0$ and $ f(\\frac {1}{2})\\geq 0$\r\n\r\n$ f(0) \\equal{} \\minus{} z^4x^2 \\plus{} x^6 \\plus{} y^6 \\plus{} z^6 \\minus{} y^4z^2 \\minus{} x^2y^4 \\minus{} x^4y^2 \\minus{} y^2z^4 \\plus{} 3x^2y^2z^2 \\minus{} x^4z^2$\r\n\r\n$ \\equal{} \\sum(x^4(x \\minus{} y)(x \\minus{} z)) \\plus{} \\sum(x^3(y \\plus{} z)(x \\minus{} y)(x \\minus{} z)) \\plus{} xyz\\sum(x(x \\minus{} y)(x \\minus{} z))\\geq 0$\r\n\r\n$ f(\\frac {1}{2}) \\equal{} \\frac {1}{4}(3x^2y^2z^2 \\plus{} xy^4z \\plus{} 2z^6 \\plus{} x^4yz \\minus{} 3x^4y^2 \\minus{} 3x^4z^2 \\plus{}$ $ 2x^3y^3 \\minus{} 3x^2y^4 \\minus{} 3y^4z^2 \\plus{} 2y^6$ $ \\plus{} 2y^3z^3 \\minus{} 3y^2z^4 \\plus{} 2x^6 \\minus{} 3z^4x^2 \\plus{} 2z^3x^3 \\plus{} yz^4x)$\r\n$ \\equal{} \\frac {1}{4}(2\\sum(x^4(x \\minus{} y)(x \\minus{} z)) \\plus{} 2\\sum(x^3(y \\plus{} z)(x \\minus{} y)(x \\minus{} z)) \\minus{} (x \\minus{} y)^2(y \\minus{} z)^2(z \\minus{} x)^2$ $ \\plus{} xyz\\sum(x(x \\minus{} y)(x \\minus{} z)) \\plus{} xyz\\sum((y \\plus{} z)(x \\minus{} y)(x \\minus{} z)))\\geq 0$", "Solution_2": "[quote=\"hedeng123\"][quote=\"vuthanhtu_hd\"][b]Problem[/b]. Let $ a,b,c$ be non-negative real numbers and $ 0 \\le k \\le \\frac {1}{2}$.Prove that\n\n$ \\prod (a \\plus{} k\\sqrt {bc}) \\ge \\prod ((k \\minus{} 1)a \\plus{} b \\plus{} c)$\n\n :)\n\nI think this problem is true and It's old problem but I donn't know author :)[/quote]\n\n\n$ a \\equal{} x^2,b \\equal{} y^2,c \\equal{} z^2$\n\n$ \\prod (a \\plus{} k\\sqrt {bc}) \\ge \\prod ((k \\minus{} 1)a \\plus{} b \\plus{} c)$\n\n$ \\Longleftrightarrow f(k) \\equal{} ( \\minus{} x^2y^4 \\plus{} xzy^4 \\plus{} 3y^2z^2x^2 \\plus{} yzx^4 \\minus{} z^4x^2 \\minus{} x^4z^2 \\minus{} y^4z^2$ $ \\minus{} z^4y^2 \\minus{} x^4y^2 \\plus{} xyz^4)k^2 \\plus{}$\n\n$ (z^4y^2 \\plus{} y^3x^3 \\plus{} y^4z^2 \\plus{} z^4x^2 \\plus{} x^3z^3 \\minus{} y^6 \\plus{} x^4y^2 \\plus{} x^4z^2 \\minus{} 6y^2z^2x^2 \\plus{} y^3z^3 \\plus{}$ $ x^2y^4 \\minus{} z^6 \\minus{} x^6)k$ \n\n$ \\minus{} z^4x^2 \\plus{} x^6 \\plus{} y^6 \\plus{} z^6 \\minus{} y^4z^2 \\minus{} x^2y^4 \\minus{} x^4y^2 \\minus{} z^4y^2 \\plus{} 3y^2z^2x^2 \\minus{} x^4z^2\\geq 0$\n\nNote\n\n$ \\minus{} x^2y^4 \\plus{} xzy^4 \\plus{} 3y^2z^2x^2 \\plus{} yzx^4 \\minus{} z^4x^2 \\minus{} x^4z^2 \\minus{} y^4z^2$ $ \\minus{} z^4y^2 \\minus{} x^4y^2 \\plus{} xyz^4$ \n\n$ \\equal{} \\minus{} (x \\minus{} y)^2(y \\minus{} z)^2(z \\minus{} x)^2 \\minus{} 2\\sum(y^2z^2(x \\minus{} y)(x \\minus{} z))$ $ \\minus{} xyz\\sum(x(x \\minus{} y)(x \\minus{} z)) \\minus{} xyz\\sum((y \\plus{} z)(x \\minus{} y)(x \\minus{} z))\\le 0$\n\nwe only prove $ f(0)\\geq 0$ and $ f(\\frac {1}{2})\\geq 0$\n\n$ f(0) \\equal{} \\minus{} z^4x^2 \\plus{} x^6 \\plus{} y^6 \\plus{} z^6 \\minus{} y^4z^2 \\minus{} x^2y^4 \\minus{} x^4y^2 \\minus{} y^2z^4 \\plus{} 3x^2y^2z^2 \\minus{} x^4z^2$\n\n$ \\equal{} \\sum(x^4(x \\minus{} y)(x \\minus{} z)) \\plus{} \\sum(x^3(y \\plus{} z)(x \\minus{} y)(x \\minus{} z)) \\plus{} xyz\\sum(x(x \\minus{} y)(x \\minus{} z))\\geq 0$\n\n$ f(\\frac {1}{2}) \\equal{} \\frac {1}{4}(3x^2y^2z^2 \\plus{} xy^4z \\plus{} 2z^6 \\plus{} x^4yz \\minus{} 3x^4y^2 \\minus{} 3x^4z^2 \\plus{}$ $ 2x^3y^3 \\minus{} 3x^2y^4 \\minus{} 3y^4z^2 \\plus{} 2y^6$ $ \\plus{} 2y^3z^3 \\minus{} 3y^2z^4 \\plus{} 2x^6 \\minus{} 3z^4x^2 \\plus{} 2z^3x^3 \\plus{} yz^4x)$\n$ \\equal{} \\frac {1}{4}(2\\sum(x^4(x \\minus{} y)(x \\minus{} z)) \\plus{} 2\\sum(x^3(y \\plus{} z)(x \\minus{} y)(x \\minus{} z)) \\minus{} (x \\minus{} y)^2(y \\minus{} z)^2(z \\minus{} x)^2$ $ \\plus{} xyz\\sum(x(x \\minus{} y)(x \\minus{} z)) \\plus{} xyz\\sum((y \\plus{} z)(x \\minus{} y)(x \\minus{} z)))\\geq 0$[/quote]\r\n\r\nThanks hedeng123 :)" } { "Tag": [ "linear algebra", "matrix" ], "Problem": "For matrices $ A \\equal{} \\left( \\begin{array}{cc} 1 & 2 \\\\\r\n\\minus{} 3 & 4 \\end{array} \\right), \\ B \\equal{} \\left( \\begin{array}{cc} 4 & 1 \\\\\r\n\\minus{} 3 & 2 \\end{array} \\right),$ Find $ (2A \\plus{} B)(2A \\minus{} B) , 4A^2 \\minus{} B^2$.\r\n\r\nQuestion from kunny:\r\n\r\nWhat should we learn from the problem?", "Solution_1": "$ 2A \\equal{}\\left(\\begin{array}{cc}2 & 4\\\\ \\minus{}6 & 8\\end{array}\\right)$\r\n\r\n$ 2A\\plus{}B\\equal{}\\left(\\begin{array}{cc}6 & 5\\\\ \\minus{}9 & 10\\end{array}\\right)$\r\n\r\n$ 2A\\minus{}B\\equal{}\\left(\\begin{array}{cc}\\minus{}2 & 3\\\\ \\minus{}3 & 6\\end{array}\\right)$\r\n\r\n$ (2A\\plus{}B)(2A\\minus{}B)\\equal{}\\left(\\begin{array}{cc}6 & 5\\\\ \\minus{}9 & 10\\end{array}\\right)\\left(\\begin{array}{cc}\\minus{}2 & 3\\\\ \\minus{}3 & 6\\end{array}\\right)$\r\n\r\n$ \\equal{}\\left(\\begin{array}{cc}\\minus{}27 & 48\\\\ \\minus{}12 & 33\\end{array}\\right)$\r\n\r\n$ A^2\\equal{}\\left(\\begin{array}{cc}1 & 2\\\\ \\minus{}3 & 4\\end{array}\\right)\\left(\\begin{array}{cc}1 & 2\\\\ \\minus{}3 & 4\\end{array}\\right)$\r\n\r\n$ A^2\\equal{}\\left(\\begin{array}{cc}\\minus{}5 & 10\\\\ \\minus{}15 & 10\\end{array}\\right)$\r\n\r\n$ 4A^2\\equal{}\\left(\\begin{array}{cc}\\minus{}20 & 40\\\\ \\minus{}60 & 40\\end{array}\\right)$\r\n\r\n$ B^2\\equal{}\\left(\\begin{array}{cc}13 & 6\\\\ \\minus{}18 & 1\\end{array}\\right)$\r\n\r\n$ 4A^2\\minus{}B^2\\equal{}\\left(\\begin{array}{cc}\\minus{}43 & 34\\\\ \\minus{}42 & 39\\end{array}\\right)$\r\n\r\n$ \\Longrightarrow (2A\\plus{}B)(2A\\minus{}B) \\ne 4A^2\\minus{}B^2$\r\n\r\n[quote]\nWhat should we learn from the problem?[/quote]\r\n\r\nAlgebraic identities aren't always true for matrices :maybe:", "Solution_2": "You seemed to make a mistake in calculation of $ 4A^2\\minus{}B^2$.", "Solution_3": "[quote]$ 4A^{2}\\minus{}B^{2}\\equal{}\\left(\\begin{array}{cc}\\minus{}43 & 34\\\\ \\minus{}42 & 39\\end{array}\\right)$[/quote]\r\n\r\nIt should have been $ \\left(\\begin{array}{cc}\\minus{}33 & 34\\\\ \\minus{}42 & 39\\end{array}\\right)$", "Solution_4": "[quote=\"picture74\"][quote]$ 4A^{2} \\minus{} B^{2} \\equal{} \\left(\\begin{array}{cc} \\minus{} 43 & 34 \\\\\n\\minus{} 42 & 39\\end{array}\\right)$[/quote]\n\nIt should have been $ \\left(\\begin{array}{cc} \\minus{} 33 & 34 \\\\\n\\minus{} 42 & 39\\end{array}\\right)$[/quote]\n\nNow that's okay. :lol: \n\n[quote=\"picture74\"]\n[quote]\nWhat should we learn from the problem?[/quote]\n\nAlgebraic identities aren't always true for matrices :maybe:[/quote]\r\n\r\nWhat would become the result of $ (2A \\plus{} B)(2A \\minus{} B)$ for matrices $ A,\\ B?$", "Solution_5": "[quote=\"kunny\"]What would become the result of $ (2A \\plus{} B)(2A \\minus{} B)$ for matrices $ A,\\ B?$[/quote]\r\nI didn't really get what you meant :(", "Solution_6": "The issue is this: $ (2A\\plus{}B)(2A\\minus{}B)$ would equal $ 4A^2\\minus{}B^2$ if $ A$ and $ B$ commuted - that is, if $ AB\\equal{}BA.$ But these two matrices don't commute and the alleged identity doesn't work.\r\n\r\nThe lesson is this: algebra as you know it does work for matrices, mostly, but the non-commutativity of multiplication is a fact of life that you're just going to have to adjust to.\r\n\r\nNon-commutativity is also why we never write $ \\frac{A}{B}$ for $ A$ and $ B$ matrices; instead we write either $ B^{\\minus{}1}A$ or $ AB^{\\minus{}1},$ whichever of those two expression we mean (and we have to choose, since they may not be equal.)", "Solution_7": "[quote=\"picture74\"][quote=\"kunny\"]What would become the result of $ (2A \\plus{} B)(2A \\minus{} B)$ for matrices $ A,\\ B?$[/quote]\nI didn't really get what you meant :([/quote]\r\n\r\nThe answer is $ 4A^2 \\minus{} 2AB \\plus{} 2BA \\minus{} B^2$." } { "Tag": [ "trigonometry" ], "Problem": "Show that $x=a+rcos\\theta, y= b+rsin\\theta$ defines a circle with $center (a,b)$ and radius $r$.", "Solution_1": "$(x-a)^{2}+(y-b)^{2}=(r\\cos x)^{2}+(r\\sin x)^{2}= r^{2}(\\sin^{2}x+\\cos^{2}x)=r^{2}$\r\nThus, $(x-a)^{2}+(y-b)^{2}=r^{2}$." } { "Tag": [ "inequalities", "Canada", "am-gm inequality", "AM-GM", "algebra" ], "Problem": "Prove that for all positive real numbers $a$, $b$, and $c$,\r\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\r\nand determine when equality occurs.", "Solution_1": "By rearrangement: $a^3 \\geq b^3 \\geq c^3$ and $\\frac{1}{bc} \\geq \\frac{1}{ac} \\geq \\frac{1}{ab}$.", "Solution_2": "[quote=\"shobber\"]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\r\n\r\n It isn't hard. :D \r\nBy AM-GM, we have \r\n$\\frac{a^3}{bc} +b+c \\geq 3a$\r\nSum up and done", "Solution_3": "Yes, the inequality is simple applications of $x^2+y^2+z^2\\geq xy+yz+zx$ for $a^2, b^2, c^2$ and $ab, bc,ca,$.\r\nWe have\r\n\\[ a^4+b^4+c^4\\geq a^2b^2+b^2c^2+c^2a^2\\geq abc(a+b+c). \\]", "Solution_4": "Also you can prove the last one by the Muirheads Inequality...\r\n\r\nDavron", "Solution_5": "Now sum cyclically using AM-GM that\n$\\frac{a^3}{2bc}+\\frac{b^3}{4ac}+\\frac{c^3}{4ab}\\ge a$", "Solution_6": "By $Cauchy-Schwarz$ and $AM-GM$, we have:\n\\[\\sum\\frac{a^3}{bc}\\ge \\frac{(a^2+b^2+c^2)^2}{3abc}\\ge \\frac{(ab+bc+ca)^2}{3abc}=a+b+c\\]", "Solution_7": "Very simple.\nBy AM-GM\n\\[\\sum( \\frac{a^3}{bc} +b+c)\\ge 3\\sum a\\iff \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]", "Solution_8": "Too easy.Using the fact that $\\sum {x^2}\\ge \\sum{xy}$ we see that\n$\\sum{a^4} \\ge \\sum{a^2b^2} \\ge \\sum{a^2bc}$.\nResult follows by dividing both sides by $abc$.\n\n\nBye...", "Solution_9": "It's trivial by chebychev, $\\sum \\frac{a^3}{bc} \\ge\\frac{a+b+c}{3} \\sum \\frac{a^2}{bc} \\ge a+b+c$\nI am learning inequalities and found chebychev very useful tool.", "Solution_10": "Inequality is equivalent to proving that $f(a,b,c) \\geq 0$ where \n\\[\nf(a,b,c)=a^4+b^4+c^4-a^2 b c-a b^2 c-a b c^2\n\\].\nIt is symmetric so we can take $a=c+x+y$ and $b=c+x$ for some $x,y \\geq 0$ and then\n\\[\nf(a,b,c)=f(c+x+y,c+x,c)=5 c^2 x^2+5 c^2 x y+5 c^2 y^2+6 c x^3+9 c x^2 y+11 c x y^2+4 c y^3+2 x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4 \\geq 0\n\\]\nDone.", "Solution_11": "[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\n\nThe following inequality is also true.\nLet $a$, $b$ and $c$ be positive numbers such that $a^{15}+b^{15}+c^{15}=3$. Prove that:\n\\[\\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq3\\]", "Solution_12": "[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\n\n$\\iff a^4+b^4+c^4\\geq a^2b^2+c^2a^2+b^2c^2\\geq abc(a+b+c) $ very simple", "Solution_13": "[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\n\nalternatively by tchebytcheff we have \n$\\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab}\\ge [\\frac{(a+b+c)}{3}](\\frac{a^2}{bc}+\\frac{b^2}{ca}+\\frac{c^2}{ab})$\nbut by cauchy\n$[\\frac{(a+b+c)}{3}](\\frac{a^2}{bc}+\\frac{b^2}{ca}+\\frac{c^2}{ab})\\ge a+b+c$\n\nas $(a+b+c)^2\\ge 3(ab+bc+ca)$", "Solution_14": "Here is another solution:\n Note that the inequality is homogeneous in $a,b,c$ so we let $abc=1$. The first observation is that, by AM-GM, $a+b+c \\geq 3\\sqrt[3]{abc} = 3$. Further, the given inequality reduces to $a^4 + b^4 + c^4 \\geq a + b + c$. This is easy, as by the AM-QM (or the power mean) inequality, we have $a^4 + b^4 + c^4 \\geq 4 \\cdot (\\frac{a+b+c}{4})^4 = (a+b+c) (\\frac{a+b+c}{3})^3 \\geq a+b+c$, and we are done. ", "Solution_15": "[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\n$$\\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab}\\geq\\sqrt[15]{3^{14}(a^{15}+b^{15}+c^{15})}$$\nis also true. \nThere is solution by hand. \n\n", "Solution_16": "[quote=shobber]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\n\nBy am-gm: $$\\sum_{\\text{cyc}} \\frac{a^3}{bc} +b+c \\geq \\sum_{\\text{cyc}} 3a \\geq \\sum_{\\text{cyc}} a$$ Equality when $a=b=c.$", "Solution_17": "Time $abc$ for both sides and we have to prove $a^4+b^4+c^4 \\geq a^2bc+b^2ac+c^2ab$ which is trivial by Muirhead", "Solution_18": "$\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]$\nImplies\n$\\[a^4+b^4+c^4\\geq a^2bc+ab^2c+abc^2\\]$\nSince $(4,0,0)$ Majorizes $(2,1,1)$ This will always hold by muirhead.\nEquality occurs when $a=b=c$", "Solution_19": "Trivial by Murihead Inequality:\n\n\nRearrange gives $$a^5bc+b^5ac+c^5ab\\geq a^3b^2c^2+a^2b^3c^2+a^2b^2c^3$$\n\nWhich finishes since $(5, 1, 1)\\succ \\ (3, 2, 2)$. Equality holdes iff $a=b=c$. \n\n[i]q.e.d.[/i]", "Solution_20": "Multiplying the inequality by $abc$ gives $$a^4+b^4+c^4 \\geq a^2bc+ab^2c+abc^2$$\nBy AM-GM:\n$$a^4+b^4+c^4 \\geq a^2b^2+a^2c^2+b^2c^2$$\nAlso by AM-GM:\n$$a^2b^2+a^2c^2+b^2c^2 \\geq a^2bc+ab^2c+abc^2$$\nTherefore,\n$$a^4+b^4+c^4 \\geq a^2bc+ab^2c+abc^2$$\nEquality is when $a=b=c$. $\\square$\n\n", "Solution_21": "[quote=Gibbenergy][quote=\"shobber\"]Prove that for all positive real numbers $a$, $b$, and $c$,\n\\[ \\frac{a^3}{bc} + \\frac{b^3}{ca} + \\frac{c^3}{ab} \\geq a+b+c \\]\nand determine when equality occurs.[/quote]\n\n It isn't hard. :D \nBy AM-GM, we have \n$\\frac{a^3}{bc} +b+c \\geq 3a$\nSum up and done[/quote]\n\nthat's what I did lol", "Solution_22": "Given that $a, b, c$ are positive reals, multiplying both sides of the inequality by $abc$ attains \n$$a^4+b^4+c^4 \\geq a^2bc+ab^2c+abc^2$$\nmeaning that\n$$\\sum_{cyc}a^4\\geq \\sum_{cyc} a^2bc$$\nsince $(4,0,0)\\succ (2,1,1)$ the inequality is true by Muirhead.", "Solution_23": "Multiply on $abc$ on both sides to get $a^4+b^4+c^4 \\ge a^2bc+a^2bc+abc^2$\nBecause $(4,0,0) > (2,1,1)$, the inequality is true", "Solution_24": "One-Liner\n\\[\\sum \\frac{a^3}{bc}=\\sum\\frac{\\frac{a^3}{bc}+\\frac{c^3}{ab}}{2}\\geq\\sum\\frac{ac}{b}=\\sum\\frac{\\frac{ac}{b}+\\frac{ab}{c}}{2}\\geq\\sum a\\]", "Solution_25": "[quote=CrazyInMath]One-Liner\n\\[\\sum \\frac{a^3}{bc}=\\sum\\frac{\\frac{a^3}{bc}+\\frac{c^3}{ab}}{2}\\geq\\sum\\frac{ac}{b}=\\sum\\frac{\\frac{ac}{b}+\\frac{ab}{c}}{2}\\geq\\sum a\\][/quote]\n\nThis is nice, but most solutions above are one-liner.\nHere is a one0line rephrasement of the most commonly discussed solution.\n\nBy rearrangement $\\frac{a^3}{bc} +\\frac{b^3}{ca}+\\frac{c^3}{ab} = \\frac{a(a^2)}{bc}+\\frac{b(b^2)}{ca}+\\frac{c(c^2)}{ab} \\ge \\frac{a(bc)}{bc} + \\frac{b(ca)}{ca}+\\frac{c(ab)}{ab} = a+b+c.$\n", "Solution_26": "Multiply both sides by $abc$, so it suffices to prove $a^4+b^4+c^4\\ge a^2bc+ab^2c+abc^2$.\n\nNotice how $a^4+a^4+b^4+c^4\\ge 4a^2bc$ by AM-GM. Summing this with its analogous inequalities proves the result, and equality occurs when $a=b=c$.", "Solution_27": "[quote=sqing]Prove that for all positive real numbers $a$, $b$, and $c$,\n$$\\frac{a^3}{bc} + \\frac{b^2c^2}{a^2} \\geq ab+c$$[/quote]\n\n[b]Remark.[/b] The claimed inequality $\\frac{a^3}{bc} + \\frac{b^2c^2}{a^2} \\geq ab+c$ is [b]not[/b] true for all positive real numbers $a,b,c$. For instance, if $\\left(a,b,c\\right)=\\left(1,\\frac{1}{2},3\\right)$, then\n\\begin{align*}\\frac{a^3}{bc} + \\frac{b^2c^2}{a^2}-( ab+c)\n=&\\frac{2}{3}+\\frac{9}{4}-\\frac{1}{2}-3\\\\\n=&-\\frac{7}{12}<0.\n\\end{align*}", "Solution_28": "Observe that $(4, 0, 0) \\succ (2,1,1)$, so from Muirhead, $a^4 + b^4 + c^4 \\ge a^2bc + b^2ca + c^2ab$, and dividing by $abc$ gives the desired result. Since Muirhead can be achieved from various applications of AM-GM, equality holds iff $a = b = c$.", "Solution_29": "Note that upon expansion we have,\n\\begin{align*}\n\\sum a^5bc \\geq \\sum a^3b^2c^2\\\\\n\\end{align*}\nThen homogenize so that $abc = 1$. Then we wish to show,\n\\begin{align*}\n\\sum a^4 &\\geq \\sum a\\\\\n\\iff \\sum a^4-a &\\geq 0\n\\end{align*}\nNow define $f(x) = x^4 - x$. We may check,\n\\begin{align*}\nf''(x) = 12x^2\n\\end{align*}\nand hence $f$ is convex everywhere. However then by Jensen's we may check,\n\\begin{align*}\nf(a) + f(b) + f(c) \\geq 3\\cdot f\\left(\\frac{a+b+c}{3} \\right)\n\\end{align*}\nNote that by AM-GM\n\\begin{align*}\n\\frac{a+b+c}{3} \\geq 1\n\\end{align*}\nFinally we may check,\n\\begin{align*}\nf([x \\geq 1]) \\geq f(1) = 0\n\\end{align*}\nIndeed note that this holds as,\n\\begin{align*}\nx^4 - x \\geq 0 \\iff x(x^3 - 1) \\geq 0 \\iff x \\in (-\\infty, 0) \\cup (1, \\infty)\n\\end{align*}\nThus we may finish as,\n\\begin{align*}\n\\sum f(a) \\geq 3 \\cdot f \\left( \\frac{a+b+c}{3} \\right) \\geq 3 \\cdot f(1) \\geq 0\n\\end{align*}" } { "Tag": [ "quadratics", "arithmetic series", "arithmetic sequence", "number theory unsolved", "number theory" ], "Problem": "Show there for any positive integer $ m$ there exist a positive integer $ N$ such that for all primes $ p > N$ there are $ m$ consecutive quadratic residues modulo $ p$.", "Solution_1": "[quote=\"hollandman\"]Show there for any positive integer $ m$ there exist a positive integer $ N$ such that for all primes $ p > N$ there are $ m$ consecutive quadratic residues modulo $ p$.[/quote]\r\n\r\nThis is an amazingly difficult problem. It has appeared on one of the USA MOSP test and no one ever had ANY idea.\r\n\r\nI can't recall the exact proof. But I can recall that it uses Van der Warden Theorem. Consider a number N so big that when we use red and blue to color numbers in range 1 to N-1, there'll always exist $ m^2$ numbers in an arithmatic sequence(like 2,7,12,17,...) that they are either all blue or all red. Now, let blue represent quadratic residue and red represent otherwise, the statement becomes that there exist $ m^2$ numbers(in arithmatic sequence) that either all quadratic residue or all otherwise.\r\n\r\nI forget all the following things, and I can't really recall why $ m^2$ is needed.", "Solution_2": "[hide=\"a finish to the above\"]\nIf $ 1,2,...,m$ are all quadratic residues, then we are done. Otherwise, suppose that there is at least one nonquadratic residue in that set. Of course, there is always at least one quadratic residue, such as $ 1$.\n\nIn the arithmetic series of $ m^2$ elements, let the common difference be $ d$. We can come up with arithmetic progression within this progression with $ m$ elements that have common difference of $ d,2d,...,md$. Among these possible differences, at least one must be a quadratic residue and at least one is a nonquadratic residue. Now, choose one of these differences $ ad$ such that $ \\left(\\frac{ad}{p}\\right)$ is the same as the legendre of any element in this arithmetic progression.\n\nThen, we can multiply each element in an arithmetic progression with difference $ ad$ by $ (ad)^{\\minus{}1}$; this will give us an arithmetic progression of difference $ 1$, meaning they are consecutive integers. Also, since $ \\left(\\frac{ad}{p}\\right)\\equal{}\\left(\\frac{(ad)^{\\minus{}1}}{p}\\right)$ all elements in this new progression will all be quadratic residues.\n[/hide]" } { "Tag": [], "Problem": "Prove that any two numbers in the following sequence are prime.\r\n\r\n$ 2 \\plus{} 1, 2^{2^1} \\plus{} 1, \\dots , 2^{2^n} \\plus{} 1$\r\n\r\nHints only, please, and hide them!\r\n\r\nThanks in advance.\r\n\r\nEDIT: Oh, fixed.", "Solution_1": "I didn't check fully, but it seems that they are not just relatively prime, but completely prime.", "Solution_2": "Is $ n\\in\\mathbb{Z^\\plus{}}$", "Solution_3": "[quote=\"hunter34\"]Is $ n\\in\\mathbb{Z^ \\plus{} }$[/quote]\r\n\r\ni think $ n$ is just integers from 0-infinity, becuz it starts aas n=0", "Solution_4": "but some terms are not prime...\r\nehh\r\ni didnt get the question", "Solution_5": "which terms aren't prime?", "Solution_6": "$ 2^{2^5}\\plus{}1\\equal{}641 \\cdot 6700417$", "Solution_7": "[quote=\"isabella2296\"]Prove that any two numbers in the following sequence are [b]relatively[/b]prime.\n\n$ 2 \\plus{} 1, 2^{2^1} \\plus{} 1, \\dots , 2^{2^n} \\plus{} 1$\n\nHints only, please, and hide them!\n\nThanks in advance.\n\n[/quote]\r\n\r\nReverted back to original.", "Solution_8": "[hide]Fermat primes and look at the +1 part it makes a the problem trivial almost[/hide]", "Solution_9": "[quote=\"fishythefish\"]I didn't check fully, but it seems that they are not just relatively prime, but completely prime.[/quote]\r\n\r\n[hide]\nNo. What about n=4, n=5, up to n=12? Check wikipedia for \"fermat numbers\"\n[/hide]", "Solution_10": "I figured it out. Thanks for your help, everyone!" } { "Tag": [ "algebra", "polynomial", "number theory", "greatest common divisor", "search", "number theory unsolved" ], "Problem": "Let $P(x) \\in Z[x]$ and $A=\\{a_{1},a_{2},... a_{n}\\}\\subset N^{*}$ .\r\n$\\forall k \\in Z, \\ \\ \\exists a_{i}\\in A$ such that $a_{i}| P(k)$ \r\nProve that: $\\exists a \\in A$ such that $a | P(x) \\forall x \\in Z$", "Solution_1": "Obviosly $a=gcd(a_{1},...,a_{n})$ work.", "Solution_2": "Rust, could you at least try to read the statement of these problems? Why on earth not taking $a=1$? After all, who said that $A$ contains the $gcd$ of its elements. By the way, the problem is taken from Leningrad olympiad and it is surely not an easy one. But it was discussed on the forum and Iura gave a solution. Use search button.", "Solution_3": "It is obviosly $gcd(a_{i})\\in A$.", "Solution_4": "Yes, very obviously that $1 \\in\\{2,3\\}$ if the polynomial is e.g. $2x$.", "Solution_5": "Ok, from now on I will just le you do whatever you want. It seems useless to try to say something, everything is too \"obviously\". Sleep in peace!" } { "Tag": [ "quadratics", "number theory theorems", "number theory" ], "Problem": "I was wondering: suppose you want to construct an equation to be solved in integers so that the sequence of solutions for one of the unknown $s_i$ has a property like: $2s_{i}+s_{i+1}=s_{i+2}$ (or any other recursive equation), does anyone know how to do this or maybe someone who can show there isn't such an equtation?", "Solution_1": "You have $s_{i+2}=2s_{i}+s_{i+1}$\r\nThe charastic equation is $x^2=x+2 \\rightarrow x^2-x-2=0$\r\n\r\n$\\text{With roots 2 and -1}$ therefore the general solution is:\r\n\r\n$s_{n}=a({2}^n)+b({-1}^n)$\r\n\r\nYou get $a$ and $b$ from your starting values", "Solution_2": "That is not what he asked for ;)", "Solution_3": "ok I know, but you misunderstood the question: what I meant is can you find a diophantic equation (quadratic or whatever) like $3n^2+3n+1=m^2$ or $2^n+n^2=3^m$ which satisfies this: the sequence of solutions for $n$ satisfies an equation like the given one, so that it becomes easy to find the next solution.\r\n \r\nI'll give an easy example: suppose you have to construct an equation so that the a sequence with all the solutions of one of the unknown satisfies $x_n=x_{n-1}-2$, then a good answer would be $x+2y=3$ because if $x_{n}$ provides a solution, then so does $x_{n}-2$ etc... This would give the sequence (...,-1,1,3,5,..) The corresponding sequence for $y_n$ would then satisfie $y_n=y_{n-1}+1$ and be (...,-2,-1,0,1,2,...).\r\n\r\nI hope it is clear now although it is maybe a weird question", "Solution_4": "Ah, it is all clear now. And much more interesting!", "Solution_5": "does it have to give all the solutions, or just some of them?", "Solution_6": "if possible all of them, but even partial solutions may be usefull", "Solution_7": "Here i found a kinda answer for you.\r\n\r\nSay were trying to find all $x$ such that the equation\r\n\r\n$ax-2y=1$ has a solution, with $y$ an integer.\r\n\r\nthen if $x=s_i$ and $x=s_{i+1}$ are solutions, so is $x=s_{i+1}+2s_i$\r\n\r\nI highly doubt its all of them, but im tired and want to go to bed...", "Solution_8": "indeed that is a nice equation (with solutions only if a is odd)\r\nit will only give all (positive) solutions if you found out that $1$ and $3$ are solutions for $x$.\r\n\r\nI think this might be generalized to all linear recursive equations, but I'm still thinking about it." } { "Tag": [ "geometry", "incenter", "circumcircle", "rectangle", "angle bisector" ], "Problem": "In $ \\Delta ABC$, $ M$ is the midpoint of $ II_{a}$. Show that $ M$ lies on the circumcircle of $ \\Delta ABC$\r\n\r\n[hide=\"hint\"]$ MA\\equal{}MB$[/hide]", "Solution_1": "[quote=\"CatalystOfNostalgia\"]In $ \\Delta ABC$, $ M$ is the midpoint of $ II_{a}$. Show that $ M$ lies on the circumcircle of $ \\Delta ABC$\n\n[hide=\"hint\"]$ MA \\equal{} MB$[/hide][/quote]\nIs $ I_a$ the excenter of $ A$ in $ \\triangle ABC$? If so, then the hint should be:\n[hide=\"Hint\"]\n$ MC \\equal{} MB$. [/hide]\nAnyways, here is the solution. \n[hide=\"Solution\"]\nLet the point of tangency with the $ A$-Excircle and $ BC$ be $ X$ and the point of tangency between the incircle and $ BC$ be $ Y$. Furthermore, let the perpindicular to $ BC$ from $ M$ meet $ BC$ at $ Z$. \n[b]Lemma: [/b]$ ZX \\equal{} ZY$\n[i]Proof:[/i] Let $ P$ be the point such that $ P$ is on $ I_aX$ extended and $ IP\\perp I_aP$ and let $ Q$ be the point so that $ Q$ is on $ IY$ extended and $ IQ\\perp I_aQ$. Since $ I_aX\\perp XY$ and $ IY\\perp XY$, we see that $ IP\\parallel XY\\parallel I_aQ$. This means that $ \\angle PXY \\equal{} 180 \\minus{} \\angle I_aPI \\equal{} 90$ and $ \\angle PIQ \\equal{} 180 \\minus{} \\angle IQI_a \\equal{} 90$, so $ PIYX$ is a rectangle as are $ XYQI_a$ and $ PIQI_a$. Thus, $ PQ$ bisects $ II_a$, so $ P$, $ M$, and $ Q$ are collinear. This means that $ M$ is the center of rectangle $ PIQI_a$, so the altitude to $ PI$ from $ M$ meets $ PI$ at its midpoint. Since $ XY\\parallel PI$, we have that the altitude from $ M$ to $ XY$ meets $ XY$ at its midpoint. Hence, $ XZ \\equal{} YZ$. \n\nNow, it is well-known that $ BX \\equal{} YC$, so $ BX \\plus{} XZ \\equal{} YZ \\plus{} YC\\implies BZ \\equal{} ZC$ and $ MZ\\perp BC$, so $ MB \\equal{} MC$. Also, $ I$ and $ I_a$ are on the bisector of $ \\angle A$, so $ II_a$ is the bisector of $ \\angle A$. Notice that $ II_a$ and the circumcircle of triangle $ ABC$ only meet at two points, which are $ A$, and say $ M'$. Notice that $ \\angle M'BC \\equal{} \\angle M'AC \\equal{} \\angle M'AB \\equal{} \\angle M'CB$, so $ M'B \\equal{} M'C$, so $ M'$ is the intersection between the angle bisector of $ \\angle A$ (or $ II_a$) and the perpindicular bisector of $ BC$. This, however, is also $ M$, so $ M \\equal{} M'$, which means that $ M$ is on the circumcircle of $ \\triangle ABC$. [/hide]" } { "Tag": [], "Problem": "For how many pairs of consecutive integers in $\\{1000,1001,1002,\\ldots,2000\\}$ is no carrying required when the two integers are added?", "Solution_1": "[hide]First, let $1abc$ represent each of the numbers in the set. Let its consecutive integer, $1abc+1$ be denoted as $1abc'$. Notice that if $a$, $b$, and $c$ are 0, 1, 2, 3, or 4, $1abc+1abc'$ requires no carrying. (125 possibilities) Additionally, there is no carrying when $c=9$and $a$ and $b$ are 0, 1, 2, 3, or 4. (25 possibilities) Finally, if $b=c=9$, then $a$ could be 0, 1, 2, 3, 4, or 9. (6 combinations) Thus, our answer is $125+25+6=\\boxed{156}$[/hide]", "Solution_2": "[hide=\"Answer\"]Looking at the first number of the pair only: for hundreds, only from $0$ to $4$ will work. For tens, we again have from $0$ to $4$. For units, we again have $0$ to $4$, but also $9$. The first condition removes $500$ numbers, and the second removes $250$. The third removes $100$, so we have $150$. However, we have oversimplified somewhat. $1999+2000=3999$, which involves no carrying, so we add one for that, and for the second condition, when the numbers ends with $99$, this will work. This adds $5$ numbers, so our answer is $\\boxed{156}$.[/hide]", "Solution_3": "I understand where the $125$ comes from when considering $[0,5]$ and the $25$ when considering $(9,0)$ and the pair $(1999,2000)$, giving a total of $151$. Where do the last $5$ come from?", "Solution_4": "$1099+1100$, $1199+1200$, $1299+1300$, $1399+1400$, and $1499+1500$.\nWe missed these because we took out the numbers whose tens digits were $9$s.", "Solution_5": "First of all if the last three digits are in $0,1,2,3,4$ we have $5$ ways for each digit, so $125$ total ways. The units digit could also be $9$, in which case the second and third digits are in $0,1,2,3,4$, for $25$ ways. Lastly, the final two digits could each be $9$ and the second digit could be $0,1,2,3,4,9$ for $6$ ways. Summing gives $\\boxed{156}$." } { "Tag": [ "algebra", "polynomial", "number theory proposed", "number theory" ], "Problem": "Determine all non-constant polynomials $ f\\in \\mathbb{Z}[X]$ with the property that there exists $ k\\in\\mathbb{N}^*$ such that for any prime number $ p$, $ f(p)$ has at most $ k$ distinct prime divisors.", "Solution_1": "Nice problem. \r\n[hide=\"Answer\"]$ f(x)\\equal{}nx^m$[/hide]\n[hide=\"Proof\"][b]Lemma[/b]\nFor each $ f\\in\\mathbb{Z}[x]$ let $ A(f)$ be the set of prime divisors of $ f(x)$ when $ x$ runs over the integers. If $ f$ has positive degree then $ A(f)$ is infinite.\n[hide=\"Proof of lemma\"]The result is clear for all $ f$ with $ f(0)\\equal{}0$. \nSuppose for the sake of contradiction that $ A(f)$ is finite for some $ f$ with nonzero independent term. \n\nLet $ P\\subseteq A(f)$ be the set of primes which divide $ f(x)$ for any integer $ x$ and let $ Q\\equal{}A\\setminus P$. \n\nLet $ b$ and $ c$ be integers such that $ bc\\equal{}f(0)$, $ (b,c)\\equal{}1$, and $ p|b$ for any $ p\\in P$. \n\nFor each $ q\\in Q$ there exists an integer $ r_q$ such that $ q\\not| f(r_q)$, because of the chinese remainder theorem there exists an integer $ d$ with $ b^2d\\equiv r_q\\mod q$ for each $ q\\in Q$. \n\nNow $ f(x)\\equal{}bc\\plus{}xg(x)$ with $ g(x)\\in\\mathbb{Z}[x]$ and $ f(b^2d)\\equal{}b(c\\plus{}bdg(b^2d))$. \n\nSince $ f(b^2d)\\equiv r_q\\mod q$ we have that $ q\\not| f(b^2d)$ for any $ q\\in Q$. Also $ (p,c\\plus{}bdg(b^2d))\\equal{}1$ for any $ p\\in P$, so $ c\\plus{}bdg(b^2d)$ does not have prime factors in $ A(f)$. \n\nAs there exist inifinitely many such integers $ d$ it is sure that for some $ d$, $ c\\plus{}bdg(b^2d)\\not\\equal{}\\pm 1$ so it does have prime factors outside $ A(f)$ and this is the contradiction.[/hide]\n\n(Returning to the problem...) Let $ f$ be one such polynomial ($ f(p)$ has at most $ k$ disctintc prime divisors) and rewrite it as $ f(x)\\equal{}x^mg(x)$, with $ g\\in\\mathbb{Z}[x]$ and $ g(0)\\not\\equal{} 0$. Suppose for the sake of contradiction that $ g$ has positive degree. We can apply the lemma to $ g$ so $ A(g)$ is inifinite. In particular we can find $ k\\plus{}1$ primes in $ A(g)$ not dividing $ g(0)$ and let $ s$ be their product. For each of these primes $ q$ there exist and integer $ r_q\\not\\equiv 0\\mod q$ such that $ q|f(r_q)$. By the chinese remainder theorem there exists $ a\\mod s$ such that $ a\\equiv r_q\\mod q$ for each of these $ q$. Since $ r_q\\not\\equal{} 0\\mod q$ we have $ (a, s)\\equal{}1$ then Dirichlet's theorem on primes in arithmetic progressions asures the existence of infinitely many primes $ p\\equiv a\\mod s$. Choose one of these $ p$ such that $ f(p)\\not\\equal{} 0$ and we have that $ s|f(p)$ so $ f(p)$ has more than $ k$ prime divisors, this is a contradiction. \n\nSo the only polynomials $ f\\in\\mathbb{Z}[x]$ with the mentioned property are $ f(x)\\equal{}nx^m$[/hide]\r\n\r\n:)" } { "Tag": [ "inequalities", "calculus", "integration", "inequalities unsolved" ], "Problem": "Hi!\r\n\r\nI'm not sure if this is the right place for asking this. My question is about the planar isoperimetric proof. Hopefully you can see the picture (equality.gif) at the end of this message.\r\n\r\nEverything else is quite clear except for the last part of the equation. More precisely the part where dz^dz* (z* = the conjugate of z) \"disappears\" and integration over omega becomes somehow (with the help of Green's theorem?) integration along omega's bounding. How does this happen?\r\n\r\nCan someone please help me?! If this isn't the right place to ask, does anyone know a better one?\r\n\r\nThank you!", "Solution_1": "I solved the problem by myself. :first:" } { "Tag": [ "algebra", "polynomial", "induction", "superior algebra", "superior algebra solved" ], "Problem": "let f,g denote single variable poly. having coeff. in ring R.(commutative with identity)\r\n 1.find all units in R[x]\r\n 2.if f nonzero and fg=0,then show there exists constant c in R satisfying\r\n cg=0.", "Solution_1": "1. http://www.mathlinks.ro/Forum/viewtopic.php?t=87441\r\n2. use induction on the degree of $f$" } { "Tag": [ "induction", "limit" ], "Problem": "Let array ${a_n}$ is defined by ${a_1} = \\frac{27}{10}, {a_{n+1}^{3}} -3{a_{n+1}}({a_{n+1}-1})-{a_n}=1 \\forall {n} > 1$\r\nProve that array has a limit and find that limit", "Solution_1": "[hide=\"almost complete solution\"]just to start...looking for bounds on $x^3-3x^2+3x+c$, $a_{n+1} \\le 3$ iff $9+c=8-a_n \\ge 0$\n\nor $a_n \\le 8$ then $a_{n+1} \\le 3$, since $a_1<8$, then by induction, $\\forall n, a_n \\le 3$\n\nthen by decartes', as long as -(a_n+1) is negative, there is no sign variation, and therefore, no negative roots, this similarly follows by induction, so we have $a_n \\in (0,3)$\n\nthis does not proof it is convergent, but it is bounded...\n\nand of course, when there is a limit, for large n, we can say that $a_n=a_{n+1}=L$, so we have a cubic...\n\n$L^3-3L^2+2L-1=0$, by rrt, there are no rational roots, and since irrational/complex roots come in pairs, but there is only one limit, so it does not have a simple exact form...(you can use cardano's method), but using a calculator approximation, we get $L \\approx 2.325$[/hide]", "Solution_2": "$a_{n + 1}^3 - 3a_{n + 1} \\left( {a_{n + 1} - 1} \\right) - a_n = 1 \\Leftrightarrow$\r\n\r\n$\\Leftrightarrow a_{n + 1}^3 - 3a_{n + 1}^2 + 3a_{n + 1} - 1 = a_n \\Leftrightarrow$\r\n\r\n$\\Leftrightarrow a_{n + 1} = 1 + \\sqrt[3]{{a_n }}$", "Solution_3": "$a_{n + 1} = 1 + \\sqrt[3]{{a_n }}\\mathop \\Leftrightarrow \\limits_{\\mathop {f\\left( x \\right) = 1 + \\sqrt[3]{x}}\\limits^ \\uparrow } a_{n + 1} = f\\left( {a_n } \\right)$\r\n\r\nthen:\r\n$\\mathop {\\lim }\\limits_{n \\to + \\infty } a_n = l$\r\n\r\nand:\r\n$l - \\sqrt[3]{l} - 1 = 0,l > 0$", "Solution_4": "[quote=\"hien\"]Let array ${a_n}$ is defined by ${a_1} = \\frac{27}{10}, {a_{n+1}^{3}} -3{a_{n+1}}({a_{n+1}-1})-{a_n}=1 \\forall {n} > 1$\nProve that array has a limit and find that limit[/quote]\r\n ${a_{n+1}^{3}} -3{a_{n+1}}({a_{n+1}-1})-{a_n}=1\\Leftrightarrow(a_{n+1}-1)^3=a_n\\Leftrightarrow$\r\n$\\Leftrightarrow(a_{n+1}-1)^3-(A-1)^3=a_n-(A-1)^3.$ All this is $\\forall A\\in\\mathbb R.$\r\nLet $A$ is root of equation $(x-1)^3=x.$\r\n$(x-1)^3=x\\Leftrightarrow x^3-3x^2+2x-1=0.$ Let $f(x)=x^3-3x^2+2x-1.$\r\nHence $f'(x)=3x^2-6x+2.$ $f'(x)=0\\Leftrightarrow x=x_{1}=\\frac{3+\\sqrt3}{3}$ or $x=x_{2}=\\frac{3-\\sqrt3}{3}.$\r\nHence $x_{max}=x_{2}$ and $f(x_{2})<0.$\r\nHence $A$ is alone root of the equation and $A>2.3$ since $f(2.3)<0.$\r\n$a_2=1+\\sqrt[3]{2.7}=2.39...>2.3.$ Hence $\\forall n \\in \\mathbb N$ $a_{n+1}=1+\\sqrt[3]{a_n}>1+\\sqrt[3]{2.3}=2.32...>2.3.$\r\nThence, $(x_{n+1}-1)^3-(A-1)^3=x_n-(A-1)^3\\Leftrightarrow$\r\n$\\Leftrightarrow(x_{n+1}-A)((x_{n+1}-1)^2+(x_{n+1}-1)(A-1)+(A-1)^2)=x_n-A.$\r\nHence, $|x_{n+1}-A|=\\frac{|x_n-A|}{(x_{n+1}-1)^2+(x_{n+1}-1)(A-1)+(A-1)^2}<$\r\n$<\\frac{|x_n-A|}{1.3^2+1.3\\cdot1.3+1.3^2}<\\frac{|x_n-A|}{2}.$\r\nId est, $\\forall n\\in\\mathbb N$ $|a_{n+1}-A|<\\frac{1}{2}\\cdot|a_n-A|.$\r\nHence, $|a_n-A|<\\frac{1}{2}\\cdot|a_{n-1}-A|<...<\\frac{1}{2^{n-1}}\\cdot|a_1-A|.$\r\nHence, $\\forall n\\in\\mathbb N$ $|a_n-A|\\leq\\frac{1}{2^{n-1}}\\cdot|a_1-A|.$\r\n$lim\\frac{|a_1-A|}{2^{n-1}}=0.$\r\nHence, $lim a_n=A,$ where $A$ is root of equation $(x-1)^3=x.$ :)" } { "Tag": [], "Problem": "A sumsquare is a 3 x 3 array of positive integers such that each row, each column and each of the two main diagonals has sum $ m$. Show that $ m$ must be a multiple of 3 and that the largest entry in the array is at most $ 2m/3\\minus{}1$.", "Solution_1": "I'll prove that $ m\\equal{}3x$, where $ x$ is the value in the central square.\r\n\r\nGiven three squares x, a, and b we can get the following:\r\n\r\n[code]\n b m-a-2b+x a+b-x\n a x m-x-a\n m-a-b whatever m-x-b\n\n[/code]\r\n\r\nLooking at the right column, $ (a\\plus{}b\\minus{}x)\\plus{}(m\\minus{}x\\minus{}a)\\plus{}(m\\minus{}x\\minus{}b)\\equal{}m$, so $ m\\equal{}3x$.\r\n\r\n\r\n\r\nSomeone else can do the second part :p", "Solution_2": "Ok then, but you pretty much already did it.\r\n\r\nSuppose the largest number is greater than or equal to $ 2m/3$. By the proof above, it can't be in the center. Pick $ b$ and $ a$ (as above) so that one of them is the largest number. If it is $ a$, then note we have $ m\\minus{}x\\minus{}a \\equal{} 2m/3\\minus{}a\\leq 0$ in the square, contradiction. If it is $ b$, then note we have $ m\\minus{}x\\minus{}b \\equal{} 2m/3\\minus{}b\\leq 0$ in the square, contradiction.\r\n\r\n(EDIT: forgot the $ \\equal{} 2m/3$ case)", "Solution_3": "I'll finish the [hide=\"second part\"]\nLazarus proved that the center entry is one third of the sum.\n\nLet $ a$ be the least entry, and $ x$ at the center. Then the entry opposite $ a$ is $ m\\minus{}x\\minus{}a\\equal{}2x\\minus{}a$. By definition $ a$ is positive, so $ a\\geq1$.\n\nWe see that $ 2x\\minus{}a\\equal{}\\frac{2m}{3}\\minus{}a\\leq\\frac{2m}{3}\\minus{}1$ since $ a\\geq1$.\n\nAnd we are done.\n[/hide]" } { "Tag": [ "function", "induction", "logarithms", "algebra", "functional equation" ], "Problem": "A function $ f: R \\rightarrow R$,not constant with $ f(1)\\equal{}e$.If $ f$ satisfy the condition $ f(x\\plus{}y)\\equal{}f(x)f(y)$ $ \\forall x,y \\in R$.Show that:\r\n$ 1)$ $ f(x\\minus{}y)\\equal{}\\frac{f(x)}{f(y)},\\forall x,y \\in R$\r\n\r\n$ 2)$ $ f(kx)\\equal{}f^k(x),\\forall k \\in Z$ and $ x\\in R$\r\n\r\n$ 3$ $ f(\\frac{1}{n}x)\\equal{}f^{\\frac{1}{n}},\\forall n \\in N^*$\r\n\r\n$ 4$ $ f(x)\\equal{}e^x,\\forall x\\in Q$", "Solution_1": "[quote=\"Dimitris X\"]A function $ f: R \\rightarrow R$,not constant with $ f(1) \\equal{} e$.If $ f$ satisfy the condition $ f(x \\plus{} y) \\equal{} f(x)f(y)$ $ \\forall x,y \\in R$.Show that:\n$ 1)$ $ f(x \\minus{} y) \\equal{} \\frac {f(x)}{f(y)},\\forall x,y \\in R$\n\n$ 2)$ $ f(kx) \\equal{} f^k(x),\\forall k \\in Z$ and $ x\\in R$\n\n$ 3$ $ f(\\frac {1}{n}x) \\equal{} f^{\\frac {1}{n}},\\forall n \\in N^*$\n\n$ 4$ $ f(x) \\equal{} e^x,\\forall x\\in Q$[/quote]\r\n\r\nUmm....... It's more like a solution than questions.", "Solution_2": "$ x \\equal{} 1$ and $ y \\equal{} 0$;\r\n\r\n$ \\Rightarrow \\, f(1) \\equal{} f(1).f(0)$\r\n\r\n$ \\Rightarrow\\, f(0) \\equal{} 1$.\r\n\r\n$ y \\equal{} \\minus{} x \\Rightarrow\\, f(0) \\equal{} f(x).f( \\minus{} x)$\r\n\r\n$ \\Rightarrow\\, 1 \\equal{} f(x).f( \\minus{} x)$\r\n\r\n$ \\Rightarrow\\, f( \\minus{} x) \\equal{} \\frac {1}{f(x)}$.\r\n\r\nFor 2):\r\n\r\nBy mathematical induction:\r\n\r\nThe basis step is\r\n\r\n$ f(2x) \\equal{} f(x \\plus{} x) \\equal{} f(x).f(x) \\equal{} f^{2}(x)$\r\n\r\nand the inductive step is straightforward.\r\n\r\nP.S. I guessed someone would beat me to it.", "Solution_3": "I think that any function that satisfies the property $ f(x\\plus{}y) \\equal{} f(x).f(y)$ can be written as $ f(x) \\equal{} a^x$ \r\n\r\nhere, $ f(1) \\equal{} e \\ \\Rightarrow \\ a \\equal{} e$\r\n\r\nso, $ f(x) \\equal{} e^x$ .\r\n\r\n\r\nRest of the proofs follow from the properties of exponential functions.", "Solution_4": "[hide]1. $ f(x \\minus{} y \\plus{} y) \\equal{} f(x \\minus{} y)f(y)$\n $ f(x \\minus{} y) \\equal{} \\frac {f(x)}{f(y)}$\n\n2. Induction using $ f(2x) \\equal{} f(x)f(x)$\n\n3. Using 2, let change $ x$ into $ \\frac {x}{k}$\n\n $ f(x) \\equal{} (f(\\frac {x}{k}))^k$\n\n\n $ (f(x))^\\frac {1}{k} \\equal{} f(\\frac {x}{k})$\n\n4. Using 2 and 3 and $ f( \\minus{} 1) \\equal{} e^{ \\minus{} 1}$[/hide]", "Solution_5": "[quote=\"Nora.91\"]I think that any function that satisfies the property $ f(x \\plus{} y) \\equal{} f(x).f(y)$ can be written as $ f(x) \\equal{} a^x$ \n[/quote]\r\n\r\nReally? But it's true over rational numbers only.......\r\n\r\n[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]\r\n\r\nAnd the proof that it's true over rational numbers is $ 1) \\to 2) \\to 3) \\to 4)$, i.e. you will do circular proof.", "Solution_6": "[quote=\"stephencheng\"][quote=\"Nora.91\"]I think that any function that satisfies the property $ f(x \\plus{} y) \\equal{} f(x).f(y)$ can be written as $ f(x) \\equal{} a^x$ \n[/quote]\n\nReally? But it's true over rational numbers only.......\n\n[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]\n\nAnd the proof that it's true over rational numbers is $ 1) \\to 2) \\to 3) \\to 4)$, i.e. you will do circular proof.[/quote]\r\n\r\n\r\nBut isn't Cauchy's functional equation : $ f(x\\plus{}y) \\equal{} f(x) \\plus{} f(y) \\ and \\ not \\ f(x).f(y)$ :?: \r\n\r\n\r\n\r\n[url=http://en.wikipedia.org/wiki/Functional_equation]This page[/url] mentions that $ f(x\\plus{}y) \\equal{} f(x).f(y)$ is satisfied by[b] all exponential functions[/b].\r\n\r\n\r\n_", "Solution_7": "[quote=\"Nora.91\"][quote=\"stephencheng\"][quote=\"Nora.91\"]I think that any function that satisfies the property $ f(x \\plus{} y) \\equal{} f(x).f(y)$ can be written as $ f(x) \\equal{} a^x$ \n[/quote]\n\nReally? But it's true over rational numbers only.......\n\n[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]\n\nAnd the proof that it's true over rational numbers is $ 1) \\to 2) \\to 3) \\to 4)$, i.e. you will do circular proof.[/quote]\n\n\nBut isn't Cauchy's functional equation : $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y) \\ and \\ not \\ f(x).f(y)$ :?: \n\n\n\n[url=http://en.wikipedia.org/wiki/Functional_equation]This page[/url] mentions that $ f(x \\plus{} y) \\equal{} f(x).f(y)$ is satisfied by[b] all exponential functions[/b].\n\n\n_[/quote]\r\nIf $ f(x \\plus{} y) \\equal{} f(x).f(y) \\Rightarrow f(x) \\equal{} a^x\\equal{}e^{cx}$ for $ f: R\\rightarrow R$ and $ f$ is not a constant.\r\n\r\nThen let $ g(x)\\equal{}In(f(x))$ we get $ g(x\\plus{}y)\\equal{}g(x)\\plus{}g(y) \\Rightarrow g(x)\\equal{}cx$ which is a wrong statement as in the link.", "Solution_8": "[quote=\"stephencheng\"]\nIf $ f(x \\plus{} y) \\equal{} f(x).f(y) \\Rightarrow f(x) \\equal{} a^x \\equal{} e^{cx}$ for $ f: R\\rightarrow R$ and $ f$ is not a constant.\n\nThen let $ g(x) \\equal{} In(f(x))$ we get $ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y) \\Rightarrow g(x) \\equal{} cx$ which is a wrong statement as in the link.[/quote]\r\n\r\nAhem... $ \\text{In}(f(x))\\equal{}\\ln {x}$.\r\n\r\n$ \\ln{x\\plus{}y}\\neq\\ln x \\plus{} \\ln y$.\r\n\r\n$ \\ln xy\\equal{}\\ln x \\plus{}\\ln y$.", "Solution_9": "[quote=\"hsiljak\"][quote=\"stephencheng\"]\nIf $ f(x \\plus{} y) \\equal{} f(x).f(y) \\Rightarrow f(x) \\equal{} a^x \\equal{} e^{cx}$ for $ f: R\\rightarrow R$ and $ f$ is not a constant.\n\nThen let $ g(x) \\equal{} In(f(x))$ we get $ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y) \\Rightarrow g(x) \\equal{} cx$ which is a wrong statement as in the link.[/quote]\n\nAhem... $ \\text{In}(f(x)) \\equal{} \\ln {x}$.\n\n$ \\ln{x \\plus{} y}\\neq\\ln x \\plus{} \\ln y$.\n\n$ \\ln xy \\equal{} \\ln x \\plus{} \\ln y$.[/quote]\r\n\r\nWhat do you mean?\r\n\r\nWe take log to $ f(x \\plus{} y) \\equal{} f(x).f(y) \\Rightarrow f(x) \\equal{} a^x \\equal{} e^{cx}$\r\n\r\n$ In(f(x\\plus{}y))\\equal{}In (f(x).f(y))\\equal{}In(f(x))\\plus{}In(f(y)) \\Leftrightarrow g(x\\plus{}y)\\equal{}g(x)\\plus{}g(y)$\r\n\r\nAlso $ f(x)\\equal{}e^{cx} \\Leftrightarrow g(x)\\equal{}cx$ ,\r\n\r\nSo $ f(x \\plus{} y) \\equal{} f(x).f(y) \\Rightarrow f(x) \\equal{} a^x \\equal{} e^{cx}$ is a statement equivalent to\r\n\r\n$ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y) \\Rightarrow g(x) \\equal{} cx$, isn't it?", "Solution_10": "1. Sorry, I thought you were using your $ In$ as notation for inverse function, and not the natural logarithm (please, use \\ln in the future).\r\n\r\n2. I obviously misunderstood the sentence\r\n\r\n[quote]Then let $ g(x)\\equal{}In(f(x))$ we get $ g(x\\plus{}y)\\equal{}g(x)\\plus{}g(y)\\Rightarrow g(x)\\equal{}cx$ which is a wrong statement as in the link.[/quote]\r\n\r\nbecause I don't see anything wrong there, since log of exp is linear identity function.", "Solution_11": "[quote=\"hsiljak\"]2. I obviously misunderstood the sentence\n\n[quote]Then let $ g(x) \\equal{} In(f(x))$ we get $ g(x \\plus{} y) \\equal{} g(x) \\plus{} g(y)\\Rightarrow g(x) \\equal{} cx$ which is a wrong statement as in the link.[/quote]\n\nbecause I don't see anything wrong there, since log of exp is linear identity function.[/quote]\r\n\r\nFrom this link\r\n[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]\r\n\r\nThere exists function $ f_1(x)$ which satisfies $ f_1(x\\plus{}y)\\equal{}f_1(x)\\plus{}f_1(y)$ and is not of the form $ cx$\r\n\r\nDefine $ g(x)\\equal{}e^{f_1(x)}$.\r\n\r\nThen $ g(x)$ satisfies $ g(x\\plus{}y)\\equal{}g(x).g(y)$ and it's not of the form $ e^{cx}$\r\n\r\n\r\nI am sorry for not presenting my idea well. :(", "Solution_12": "Now I get it :wink: You were showing the equivalence between Cauchy's equation and the Cauchy-type equation such as one given in the O.P.'s problem. Nice." } { "Tag": [ "geometry", "analytic geometry" ], "Problem": "Find the area of the region enclosed by the line $3x+4y=12$ and the coordinate axes.", "Solution_1": "[hide]\nThe intercepts are 3 and 4.\n\n$\\text{Area}=\\frac{1}{2}\\cdot 3\\cdot 4= \\boxed{6}$. [/hide]", "Solution_2": "Using double intercept form:\r\n$x/4+y/3=1$\r\nyou get the intercepts of $3 and 4$\r\nSince it's a triangle\r\n$3*4/2$\r\n$\\boxed{=6}$\r\nhope I did the box correctly..." } { "Tag": [ "Euler" ], "Problem": "Se dau 3 cercuri de aceeasi raza 2 cate 2 secante, ce au in comun punctul M. Fie A;B;C celelalte puncte de intersectie, iar O' O\" si O''' centele cercurilor.\r\nSa se demonstreze ca dreptele lui Euler ale triunghiurilor O'O\"O''' si ABC coincid", "Solution_1": "nu vine nimeni cu o solutie?", "Solution_2": "In forma asta, pur si simplu nu e adevarat. Poate trebuie ca cercurile sa aiba raze egale?", "Solution_3": "Ai dreptate grobber cercurile sunt de aceeasi raza. imi cer scuze. Am facut modificarea in enunt. Subiectul este dat la examenul de grad II in 1997", "Solution_4": "[hide=\"Simpatica problema, dar cam usurica.\"]Se arata ca punctul $M$ este simultan centrul cercului circumscris triunghiului $OO'O''$ si ortocentrul triunghiului $ABC$. Mai departe, se considera ortocentrul N al triunghiului $OO'O''$ si se arata usor ca N este si centrul cercului circumscris triunghiului $ABC$. De aici, concluzia este imediata.[/hide]\n[hide=\"Observatie.\"]Enuntul problemei trebuia sa spuna ca centrele $O, O', O''$ ale cercurilor date nu trebuie sa fie situate in varfurile unui triunghi echilateral.[/hide]" } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra solved" ], "Problem": "It's well-known that a group can't be the reunion of 2 subgroups different from itself. Here's a nice problem (even if it's kind of easy :D ):\r\n\r\nProve that a field K (commutative or not) can't be the reunion of 3 subfields different from itself.\r\n\r\n[i]Romanian NMO, final round, 1987, 12'th grade[/i]", "Solution_1": "Maybe you could use another problem I posted somewhere around here (\"Proposed problems\") and which you people disregarded as well... :D", "Solution_2": "Suppose it is possible to have K as the union of K1, K2, K3.\r\nThen K1 not included in K2 U K3 and same for the others (otherwise K would be the union of only two subfields)\r\n\r\nTake x in K1-(K2 U K3), y in K2-(K1 U K3)\r\nx<>0, y<>0 () is in all subfields)\r\nxy not in K1 or K2 so xy in K3\r\nx*xy=x 2 y not in K1 or K3 so x2y in K2\r\ny in K2 so x 2 in K2\r\nx+y not in K1 or K2 so x+y in K3\r\nx(x+y) not in K1 or K3 so x(x+y)=x 2 +xy in K2\r\nx 2 in K2 so xy in K2\r\ny in K2 so x in K2, false\r\n\r\nI suppose this is all. Correct me if I'm wrong. :)", "Solution_3": "Good job :) blondu I moved ot to solved problems", "Solution_4": "Right! \r\nChanged it now.\r\nThx for pointing it out. :)" } { "Tag": [ "function", "probability and stats" ], "Problem": "If $ X$ and $ Y$ are iid r.v's, what can we say about $ E[X.1_A]$ and $ E[Y.1_A]$ for $ A\\in \\sigma(X,Y)$?", "Solution_1": "You cannot say anything deep if you don't make assumptions on your set $ A$\r\nthose expectations are not even equal ( take $ A\\equal{}(X\\in[a,b];Y\\in[c,d])$ for suitable $ a,b,c,d$)\r\n\r\nso what is your point ?", "Solution_2": "Can't we have any relationship between $ E[X.1_A]$ and $ E[Y.1_A]$ when $ X$ and $ Y$ are iid?", "Solution_3": "no, because there are a lot of events inside $ \\sigma(X,Y)$ where $ X$ and $ Y$ can be very different.", "Solution_4": "I think, as siiz said, $ E[X.1_A] \\equal{} E[Y.1_A]$ for events $ A \\equal{} \\{X \\plus{} Y\\le a\\}$ and these events form a $ \\pi$-system which generates the $ \\sigma$- field $ \\sigma(X\\plus{}Y)$", "Solution_5": "oh! now you are talking about $ \\sigma(X\\plus{}Y)$, which is completely different of $ \\sigma(X,Y)$ ..", "Solution_6": "[quote=\"limsup\"]I think, as siiz said, $ E[X.1_A] \\equal{} E[Y.1_A]$ for events $ A \\equal{} \\{X \\plus{} Y\\le a\\}$ and these events form a $ \\pi$-system which generates the $ \\sigma$- field $ \\sigma(X \\plus{} Y)$[/quote]\r\nFurthermore, we can easily show that\r\n\\[ E(X|\\phi(X,Y)) \\equal{} E(Y|\\phi(X,Y)) \\equal{} 0.5\\phi(X,Y)\r\n\\]\r\nfor all symmetric Borel function $ \\phi$ such that $ \\phi(X,Y)$ is integrable. Hence $ E[X.1_A] \\equal{} E[Y.1_A]$ for any $ A$ in sigma algebra generated by the family of $ \\phi(X,Y)$, where $ \\phi$ satisfies above requirements.", "Solution_7": "[quote=\"siiz\"][quote=\"limsup\"]I think, as siiz said, $ E[X.1_A] \\equal{} E[Y.1_A]$ for events $ A \\equal{} \\{X \\plus{} Y\\le a\\}$ and these events form a $ \\pi$-system which generates the $ \\sigma$- field $ \\sigma(X \\plus{} Y)$[/quote]\nFurthermore, we can easily show that\n\\[ E(X|\\phi(X,Y)) \\equal{} E(Y|\\phi(X,Y)) \\equal{} 0.5\\phi(X,Y)\n\\]\n[/quote]\r\nI am not completely sure about the last assertion. I think for example that\r\n\\[ E[X | X\\plus{}Y] \\equal{} E[X | 100(X\\plus{}Y)] \\equal{} \\frac{X\\plus{}Y}{2} \\neq 50(X\\plus{}Y)\\]", "Solution_8": "You are right, as usual :D\r\nOk, let say it again:\r\n\\[ E(X|\\phi(X,Y)) \\equal{} E(Y|\\phi(X,Y))\r\n\\]\r\nfor all symmetric Borel function $ \\phi$. Hence $ E[X.1_A] \\equal{} E[Y.1_A]$ for any $ A$ in sigma algebra generated by the family of $ \\phi(X,Y)$, where $ \\phi$ satisfies above requirements.\r\nI have made so many mistakes recently, not only on this forum :(" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Find all $ x\\in N$ for $ x^2\\plus{}6x\\plus{}2008$ is a perfect square number.", "Solution_1": "[quote=\"thanhnam2902\"]Find all $ x\\in N$ for $ x^2 \\plus{} 6x \\plus{} 2008$ is a perfect square number.[/quote]\r\n\r\n$ y^2\\equal{}x^2\\plus{}6x\\plus{}2008\\equal{}(x\\plus{}3)^2\\plus{}1999$\r\n$ \\Leftrightarrow (y\\plus{}x\\plus{}3)(y\\minus{}x\\minus{}3)\\equal{}1999$\r\n\r\nAs $ 1999$ is prime, $ \\begin{cases}y\\plus{}x\\plus{}3\\equal{}1999 \\\\ y\\minus{}x\\minus{}3\\equal{}1 \\end{cases}$\r\n\r\ngives $ x\\equal{}996$ as the only solution." } { "Tag": [ "puzzles" ], "Problem": "The players of a champion football team are to be given medals by their club managers. Including the reserves, the team of 15 players stand in a line along the pitch with one meter between them each. The manager will begin with any player and then continue presenting the medals to the others. If this procedure is to be completed using the shortest route, then beginning with player number 1 and continuing in line until player number 15 is the most appropriate. (In this case 14 meters would be covered). What will the total distance be if the same procedure is to be completed using the longest possible route for n players?", "Solution_1": "I have no idea if\r\n[hide=\"this\"]\nthe coach starts at position 1 and goes to 15, then 2, then 14, and so on, thus covering 14+13+12+...+1 meters or 105 meters\n[/hide]\r\nis right.", "Solution_2": "Well, $14+13+12+...+2+1=105$...", "Solution_3": ":rotfl: Wow--just wow. :o What a bad start. Problem fixed.\r\n\r\nSo is it right now?", "Solution_4": "I think it's right. At least I can't find a longer path... :P", "Solution_5": "Well taking a smaller example, with 1 2 3 4 5 players, lined up like that, then your path would be to go in this order: 15243 for a total distance of 4+3+2+1=10.\r\n\r\nConsider this path instead: 25143 for a total distance of 3+4+3+1=11.\r\n\r\nI don't know what the longest possible path is but I'll think about it some more, but I think by considering this smaller path, it shows that there is likely a longer path than the 1, 15, 2, 14, 3, 13, ... path.", "Solution_6": "He can wallk through path between men number\r\n1,2 atmost 2 times as well as 14,15\r\n2,3 atmost 4 times as well as 13,14\r\n3,4 atmost 6 times as well as 12,13\r\n4,5 atmost 8 times as well as 11,12\r\n5,6 atmost 10 times as well as 10,11\r\n6,7 atmost 12 times as well as 9,10\r\n7,8 atmost 14 times as well as 8,9\r\nwhich is 2(2+4+6+8+10+12+14) = 112 m because each path length is 1.\r\nbut he cannot walk 112 because of the beginning point. <= Sorry for unclear explanation. It's hard to expain.\r\n\r\nSo the he can walk at most\r\n112 - 1 = 111 m\r\n\r\nAnd the path 8,1,15,2,14,3,13,4,12,5,11,6,10,7,9 is 111 m" } { "Tag": [ "linear algebra", "matrix", "superior algebra", "superior algebra unsolved" ], "Problem": "I have theorem:\r\nLet $ A$ be an algebra with unity. If $ a$ contains matrix units $ e_{ij}, i,j \\equal{} 1,2....,n$ then A is isomorfic $ M_{n}(e_{tt}Ae_{tt})$ for each $ t \\equal{} 1,...,n.$\r\n\r\n We can write $ a_{ij} \\equal{} e_{ti}ae_{jt}$ and $ \\phi : A \\minus{} \\minus{} > M_{n}(e_{tt}Ae_{tt})$ where $ \\phi (a) \\equal{} (a_{ij})$ I don't know why $ \\phi$ is surjective.", "Solution_1": "[quote=\"wiosna\"]I have theorem:\nLet $ A$ be an algebra with unity. If $ a$ contains matrix units $ e_{ij}, i,j \\equal{} 1,2....,n$ then A is isomorfic $ M_{n}(e_{tt}Ae_{tt})$ for each $ t \\equal{} 1,...,n.$\n\nWe can write $ a_{ij} \\equal{} e_{ti}ae_{jt}$ and $ \\phi : A \\minus{} \\minus{} > M_{n}(e_{tt}Ae_{tt})$ where $ \\phi (a) \\equal{} (a_{ij})$ I don't know why $ \\phi$ is surjective.[/quote]\r\n\r\nIf $ b\\equal{}(e_{tt}a_{ij}e_{tt}) \\in M_n(e_{tt}Ae_{tt}),$ then let $ a\\equal{}\\sum_{i,j}e_{it}a_{ij}e_{tj}.$ It's easy to show that $ \\phi(a)\\equal{}b.$" } { "Tag": [], "Problem": "When $ 0.\\overline{36}$ is expressed as a common fraction, what is the sum of the numerator and denominator?", "Solution_1": "To find the fraction N that equals $ 0.\\overline{36}$, you solve:\r\n\r\n$ 100N \\equal{} 36.\\overline{36}$\r\n$ 1 \\quad N \\equal{} 00.\\overline{36}$.\r\n\r\n(so that the repeating part will cancel out)\r\n\r\nSubtracting the second equation from the first,\r\n$ 99N \\equal{} 36 \\implies N \\equal{} \\frac{36}{99} \\implies N\\equal{} \\frac{4}{11}$.\r\nThus, the sum of the numerator and denominator is $ 4 \\plus{} 11 \\equal{} 15$." } { "Tag": [ "inequalities", "absolute value", "combinatorics proposed", "combinatorics" ], "Problem": "Suppose $x_1,...,x_n$ are real numbers and $-1 \\leq x_i \\leq 1$ and $\\sum x_i =0$\r\nProve there is a permutation $\\sigma$ one the ${ 1,...,n \\}}$ that for $1 \\leq p \\leq q \\leq n$ we have\r\n\\[\\mid x_{\\sigma (p)}+...+x_{\\sigma (q)} \\mid \\leq 2 - \\frac 1n\\]\r\nProve we can't use $2 - \\frac 4n$ instead of $2- \\frac 1n$\r\n____________________________________\r\nNo pain, no gain", "Solution_1": "i'm sorry but I don't understant the pb ?? Could you reformulate it please ..", "Solution_2": "I think I have explained enough.\r\nWe have n real numbers between 1 and -1 with sum 0.\r\nProve we can put tem in a row that absolute value of sum of any block is not bigger than $2-\\frac 1n$", "Solution_3": "Hey people solve this problem ;) ;) ;)", "Solution_4": "I'll reformulate it for you so pbornstein can solve it ;-)\r\n\r\n[quote=\"Omid Hatami\"]Consider $x_1,...,x_n \\in \\mathbb{R}$ and $-1 \\leq x_i \\leq 1$ and $\\displaystyle \\sum^n_{i=1} x_i = 0$\n\n1. Show that there exists a permutation $\\sigma$ of the numbers $1 \\rightarrow n$, so that for each $p,q \\in \\mathbb{N}$ for which $1 \\leq p \\leq q \\leq n$, we have $\\mid x_{\\sigma (p)}+...+x_{\\sigma (q)} \\mid \\leq 2 - \\frac 1n$\n\n2. Prove that the above statement is false if we replace $2- \\frac 1n$ by $2 - \\frac 4n$.[/quote]", "Solution_5": "Well why didn't you understand what I said ?", "Solution_6": "Why me? It was Alekk asking :( \r\n\r\nAnyway, I think that this one has been proposed on ML before, maybe on the inequalities forum.\r\nIt also appear in the book from Andreescu and Feng so...\r\n\r\nPierre.", "Solution_7": ":P woops\r\n\r\nsorry pierre, I saw the french flag without avatar and thought it was you :blush: \r\n\r\n\r\nshould have known the post was to short :lol:", "Solution_8": "Don't worry, Peter, even Valentin made this kind of mistake before :D (in the days when belenos had his French flag there for a time): [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=11271]click here[/url] ;) .", "Solution_9": "It passed 3 months and nobody has solution for this problem.I'll send my solution soon." } { "Tag": [ "function", "AMC", "AIME", "number theory", "relatively prime", "arithmetic sequence" ], "Problem": "This is a pretty fun problem. \r\n\r\nA sequence is defined by $a_1=1995$, $a_{n+1}=\\frac{a_n}{a_n+1}$ for all $n\\geq 1$. If $a_{1995}=\\frac{m}{n}$, where $m$ and $n$ are relatively prime integers, find the remainder when $m+n$ is divided by $1000$.\r\n\r\nEdit: This is my $29^2$ post! :D", "Solution_1": "[hide]$a_{n+1}=\\frac{a_n}{a_n+1}$\n$a_{n+2}=\\frac {\\frac{a_n}{a_n+1}}{\\frac{a_n}{a_n+1}+1}=\\frac{a_n}{2a_n+1}$\n$a_{n+3}=\\frac {\\frac{a_n}{2a_n+1}}{\\frac{a_n}{2_n+1}+1}=\\frac{a_n}{3a_n+1}$\n\\vdots\n$a_{n+1994}=\\frac {\\frac{a_n}{1993a_n+1}}{\\frac{a_n}{1993a_n+1}+1}=\\frac{a_n}{1994a_n+1}$\n\nfor $n=1$, $a_n=1995$ so $\\frac{1995}{1994*1995+1}$\n$1995+1994(1995)+1\\equiv (-5)+(-6)(-5)+1\\equiv \\boxed{26}\\mod 1000$[/hide]\r\n\r\nedit: i'm an idiot >_<", "Solution_2": "This isn't from 1995 AIME? It looks exactly like an AIME problem, and $a_1=1995$... out of curiosity, what's the source?", "Solution_3": "[quote=\"eryaman\"]This is a pretty fun problem. \n\nA sequence is defined by $a_1=1995$, $a_{n+1}=\\frac{a_n}{a_n+1}$ for all $n\\geq 1$. If $a_{1995}=\\frac{m}{n}$, where $m$ and $n$ are relatively prime integers, find the remainder when $m+n$ is divided by $1000$.\n\nEdit: This is my $29^2$ post! :D[/quote]\r\n\r\nWe have $a_{n+1}=\\frac{a_n}{a_n+1}$ \r\n\r\n$\\Rightarrow \\frac{1}{{a_{n + 1} }} = \\frac{1}{{a_n }} + 1$\r\n\r\n$\\Rightarrow \\frac{1}{{a_{n + 1} }} - \\frac{1}{{a_n }} = \\frac{1}{{a_n }} - \\frac{1}{{a_{n - 1} }} = \\frac{1}{{a_{n - 1} }} - \\frac{1}{{a_{n - 2} }} = ....1$\r\n\r\nHence the numbers $\\frac{1}{{a_{n + 1} }},\\frac{1}{{a_n }},\\frac{1}{{a_{n - 1} }},\\frac{1}{{a_{n - 2} }}.....$ form an Arithmetic Progression \r\n\r\n$\\Rightarrow \\frac{1}{{a_{1995} }} = \\frac{1}{{a_1 }} + 1994$\r\n\r\n$\\Rightarrow \\frac{1}{{a_{1995} }} = \\frac{1}{{1995}} + 1994 = \\frac{{1995.1994 + 1}}{{1995}}$\r\n\r\n$\\Rightarrow a_{1995} = \\frac{{1995}}{{1995.1994 + 1}} = \\frac{m}{n}$\r\n\r\nSo, $m + n = 1995 + 1995.1994 + 1 = (1995)^2 + 1 = (2000 - 5)^2 + 1$.\r\n\r\nThe remainder when $m+n$ is divided by $1000$ is $26$", "Solution_4": "[quote=\"JesusFreak197\"]This isn't from 1995 AIME? It looks exactly like an AIME problem, and $a_1=1995$... out of curiosity, what's the source?[/quote]Actually it's based off an old Math League question, I modified it some though, in addition to making it fit the AIME format.\r\n\r\nEdit: Oh and $26$ is the correct answer." } { "Tag": [ "linear algebra", "matrix", "algebra", "polynomial", "linear algebra unsolved" ], "Problem": "A matrix in M(n,C) such that \r\n\r\n$tr(A)=0$\r\n$tr(A^2)=0$\r\n.............\r\n$tr(A^{n-1})=0$\r\n$tr(A^n)=k \\neq 0$\r\n\r\nprove that A is diagonalisable", "Solution_1": "Suppose the eigenvalues are not disinct: \r\nThere are $m_1$ eigenvalues equal to $a_1$,$m_2$ eigenvalues equal to $a_2$, and .., ,$m_r$ eigenvalues equal to $a_r$, with $m_1+..+m_r=n$, $r 0$, \u03bd.\u03b4.\u03bf. \r\n\r\n$ \\frac {(ab)^2}{a \\plus{} b} \\plus{} \\frac {(bc)^2}{b \\plus{} c} \\plus{} \\frac {(ca)^2}{c \\plus{} a}\\leq\\frac {a^3 \\plus{} b^3 \\plus{} c^3}{2}$\r\n\r\n\r\n\u0392\u03ac\u03b6\u03c9 \u03bc\u03b9\u03b1 \u03c0\u03c1\u03cc\u03c7\u03b5\u03b9\u03c1\u03b7 \u03bb\u03cd\u03c3\u03b7 :\r\n\r\n\r\n$ a \\plus{} b\\geq 2\\sqrt {ab}$ \r\n\r\n\u0388\u03c4\u03c3\u03b9\r\n\r\n$ RHS\\leq\\frac {(ab)^2}{2\\sqrt {ab}} \\plus{} \\frac {(bc)^2}{2\\sqrt {bc}} \\plus{} \\frac {(ca)^2}{2\\sqrt {bc}} \\equal{} \\frac {ab\\sqrt {ab} \\plus{} bc\\sqrt {bc} \\plus{} ca\\sqrt {ca}}{2}$\r\n\r\n\r\n\u038c\u03bc\u03c9\u03c2\r\n\r\n$ \\frac{a^3 \\plus{} b^3 \\plus{} c^3}{2} \\equal{} \\frac{(a\\sqrt {a})^2 \\plus{} (b\\sqrt {b})^2 \\plus{} (c\\sqrt {c})^2}{2}$\r\n\r\n\u03ba\u03b1\u03b9 \u03b1\u03c0\u03cc \u03c4\u03b7\u03bd $ x^2 \\plus{} y^2 \\plus{} z^2\\geq xy \\plus{} yz \\plus{} zx$\r\n\r\n\u03c0\u03c1\u03bf\u03ba\u03cd\u03c0\u03c4\u03b5\u03b9 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03cd\u03bc\u03b5\u03bd\u03bf .\r\n\r\n\r\n\u03a7\u03ac\u03bd\u03c9 \u03ba\u03ac\u03c0\u03bf\u03c5 ;\r\n\r\n\r\n\r\n\u03a3\u03c4\u03ad\u03bb\u03b9\u03bf\u03c2", "Solution_1": "\u038c\u03c7\u03b9 \u03c3\u03c9\u03c3\u03c4\u03cc\u03c2 \u03b5\u03af\u03c3\u03b1\u03b9. :wink: \u039a\u03b9 \u03b5\u03b3\u03ce \u03c4\u03b7\u03bd \u03b5\u03af\u03b4\u03b1 \u03c4\u03b7\u03bd \u03ce\u03c1\u03b1 \u03c4\u03bf\u03c5 \u03b4\u03b9\u03b1\u03b3\u03c9\u03bd\u03b9\u03c3\u03bc\u03bf\u03cd \u03b1\u03c0\u03cc \u03ad\u03bd\u03b1 \u03bc\u03b9\u03ba\u03c1\u03cc \u03c0\u03bf\u03c5 \u03ba\u03b1\u03b8\u03cc\u03c4\u03b1\u03bd \u03b4\u03af\u03c0\u03bb\u03b1 \u03ba\u03b1\u03b9 \u03c4\u03b7\u03bd \u03ad\u03bb\u03c5\u03c3\u03b1 \u03c3\u03b5 2 \u03bb\u03b5\u03c0\u03c4\u03ac \u03c0\u03c1\u03b9\u03bd \u03b1\u03c3\u03c7\u03bf\u03bb\u03b7\u03b8\u03ce (\u03b1\u03c0\u03bf\u03c4\u03c5\u03c7\u03ce\u03c2) \u03bc\u03b5 \u03c4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03c4\u03c9\u03bd \u03bc\u03b5\u03b3\u03ac\u03bb\u03c9\u03bd. :rotfl: \u0389\u03c4\u03b1\u03bd \u03cc\u03bd\u03c4\u03c9\u03c2 \u03b5\u03cd\u03ba\u03bf\u03bb\u03b1 \u03c4\u03b1 \u03b8\u03ad\u03bc\u03b1\u03c4\u03b1 \u03c4\u03c9\u03bd \u03bc\u03b9\u03ba\u03c1\u03ce\u03bd...", "Solution_2": "\u0395\u03b3\u03c9 \u03c4\u03b7\u03bd \u03b5\u03bb\u03c5\u03c3\u03b1 \u03c3\u03b5 1 \u03b5\u03bd\u03b1 \u03bb\u03b5\u03c0\u03c4\u03bf :lol: \r\n\r\n\u03a0\u03b1\u03bd\u03c4\u03c9\u03c2 \u03c4\u03b7\u03bd \u03bb\u03c5\u03c3\u03b7 \u03b5\u03b4\u03c9\u03c3\u03b5 \u03bf r_boris \u03b5\u03b4\u03c9 [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1102051#1102051[/url]", "Solution_3": "\u0393\u03b9\u03b1 \u03b5\u03c3\u03b1\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03b1 \u03b5\u03bd\u03b1 \u03c0\u03bf\u03bb\u03c5 \u03b5\u03c5\u03ba\u03bf\u03bb\u03bf \u03b8\u03b5\u03bc\u03b1, \u03b5\u03bc\u03b5\u03bd\u03b1 \u03bf\u03bc\u03c9\u03c2 \u03c0\u03bf\u03c5 \u03c4\u03c9\u03c1\u03b1 \u03b1\u03c1\u03c7\u03b9\u03c3\u03b1 \u03bd\u03b1 \u03b1\u03c3\u03c7\u03bf\u03bb\u03bf\u03c5\u03bc\u03b5 \u03bc\u03b5 \u03b5\u03be\u03c9\u03c3\u03c7\u03bf\u03bb\u03b9\u03ba\u03b1 \u03bc\u03b1\u03b8\u03b7\u03bc\u03b1\u03c4\u03b9\u03ba\u03b1 \u03bc\u03bf\u03c5 \u03c6\u03b1\u03bd\u03b7\u03ba\u03b5 \u03ba\u03b1\u03bd\u03bf\u03bd\u03b9\u03ba\u03bf \u03bf\u03c0\u03c9\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03bf 4\u03bf \u03b8\u03b5\u03bc\u03b1 \u03c4\u03c9\u03bd \u03bc\u03b9\u03ba\u03c1\u03c9\u03bd :huh: . \u039f\u03c1\u03b9\u03c3\u03c4\u03b5 \u03b7 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03b3\u03b9\u03b1 \u03c4\u03bf 3\u03bf \u03b8\u03b5\u03bc\u03b1:\r\n\r\n\u0397 \u03b1\u03bd\u03b9\u03c3\u03c9\u03c4\u03b7\u03c4\u03b1 $ xy(x \\plus{} y) \\leq x^3 \\plus{} y^3 (1)$ \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 2 \u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 $ x, y$ \u03b1\u03c6\u03bf\u03c5 \u03b3\u03b9\u03b1 $ (x \\plus{} y)$ \u03b4\u03b9\u03b1\u03c6\u03bf\u03c1\u03bf \u03c4\u03bf\u03c5 \u03bc\u03b7\u03b4\u03b5\u03bd \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b9\u03c3\u03bf\u03b4\u03c5\u03bd\u03b1\u03bc\u03b7 \u03bc\u03b5 \u03c4\u03b7\u03bd $ (x \\minus{} y)^2 \\geq 0$, \u03b5\u03bd\u03c9 \u03b3\u03b9\u03b1 $ x \\plus{} y \\equal{} 0$ \u03b9\u03c3\u03c7\u03c5\u03b5\u03b9 \u03b7 \u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1.\r\n\r\n\u0391\u03ba\u03bf\u03bc\u03b1 \u03b1\u03c0\u03bf \u03c4\u03b7\u03bd \u03b1\u03bd\u03b9\u03c3\u03c9\u03c4\u03b7\u03c4\u03b1 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b9\u03c4\u03b9\u03ba\u03bf\u03c5 - \u03b3\u03b5\u03c9\u03bc\u03b5\u03c4\u03c1\u03b9\u03ba\u03bf\u03c5 \u03bc\u03b5\u03c3\u03bf\u03c5 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 $ (x \\plus{} y)^2 \\geq 4xy (2)$, \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 $ x, y$ \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 $ xy > 0$ \u03ba\u03b1\u03b9 $ (x \\plus{} y) > 0$, \u03b1\u03c1\u03b1 \u03ba\u03b1\u03b9 $ \\frac {xy}{x \\plus{} y} > 0$, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03c0\u03bf\u03bb\u03bb\u03b1\u03c0\u03bb\u03b1\u03c3\u03b9\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 2 \u03bc\u03b5\u03bb\u03b7 \u03c4\u03b7\u03c2 (2) \u03bc\u03b5 $ \\frac {xy}{x \\plus{} y}$ \u03bb\u03b1\u03bc\u03b2\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5 \r\n$ xy(x \\plus{} y) \\geq 4xy{\\frac {xy}{x \\plus{} y}} < \\equal{} > xy(x \\plus{} y) \\geq \\frac {4(xy)^2}{x \\plus{} y} (3)$\r\n\r\n\u0391\u03c0\u03bf (1) \u03ba\u03b1\u03b9 (3) \u03c0\u03c1\u03bf\u03ba\u03b5\u03b9\u03c0\u03c4\u03b5\u03b9 \u03bf\u03c4\u03b9 $ \\frac {4(xy)^2}{x \\plus{} y} \\leq x^3 \\plus{} y^3$ \u03b3\u03b9\u03b1 \u03ba\u03b1\u03b8\u03b5 2 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 $ x,y$. \u0395\u03c4\u03c3\u03b9 \u03b3\u03b9\u03b1 \u03c4\u03bf\u03c5\u03c2 \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c5\u03c2 $ a, b, c$ \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 $ \\frac {4(ab)^2}{a \\plus{} b} \\leq a^3 \\plus{} b^3 (4) , \\frac {4(ac)^2}{a \\plus{} c} \\leq a^3 \\plus{} c^3 (5)$ , \u03ba\u03b1\u03b9 $ \\frac {4(bc)^2}{b \\plus{} c} \\leq b^3 \\plus{} c^3 (6)$\r\n\r\n\u039c\u03b5 \u03c0\u03c1\u03bf\u03c3\u03b8\u03b5\u03c3\u03b7 \u03ba\u03b1\u03c4\u03b1 \u03bc\u03b5\u03bb\u03b7 \u03c4\u03c9\u03bd $ (4), (5)$ \u03ba\u03b1\u03b9 $ (6)$, \u03bb\u03b1\u03bc\u03b2\u03b1\u03bd\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ \\frac {4(ab)^2}{a \\plus{} b} \\plus{} \\frac {4(ac)^2}{a \\plus{} c} \\plus{} \\frac {4(bc)^2}{b \\plus{} c} \\leq 2(a^3 \\plus{} b^3 \\plus{} c^3) (7)$\r\n \r\n\u03ba\u03b1\u03b9 \u03b4\u03b9\u03b1\u03b9\u03c1\u03c9\u03bd\u03c4\u03b1\u03c2 \u03ba\u03b1\u03b9 \u03c4\u03b1 2 \u03bc\u03b5\u03bb\u03b7 \u03c4\u03b7\u03c2 $ (7)$ \u03bc\u03b5 \u03c4\u03bf 4 \u03b5\u03c7\u03bf\u03c5\u03bc\u03b5:\r\n\r\n$ \\frac {(ab)^2}{a \\plus{} b} \\plus{} \\frac {(ac)^2}{a \\plus{} c} \\plus{} \\frac {(bc)^2}{b \\plus{} c} \\leq \\frac {a^3 \\plus{} b^3 \\plus{} c^3}{2}$\r\n\r\n\u0395\u03b9\u03bd\u03b1\u03b9 \u03c3\u03c9\u03c3\u03c4\u03b7 \u03b7 \u03bb\u03c5\u03c3\u03b7 \u03b7 \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03c0\u03bf\u03c5\u03b8\u03b5\u03bd\u03b1 \u03ba\u03b1\u03bd\u03b5\u03bd\u03b1 \u03bb\u03b1\u03b8\u03bf\u03c2;" } { "Tag": [ "blogs", "MATHCOUNTS" ], "Problem": "i am trying to train my school for mathcounts since my teachers arnt doing much so im making a powerpoint and although i might not complete it this year ill definetly finish it over summer tpo have it ready for next year and if anyone reading this is from longfellow i will make sure u never get first again at mathcounts >:O (dont worry im not psychotic).\r\nhttp://www.artofproblemsolving.com/Forum/weblog.php?w=1211", "Solution_1": "It's really hard to read your post if it's only one long, run-on sentence and every word is shortened.\r\n\r\nWhat's the powerpoint about?", "Solution_2": "See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779[/url]. If you insist on making more posts of this quality, I'll probably just start deleting them.", "Solution_3": "[quote]I want to help my school train for MathCounts. My teachers aren't doing too much, so I want to make a PowerPoint. I might not finish it this year, but I'll definitely finish it by the end of the summer and have it ready for next year.[/quote]\r\n\r\nWhat do you think you'll accomplish by making a PowerPoint? I can see that you don't have much of a sense of direction because you did not tell us even vaguely what your PowerPoint will be about, or the goals of your PowerPoint, nor have you asked us about either of these.\r\n\r\n[b]If you spend that same amount of time preparing for MathCounts yourself, you will do much better, thereby improving your team's total score. You will have a much more significant effect on your team's score by doing this than than by putting together a PowerPoint without a sense of direction.[/b]", "Solution_4": "Firstly, what do you intend to accomplish by posting this in this forum?\r\n\r\nIt looks like the PowerPoint presentation is supposed to be used for instruction, or something of an all-things-MATHCOUNTS reference/tutorial. I will assume this in the remainder of this post.\r\n\r\n[quote=\"peregrinefalcon88\"]i am trying to train my school for mathcounts[/quote]\r\n\r\nIt seems more time was put into the creating PowerPoint than into creating this topic. I hope your efforts won't be wasted... at the very least teaching others forces the teacher to have a thorough understanding, and can enhance a teacher's skill in a subject through new perspectives. However, there are quite a few things which should be taken into consideration. Creating a comprehensive curriculum for any subject is huge task, much bigger than I first thought (I once tried to do something similar). Then to teach it is yet another big endeavor - one that tends to fail when the targeted students lack motivation. People spend their entire lives doing this, mind you. Before proceeding, think of what you ultimately want to do with this, and if it is feasible.", "Solution_5": "People I didnt tell u what it was about so you'd be curious and go figure it out yourself. JBL i'm sorry about the post I didn't know it was wrong but I just wanted to see how I was doing on the powerpoint. TZF its nice to make a ppt because its relaxing and gives me something to present as opposed to standing there just talking and when I did the bases persentation it worked out very good. I like to make my powerpionts look fancy so i can have some fun while making them and people think they're really cool.", "Solution_6": "[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding yours.[/color]\r\n\r\nI aw your power point...but there are a lot of spelling mistakes... (sumation ?)", "Solution_7": "As I feared, the OP's powerpoint skills exceed his mathematical skills :(. For instance, slide 18 currently reads\r\n[quote]\nTo find the sum of an arithmetic series such as you simply add the first and last terms and multiply by the number of terms there are. For example to sum 5, 10, 15, 20, 25, \u2026\u2026\u2026 355 we have (5+355)(71) \n[/quote]\r\nin big black letters on the nice red background :P. \r\n\r\nSo, alas, I have to agree with TZF.", "Solution_8": "well i cant say that i really like making the powerpoint or that im good at grammar but i just make it to try and help out my school because im the only aopser and im trying to get others to become aopsers also and fedja thank you i see i forgot to divide by two in that part of the ppt (which is why i would like people to review it with me) but if we could all continue this thread on my blog comments i think it would be better because i dont want to keep returning to this forum thread after the mod said i wasnt supposed to post this here" } { "Tag": [ "algebra", "polynomial", "algorithm", "calculus", "superior algebra", "superior algebra solved" ], "Problem": "Here are some problems from the GM 3/2000 :\r\n\r\n\r\n1) Compute all the matrices A in M_2(Z) with the property that there exists n natural n>=2 so that:\r\nA^(n+1)-nA=B.\r\n\r\nwhere B=\r\n2 2\r\n2 2 \r\n\r\n2)Let f be in R[x] a polinomial (the polinomial is not constant). Prove the equivalence if the statements:\r\ni) f doesn't have real roots;\r\nii)for any A in M_2(R), det(f(A))=0 iff f(A)=O2.\r\n\r\n3)Compute the det:\r\nD=\r\n1/k 1/(k+1) ... 1/n\r\n1/(k+1) 1/(k+2) ... 1/(n+1)\r\n............................\r\n1/n 1/(n+1) ... 1/(2n-k)\r\nwhere k and N are natural, k<=n.", "Solution_1": "For the third:\r\n\r\nLook at this determinant:\r\ndet(A) where A=(a_i,j) and a_i,j=1/(x_i+y_j). It is easier to compute this det. It can be computed using polynomial methods, but it does not have a nice form. It belongs to Cauchy. If you can't compute it look for the solution in Knuth's \"The Art of Computer Programming - Fundamental Algorithms\" (it has a lot of math in it).", "Solution_2": "For the first.\r\n\r\nAB=BA. By writing A=\r\nx y \r\nz t\r\nwe get x=t and y=z, so A has the form\r\nx y\r\ny x\r\nIt is easy to see that A k has the form\r\nx_k y_k\r\ny_k x_k\r\nand \r\nx_(k+1)=x*x_k+y*y_k \r\ny_(k+1)=y*x_k+x*y_k\r\nDenote s_k=x_k+y_k and s=x+y.\r\nObviously s_k=s k.\r\n\r\nOur problem turns to finding all x,y such that \r\nx_(n+1)-nx=2\r\ny_(n+1)-ny=2.\r\nBy adding these two we get s n+1-ns=4. In integers s and n with n \\geq 2, this equation has the solutions: s=-1 n=3 and s=2 n=2.\r\n\r\nIf s=-1 then y=-1-x and x_(k+1)=x*x_k+(-1-x)*((-1) k-x_k)=(2x+1)*x_k+(-1) k+1+x.\r\nSince n=3 we must have x_4=3x+2. But x_4=(2x+1)*x_3-1+x, so\r\n(2x+1)*x_3=2x+3. This should be easy to solve.\r\nDo smth similar 4 s=n=2.", "Solution_3": "for the third one\r\nlet A \\in Mp(R).if A=(a_ij) with a_ij=1/(x_i+y_j)\r\nwe have that \r\ndet(A)=V(x1,x2,...,xp)*V(y1,y2,...,yp)/ \\pi {i=1,j=1}(xi+yj), where V(x,y,z,...) is a van der Mont determinant... \r\nwhich is not hard to proove... but it's a lot of calculus :D" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "let $P(x)$ be a polynomial with degree $3n$ such that:\r\n$P(0)=P(3)=...=P(3n)=2$\r\n$P(1)=P(4)=...=P(3n-2)=1$\r\n$P(2)=P(5)=...=P(3n-1)=0$\r\nand we know $P(3n+1)=730$\r\nfind all possible values of $n$\r\nIt is for years ago but i have got stuck with this. help is desparetly needed.", "Solution_1": "Please solve this \r\n\r\ni have done what ever i can but i dont know, how on the earth can we find out the $n$?\r\n\r\nCan someone find the solution for it in usa olympiad archives or something?", "Solution_2": "huh ? :huh: \r\nLets take n=1 then P(3*1+1)=730\r\nNow take n=2 then P(3*2-2)=1\r\nBut 1 =/= 730 ...\r\n\r\nAm I missing something ?", "Solution_3": "You have shown that $n \\neq 2$.\r\n\r\nhttp://www.kalva.demon.co.uk/usa/usa84.html", "Solution_4": "I don't quite understand the problem :D \r\n(but to say true I don't give a d*** )" } { "Tag": [ "inequalities", "logarithms" ], "Problem": "Engel mistakenly asserts the identity\r\n\r\n$ \\sum_{k \\equal{} 1}^{\\infty} \\frac {1}{F_k} \\equal{} 4 \\minus{} \\phi$\r\n\r\nwhere $ F_n$ is the Fibonacci sequence and $ \\phi$ is the golden ratio. The value of this sum is in fact [url=http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html]not known[/url] (or at least not known in a form this neat). However, can you figure out a reciprocal identity involving the Fibonacci numbers whose value is actually $ 4 \\minus{} \\phi$?\r\n\r\n(I do not know the answer. I also have no idea why such a blatant error is in Engel.)", "Solution_1": "Engel almost certainly means the so-called \"Millin series\":\r\n[url]http://mathworld.wolfram.com/MillinSeries.html[/url]\r\n\r\nWhile we are on the subject, here is another error I found. On page 184, Engel asks to prove the inequality\r\n\\[ \\sqrt {xy} < \\frac {x \\minus{} y}{\\ln x \\minus{} \\ln y} < \\frac {x \\plus{} y}{2},\r\n\\]\r\nwhere $ x > y > 0$.\r\n\r\nIn the solution on page 197, using a geometric argument, Engel derives the inequality\r\n\\[ \\frac {2(x \\minus{} y)}{x \\plus{} y} < \\ln x \\minus{} \\ln y,\r\n\\]\r\nwhich gives the right inequality. However, he also derives the inequality\r\n\\[ \\frac {x \\minus{} y}{\\sqrt {xy}} < \\ln x \\minus{} \\ln y,\r\n\\]\r\nwhich is the opposite of what he wants! So my challenge to anyone is to fix the argument.", "Solution_2": "I think this works to prove the left inequality . . .\r\n[hide]\nFirst, we find the minimum of \n\n$ \\sqrt{z}-\\sqrt{\\frac{1}{z}}-\\ln(z)$.\n\nDifferentiating, we want\n\n$ \\frac{1}{2}z^{-1/2}+\\frac{1}{2}z^{-3/2}-\\frac{1}{z}=0$,\n\nso we can multiply through by $ 2z^{3/2}$ to get\n\n$ z-2\\sqrt{z}+1=0$,\n\nwhich clearly has a double root at $ 1$. The original expression is equal to $ 0$ at $ z=1$, so $ 0$ is its minimum. We see that the expression is $ >0$ for $ z>1$. Thus, substituting $ \\frac{x}{y}$ for $ z$, we find that for $ x>y$, \n\n$ ln(\\frac{x}{y})<\\sqrt{\\frac{x}{y}}-\\sqrt{\\frac{y}{x}}$.\n\nSo\n\n$ \\ln x-\\ln y< \\frac{x-y}{\\sqrt{xy}}$\n\nas desired?\n[/hide]" } { "Tag": [], "Problem": "In 4 liter of water there are 200 gr $ NaCl$ .What is the molar concentration of $ NaCl$ in the water?\r\n\r\nThe molar concentration is the number of moles in 1 liter", "Solution_1": "Well $ \\mbox{Molarity}\\equal{}\\frac{\\mbox{mols}}{\\mbox{Volume}}$ if I can recall correctly. I don't have a periodic table in front of me so I don't know the molar mass of $ \\mbox{NaCl}$ but once you know it, you just divide grams by molar mass to get mols and then divide by the 4 liters.", "Solution_2": "For this I'm only going to use $ 5$ significant figures to describe the numbers. According to my periodic table, $ \\text{1 mol of NaCl \\equal{} 58.442 g}$.\r\n$ \\text {200 g of NaCl}\\times{\\frac {\\text{1 mol}}{\\text{58.442 g of NaCl}}} \\equal{} \\text {3.4222 mols of NaCl}$\r\nEven though this can be simplified earlier, we now divide $ \\text{3.4222 mols of NaCl}$ by $ \\text{4 liters}$, giving us the molar concentration of $ \\text{0.85555 mols of NaCl per liter}$ of water." } { "Tag": [ "probability", "geometry", "trigonometry", "rectangle", "inequalities", "LaTeX", "Support" ], "Problem": "The Mock AMC contest will take place on Friday, July 23rd, at 3:00 PM Eastern time. If you cannot make this date, the problems will still be left up in this forum, so you may take the contest whenever you want, albeit \"unofficially\". You can PM me your answers and I will send you back your score (or you can check your score on your own after consulting the answers), but unfortunately I will not be able to add it to the high score list, should it qualify.\r\n\r\nThe rules of this contest are the same as those of the AMC 10/12. The test contains 25 multiple-choice questions in approximate increasing order of difficulty. Each question has five answer choices, and only one of these is the correct answer. You will have 75 minutes to complete the test. Each correct answer is worth 6 points, each skipped question is worth 2.5 points, and each incorrect answer is worth 0 points. Therefore, it is [i]not[/i] to your advantage to guess on a question unless you have eliminated a significant number of answer choices. [url=http://www.collegeboard.com/counselors/hs/sat/registration/calc.html\\]Any calculator acceptable for use on the SAT[/url] may be used for this contest. You may not use outside resources such as books, the internet, etc., but you may use graph paper, a ruler, compass, and a protractor.\r\n\r\nAt 3:00 PM Eastern time, I will post the problems in an announcement entitled \"Problems of the Mock AMC Contest\" in the AMC forum. The contest will end at precisely 4:15 PM Eastern time. In case your clock is a bit off, set a timer for 75 minutes and start it at the time that the questions are posted. Immediately after the contest is over, please PM me your answers (no solutions are necessary; all you need to enter on the PM is the letter corresponding your answer). The best way to do this would be to open a new PM before the contest and type in the following:\r\n\r\n1.\r\n2.\r\n3.\r\n\r\n...all the way to 25. As you work on the contest, enter your answers on the PM. That way, as soon as 75 minutes are up, you will have all your answers ready to be sent. I will allow a few minutes' buffer time for slow modems, etc., but not more than that, so please be prompt!\r\n\r\nAfter I have graded the contest and sent back individual scores, I will post (in spoiler) the correct answers on the forum, and solutions may be discussed (but please do not discuss solutions in the thread with the problems, so as not to ruin the contest for the unofficial test takers). I will also post a list of high scorers (all scorers of 100+ or the top 10 scorers, whichever would have more) on the forum. \r\n\r\nIf you have any questions, feel free to ask! Good luck to everyone! ;)", "Solution_1": "I assume, that next to the 1, for example, there will be (a) or (b) ..... (e)?", "Solution_2": "Yes, when 75 minutes are up you should have the PM filled out with a, b, c, d, or e next to all the questions you wish to answer.\r\n\r\nHere is a practice question:\r\n\r\nA weighted coin has a [tex]\\frac{2}{5}[/tex] chance of coming up heads and a [tex]\\frac{3}{5}[/tex] chance of coming up tails. Given that the probability of getting exactly 4 heads out of 5 tosses is [tex]\\frac{m}{n}[/tex], with m and n relatively prime positive integers, find [tex]m+n[/tex]. \r\n\r\n(A) 141 (B) 173 (C) 673 (D) 1202 (E) 1242", "Solution_3": "JSRosen3 wrote:Here is a practice question:\n\nA weighted coin has a [tex]\\frac{2}{5}[/tex] chance of coming up heads and a [tex]\\frac{3}{5}[/tex] chance of coming up tails. Given that the probability of getting exactly 4 heads out of 5 tosses is [tex]\\frac{m}{n}[/tex], with m and n relatively prime positive integers, find [tex]m+n[/tex]. \n\n(A) 141 (B) 173 (C) 673 (D) 1202 (E) 1242\n\n\n\nSpoiler:\n\n\n\n[hide]\n\nC.\n\n\n\n2/5*2/5*2/5*2/5*3/5+2/5*2/5*2/5*3/5... there are 5 ways to order 2/5*2/5*2/5*2/5*3/5, so the probability is 5*2/5*2/5*2/5*2/5*3/5 = 48/625. 48 + 625 = 673.[/hide]", "Solution_4": "Approximately what number would that practice question be on the test?", "Solution_5": "Probably about #11 or #12. Hard to say for sure.\r\n\r\nAnd yes, Rep, there will be plenty of harder questions.", "Solution_6": "[hide]\n\n\n\nC)\n\n\n\n(2/5)^4 * 3/5 * C(5,4)\n\n\n\n[/hide]", "Solution_7": "Another few sample questions:\r\n\r\n(approx. difficulty: #4) If [tex]2^{1000}-2^{999}-2^{998}+2^{997}=q*2^{997}[/tex], find the value of [tex]q[/tex].\r\n\r\n(A) 1 (B) 2 (C) 3 (D) 4 (E) 5\r\n\r\n\r\n(approx. difficulty: #13) Find the largest integer [tex]n[/tex] with [tex]02 there exist exactly k integers a such that:\r\n\r\n$k|\\varphi(p^{a})$\r\n\r\nand\r\n\r\n$a^{k}\\equiv 1 \\mod{p^{a}}$\r\n\r\nI apolgise if I have misunderstood the question, just wanted to make sure", "Solution_2": "Not $k$ integers, but $k$ different residues $b \\mod p^{a}$ with $b^{k}\\equiv 1 \\mod p^{a}$ (he just used $a$ twice).\r\nThis follows from the existence of primitive roots $\\mod p^{a}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=55473 ) for example.", "Solution_3": "Hi Fabian, nice to see you on ML!\r\n\r\nWell, I assume that the $a$ in $p^{a}$ is another $a$ then the $a$ in $a^{k}$, otherwise the problem would not make much sense :D\r\n\r\nLet's rephrase the problem:\r\n[color=brown]Let $p$ be an odd prime and $a$ be a positive integer. Furthermore let $k$ be a divisor of $\\varphi(p^{a})$. Then the congruence $x^{k}\\equiv 1 \\pmod{p^{a}}$ has exactly $k$ pairwise incongruent solutions in $x$.[/color]\r\n\r\nProof: Let $g$ be a primitive root modulo $p^{a}$. Then $x\\equiv g^{y}$ for some $y \\in \\{1,\\ldots,\\varphi(p^{a})\\}$. \r\nWe have\r\n$x^{k}\\equiv 1 \\pmod{p^{a}}\\Leftrightarrow g^{yk}\\equiv g^{\\varphi(p^{a})}\\pmod{p^{a}}\\Leftrightarrow yk\\equiv \\varphi(p^{a}) \\pmod{\\varphi(p^{a})}$.\r\nHence the number of solutions $x \\mod p^{a}$ is equal to the number of solutions in $y\\mod{\\varphi(p^{a})}$. But $yk\\equiv \\varphi(p^{a}) \\pmod{\\varphi(p^{a})}$ has exactly $k$ solutions in $y$ since $gcd(k, \\varphi(p^{a}))=k$ and $k\\mid \\varphi(p^{a})$ by definition.\r\n$\\Box$\r\n\r\nHere we use the fact that $ax\\equiv b \\pmod{c}$ has solutions in $x$ iff $d=gcd(a,c)\\mid b$ and that if it has solutions, the number of solutions is $d$.\r\n\r\nEdit: ZetaX was faster.....once again :P", "Solution_4": "danke recht sch\u00f6n, yimin. ja, hab mich tats\u00e4chlich auf mathlinks verirrt.\r\nalso, bau das n\u00e4mlich in mein spezialgebiet ein und bin auf den beweis nicht draufgekommen.\r\ndann, gute nacht", "Solution_5": "[quote=\"Fabian May\"]danke recht sch\u00f6n, yimin. ja, hab mich tats\u00e4chlich auf mathlinks verirrt.\nalso, bau das n\u00e4mlich in mein spezialgebiet ein und bin auf den beweis nicht draufgekommen.\ndann, gute nacht[/quote]\r\n\r\nEnglish here please. If not, use PM's." } { "Tag": [], "Problem": "This is from Singapore Math:\r\n\r\nPeter and Paul each had an equal amount of money. Each day Peter spent 36 dollars\r\nand Paul spent 48 dollars. When Paul used up all his money, Peter still had 240 dollars left. \r\nHow much money did each of them have at first?\r\n\r\nI just stated that Peter saved 12 bucks everyday compared to Paul, and hence, Paul spent money for 20 days, and 20 x 48 is 960. I was just wondering if I can only make this assumption/statement because they started off with the same amount. Also, any other ways to solve?", "Solution_1": "Yes. That's correct. You can only make that assumption when you have the same starting amount." } { "Tag": [ "function" ], "Problem": "I have this one math problem that my friend gave to me as a joke about an hour ago.\r\n\r\nWhat is the gradient of the function y = [2x(x^2+3x-4)+4x-4]/[x-1] when x = 2?\r\n\r\nSo far I think I got this, but it's probably wrong.\r\n\r\n[(4x^8+12x-16)+8-4]\r\n[(520)+8-4]\r\n[525] = y\r\n\r\nI need someone to help me out with the answer, and the steps you took to get it.", "Solution_1": "[quote=\"LJ6580\"]\nWhat is the gradient of the function y = [2x(x^2+3x-4)+4x-4]/[x-1] when x = 2?\n\n[(4x^8+12x-16)+8-4]\n[(520)+8-4]\n[525] = y\n[/quote]\r\nYour $ 4x^8$ term should be $ 4x^2$ instead.\r\nAnswer:\r\n$ \\frac{2x(x^2\\plus{}3x\\minus{}4)\\plus{}4x\\minus{}4}{x\\minus{}1}\\equal{} \\frac{(2x)(x\\plus{}4)(x\\minus{}1)\\plus{}4(x\\minus{}1)}{x\\minus{}1}\\equal{} 2x(x\\plus{}4)\\plus{}4\\\\\r\n2(2)(2\\plus{}4)\\plus{}4\\equal{}4\\cdot6\\plus{}4\\equal{}28$", "Solution_2": "How do I get the EXACT gradient, not the average?" } { "Tag": [ "function", "limit", "calculus", "calculus computations" ], "Problem": "Can anyone help me set up this problem. \r\n\r\nA pond initially contains 1,000,000 gal of water and an unknown amount of undesirable chemical. Water containing 0.01 gram of this chemical per gallon flows into the pond at a rate of 300 gal/hr. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. \r\n\r\na). Write a differential equation for the amount of chemical in the pond at any time. I am confused with the units, including gal and grams. Am I supposed to work with grams or concentration? \r\n\r\nAny help would be greatly appreciated. THx", "Solution_1": "My suggestion: make your main function, $y(t),$ be the mass of the the contaminant in the whole pond, measured in grams.\r\n\r\nThe concentration in the pond at any time is then $10^{-6}y(t),$ with the units of measurement there being grams per gallon. (Yes, that's a truly weird unit of concentration - but I've actually seen it. Just go along with it.)\r\n\r\nNow, what happens during a small interval of time $\\Delta t?$ (Use hours as the unit of measurement for $\\Delta t.$)\r\n\r\nThe amount of the chemical flowing into the pond is $300\\cdot 0.01\\cdot \\Delta t.$ The units of measurement there are\r\n\r\n$\\frac{\\text{gal}}{\\text{hr}}\\cdot\\frac{\\text{gram}}{\\text{gal} }\\cdot \\text{hr}=\\text{gram}$\r\n\r\nNow, what is the amount flowing out of the pond? The volume of water flowing out is $300\\cdot\\Delta t.$ The concetnration varies with time, but taking $\\Delta t$ to be a very short interval, we approximate it as $10^{-6}\\cdot y(t).$ Then the amount flowing out is approximately $10^{-6}\\cdot y(t)\\cdot 300\\cdot\\Delta t.$ Once again the units work out to be grams. So:\r\n\r\n$y'(t)\\cdot\\Delta t\\approx 300\\cdot 0.01\\cdot \\Delta t-10^{-6}\\cdot y(t)\\cdot 300\\cdot\\Delta t$\r\n\r\n$y'(t)=3-(3\\times 10^{-4})y(t).$\r\n\r\nI'll leave you to solve that equation yourself. It is both linear and separable and hence can be solved in more than one way.\r\n\r\nTwo questions:\r\n\r\n1. What is the steady-state condition of the pond? In particular, what is $\\lim_{t\\to\\infty}y(t)$ (or the concentration you could derive from that)? Hint: the \"common sense\" value is the correct value, and could be seen in this autonomous differential equation without bothering to find the solution.\r\n\r\n2. What is the time scale? Here's an interesting alternative: imagine setting up a movable impermeable membrane between the inlet and the outlet so that the input solution never mixes with the original contents of the pond at all. It would take $\\frac{10^6}{300}$ hours to completely replace the original contents of the pond. So that - about 333 hours - is some kind of fundamental time scale here. Now for my question: for the mixing process, what is the half-life? (Yes, that is a meaningful question, if you look at it correctly.)" } { "Tag": [ "geometry", "trapezoid", "circumcircle", "geometry open" ], "Problem": "In trapezoid $ ABCD$, $ AD\\parallel BC$. $ K$ on $ AB$ and $ L$ on $ CD$, so that $ \\angle ABL \\equal{} \\angle CDK$. Prove that $ \\angle ALB \\equal{} \\angle CKD$.\r\n\r\nI've posted this problem in Pre-Olympiad forum, but since no one are keen to answer my problem, I decided to put it here\r\n\r\nThanks", "Solution_1": "I think that the problem is not true in general.\r\n\r\nBecause of $ \\angle ABL \\equal{} \\angle CDK,$ we have that the circumcircles $ (O),\\ (O'),$ of the triangles $ \\bigtriangleup BKL,\\ \\bigtriangleup DKL$ respectively, are congruent.\r\n\r\nIt is easy to show that $ \\angle ALB \\equal{} \\angle CKD,$ in the particular case when the vertices $ C,\\ A,$ of the given trapezium $ ABCD,$ lie on the circles $ (O),\\ (O'),$ respectively.\r\n\r\nKostas Vittas." } { "Tag": [ "algebra", "polynomial", "algebra open" ], "Problem": "P(x) is a polinomyal with integers coefficients such that $P(n)>n$ ; $n \\in Z$ , for each positive integer m, there's a term of the serie:\r\n\r\n$P(1),P(P(1)),P(P(P(1))),...$\r\n\r\nwhich is divisible by m.\r\nShow that $P(x)= x+1$", "Solution_1": "$f(x)=P(x)-x$ is a polynomial of the same degree $d$ as $P(x)$. \r\n\r\nPut $N=P(P(P(\\dots(P(a))\\dots)), m=f(N)$, where $P$ is applied $k$ times. Then $P(N)$ equals $N$ modulo $m$, so $P(P(n))$ equals $N$ modulo $m$, etc. Hence our sequence contains at most $k$ different residues modulo $m$. But $m>k$ for large $k$, the onliest exception is $P(x)=x+1$.", "Solution_2": "\u00bfPuedes traducirlo al espa\u00f1ol???? >>>>>Esto para Mario-Peru\r\nYo no se mucho ingles y no entiendo muy bien la solucion :blush: \r\n\r\nPor favor :oops:", "Solution_3": "$f(x)=P(x)-x$ es un polinomio del mismo grado $d$ que $P(x)$. \r\n\r\n\r\nPonemos $N=P(P(P(\\dots(P(a))\\dots)), m=f(N)$, donde $P$ es aplicado $k$ veces. Luego $P(N)$ es congruente con $N$ en m\u00f3dulo $m$, entonces $P(P(n))$ es congruente con $N$ en m\u00f3dulo $m$, etc. Entonces nuestra sucesi\u00f3n contiene como m\u00e1ximo $k$ residuos diferentes en m\u00f3dulo $m$. Pero $m>k$, para $k$ mayor, la \u00fanica excepci\u00f3n es $P(x)=x+1$.", "Solution_4": "Gracias,no se mucho ingles ,recien estoy aprendiendo algo..... :blush:" } { "Tag": [], "Problem": "Can anyone post solutions to Round 4?", "Solution_1": "Here is my solution to the second problem:\r\n\r\nVerify that the number of such sequences is the number of paths from the leftmost $ 0$ to the rightmost $ 0$ in the following figure:\r\n\r\n[code]\n 0 0 0 0 . . . . . 0 0 \n 1 1 1 . . . . . 1 1 \n 2 2 2 . . . . 2 2 \n 3 3 . . . 3 3 \n 4 4. . . 4 4 \n 5 . . . 5 \n . . . \n n [/code] \n\n\nwhere from each of the entries we may reach the next column either to the top-right or to the bottom-right entry. Here the $ i$th column contains all the possible values of $ a_i$.\n\nConsider the following configuration of the above figure:\n\n\n[code]\n 0\n 0 1\n 0 1 2\n . \n . \n 0 1 2 . . .n-2\n 0 1 2 3 . . .n-1\n0 1 2 3 4 . . . n\n[/code]\r\n\r\nwhere each of the numbers represents a lattice point in the region $ 0\\le y\\le x\\le n$. Hence our problem is equivalent to finding the number of lattice paths from $ (0,0)$ to $ (n,n)$ which do not cross the line $ y \\equal{} x$ and from a point we may go either up or right. This is the $ n$-th Catalan number $ \\frac {1}{n \\plus{} 1}\\binom{2n}{n}$, which is well known. Q.E.D.", "Solution_2": "Yeah, that was all I got as well (with the same solution via the standard Catalan path count), though admittedely I was more focussed on other things (school exams and music stuff) the last couple of weeks.", "Solution_3": "Would the marks have come from showing it is equivalent to the standard catalan path, or would the marks be for the proof that they count to catalan numbers?", "Solution_4": "I messaged Valentin about this, and he assured me that just showing it was equivalent to the standard Catalan (going up in lattice points such that x>=y) and then quoting the formula was enough." } { "Tag": [ "absolute value" ], "Problem": "Okay, so I got x = -17, which makes the statement true (at least it seems to me).\r\n\r\nx - |x+4| = \u00b130\r\nx+4 = -(x + 30)\r\nx+4 = -x-30\r\nx = -17\r\n\r\n-----\r\n\r\nx - |x+4| = \u00b130\r\nx+4 = -(x - 30)\r\nx+4 = -x+30\r\nx = 13\r\n\r\n13 does not make the statement true. :| Why is that? Can someone explain this please? Also, is -17 the only solution?", "Solution_1": "[quote=\"I Hate Watermelons\"]Okay, so I got x = -17, which makes the statement true (at least it seems to me).\n\nx - |x+4| = \u00b130\nx+4 = -(x + 30)\nx+4 = -x-30\nx = -17\n\n-----\n\nx - |x+4| = \u00b130\nx+4 = -(x - 30)\nx+4 = -x+30\nx = 13\n\n13 does not make the statement true. :| Why is that? Can someone explain this please? Also, is -17 the only solution?[/quote]\r\n\r\n\r\n$ |x|$----> if $ x\\geq 0$---> $ |x|\\equal{}x$\r\n ----> if $ x<0$ ---> $ |x|\\equal{}\\minus{}x$\r\n\r\n\r\n$ |x \\minus{} |x\\plus{}4\\parallel{} \\equal{} 30$\r\n\r\na) $ x\\plus{}4\\geq 0$-----> $ x\\geq \\minus{}4$\r\n$ |x \\minus{} (x\\plus{}4)| \\equal{} 30$---> $ |x \\minus{} x\\minus{}4| \\equal{} 30$-----> $ |4| \\equal{} 30$ ----> $ \\phi$\r\n\r\n$ x\\plus{}4\\leq 0$-----> $ x\\leq \\minus{}4$\r\n$ |x \\plus{} (x\\plus{}4)| \\equal{} 30$----> $ |2x\\plus{}4| \\equal{} 30$---> $ x\\leq \\minus{}4$---> $ \\minus{}(2x\\plus{}4)\\equal{}30$----> $ \\minus{}2x\\minus{}4\\equal{}30$----> $ x\\equal{}\\minus{}17$", "Solution_2": "yes, -17 is the only solution.\r\n\r\nif x>=-4, the value of x-|x+4| is always -4, and absolute value of that is 4.\r\n\r\nwhich doesnt equal 30.\r\n\r\nthus, we know that 13, which is larger than -4, is not the solution." } { "Tag": [ "algebra unsolved", "algebra", "Function equations" ], "Problem": "Note: Two problems below is not \"unsolved problem\" :arrow: \r\n\r\n[b]Problem 1[/b]: Find all $f: R\\to R$ sastifying $f(x^{2}+f(y))=xf(x)+y$\r\n\r\n[b]Problem 2[/b]: Find all $f: R\\to R$ sastifying $f(xf(x)+f(y))=(f(x))^{2}+y$\r\n\r\nEnjoy!", "Solution_1": "1)Firstly, by putting $x=0$ we get $f(f(y))=y$ and $f(x)=f(y)\\Rightarrow f(f(y))=f(f(x))\\Rightarrow x=y$. \r\nThis means $xf(x)=f(f(x))f(x) \\rightarrow f(x^{2}+f(0))=f(f(x)^{2}+f(0))\\Rightarrow$\r\n$\\Rightarrow x^{2}=f(x)^{2}\\Rightarrow |x|=|f(x)|$.\r\nSo $\\exists \\sigma: \\mathbb{R}\\to \\{1,\\;-1\\}\\;/\\;f(x)=\\sigma(x)x$. Notice in any case $f(0)=0$. From inyectivity and by letting $y=0$ respectively we get $\\sigma(-x)=\\sigma(x)$ and $\\sigma(x)=\\sigma(x^{2})$. Now suppose $\\exists\\; z>y\\in \\mathbb{R}^{+}\\;/\\;\\sigma(z)\\neq \\sigma (y)$; then by letting $x=\\sqrt{z-f(y)}$ we have \r\n$xf(x)+y=\\sigma(x)x^{2}+y =f(x^{2}+f(y))=f(z)=\\sigma(z)z=$\r\n$=\\sigma(z)x^{2}+\\sigma(y)\\sigma(z)y=\\sigma(z)x^{2}-y \\Rightarrow 2y=x^{2}(\\sigma (z)-\\sigma(x^{2}))$; here $y>0$ implies $2y=2x^{2}\\Rightarrow y=z-f(y)$; now, since $z>y$ we have actually $2y=z$; however $\\sigma (y^{2})=\\sigma(y)\\neq \\sigma (2y)=\\sigma (4y^{2})$ but since $(2x)^{2}>x^{2}>0$ repeating the reasonament we reach $2x^{2}=4x^{2}\\rightarrow 2=4$, a contradiction. So $\\forall x,y\\in \\mathbb{R}^{+}\\;\\sigma(x)=\\sigma(y)$, which means either $\\forall x\\in \\mathbb{R}\\;f(x)=x$ either $\\forall x\\in \\mathbb{R}f(x)=-x$.\r\n2) It seems it is indeed almost the same as the previous one, once you have noticed biyectivity. If you let $A=f^{-1}(0)$, by letting $x=A$ you get $f(f(y))=f(Af(A)+f(y))=f(A)^{2}+y=y$. And then you can make directly, I think, the previous substitution $x$ by $f(x)$ or realize this is actually the same as equation 1) using the fact $f^{-1}(x)=f(x)$.\r\n[size=59]But this is wrong.[/size]" } { "Tag": [ "function" ], "Problem": "I don;t know if it difficult, but it quite interesting. any explanation?\r\n 1/89 = 0. 0 1 1 2 3 5 955056... \r\n 1/9899 = 0.00 01 01 02 03 05 08 13 21 34 55 904636...\r\n 1/998999 = 0.000 001 001 002 003 005 008 013 021 034 055 089 144 233 377 \r\n 610 9885... \r\n 1/99989999 = 0.0000 0001 0001 0002 0003 0005 0008 0013 0021 0034 0055 \r\n 0089 0144 0233 0377 0610 0987 1597 2584 4181 67660947...", "Solution_1": "The Fibonacci Series has the generating function\r\n\r\n$\\frac{x}{1-x-x^{2}}= \\sum_{k=0}^{\\infty}F_{k}x^{k}$\r\n\r\nPlug in $x = \\frac{1}{10^{k}}$. ;)" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "angle bisector", "perpendicular bisector", "geometry unsolved" ], "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14044[/img]\r\n\r\n :(", "Solution_1": "$ AH$ is the A-altitude line, perpendicular to $ AH$ through $ T_a$ is the $ a \\equiv BC$ sideline, cutting $ AH$ at the foot $ D$ of the A-altitude. $ O$ is the circumcenter, $ AO, AH$ are isogonals with respect to the angle $ \\angle CAB$ $ \\Longrightarrow$ $ AT_a$ bisects the angle $ \\angle OAH.$ Let $ M_a$ be the midpoint of $ BC.$ Then $ OM_a \\equal{} \\frac{AH}{2}.$ Let $ K$ be the midpoint of $ AH.$ Reflect $ h_a \\equiv AH \\equiv AD$ in angle bisector $ AT_a$ into $ o_a.$ Reflect $ K$ in the perpendicular bisector of $ AD$ into $ K'.$ Parallel $ k'$ to $ a \\equiv T_aD$ through $ K'$ cuts $ o_a$ at $ O.$ Circle $ (O)$ with radius $ OA$ cuts $ a \\equiv T_aD$ at $ B, C.$", "Solution_2": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14053[/img]\r\n\r\n :thumbup:" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find continuous$ F: \\mathbb{R} \\mapsto \\mathbb{R}$ satisfying\r\nfor all$ x_1, x_2 \\in \\mathbb{Q}$,$ x_1 \\minus{} x_2 \\in \\mathbb{Q}$ then $ f(x_1) \\minus{} f(x_2) \\in \\mathbb{Q}$", "Solution_1": "Let $ g_a(x)\\equal{}f(x\\plus{}a)\\minus{}f(x),a\\in Q$. Because $ g_a(x)$ continiosly and $ g_a(x)\\in Q \\forall x$ we get $ g_a(x)\\equiv b\\in Q$\r\nTherefore $ f(x\\plus{}na)\\equal{}f(x)\\plus{}nb, f(x\\minus{}na)\\equal{}f(x)\\minus{}nb$.\r\nLet $ a_1,a_2,b_1,b_2\\in Q$, suth that $ g_{a_i}\\equiv b_i$.Exist $ n_i\\in Z$, suth that $ n_1a_1\\equal{}n_2a_2\\to n_1b_1\\equal{}n_2b_2$.\r\nTherefore $ \\frac{b_i}{a_i}\\equiv c\\in Q$ and $ f(x\\plus{}a)\\minus{}f(x)\\equal{}ac, forall a\\in Q,x\\in R$. Therefore $ f(r)\\equal{}f(0)\\plus{}rc, \\forall r\\in Q$,\r\nby continiosly $ f(x)\\equal{}cx\\plus{}d, c\\in Q,d\\in R$.", "Solution_2": "[quote=\"Rust\"]Let $ g_a(x) \\equal{} f(x \\plus{} a) \\minus{} f(x),a\\in Q$. Because $ g_a(x)$ continiosly and $ g_a(x)\\in Q \\forall x$ we get $ g_a(x)\\equiv b\\in Q$\n[/quote]\r\nSorry, Rust But I can't understand this line :blush: .Can you explain more?", "Solution_3": "$ g_a(x)$ is continiosly. If $ g_a(x_1)\\equal{}b_1,g_a(x_2)\\equal{}b_2>b_1$, then for any $ b\\in (b_1,b_2)$ exist x, suth that $ g_a(x)\\equal{}b$.\r\nIf $ g_a(x)$ is not constant, then we can chose $ b\\in (b_1,b_2)$ - irrational. Therefore for any rational a $ g_a(x)\\equiv b\\in Q$." } { "Tag": [ "function", "algebra", "domain", "integration", "complex analysis" ], "Problem": "Let $ c$ be a simple closed curve, and $ u$ be a function defined on a domain $ D$ disjoint from $ c$. $ u$ is continuous on $ c$.\r\ndefine a new function \\[ h(z)\\equal{}\\int_c \\frac{u(s)}{s\\minus{}z}ds,\\indent z\\in D\\]\r\n\r\nShow that $ h$ is an analytic function.", "Solution_1": "We can see easily that $ h'(z)\\equal{}\\int_{C}\\frac{u(z)}{(s\\minus{}z)^2}ds$. So $ h$ is analytic." } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ a_1,a_2,a_3$ are positive real numbers such that $ a_1\\plus{}a_2\\plus{}a_3\\equal{}1$.\r\nProve that: $ \\frac{a_1^2}{a_1\\plus{}a_2}\\plus{}\\frac{a_2^2}{a_2\\plus{}a_3}\\plus{}\\frac{a_3^2}{a_3\\plus{}a_1} \\ge \\frac{1}{2}$", "Solution_1": "this is directly from Cauchy-Schwarz. :lol: \r\n$ {[(a_{1} + a_{2}) + (a_{2} + a_{3}) + (a_{3} + a_{1})]\\left(\\frac {a_{1}^{2}}{a_{1} + a_{2}} + \\frac {a_{2}^{2}}{a_{2} + a_{3}} + \\frac {a_{3}^{2}}{a_{3} + a_{1}}\\right)\\ge{(a_{1} + a_{2} + a_{3})^2}}$\r\n\r\n$ \\Rightarrow \\frac{a_{1}^{2}}{a_{1}+a_{2}}+\\frac{a_{2}^{2}}{a_{2}+a_{3}}+\\frac{a_{3}^{2}}{a_{3}+a_{1}}\\ge\\frac{(a_{1}+a_{2}+a_{3})^{2}}{2(a_{1}+a_{2}+a_{3})}=\\frac{1}{2}$", "Solution_2": "[quote=\"earldbest\"]this is directly from Cauchy-Schwarz. :lol: \n$ \\frac {a_{1}^{2}}{a_{1} \\plus{} a_{2}} \\plus{} \\frac {a_{2}^{2}}{a_{2} \\plus{} a_{3}} \\plus{} \\frac {a_{3}^{2}}{a_{3} \\plus{} a_{1}}\\ge\\frac {(a_{1} \\plus{} a_{2} \\plus{} a_{3})^2}{2(a_{1} \\plus{} a_{2} \\plus{} a_{3})}$[/quote]\r\n- Can you explain me step by step,couse i`m having trouble to get it,thanks :)", "Solution_3": "$ {[(a_{1}+a_{2})+(a_{2}+a_{3})+(a_{3}+a_{1})]\\left(\\frac{a_{1}^{2}}{a_{1}+a_{2}}+\\frac{a_{2}^{2}}{a_{2}+a_{3}}+\\frac{a_{3}^{2}}{a_{3}+a_{1}}\\right)\\ge{(a_{1}+a_{2}+a_{3})^{2}}}$ is from cauchy-schwarz - $ (x^2 +y^2+z^2)(p^2+q^2+r^2) \\ge (xp+yq+zr)^2$ \r\n\r\nThen, $ (a_1 +a_2 )+(a_2 +a_3 )+(a_3 +a_1 ) =2 (a_1 +a_2 + a_3 )=2$ is definite :!:" } { "Tag": [ "topology", "parameterization", "analytic geometry", "calculus", "derivative", "trigonometry", "advanced fields" ], "Problem": "Can someone tell me how to embed the Klein bottle into R^4???\r\nI know the intuition part...but I'm have no idea on how to figure out the parametrization of it in R^4. \r\n\r\nThanks alot", "Solution_1": "Represent the Klein bottle as $ \\mathbb{R}^2/G$, where $ G$ is the group generated by $ (x,y)\\mapsto (x\\plus{}1,y)$ and $ (x,y)\\mapsto (\\minus{}x,y\\plus{}1)$.\r\n\r\nAn embedding into $ \\mathbb{R}^4$, represented as $ \\mathbb{C}^2$:\r\n$ f(x,y)\\equal{}(e^{\\pi iy}\\sin(2\\pi x),e^{2\\pi iy}\\cos(2\\pi x))$\r\n\r\nSymmetries of the first coordinate: $ (x,y)\\mapsto (x\\plus{}1,y),(x,y)\\mapsto (\\minus{}x,y\\plus{}1)$, $ (x,y)\\mapsto (\\frac12\\minus{}x,y)$. Symmetries of the second coordinate: $ (x,y)\\mapsto (x\\plus{}1,y)$, $ (x,y)\\mapsto (\\minus{}x,y\\plus{}1)$, $ (x,y)\\mapsto (\\minus{}x,y)$.\r\nThe intersection is exactly the desired group.\r\n\r\nWe need both coordinates to avoid zero derivatives with respect to $ y$; $ \\sin 2\\pi x$ and $ \\cos 2\\pi x$ are never both zero.\r\n\r\nThe symmetries are the key to the construction; we need to build it with periodic functions." } { "Tag": [ "algebra", "polynomial", "AMC" ], "Problem": "1990 AHSME #30\r\n\r\nIf R_n = 1/2*(an + bn) where a = 3 + 2 :rt2:, b = 3 - 2 :rt2: , and n = 0, 1, 2, ...., then R_12345 is an integer. Its units digit is\r\n\r\n(A) 1\r\n(B) 3\r\n(C) 5\r\n(D) 7\r\n(E) 9", "Solution_1": "mathfanatic wrote:1990 AHSME #30\n\nIf R_n = 1/2*(an + bn) where a = 3 + 2 :rt2:, b = 3 - 2 :rt2: , and n = 0, 1, 2, ...., then R_12345 is an integer. Its units digit is\n\n(A) 1\n(B) 3\n(C) 5\n(D) 7\n(E) 9\n\n[hide]Yuck. This needs to be made more beautified.\n\na=3+2 :rt2:\n\na :^2: = 9 + 8 + 12 :rt2: = 17 + 12 :rt2:\n\na :^3: = 27 + 16 :rt2: + 72 + 54 :rt2: = 99 + 70 :rt2:\n\n\n\nb=3-2 :rt2:\n\nb :^2: = 9 + 8 - 12 :rt2: = 17 - 12 :rt2:\n\nb :^3: = 27 - 54 :rt2: + 72 - 16 :rt2: = 99 - 70 :rt2:\n\n\n\nSo it's pretty obvious that the radicals cancel, and the integer terms are the same. And then we know that (1/2)(a^n + b^n) is the same as the integer part of a^n. (where I'm using \"integer part\" to mean the term without the radical)\n\nn[tab]units digit of integer part of a^n\n\n0[tab]1\n\n1[tab]3\n\n2[tab]7\n\n3[tab]9\n\n4[tab]7\n\n5[tab]3\n\n6[tab]1\n\n7[tab]3\n\n8[tab]7\n\n9[tab]9\n\n10[tab]7\n\n11[tab]3\n\n12[tab]1\n\nYeah, I used a calculator for some of that, but it's AMC so that's okay. It repeats in groups of 6. 12345 mod 6 :equiv: 3 so the answer is 9. E amirighthuhuhuhuh?[/hide]", "Solution_2": "It might help to note [hide]the recurrence relation Rn+2 = 6 Rn+1 - Rn. That equation works because a and b are roots of the polynomial x2 - 6x + 1.[/hide]", "Solution_3": "Matt is [hide]correct[/hide].", "Solution_4": "My answer is 1. But, what was the time limit?" } { "Tag": [], "Problem": "Let [tex]r_1,r_2,r_3,r_4[/tex] be the radii of four mutually externally\r\ntangent circles. \r\n\r\nProve that [tex]\\displaystyle\\[\\sum_{i=1}^4 \\frac{2}{r_1^2} = \\left( \\sum_{i=1}^4 \\frac{1}{r_i} \\right)^2.\\]\r\n[/tex]\r\n\r\nDoes anyone have a solution?", "Solution_1": "[quote=\"WarpedKlown1335\"](Descartes' circle theorem)\nLet $r_1,r_2,r_3,r_4$ be the radii of four mutually externally\ntangent circles. Prove that\n[tex]\\[\n\\sum_{i=1}^4 \\frac{2}{r_1^2} = \\left( \\sum_{i=1}^4 \\frac{1}{r_i} \\right)^2.\n\\][/tex]\n\nI know the theorem but I'm looking for a proof of it.[/quote]", "Solution_2": "You can find something [url=http://www.mathlinks.ro/viewtopic.php?t=352&highlight=descartes]here[/url]." } { "Tag": [ "geometry", "geometric transformation", "reflection", "algebra", "polynomial", "number theory unsolved", "number theory" ], "Problem": "Sorry this isn't really a contest problem per se, but this seemed like a good place to bring it up:\r\n\r\nLet 0 < a, b < 1. Evaluate this expression:\r\n\r\na + b*(a + b*(...))\r\n\r\nwhere the ellipsis indicates (infinite) recursion. The expression has a finite value e.g. when a = (1-r) and b = r^2 with 0 < r < 1, in which case you can think of it as an expression of the amount of light which escapes from a thin gap of air between two mirrors. Things get even more interesting when you add more mirrors :-)", "Solution_1": "Well, assuming it converges (which is probably easy to prove with those bounds on a and b, but I'm not worried about that part), then we just have $a+bx = x$, so $x = \\frac{a}{1-b}$ is the answer.", "Solution_2": "Er... right :blush: \r\n\r\nHere's the more general problem:\r\n\r\nSay you shine a flashlight at a stack of n partial mirrors, where each mirror reflects r of any light which falls on it, and lets 1-r through. How much of the light from the flashlight makes it through all n mirrors? Keep in mind that the mirrors reflect light in both directions.\r\n\r\nIt's pretty simple to solve for small n by mapping out the (2n-1) \"states\" of a light particle as it bounces around in the mirrors, e.g. for n = 2 you have:\r\n\r\n0_right = (1-r)*1_right + r*0\r\n1_right = (1-r)*2_right + r*2_left\r\n1_left = (1-r)*0 + r*1_right\r\n2_right = (1-r) + r*1_left\r\n2_left = (1-r)*1_left + r*2_right\r\n\r\n0_right is the state of a unit of light heading into the first mirror from the outside. 1_left is in between the first mirror and the second, heading back out the \"wrong way\", and 2_right is one step away from the other side of the mirrors. Solving for 0_right gives you the answer after some hairy algebra, but I haven't found the general solution for any number of mirrors.\r\n\r\nThe idea for this problem came from an exercise in an optics textbook, which gave an answer of (1-r)^n. That's actually the amount of light which is transmitted directly through all n mirrors, but it ignores the light which bounces around between the mirrors before escaping.", "Solution_3": "Oops. I meant n=3 for the example.", "Solution_4": "Hmm. I could solve \"How much light gets through mirror number k\" if there are an infinite number of mirrors by setting up a double recursion, but that doesn't quite work when there are a finite number as the recursion gets stuffed up by the end terms.. still thinking.", "Solution_5": "This is the answer I found empirically by looking at answers for N = 0,1,2,3,4,5:\r\n\r\n\\[ \\frac{1-r}{(N-1)r+1} \\]\r\n\r\nNo idea about any proof, though.", "Solution_6": "Ahha. Yep, that works, you can prove it by induction.", "Solution_7": "OK, It can be done inductively. Because no light is ever absorbed, the reflectivity of N mirrors with a reflectivity of r is $1 - \\frac{1-r}{(N-1)r+1}$.\r\n\r\nSo these N mirrors are equivalent to [b]one[/b] mirror with a reflectivity of $1 - \\frac{1-r}{(N-1)r+1}$. We can find the transmittivity of N+1 mirrors by solving a situation with two mirrors; one with reflectivity of r, and the other with reflectivity of $1 - \\frac{1-r}{(N-1)r+1}$. The total transmittivity of the whole thing turns out to be $\\frac{1-r}{Nr+1}$, completing the inductive step.", "Solution_8": "That's very neat. Out of curiosity, how did you arrive at the answer? I worked out the solution for a few n but always ended up with a quotient of polynomials which I couldn't simplify any further.", "Solution_9": "Well, I , uh, [size=75]used Maple[/size]. But it might have also helped that I actually worked with transmittivity instead of reflectivity, since that gives a numerator of just -t instead p-1. I'm guessing that'd make it a lot easier to find any common factors in the numerator and denominator." } { "Tag": [ "geometry", "trapezoid", "calculus", "integration", "modular arithmetic" ], "Problem": "(1) A regular hexagon has side length 10. What is the sum of the lengths of all the diagonals for the hexagon? Express your answer in simplest radical form.\r\n\r\n\r\n[hide=\"My Answer\"]$ 60 \\plus{} 72\\sqrt{3}$[/hide]\n\n[hide=\"The Answer\"]Hm...I can't read it :P (No, I'm not kidding.)[/hide]\n\n(2)How many different arithmetic sequences are there with all the following properties: (1) The first term is 31, (2) The last term is 61, (3) The common difference is a whole number and (4) The total number of terms is at least 3? \n\n\n[hide]7[/hide]\r\n\r\n(3) An equilateral triangular region of side length 2 cm is cut form a corner of an equilateral triangular region with side length 6 cm. What is the number of square centimeters in the area of the resulting trapezoid?\r\n\r\n(4)Jill has two watches. One gains 2 seconds per hour and the other loses 1 seconds every 2 hours. If the clocks are set at the right time at 8:00 AM, at what time will the clocks be exactly one hour apart?\r\n\r\n(5) the 9-digit number $ abbababa3$ is a multiple of 99 for some pair of digits $ a$ and $ b$. What is $ b \\minus{} a$", "Solution_1": "[hide=\"1\"]\n$ 3\\cdot20 \\plus{} 6\\cdot10\\sqrt3 \\equal{} \\boxed{60 \\plus{} 60\\sqrt3}$.\n\nAs a check, this encompasses all $ 3 \\plus{} 6 \\equal{} 9 \\equal{} \\dbinom62 \\minus{} 6$ diagonals.\n[/hide]\n[hide=\"2\"]\nOur sequence is $ 31,31 \\plus{} d,\\ldots,61 \\equal{} 31 \\plus{} nd$.\n\nSince $ n$ must be integral, $ n|(61 \\minus{} 31 \\equal{} 30)$ ($ 30 \\equal{} 2\\cdot3\\cdot5$), which means that there are $ 2\\cdot2\\cdot2 \\equal{} 8$ possible values for $ n$.\n\nHowever, $ n \\equal{} 1$ is invalid, since it produces the sequence $ 31,61$, with only two terms.\n\nOur final answer is $ 8 \\minus{} 1 \\equal{} \\boxed7$.\n[/hide]\n[hide=\"3\"]\nThe area of an equilateral triangle of side $ s$ is $ \\frac {s^2\\sqrt3}4$, so the answer is\n\n$ \\frac {\\sqrt3}4(6^2 \\minus{} 2^2) \\equal{} \\boxed{8\\sqrt3}$.\n[/hide]\n[hide=\"4\"]\nEvery hours, the two clocks get $ 2 \\plus{} \\frac12 \\equal{} \\frac52$ seconds apart.\n\nThere are $ 60\\cdot60 \\equal{} 3600$ seconds in an hour, so it will take $ \\frac {3600}{\\frac52} \\equal{} 1440$ hours.\n\n$ 1440\\equiv0\\pmod{24}$ and $ 1440 \\equal{} 60\\cdot24$, so this will happen after $ 60$ days at $ 8: 00$ AM.\n[/hide]\n[hide=\"5\"]\nThis number must be divisible by $ 9$ and $ 11$.\n\nTo be divisible by $ 11$, we must have\n\n$ 3 \\minus{} a \\plus{} b \\minus{} a \\plus{} b \\minus{} a \\plus{} b \\minus{} b \\plus{} a \\equal{} 3 \\minus{} 2a \\plus{} 2b\\equiv0\\pmod{11}$, or\n\n$ 2b \\minus{} 2a\\equiv \\minus{} 3\\equiv8\\pmod{11}\\implies b \\minus{} a\\equiv4\\pmod{11}$.\n\nSince $ 0\\leq a,b\\leq9$, we can see that $ b \\minus{} a \\equal{} \\boxed4$ is the only possible answer.\n[/hide]" } { "Tag": [ "email", "ratio", "articles", "geometric series" ], "Problem": "Hello, im am writing to ask if there is a summation formula for \r\n1^n +2^n +3^n + . . . + k^n =?\r\nI do not know how to find it, i would appreciate any help that includes proof and formula. Thanks\r\nPlease reply as fast as possible. \r\n\r\nBen\r\n\r\nPD: My email is benjalizana@gmail.com and msn benjamin_lizana@hotmail.com\r\n Aim's Screen Name: benjalizana", "Solution_1": "[quote=\"benjalizana\"]Hello, im am writing to ask if there is a summation formula for \n1^n +2^n +3^n + . . . + k^n =?\nI do not know how to find it, i would appreciate any help that includes proof and formula. Thanks\nPlease reply as fast as possible. \n\nBen\n\nPD: My email is benjalizana@gmail.com and msn benjamin_lizana@hotmail.com\n Aim's Screen Name: benjalizana[/quote]\r\n\r\nI know the formulas for $n=1,2,3$:\r\n$n = 1 \\quad \\frac{n(n+1)}{2}$\r\n\r\n$n = 2 \\quad \\frac{n(n+1)(2n+1)}{6}$\r\n\r\n$n = 3 \\quad \\left( \\frac{n(n+1)}{2}\\right)^{2}$\r\n\r\nI don't know if someone could generalize this... They are all fairly simple to prove with induction. But, I don't know how to derive them.", "Solution_2": "[quote=\"benjalizana\"]Hello, im am writing to ask if there is a summation formula for \n1^n +2^n +3^n + . . . + k^n =?\nI do not know how to find it, i would appreciate any help that includes proof and formula. Thanks\nPlease reply as fast as possible. \n\nBen\n\nPD: My email is benjalizana@gmail.com and msn benjamin_lizana@hotmail.com\n Aim's Screen Name: benjalizana[/quote]\r\n\r\nBleh: http://en.wikipedia.org/wiki/Summation#Identities", "Solution_3": "See this pdf on the [url=http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf]sum of the $k^{th}$ powers[/url]. See page 45 to skip directly to your question.", "Solution_4": "yes i found it up to n=4 but cant find a relation :S Help Please", "Solution_5": "The formula was on the link I sent you...", "Solution_6": "easy \r\n$\\sum_{i=1}^{n}n^{n}=\\frac{n^{n}-1}{n-1}$ or $\\frac{n^{r}-1}{r-1}$", "Solution_7": "[quote=\"binomial_4eva\"]easy \n$\\sum_{i=1}^{n}n^{n}=\\frac{n^{n}-1}{n-1}$ or $\\frac{n^{r}-1}{r-1}$[/quote]\r\n\r\nIsn't that for geometric series? That wouldn't work here...", "Solution_8": "i think ur rite but no wait i think i am rite coz the formula for this sequence is $n^{n}$ isnt it ? so therefore its the sum of the geometric seq. $U_{n}=a(r)^{n-1}$ $s_{\\infty}=\\frac{1-r^{n}}{1-r}$ with $a=1$ $r=n-1$", "Solution_9": "right, the thing is its not n to the power of n, but n to any power different from n, lets say K.. The formula in that pdf.. or on wikipedia seems to be the answer, but i cant really understand the proof behind it; how and why does newtons binomial theory apply to this problem?? how does \"Bk\", the kth Bernoulli number relate to this problems solution..also in that pdf page 46 i do not understand what \"fk(n + m)\" Please help as fast as posible.\r\n\r\nThanks", "Solution_10": "[quote=\"binomial_4eva\"]i think ur rite but no wait i think i am rite coz the formula for this sequence is $n^{n}$ isnt it ? so therefore its the sum of the geometric seq. $U_{n}=a(r)^{n-1}$ $s_{\\infty}=\\frac{1-r^{n}}{1-r}$ with $a=1$ $r=n-1$[/quote]\r\n\r\nThis:\r\n$1^{1}, 2^{2}, 3^{3}, ..., N^{N}$\r\nis [b]NOT[/b] a geometric series!\r\n\r\nThere is no common ratio for this. In a geometric series (strictly speaking, sequence), there is always a common ratio. For example,\r\n$1, 2, 4, 8, 16, ...$\r\nhas a common ratio of $2$. \r\nIf you still don't believe me, look at this:\r\n$2^{2}/1^{1}= 4$\r\n$3^{3}/2^{2}= 27/4$\r\n$4^{4}/3^{3}= 256/27$\r\nClearly, there is no common ratio.", "Solution_11": "[quote=\"vishalarul\"][quote=\"binomial_4eva\"]i think ur rite but no wait i think i am rite coz the formula for this sequence is $n^{n}$ isnt it ? so therefore its the sum of the geometric seq. $U_{n}=a(r)^{n-1}$ $s_{\\infty}=\\frac{1-r^{n}}{1-r}$ with $a=1$ $r=n-1$[/quote]\n\nThis:\n$1^{1}, 2^{2}, 3^{3}, ..., N^{N}$\nis [b]NOT[/b] a geometric series!\n\nThere is no common ratio for this. In a geometric series (strictly speaking, sequence), there is always a common ratio. For example,\n$1, 2, 4, 8, 16, ...$\nhas a common ratio of $2$. \nIf you still don't believe me, look at this:\n$2^{2}/1^{1}= 4$\n$3^{3}/2^{2}= 27/4$\n$4^{4}/3^{3}= 256/27$\nClearly, there is no common ratio.[/quote]\r\nBut still that isn't what the problem is asking for.\r\nhttp://mathworld.wolfram.com/PowerSum.html", "Solution_12": "[quote=\"benjalizana\"]..also in that pdf page 46 i do not understand what \"fk(n + m)\" Please help as fast as posible.[/quote]\r\n\r\nThe notation you're asking about is mentioned in the first couple of pages. Our specific series is mentioned on page 7. $f_{k}(n)$ denotes a sum of consecutive integers from $1$ to $n$, each of which are raised to the power $k$. For example, $f_{2}(5) = 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}$.", "Solution_13": "From Wikipedia:\r\n$\\sum_{i=0}^{n}i^{p}= \\frac{(n+1)^{p+1}}{p+1}+\\sum_{k=1}^{p}\\frac{B_{k}}{p-k+1}{p\\choose k}(n+1)^{p-k+1}$\r\n\r\n where Bk is the kth Bernoulli number.", "Solution_14": "Here's the exact article: http://en.wikipedia.org/wiki/Faulhaber%27s_formula" } { "Tag": [ "IMO" ], "Problem": "hey guys i'm sorry to bother you with this thing, but i just can't find a topic i've been looking for. it's the one where one posts a problem of an IMO prep level, and the one who solves it - posts the next.\r\nthanks a lot!", "Solution_1": "You may refer to the following link. Hope that helps.\r\n\r\nhttp://www.mathlinks.ro/viewtopic.php?t=288069\r\n\r\nRegards,", "Solution_2": "Very funny. \r\n\r\nHe wants IMO problems, not MATHCOUNTS...\r\nI don't know if it exists, but look at Olympiad section, you'll find many problems..." } { "Tag": [], "Problem": "Could someone please tell when the JBMO 2007 will be hosting ?\r\nAlso there is a site for this ?Thank you", "Solution_1": "The jBMO 2007 is held in Shumen, Bulgaria, on 25-30 June (these days). The official site of the competion is [url]http://www.jbmo.eu/[/url]." } { "Tag": [ "algebra", "polynomial", "number theory proposed", "number theory" ], "Problem": "Let$P(x)$ be a polynomial,such that : \r\n $P(x)\\equiv a \\mod(x-a)$\r\n$P(x)\\equiv b \\mod(x-b)$\r\n$P(x)\\equiv c \\mod(x-c)$\r\nfind$d$ such that : $P(x)\\equiv d \\mod(x-a)(x-b)(x-c)$", "Solution_1": "Isn't it obviously $x$?\r\nOf course if $a,b,c$ are not pairwise unequal, you can not determine, but in the other case it is obviously $x$." } { "Tag": [ "puzzles" ], "Problem": "A man is standing on the platform of a train station. As soon as a train pulls out of the station, he visualises a lady aboard it waving. He runs over to the train, gets on it and kills the lady. Thereafter, he gets off the train. Then, he goes to the nearest polce station and confesses to murdering the lady. \r\n\r\nHowever, he is able to walk away from the police station without consequence. \r\n\r\nHow is this possible?", "Solution_1": "[hide]\n\nIf he only \"visualises\" the lady, then she isn't real and he's insane?\n\n [/hide]", "Solution_2": "i guess its pretty clear that the woman was most wanted. then, the man confesses and the police dont give him a reward for kiling her, but let the man walk away safe.", "Solution_3": "The solution to the puzzle is as follows:\r\n\r\n[hide]The lady was the man's wife. Years before she faked her own death and framed her husband. Her husband was convicted and served time for murder. In the country where this occurred, a person can not be prosecuted for the same crime twice. Since he had already served his time, he could not be sentenced again.[/hide]", "Solution_4": "Hey people. you seem to misunderstand these riddles:\r\nyou shall not show how cool you are (by giving what you heard whereever else), but you should ask no-or-yes-questions which the author is supposed to answer.", "Solution_5": "[quote=\"K Sengupta\"]The solution to the puzzle is as follows:\n\n[hide]The lady was the man's wife. Years before she faked her own death and framed her husband. Her husband was convicted and served time for murder. In the country where this occurred, a person can not be prosecuted for the same crime twice. Since he had already served his time, he could not be sentenced again.[/hide][/quote]\r\n\r\nK Sengupta, just a question..if that is the actual solution, how are we supposed to get that kinda far fethched answer from the puzzle and teh information that was provided to us?? just wondering, cuz don't riddles have to some contexual clues to obtain the answer? i don't mean to be disrespectful, i was just simply curious... :?: :maybe:", "Solution_6": "As I already wanted to point out, this is a lateral (or however it's called in english), thus normally all are allowed to ask the person who gave the riddle questions; as a restriction, these questions have to have \"yes\" or \"no\" as only answers (possibly allowing finite casechecking, too, since that could always be realised by yes-no-questions).", "Solution_7": "That's a very good point. There should be some kind of rule saying if you already know the answer, don't spoil the problem for everyone else.", "Solution_8": "[hide]The lady is a terrorist/criminal/dangerous maniac!! lol :D [/hide]\nor\n[hide]the lady isn't real..she's just his imagination...and the police have him registered as a mental asylum patient[/hide]\r\ndunno yaar..." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME", "email", "ARML", "HMMT" ], "Problem": "[url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamoqual.shtml[/url]\r\n\r\nI guess here is the preliminary list. A lot from my state this year, but it seems TJHSST and Vestivia Hills didn't get as many as they usually do... Exeter had like a freakin WALL of qualifiers :P Stuy did pretty awesome as usual....\r\n\r\nGood luck to all on USAMO!! :)", "Solution_1": "Thanks!!\r\n\r\nWow, I think I have met everyone from my state that made USAMO.", "Solution_2": "Wow, you're right, it's already up! Congratulations to all qualifiers! Good luck on USAMO!\r\n\r\nSummary:\r\n\r\n11/12 Index = 204.0\r\n\r\nFloor = 6\r\n\r\n10 or below index (with floor) = 202.5\r\n\r\n[b]NOTE[/b]: Some people who took the AIME II are not on the list YET but will qualify. That's why it is \"preliminary.\"", "Solution_3": "Yay I'm on the list. Which means I made less than 3 bubbling mistakes on the AIME, if any. :P \r\nI recognize many names...lol. If I counted correctly, there are around 89 people from California who qualified. Yee.", "Solution_4": "Heh, to think that for the last few weeks I've been thinking that my 216 index was borderline . . . \r\n\r\nAnyway, congratulations to everyone who qualified (including me :D )!", "Solution_5": "I think it's pretty awesome that the only states with more qualifiers than MA are CA (obviously), and NY, which only has like 3 more. Maybe there are other states that I misssed, but I'm pretty sure those are the only ones.", "Solution_6": "I should have qualified (AMC 10:126, AIME II:9), but I'm not on the list. This might be because on my AIME answer form, I put that my AMC 12 score was a 109.5.\r\n\r\nAlso, I know that the forms for my school have arrived, for I can see that one person from my school has already made USAMO.", "Solution_7": "Or it might be that\r\n\r\n[quote=\"AMCDirector\"]There are still a few (200 or less) AIME II answer forms to come in yet and be scored[/quote]", "Solution_8": "[quote=\"knexpert\"]I should have qualified (AMC 10:126, AIME II:9), but I'm not on the list. This might be because on my AIME answer form, I put that my AMC 12 score was a 109.5.[/quote]\r\n\r\nFor you and anyone else who believes s/he may have qualified,\r\n\r\nPlease email the following information to amcinfo@maa.org after April 13th:\r\n\r\n1) Your name.\r\n2) Your school's name and CEEB number. If you took the contests at more than one location, include both.\r\n3) Your address.\r\n\r\nAs has been mentioned before the score you bubble on the answer form is just an initial identity check; it does not serve as the qualification score.\r\n\r\nAs E^(pi*i)=-1 says, we may not have received your school's answer forms, yet, and therefore you may appear later this week.\r\n\r\nElizabeth Claassen\r\nAMC Senior Accounting Clerk\r\n(and resident number cruncher)", "Solution_9": "Woah, this list looks so loooong!", "Solution_10": "Crap. My index=202, so I was two points off.", "Solution_11": "Actually, you were .5 points off. Hope that makes you feel somewhat better, Eric. :) \r\n\r\nThen again, it might not. :maybe: \r\n\r\nBut just feel good about the fact that you're still one of the most pwnzinating ARML kids in GA. :D", "Solution_12": "No, I'm a junior. 11/12 index was 204.", "Solution_13": "Hmm I wonder which state had the most qualifiers? I think it was CA...IL had a largeish number but not that many.", "Solution_14": "[quote=\"calc rulz\"][b]NOTE[/b]: Some people who took the AIME II are not on the list YET but will qualify. That's why it is \"preliminary.\"[/quote]\r\n\r\nany description of those people?\r\n\r\njust the 200 answer sheets who havn't been scored yet?", "Solution_15": "144/9. On to blue mop.\r\n\r\nI need remedial additon.", "Solution_16": "[quote=\"pythag011\"]144/9. On to blue mop.\n\nI need remedial additon.[/quote]\r\nWhoa...6th grader aiming for MOP...I wish you luck.", "Solution_17": "Finally in, in my last chance. \r\n\r\n(I'm the only one from IL outside Chicago, other than Alex Zhai)", "Solution_18": "haha, what a true illinoiser.\r\n\"in Chicago\" meaning within two hours driving distance :P \r\nyay only freshman from il. no eighth graders this year. troubling.", "Solution_19": "I count as [i]at least [/i] half Illinois-ese, Aryth. So 1.5 Illinois freshmen this year =P.", "Solution_20": "[quote=\"Aryth\"]haha, what a true illinoiser.\n\"in Chicago\" meaning within two hours driving distance :P \nyay only freshman from il. no eighth graders this year. troubling.[/quote]\r\n\r\nEh, it's all the same up there when you live down near St. Louis like I do.", "Solution_21": "When do the official lists come out??? (if they change something)\r\n\r\nWhen are you officially imformed through email?", "Solution_22": "[quote=\"firecricket91\"]When are you officially imformed through email?[/quote]\r\n\r\nYou'll be emailed by tomorrow, according to the announcement.", "Solution_23": "it's unfair!!! \r\nif a junior took amc12 and he/her index was 203,he/she would not qualify to usamo! but the one who took amc10 (and his/her index was 203 too) is qualified !!! :mad:", "Solution_24": "[quote=\"gpl_coll4ever\"]it's unfair!!! \nif a junior took amc12 and he/her index was 203,he/she would not qualify to usamo! but the one who took amc10 (and his/her index was 203 too) is qualified !!! :mad:[/quote]\r\n\r\nThat is false no matter how you look at it. A \"junior\" meaning 11th grade can not take the AMC 10. If you meant an underclassman (10th grade or below), then a 203 index qualified regardless of which test they took. It is unfair because it's easier to get a 203 with the AMC10 but not because of what you said.", "Solution_25": "i am so sad.\r\ni am in 10 grade , took amc12a(108) and 12b(118.5) at two different schools, and took aime at my high school where i got a lower score. so they added my aime score(9) to 108, and i am not qualified to usamo.", "Solution_26": "[quote=\"gpl_coll4ever\"]i am so sad.\ni am in 10 grade , took amc12a(108) and 12b(118.5) at two different schools, and took aime at my high school where i got a lower score. so they added my aime score(9) to 108, and i am not qualified to usamo.[/quote]\r\n\r\nYou will still qualify. The score reports sent to the schools will typically report scores from one school - not those taken at other schools. The AMC folks have no easy way of knowing how to match your AIME score to the 12b score unless you reported the higher score on the AIME information form. If you aren't on the qualifier list posted on the AMC website yet, call the AMC office and explain the situation.", "Solution_27": "i am very happy to hear that. \r\na teacher told me i had to try next year.\r\nthank you very much!!!", "Solution_28": "gt59 wrote:\r\n[quote] If you aren't on the qualifier list posted on the AMC website yet, call the AMC office and explain the situation.[/quote]\n\ngpl_coll4ever (and others in the same circumstances): Remember that the AMC has requested that you [b]WAIT until AFTER April 13 before you contact [/b]them to inquire about this.\n\nAnd the AMC has requested that you [b]email [/b]them, NOT call them.\n\nSee excerpt from Dave Patrick's posting below:\n\n[quote]Elizabeth Claassen of the AMC posts the following information for those people who think that they should be on the USAMO qualifiers list, but are not listed:\n[quote]\nFor you and anyone else who believes s/he may have qualified,\n\nPlease email the following information to amcinfo@maa.org [b]after April 13th[/b]:\n\n1) Your name.\n2) Your school's name and CEEB number. If you took the contests at more than one location, include both.\n3) Your address.[/quote]\nI have added the emphasis on \"after April 13th\". They are still processing some scores and rechecking some others. Please let them finish this task before contacting them. There is still plenty of time to get properly registered for the USAMO if indeed you qualified but are not yet listed.[/quote]", "Solution_29": "Only freshman in Ohio... :thumbup: Hope to make it to Red MOP" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "AIME" ], "Problem": "Rumors say that MOP is expanding, who's got the details. Post 'em here please.\r\n\r\nThank you,\r\nIW.OAV", "Solution_1": "My information is limited, but yes, the current plan is to expand MOP to about 60 people in 2004.", "Solution_2": "I heard that MOP in 2002, the large one was for everyone who wasn't a senior and seniors who did very well. Why is that the case? I am a senior and if the chances of making it to MOP were made a bit easier to achieve then that will motivate me to work harder. If MOP is expanded, does it make it easier for anyone to get there or usually non-seniors? Ok thanks", "Solution_3": "I believe 2002 was because of excessive sponsorship. I'm guessing that any further expansion will not be open to seniors, based upon the 2002 case and that under the current system the non-winners who attend are non-senior. Reason for this is probably because while it makes sense to accept all 12 potential IMO-quals for that year, any seniors who are not USAMO winners can't qualify for IMO the following year and thus it doesn't help the US team's future to accept nonwinner seniors, whereas students from lower grades can someday become a member of the team... thus it makes more sense to give them adequate training...", "Solution_4": "[quote=\"Ripper1\"]I heard that MOP in 2002, the large one was for everyone who wasn't a senior and seniors who did very well. Why is that the case? I am a senior and if the chances of making it to MOP were made a bit easier to achieve then that will motivate me to work harder. If MOP is expanded, does it make it easier for anyone to get there or usually non-seniors? Ok thanks[/quote]\r\n\r\nIn 2002, the program was expanded. But it didn't include [i]everyone[/i] who was a senior, only the top 120 or so, I believe. In order to qualify for MOP, you had to be in the top 12 (senior or not) or you had to be a non-senior (preferably a young student) with a medium-rate score.\r\n\r\nIf MOP expands next year, it won't be to the same magnitude as in 2002.", "Solution_5": "In 2002, the top 12 were invited, and then every non-senior who took the USAMO was also invited, regardless of score. So some of the people at MOP 2002 got 0s on the USAMO.", "Solution_6": "oh sigh...if only they expanded that much this year", "Solution_7": "Oh...I didn't know that. Oy, if only they'd do that again...", "Solution_8": "Hmm, I need to find a way to shave a couple years off my age.", "Solution_9": "How do you \"shave\" off a couple of years? :) \r\nMaybe :evil: can help you... :)", "Solution_10": "So does anyone have more updated info about this MOP expansion? We're a ways into the school year, but the AMC website has not stated any expansion of the summer program.", "Solution_11": "I am a member of an advisory committee to the CAMC and while we all agreed that the expanded 2002 MOP was a good thing that we'd like to see repeated, this did not seem likely at our meeting last March. As other people have indicated, the large 2002 MOP was due to very generous donations that year, primarily one large donation from Akamai if I remember correctly, and the current economic climate made us feel that such donations were unlikely any time soon. For those who have heard \"rumors\" of an expanded MOP this year (Eve, IWoAV), where are you hearing them? I'm very interested in learning more about this. I agree with the general opinion that most of the expansion would be for non-seniors who might end up on future IMO teams.", "Solution_12": "I corresponded with Steve Dunbar, the AMC Director, about this around 3 months ago. He indicated that they received more funding from Akamai & would be expanding MOP next summer by \"about 30\". (TomC - I'm guessing this all came to fruition after that meeting in March.) The selection criteria were not set at the time, but he suggested they would be included with the AMC-10/12 packets that your schools should receive this month.", "Solution_13": "Hi -- I'm here to answer a few questions and put to rest some rumors.\r\n\r\nYes, MOSP 2004 will have room for about 60 students, an expansion of \"about 30\" from our 2003 size of \"about 30.\"\r\n\r\nRichard is correct, this did come to fruition in about June-July of last summer, several months after the AMC Advisory Board meeting, and the provisional advice and input of the Advisory Board at that meeting was used in crafting the agreement during summer 2003.\r\n\r\nFinal decisions about the composition of the MOSP 2004, (i.e. who will be invited and under what criteria) have not been made at this moment, although a proposal has been circulated to our policy-proposing MAA Committee on the American Mathematics Competitions (CAMC) and the (financially advising and personnel managing) MAA Executive Committee. The proposal should be discussed face-to-face at the Joint Mathematics Meetings in January and finalized shortly thereafter. I will publicize the decisions through our website, and in the AIME-USAMO Teachers' Manual.\r\n\r\nThe reasons for inviting only non-seniors (except for the 12 USAMO Finalists) given by complex zeta, mathfanatic and J. Lin are essentially correct.\r\n\r\nI would be interested in hearing ideas form this group about what should be the criteria for inviting \"about 30\" more students to MOSP 2004. Remember that the decision making rests with several committees, and that not all your suggestions may be accommodated for practical and financial reasons which are beyond the scope of this forum!\r\n\r\nFinally, please remember that our decision-making and planning cycle takes place on an (annual) timescale which is much longer than a volatile economic time scale (~3 months) which in turn is much longer than the information time scale of the Internet (~minutes). That can be frustrating for everyone, including me!\r\n\r\nSteve Dunbar, AMC Director", "Solution_14": "wow, thats cool, though im definitely not making this year... maybe in my junior year if it stays this size, i have a chance.\r\n\r\nthats too bad for seniors... but i guess that they've had enough tries already :P", "Solution_15": "Maaaaaan, what a rip...\r\n\r\nThere goes my chance at MOP\r\n\r\nThanks a lot Jongmin Baek, pffft.\r\n\r\n:P", "Solution_16": "What did Jongmin do? He's a good friend of mine, by the way." } { "Tag": [ "AMC", "AIME", "AMC 10", "AMC 10 B" ], "Problem": "Is it possible [b]not[/b] to qualify for the AIME with a 120. On the 2003 AMC10B, the AIME qualification score was [b]121+[/b], why is that? If I make a 120 in 2010, will I be guaranteed to move on to the AIME?", "Solution_1": "I depends on the how easy/hard the AMC was that year. But you can rest assured that the cutoff score won't deviate more than $ 5$ points from $ 120$.", "Solution_2": "Back in 2003, there was no rule that a 120 on the AMC 10 was guaranteed to make AIME. Now there is such a rule.", "Solution_3": "If you take the AMC 10 practice test for that year, you may find that that test was relatively easy. (Like I did. :))", "Solution_4": "This is a somewhat related thread:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=321425", "Solution_5": "RaviB is correct, I checked the AMC 10s and from 2000-2003 the AIME qualification level from the AMC 10 was set at the top 1%. After 4 years of experience the contest creation committee members felt that we had enough experience in setting the test that we could use a fixed level of 120 (or the top 1%, whichever is lower.)\r\n\r\nIf you have the 21st Century CD, look at the fine print on the covers of the AMC 10s for those years to see for yourself.\r\n\r\n-- Steve Dunbar\r\nMAA Director, American Mathematics Competitions" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "let a,b,c>0 and a+b+c=1 prove that this inequality \r\n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $", "Solution_1": "We have : \r\n\r\n $\\frac{a^2}{b+c}+\\frac{b^2}{a+c}+\\frac{c^2}{c+a} +1= \\frac{a}{b+c}\r\n+\\frac{b}{a+c}+\\frac{c}{a+b}$\r\n\r\nSo our inequality is equal to: \r\n\r\n $\\frac{a+b}{b+c}+\\frac{b+c}{c+a}+\\frac{c+a}{a+b} \\geq 3$\r\n\r\nand it is easy from AM-GM.", "Solution_2": "here is a stronger form with condition $a+b+c \\ge \\frac{1}{5}$ with Perfect proof of darij .\r\n [url]http://www.mathlinks.ro/Forum/topic-51000.html[/url]\r\nAs the source you wrote :My Book ,quite enjoying yourself with these things ha ? :| :|", "Solution_3": "$f(x)=\\frac{x}{1-x}$ and $g(x)=\\frac{1}{1-x}$ are convex on $(0; 1)$ and $a+b+c=1$ so from Jensen inequality:\r\n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} = \\frac{a^2+b}{1-a}\\ + \\frac{b^2+c}{1-b}\\ + \\frac{c^2+a}{1-c} = (\\frac{a^2}{1-a}\\ + \\frac{b^2}{1-b}\\ + \\frac{c^2}{1-c}) + (\\frac{b}{1-a}\\ + \\frac{c}{1-b}\\ + \\frac{a}{1-c}) = (a\\frac{a}{1-a}\\ + b\\frac{b}{1-b}\\ + c\\frac{c}{1-c}) + (b\\frac{1}{1-a}\\ + c\\frac{1}{1-b}\\ + a\\frac{1}{1-c}) = (af(a) + bf(b) + cf(c)) + (bg(a) + cg(b) + ag(c)) \\geq\r\n\\frac{a^2+b^2+c^2}{1-(a^2+b^2+c^2)} + \\frac{1}{1-(ab+bc+ca)} \\geq \\frac{a^2+b^2+c^2}{1-(ab+bc+ca)} + \\frac{1}{1-(ab+bc+ca)} = \\frac{a^2 + b^2 + c^2 + 1}{1-(ab+bc+ca)}=\\frac{a^2 + b^2 + c^2 + (a+b+c)^2}{(a+b+c)^2-(ab+bc+ca)}=\\frac{2(a^2+b^2+c^2+ab+bc+ca)}{a^2+b^2+c^2+ab+bc+ca}=2$", "Solution_4": "\\[\\sum\\frac{a^2+b}{b+c}=\\sum\\frac{a^2+ab+b(b+c)}{b+c}=\\sum\\frac{a(a+b)}{b+c}+\\sum{a}=\\\\\\sum\\frac{(a+b)-(a+b)(b+c)}{b+c} +1=\\sum\\frac{a+b}{b+c} -(\\sum{a+b}) +1=\\\\\\sum\\frac{a+b}{b+c}-1\\geq 3-1=2\\] where the last inequality holds by AM-GM", "Solution_5": "This problem is in Old and New ineqs . \r\n The solution showed in that book is not nice (it's just my idea) . This one is nicer :\r\n $ \\frac{a^2+b}{b+c}+a = \\frac{a^2+b}{1-a}+a = \\frac{a+b}{b+c} $ \r\n It's enough to prove : $ \\frac{a+b}{b+c}+\\frac{b+c}{c+a}+\\frac{c+a}{a+b} \\geq 3 $ , which is true by AM-GM", "Solution_6": "[quote=\"nttu\"]This problem is in Old and New ineqs . \n The solution showed in that book is not nice (it's just my idea) . This one is nicer :\n $ \\frac{a^2+b}{b+c}+a = \\frac{a^2+b}{1-a}+a = \\frac{a+b}{b+c} $ \n It's enough to prove : $ \\frac{a+b}{b+c}+\\frac{b+c}{c+a}+\\frac{c+a}{a+b} \\geq 3 $ , which is true by AM-GM[/quote] This is exactly [i]xtar[/i]'s solution. ;)", "Solution_7": "[quote=\"nttu\"]This problem is in Old and New ineqs . \r\n What's the book??", "Solution_8": "because the ineq is homogenous we can give the problem without the condition $a+b+c=1$ Sure?", "Solution_9": "It's not homogeneous.", "Solution_10": "I was wrong sorry :oops:", "Solution_11": "[quote=\"ALGEBRA\"]let a,b,c>0 and a+b+c=1 prove that this inequality \n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $[/quote]\r\nAfter taking common denominator, making cross product and cancelling common terms on LHS and RHS the inequality reduces to a^3+b^3+c^3>=a*(b^2)+b*(c^2)+c(a^2)\r\nwhic follows directly from rearrangement inequality..Namely, suppose a>=b>=c\r\n(a^2)*a+(b^2)*b+(c^2)*c>=(a^2)8c+(b^2)*a+(c^2)*b", "Solution_12": "[quote=\"ALGEBRA\"]let a,b,c>0 and a+b+c=1 prove that this inequality \n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $[/quote]\r\nAfter taking common denominator, making cross product and cancelling common terms on LHS and RHS the inequality reduces to a^3+b^3+c^3>=a*(b^2)+b*(c^2)+c(a^2)\r\nwhic follows directly from rearrangement inequality..Namely, suppose a>=b>=c\r\n(a^2)*a+(b^2)*b+(c^2)*c>=(a^2)8c+(b^2)*a+(c^2)*b", "Solution_13": "[quote=\"Makaveli\"][quote=\"ALGEBRA\"]let a,b,c>0 and a+b+c=1 prove that this inequality \n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $[/quote]\nAfter taking common denominator, making cross product and cancelling common terms on LHS and RHS the inequality reduces to a^3+b^3+c^3>=a*(b^2)+b*(c^2)+c(a^2)\nwhic follows directly from rearrangement inequality..Namely, suppose a>=b>=c\n(a^2)*a+(b^2)*b+(c^2)*c>=(a^2)8c+(b^2)*a+(c^2)*b[/quote]\r\n No, I think you can't suppose $ a \\geq b \\geq c $ \r\n This ineq $ a^3+b^3+c^3 \\geq ab^2+bc^2+ca^2 $ can be easily solved by AM-GM : \r\n $ a^3+b^3+c^3 \\geq 3ab^2 $", "Solution_14": "I have a question . which site I can see the problem of Romanian mathmatic olympiad and the problem of Romanian olympiad team?", "Solution_15": "well , you can find it in \"www.ajorza.org\"", "Solution_16": "[quote=\"ALGEBRA\"]let a,b,c>0 and a+b+c=1 prove that this inequality \n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $[/quote]\nExpanding gives $\\sum_{cyc}(a^4+a^3b-a^2b^2-a^2bc)\\geq0,$ which is obviously true.", "Solution_17": "The book referred to earlier is either [i]Old and New Inequalities [/i]by T. Andreescu, V. Cirtoaje, G. Dospinescu, and M. Lascu or [i]Old and New Inequalities, Volume 2[/i] by Vo Quoc Ba Can and Cosmin Pohoata. Both are published by GIL Publishing House, Zalau, Romania (2004, 2008, respectively)", "Solution_18": "[quote=\"ALGEBRA\"]let a,b,c>0 and a+b+c=1 prove that this inequality \n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $[/quote]\n\nwhere did I go wrong here?\n\nby cauchy, $\\sum_{cyc} \\dfrac{a^2+b}{b+c} = \\sum_{cyc} \\dfrac{a^2}{b+c} + \\dfrac{b^2}{b^2+bc} \\ge \\dfrac{(a+b+c)^2}{2(a+b+c)} + \\dfrac{(a+b+c)^2}{a^2 + b^2 + c^2 + ab + bc + ac} = \\dfrac{1}{2} + \\dfrac{1}{(a+b+c)^2 - ab - bc - ca}= \\dfrac12 + \\dfrac{1}{1-ab-bc-ca}$\n\nWhat remains to be proven is that $ab + bc + ca \\ge \\dfrac13$ if $a+b+c = 1$ which is ... false? because by rearrangement, $a^2 + b^2 + c^2 \\ge ab + bc + ca$ so $(a+b+c)^2 \\ge 3ab + 3bc + 3ca$ or $\\dfrac{1}{3} \\ge ab + bc + ca$", "Solution_19": "Dear AkshajK, that is because i think the inequality $ \\sum \\frac{b}{b+c} \\ge \\frac{3}{2}$ isn't true for all positive numbers", "Solution_20": "[quote=\"AkshajK\"][quote=\"ALGEBRA\"]let a,b,c>0 and a+b+c=1 prove that this inequality \n$\\frac{a^2+b}{b+c}\\ + \\frac{b^2+c}{c+a}\\ + \\frac{c^2+a}{a+b} \\geq\\ 2 $[/quote]\n\nwhere did I go wrong here?\n\nby cauchy, $\\sum_{cyc} \\dfrac{a^2+b}{b+c} = \\sum_{cyc} \\dfrac{a^2}{b+c} + \\dfrac{b^2}{b^2+bc} \\ge \\dfrac{(a+b+c)^2}{2(a+b+c)} + \\dfrac{(a+b+c)^2}{a^2 + b^2 + c^2 + ab + bc + ac} = \\dfrac{1}{2} + \\dfrac{1}{(a+b+c)^2 - ab - bc - ca}= \\dfrac12 + \\dfrac{1}{1-ab-bc-ca}$\n\nWhat remains to be proven is that $ab + bc + ca \\ge \\dfrac13$ if $a+b+c = 1$ which is ... false?[/quote]\n\n[b]AkshajK[/b], because your estimations were weak." } { "Tag": [ "combinatorial geometry", "combinatorics proposed", "combinatorics" ], "Problem": "Do there exist two disjoint infinite sets A and B of points in the plane such that: (1) no three points in the union of A and B are collinear; (2) the distance between any two points in the union of A and B is at least 1; (3) given any three points in B, there is a point of A in the triangle formed by the three points; (4) given any three points in A, there is a point of B in the triangle formed by the three points? A triangle means all points in or on the triangle (the vertices, all points on the sides of the triangle and all points in the interior of the triangle).", "Solution_1": "any solution?/??/", "Solution_2": "You simply need to find 5 points in A such that their convex hull does not contain any other points from A. Since A is discrete this is not difficult.\r\nWhen you have this points suppose the convex hull is 5-gon. Divide it into 3 disjoint triangles. You have point in B in each of this triangles. This gives you triangle formed by three points from B. So there is point in A in it. Contradiction.\r\nIt is the same when convex hull is quadrilateral or triangle." } { "Tag": [ "probability" ], "Problem": "4. Whats six times one-seventh?\r\nThis is the official number four in the chapters.\r\nHow many seconds did it take you to do that?", "Solution_1": "Yeah, I was appaled. \r\n\r\nIt took me 20 seconds because I was searching for some pitfall, I really didn't believe there was a problem THAT easy.", "Solution_2": "I disagree, I believe #3 on the Chapter Team Round was easier, I spent 2 minutes on that looking for pitfall, however, I just moved past this problem as there's no extra words that could've trapped you.\r\n\r\nLink to 2006 Chapter Team Round:\r\nhttp://www.mathcounts.org/webarticles/articlefiles/510-06CTeam.pdf", "Solution_3": "yes, i agree with Ignite168 on this one. Our strategy was to assign everyone on our team 2 problems and do the last two together and i was assigned #3. :D thats how i remember it", "Solution_4": "Yeah, that was one of them too, it was hard because it was so easy, I got confused for a minute.\r\n\r\nI still have to say 6* 1/7 is easier, it tells you the answer... And less reading.", "Solution_5": "i say #4, 6*1/7 was easier. i have to be the slowest reader here and i hate \"interpreting\" graphs. i remember it. didn't take long.", "Solution_6": "In 2005 problem #1 on chapter was to find the distance between cities A and C given a mileage chart for cities A,B,C,D. \r\n\r\n#1 at state was the same mileage chart except calculate the probability that two cities are more than 7000 mi apart.", "Solution_7": "I have to say that it is sadly pathetic...", "Solution_8": "[quote=\"kyyuanmathcount\"]Yeah, I was appaled. \n\nIt took me 20 seconds because I was searching for some pitfall, I really didn't believe there was a problem THAT easy.[/quote]\r\nSame here. :D :D :D", "Solution_9": "I didn't even think when I did that problem. :spider: :icecream:", "Solution_10": "#3 of the team competition was a piece of cake! :icecream:", "Solution_11": "I read it 3 times to check for any tricks [hide=Will put something in here after we are allowed to discuss state][/hide]" } { "Tag": [ "function", "algebra", "domain", "logarithms" ], "Problem": "f(xy/3) = f(x/3) + f(y) then find f(101)=?", "Solution_1": "[quote=\"mathservant\"]f(xy/3) = f(x/3) + f(y) then find f(101)=?[/quote]\r\nWhat is the domain of $f$?\r\nIf $f$ is defined on the whole set of reals, then let $x=0$. We get $f(0)=f(0)+f(y)\\Rightarrow f(y)=0\\forall y\\in\\mathbb{R}\\Rightarrow f(101)=0$\r\nIf $f$ is only defined on positive reals, then by letting $x=3$, we get $f(y)=f(1)+f(y)\\Rightarrow f(1)=0$. Substituting $a=\\frac{x}{3}$ and $b=y$ in the given condition yields $f(ab)=f(a)+f(b)$. Since $f$ is continous at one point, it is continous everywhere, and so the solution to this Cauchy equation is $f(x)=c\\log x$ where $c$ is a constant. Therefore $f(101)=c\\log 101$. I think we don't have enough conditions to work out $c$ here... :(" } { "Tag": [ "geometry" ], "Problem": "ABC is an equilateral triangle with sides equal to 2 cm. Line BC is extended its own length to D, and E is the midpoint of Line AB. Line ED meets Line AC at F. Find the area of quadrilateral BEFC in square centimeters in simplest radical form.", "Solution_1": "[hide]Draw segment AD. We now have triangle ABD with medians AC and DE drawn. Now draw the thrid median from B to side AD at point Z. We have all medians drawn, and the triangle is divided into six triangles of equal area. So to find the area of BEFC, we need to find the area of two of the equal area triangles, or 2/3 of the area of ABC. \\[\\sqrt{3}\\times \\frac{2}{3}=\\boxed{\\frac{2\\sqrt{3}}{3}}.\\] [/hide]" } { "Tag": [ "function", "Ross Mathematics Program", "logarithms", "calculus", "derivative", "inequalities" ], "Problem": "the followings are the standard way of proving [code]e^pi>pi^e[/code] :\n\n1) By considering the maximality of the functions [code]x^(1/x)[/code] or [code]x/ln(x)[/code].\n2) Another indirect proof given by Ross Honsberger in his book \"Mathematical Morsels\".\n\nI am required at least 3 more independent proofs of the above inequality. Is there way to prove it directly from\n[code]e^pi=1+pi+(pi^2/2!)+(pi^3/3!)+....[/code] ?\r\n\r\nCan anyone help ? :!: :( :( :?:", "Solution_1": "hello, let $ f(x)=\\frac{x}{\\ln(x)}$ then it is easy to show that $ f(x)$ has a global minimum ar $ x=e$, so we get since $ \\pi>e$ implies ${ \\frac{\\pi}{\\ln(\\pi)}}>e=f(e)$.\r\nSonnhard.", "Solution_2": "[quote=\"Dr Sonnhard Graubner\"]hello, let $ f(x) = \\frac {x}{\\ln(x)}$ then it is easy to show that $ f(x)$ has a global minimum ar $ x = e$, so we get since $ \\pi > e$ implies ${ \\frac {\\pi}{\\ln(\\pi)}} > e = f(e)$.\nSonnhard.[/quote]\r\n\r\nI think that's the method Shanku is referring to in #1.\r\n\r\nThis is probably not a completely independent proof, but here goes:\r\n\r\nWe can prove the result that $ e^x > x^e$ for $ x > e$. There exists $ k > 1$ such that $ x = ke$; we therefore must prove that $ e^{ke} = (e^e)^k > (ke)^e = (k^e)(e^e)$. This is the same as requiring $ (e^e)^{k - 1} = (e^{k - 1})^e > k^e$; taking the ln of both sides, $ e^{k - 1} > k$. Substituting $ g = k - 1$, we require $ e^g > g + 1$. We know that $ e^0 = 0 + 1$, so by comparing derivatives, it is easy to prove that the inequality $ e^g > g + 1$ holds for $ g > 0$. So $ e^x > x^e$ for $ x > e$, as needed. Since $ \\pi > e$ this concludes the proof.", "Solution_3": "Yes, Sonnhard's proof is the one I've already referred.\r\nThanx mqtrinh, though not entirely independent, I can consider your approach as a new proof because of the methodology. Pl keep me posted in case any other is found.", "Solution_4": "See http://www.mathlinks.ro/viewtopic.php?t=315059", "Solution_5": "Nothing New Andrew.\r\nI wonder whether these are the only ways to prove it." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "let {a(n)} satisfies:\r\n1.a(0)=a (a is an integer,a>=2)\r\n2.if a(n) is a square number then a(n+1)=[a(n)]^(1/2)\r\n3.if a(n) is not a square number then a(n+1)=a(n)+b.Here b is a odd prime.(a,b)=1\r\n\r\nprove or disprove that we can find a number M such that for all n>M we have a(n+1)=a(n)+b", "Solution_1": "Are you sure? Try $a=2$ and $b=127$.", "Solution_2": "ok,a(0)=2,a(1)=2+127=129,a(2)=256\r\na(3)=16\r\na(3+k)=16+127k\r\nput 16+127k=u^2 then 127|(u-4)(u+4)\r\n127 is an odd prime.so the min u is 123\r\nk=119\r\na(122)=123^2\r\na(123)=123\r\na(123+k)=123+127k\r\nput 123+127k=u^2 127|u^2+4\r\nbut (-4/127)=-1 because (-4)^63=1 (mod 127)\r\nso there exists an M=123 satisfies it.", "Solution_3": "a(3)=16, a(4)=4, a(5)=2. Not?", "Solution_4": "Oh,sorry.A mistake......\r\nand what about \"when a b are both odd primes\"?\r\njust a guessing of my friend." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Hi I am new to Mathcounts and how do the problems in the school handbook compare to the actual competition problems for school, chapter, states, and nationals?\r\nLike warmups 1- x are school level etc. etc. . If the problem are even similar at all.", "Solution_1": "The warmups are somewhat similar to the school round, perhaps a bit harder, and the workouts have problems in the state difficulty.\r\n\r\nOf course, some problems vary in difficulty, and workout problems could get kinda hard...maybe about easy national level. The problems do resemble the actual test problems in a way.", "Solution_2": "It really depends. This year's handbook seemed a bit easier than prior years. But I'd say:\r\nWarm ups=School Sprint\r\nWorkouts=Chapter Target", "Solution_3": "Let's say if a student mastered all problems in 2010 handbook.\r\nWill he/she do well in School, Chapter or State round?\r\n\r\nHow much School/Chapter/State contest problems in a year, say 2008, differ from 2008 handbook problems?", "Solution_4": "One should do both this year's along with those of previous years to get a firmer grip on the real School/Target Rounds. i think that the 'Warm-Ups' and 'Work-outs' are pretty similar to the real competition.", "Solution_5": "This varies of course ... but I think it is a good question. I wonder myself sometimes how to best use the handbooks to simulate competitions (as there are many more handbook problems than competition sets).\r\n\r\nI use the later Workouts (6-9) to simulate state-level team rounds, and I think in some cases they are more difficult (depends on the year ... handbooks and competitions vary widely sometimes). There are national level problems within the later Warm-Ups and Workouts (actually taken from national sets, for example, Workout 7 from this year's handbook contains a previous year's state team round problem).\r\n\r\nI would say that Warm-Ups 12-15 are similar to problems 15-25 on a state sprint round, while WU 16-18 are like the middle-pages of a national sprint.\r\n\r\nIf you are going to simulate a state-level sprint using the handbook, I would say ... take Warm-Ups 6, 12, and 18 ... and try do them all in 40 minutes. For chapter, take 4, 8, and 12 ... something like that.\r\n\r\nI am interested in hearing what students perceive as the difficulty of the competitions versus the handbook problems." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Given two numbers m and n, such that $2n \\geq n \\geq 0$. Then show that $2^{2n+2} + 2^{m+2} +1$ \r\nis a perfect square if and only if $m=n$.\r\n\r\nDavron", "Solution_1": "By hypothesis the given number is an odd square. Say it is (2k+1)^2. Setting these as equals and simplifying yields\r\n 2^(2n)+2^m=k(k+1) \r\nNow if 2n>m We have (2^m)(2^(2n-m)+1)=k(k+1) One of the factors on each side is odd while the other is even. If k is even then k=2^m and k=2^(2n-m) and m=n. If k is odd then k>k+1 (RAA).\r\nIf m>2n We have once again k is even, k=2^(2n), and k=2^(m-2n) then k^2=2^m. Thus 4n=m. Then the given square is (2^(2n+1)+1)^2. I think 2n>m may be required for the given result" } { "Tag": [], "Problem": "", "Solution_1": "", "Solution_2": "", "Solution_3": "", "Solution_4": "BOOOOOOOOOOOOOOOOO NO MATH" } { "Tag": [ "function", "integration", "algebra unsolved", "algebra" ], "Problem": "Does there exits a function $f: \\ (0;+\\infty)\\rightarrow (0;+\\infty)$ such that $\\exists f'(x) \\ \\forall x>0$ and $f'(x)\\ge f(f(x)) \\ \\forall x>0$", "Solution_1": "Hope this is not another proof of me being mentally retarded.\r\nSince $f'(x)\\geq f(f(x))>0$, if $f$ exists, it is monotonically increasing. If it was bounded then $f$ would tend (how do you say this in English?) to a limit, $c>0$ as $x$ grows to $\\infty$; in addition for fixed $\\delta$, $lim_{x\\to \\infty}\\frac{f(x+\\delta)-f(x)}{\\delta}=0$, and since letting $g(x)=y|[f'(y)=min\\{f'(z)| z\\in[x,x+\\delta]\\}]$, we would have that $f'(g(x))<\\frac{f(x+\\delta)-f(x)}{\\delta}$, we would get\r\n$0=lim_{x\\to \\infty}\\;f'(g(x))\\geq lim_{x\\to \\infty}f(f(g(x)))=lim_{x\\to \\infty}f(f(x))=f(c)>0$ \r\nwhich is a non-sense. So it would be unbounded and therefore there would be an $X$ such that $f(f(x))\\geq 1+\\epsilon \\;\\; \\forall x\\geq X$, and so, such that $f'(x)\\geq 1+\\epsilon\\;\\; \\forall x\\geq X$ for some $\\epsilon>0$ (it has no relation with the previous one). This means that $[f(x)-x]'=f'(x)-1\\geq \\epsilon>0 \\;\\;\\;\\forall x\\geq X$, so the function $f(x)-x$ would be unbounded; in particular there must exist a $Z$ such that $f(x)\\geq x+1 \\;\\;\\forall x\\geq Z$. But then $\\forall x\\geq Z$ we would have,\r\n$f(x+1)=f(x)+\\int_{x}^{x+1}f'(t)dt\\geq f(x)+\\int_{x}^{x+1}f(f(t))dt\\geq$\r\n$\\geq f(x)+\\int_{x}^{x+1}f(f(x))dt=f(x)+f(f(x))\\geq f(x)+f(x+1)>f(x+1)$, \r\nso the funtion doesn't exist." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "in every triangle with a,b,c sides, prove that:\r\n1/a+b-c +1/b+c-a +1/a+c-b>=1/a +1/b +1/c", "Solution_1": "Use Cauchy.So easy.", "Solution_2": "nice inequality:\r\nput: $ x=\\frac{1}{a+b-c}$ and $ y=\\frac{1}{a-b+c}$ and $ z=\\frac{1}{-a+b+c}$\r\n\r\n$ \\longleftrightarrow \\frac{1}{a}=\\frac{2xy}{x+y}$ ;$ \\frac{1}{b}=\\frac{2xz}{x+z}$ and $ \\frac{1}{c}=\\frac{2yz}{y+z}$\r\nthe inequality is equivalent has:\r\n$ S=\\frac{2xy}{x+y}+\\frac{2xz}{x+z}+\\frac{2yz}{y+z}\\leq x+y+z$\r\nwe know that: $ (x+y)^2\\geq 4xy$\r\n$ \\longleftrightarrow \\frac{2xy}{x+y}\\leq\\frac{1}{2}(x+y)$\r\nand $ \\longleftrightarrow \\frac{2yz}{y+z}\\leq\\frac{1}{2}(y+z)$\r\nand $ \\longleftrightarrow \\frac{2zx}{z+x}\\leq\\frac{1}{2}(z+x)$\r\nby summation:\r\n$ S\\leq x+y+z$", "Solution_3": "[quote=\"behdad.math.math\"]In every triangle with $ a,b,c$ sides, prove that\n\n$ \\frac{1}{b+c-a} +\\frac{1}{c+a-b}+\\frac{1}{a+b-c }\\geq\\frac{1}{a} +\\frac{1}{b} +\\frac{1}{c}.$[/quote]$ \\sum{\\frac{1}{b+c-a}}-\\sum{\\frac{1}{a}}=\\frac{1}{2}\\sum{\\left(\\frac{1}{c+a-b}+\\frac{1}{a+b-c }\\right)-\\sum{\\frac{1}{a}}=\\sum\\frac{(b-c)^2}{a(c+a-b)(a+b-c)}\\geq0.}$", "Solution_4": "note that $ a\\plus{}b\\minus{}c,b\\plus{}c\\minus{}a,c\\plus{}a\\minus{}b>0$\r\n\r\nThe rest follows. :D" } { "Tag": [], "Problem": "Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?\r\n(A) 45\r\n(B) 48\r\n(C) 50\r\n(D) 55\r\n(E) 58", "Solution_1": "[hide] $x$ is the distance to his workplace. $y$ is how long it should take for him to be on time. \n$\\frac{x}{40}=y+3$\n$\\frac{x}{60}=y-3$\n$40y+120=60y-180$\n$y=15$\n$x=60\\cdot12=720$\n$\\frac{720}{15}=\\boxed{B)48}$[/hide]", "Solution_2": "[hide=\"Another method\"] In this type of problem we can use harmonic mean. $\\frac{1}{\\frac{\\frac{1}{40}+\\frac{1}{60}}{2}}=\\frac{2}{\\frac{3}{120}+\\frac{2}{120}}=\\frac{2}{\\frac{5}{120}}=2\\cdot\\frac{120}{5}=\\boxed{48}$[/hide]", "Solution_3": "[hide=\"maybe simpler\"]\nsince going at 40 mph and 60 mph get him to work the same amount of time late as the other does early, then the desired speed is the harmonic mean of 40 and 60.\nharmonic mean is $\\frac{2ab}{a+b}$ or in our case, $\\frac{2\\times60\\times40}{60+40}=\\fbox{48}$[/hide]", "Solution_4": "[quote=\"lotrgreengrapes7926\"][hide=\"Another method\"] In this type of problem we can use harmonic mean. $\\frac{1}{\\frac{\\frac{1}{40}+\\frac{1}{60}}{2}}=\\frac{2}{\\frac{3}{120}+\\frac{2}{120}}=\\frac{2}{\\frac{5}{120}}=2\\cdot\\frac{120}{5}=\\boxed{48}$[/hide][/quote]\r\n\r\nWhat's a harmonic mean?", "Solution_5": "Why would it be the harmonic mean and not some other kind?", "Solution_6": "sapphyre571, yours is almost exactly like mine. \r\n\r\nThe harmonic mean is the reciprocal of the arithmetic mean of the reciprocals of the numbers. For example, $\\frac{1}{a}$ and $\\frac{1}{b}$ are the 2 reciprocals. The arithmetic mean of those are $\\frac{\\frac{1}{a}+\\frac{1}{b}}{2}=\\frac{\\frac{a+b}{ab}}{2}=\\frac{a+b}{2ab}$. Then, the reciprocal of that is $\\frac{2ab}{a+b}$.\r\n\r\nI don't know exactly why it works, but it almost always does in problems involving rates.", "Solution_7": "Does it always work?\r\nHow do you use use it (like know which numbers to average)?", "Solution_8": "[quote=\"laughinghead505\"]Why would it be the harmonic mean and not some other kind?[/quote]\r\n\r\nI don't know.\r\n\r\nIt is by name it is called harmonic mean. you can call it whatever you remember it by, and this would help with average speed problems.", "Solution_9": "[hide]B[/hide]", "Solution_10": "[hide] (rate)(time) = (distance)\n\nlet $T$ be the time it takes to get to work in minutes.\n\n$40(T+3) = 60(T-3) \\Rightarrow T=15$\n\nlet $x$ be the desired speed. then $40(18) = 15x$\ngiving $x = 48$\nMaking the answer $\\boxed{B}$[/hide]", "Solution_11": "The simplest problem you can use harmonic mean on is \"I drove at $x$ mph to point A and drove at $y$ mph on the way back. What was my average speed?\" For other problems, you may need to adjust and use it differently.", "Solution_12": "No, I meant why would it be the harmonic mean instead of, say the arithmetic or geometric?" } { "Tag": [], "Problem": "Is it possible to use Kile on a Windows XP platform?\r\nIf possible, how should one install it?", "Solution_1": "I just downloaded it from sourceforge.\r\nIt is an archive of thousands of files, none of them executable.\r\nHow do I use it?", "Solution_2": "It looks like you need KDE (or at least QT) to run Kile. KDE is a desktop environment for Unix-based operating systems. You might be able to get KDE to work in Windows using Cygwin, if you really want to use Kile, but I'd suggest finding a different editor." } { "Tag": [ "inequalities", "function", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Prove that for every continous function $f: [0,1]\\to R$ the following inequality holds:\r\n$\\frac{1}{4}\\int_{0}^{1}f^{2}(x)dx+2\\left ( \\int_{0}^{1}f(x)dx \\right )^{2}\\geq 3 \\int_{0}^{1}f(x)dx \\int_{0}^{1}xf(x)dx$", "Solution_1": "This really is the typical Cezar Lupu inequality.\r\n\r\nTrying some functions we discover that for $f(x) \\equiv 3x-1$ we have equality in\r\n\\[\\int_{0}^{1}f^{2}(x) \\, dx+8 \\left( \\int_{0}^{1}f(x) \\, dx \\right)^{2}\\geq 12 \\int_{0}^{1}f(x) \\, dx \\int_{0}^{1}x f(x) \\, dx . \\]\r\nSo, let's apply an inequality. But what inequality should that be? You've guessed it, it's Cauchy:\r\n\\[\\int_{0}^{1}f^{2}(x) \\, dx \\int_{0}^{1}\\left( x-\\frac13 \\right)^{2}\\, dx \\geq \\left( \\int_{0}^{1}f(x) \\cdot \\left( x-\\frac13 \\right) \\, dx \\right)^{2}, \\]\r\ni.e.\r\n\\[\\int_{0}^{1}f^{2}(x) \\, dx \\geq \\left( \\int_{0}^{1}f(x) \\cdot \\left( 3x-1 \\right) \\, dx \\right)^{2}. \\]\r\nThus, it suffices to prove that\r\n\\[\\left( \\int_{0}^{1}f(x) \\cdot \\left( 3x-1 \\right) \\, dx \\right)^{2}\\geq 12 \\int_{0}^{1}f(x) \\, dx \\int_{0}^{1}x f(x) \\, dx-8 \\left( \\int_{0}^{1}f(x) \\, dx \\right)^{2}\\]\r\n\r\n\\[\\iff 9 \\left( \\int_{0}^{1}x f(x) \\, dx \\right)^{2}-6 \\int_{0}^{1}x f(x) \\, dx \\int_{0}^{1}f(x) \\, dx+\\left( \\int_{0}^{1}f(x) \\, dx \\right)^{2}\\geq \\]\r\n\r\n\\[\\geq 12 \\int_{0}^{1}f(x) \\, dx \\int_{0}^{1}x f(x) \\, dx-8 \\left( \\int_{0}^{1}f(x) \\, dx \\right)^{2}\\]\r\n\r\n\\[\\iff 9 \\left( \\int_{0}^{1}x f(x) \\, dx \\right)^{2}+9 \\left( \\int_{0}^{1}f(x) \\, dx \\right)^{2}\\geq 18 \\int_{0}^{1}f(x) \\, dx \\int_{0}^{1}x f(x) \\, dx , \\]\r\nwhich is true by $AM-GM$.\r\n\r\nI hope I haven't made any stupid mistakes!", "Solution_2": "I think the solution is correct. Thanx a lot perfect_radio! This bloody problem kept annoying me since the olympiad :) !", "Solution_3": "Here is a similar but slightly different approach. Note that $1$ and $x-1/2$ are orthogonal as elements of $L^{2}[0,1]$. So, let $f=a+b(x-1/2)+g$ where $g$ is orthogonal to both $1$ and $x-1/2$. Then $\\int_{0}^{1}f^{2}\\ge a^{2}+\\frac{b^{2}}{12}$, $\\int_{0}^{1}f=a$ and $\\int_{0}^{1}xf=\\frac{a}{2}+\\frac{b}{12}$, which brings us to the inequality\r\n\\[a^{2}+\\frac{b^{2}}{12}+8a^{2}\\ge 12 a\\left(\\frac{a}{2}+\\frac{b}{12}\\right)\\]\r\nor, which is the same,\r\n\\[3a^{2}+\\frac{b^{2}}{12}-ab = 3\\left(a-\\frac{b}{6}\\right)^{2}\\ge 0\\,..\\]\r\nIt is clear from here that, to get equality, we must have $b=6a$, $g\\equiv 0$, so $f(x)=a(1+6(x-1/2))=a(6x-2)$.\r\n\r\nIn general, when you see an inequality involving a few linear functionals of $f$ and $\\int f^{2}$, it may be helpful to use some orthogonal system that is rich enough to allow you to express all the involved linear functionals in terms of the Fourier coefficients of $f$ with respect to this system. Then, instead of an inequality for functions, you'll get one for numbers." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be positive numbers such that: $ abc \\equal{}1$.Prove that:\r\n$ \\frac{\\sqrt{a^2\\plus{}ab\\plus{}2b^2}}{c^3\\plus{}3}\\plus{}\\frac{\\sqrt{b^2\\plus{}bc\\plus{}2c^2}}{a^3\\plus{}3}\\plus{}\\frac{\\sqrt{c^2\\plus{}ca\\plus{}2a^2}}{b^3\\plus{}3} \\ge \\frac{3}{2}$", "Solution_1": "[quote=\"quykhtn-qa1\"]Let $ a,b,c$ be positive numbers such that: $ abc \\equal{} 1$.Prove that:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^3 \\plus{} 3} \\ge \\frac {3}{2}$[/quote]\r\nI have an nice solution for this inequality.But with the following inequality:\r\n$ \\frac{\\sqrt{a^2\\plus{}ab\\plus{}2b^2}}{c^4\\plus{}4}\\plus{}\\frac{\\sqrt{b^2\\plus{}bc\\plus{}2c^2}}{a^4\\plus{}4}\\plus{}\\frac{\\sqrt{c^2\\plus{}ca\\plus{}2a^2}}{b^4\\plus{}4} \\ge \\frac{6}{5}$\r\nI don't know it 's true or false? :(\r\n .I hope you can help me.Thanks :)", "Solution_2": "[quote=\"quykhtn-qa1\"]Let $ a,b,c$ be positive numbers such that: $ abc \\equal{} 1$.Prove that:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^3 \\plus{} 3} \\ge \\frac {3}{2}$[/quote]\r\nDear quykhtn-qa1 !\r\nCan you give some hint with this inequality,thanks.", "Solution_3": "[quote=\"quykhtn-qa1\"][quote=\"quykhtn-qa1\"]Let $ a,b,c$ be positive numbers such that: $ abc \\equal{} 1$.Prove that:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^3 \\plus{} 3} \\ge \\frac {3}{2}$[/quote]\nI have an nice solution for this inequality.But with the following inequality:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^4 \\plus{} 4} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^4 \\plus{} 4} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^4 \\plus{} 4} \\ge \\frac {6}{5}$\nI don't know it 's true or false? :(\n .I hope you can help me.Thanks :)[/quote]\r\n\r\nThe second is also correct :wink:", "Solution_4": "[quote=\"quykhtn-qa1\"]Let $ a,b,c$ be positive numbers such that: $ abc \\equal{} 1$.Prove that:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^3 \\plus{} 3} \\ge \\frac {3}{2}$[/quote]\r\nIt's easy because $ \\sqrt{a^2\\plus{}ab\\plus{}2b^2}\\geq\\frac{3a\\plus{}5b}{4}$\r\nand $ \\sum_{cyc}{\\frac {a}{c^3 \\plus{} 3}}\\geq\\frac {3}{4}$\r\nand $ \\sum_{cyc}{\\frac {b}{c^3 \\plus{} 3}}\\geq\\frac {3}{4}$ (It's old and easy by CS) :P", "Solution_5": "[quote=\"zaizai-hoang\"][quote=\"quykhtn-qa1\"][quote=\"quykhtn-qa1\"]Let $ a,b,c$ be positive numbers such that: $ abc \\equal{} 1$.Prove that:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^3 \\plus{} 3} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^3 \\plus{} 3} \\ge \\frac {3}{2}$[/quote]\nI have an nice solution for this inequality.But with the following inequality:\n$ \\frac {\\sqrt {a^2 \\plus{} ab \\plus{} 2b^2}}{c^4 \\plus{} 4} \\plus{} \\frac {\\sqrt {b^2 \\plus{} bc \\plus{} 2c^2}}{a^4 \\plus{} 4} \\plus{} \\frac {\\sqrt {c^2 \\plus{} ca \\plus{} 2a^2}}{b^4 \\plus{} 4} \\ge \\frac {6}{5}$\nI don't know it 's true or false? :(\n .I hope you can help me.Thanks :)[/quote]\n\nThe second is also correct :wink:[/quote]\r\nI think,if second true.It's very strong! :) \r\ncan you post your solution?thank you!" } { "Tag": [ "summer program", "MathPath", "pigeonhole principle", "Duke", "college", "Stanford" ], "Problem": "Besides MathPath, what are some of theo ther good MS programs?", "Solution_1": "For middle school students, I highly recommend attending a program such as \r\n\r\nCenter for Talented Youth \r\n\r\ninstead of a program like MathPath. \r\n\r\nI don't really think it's good to pigeonhole students at THAT young of an age.\r\n\r\n(Disclaimer Yes, I do work for CTY, so there is a bias element..)", "Solution_2": "the one from Johns Hopkins?? \r\n\r\nwhat about Stanford's MS institute?", "Solution_3": "I think Stanford's one is great as well (JHU, Duke and Stanford all collaborate together)." } { "Tag": [ "calculus", "derivative", "function", "induction", "algebra", "functional equation", "real analysis" ], "Problem": "Find all fuctions $f: \\mathbb{R}\\to \\mathbb{R}$ continuous at $0$ and satisfying for all real $x,y$ the following equation\r\n\\[f(x+y)=f(x)+f(y)+xy(x+y) \\]\r\n[hide=\"the case when there exists $f'(0)$\"]Let $c=f'(0)$, then there exists $\\exists f'(x)=x^{2}+c \\Longrightarrow f(x)=\\frac{x^{3}}{3}+cx$ is the only solution. What if the derivative at 0 does not exist :?: [/hide]", "Solution_1": "Since $f(0)=0,\\ f(x)=\\frac{1}{3}x^{3}.$", "Solution_2": "why kunny? \r\nwhat;s wrong with my answer?\r\nand by the way how do I have to put the text to be hiden?", "Solution_3": "Plugging $x=y=0$ into the given functional equation gives the desired answer. This problem is made by the following equality.\r\n\r\n$\\boxed{(x+y)^{3}=x^{3}+y^{3}+3xy(x+y)}\\Longleftrightarrow \\frac{1}{3}(x+y)^{3}=\\frac{1}{3}x^{3}+\\frac{1}{3}y^{3}+xy(x+y).$", "Solution_4": "\\[f(x+y)=\\frac13 \\cdot (x+y)^{3}+c(x+y)= \\frac13 x^{3}+cx+\\frac13 y^{3}+cy+xy(x+y)= f(x)+f(y)+xy(x+y) \\]\r\nSO $f(x)=\\frac13x^{3}+cx$ is solution.\r\n\r\nBut how do we solve the problem if the [color=red]fuction is continuous in one point 0 and you dont know whether there exists or not[/color] $f'(0)$ :|", "Solution_5": "Define $g(x)$ so that $f(x)=\\frac{x^{3}}3+g(x).$ Then the functional equation reduces to:\r\n\r\n$g(x+y)=g(x)+g(y).$\r\n\r\nThis is a fairly famous problem, and it is known that the only continuous solutions are of the form $g(x)=cx,$ although there do exist discontinuous solutions (the \"constuction\" of such monsters requires the Axiom of Choice.)\r\n\r\nLet $c=g(1).$ By induction, $g(n)=cn.$ By another induction, $g\\left(\\frac{m}{n}\\right)=c\\frac{m}{n}.$\r\n\r\nNow let $x$ be any real number, and find a sequence $q_{n}$ of rational numbers which tends to $x.$ \r\n\r\n$g(x)=g(q_{n})+g(x-q_{n})=cq_{n}+g(x-q_{n}).$\r\n\r\nTak the limit. $cq_{n}$ tends to $cx$ while $g(x-q_{n})$ tends to $g(0)=0$ because $g$ is presumed continuous at zero. We get that $g(x)=cx.$\r\n\r\nSince $g(x)=cx,$ we have that $f(x)=\\frac{x^{3}}3+cx.$", "Solution_6": "Thank you!\r\nI do had to figure it out from my previous post :oops:" } { "Tag": [ "calculus", "derivative", "analytic geometry", "graphing lines", "slope", "calculus computations" ], "Problem": "Find the equation of the line tangent to $ y\\equal{}f(x)$ at $ x\\equal{}1$, where $ f(x)\\equal{}6e^{5x} \\plus{} e^{\\minus{}x^2}$?\r\n\r\n[hide=\"what I got\"]The derivative of the equation above:\n$ \\frac{df(x)}{dx}\\equal{}30e \\minus{} \\frac{2}{e}$\n\nPlugging in $ x\\equal{}1$ to $ f(x) \\rightarrow 6e^4$.\n\nSo I set-up point slope form and got\n$ y \\equal{} (x\\minus{}1)(30e \\minus{} \\frac{2}{e}) \\plus{} 6e^4$\n\nI got a feeling that it is wrong...[/hide]", "Solution_1": "[quote=\"jeez123\"]Plugging in $ x \\equal{} 1$ to $ \\boxed{f(x) \\rightarrow 6e^4}$.\n\nSo I set-up point slope form and got\n$ y \\equal{} (x \\minus{} 1)(30e \\minus{} \\frac {2}{e}) \\plus{} \\boxed{6e^4}$\n\nI got a feeling that it is wrong...[/hide][/quote]\r\n\r\n :idea:", "Solution_2": "[quote=\"jeez123\"]Find the equation of the line tangent to $ y \\equal{} f(x)$ at $ x \\equal{} 1$, where $ f(x) \\equal{} 6e^{5x} \\plus{} e^{ \\minus{} x^2}$?\r\n\r\n\r\n$ f'(x) \\equal{} 30e^{5x} \\minus{} 2xe^{\\minus{}x^{2}}$ \r\n\r\n$ f'(1) \\equal{} 30e^{5} \\minus{} 2e^{\\minus{}1}$\r\n\r\nHence: $ m \\equal{} 30e^{5} \\minus{} 2e^{\\minus{}1}$\r\n $ x_{0} \\equal{} 1$\r\n $ y_{0} \\equal{} 6e^{5} \\plus{} e^{ \\minus{} 1}$\r\n $ y \\minus{} y_{0} \\equal{} m(x \\minus{} x_{0})$" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let $A_1B_1C_1$ be an acute triangle. $A , B , C$ are respectively belong to segments $B_1C_1 , C_1A_1 , A_1B_1$ such that : $\\angle BAC = \\angle B_1A_1C_1$ , $\\angle ABC = \\angle A_1B_1C_1$ , $\\angle BCA = \\angle B_1C_1A_1$ . Let $(O)$ be the circumcircle of $\\triangle A_1B_1C_1$ . $H , H_1$ are respectively the orthocenters of $\\triangle ABC$ , $\\triangle A_1B_1C_1$ . Prove that : $OH =OH_1$", "Solution_1": "I found that : $(A_1BC) , (B_1AC)$ and $(C_1AB)$ have a common point namely $H$ . We also have $\\R(A_1BC) =\\R(AB_1C) =\\R(ABC_1) =\\R(ABC)$ (it can be prove by applying sin law ) \r\n And it's easy to get $H$ is the circumcenter of $(A_1B_1C_1)$ \r\n Could you please help me to finish it ?" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Call a super-integer an infinite sequence of decimal digits: $\\ldots d_n \\ldots d_2d_1$. \n\n(Formally speaking, it is the sequence $(d_1,d_2d_1,d_3d_2d_1,\\ldots)$ )\n\nGiven two such super-integers $\\ldots c_n \\ldots c_2c_1$ and $\\ldots d_n \\ldots d_2d_1$, their product $\\ldots p_n \\ldots p_2p_1$ is formed by taking $p_n \\ldots p_2p_1$ to be the last n digits of the product $c_n \\ldots c_2c_1$ and $d_n \\ldots d_2d_1$. \nCan we find two non-zero super-integers with zero product?\n (a zero super-integer has all its digits zero)", "Solution_1": "$2^n(mod \\ 10^n), 5^n (mod \\ 10^n)$.", "Solution_2": "This problem is probably from some older ISL or something like hat since it was used in a german TST some years ago.\r\nThe solution of Rust is incorrect since it doesn't give such sequence (or at least I don't see how it shall work).\r\n\r\nThe idea behind this is to work in the ring $\\mathbb{Z}_2 \\times \\mathbb{Z}_5$ (being product of two rings of $p$-adic integers).\r\nWe are looking for some solution to $xy=0$ other than the trivial ones. Such solution exists, e.g. $(0,1) \\cdot (1,0) = (0,0)$ in the ring above.\r\nSo taking some $x \\equiv 1 \\mod 5^n$ , $x \\equiv 0 \\mod 2^n$ and inversly $y$ will give us the result, and they will fulfill all we want (they are unique determined $\\mod 10^n$, thus by highering $n$ the last digits will remain the same)." } { "Tag": [ "function", "induction", "trigonometry", "logarithms" ], "Problem": "If $ f(x)$ is a real valued function of the real variable $ x$, and $ f(x)$ is not identically zero, and for all $ a$ and $ b$ we have $ f(a \\plus{} b) \\plus{} f(a \\minus{} b) \\equal{} 2f(a) \\plus{} 2f(b)$, then for all $ x$ and $ y$\r\n(A) $ f(0) \\equal{} 1$\r\n(B) $ f( \\minus{} x) \\equal{} \\minus{} f(x)$\r\n(C) $ f( \\minus{} x) \\equal{} f(x)$\r\n(D) $ f(x \\plus{} y) \\equal{} f(x) \\plus{} f(y)$\r\n(E) There is a positive number $ T$ such that $ f(x \\plus{} T) \\equal{} f(x)$\r\n\r\nThanks in advance!", "Solution_1": "[hide=\"solution\"]\n$ a\\equal{}b\\rightarrow f(2b)\\plus{}f(0)\\equal{}4f(b)$\n$ a\\equal{}\\minus{}b\\rightarrow f(0)\\plus{}f(\\minus{}2b)\\equal{} 2f(\\minus{}b)\\plus{}2f(b)$\n$ f(2b)\\minus{}f(\\minus{}2b)\\equal{}2f(b)\\minus{}2f(\\minus{}b)$\n$ f(2b)\\minus{}2f(b)\\equal{}f(\\minus{}2b)\\minus{}2f(\\minus{}b)$\nThus, $ f(b)\\equal{}f(\\minus{}b)$, or C.[/hide]", "Solution_2": "What kind of function would satisfy that relationship?", "Solution_3": "[hide=\"Discussion\"] First of all, there is a more direct way to get the answer: substitute $ b \\equal{} \\minus{}b$ on the LHS and you get $ f(a \\minus{} b) \\plus{} f(a \\plus{} b) \\equal{} 2f(a) \\plus{} 2f(\\minus{}b)$, which we already know is equal to $ 2f(a) \\plus{} 2f(b)$.\n\nOne solution is $ f(x) \\equal{} cx^2$. Are there others? [/hide]", "Solution_4": "t0rajir0u, to add on to what you said, substituting $ b \\equal{} a$ yields $ f(0) \\plus{} f(2a) \\equal{} 4 f(a)$. Substituting $ a \\equal{} 0$ yields $ 2f(0) \\equal{} 4 f(0)$, implying that $ f(0) \\equal{} 0$, so $ f(2a) \\equal{} 4 f(a)$, so $ f(a) \\equal{} f(2a) / 4$. \r\n\r\nIf we let $ f(1) \\equal{} c$, it follows from induction that $ f(2^k) \\equal{} 4^k c$ for all integers $ k$ (positive and negative.)\r\n\r\nNot sure where else we can go from there... other than it's even.", "Solution_5": "Rephrase that as, if $ x$ is a power of $ 2$, then $ f(x) \\equal{} cx^2$. It is then natural to see if this relationship also holds true if $ x$ is not necessarily a power of $ 2$.", "Solution_6": "I wouldn't spend too much time. I'm almost positive that the same kind of noncontinuous solutions that the Cauchy equation has will show up here. But I think you can get $ f(x) \\equal{} cx^2$ if the function is defined on rationals without too much trouble.", "Solution_7": "[quote=\"TZF\"]Rephrase that as, if $ x$ is a power of $ 2$, then $ f(x) \\equal{} cx^2$. It is then natural to see if this relationship also holds true if $ x$ is not necessarily a power of $ 2$.[/quote]\r\nThis is not at all the case, and you don't even need Cauchy-type solutions to see this. Consider the function $ f(x) \\equal{} c \\left( \\cos (2 \\pi \\log_2 |x| ) \\right)^2$." } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "Evaluate\r\n\r\n$ \\int\\frac{x^{3n-1}-x^{n-1}}{x^{4n}+1}dx$", "Solution_1": "Observe that $ x^{4n}+1 = (x^{2n}-\\sqrt{2}x^{n}+1)(x^{2n}+\\sqrt{2}x^{n}+1)$.\r\n\r\nThe rest is easy.\r\n\r\nThe result I obtained is:\r\n\r\n\\[ \\frac{\\sqrt{2}}{4n}\\left( \\ln(x^{2n}-\\sqrt{2}x^{n}+1)-\\ln(x^{2n}+\\sqrt{2}x^{n}+1) \\right)\\]", "Solution_2": "show your work even though the answer is right", "Solution_3": "can someone work this fully work this out?", "Solution_4": "[quote=\"mdk\"]Ascertain the indefinite integral $ I=\\int\\frac{x^{3n-1}-x^{n-1}}{x^{4n}+1}\\ dx$[/quote]\r\n[color=darkblue][b]Proof.[/b] Observe that $ I=\\int\\frac{x^{n-1}\\left(x^{2n}-1\\right)}{x^{4n}+1}\\ dx=\\frac{1}{n}\\cdot\\int\\frac{\\left(x^{n}\\right)^{2}-1}{\\left(x^{n}\\right)^{4}+1}\\cdot\\left(x^{n}\\right)'\\ dx$. Using the substitution $ y=x^{n}$ obtain $ I=\\frac{1}{n}\\cdot K\\left(x^{n}\\right)+\\mathcal C$, where\n$ K=\\int\\frac{y^{2}-1}{y^{4}+1}\\ dy=\\int\\frac{1-\\frac{1}{y^{2}}}{y^{2}+\\frac{1}{y^{2}}}\\ dy=\\int\\frac{\\left(y+\\frac{1}{y}\\right)'}{\\left(y+\\frac{1}{y}\\right)^{2}-2}\\ dy$. Using the substitution $ z=y+\\frac{1}{y}$ obtain $ K=L\\left(y+\\frac{1}{y}\\right)+\\mathcal C$, where\n$ L=\\int\\frac{1}{z^{2}-2}\\ dz=\\frac{1}{2\\sqrt 2}\\ln\\left|\\frac{z-\\sqrt 2}{z+\\sqrt 2}\\right|+\\mathcal C$. Therefore, $ K=\\frac{\\sqrt 2}{4}\\cdot \\ln\\frac{y^{2}-y\\sqrt 2+1}{y^{2}+y\\sqrt 2+1}+\\mathcal C$ and $ \\boxed{I=\\frac{\\sqrt 2}{4n}\\cdot\\ln\\frac{x^{2n}-x^{n}\\sqrt 2+1}{x^{2n}+x^{n}\\sqrt 2+1}+\\mathcal C}$.\n\n[b]Remark.[/b] Can apply directly the substitution $ t=x^{n}+\\frac{1}{x^{n}}$ and obtain $ I=\\frac{1}{n}\\cdot\\int \\frac{1}{\\left(x^{n}+\\frac{1}{x^{n}}\\right)^{2}-2}\\cdot\\left(x^{n}+\\frac{1}{x^{n}}\\right)'\\ dx$, i.e. $ I=\\frac{\\sqrt 2}{4n}\\cdot\\ln\\frac{x^{2n}-x^{n}\\sqrt 2+1}{x^{2n}+x^{n}\\sqrt 2+1}+\\mathcal C$.[/color]", "Solution_5": "[quote=\"Virgil Nicula\"]\n[b]Proof.[/b] Observe that $ I=\\int\\frac{x^{n-1}\\left(x^{2n}-1\\right)}{x^{4n}+1}\\ dx=\\frac{1}{n}\\cdot\\int\\frac{\\left(x^{n}\\right)^{2}-1}{\\left(x^{n}\\right)^{4}+1}\\cdot\\left(x^{n}\\right)'\\ dx$. Using the substitution $ y=x^{n}$ obtain $ I=\\frac{1}{n}\\cdot K\\left(x^{n}\\right)+\\mathcal C_{1}$, where\n$ K=\\int\\frac{y^{2}-1}{y^{4}+1}\\ dy=\\int\\frac{1-\\frac{1}{y^{2}}}{y^{2}+\\frac{1}{y^{2}}}\\ dy=\\int\\frac{\\left(y+\\frac{1}{y}\\right)'}{\\left(y+\\frac{1}{y}\\right)^{2}-2}\\ dy$. Using the substitution $ z=y+\\frac{1}{y}$ obtain $ K=L\\left(y+\\frac{1}{y}\\right)+\\mathcal C_{2}$, where\n$ L=\\int\\frac{1}{z^{2}-2}\\ dz=\\frac{1}{2\\sqrt 2}\\ln\\left|\\frac{z-\\sqrt 2}{z+\\sqrt 2}\\right|+\\mathcal C_{3}$. Therefore, $ K=\\frac{\\sqrt 2}{4}\\cdot \\ln\\frac{y^{2}-y\\sqrt 2+1}{y^{2}+y\\sqrt 2+1}+\\mathcal C_{2}+\\mathcal C_{3}$ and $ \\boxed{I=\\frac{\\sqrt 2}{4n}\\cdot\\ln\\frac{x^{2n}-x^{n}\\sqrt 2+1}{x^{2n}+x^{n}\\sqrt 2+1}+\\mathcal C_{1}+\\mathcal C_{2}+\\mathcal C_{3}}$.\n[/quote]\r\n\r\nnice ! :coolspeak:" } { "Tag": [ "inequalities" ], "Problem": "I really want to know how to prove Holder ineqality\u2026\u2026This maybe the most simple one:\r\n\u6709\u5b9e\u6570\u5217$a_1,a_2,a_3...a_n$\u4e0e$b_1,b_2,b_3...b_n$\u4ee5\u53ca$\\frac{1}{p}+\\frac{1}{q}=1$\r\n\u6c42\u8bc1:$({a_1}^p+{a_2}^p+...+{a_n}^p)^{\\frac{1}{p}}({b_1}^q+{b_2}^q+...+{b_n}^q)^{\\frac{1}{q}} \\geq |a_1b_1+a_2b_2+...+a_nb_n|$", "Solution_1": "One approach is to use Young's inequality.", "Solution_2": "As mentioned above, Young's inequality can be used.\r\nIf p,q are positive such that $\\frac{1}{p}+\\frac{1}{q}=1$ and a,b are positive, then $\\frac{a^p}{p}+\\frac{b^q}{q} \\ge ab$\r\nProof: This is directly true by using the fact that ln is concave.\r\n\r\nNow, without loss of generality, assume $\\sum a^p = 1$ and $\\sum b^q=1$, then by young's inequality,\r\n$\\sum a_1b_1 \\le \\sum \\frac{a_1^p}{p} + \\frac{b_1^q}{q} = \\frac{1}{p}+\\frac{1}{q} = 1$" } { "Tag": [ "probability", "Support" ], "Problem": "when you get a problem wrong/right it will show u a solution that you get to rate. are there multiple DIFFERENT solutions for each problem?", "Solution_1": "I don't think there are different solutions that they show, but most problems could be solved a unique way than the solutions provided.", "Solution_2": "There are a lot of ways to solve most probability problems, and the solution usually states the easiest method for the specific problem.", "Solution_3": "If you find an easier solution, you can post it on the forum too! :)", "Solution_4": "[quote=\"007math\"]I don't think there are different solutions that they show, but most problems could be solved a unique way than the solutions provided.[/quote]\r\n\r\nyeah they could \r\ndifferent solutions would be easier to understand for different ppl", "Solution_5": "If you believe that there is a better solution or a solution that can be better understood, you could always suggest it on the forum, and maybe an administrator would consider it as an alternate solution.", "Solution_6": "We have multiple solutions on the \"to be added\" list for Alcumus. I would love to see this as well since not only may one solution be easier than another, it's helpful to understand how you can solve problems in different ways." } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "modular arithmetic", "algebra", "polynomial", "limit" ], "Problem": "It is a nice and famous problem (given at 1995 USAMO and, I believe, some Russian olympiad before this): let $f : \\mathbb{N} \\to \\mathbb{Z}$ be such that $f(x+m) \\equiv f(x) \\pmod{m}$ for all $x,m \\in \\mathbb{N}$. If $f(n)$ is bounded by a polynomial (that is, there are constants $C,k$ such that $|f(n)| < Cn^k$ for all $n$), then $f(n)$ [b] is[/b] a polynomial (that is, there exists a polynomial $Q \\in \\mathbb{Q}[x]$ such that $f(n) = Q(n)$ for all $n$).\r\n\r\nNow, I propose a stronger result: it suffices that $f$ is bounded [i] exponentially[/i]!\r\n\r\n[b]Claim: [/b] Let $f : \\mathbb{N} \\to \\mathbb{Z}$ be such that $f(x+m) \\equiv f(x) \\pmod{m}$ for all $x,m \\in \\mathbb{N}$. Suppose that there exists a constant $A > 0$ such that $|f(n)| < A^n$ for all $n$. Then, $f(n)$ must be a polynomial in $n$.\r\n\r\n--Vesselin", "Solution_1": "I think this is true if and only if $Ae$ immediately: construct $a_n$ recurrently: take $a_{n+1}=b_{n+1}+l_nA^n$ for all $A 1$ for the region $ A$ to exist.\r\n\r\nThe volume of the cylinder formed when the total rectangular region ($ A \\plus{} B$) is revolved about the $ x$-axis is $ \\pi k^2 \\sqrt(k \\minus{} 1)$.\r\n\r\nThus, the sume of the volume generated by $ A$, $ V_A$, and the volume generated by $ B$, $ V_B$, is\r\n\r\n$ V_A \\plus{} V_B \\equal{} \\pi k^2 \\sqrt {k \\minus{} 1}$\r\n\r\nThus, when $ V_A \\equal{} V_B$ (which the problem asks for), $ V_B \\equal{} \\frac {\\pi}{2} k^2 \\sqrt {k \\minus{} 1}$.\r\n\r\n$ V_B$ can be found using the disk method:\r\n\r\n$ V_B \\equal{} \\pi \\int_0^{\\sqrt {k \\minus{} 1}} (x^2 \\plus{} 1)^2 \\ dx$\r\n\r\nThis can be evaluated to get\r\n\r\n$ V_B \\equal{} \\frac {\\pi}{15} \\sqrt {k \\minus{} 1} \\ (3k^2 \\plus{} 4k \\plus{} 8)$\r\n\r\nUsing the two expressions for $ V_B$, solve for $ k$.\r\n\r\n$ \\frac {\\pi}{15} \\sqrt {k \\minus{} 1} \\ (3k^2 \\plus{} 4k \\plus{} 8) \\equal{} \\frac {\\pi}{2} k^2 \\sqrt {k \\minus{} 1}$\r\n\r\nSolving this yields a few extraneous solutions, leaving a final answer of \r\n\r\n$ k \\equal{} \\frac {4}{9} \\plus{} \\frac {4 \\sqrt {10}}{9}$", "Solution_2": "Correct. :)" } { "Tag": [ "inequalities", "trigonometry", "geometry", "circumcircle", "inradius", "logarithms", "function" ], "Problem": "Let [tex]A,B,C[/tex] be the angles of a triangle, Prove that\r\n\r\n[tex]\\sin{A/2}\\sin{B/2}\\sin{C/2}\\leq1/8[/tex]\r\n\r\n(This is from p.6 of Kiran Kedlaya's MOP notes on inequalities)\r\n\r\nNote: This inequality implies that [tex]r\\leq{1/2R}[/tex], where [tex]R[/tex] is the circumradius of [tex]ABC[/tex] and [tex]r[/tex] is the inradius of [tex]ABC[/tex]. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=86151#86151]here[/url] for a proof of the identity [tex]\\sin{A/2}\\sin{B/2}\\sin{C/2}=r/4R[/tex]", "Solution_1": "[hide]Taking ln of both sides and using a property of logs we get [tex]\\ln\\sin A/2 + \\ln\\sin B/2 + \\ln\\sin C/2 \\leq 3(\\ln 1/2)[/tex]. Note that the function [tex]f(x) = \\ln\\sin(x) [/tex] has second derivative [tex] -1/\\sin(x)^2[/tex] which is always negative, so that the function is concave. Then we apply Jensen's Inequality to the function to get [tex] \\ln\\sin A/2 + \\ln\\sin B/2 + \\ln\\sin C/2 \\leq 3( \\ln(A+B+C)/6) = 3(\\ln 30) = 3(\\ln 1/2)[/tex], which is what we wanted to show.[/hide]", "Solution_2": "Nice proof. There is an elementary proof which doesn't require knowing the concavity of the function [tex]\\ln\\sin x[/tex] (which requires calculus to prove). Actually, [tex]1-8\\sin A/2\\sin B/2\\sin C/2[/tex] can be written as a sum of squares.", "Solution_3": "Dont post the second proof yet, Im going to try it. Does the concavity of ln(sin(x)) really require calc? Of course its very easy with calc. Is there another method?", "Solution_4": "[quote=\"tetrahedr0n\"]Does the concavity of ln(sin(x)) really require calc? Of course its very easy with calc. Is there another method?[/quote]\r\n\r\nI have no idea.", "Solution_5": "Concavity is essentially a property of the derivatives -- the only other way to define concavity that I know of is in terms of Jensen's inequality, which obviously isn't helpful in this case.", "Solution_6": "There are other ways to prove concavity. In fact you can prove concavity for things that aren't even differentiable (though concavity implies continuity).\r\n\r\nThe basic definition of concavity is a function for which t*f(a) + (1-t)*f(b) :le: f(t*a + (1-t)*b) for any a and b and any t betwen 0 and 1. [It turns out that this stronger statement is equivalent to just letting t be any fixed number between 0 and 1 (say 1/2 for instance)].\r\n\r\nFor most functions, proving that it is concave without using calculus is not easy. But it is possible.", "Solution_7": "Gauss202 -- Can you give an example of a function for which is it not sufficient to let t=1/2? (That is, can you come up with a function that would appear to be concave if you were only to check t=1/2 that it not actually concave?)", "Solution_8": "Yes. If f is not continuous then you can define it to be non-concave on the span of some basis element of R over Q, while keeping it concave on the span of the other bases. So I guess I should say - if f is continuous then you can take t to be some fixed number between 0 and 1 (such as 1/2). Good point!" } { "Tag": [ "algebra", "polynomial", "quadratics", "superior algebra", "superior algebra unsolved" ], "Problem": "Show that $ \\bar{\\mathbb{Q}}$ is not a finite extension of $ \\mathbb{Q}$.", "Solution_1": "Roots of $ x^n\\minus{}2$.", "Solution_2": "Should I prove that the roots are linearly independent over $ \\mathbb{Q}$?", "Solution_3": "They aren't; $ \\sqrt{2} \\plus{} (\\minus{}\\sqrt{2}) \\equal{} 0$.\r\n\r\nWhat you want to do is just add one of the roots and find the dimension.", "Solution_4": "I actually meant the positive roots, my mistake.I think those are independent.\r\n\r\nBut I see how it's easier just to adjoin a root.\r\n\r\nI have another question, not exactly on the same subject but close enough.\r\n\r\nSay you have $ F$ a field and $ E_1$, $ E_2$ finite extensions of that field. Will one extension be necessarily a subfield of the other or at least contain a subfield isomorphic to the other? I think the answer is yes but I cannot prove it.", "Solution_5": "$ F \\equal{} \\mathbb{Q}, E_1 \\equal{} \\mathbb{Q}[ \\sqrt {2} ], E_2 \\equal{} \\mathbb{Q}[ \\sqrt {3} ]$. Generally, field extensions form a [url=http://en.wikipedia.org/wiki/Lattice_(order)]lattice[/url], not a chain.", "Solution_6": "1.Would this mean that if extensions $ E_1$ and $ E_2$ would have different dimensions over $ F$ then one would be the subfield of the other?\r\n2. Actually this question arose from the following exercise: Argue that every finite field extension $ E$ of $ \\mathbb{R}$ is either $ \\mathbb{R}$ or is isomorphic to $ \\mathbb{C}$. \r\n\r\nI solved the problem using uniqueness of algebraic closure but I still can't answer the above question which in this case would be: If I assume that $ [E: \\mathbb{R}]\\equal{}3$ does this imply that $ E\\geq \\mathbb{C}$?", "Solution_7": "$ E_2 \\equal{} \\mathbb{Q}[x]/(x^3 \\minus{} x \\minus{} 1)$. Lattices don't behave like that, either. The standard example to keep in mind is the lattice of subsets of a given set: two sets can have different cardinalities without one being the subset of the other. If you're more number-theoretically inclined, think of the lattice of divisors of a given positive integer.\r\n\r\nThe given property of $ \\mathbb{R}$ is a consequence of the fundamental theorem of algebra. Every finite field extension of $ \\mathbb{R}$ is of the form $ \\mathbb{R}[x]/(P(x))$ for some polynomial $ P$ which is irreducible over $ \\mathbb{R}$ if and only if $ P$ is a quadratic with negative discriminant or linear; in other words, $ \\mathbb{R}$ can't have any finite field extensions of degree greater than $ 2$. Now show that the extensions of degree $ 2$ are all isomorphic to $ \\mathbb{C}$. \r\n\r\nIn any case, this is a very special property of $ \\mathbb{R}$: the lattice of finite field extensions of $ \\mathbb{Q}$ is very complicated indeed." } { "Tag": [ "function", "algebra", "domain", "induction", "limit", "algebra unsolved" ], "Problem": "f(f(f(x)))+f(f(x))+f(x)=3n f(x)=?", "Solution_1": "Who is n, and who is de domain and image of the function?", "Solution_2": "I think $f: N\\to N$. In this case it is very easy $f(n)=n$.\r\nConsider n=1 $f(f(f(n)))+f(f(n))+f(n)\\ge 3$ if is equality then f(1)=1. If f(2)=a>2, then f(a) or f(f(a))=1. It give $ffff(a)+fff(a)+ff(a)<3f(a)$ contradition. By induction $f(n)=n \\ \\forall n\\in N$.", "Solution_3": "I have prove that if the condition is $f: R^{+}\\to R^{+}$then the solution is also $f(x)=x$\r\nfirst,$f(x)<3x$\r\nthen $f(x)+ff(x)+fff(x)<(1+3+9)f(x)$then $f(x)>\\frac{1}{13}x$\r\n$\\dots \\dots \\dots$\r\nwe can find two sequence $a_{n},b_{n}$in this way such that $a_{n}x\\leq f(x)\\frac{1}{13}x$\n[/quote]\r\nI think it is not true.", "Solution_5": "Thank you for pointing out my mistake. :oops: \r\nBut I think the solution can be modified :\r\nfirst,$f(x)<3x$ $ff(x)<3x$\r\nthen $f(x)+ff(x)+fff(x)<(1+3+3)f(x)$then $f(x)>\\frac{1}{7}x$\r\nthen$3x=f(x)+ff(x)+fff(x)>\\frac17x+\\frac87ff(x)$ $ff(x)<\\frac52x$\r\n$\\dots \\dots \\dots$\r\nwe can find two sequence $a_{n},b_{n},c_{n}.d_{n}$in this way such that $a_{n}x\\leq f(x)\\frac{1}{7}x$\nthen$3x=f(x)+ff(x)+fff(x)>\\frac17x+\\frac87ff(x)$ $ff(x)<\\frac52x$\n$\\dots \\dots \\dots$\n[/quote]\r\nthen $3x=f(x)+ff(x)+fff(x)<7f(x)\\Longrightarrow f(x)>\\frac{3}{7}x$.\r\nLet $a_{1}=\\frac{3}{7},b_{1}=3$ It is easy to show if $a_{n}xa_{1}$ Because $a_{n}$ increase, $b_{n}$ decrease we have $\\lim_{n}a_{n}=a,\\lim_{n}b_{n}=b$. It give $b=\\frac{3}{1+a+a^{2}}, a=\\frac{3}{1+b+b^{2}}, a\\le 1\\le b$. It is easy to chek, that a=b=1.", "Solution_7": "To rust,thank you for the complete solution. :) \r\nI guess if the condition is $f: R\\to R$ and $f(x)$is continuous,there is also exactly one solution$f(x)=x$,(there is a another similar well-known problem if $f: R\\to R$,and f is continuous,and $ff(x)+f(x)=2x$then f(x)=x)\r\nBut I can't solve this supposition,Would you like to have a try?", "Solution_8": "[quote=\"Hawk Tiger\"] if $f: R\\to R$,and f is continuous,and $ff(x)+f(x)=2x$then f(x)=x)\n[/quote]\r\nIt is not true. For example f(x)=-2x is solution.", "Solution_9": "Oh,I am wrong again. :D ,it should be $ff(x)+x=2f(x)$ then $f(x)=x$\r\nSorry." } { "Tag": [ "trigonometry", "geometry", "perimeter", "trig identities", "Law of Sines" ], "Problem": "Consider triangle $ ABC$. If $ AB\\equal{}5$ and $ AC\\equal{}8$ and $ sin C\\equal{}0.5$, then what is the value of $ cos B$, given that triangle $ ABC$ is obtuse?\r\n$ A. \\minus{}\\frac{3}{5}$\r\n$ B. \\minus{}\\frac{4}{5}$\r\n$ C. \\frac{4}{5}$\r\n$ D. \\frac{3}{5}$\r\n[hide=\"Why doesnt this work?\"]\nUsing law of sines, we have \n$ \\frac{sin B}{8}\\equal{}\\frac{sin C}{5}$.\n$ 5sinB\\equal{}4$ ==> $ sin B\\equal{}\\frac{4}{5}$, so $ cos B\\equal{}\\frac{3}{5}$. Now because the problem gave that the triangle is obtuse, one would think to make this answer negative, since cos in the 2nd quadrant is negative. However we do not know the length of the third side, and since there can be only one obtuse angle in a triangle, we don't know if that's longer then 8. So using law of cosines...\n$ 25\\equal{}64\\plus{}BC^{2}\\minus{}2(8)(BC)(cos30)$\n$ 25\\equal{}64\\plus{}BC^{2}\\minus{}8\\sqrt{3}BC$\n$ BC^{2}\\minus{}8\\sqrt{3}BC\\plus{}39\\equal{}0$\n$ \\frac{8\\sqrt{3}\\pm\\sqrt{192\\minus{}4(1)(39)}}{2}$\n$ \\frac{8\\sqrt{3}\\pm\\sqrt{36}}{2}$\n$ 4\\sqrt{3}\\plus{}3$, which is indeed greater than 8 so the obtuse angle doesn't belong to 8, meaning the angle B is acute, so cosine would be positive.[/hide]", "Solution_1": "whats wrong with ur problem? it just shows that angle B is acute=36.87, so leaving angle A the obtuse angle", "Solution_2": "You don't need to use law of cosines.\r\nSimply, if $ \\sin B \\equal{} \\frac {4}{5}$, just use the famous identity $ \\sin^2 B \\plus{} \\cos^2 B \\equal{} 1$.\r\n\r\n$ \\sin^2 B \\plus{} \\cos^2 B \\equal{} 1$\r\n$ \\Leftrightarrow \\left( \\frac {4}{5}\\right)^2 \\plus{} \\cos^2 B \\equal{} 1$\r\n$ \\Leftrightarrow \\cos^2 B \\plus{} \\frac {16}{25} \\minus{} 1 \\equal{} 0$\r\n$ \\Leftrightarrow \\cos^2 B \\minus{} \\frac {9}{25} \\equal{} 0$\r\nBy factoring,\r\n$ \\Leftrightarrow \\left( \\cos B \\minus{} \\frac {3}{5}\\right) \\left(\\cos B \\plus{} \\frac {3}{5} \\right) \\equal{} 0$\r\nSo, the value of $ \\cos B$ is $ \\frac {3}{5}$ or $ \\minus{} \\frac {3}{5}$. But since $ \\angle{B}$ is obtuse, it can be concluded that $ \\cos B \\equal{} \\minus{} \\frac {3}{5}$. Finally, $ A$ is the answer.", "Solution_3": "I have another problem about triangle.\r\n\r\nThe perimeter of triangle $ PQR$ is $ 15$ centimeters. $ PQ$ is $ 2$ centimeters longer than $ QR$. If $ \\angle{Q}\\equal{}\\frac{2\\pi}{3}$, find the length of each side of this triangle.", "Solution_4": "You know , you could just have drawn the triangle and all of it would have been clear. The length of the 3rd side is \r\n3+ 4*3^0.5.\r\nAnd so because B=37degrees. cos37 is 3/5." } { "Tag": [ "algorithm" ], "Problem": "I do not know what they are called but something like this:\r\n\r\nFind integers $m$ and $n$ such that $\\frac{1}{m}+\\frac{1}{n} = \\frac{1}{7}$\r\n\r\n\r\nMy answer is that m and n both equal 14. That seems to easy and I don't know if they have to be separate integers. \r\nI recall reading that, for similar problems, there is a way to solve these by starting with $\\frac{1}{2}$ or something like that.\r\n\r\nIf anyone knows what I'm talking about (the name or the method of solving), I'd appreciate a little help\r\n\r\nThanks", "Solution_1": "The greedy algorithim can be helpful. I don't remember it off the top of my head though.", "Solution_2": "I researched the Greedy Algorithm and that led me to Egyptian fractions, which is what I think we are dealing with. But I am still not understanding the steps to how to solve these equations. \r\n\r\nThanks for the lead, though.", "Solution_3": "If the numbers don't have to be distinct than this is trivial since\r\n$\\frac{1}{x} + \\frac{1}{y} = \\frac{1}{N}$\r\nhas a solution where $x = y = 2N$ since we get\r\n$\\frac{1}{x} + \\frac{1}{x} = \\frac{2}{x} = \\frac{1}{N}$\r\n\r\nDid you want the solution for any number of variables instead??", "Solution_4": "[quote=\"vinny919\"]Did you want the solution for any number of variables instead??[/quote]\r\n\r\nYes.", "Solution_5": "Assuming that each number has to be distinct, you can use the greedy algorithm. While the algorithm might not give you an unique solution, it should give you one. Also, you cannot really control how many fractions are in the end result using this algorithm (as far as I know), although you might be able to split another fraction up a bit more.\r\n\r\nIn your example:\r\n\r\n$\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{7}$\r\n\r\nThe greatest fractions smaller than $\\frac{1}{7}$ that is of the form $\\frac{1}{n}$ where $n\\geq1$ is $\\frac{1}{8}$. Subtracting $\\frac{1}{8}$ from $\\frac{1}{7}$ gives us the other fraction as $\\frac{1}{56}$.", "Solution_6": "$\\frac {1}{x} + \\frac{1}{y} = \\frac 17$\r\n\r\n$\\frac{x+y}{xy} = \\frac 17$\r\n\r\n$7x + 7y = xy$\r\n\r\ncan't you do that?", "Solution_7": "A [hide=\"General Solution\"]$\\frac 1x+\\frac1y=\\frac1N$\n\n$\\frac yy \\cdot \\frac 1x+\\frac xx \\cdot \\frac1y=\\frac 1N$\n\n$\\frac{y}{xy}+\\frac{x}{xy}=\\frac 1N$\n\n$\\frac{y+x}{xy}=\\frac 1N$\n\n$N(x+y)=xy$\n\n$Nx+Ny=xy$\n\n$Nx=xy-Ny$\n\n$Nx=(x-N)y$\n\n$y=\\frac{Nx}{N-x}$[/hide]", "Solution_8": "We can rewrite $\\frac{7}{m}+\\frac{7}{n}=1$, we are to imagine the form of intercept to equation of the line.", "Solution_9": "[quote=\"ankur87\"]I do not know what they are called but something like this:\n\nFind integers $m$ and $n$ such that $\\frac{1}{m}+\\frac{1}{n} = \\frac{1}{7}$\n\n\nMy answer is that m and n both equal 14. That seems to easy and I don't know if they have to be separate integers. \nI recall reading that, for similar problems, there is a way to solve these by starting with $\\frac{1}{2}$ or something like that.\n\nIf anyone knows what I'm talking about (the name or the method of solving), I'd appreciate a little help\n\nThanks[/quote]\r\n\r\n[/hide=Solution]\r\n1/m + 1/n = 1/7\r\n\r\n7m + 7n = mn\r\nmn - 7m - 7n = 0\r\nmn - 7m - 7n + 49 = 49\r\n(m - 7)(n - 7) = 49\r\n\r\nYeah.\r\n[/hide]\r\n[/hide]", "Solution_10": "[quote=\"Treething\"][quote=\"ankur87\"][hide]I do not know what they are called but something like this:\n\nFind integers $m$ and $n$ such that $\\frac{1}{m}+\\frac{1}{n} = \\frac{1}{7}$\n\n\nMy answer is that m and n both equal 14. That seems to easy and I don't know if they have to be separate integers. \nI recall reading that, for similar problems, there is a way to solve these by starting with $\\frac{1}{2}$ or something like that.\n\nIf anyone knows what I'm talking about (the name or the method of solving), I'd appreciate a little help\n\nThanks[/hide][/quote]\n\n[hide=\"Solution\"]\n1/m + 1/n = 1/7\n\n7m + 7n = mn\nmn - 7m - 7n = 0\nmn - 7m - 7n + 49 = 49\n(m - 7)(n - 7) = 49\n\nYeah.\n[/hide]\n[/quote]In general the equation $\\frac1x+\\frac1y=\\frac1n$ for a given $n$ and positive integer unknowns $x,y$ has as many different solutions as factors has $n$ (up to permutation of the terms). The solution given by Treething is easely generalizable $\\frac1x+\\frac1y=\\frac1n\\Rightarrow n(x+y)=xy \\Rightarrow xy-(x+y)+n^2=(x-n)(y-n)=n^2$. If we assume that $x\\leq y$, then $(x-n)$ is a factor of $n$.", "Solution_11": "[quote=\"ankur87\"]I do not know what they are called but something like this:\n\nFind integers $m$ and $n$ such that $\\frac{1}{m}+\\frac{1}{n} = \\frac{1}{7}$\n[/quote]\r\n\r\nI think we call such a fraction as Egyptian Fraction.\r\n\r\nBy the procedure explained by some guys have explained, the answer is $(m,n)=(8,56),(14,14),(56,8),(6,-42),(-42,6)$" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $x_i>0$ and $x_iy_i-z_i^2>0$ for $i<0$ then \r\n$\\frac{n^3}{(\\sum_{i=1}^nx_i)(\\sum_{i=1}^ny_i)-(\\sum_{i=1}^nz_i)^2}\\leq\\sum{\\frac{1}{x_iy_i-z_i^2}}$", "Solution_1": "[quote=\"silouan\"]If $x_i>0$ and $x_iy_i-z_i^2>0$ for $i<0$ then \n$\\frac{n^3}{(\\sum_{i=1}^nx_i)(\\sum_{i=1}^ny_i)-(\\sum_{i=1}^nz_i)^2}\\leq\\sum{\\frac{1}{x_iy_i-z_i^2}}$[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=59017" } { "Tag": [], "Problem": "So I was browsing through a Bleach fansite forum one day and stumbled upon this interesting exchange: (May contain spoilers regarding the actual manga)\r\n\r\nhttp://forums.bleachportal.net/showpost.php?p=3618592&postcount=278\r\nhttp://forums.bleachportal.net/showpost.php?p=3618628&postcount=279\r\nhttp://forums.bleachportal.net/showpost.php?p=3620430&postcount=357\r\nhttp://forums.bleachportal.net/showpost.php?p=3620642&postcount=360\r\nhttp://forums.bleachportal.net/showpost.php?p=3620880&postcount=369 <--- Interesting tidbit\r\nhttp://forums.bleachportal.net/showpost.php?p=3620899&postcount=371\r\nhttp://forums.bleachportal.net/showpost.php?p=3620914&postcount=372 ...and the rest of them shuts up after this.\r\n\r\nPost other stupidity you found on the internet here! No Bash/QDB/FML or similar unless it's work-safe.", "Solution_1": "Thank you for linking us to this very interesting discussion on your favorite anime message board. I, for one, had no idea that the internet was full of such ill-informed persons!", "Solution_2": "You'd be surprised.", "Solution_3": "You should see the flame wars that erupt about whether 0.9999999999...=1", "Solution_4": "http://www.casiocalc.org/?showtopic=1757&st=0\r\n\r\nI randomly googled this and got the above...\r\n\r\nFeeling an urge to split my hair as going through their discussion...", "Solution_5": "Why bother? Anybody who isn't curious enough to look it up themselves obviously isn't going to be doing anything in life that requires knowing that 0.9999... = 1. So if your goal is to improve people's general understanding of mathematics, then trying to explain it to these folks isn't a good use of your time.\r\n\r\nIf that's not your goal, then stop bashing your head over it -- as a matter of fact, stop thinking about it. Go do something more useful (or at least something fun)." } { "Tag": [], "Problem": "If the lengths of a kite's diagonals are 6 cm and 12 cm, can one of the kite's sides be twice as long as another? Prove your answer.", "Solution_1": "If the sides are $ a$ and $ 2a$, and if, for example, the diagonals bisect each other at right angles into two equal line segments of length $ 3$ and two line segments of lengths $ b$ and $ 12 \\minus{} b$, then by Pythagoras,\r\n\r\n$ a^{2} \\equal{} b^{2} \\plus{} 9$ and $ 4a^{2} \\equal{} (12 \\minus{} b)^2 \\plus{} 9$.\r\n\r\nMultiplying both sides of the former equation by $ 4$ and substituting into the latter, we get (I think)\r\n\r\n$ b^{2} \\plus{} 8b \\minus{} 39 \\equal{} 0$ which has a real positive solution.\r\n\r\nOn the other hand, if the diagonal are divided into segments of lengths $ 6 \\plus{} 6$ and $ b \\plus{} (6\\minus{}b)$, then we get the equation $ b^{2} \\plus{} 4b \\plus{}24 \\equal{}0$ (I think) which does not have real solutions, so there is only one possible kite under the given conditions.", "Solution_2": "Nicely done, thanks." } { "Tag": [ "modular arithmetic" ], "Problem": "Prove that $ 2222^{5555}\\plus{}5555^{2222}$ is divisible by 7.", "Solution_1": "[quote=\"fibointepi\"]Prove that $ 2222^{5555} \\plus{} 5555^{2222}$ is divisible by 7.[/quote]\r\n\r\n\r\n$ 2222^{5555}\\equiv 3^{5555}\\equiv 3^{2}\\cdot(3^{3})^{1851}\\equiv9\\cdot(\\minus{}1)^{1851}\\equiv\\minus{}9\\equiv\\minus{}2\\equiv5(7)$\r\n\r\n$ 5555^{2222}\\equiv 4^{2222}\\equiv 4^{2}\\cdot(4^{3})^{740}\\equiv16\\cdot1^{740}\\equiv16\\equiv2(7)$\r\n\r\n$ 2222^{5555} \\plus{} 5555^{2222}\\equiv2\\plus{}5\\equiv0(7)$", "Solution_2": "[hide]First, $ 2222\\equiv3\\pmod{7}$, and $ 5555\\equiv4\\pmod{7}$. So it becomes $ 3^{5555}+4^{2222}$ in modulo 7. From Fermat's, we have $ 3^6$ and $ 5^6$ are both 1 (mod 7), so it reduces to $ 3^5+4^2\\pmod{7}\\rightarrow{259\\pmod{7}\\equiv0\\pmod{7}}$.[/hide]", "Solution_3": "This problem has already been posted. See\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=180357 .\r\n :)" } { "Tag": [ "geometry proposed", "geometry" ], "Problem": "Let ABC be a triangle such each interior angle is $ \\leq \\frac {\\pi}{2} $.\r\n\r\nThe points of the triangle sides are colored with $ red $ and $ blue $ .\r\n\r\nProve that there are 3 points M,N,P colored with the same color, such that the triangle MNP is right .\r\n\r\n :cool: Do you know the source of this problem ?! :cool:", "Solution_1": "Let A'B'C' the inscribed triangle in ABC s.t. A'B' is ortogonal to BC, B'C' is ortogonal to AC and A'C' is ortogonal to AB. (*)\r\n\r\nTwo of tre points in {A',B',C'} must have the same colour X. Let's suppose those are A' and B'. If B or C are of the same colour of A' we fisish the proof. If B and C have different colour wrt those of B' and A' we considere the ortogonal projection of C' on BC and call it C\". If C' has colour Y=/=X then if C\" is coloured X then B'A'C\" is the triangle MNP; if C\" is Y then C'C\"B is MNP.\r\n\r\nIf instead C' has colour X too and A has colour Y (the case A Y is obvious) consider the projection A\" of A on BC and analise the cases.\r\n\r\nPS\r\nSorry for my pour english. I hope at least the idea is clear.\r\n\r\n(*) could be a simple but nice exercise to construct such a triangle." } { "Tag": [ "counting", "derangement", "probability" ], "Problem": "This is some FIITJEE test sum, I think his answer is wrong and as usual, I can't understand head or tail of the solution.\r\n\r\nFind the number of tuples (x1,x2,x3,x4,x5) such that $ x_1x_2x_3x_4x_5 \\equal{} 2310$\r\n\r\n.\r\n\r\n[hide]I think the answer given was $ 16.5^5$[/hide]\r\n\r\nP.S: Check this out. Another FIITJEE question. http://www.mathlinks.ro/viewtopic.php?p=1095007#1095007", "Solution_1": "Its correct only. What didnt you understand? :maybe:", "Solution_2": "[hide]Think negative[/hide]", "Solution_3": "If it helps, i did that fiitjee derangements sum by classical fraud :D\r\nObviously, the total number of cases (which is the denominator in the probability) is $ \\frac {11!}{2!2!2!}$ \r\nIn the choices given, all the denominators were $ 11!$ only. So, obviously, the numerator should be a multiple of 8, which was true only in one case. \r\n\r\nQ.E.D. :D" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Prove that $\\mathop{\\lim }_{n \\to+\\infty }\\left|{y_{n}}\\right| = \\left| L \\right|$ when you know that $\\mathop{\\lim }_{n \\to+\\infty }y_{n}= L$. Why the inverse sentence is not a theorem, for $L \\ne 0$?", "Solution_1": "Take a look at $y_{n}=(-1)^{n}$.", "Solution_2": "I believe the forward statement just follows from the fact that $| |L|-|y_{n}| | \\le |L-y_{n}|$ by the triangle inequality." } { "Tag": [ "linear algebra", "matrix", "vector" ], "Problem": "Let $ A$ be a $ n\\times n$ matrix of numbers and $ \\mathbf x$ a $ n\\times1$ matrix of vectors. ($ \\mathbf x\\equal{}[\\mathbf v_1\\,\\cdots\\,\\mathbf v_n]^T$)\r\nDoes $ A\\mathbf x\\equal{}\\mathbf x$ imply $ A\\equal{}I$?", "Solution_1": "No. All $ A\\mathbf{x}\\equal{}\\mathbf{x}$ for a single nonzero vector $ \\mathbf{x}$ implies is that $ 1$ is an eigenvalue of $ A.$\r\n \r\nSimple example: $ A\\equal{}\\begin{bmatrix}1&\\minus{}4\\\\0&7\\end{bmatrix},\\quad\r\n\\mathbf{x}\\equal{}\\begin{bmatrix}1\\\\0\\end{bmatrix}.$", "Solution_2": "Here I mean $ \\mathbf x\\equal{}\\begin{bmatrix}\r\n\\mathbf v_1 \\\\\r\n\\vdots\\\\\r\n\\mathbf v_n\\\\\r\n\\end{bmatrix},$ where $ \\mathbf v_1, \\cdots, \\mathbf v_n$ are nonzero vectors, not numbers.", "Solution_3": "All right, $ A\\equal{}\\begin{bmatrix}1&\\minus{}4\\\\0&7\\end{bmatrix},\\quad \\mathbf{x}\\equal{}\\begin{bmatrix}1&1\\\\0&0\\end{bmatrix}.$\r\n\r\nIf $ X$ is invertible, then of course $ AX\\equal{}X$ would imply $ A\\equal{}I.$ If $ X$ is not invertible, then no.", "Solution_4": "[quote=\"ifai\"]Here I mean $ \\mathbf x \\equal{} \\begin{bmatrix} \\mathbf v_1 \\\\\n\\vdots \\\\\n\\mathbf v_n \\\\\n\\end{bmatrix},$ where $ \\mathbf v_1, \\cdots, \\mathbf v_n$ are nonzero vectors, not numbers.[/quote]\r\nOK. And how about if $ \\mathbf v_1, \\cdots, \\mathbf v_n$ are vectors in general vector space, or they are column vectors?", "Solution_5": "Don't go there- you're going to have trouble even defining the product. As written, it only makes sense if the $ v_i$ are row vectors.\r\n\r\nThe key condition is linear dependence/independence. If the $ v_i$ are linearly independent, $ A\\equal{}I$ is the only solution. If the $ v_i$ are dependent, there are other solutions." } { "Tag": [ "geometry", "algorithm", "abstract algebra", "number theory", "superior algebra", "superior algebra theorems" ], "Problem": "Through studying Topics in Algebra by Herstein, some online sources, and occasionally this forum, I've figured out most of the basics in abstract algebra, and I like the subject. I was wondering if anyone could recommend a book/textbook with which I could continue learning about it and preferably doesn't waste too much time on the very basics. Some specific topics I have interest in learning are algebraic geometry, category theory, and algebraic number theory.\r\n\r\nThanks.", "Solution_1": "A great introduction to some of the ideas in algebraic geometry is [url=http://www.amazon.com/Ideals-Varieties-Algorithms-Computational-Undergraduate/dp/0387946802]Cox, Little, and O'Shea[/url]. The book is written very accessibly and puts a great deal of emphasis on motivation and avoiding \"dryness\" but still covers its subject thoroughly. (It doesn't even assume knowledge of abstract algebra - nevertheless, there is a great deal to learn from it.)", "Solution_2": "I would recommend [i]Introduction to Commutative Algebra[/i] by Atiyah and Macdonald. That book has great problems, just like Herstein, and it continues right where Herstein left off. The material is commutative ring theory, so it is very applicable to algebraic number theory and algebraic geometry. I also like the homological algebra in the modules section, which is very useful for later math. (Another benefit is that, only from what I've heard, if you do all of the problems in the book, you'll have an understanding of schemes, which are an important but complicated idea in algebraic geometry.)\r\n\r\nUnfortunately, the book doesn't have tons of geometric intuition, so if that's what you're looking for, you might want to try [i]Commutative Algebra[/i] by Miles Reid." } { "Tag": [ "function", "integration", "limit", "calculus", "real analysis", "advanced fields", "advanced fields unsolved" ], "Problem": "I want to prove the following statement:\r\n\r\nLet $ u$ be harmonic in $ \\mathbb{D}$ and suppose that there exists a $ C > 0$ such that\r\n$ \\int_0^{2\\pi} |u(re^{it})| \\frac {dt}{2\\pi} \\leq C$\r\nfor all $ 0 < r < 1$. Then there exists a function of bounded variation $ \\mu$ defined on $ [0,2\\pi]$ whose total variation is less than $ C$ such that\r\n$ u(z) \\equal{} \\int_0^{2\\pi} \\frac {1 \\minus{} |z|^2}{|e^{it} \\minus{} z|^2} \\, d\\mu$\r\n\r\nThere is a hint to use Helly's selection theorem, but I don't get it. I am unfamiliar with the theorem, and I don't see how to apply it.\r\n\r\nI am supposing the proof should run like this (please note I am guessing wildly on things, so please don't take my steps for true):\r\nIf for $ 0 < r < 1$ we let $ u_r(z) \\equal{} u(rz)$ then $ u_r$ are harmonic in $ |z| < \\frac {1}{r}$. So we know that for $ z \\in \\mathbb{D}$\r\n$ u_r(z) \\equal{} \\frac {1}{2\\pi}\\int_0^{2\\pi} \\frac {1 \\minus{} |z|^2}{|e^{it} \\minus{} z|^2}u_r(e^{it}) \\, dt$\r\nand for fixed $ z$, $ \\lim_{r\\to 1} u_r(z) \\equal{} u(z)$\r\n\r\n\r\nIf we let $ \\mu_r(t) \\equal{} \\frac {1}{2\\pi}\\int_0^{t} u(re^{is}) \\, ds$ then they satisfy the criterions for Helly's theorem, so there is a $ \\mu$ of bounded variation and a sequence $ (r_n) \\in (0,1)$ such that $ r_n \\to 1$ and for continuous functionals $ T \\in \\mathbb{C}^*$ we have $ T(\\mu_{r_n}(t)) \\to T(\\mu(t))$ as $ n \\to \\infty$ \r\n\r\n\r\nBased on a proof of a similar statement for $ p > 1$ where we can apply reflexivity, I am guessing that this should imply that\r\n$ \\lim_{n\\to \\infty} \\int_0^{2\\pi} v \\, d\\mu_{r_n} \\equal{} \\int_0^{2\\pi} v \\, d\\mu$ (*)\r\nAt least for $ v$ the Poisson kernel. I don't seem to quite get this, since there is a $ t$ involved in the weak convergence statement above, while for a continuous $ v$, I would like to look at the functional $ T(\\mu) \\equal{} \\int_0^{2\\pi} v \\, d\\mu$.\r\n\r\n\r\nSince my $ \\mu_r$ are actually absolutely continuous, we have\r\n$ \\frac {1}{2\\pi}\\int_0^{2\\pi} v \\, u_r(e^{it}) \\, dt \\equal{} \\int_0^{2\\pi} v \\, d\\mu_{r}$\r\nand this would complete the proof, modulo that I can't show (*)\r\n\r\n\r\nSorry for the mess, but when I started this post I didn't know what I was doing at all. I think I am close now. Please someone help me fix this. Meanwhile I will try to understand it better, and probably edit my post a million more times.", "Solution_1": "You are on the right track. All you need to do is to integrate by parts and express $ \\int P(z,t)\\,d\\mu_r(t)$ as $ \\int Q(z,t)\\mu_r(t)\\,dt\\plus{}c\\mu_r(2\\pi)$, thus showing that pointwise convergence of $ \\mu_r$ to $ \\mu$ implies the convergence of the integrals. Then you can integrate by parts back to return to the integral with respect to $ d\\mu$. :)\r\n\r\nAs a side remark, I find the language using integrals with respect to $ d\\mu$ where $ \\mu$ is a function of bounded variation rather archaic. The language of signed measures, weak convergence, Riesz theorem and weak compactness of the unit ball in the dual space seems to be much more convenient. Of course, this remark applies only if you are familiar with Lebesgue integration.", "Solution_2": "Thanks a lot. I was a little suspicious of the integration by parts first, but combining some theorems on Lebesgue-Stieltjes integrals and some on Riemann-Stieltjes integrals I have meticulously checked that yes, it is ok, although in one step you have a Lebesgue integral.\r\n\r\nHow come integration by parts is such a powerful tool also in \"abstract\" mathematics (as opposed to what americans refer to as calculus)? It's often used to get integrands with better convergence properties, although I am not fully used to such techniques, and am often mesmerized by their application. One example that comes to mind is when you have a periodic function $ f$ and look at integrals of $ f(nx)$ - I don't remember the exact formulation. I had such a question on a take-home exam and I provided a whopping four pages of approximation by simple functions and use of some strong convergence theorems, and then I was shown a three-line solution with integration by parts.\r\n\r\nOne more question: In the statement of Helly's theorem, there is a weak convergence statement, that $ T(f_k(t)) \\to T(f(t))$ for continuous functionals. In this example, the relevant property was pointwise convergence; but in fact, isn't pointwise convergence [b]equivalent[/b] to the weak convergence? So what is the point of weak convergence? :maybe:", "Solution_3": "[quote=\"Kalle\"]In this example, the relevant property was pointwise convergence; but in fact, isn't pointwise convergence [b]equivalent[/b] to the weak convergence? So what is the point of weak convergence? :maybe: \n[/quote]\r\nTry to do everything in $ \\mathbb R^3$ instead of $ \\mathbb R^2$ where the integral is taken over the 2-dimensional sphere. Where will be the Stiltjes integral, integration by parts, and Helly's theorem now? The result is exactly the same nevertheless ;)." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "$ x_1^2\\plus{}x_2^2\\plus{}x_3^2\\plus{}x_4^2\\equal{}\\frac{1}{2}$\r\nFind the min of $ (x_1\\minus{}2x_2\\plus{}x_3)^2\\plus{}(x_2\\minus{}2x_3\\plus{}x_4)^2\\plus{}(x_2\\minus{}2x_1)^2\\plus{}(x_3\\minus{}2x_4)^2$", "Solution_1": "please help us everyone!!!!!! we must have the solution before Monday (19/10) :wallbash: My teacher will check our exercise on Monday!" } { "Tag": [], "Problem": "The Randomest Question ever. What type of gum do you like the most? YAY", "Solution_1": "trident splash!!", "Solution_2": "Wrigley's Doublemint", "Solution_3": "I don't normally chew gum, but I'd have to say that the one my parents seem to prefer is Excel.\r\n\r\nPersonally, I like Trident Splash.", "Solution_4": "Gum makes my tongue super-sensitive.\r\n\r\nLemons FTW.\r\n\r\nI hate gum.", "Solution_5": "darn, you must be pretty beast if you chew lemons\r\nanyways, 5 cobalt if you don't count price, trident otherwise", "Solution_6": "I don't really chew gum that much (braces), but I like Orbit Bubblemint. I also like Wrigley's Fruit chewed at the same time with those little Wonka Nerds", "Solution_7": "trident tropical twist!!! ;]]", "Solution_8": "five and orbit.", "Solution_9": "five and orbit.", "Solution_10": "[quote=\"shelly32494\"]trident tropical twist!!! ;]][/quote] me too!!! :D :D :D", "Solution_11": "five solstice.", "Solution_12": "[quote=\"jonathanchou711\"]five solstice.[/quote]\r\n\r\nmakes me feel sick... but five itself isnt bad", "Solution_13": "WRIGLEY'S SPEARMINT! YEAH!", "Solution_14": "Everyone's choices- taste or long-lasting?", "Solution_15": "I would prefer healthy, then long lasting, then taste, in that order.", "Solution_16": "me2/.... :lol:", "Solution_17": "er. bothhhh.", "Solution_18": "You know that kind of new 5 gum zing?\r\n\r\nIt tastes weird.. It's kind of like those doctor's gum where they give you gum that tastes minty and after a while it becomes all crumbly and you can swallow it! That's the same with zing. It becomes swallowable and dry.", "Solution_19": "that's what she said", "Solution_20": "[quote=\"ra5249\"]that's what she said[/quote]\r\n\r\n-___-;;;", "Solution_21": "@ OP's title: gum in Spanish is chicle\r\nchiste = joke\r\n\r\nI assume that's what you were going for, since I searched for and couldn't find any brand of gum called \"Chiste\"", "Solution_22": ":o and :rotfl: to ra's comment on \"swallowable and dry\"\r\n\r\nand yes that's espanol for gum, i taKE espanol. and i was working on the hw. But not now cuz i have SUMMER! and no, im not hispanic........." } { "Tag": [ "function", "puzzles" ], "Problem": "Find first \r\n(1) decimal digit of 2^987654321 \r\n(2) two decimal digits of 2^987654321 \r\n(3) three decimal digits of 2^987654321 \r\n(4) n decimal digits of 2^98765321 (represent as function of n)", "Solution_1": "Do not post the same post in different forums. This is already in the intermediate forum. Mods, please delete this.", "Solution_2": "Yes, he is right. There is no need to double post. If you want to participate on the discussion of this thread, go to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36661]the thread on intermediate forum[/url]. This is locked... and to be deleted." } { "Tag": [ "Divisibility", "modular arithmetic", "number theory", "IMO Shortlist" ], "Problem": "Let $b,n > 1$ be integers. Suppose that for each $k > 1$ there exists an integer $a_k$ such that $b - a^n_k$ is divisible by $k$. Prove that $b = A^n$ for some integer $A$.\n\n[i]Author: Dan Brown, Canada[/i]", "Solution_1": "let $ k\\equal{}b^2$:\r\n$ b^2|b\\minus{}a^n_k \\Rightarrow a^n_k\\equal{}b(bx\\plus{}1)$ but $ gcd(b , bx\\plus{}1)\\equal{}1$ therefore $ b\\equal{}A^n$ for some integer $ A$. :D", "Solution_2": "[quote=\"orl\"]Let $ b,n > 1$ be integers. Suppose that for each $ k > 1$ there exists an integer $ a_k$ such that $ b \\minus{} a^n_k$ is divisible by $ k.$ Prove that $ b \\equal{} A^n$ for some integer $ A.$\n\n[i]Author: unknown author, Canada[/i][/quote]\r\n[hide=\"Solution\"]\nAssume that $ b$ has a prime factor, $ p$, so that $ p^{xn \\plus{} r}\\parallel{}b$ with $ 0 < r < n$. Then, we can let $ k \\equal{} p^{xn \\plus{} n}$. It follows that $ b\\equiv a_k^n\\bmod p^{xn \\plus{} n}$. Since $ p^{xn \\plus{} r}\\parallel{}b$, we see that $ p^{xn \\plus{} r}\\parallel{}a_k^n$. Then, $ p^{x \\plus{} \\frac {r}{n}}\\parallel{}a_k^n$, which is a contradiction since $ \\frac {r}{n}$ is not an integer. Thus, we have a contradiction, so $ r \\equal{} 0$, which means that only $ n$th powers of primes fully divide $ b$, so $ b$ is an $ n$th power. [/hide]", "Solution_3": "It may be interesting to know that if $ 8 \\nmid n$, then it suffices to consider primes $ k$ only.\r\nMore generally (more or less the Grunwald-Wang-theorem):\r\n\r\nLet $ n,b$ be a integers, $ n > 0$. If for all primes $ p$ there is an $ a_p$ such that $ p|b \\minus{} a_p^n$, then the following holds:\r\na) If $ 8 \\nmid n$, then $ b \\equal{} a^n$ for some integer $ a$.\r\nb) If $ 8|n$ then $ b \\equal{} a^{\\frac n2}$ for some integer $ a$.\r\n\r\nThis was posted and partially solved before.\r\n\r\nFor those knowing algebraic number theory, this can be generalised even more (those don't knowing algebraic number theory, please stop here :D ):\r\n\r\nLet $ K$ be a number field, $ b \\in K$, $ n \\equal{} m \\cdot 2^s > 0$ an integer ($ m$ odd).\r\nIf $ b$ is a $ n$-th power in all but finitely many completions $ K_v$ (where $ v$ runs through the primes of $ K$), then:\r\na) If $ K[\\zeta_{2^s}]|K$ is cyclic, then $ b \\equal{} a^n$ for some $ a \\in K$.\r\nb) Otherwise, $ b \\equal{} a^{\\frac n2}$ for some $ a \\in K$.\r\n\r\nProving this is some work in (pre)global class field theory (the main step being to show that if $ L|K$ has all but finitely many primes totally split, then $ K \\equal{} L$).", "Solution_4": "Note that a) indeed requires $ 8\\nmid n$:\r\nLet $ b \\equal{} 16$, $ n \\equal{} 8$. We have $ x^8 \\minus{} 16 \\equal{} (x^2 \\minus{} 2)(x^2 \\plus{} 2)(x^2 \\minus{} 2x \\plus{} 2)(x^2 \\plus{} 2x \\plus{} 2)$. Of course one of the numbers $ \\minus{} 1, \\minus{} 2,2$ is a quadratic residue $ \\mod p$. It means that for every prime $ p$ one of the equations $ (x \\plus{} 1)^2 \\plus{} 1 \\equal{} 0$, $ x^2 \\plus{} 2 \\equal{} 0$, $ x^2 \\minus{} 2 \\equal{} 0$ has a solution $ \\mod p$ and so does $ x^8 \\minus{} 16 \\equal{} 0$. And $ 16\\neq A^8$.\r\n\r\n[Moderator edit: See also http://www.mathlinks.ro/viewtopic.php?t=4874 for this counterexample.]", "Solution_5": "I also didn't see it at first, but $ p$ is not necessarily prime. Using this, it gets really easy (using valuations).\r\nBut you will have a lot of fun to show that if we require that $ 8 \\nmid n$, then we need only to look at primes $ p$.", "Solution_6": "[b]Solution[/b]\n\nAssume that there is no $A$ such that $b=A^n$. Then there must exist a prime number $p$ such that, if $p^a \\| b$, then $n \\not | a$. Assume now that $mn < a < (m+1)n$ for some $m \\in \\mathbb{N}$. Taking $k=p^{(m+1)n}$ yields that\n\n\\[b \\equiv a_{k}^n \\pmod{p^{(m+1)n}}\\]\nWhich implies that $p^{a} |a_{k}^n$ and, since $a_{k}^n$ is a perfect $n$th power, that $p^{(m+1)n} | a_{k}^n$. Hence $b \\equiv 0 \\pmod{p^{(m+1)n}}$ and $n|a$ which is a contradiction. $\\blacksquare$", "Solution_7": "Does anyone have a solition with looking at a prime divisor of $n$?", "Solution_8": "[quote=ali666]let $ k\\equal{}b^2$:\n$ b^2|b\\minus{}a^n_k \\Rightarrow a^n_k\\equal{}b(bx\\plus{}1)$ but $ gcd(b , bx\\plus{}1)\\equal{}1$ therefore $ b\\equal{}A^n$ for some integer $ A$. :D[/quote]\n\nCan you explain more detailed?", "Solution_9": "A different [u]Solution[/u]\nwrite $b$ as $b=B^n \\times j$ where $gcd(B,j)=1$ now if $j=1$ then we are done.\nSuppose $j$ is not equal to $1$ then consider a prime which divides $j$ and consider $k=p^n$\nnow since $p|j$ then $p|a_k$ this implies $p^n|{a_k}^n$ implies $p^n|j$ but then it follows that $p^{n}|B$ by definition of B.\nSo this a contradiction and hence we are infact done.", "Solution_10": "hehe my solution if exist p: Vp(b)=r.n+s ( 0 r then Vp(ak) >= r+1 then Vp(b-ak^n) = rn+s < r(n+1)", "Solution_11": "in this problem b and n are not fixed right? they can vary, just like k? ", "Solution_12": "Let $b=p^{an+c}l$ where p is prime and $(p,l)=1$ and $1\\leq c\\leq n-1$. Let $k=p^{an+n}$, $k|b-a_k^n,=>v_p(k)\\leq v_p(b-a_k^n)<=>an+n\\leq v_p(b-a_k^n)=min\\left\\{v_p(b),v_p(a_k^n)\\right\\}\\leq an+c$ . Which is plainly a contradiction, thus $b=m^n$", "Solution_13": "[quote=vjdjmathaddict]A different [u]Solution[/u]\nwrite $b$ as $b=B^n \\times j$ where $gcd(B,j)=1$ now if $j=1$ then we are done.\nSuppose $j$ is not equal to $1$ then consider a prime which divides $j$ and consider $k=p^n$\nnow since $p|j$ then $p|a_k$ this implies $p^n|{a_k}^n$ implies $p^n|j$ but then it follows that $p^{n}|B$ by definition of B.\nSo this a contradiction and hence we are infact done.[/quote]\n Wrong: If you take B to be the max n-th power that divides b, then not necessarily GCD(B,j)=1, but if you take by construction that GCD(B,j) , B and j are clearly unique, but not necessarily all n-th power that divides b go to B. Take for instanceb=(2^n)*(3^(n+1)), then B=2^n and j=3^(n+1) are the only air (B,j), but 3^n divides j, and therefore no contradiction\n", "Solution_14": "Nice problem. Hope, my solution is not the same as anyone else's :blush: \n\n$\\rightarrow$. Suppose there exists a prime $\\text{p}$ such that $b = p^{\\alpha n + e_0}x , 1 \\le e_0 < n$.\nWhere, $\\text{gcd} ( x,p)=1$ and $\\text{v}_p(b)$ is not congruent to $0 \\pmod{n}$.\n\nNow, let $k= p^{n (\\alpha +1)}$. Now, $\\text{v}_p(b-a_k^n)$ must be greater than or equal to $\\text{v}_p(k) = n(\\alpha + 1)$.\n\n$\\blacksquare$ [b]Case 1.[/b]: $\\text{v}_p(a_k) \\le \\alpha$.\n$\\implies \\text{v}_p(b)$$ \\ge \\text{v}_p(a_k^n)$$ \\implies \\text{v}_p(b-a_k^n) =\\text{v}_p(a_k^n) \\le \\alpha $\n[i]contradiction.[/i]\n\n$\\blacksquare$ [b]Case 2.[/b]: $\\text{v}_p(a_k) \\ge \\alpha +1$.\n$\\implies \\text{v}_p(a_k^n) \\ge \\text{v}_p(b) \\implies \\text{v}_p(b-a_k^n) = \\text{v}_p(b) < n(\\alpha + 1)$\n[i]contradiction.[/i]\n\nHenceforth, we conclude that $e_0 \\equiv$$ 0 \\pmod {n} \\implies b=m^n$. ", "Solution_15": "Assume that $n \\nmid v_p(b)$ for some prime $p$. Then let $v_p(b)=na+r$ with $0 < r < n$ and $\\frac{b}{p^{v_p(b)}}=c$. Then taking $k=p^{n(a+1)}$ and $a_k=m$ for simplicity, we get that $$p^{n(a+1)} \\mid p^{na+r}c-m^n.$$ This obviously implies that $p^a \\mid m$. Let $\\frac{m}{p^a}=o$. We then get that $$p^n \\mid p^rc-o^n.$$ Since $r>0$, we get that $p \\mid o$. Thus $$p^n \\mid p^rc-p^n(\\frac{o}{p})^n \\implies p^n \\mid p^rc,$$ contradiction since $rv_p(b)$.\n\nThen, we have\n$$a_{p^{kn}}^n\\equiv b\\pmod{p^{kn}}$$\n\nSince $v_p(b)v_p(b)$. Then we obtain $n\\mid v_p(b)$. $\\blacksquare$ (same solution as #50)", "Solution_24": "Suppose not. Then there exists $p$ such that $n\\nmid \\nu_p(b)$. Let $\\nu_p(b) = t$. Setting $k = p^{t+1}$ gives $\\nu_p(c^n) = n\\nu_p(c) = \\nu_p(b)$ for some $c$, which is impossible. ", "Solution_25": "Take $k=b^2$ so that $a_{b^2}^n \\equiv b \\pmod{b^2}$. Then, $a_{b^2}^n = b(bj+1)$ for some nonnegative integer $j$, but $(b, bj+1)=1$, so $b=A^n$ for some integer $A$, as desired. ", "Solution_26": "The main idea of the problem is the assertion $k=b^2$.\nThen for every prime $p$ dividing $b$, we clearly need $n\\mid\\nu_p(b)$. $\\blacksquare$", "Solution_27": "$k=b^2$ implies that $a_{b^2}^n = b(1-cb)$ for some $c$. However, for all $p \\mid b$, $\\nu_p(b(1-cb)) = \\nu_p(b)$ and we're done since $n \\mid \\nu_p(b(1-cb))$. $\\square$", "Solution_28": "Another one liner?\n\nThe assertion $$k=b^2\\implies a_{b^2}^n = b(1-kb)\\implies n\\cdot v_pa_{b^2}=v_pa_{b^2}^n=v_p(b(1-kb))=v_pb\\forall p\\mid b\\implies n\\mid v_pb,$$ which is just the problem rephrased since each exponent of a prime in b must be a multiple of n. $\\blacksquare$", "Solution_29": "Consider a $k$ such that for every prime $p$ dividing $b, \\nu_p(k)>\\nu_p(b)$ (here $k=b^2$ works easily), note that $k\\mid b-a_k^n$, if $\\nu_p(b)=n\\nu_p(a_k)$ we are done, so assume the contrary then\n\\[\\nu_p\\left(b-a_k^n\\right)=\\min\\{\\nu_p(b),n\\nu_p(a_k)\\}>\\nu_p(k)\\]\nWhich is impossible as $\\nu_p(b)<\\nu_p(k)$" } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "This year the in a college has members from all 4 classes - 9 freshmen, 11\r\nsophomores, 18 juniors and 22 seniors.\r\na. The photographer wants to line up the students in one line for a photograph so that all of the students in each class will be together. How many ways can this be done? Can anyone solve this problem?", "Solution_1": "treat each \"class group\" as a block. It really doesn't matter how many students are in each class, if I'm reading the question correctly. It's basically finding the number of permutations of 4 distinct \"objects\".", "Solution_2": "Here we are not considering the order. We are concerned with just lining them up . I thought this is a combination problem not permutation. Try once again and solve it." } { "Tag": [], "Problem": "Let $t_j=\\frac{j(j+1)}{2}$, and $S_n=\\displaystyle\\sum_{j=1}^{n} t_j$. Find the smalles possible value of $n+k$ if $S_{n+k}-S_n>2007$", "Solution_1": "this uses a classic result, first notice that $t_j=\\binom{j+1}{2}$\r\n\r\nit can be shown through pascal's identity, i believe that the specific name is the vandermonde identity, but anyway, you get\r\n\r\n$S_n=\\binom{n+2}{3}$\r\n\r\nclearly, we want to minimize $n$ since we are subtracting that, so the last binomial coefficient is defined when $n=1$, then\r\n\r\n$\\binom{k+3}{3}>2008$\r\n\r\nchecking a few values, $k=24$, so $n+k=\\boxed{25}$", "Solution_2": "Ok, I see. Your solution seems to be right. My solution had an error.\r\nI made this up for an exam without a calculator. Do you think the last step is too much computation (3mins/problem avg.)." } { "Tag": [ "LaTeX" ], "Problem": "Find all prime numbers p and q such that p divides q+6 and q divides p+7", "Solution_1": "bnick, you need better titles for your posts. (LaTeX is nice, too.)\r\n\r\n[quote=\"bnick17\"]Find all prime numbers $p$ and $q$ such that $p$ divides $q+6$ and $q$ divides $p+7$[/quote]\r\n[hide=\"Boring solution\"]\nNote that if $q = 2$ then $p \\mid 8$ and so $p = 2$, but $2 \\not\\mid 9$, a contradiction. Thus $q$ is odd and so $p \\mid q+6$ is odd as well. Thus $p+7$ is even so in fact $q$ divides $\\frac{p+7}{2}$.\n\nCase 1: $p = q+6$. Then $q$ divides $p+7 = q+13$ so $q = 13$, $p = 19$.\n\nCase 2: $p \\neq q+6$, so in particular $p \\leq \\frac{q+6}{2}$. Then $q \\leq \\frac{\\frac{q+6}{2}+7}{2}= \\frac{q+20}{4}$ so $q \\leq 5$ and $p \\leq 5$. There are only 4 cases to check, and none of them work.[/hide]\r\n\r\nAnyone have anything cleverer?", "Solution_2": "[hide=\"this also requires some casework\"]\nClearly $p\\neq q$ (otherwise $p$ divides both $p+7$ and $p+6$), so $p$ and $q$ are relatively prime. Now observe that\n\\[p|(6p+7q+42),\\ q|(6p+7q+42)\\Rightarrow pq|(6p+7q+42)\\]\nLet $6p+7q+42=kpq$. If $k$ is even, then $q=2$. This means $p|8$ so $p=2$, but this fails because $2\\not| 9$. Similarly, if $k$ is divisible by 3, then $q=3$. This means $p|9$ so $p=3$, but this fails because $3\\not| 10$. Therefore either $k=1$ or $k\\geq 5$.\n\nIf $k=1$, then $6p+7q+42=pq$ implies\n\\[(p-7)(q-6)=84\\]\nBoth factors are positive (if both negative, product less than 84). Then $q>6$ and $p>7$, so $q-6$ can't be divisible by 2 or 3, $p-7$ can't be divisible by 7. The only case that remains is $p-7=12,q-6=7$ which leads to $(19,13)$.\n\nIf $k\\geq 5$, then $6p+7q+42\\geq 5pq$ implies\n\\[(5p-7)(5q-6)\\leq 252\\]\nAs shown before, $q=2$ and $q=3$ are bad, so $q\\geq 5$. This means $p\\leq \\frac{252}{19}<18\\Rightarrow p\\leq 3$. In either case, $p|6$ so $p=q$, but this is a contradiction!\n\nTherefore $\\boxed{(19,13)}$ only solution.\n[/hide]" } { "Tag": [], "Problem": "If you're interested in participating in COP 4th, please sign up here by putting \"I'm in\" or \"Yes!\" or et cetera.\r\n\r\nThanks.", "Solution_1": "I'm in! :D", "Solution_2": "Hurray! Another COP. [b][i][u]I'm in[/u][/i][/b]. [b][i][u]Yes[/u][/i][/b].", "Solution_3": "I'll join", "Solution_4": "Wow, endless problem supply. I'm in. :P", "Solution_5": "I'll do it. Can you have one or more of tokenadult, rcv, or frost13 check the problems for clarity so we don't have to ask questions?", "Solution_6": "I'll try to do it, if I can remember that is.", "Solution_7": "I'm in at least through the proof-free rounds...", "Solution_8": "What's this?? When does it take place?", "Solution_9": "[quote=\"Silverfalcon\"]If you're interested in participating in COP 4th, please sign up here by putting \"I'm in\" or \"Yes!\" or et cetera.\n\nThanks.[/quote]\r\n\r\nw/e i'll do it", "Solution_10": "To RC-7th:\r\n\r\nI'll try to do it.. But often times, I do not want the problems to be out while it's in process so I rarely do that. Of course, making problems and realizing the mistakes make me a better problem writer though. :) I remember writing the my former competition and had huge mistakes while the mistakes in here are usually minor.\r\n\r\nTo shobber:\r\n\r\nPlease read the stickied thread \"About COP (Contest of Opportunity)\" on this forum.\r\n\r\nThank you.", "Solution_11": "I'll do it.", "Solution_12": "Is there going to be a COPSD? If there is, i'll be joining that instead.", "Solution_13": "Done.\r\n\r\nIt's on the Basics Forum now. :) :D", "Solution_14": "$Sign-Up$", "Solution_15": "im in i guess", "Solution_16": "Can you do both the COP and COPSD? If so, I'm in, else I'll just do the COP.", "Solution_17": "I didn't sumbit the last one.................. I gotta do this one to make up for the stupidity. :lol:", "Solution_18": "I'm in as well :D", "Solution_19": "[quote=\"solafidefarms\"]Can you do both the COP and COPSD? If so, I'm in, else I'll just do the COP.[/quote]\r\n\r\nIf you're middle schooler, yes.\r\n\r\nIf you're graduated from middle schooler, sorry no. :lol:", "Solution_20": "Heck, I'll give it a shot.", "Solution_21": "what's the difference between\r\n\r\nCOPSD and COP?/\r\n\r\nand I'm in for both ...just notify me\r\n....\r\n\r\nI'm kinda busy these days, so I'm not on every often times", "Solution_22": "[quote=\"Silverfalcon\"]If you're interested in participating in COP 4th, please sign up here by putting \"I'm in\" or \"Yes!\" or et cetera.\n\nThanks.[/quote]\r\n\r\nEt cetera.", "Solution_23": "I'd like to participate in the COP.", "Solution_24": "[quote=\"shinwoo\"]what's the difference between\n\nCOPSD and COP?/\n[/quote]\r\n\r\nI'm pretty sure COPSD is for middle schoolers only", "Solution_25": "Yes.\r\n\r\nExactly. :)", "Solution_26": "I 'm in :lol:\r\n(this is the first post for me)\r\nWhat is middle schooler's age? (I'm 14 :D )", "Solution_27": "You just have to still be in middle school :)", "Solution_28": "Or your country's equivalent, if applicable.", "Solution_29": "i'm in", "Solution_30": "I'm In!\r\n\r\nUm and Yes!", "Solution_31": "$yes$", "Solution_32": "i'll join", "Solution_33": "mee three, yes i want in", "Solution_34": "$\\text{ count me in!!!! }$", "Solution_35": "I'm proud to announce that the fourth COP will start in few weeks.\r\n\r\nBe ready!\r\n\r\nI'll post a thread when problems are ready.\r\n\r\nP.S. This is same for COPSD.", "Solution_36": "I'm in too.", "Solution_37": "$\\text{I AM IN!!!!}$", "Solution_38": "Sign me in. If is not to late :)" } { "Tag": [], "Problem": "Hi, I was wondering if anyone knows of any (possibly online) math contests that don't run during the school year? Just trying to plan out my summer a little, see if there's anything of the sort to do...", "Solution_1": "Mu Alpha Theta National Convention, but the registration is already over. and it is pretty expensive ( I think 500 dollars for registration and sleeping and stuff) but it goes for a week and it is pretty well-organized and worked out. This year it is in Hawaii(which is why iam going despite hating the competition itself).\r\n\r\nOther than that, I dunno." } { "Tag": [ "function", "algebra", "functional equation", "algebra unsolved" ], "Problem": "Find all functions $f: \\mathbb{Q} \\mapsto \\mathbb{C}$ satisfying\r\n\r\n(i) For any $x_1, x_2, \\ldots, x_{1988} \\in \\mathbb{Q}$, $f(x_{1} + x_{2} + \\ldots + x_{1988}) = f(x_1)f(x_2) \\ldots f(x_{1988})$.\r\n\r\n(ii) $\\overline{f(1988)}f(x) = f(1988)\\overline{f(x)}$ for all $x \\in \\mathbb{Q}$.", "Solution_1": "[quote=\"orl\"]Find all functions $f: \\mathbb{Q} \\mapsto \\mathbb{C}$ satisfying\n\n(i) For any $x_1, x_2, \\ldots, x_{1988} \\in \\mathbb{Q}$, $f(x_{1} + x_{2} + \\ldots + x_{1988}) = f(x_1)f(x_2) \\ldots f(x_{1988})$.\n\n(ii) $\\overline{f(1988)}f(x) = f(1988)\\overline{f(x)}$ for all $x \\in \\mathbb{Q}$.[/quote]\r\n\r\nat the beginning, we suppose $f(x)\\neq0$ for all $x \\in \\mathbb{Q}$.\r\n\r\nby $\\overline{f(1988)}f(x) = f(1988)\\overline{f(x)}$ we find $\\frac{f(x)}{\\overline{f(x)}}=\\frac{f(1998)}{\\overline{f(1998)}}\\longrightarrow\\frac{f(x)}{\\overline{f(x)}}=\\frac{f(y)}{\\overline{f(y)}}=$costant for all x,y.\r\nThis means that, if $f(x)=p(x)e^{i\\theta(x)}$, $\\frac{p(x)e^{i\\theta(x)}}{p(x)e^{-i\\theta(x)}}=e^{2i\\theta(x)}=$costant, so $2i\\theta(x)=$costant $(mod$ $2\\pi)$, $\\theta(x)=$costant $(mod$ $\\pi)$: if we make $p(x)$ vary in the positive and negative reals, that means that $\\theta(x)$ is costant.\r\n\r\nnow consider the first relation: if we put $x_1=x_2=\\ldots=x_{1998}=0$, we have $f(0)^{1998}=f(0)$, so f(0) is one of the 1997 complex 1997-roots of the unity, $f(0)=e^{\\frac{2ki\\pi}{1997}}$ (and this is $\\theta(x)$ for all x)\r\n\r\nnow let's put $x_1=x_2=\\ldots=x_{1998}=a$, and then we have $f(a)^{1998}=f(1998a)$\r\nagain, let's put $x_3=x_4=\\ldots=x_{1998}=a$, $x_1=a+b$, $x_2=a-b$ , from which we have $f(a)^{1996}f(a+b)f(a-b)=f(1998a)$.\r\nFrom this two relations we have $f(a)^2=f(a+b)f(a-b)$, changing\r\n$f(\\frac{x+y}{2})^2=f(x)f(y)$,for all x,y.\r\nThis means also that $f(\\frac{x+y}{2})^2=f(x+y)f(0)$, and then\r\n$f(x+y)f(0)=f(x)f(y)$.\r\n\r\nWith the substitution $f(x)=p(x)e^{i\\theta(x)}=p(x)e^{\\frac{2ki\\pi}{1997}}$ we have\r\n$p(x)e^{\\frac{2ki\\pi}{1997}}p(y)e^{\\frac{2ki\\pi}{1997}}=p(x+y)e^{\\frac{2ki\\pi}{1997}}e^{\\frac{2ki\\pi}{1997}}$\r\n$p(x)p(y)=p(x+y)$, with $p(x)$ function $p: \\mathbb{Q} \\mapsto \\mathbb{R}$ and $p(0)=1$.\r\n\r\nThis functional equation is trivial: the solutions are $p(x)=a^x$ for $a>0$, from which $f(x)=a^xe^{\\frac{2ki\\pi}{1997}}$ ($k\\in\\mathbb{Z}/_{1997}$).\r\nIt can be easily seen that these functions satisfy the two text relations.\r\n\r\nNow suppose that there is an y such that $f(y)=0$.\r\nin the first relation, if we put $x_1=y$, $x_2=z-y$, $x_3=x_4=\\dots=x_{1998}=0$, we see that\r\n$f(z)=f(y)f(z-y)f(0)^{1996}=0$ for all $z\\in\\mathbb{Q}$.\r\nso $f(x)=0$ is another solution. this ends the demonstration.\r\n\r\nhi everyone\r\n\r\nEDIT: exscuse me, i treated 1988 as 1998.... :D :D" } { "Tag": [ "geometry open", "geometry" ], "Problem": "It is trivial to dissect a triangle into $ n$ triangles for any $ n\\ge 2$.\r\nSame is true for convex quadrilaterals. It is not hard to show that every \r\nconvex pentagon can be dissected (partitioned) into $ 6$ convex pentagons. \r\nWhat about the general case?\r\n\r\n[b]Problem.[/b] Is it true that every convex $ n$-gon can be dissected into\r\nfinitely many convex $ n$-gons, for every $ n\\ge 3$?\r\n\r\nThe problem is probably not new, but one of my students asked me the question\r\nand I do not know the answer. Any reference is welcome.\r\n\r\nDan", "Solution_1": "nice question, i will be back with a solution if i figured this out :)" } { "Tag": [], "Problem": "Apparently, KoMaL for this year starts in September. At least that's what the web site said.\r\n\r\nSo, how does turning in solutions work? I signed up in July, but this is the first time I've done it.", "Solution_1": "You have to contact them directly. I have no idea what their regulation are." } { "Tag": [ "email", "search" ], "Problem": "Kung Fu Madness, does anyone play this game?\r\n\r\nClick [url=http://www.kungfumadness.com/PlayerCreate.cfm?Friend=2583]Here[/url] to create an account.\r\n\r\nI'm a black belt BTW, my account = sin4pi.", "Solution_1": "I made an account. My name is DarkKnight. Orange belt. :P", "Solution_2": "wtf????? I didnt get an email with my password. I think i made like 5 accounts in like 5 different email accounts but still havnt gotten any password", "Solution_3": "I'd like to start a dojo if enough AOPSers sign up.\r\n\r\nBut I need at least 6 or so more first.", "Solution_4": "you can count me in. I'm a purple belt and my name is elvenchamp. what do dojo do though?\r\n\r\nelvenchamp777", "Solution_5": "[quote=\"elvenchamp777\"]you can count me in. I'm a purple belt and my name is elvenchamp. what do dojo do though?\n\nelvenchamp777[/quote]\r\n\r\nI have three acounts so that makes 5. And i know a guy that would be willing to join. I think hes a second dan. My accounts are brown belt, brown belt and second dan. working up one of my brown belts and using the other to get some money because i don't care about that one's record. To find my accounts search DarkKnight.", "Solution_6": "I'm black now :D and is there some other way to get money cause I'M SOOOO POOR lol and how do you transfer money from one account to another?\r\n\r\nelvenchamp777", "Solution_7": "[quote=\"elvenchamp777\"]I'm black now :D and is there some other way to get money cause I'M SOOOO POOR lol and how do you transfer money from one account to another?\n\nelvenchamp777[/quote]\r\n\r\nTo transfer money to another account, go to the gold ledger and click send gold to friend. I made a dojo with three people i've fought and 2 accounts that are me. Please message me to join. It's called Arcane Vault. I've sent a message to you and lucky707 asking you to join my dojo but you will have to leave the one you are currently in. Remember to check your mail. ;)", "Solution_8": "Guiseppe - Black Belt", "Solution_9": "Could ya invite WindSlicer to the clan? Thanks!\r\n\r\nI think I'm Green Belt workin' my way up :P", "Solution_10": "[quote=\"miraculouspostmaster\"]Guiseppe - Black Belt[/quote]\r\n\r\nWhy did you leave our dojo after you joined?" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Let $ m_a,m_b,m_c$ the medians and $ r_a,r_b,r_c$ the exradii of an scalene triangle $ ABC$. Find the minimum value of\r\n\r\n$ P\\equal{}\\sum\\frac{m_a}{r_a}$", "Solution_1": "[quote=\"Ligouras\"]Let $ m_a,m_b,m_c$ the medians and $ r_a,r_b,r_c$ the exradii of an scalene triangle $ ABC$. Find the minimum value of\n\n$ P \\equal{} \\sum\\frac {m_a}{r_a}$[/quote]\r\nA computer program shows that $ P_{min}\\equal{}3.$", "Solution_2": "[quote=\"Ligouras\"]Let $ m_a,m_b,m_c$ the medians and $ r_a,r_b,r_c$ the exradii of an scalene triangle $ ABC$. Find the minimum value of\n\n$ P \\equal{} \\sum\\frac {m_a}{r_a}$[/quote]\r\n\r\n$ \\frac{h_a}{r_a}\\equal{} \\frac{2(s\\minus{}a)}{a}$ , $ s\\equal{}\\frac{1}{2}(a\\plus{}b\\plus{}c)$\r\n\r\n$ \\sum\\frac {m_a}{r_a} \\ge \\sum\\frac {h_a}{r_a}\\equal{}2\\sum\\frac {s}{a}\\minus{}6 \\ge 3$" } { "Tag": [ "logarithms" ], "Problem": "[size=150]solve to 3 s.f [/size]\r\n\r\ne^1-x = 17.5\r\n3-2e^x-1 = 0\r\n\r\nif someone could show me how to do them that would very helpful\r\nthank you\r\nMAGIC MAN", "Solution_1": "Am I reading correctly:\r\n$ e^{1\\minus{}x}\\equal{}17.5$ and $ 3\\minus{}2e^{x\\minus{}1}\\equal{}0$\r\nI'm assuming these are two separate problems so, for the first one, we can decompose the exponent into $ e\\cdot\\frac{1}{e^{x}}\\equal{}17.5$. We can multiply both sides by $ e^{x}$ and divide by 17.5 to obtain $ \\frac{e}{17.5}\\equal{}e^{x}$. Take the natural log of both sides to get $ x\\equal{}\\ln\\frac{e}{17.5}$. Using properties of logs, we can rewrite $ x\\equal{}\\ln{e}\\minus{}\\ln{17.5}$. Since the first natural log is 1, our final answer is $ x\\equal{}1\\minus{}\\ln{17.5}$.\r\n\r\nFor number 2, it's basically the same deal. Move the 3 over and divide both sides by 2 to obtain $ e^{x\\minus{}1}\\equal{}\\frac{3}{2}$. From there, we follow the same procedure as the first problem.", "Solution_2": "1. $ e \\minus{} x \\equal{} 17.5$?\r\n2. $ 3 \\minus{} 2e^{x \\minus{} 1} \\equal{} 0$\r\nThese the questions? I probably misread so no need to look at my corrections, but\r\n[b]Please[/b] refer to the link provided in the hint.\r\n[hide=\"Hint\"]\nFor the first question, just subtract $ e$ from both sides and divide by $ \\minus{} 1$.\nFor the second one, subtract both sides by $ 3$, divide by $ \\minus{} 2$, then use logarithms. (If you're having trouble with logs, refer to this link,[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175427[/url], and look how stimnc did it).[/hide]", "Solution_3": "[quote=\"Magic Man\"][size=150]solve to 3 s.f [/size]\n\ne^1-x = 17.5\n3-2e^x-1 = 0\n\nif someone could show me how to do them that would very helpful\nthank you\nMAGIC MAN[/quote]\r\n1. 1-x=ln 17.5, so x=1-ln 17.5\r\n2. 2e^(x-1)=3, so e^(x-1)=3/2 and x-1=ln 1.5, so x=1+ln 1.5" } { "Tag": [ "AMC", "AIME", "MATHCOUNTS", "search", "ARML", "geometry", "function" ], "Problem": "(Mods--I'm not sure if here is the best place to post this; please feel free to move this thread to a more appropriate location.)\r\n\r\nNext year, we will try to start a math club at my school, and I would appreciate any suggestions anyone has. \r\n\r\nFirst, I\u2019d like to explain a little about my school and school district. My school is really heavy on humanities/liberal arts and quite lighter on math/science; in fact, the only math competition we offer is basically the AMCs. We\u2019ve only had five AIME qualifications in the history of our school district, and this year, we sent out personal invitations for the AMCs in addition to announcements, but only 7-8 people showed up (some with the motivation of missing class). \r\n\r\nMy school district does, however, have many strong math students; the middle school MathCounts program is quite successful, for example. Unfortunately, MathCounts ends in 8th grade, and there is no math program to hold up MathCounts alums. Sadly, by sophomore and junior years, student participation in activities differentiates so much, and interest in mathematics (and academic competitions in general) declines substantially. \r\n\r\nI strongly believe that, especially in the lower grades, there is interest in mathematics that can be harnessed. For this reason, I'm planning to help spearhead a movement to establish a running math program in the high school. We already have a boost with a supportive administration and substantial funds available from a grant proposal I wrote specifically for this purpose. Additionally, I know probably we\u2019ll need to get an enthusiastic coach (I\u2019m planning to talk to a teacher), and most importantly, members. \r\n\r\nI\u2019m wondering about other suggestions/ideas. For example: What is the best way to recruit new members? What kinds of competitions should we participate in? Should they involve lots of travel or be stay-at-home type contests? How many should we do? For a new club, what general structure/schedule would be best to follow? Weekly meetings? How would our meetings be structured, anticipating limited membership and relatively inexperienced problem solvers? What kinds of books/resources should we utilize? Most importantly, how would we best sustain interest in members and develop critical mass? How does math club work at your school? \r\n\r\nThanks in advance. This actually isn\u2019t the first time I\u2019ve tried to help start a math club. We didn\u2019t have a great start last time, so I\u2019m really hoping this try will be successful. :)", "Solution_1": "Ah!!! It's a curse when we write up a long thread and no one answers it. Maybe people just like small blocks of text and such... For one thing, I don't know the answer to all your questions and it might be a turn off for some people to try to answer all of them :(\r\n\r\nThe main thing is to have your team do problem sets regurlarly. Consider going over the problems and/or giving out the solutions afterward. Also consider giving awards and having rankings as incentives for members to study. That's how TJ http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=477000247&t=199100\r\n\r\ndoes it. It motivates a few smart kids to work... meh.", "Solution_2": "The most important thing is to have a supportive sponsor for your club, and it sounds like you have that. Having money is also good. I am shocked that people are giving you money; most clubs have to raise their own money.\r\n\r\nSome simple suggestions, based on the fact that you're trying to build math competition culture out of nothing:\r\n\r\nTry to compete in whatever local competitions you can. Even if the quality isn't that great, they tend to be more fun as a social activity than tests that you take at your own school (like AMC). It also builds camaraderie as well as rivalry with other schools. After all, you're probably not going to bring a team to ARML and bring down the big boys, but you can certainly try to take on Shelbyville. Unfortunately, a lot of areas don't have local math competitions. You might even try non-math competitions just to get people interested in competing. Stay-at-home contests are also worth doing and require minimal effort, but I don't think they're your key to success.\r\n\r\nIn general, try to compete as much as possible. Practice is boring. Competing is fun. Most people only practice because it leads to winning. People have to understand the desirability of winning before they can understand the point of practice. \r\n\r\nTry to organize at least one big trip per year, perhaps to a regional math competition.\r\nEveryone likes field trips, especially overnight ones.\r\n\r\nTry to convince your math teachers to offer extra credit for math competition participation. (Note that I said participation rather than performance.)\r\n\r\nInstead of inviting everyone to take the AMC, try to convince your math teachers to simply cancel all advanced math classes and just force everyone to take the AMC by default. Or if you can't do that, at least get them to offer extra credit. \r\n\r\nTell people that being on the math team and winning math awards looks really good on your college application.\r\n\r\nI respect what the previous poster wrote, but you should know that what works at TJ is probably not going to work for you. Maybe if you said where your high school is, others might have more specific advice for you. I wish you the best of luck.", "Solution_3": "Here's what I learned from 5 minutes with the search function:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=703006746&t=36197\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=703006746&t=50282\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=703006746&t=91091\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=703006746&t=109466\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=703006746&t=167763\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=703006746&t=176025\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=148210637&t=9792\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=148210637&t=48507\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=148210637&t=81905", "Solution_4": "I'm sure JBL's post would answer all of your questions, but I could offer a little help. If your school system isn't too focused on math/science, take the people you can and ask what afternoons they have free. You can base a schedule off that. Also, check with the parents and print up fliers if they want their child to go to faraway contests or stay-at-home ones. \r\n\r\nEntirely based on the people. Like the Articles of Confederation :P \r\n\r\nFor those younger students who have no reason or will to join, try doing a separate date for those younger students. If you plan something like Jeopardy, Are you Smarter than a ____ Grader, or a Deal-Or-No-Deal Math game it will certainly attract most people. Even a \"lottery\" with prizes such as difficult problem solving flash cards could score a few people.", "Solution_5": "For contests, emphasize the AMC's, and have a couple teams take the iTest. The iTest is free, and it can be found here: http://www.theitest.com/\r\n\r\nOnce the club has grown in size, you might want to try some bigger competitions, like the ARML Team Selection Test(or equivalent), HMMT, and Princeton. Some people are against driving(especially long distance), so car pooling would be a good idea for the ones that are far away. If you meat weekly at say dinner time, you might want to bring a pizza or two; my Math Circle does that(pizza money is needed, like 10 dollars a month).\r\n\r\nMy math circle sends out problems on our yahoo group (AMC 12s, AHSMEs, AIMEs, HMMTs, ARMLs, you get it), and every week we meet (at dinner time, :roll: ), work for an hour, have pizza, and get to working on problems. In the last 45 minutes of the meeting, we usually do either a relay round, a guts round, or work. After three hours are up, people clean up and drive home.\r\n\r\nA way to get people to come is to make a member list, and have everyone check their name off when they come. You establish a point system: +3 points for going to a meeting, -2 points for missing a meeting, +100 points for going to a contest, +50 points for competing in the AMCs, +100 points for scoring high/advancing to the next competition, etc.\r\n\r\nAlso recommend AoPS and it's books, ACoPS, and MAA books.\r\n\r\nMy MathCounts group has a coach who hands out problems to groups of students, and assistant coaches for each group of students. If you don't know a problem, it can be explained by the coach, or another student. The coaches are people who are in high-school or who have participated in MathCounts competitions for three years.", "Solution_6": "I strongly second everything that yenlee said. The 1 or 2 hardcore kids might be motivated by your having an ACoPS or AoPS book in the sponsoring teacher's classroom, and they might be motivated by the once-a-year AMC and opportunity to qualify for AIME. But if you really want a good number of people to come, if you want to build an actual program, you need: \r\n\r\n- accessible things (reasonably easy contests)\r\n- frequent things (look up Math League for whatever state you're in, Mandelbrot, NAT, other periodic contests) \r\n- travel (people LOVE going to contests together)\r\n- motivation (extra credit for math classes, [i]just for participation[/i], is one of the best ways to do this, as yenlee said)\r\n\r\nYou could also use your funds to provide some extra motivation, like: \r\n\r\n- randomly distributed prizes for people who participate in (say) AMC, or all 6 Math League tests (again participation only)\r\n- keep overall records for people in the club (rankings) and give prizes for that. (this can be a mix of participation and performance)\r\n- provide snacks after the in-school tests (try to make sure that kids who don't take the test can't steal the food)", "Solution_7": "your math club's success will depend on the people in it. not the number of people, the people themselves. focus on getting together your group of people who are interested in math(teachers too)--everything else will follow.", "Solution_8": "Thank you so much to everyone who helped. I'll help implement your suggestions next year.", "Solution_9": "Make it accessible. I HATE WAKING UP IN THE MORNING!!! but you can do this once kids get serious. start rewarding. i know its cliche but it works\r\ntake in-school tests and allow teamwork\r\npeople love working together\r\n\r\nif you want to win though....\r\npractice hard... WITH PIZZA (of course make them pay a little, but not so much that it is a lousy deal)", "Solution_10": "Our conditions are exactly similar to the OPs conditions except \"parents and teachers\" despise olympiads and math :mad: \r\n\r\nWhat Xevarion said is [b]completely[/b] right. My teacher *tried* to start math discussion club out of our school. But all in vain.\r\n\r\nHere is our story:\r\nFirst day, he had around 150 students. But within weeks it crumbled downed to 25. Finally after two months, it stabilized to 2 students. At the regional olympiad, no one turned up except one!(Thats me by the way) and I did not get selected! \r\n\r\nSo at the end of the day, Sir and I analyzed the reason for failure. The reasons were:\r\n\r\n1)No funds. He could not provide good study materials.\r\n\r\n2)No schools appreciate learning maths as a hobby! They all turned down offers made by sir. In fact some teachers even advised students not to join the club. They believed the students could invest their time in something more fruitful like \"textbook\" cramming.\r\n\r\n3)When Sir arranged a talk by good mathematicians in the country, only 20 people turned up. Because everyone was afraid they will not understand what they say. Their parents told them it will be a waste of money.\r\n\r\nSo my Sir is [b]still[/b] fighting this losing battle. I believe if anyone had funded the venture and parents or teachers were supportive, the students would automatically gain interest.\r\n\r\nSo I believe the most important ingredient in the recipe is support. If you have financial support, you can have good resources, if you have school support, you can get students and if you have parents and faculty support, your students can get good motivation.\r\n\r\nAnyway these are my thoughts because of the bad experience I had. It was never meant to discourage you. I really hope and wish that your math club becomes successful", "Solution_11": "[quote=\"k8reindeer\"]your math club's success will depend on the people in it. not the number of people, the people themselves. focus on getting together your group of people who are interested in math(teachers too)--everything else will follow.[/quote]\r\n\r\nIt also depends on the number of people to soem extent. You would want a math club to stay alive... with let's say 10 people.", "Solution_12": "eh, to be honest, i don't participate in math club at all\r\n\r\nit's not really interesting, and none of my friends are in it, so it's like \"why bother going?\"\r\n\r\n\r\nwith that said, there has to be some sort of motivational impetus to get kids interested in it. like i mean if you guys want a core group of really smart kids who get together and do high-tier competitions, that's great. but the annoying truth is that - at least where i live - there aren't that many ridiculously intelligent kids to do those things. so if you want an active math club up and running, you have to get a way of getting more \"normal\" kids to join, compete, and be motivated. it's definitely not easy, because socially, math has a pretty negative stigma attached to it, and...\r\n\r\nbasically, good luck.", "Solution_13": "One good way to attract more math people is bringing food...proven by bookaholic.", "Solution_14": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding your own.[/color]\r\n\r\nDisproven by ccy. His club failed worse than cube club", "Solution_15": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding your own.[/color]\r\n\r\nBut his club was web design, and about 2 people in our school are interested in that. Contrastingly, we have a whopping 14 people interested in math.", "Solution_16": "[quote=\"Xevarion\"]- travel (people LOVE going to contests together)[/quote]\r\n\r\nLet me emphasize this. Whenever we go to a contest that needs travel(HMMT, Princeton, National MC, you name it), we love playing games. Chess, cards, Contact, and even some \"What now?\", are great to play as a group.\r\n\r\nFor card games, Egyptian Rat Screw, Presidents, Mao, and Set are great card games.", "Solution_17": "[quote=\"serialk11r\"]One good way to attract more math people is bringing food...proven by bookaholic.[/quote]\r\n\r\nI would definitely agree with this; even if not all of them are interested in math, it gets more people to at least try it out and experience it before they judge the competitions.\r\n\r\nI would add Mafia to that list of games to play, especially if you're going to ARML. Mao is fun (and one of my favorites), but I wouldn't recommend it as a starting game because it tends to annoy a great deal of people, especially those that haven't played it extensively before (I know that the first time I played it at ARML, I was wondering what in the world was going on because people just kept giving me random instructions, and then giving me cards when I didn't follow them :lol: )\r\n\r\nUmmm...to be honest, sponsors, as great as they are, are overrated. If you have a strong math club program, I have found that parents are quite willing to donate and support their kids (and it adds up fairly quickly). It's great that you have a sponsor, and it may give you a lot more options, but focus on the MATH. That's what keeps the core group interested.\r\n\r\nCompetitionwise, I agree with yenlee that local ones are a lot more fun. If all you're doing are AMC's, a fair number of people are probably going to just get bored and give in. Even if you might not be vying for a top spot, sending a team to ARML would still be a great idea. Although it might not have the toughest math questions (at least compared to USAMO), it's a social competition, and I've found that it tends to be one of the funnest ones, especially if the timing of your plane flights gives you some free time as a team before/after the actual competition.\r\n\r\n-piggy *oink* *oink*", "Solution_18": "yuck...long block of texts...\r\n\r\n\r\nhere's some practical advice:\r\n\r\nhave food during practice\r\n\r\nmake it fun, don't just make them do problems, have them work on problems together, or even try to race to see which group can solve it the fastest\r\n\r\n\r\ngo to lots of competitions, especially far off ones, staying in hotel and traveling is awesome.\r\n\r\nhave food\r\n\r\ntry to let the parents know about this, they can force their kids to go since i doubt many kids will actually like doing math...\r\n\r\nhave 1 or 2 meetings per week\r\n\r\n\r\nhope this helps, i know i'm not very good at giving advise, but meh..i'm doing the best i can." } { "Tag": [ "inequalities", "geometry proposed", "geometry" ], "Problem": "Let AD,BE,CF be the medians of $\\triangle ABC$. Let k be the number for which $a>k|b-c|,b>k|c-a|,c>k|a-b|$ such that $AB\\cdot BC\\cdot CA>AD\\cdot BE\\cdot CF$. Determine the minimum value of $k$.", "Solution_1": "still no reply..... :( \r\nMaybe it'd better for me to post it in the Inequality Forum." } { "Tag": [ "linear algebra", "matrix", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ G$ be the set of all $ 3 \\times 3$ real matrices of determinant $ 1$ of the form\r\n\\[ \\begin{pmatrix} a & 0 & 0 \\\\ b & x & y \\\\ c & z & w \\end{pmatrix}.\\]Verify that $ G$ is a group. Find a homomorphism from $ G$ onto the group $ GL_2(\\mathbb{R})$ of all non-singular $ 2 \\times 2$ real matrices, and find its kernel.", "Solution_1": "G is block lower triangular, so you can multiply block wise. For the homomorphism just take the bottom-right block.\r\n\r\nThe problem is basically identical to taking G={[a,0;b,x] : ax=1} and finding a homomorphism to GL(1,R) by sending [a,0;b,x] to x. In you case \"x\" is taken from the ring M_2(R) and b from the bi-module R^2, but all of the expressions used when a,b,x are in R still work." } { "Tag": [ "Ring Theory", "superior algebra", "superior algebra solved" ], "Problem": "[color=green]In each of the following cases, determine if S is a subring of R.\n(a) R = Z[size=75]9[/size], S = { [0], [3], [6] }.\n(b) R = Q \u00d7 Q, S = { (r, s) : r, s in Q, r <= s }.\n(c) R = Q, S = { r/s : r, s in Z, s is odd }[/color]\r\n\r\nThanks very much!!! :D", "Solution_1": "[quote=\"suedenation\"][color=green]In each of the following cases, determine if S is a subring of R.\n(a) R = Z[size=75]9[/size], S = { [0], [3], [6] }.\n(b) R = Q \u00d7 Q, S = { (r, s) : r, s in Q, r <= s }.\n(c) R = Q, S = { r/s : r, s in Z, s is odd }[/color]\n\nThanks very much!!! :D[/quote]\r\n\r\n1. Get some sleep. \r\n2. grab your algebra book and check if you know the definitions well.\r\n\r\nThese steps are meant to rid you of confusion.\r\n\r\n3. (a) Make the table for addition and multiplication for $\\mathbb Z_9$ and verify that $S$ is an additive subroup stable under multiplication.\r\n4. (b) The operations on $R$ are the component-wise addition and multiplication. Check that $S$ is not an additive subgroup. If $(r,s)\\in S$ and $r l:=nops(u);\r\n> s:=1:\r\n> for i from 1 to nops(u) do\r\n> s:=s*u[i]:\r\n> od:\r\n> s;", "Solution_2": "And you may remove that 3rd line ( l:=nops(u); )." } { "Tag": [ "quadratics", "algebra", "quadratic formula" ], "Problem": "You are given an equation x^2 - 4x + 6k = 0. If the difference of the squares of its roots is 8, find the value of k.", "Solution_1": "[hide]\n\n\n\nsolving for the roots, in terms of k, using quadratic formula gives\n\n\n\nx=2 :pm: :sqrt: (4-6k)\n\n\n\nsquaring each root and subtract smaller from the larger returns\n\n\n\n8 :sqrt: (4-6k)\n\n\n\n\n\nsince difference of the squares is 8\n\n\n\n8 :sqrt: (4-6k) = 8\n\n:sqrt: (4-6k) =1\n\n4-6k = 1\n\nk=1/2, and the two roots are 3 and 1 [/hide]", "Solution_2": "Yes, this is correct.\n\n\n\nNow try to find another way to do this.\n\n\n\nHint: [hide]Sum of roots and Product of roots[/hide]", "Solution_3": "I did it that way the first time, but if it's x^2-4x+6k, you know that it is in the form of (x-2)^2 or (x-3)(x-1), and the latter is the only one with diff of squares as 8.\r\n\r\nI like the viete-ish method better, though, and I'll let someone else show it", "Solution_4": "Nice one.\r\n\r\nits works if the roots are rational. like if they aren't... but in this problem, they are.", "Solution_5": "here it is, i think\n\n[hide]\n\n\n\ndiff of squares = 8. call roots m, n\n\n\n\n(m-n)(m+n) = 8\n\n4 (m - n) = 8\n\nm - n = 2\n\n\n\nnow to find m - n...\n\n(m-n) :^2: = m :^2: - 2mn + n :^2:\n\n(m-n) :^2: = (m+n) :^2: - 4mn\n\n(m-n) :^2: = 16 - 4(6k)\n\n\n\nm-n = 2 :sqrt: (4 - 6k)\n\n2 = 2 :sqrt: (4 - 6k)\n\n1= 4 - 6k\n\nk= 1/2\n\n[/hide]", "Solution_6": "Nice.", "Solution_7": "you know that $m+n=4$ (sum of roots) and $m-n=2$ so you just solve the system and get $k$ through the product mn.", "Solution_8": "Please do not reveal old threads. \r\n\r\nI don't mind this happening if the problem wasn't answered by anyone but like in this case, please don't.\r\n\r\nLocked." } { "Tag": [ "inequalities", "geometric inequality", "inequalities proposed" ], "Problem": "Hello,\r\n\r\nLet $x_1,x_2,\\cdots,x_n$ for $n\\geq 3$ be positive real numbers. Prove that:\r\n\\[\r\n{{1}\\over{x_2+x_3+\\cdots+x_n}}+{{1}\\over{x_1+x_3+\\cdots+x_n}}\r\n+\\cdots+{{1}\\over{x_1+x_2+\\cdots+x_{n-1}}}<\\left({{x_1+x_2+\\cdots+x_n}\\over{x_1x_2\\cdots x_n}}\\right)^{{{1}\\over{n-1}}}.\r\n\\]", "Solution_1": "Use the arithmetic geometric inequality for the numerators in the left hand side and then the pOWER mean inequality. I do not have a pper, but I think it works.", "Solution_2": "just as he said it works !\r\nreally:)", "Solution_3": "Well, guess where I have found it: in Komal Contest set of problems for September! So, be careful, rules are very strict!" } { "Tag": [ "inequalities", "logarithms", "blogs", "function", "calculus" ], "Problem": "Consider the following series:\r\n\r\n$S_n = 1+\\frac{1}{2}+\\cdots+\\frac{1}{n}$\r\n\r\n$T_n = S_1+S_2+\\cdots+S_n$\r\n\r\n$U_n = \\frac{T_1}{2}+\\frac{T_2}{3}+\\cdots+\\frac{T_n}{n+1}$.\r\n\r\nProve that $T_n+\\ln{(n+1)} > U_n+n$.\r\n\r\n[hide=\"hint\"]Reading the most recent post of my [url=http://wangsblog.com/jeffrey/]blog[/url] will probably help a lot... [/hide]", "Solution_1": "Let $F : \\mathbb{N} \\rightarrow \\mathbb{N}$, the function $F(n) = T_n + ln(n+1) > U_n + n$.\r\n\r\n[b]Lemma 1.[/b] $T_n = (n+1)(S_{n+1} - 1)$. Proof. In $(n+1)(1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n+1})$ we overcount the number of 1's by 1, $\\frac{1}{2}$'s by 2, and the number of $\\frac{1}{k}$'s by k, so in total we overcounted by 1, $n+1$ times. Result follows.\r\n\r\n--\r\n\r\n[b]Lemma 2.[/b] $F(n+1) - F(n) > 0$. Its equivalent to\r\n\r\n$(T_{n+1} - T_n) + (\\ln (n+2) - \\ln (n+1) ) + (U_n - U_{n+1}) + (n - (n+1)) > 0$.\r\n\r\n$\\Leftrightarrow S_{n+1} + \\ln \\frac{n+2}{n+1} > \\frac{T_{n+1}}{n+2} + 1$\r\n\r\nand using lemma 1, $\\Leftrightarrow \\ln \\frac{n+2}{n+1} > S_{n+2} - S_{n+1}$\r\n\r\nor $\\Leftrightarrow \\ln (\\frac{n+2}{n+1})^{n+2}) > 1$, but $(\\frac{n+2}{n+1})^{n+2} > e$ where the result follows. (e is the limit when n gets big, and the function is trivially decreasing using calculus)\r\n\r\n-----------\r\n\r\nWe can verify $F(1) > 0$. Then $F(n+1) > F(n) > \\dots > F(1) > 0$, for all $n$. Result follows." } { "Tag": [ "LaTeX" ], "Problem": "Can someone tell me what does \\null command do? Some example?", "Solution_1": "\\null is a TeX command and is shorthand for \\hbox{} which is an empty horizontal box. You'll find the details on this in Donald Knuth's original TeX manual, [url=http://www.amazon.co.uk/TeXbook-Donald-E-Knuth/dp/0201134489/ref=sr_1_1?ie=UTF8&s=books&qid=1240909879&sr=8-1]The TeXbook[/url]. It is not easy reading and is really for those who want to dig deep into the system.\r\n\\hbox{} is also mentioned in Victor Eijkhout's freely available book [url=http://www.eijkhout.net/texbytopic/texbytopic.html]TeX by Topic[/url] which you can download from there. I notice that Victor was one of the people who replied to one of your queries on comp.text.tex." } { "Tag": [ "linear algebra", "matrix", "vector", "real analysis", "real analysis unsolved" ], "Problem": "(a) if $ H(p) \\equal{} \\frac {1}{r}|p|^r$ show $ L(q) \\equal{} \\frac{1}{s}|q|^s$\r\nis the legendre transform of H when $ \\frac {1}{r} \\plus{} \\frac {1}{s} \\equal{} 1$\r\n\r\n(b) if $ H(p) \\equal{} \\frac {1}{2}\\sum_{i,j \\equal{} 1}^n a_{ij}p_ip_j \\plus{} \\sum_{i \\equal{} 1}^n b_ip_i$ where $ A \\equal{} (a_{ij})$ is symm positive def matrix and b is a vector in R^n what is $ L \\equal{} H^*$?\r\n\r\n(c) define $ \\partial H (p) \\equal{} \\{ q: H(r)\\ge H(p) \\plus{} q\\cdot (r \\minus{} p) \\forall r\\in \\mathbb{R} \\}$ and show\r\n$ q \\in \\partial H(p)$ iff $ p \\in \\partial H^* (q)$ iff $ p\\cdot q \\equal{} H(p) \\plus{} H^*(q)$\r\n\r\ni hate you hopf-lax!", "Solution_1": "I can't understand the text. Could you rewrite it?" } { "Tag": [ "inequalities", "LaTeX" ], "Problem": "can someone give a nive solution to these inequilities please.\r\n\r\n1) $ \\frac{(x\\minus{}2)(x\\plus{}3)(x\\minus{}4)}{(x\\minus{}1)}<0$\r\n\r\n2) $ \\frac{2}{x\\plus{}5}>x\\plus{}4$\r\n\r\n3) $ \\frac{x(x\\minus{}2)}{2x\\minus{}5}\\geq3$\r\n\r\n4) $ \\frac{x(x\\plus{}5)}{x\\minus{}4}>\\minus{}2$\r\n\r\n5) $ \\frac{x\\plus{}3}{x\\minus{}1} \\geq \\frac{x\\minus{}1}{x\\plus{}4}$\r\n\r\n6) $ \\frac{x\\minus{}1}{x\\minus{}3}<\\frac{x\\plus{}2}{x\\minus{}5}$\r\n\r\n\r\ni suck at inequilities.", "Solution_1": "I have never solve inequations of root more than 1. But i think I know.....\r\n \r\n1)$ \\frac {(x - 2)(x + 3)(x - 4)}{(x - 1)} < 0$ for all $ x\\neq{1}$\r\n\r\n$ x_1 < 2$\r\n$ x_2 < - 3$\r\n$ x_3 < 4$\r\n\r\n2)$ \\frac {2}{x + 5} > x + 4$ for all $ x\\neq{ - 5}$\r\n$ 2 > (x + 4)(x + 5)\\Leftrightarrow{2 > x^{2} + 9x + 20}$\r\n$ 0 > x^{2} + 9x + 18\\Leftrightarrow{0 > (x + 3)(x + 6)}$\r\n\r\n$ x_1 < - 3$\r\n$ x_2 < - 6$", "Solution_2": "[quote=\"Elio (n)\"]I have never solve inequations of root more than 1. But i think I know.....\n \n1)$ \\frac {(x - 2)(x + 3)(x - 4)}{(x - 1)} < 0$ for all $ x\\neq{1}$\n\n$ x_1 < 2$\n$ x_2 < - 3$\n$ x_3 < 4$\n\n2)$ \\frac {2}{x + 5} > x + 4$ for all $ x\\neq{ - 5}$\n$ 2 > (x + 4)(x + 5)\\leftrightarrow{2 > x^{2} + 9x + 20}$\n$ 0 > x^{2} + 9x + 18\\leftarrow{0 > (x + 3)(x + 6)}$\n\n$ x_1 < - 3$\n$ x_2 < - 6$[/quote]\r\n\r\nSay what?\r\n\r\nbinomial, have you heard of a \"sign-chart\", they're used sometimes to solve inequalities with lots of roots which get messy, Try drawing one up for each problem,\r\n\r\nIf you haven't, post again and I'll draw you up one.\r\n\r\nEDIT: Never mind, here is an example of a relatively simple one from my C1 notes\r\n\r\n\r\nSolve $ (x-4)(x-1) > 0$\r\n\r\n\\begin{tabular}{|c\\parallel{}c|c|c|c|c|}\n\\hline\n& & CV & & CV & \\\\ \\hline\nx & 0 & 1 & 3 & 4 & 5 \\\\ \\hline \\hline\nx-4 & - & - & - & 0 & + \\\\ \\hline\nx-1 & - & 0 & + & + & + \\\\ \\hline \\hline\nP & + & 0 & - & 0 & +\\\\\n\\hline\n\\end{tabular}\r\n\r\nFrom this it is clear that when the product is positive, it evaluates to a number greater than $ 0$ and therefore, we can say that, $ x<1 \\mbox{ or } x>4$\r\n\r\nFor the record, that was just an excuse to show off some latex which I think look cool", "Solution_3": "$ 1)$\r\n\r\nThe roots are at $ \\minus{}3, 2, 4$. It is undefined at $ 1$. \r\n\r\nCase $ x < \\minus{}3$: all results will be greater than 0. \r\nCase $ \\minus{}3 < x < 1$: All results will be less than 0. \r\nCase $ 1 < x < 2$: All results will be greater than 0. \r\nCase $ 2 < x < 4$: All results will be less than 0. \r\nCase $ x > 4$: All results will be greater than 0. \r\n\r\nHence the solution is $ \\minus{}3 < x < 1 OR 2 < x < 4$. \r\n\r\nFor the rest of the equalities, solve them similarly. It may help to put all the terms onto one side into one fraction.", "Solution_4": "Yeah, I agree with Zhero...\r\nBut when you move it over, subtract instead of cross-multiplying\r\nCross-multiplying gives you a wrong answer.\r\nYou first find when the term=0\r\nYou get x=-3, 1, 2, 4\r\nYou just make a number line\r\n\r\n<-----|-----------|------|----------|---->\r\nWhere you plot -3, 1, 2, and 4\r\n\r\n\r\nAnd you can plug in a number from each interval to check...\r\nJust do the same with the rest.\r\nRemember when subtracting to put it all on a common denominator" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Show that $\\forall x,y,z \\geq 0$ we have:\r\n$ \\sqrt{xy}+ \\sqrt{yz} + \\sqrt{zx} \\geq 2 \\sqrt[4]{xyz(x+y+z)}$. \r\nWhen does the equality take place?", "Solution_1": "Put a 2 =yz b 2 =xz c 2 =xy and the inequality becomes\r\n\r\n\r\n (a+b+c)^4 \\geq 16((ab) 2 +(bc) 2 +(ca) 2 )\r\n\r\nsice it is homogeneus you can choose a+b+c=1 and so we have\r\n\r\n (ab) 2 +(bc) 2 +(ca) 2 \\leq 1/4 2 \r\n\r\nthis is the well known HARAZI'S INEQUALITY already discussed many times in this forum.", "Solution_2": "It seems that in a short period I will be able to write an article with the applications of so called Harazi inequality. It's funny." } { "Tag": [ "geometry" ], "Problem": "Congratulations!", "Solution_1": "YAYAYAYAYYAYAYAYAY!\r\ngo cynthia! :D", "Solution_2": "[quote=\"turak\"]YAYAYAYAYYAYAYAYAY!\ngo cynthia! :D[/quote]\r\n\r\nYou are next. :)", "Solution_3": "wow ty and IT WOULD BE SO AWESOME IF WE ALL MET UP AT MOP NEXT YEAR!!!!!!!!", "Solution_4": "yeah! (not going to happen) :D", "Solution_5": "YAY Cynthia!!!\r\n\r\nSorry, I was in Wyoming. I would have congratulated you sooner.\r\n\r\nMaybe if I learn how to solve not-so-challenging geometry problems I'll see you at MOP in 2010. Maybe turak too if she deletes her parenthetical.", "Solution_6": "you're late :P\r\n\r\nturak refuses to delete her parenthetical\r\n\r\nturak hates those geometry problems too :(" } { "Tag": [ "calculus", "integration" ], "Problem": "How many integers $ x$ satisfy the equation\r\n\\[ (x^2 \\minus{} x \\minus{} 1)^{x \\plus{} 2} \\equal{} 1\r\n\\]$ \\textbf{(A)}\\ 2 \\qquad \\textbf{(B)}\\ 3 \\qquad \\textbf{(C)}\\ 4 \\qquad \\textbf{(D)}\\ 5 \\qquad \\textbf{(E)}\\ \\text{none of these}$", "Solution_1": "[hide=\"Solution\"]Suppose $ x^2\\minus{}x\\minus{}1\\equal{}1$\nWe get solutions $ x\\equal{}2$ and $ x\\equal{}\\minus{}1$.\nIf instead $ x^2\\minus{}x\\minus{}1\\equal{}\\minus{}1$, we get $ x\\equal{}1$ and $ x\\equal{}0$. But $ x\\equal{}1$ is not a valid solution because it would make the expression $ \\minus{}1$ instead of $ 1$.\nFinally, suppose $ x\\plus{}2\\equal{}0$. We get the solution $ x\\equal{}\\minus{}2$.\nSo there are $ \\boxed{\\textbf{(C)}\\ 4}$ solutions, $ x\\equal{}\\minus{}2,\\minus{}1,0,2$.[/hide]" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c,d$ be positive real numbers such that $ ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd \\equal{} 6$. Prove that\r\n\\[ \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2} \\ge 1\r\n\\]\r\n:)", "Solution_1": "[quote=\"can_hang2007\"]Let $ a,b,c,d$ be positive real numbers such that $ ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd \\equal{} 6$. Prove that\n\\[ \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2} \\ge 1\n\\]\n:)[/quote]\r\n\r\nAnother my nice inequality (but easy) \r\nLet $ a,b,c,d$ be positive real numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 4$. Prove that\r\n\\[ 1\\ge \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2}\r\n\\]", "Solution_2": "[quote=\"zaizai-hoang\"][quote=\"can_hang2007\"]Let $ a,b,c,d$ be positive real numbers such that $ ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd \\equal{} 6$. Prove that\n\\[ \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2} \\ge 1\n\\]\n:)[/quote]\n\nAnother my nice inequality (but easy) \nLet $ a,b,c,d$ be positive real numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 4$. Prove that\n\\[ 1\\ge \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2}\n\\]\n[/quote]\r\n yes,exactly it's easy .\r\n$ \\sum \\frac {1}{((1 \\plus{} \\frac {1}{a^2})^2} \\equal{} \\sum \\frac {a^4}{(1 \\plus{} a^2)^2} \\leq \\frac {1}{4} \\sum 4a^2 \\equal{} 1$", "Solution_3": "[quote=\"Hong Quy\"][quote=\"zaizai-hoang\"][quote=\"can_hang2007\"]Let $ a,b,c,d$ be positive real numbers such that $ ab \\plus{} ac \\plus{} ad \\plus{} bc \\plus{} bd \\plus{} cd \\equal{} 6$. Prove that\n\\[ \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2} \\ge 1\n\\]\n:)[/quote]\n\nAnother my nice inequality (but easy) \nLet $ a,b,c,d$ be positive real numbers such that $ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} d^2 \\equal{} 4$. Prove that\n\\[ 1\\ge \\frac {1}{\\left(1 \\plus{} \\frac {1}{a^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{b^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{c^2}\\right)^2} \\plus{} \\frac {1}{\\left(1 \\plus{} \\frac {1}{d^2}\\right)^2}\n\\]\n[/quote]\n yes,exactly it's easy .\n$ \\sum \\frac {1}{((1 \\plus{} \\frac {1}{a^2})^2} \\equal{} \\sum \\frac {a^4}{(1 \\plus{} a^2)^2} \\leq \\frac {1}{4} \\sum 4a^2 \\equal{} 1$[/quote]\r\n\r\nOMG, I don't know it is very easy like that :P" } { "Tag": [ "calculus", "integration" ], "Problem": "Determine the number of positive integral solutions of the equation\r\na2-7a+b2-7b+2ab=0. Sorry, the 2's are suppose to be squares.", "Solution_1": "[quote=\"king_23\"]Sorry, the 2's are suppose to be squares.[/quote]\r\n\r\nFor exponents, you can use the ^ sign. Alternatively, you can also type your expression in LaTeX.", "Solution_2": "[hide]\n$a^2 + 2ab + b^2 - 7(a+b)=(a+b)(a+b) - 7(a+b) = (a+b-7)(a+b)=0$\n\nThe factor $(a+b)$ will never be $0$ for positive integer values of $a$ and $b$, so we only look at the factor $(a+b-7)$\n\nThe $6$ ordered pair of solutions are then $(6, 1)$, $(5, 2)$, $(4, 3)$ and the pairs where the order is reversed.\n[/hide]" } { "Tag": [ "\\/closed" ], "Problem": "i just went to memberlist to look up someone and:\r\n\r\nhttp://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=29314\r\n\r\n :P", "Solution_1": "Something similar happened earlier this week. It should be fixed soon.", "Solution_2": "But he's a moderator :!:", "Solution_3": "[quote=\"#H34N1\"]But he's a moderator :!:[/quote]\r\n\r\nHe's not a mod, he just editted his avatar so it says \"Moderator.\" \r\n\r\nThis is just a glitch with the post counts.", "Solution_4": "No, he did not change his avatar, that's part of the bug.\r\nBut it should be fixed soon, as I don' know if it may be harmful (I think not, but who knows...).", "Solution_5": "[quote=\"ZetaX\"] as I don' know if it may be harmful (I think not, but who knows...).[/quote]\r\n\r\nHe shouldn't be, as he's not on the [url=http://www.artofproblemsolving.com/Forum/memberlist.php?level=2]mod list[/url].", "Solution_6": "He only has 2 posts. I doubt he is hacking; hackers don't usually contribute to the forum and then hack. (I think.)\r\nThis is odd though. \r\n :| :huh:", "Solution_7": "Moderators don't have any *real* power, anyways. I honestly don't know why people think they do.", "Solution_8": "That's 1=2, we can just ask him about it? :huh: (1=2's not a moderator, is he?)", "Solution_9": "[quote=\"azjps\"]That's 1=2, we can just ask him about it? :huh: (1=2's not a moderator, is he?)[/quote]\r\n\r\nGood guess. He is my dad, who is not a hacker.\r\n\r\nShould be a glitch. He only has a couple hundred posts, not a couple million.\r\n\r\nAnd he is indeed a moderator.", "Solution_10": "The memberlist display doesn't affect anything. For some reason, some moderation process puts it off count, however it can always be quickly reevaluated by actually counting one's post so don't worry too much about it for now.", "Solution_11": "Found the reason, case closed." } { "Tag": [ "\\/closed" ], "Problem": "When I rate someone else's post, before I do, the red X which cancels the rating is on the screen, but when I rate, it disappears. Shouldn't it be the other way around? And also, it disappears only for that person.", "Solution_1": "A few observations:\r\n\r\n1) You can't click on the red X if you didn't vote (it looks faded). As to be expected.\r\n\r\n2) If you opened the topic without rating a particular post (regardless if you have voted before and retracted, or not), and then vote on that post, the red X will disappear.\r\n\r\n3) You can still change you vote while on topic, but the X will not reappear.\r\n\r\n4) If you opened a topic having rated a particular post, the red X should appear, not appearing faded. Clicking it retracts the vote, and the X will again look faded. However, this time if you revote, the X will not disappear.", "Solution_2": "Thanks! So if I want to cancel rating, I should get out of and then get in the topic to do so..." } { "Tag": [], "Problem": ":D Please Help- I'm not sure if this is the right forum- I have a problem that I think that I have worked out and I would like to see the equation for. The question is- assuming that a NC license plate has Three letters, and four numbers, how many possible combinations are there? I have come up with 6,580,454,400. Is this correct? If not, please explain. Thanks- Mike", "Solution_1": "$26^3\\cdot 10^4\\cdot\\binom{7}{3}=\\boxed{6,151,600,000}$", "Solution_2": "It depends whether letters and/or numbers can repeat, and whether you have to have a letter or number in a specific spot or not. I seem to get that if both numbers and letters can repeat and you don't have to have a number or letter in specific spot (eg: letter, number, letter, ...) there are $6,151,600,000$ different license plates.", "Solution_3": "What's the 7 choose 3 for?\r\n\r\nEDIT: Ahhh, for the order? But still, why 7 and why 3?\r\nEDIT2: Is that because you'll have $\\frac{7!}{3!4!}$ Meaning you have 7 slots, 3 to choose letters, 4 to choose numbers?", "Solution_4": "[quote=\"WindSlicer\"]EDIT2: Is that because you'll have $\\frac{7!}{3!4!}$ Meaning you have 7 slots, 3 to choose letters, 4 to choose numbers?[/quote]Yes.\r\n\r\n500th post!! :!: :D :D :thumbup: :clap: :clap2:", "Solution_5": "I did 9P4 and got 3,024\r\nthen 26P3 adn got 15,600\r\nthen mulitplyed and got 47,174,400\r\n\r\nIf im wrong tell me why :?", "Solution_6": "It depends whether the character that looks like a big circle is a 0 (a number) or a O (a letter) or whether there are 2 separate characters or whether that character is omitted. Also, it matters whether licence plate characters always follow a particular pattern. (In New York, for example, we always have letter-letter-letter-number-number-number-number.)", "Solution_7": "chances are the question's asking for any license plate with three letters in the first 3 slots and then 4 numbers in the last 4 slots. Then the answer would be:\r\n\r\n26^3*10^4=175760000", "Solution_8": "[quote=\"ineedmathhelp\"]chances are the question's asking for any license plate with three letters in the first 3 slots and then 4 numbers in the last 4 slots. Then the answer would be:\n\n26^3*10^4=175760000[/quote]\r\n\r\nBut you also have to multiply that by 7!/(3!x4!) Damn I need some latex....anyway, its because you have 7 slots, 3 letters, 4 numbers.", "Solution_9": "[quote=\"Iversonfan2005\"][quote=\"ineedmathhelp\"]chances are the question's asking for any license plate with three letters in the first 3 slots and then 4 numbers in the last 4 slots. Then the answer would be:\n\n26^3*10^4=175760000[/quote]\n\nBut you also have to multiply that by 7!/(3!x4!) - I need some latex....anyway, its because you have 7 slots, 3 letters, 4 numbers.[/quote]\r\n\r\nBut Iversofan2005 just said that he is assuming that all license plates in NC follow a general pattern: lllnnnn." } { "Tag": [ "\\/closed" ], "Problem": "It won't let me log off. BTW I have remember password.", "Solution_1": "Do you want to give some more detail? Like what happens when you try?\r\n\r\nand *please* can you stop using caps (subject of this thread). Its considered extremely rude..", "Solution_2": "I click logout and it won't let me log out because after i click i go back to the forum which is still signed in. I use Internet Explorer six and this also happened at the library (i hope no body goes to AOPS)." } { "Tag": [ "linear algebra", "matrix", "limit", "inequalities", "linear algebra unsolved" ], "Problem": "How do you prove that every matrix can be written as a limit of diagonalizable matrices ?\r\n\r\nthanks", "Solution_1": "How do you write $\\begin{pmatrix}0 &1\\\\ 0 & 0\\end{pmatrix}$ as a limit of diagonalizable matrices?", "Solution_2": "$\\left( \\begin{matrix}a & 1 \\\\ 0 &-a \\end{matrix}\\right)$, where $a \\to 0$.\r\n\r\nIt has been discuseed on the forum before; one needs only to show that the matrices with distinct eigenvalues are dense (via Schur's Theorem).\r\n[Supposing that the matrices are over $\\mathbb C$.]", "Solution_3": "OK, thanks for the reply, I try to prove it for triangular matrices.\r\n\r\nThe eigenvalues of a triangular matrix are the diagonal elements.\r\nIf they are all distinct then the matrix is diagonalizable and we are done. If not there are repeated elements and create the matrix $A_{\\epsilon}$ as follows :\r\n\r\n$[A_{\\epsilon}]_{ij}= [A]_{ij}$ for $i \\neq j$ (leave the non diagonal elements unchanged) and add $\\epsilon$, $2 \\epsilon$... to each of the repeated elements. (for example if an element is repeated three times, instead of it put element + $\\epsilon$, element + $2 \\epsilon$ and element + $3 \\epsilon$ in the three places where this element was).\r\n\r\nNow clearly $A_{\\epsilon}$ is diagonalisable for $\\epsilon$ (except at most a finite number of $\\epsilon$) and $\\lim_{\\epsilon \\to 0}A_{\\epsilon}= A$\r\n\r\nDoes this proof work ? I'll try to use Schur to extend it in general" } { "Tag": [ "modular arithmetic", "inequalities", "number theory unsolved", "number theory" ], "Problem": "Let $ p$ be an odd prime. Prove that $ \\sum_{k=1}^{p-1}k^{2p-1}\\equiv \\frac{p(p+1)}{2}(\\text{mod}p^{2})$\r\n\r\n[hide=\"My solution (but I think I am wrong)\"]\n\nWe must prove $ 4\\sum_{k=1}^{p-1}k^{2p-1}\\equiv 2p \\pmod{p^{2}}$\n\n$ k^{2p-1}+(p-k)^{2p-1}\\equiv (2p-1)p\\cdot k^{2p-2}\\pmod{p^{2}}$\n\nso $ (2p-1)(p)(\\sum_{k=1}^{p-1}k^{2p-2})\\equiv 2p \\pmod{p^{2}}\\leftrightarrow \\sum_{k=1}^{p-1}k^{2p-2}\\equiv-2\\pmod{p^{2}}$\n\nBut I think the last equality is not true..[/hide]\r\n\r\nCan anyone help me to find my mistake and solve the problem?\r\nthx..\r\n\r\nSorry if this problem posted before..\r\n\r\nBtw why I can not hide my message?", "Solution_1": "[quote=\"Yosh...\"]Btw why I can not hide my message?[/quote]\r\n\r\ni had this problem sometimes before. i think it is because of using the ' symbol in the BBCode:[Hide=My solution (but I think [u]I'm[/u] wrong)]\r\n\r\n[Moderator edit: Yes, this is the reason. Edited first post.]", "Solution_2": "[hide=\"I hope my help is useful\"]\n\nActually $ 2\\sum_{k=1}^{p-1}k^{2p-1}=\\sum_{k=1}^{p-1}k^{2p-1}+(p-k)^{2p-1}$ and not $ 4\\sum_{k=1}^{p-1}k^{2p-1}=\\sum_{k=1}^{p-1}k^{2p-1}+(p-k)^{2p-1}$\n\nSo at the end the inequality you get is:\n\n$ \\sum_{k=1}^{p-1}k^{2p-2}\\equiv-1\\pmod{p}$ (and not $ \\pmod{p^{2}}$ because you have divided by p)\n\nBut this last inequality is true, do you see why?[/hide]", "Solution_3": "If i am true then it's a problem from Canada MO." } { "Tag": [ "geometry", "rectangle", "symmetry" ], "Problem": "Two vertices of a rectangle are $ (\\minus{}3,1)$ and $ (\\minus{}3, \\minus{}3)$. The line\nwith equation $ x \\equal{} 1$ is a line of symmetry for this rectangle. How many square units are in the area of the rectangle?", "Solution_1": "The width of the rectangle is the distance from (-3,1) to (-3,-3), which is 4. Since x=1 is the symmetry the length is twice the distance from x=-3 to x=1, or 8. So, the area is 4*8=32" } { "Tag": [ "USAMTS" ], "Problem": "The original statement of Problem 2 of Round 2 did not make explicit that c and d must be positive integers. The question has been restated to include this information. Thanks to those of you who pointed this out.", "Solution_1": "Also, can we use calculators for computation (ex. working out something*sqrt(2)), or does the entire solution have to be by hand? That is, is the purpose of the calculator/computer restriction just to prevent us from writing scroll/test programs, or are we supposed to, if our solution needs it, find sqrt(2) to some number of decimal places and then multiply by hand?", "Solution_2": "We think the question is unambiguous on this point: \"Without using a calculator or computer\".", "Solution_3": "Okay, sorry. Thanks.", "Solution_4": "number 4 says the quadrilateral must intersect 4 of the 9 points. does this mean EXACTLY 4 of the 9 points or can it intersect 4 or more points?\r\n\r\nmaybe I'm missing something.", "Solution_5": "Read the question carefully; we think it is unambiguous on this point.", "Solution_6": "Folks, you really *must stop* asking specific questions about these problems! The rules prohibit *any discussion* of any pending problem. This includes publicly asking specific questions about any pending problem!\r\n\r\nI think Mr. Rusczyk has been kind so far. But I also think he would be within the rules to disqualify any contestant who publicly asks a specific question.\r\n\r\nIf you have a question, please ask the contest administrators via PM. It is exceedingly rare that a USAMTS problem is sufficiently ambiguous to require clarification. So, chances are high you will be told \"the question is clear, as written\". If you still believe the question is unclear or ambiguous, include a statement, with your submission, clearly explaining your reasoning. Based on e-mails I have exchanged with the previous contest administrators and graders, I have every confidence the current graders will consider your explanation, and you could even earn a \"commendation\" if you find something the question-writers overlooked.\r\n\r\nI am only a coach, so I don't speak for USAMTS. But please preserve the integrity of the contest." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all the pairs of positive integers $ (a,b)$ such that $ a^2 \\plus{} b \\minus{} 1$ is a power of prime number $ ; a^2 \\plus{} b \\plus{} 1$ can divide $ b^2 \\minus{} a^3 \\minus{} 1,$ but it can't divide $ (a \\plus{} b \\minus{} 1)^2.$", "Solution_1": "it has been posted, search....(it is China TST but i don't know the year).", "Solution_2": "[quote=\"shoki\"]it has been posted, search....(it is China TST but i don't know the year).[/quote]\r\nExactly it's a problem in China TST 2007 but I don't see any solution. :maybe: This problem seem to be easy, however I can't solve it. :oops:", "Solution_3": "I'm really sorry :( but anybody has solution for it? :)", "Solution_4": "i\u00b4ve solved a very similar problem at st. petesburgs olympiad, but i don\u00b4t remember the year... :|" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "If $ x,y,z$ are positive reals such that $ x \\plus{} y \\plus{} z \\plus{} xyz \\equal{} 4$, prove that:\r\n$ \\frac{x}{\\sqrt{y\\plus{}z}} \\plus{} \\frac{y}{\\sqrt{z\\plus{}x}} \\plus{} \\frac{z}{\\sqrt{x\\plus{}y}} \\ge \\frac{x \\plus{} y \\plus{} z}{\\sqrt{2}}$.", "Solution_1": "[quote=\"Raja Oktovin\"]If $ x,y,z$ are positive reals such that $ x \\plus{} y \\plus{} z \\plus{} xyz \\equal{} 4$, prove that:\n$ \\frac {x}{\\sqrt {y \\plus{} z}} \\plus{} \\frac {y}{\\sqrt {z \\plus{} x}} \\plus{} \\frac {z}{\\sqrt {x \\plus{} y}} \\ge \\frac {x \\plus{} y \\plus{} z}{\\sqrt {2}}$.[/quote]\r\nSee here:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=127956" } { "Tag": [ "blogs" ], "Problem": "its awesome. if u like to play it, check out my blog", "Solution_1": "I really like to listen to Coltrane recordings of jazz saxophone. That's a great solo instrument.", "Solution_2": "I used to play Alto Sax for about 4 years. Great intrument. \r\n\r\nI LOVE KENNY G!", "Solution_3": "I don't play the saxophone right now, but I might start to play soon. My brother plays it. :)", "Solution_4": "I don't play sax at all. I play the violin.", "Solution_5": "I play the sax.", "Solution_6": "No sax! Piano.\r\n\r\nRofl, my friend likes a girl whos last name is sax...", "Solution_7": "I play.\r\n\r\n[quote=\"appillai\"]I used to play Alto Sax for about 4 years. Great intrument. \n\nI LOVE KENNY G![/quote]\r\n\r\nUg. No.", "Solution_8": "I play the bari sax. But in high school, I will switch to the tenor or alto sax. The bari is just too big", "Solution_9": "hmm, just curious, did u start on bari?", "Solution_10": "I play tenor, mainly but started on alto. I think most people (if not all) did at my school.", "Solution_11": "[quote=\"buzzer11\"]I play tenor, mainly but started on alto. I think most people (if not all) did at my school.[/quote].\r\n\r\nyeah, thats how it is at my school too. btw, very creative with ur signature." } { "Tag": [ "function", "analytic geometry", "geometry", "geometric transformation", "reflection", "graphing lines", "slope" ], "Problem": "Im stuck with this problem:\r\nIf p,r, and s are three different prime numbers greater than 2, and n= p x r x s, how many positive factors including 1 and n does n have?", "Solution_1": "well the only thing that will divide that product is combinations of products of the prime numbers that make it up. (i'm assuming you want a hint, not the solution) i'm not sure if it is clear of what i'm saying either.", "Solution_2": "[quote=\"Silentfyre\"]Im stuck with this problem:\nIf p,r, and s are three different prime numbers greater than 2, and n= p x r x s, how many positive factors including 1 and n does n have?[/quote]\r\n\r\nI suppose that the x is multiplictation. It's pretty easy to count the factos. There's also a formula where you add one two the exponents and multiply them together to get the number of factors (easily provable). The answer is (1+1)(1+1)(1+1) = 8.", "Solution_3": "I also have another problem that i need to see how to solve:\r\nh(t)= c - (d - 4t)square\r\n\r\nAt time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds are given by the function h above, in which c and d are positive constants, If the ball reached its maximium height of 106 at time t=2.5, what was the height, in feet, of the ball at time t=1?", "Solution_4": "[quote=\"Silentfyre\"]Im stuck with this problem:\nIf p,r, and s are three different prime numbers greater than 2, and n= p x r x s, how many positive factors including 1 and n does n have?[/quote]\r\n\r\n\r\n1)n\r\n2)1\r\n3)pr\r\n4)ps\r\n5)rs\r\n6)p\r\n7)r\r\n8)s\r\n\r\nprs is the same as n, and order doesn't matter..so 8.", "Solution_5": "[quote=\"Silentfyre\"]I also have another problem that i need to see how to solve:\nh(t)= c - (d - 4t)square\n\nAt time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds are given by the function h above, in which c and d are positive constants, If the ball reached its maximium height of 106 at time t=2.5, what was the height, in feet, of the ball at time t=1?[/quote]\r\n\r\nThe height was $70$.", "Solution_6": "[i][edit: Shobhit caught his mistake :P ][/i]\r\n\r\nFor the second one, from the information given we know that $h(0)=6$ and $h(2.5)=106$, that is:\r\n\r\n\\begin{eqnarray}\r\nh(0) &=& c-d^2 = 6\\\\\r\nh(2.5) &=& c-(d-10)^2 = (c-d^2)+20d-100= 106\r\n\\end{eqnarray}\r\n\r\nIf we plug $c-d^2=6$ into $(2)$ we get\r\n\r\n\\[6+20d-100=106\\Rightarrow d=10\\]\r\n\r\nand thus $c=106$. So we have $h(t)=106-(10-4t)^2$. Plug in $t=1$ and we get $h(1)=106-(10-4)^2=\\boxed{70.}$", "Solution_7": "[quote=\"joml88\"][i][edit: Shobhit caught his mistake :P ][/i]\r\n\r\n\r\n\r\nYes, actually it would be easier if you just plug in numbers. Just say that $n=105$ by letting $p=3,r=5,s=7$ so the factors you have are $105,1,5,3,7,21,35,15$.", "Solution_8": "Getting stuck with this one:\r\n\r\nIf k, n, x, and y are positive numbers satisfying x to the -4/3 = k to the -2 and y to the \r\n4/3 = n to the second, what is (xy) to the -2/3 in terms of n and k?\r\n\r\na. 1/nk\r\nb. n/k\r\nc. k/n \r\nd. nk\r\ne. 1", "Solution_9": "So we have $x^{-4/3}=k^{-2}$ and $y^{4/3}=n^2$ and we want $(xy)^{-2/3}$.\r\n\r\nThe first equation can be simplified as \r\n\r\n\\[x^{-4/3}=k^{-2}\\Rightarrow (x^{-2/3})^{2}=(k^{-1})^2\\Rightarrow x^{-2/3}=k^{-1}.\\] \r\n\r\nThe second \r\n\r\n\\[y^{4/3}=n^2\\Rightarrow (y^{2/3})^2=(n)^2\\Rightarrow y^{2/3}=n\\Rightarrow y^{-2/3}=n^{-1}.\\] \r\n\r\nThus \\[(xy)^{-2/3}=x^{-2/3}\\cdot y^{-2/3}=k^{-1}\\cdot n^{-1}=\\frac{1}{nk}.\\] [b]A[/b]", "Solution_10": "Last few questions:\r\n\r\nLet [x] be defined as [x]= x square - x for all values of x. If [a] = [a-2]. what is the value of a?\r\n\r\n\r\nIn the xy-coordinate plane, line m is the reflection of line L about the x-axis. If the slope of line m is -4/5, what is the slope of line L?\r\n\r\n\r\n\r\nI was also wondering how the graph of f would look like if the function f is defined by \r\nf(x) = x square + bx + c, where b and c are constants.\r\n\r\nThanks.", "Solution_11": "the graph of $ax^2+bx+c$ would be parabola (u-shaped thingie)\r\n\r\nand I think line L should have slope of $4/5$\r\n\r\nAnd I'm a little confused on the first but since $[a]=[a-2]$, then $a^2-a=(a-2)^2-(a-2)$, thus $a^2-a=a^2-4a+4-a+2$, so $4a=6$ and $a=3/2$." } { "Tag": [ "trigonometry" ], "Problem": "B\u03c1\u03b5\u03b9\u03c4\u03b5 \u03c4\u03bf \u03bc\u03b5\u03b3\u03b9\u03c3\u03c4\u03bf \u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c3\u03c4\u03b1\u03c3\u03b7\u03c2\r\n$ \\frac {[(4 \\plus{} \\sin x)y^4 \\plus{} 4y \\plus{} 3\\sin x \\plus{} 12]^2}{(2y^4 \\plus{} 4y \\plus{} 6)[(\\sin x \\plus{} 2)y^4 \\plus{} 3\\sin x \\plus{} 6]}$.\r\n\u039a\u03b1\u03bb\u03bf \u03a6\u03b8\u03b9\u03bd\u03bf\u03c0\u03c9\u03c1\u03bf \u03ba\u03b1\u03b9 \u03ba\u03b1\u03bb\u03b5\u03c2 \u03c3\u03c0\u03bf\u03c5\u03b4\u03b5\u03c2 \u03c3\u03c4\u03bf\u03c5\u03c2 \u03b5\u03c0\u03b9\u03c4\u03c5\u03c7\u03bf\u03bd\u03c4\u03b5\u03c2 \u03c4\u03bf\u03c5 forum", "Solution_1": "Sil \u03c6\u03c4\u03b9\u03b1\u03be\u03b5 \u03c4\u03bf typo \u03c4\u03bf\u03c5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03b7 \u03c0\u03c1\u03bf\u03c3\u03b8\u03b5\u03c4\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c3\u03c4\u03b7\u03bd \u03b1\u03b3\u03ba\u03c5\u03bb\u03b7 \u03c4\u03bf\u03c5 \u03b1\u03c1\u03b9\u03b8\u03bc\u03b7\u03c4\u03b7 \u03c4\u03bf $ \\+3sinx$ .Sorry me ta typo..............", "Solution_2": "Done ! \u03a0\u03c1\u03bf\u03c3\u03b8\u03ad\u03c4\u03c9 \u03ba\u03ac\u03c4\u03b9 \u03b1\u03ba\u03cc\u03bc\u03b1 . \u0392\u03c1\u03b5\u03af\u03c4\u03b5 \u03ba\u03b1\u03b9 \u03c4\u03bf \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf \u03c4\u03b7\u03c2 \u03c0\u03b1\u03c1\u03b1\u03c0\u03ac\u03bd\u03c9 \u03c0\u03b1\u03c1\u03ac\u03c3\u03c4\u03b1\u03c3\u03b7\u03c2 \u03ba\u03b1\u03b8\u03ce\u03c2 \u03ba\u03b1\u03b9 \u03b3\u03b9\u03b1 \u03c0\u03bf\u03b9\u03b1 \u03b6\u03b5\u03cd\u03b3\u03b7 $ (x,y)$ \u03b5\u03c0\u03b9\u03c4\u03c5\u03b3\u03c7\u03ac\u03bd\u03b5\u03c4\u03b1\u03b9 :wink:\r\n\u03a4\u03ad\u03bb\u03bf\u03c2 \u03b4\u03b5\u03af\u03be\u03c4\u03b5 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ y$ \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03ad\u03bd\u03b1 $ x$ \u03ce\u03c3\u03c4\u03b5 \u03bd\u03b1 \u03ad\u03c7\u03bf\u03c5\u03bc\u03b5 \u03b5\u03bb\u03ac\u03c7\u03b9\u03c3\u03c4\u03bf .", "Solution_3": "\u03a4o \u03b4\u03b9\u03cc\u03c1\u03b8\u03c9\u03c3\u03b1 \u03b1\u03bb\u03bb\u03ac \u03b7 \u03b1\u03bc\u03c6\u03b9\u03b2\u03bf\u03bb\u03af\u03b1 \u03b5\u03be\u03b1\u03ba\u03bf\u03bb\u03bf\u03c5\u03b8\u03b5\u03af \u03bd\u03b1 \u03b9\u03c3\u03c7\u03cd\u03b5\u03b9", "Solution_4": "\u0397 \u03ac\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b4\u03b5\u03bd \u03c3\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ x$ \u03c5\u03c0\u03ac\u03c1\u03c7\u03b5\u03b9 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc $ y$. \u03a3\u03bf\u03c5 \u03b6\u03b7\u03c4\u03ac\u03b5\u03b9 \u03bd\u03b1 \u03b4\u03b5\u03af\u03be\u03b5\u03b9\u03c2 \u03cc\u03c4\u03b9 \u03b3\u03b9\u03b1 \u03ba\u03ac\u03b8\u03b5 $ y$ \u03c5\u03c0\u03b1\u03c1\u03c7\u03b5\u03b9 $ x$ (\u03c3\u03af\u03b3\u03bf\u03c5\u03c1\u03b1 \u03cc\u03c7\u03b9 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc).", "Solution_5": "\u039d\u03b1\u03b9 \u03c0\u03c1\u03ac\u03b3\u03bc\u03b1\u03c4\u03b9 . \u0394\u03b5\u03bd \u03b5\u03bd\u03bd\u03bf\u03cd\u03c3\u03b1 \u03bc\u03bf\u03bd\u03b1\u03b4\u03b9\u03ba\u03cc . \u03a3\u03c5\u03b3\u03b3\u03bd\u03ce\u03bc\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03b5\u03bb\u03b5\u03b9\u03c0\u03ae \u03b4\u03b9\u03b1\u03c4\u03cd\u03c0\u03c9\u03c3\u03b7 . (\u0397 \u03bb\u03cd\u03c3\u03b7 \u03bc\u03bf\u03c5 \u03bc\u03bf\u03b9\u03ac\u03b6\u03b5\u03b9 \u03bc\u03b5 \u03c4\u03bf\u03c5 \u03ba .\u039c\u03c0\u03cc\u03c1\u03b7)", "Solution_6": "O.K. Rodolfe .Kanoyme peripoy ta idia me alles synarthseis kai briskv ta akrotata dixvs diaforiko logismo." } { "Tag": [ "integration", "real analysis", "limit", "real analysis unsolved" ], "Problem": "Suppose that $ f$ is integrable on $ [ \\minus{} \\pi, \\pi]$. Choose $ \\epsilon > 0$, can we find a $ \\delta$ so that $ |\\int_{ \\minus{} \\delta}^{\\delta}f\\ dx\\ | < \\epsilon$ ?", "Solution_1": "yes, absolute continuity of Lebesgue integral. \r\nWe can find $ \\delta > 0$ such that for all $ E \\subseteq [ \\minus{} \\pi,\\pi]$ with $ m(E) < \\delta$ it is $ \\int_E |f|dx < \\epsilon$", "Solution_2": "This is a more specific than absolute continuity and has a slightly easier proof than the general case.\r\n\r\nDefine $ g_{\\delta}(x)\\equal{}f(x)\\mathbf{1}_{[\\minus{}\\delta,\\delta]}(x)\\equal{}\\begin{cases}f(x)& x\\in[\\minus{}\\delta,\\delta]\\\\0&\\text{otherwise}\\end{cases}.$\r\n\r\nThen $ g_{\\delta}(x)\\to 0$ [i]a.e.[/i] as $ \\delta\\to 0.$ Since we also have that $ |g_{\\delta}|\\le |f|$ everywhere and $ |f|$ is integrable, then the Dominated Convergence Theorem applies and \r\n\r\n$ \\lim_{\\delta\\to0}\\int d_{\\delta}\\equal{}\\lim_{\\delta\\to 0}\\int_{\\minus{}\\delta}^{\\delta}f\\equal{}0.$" } { "Tag": [ "modular arithmetic" ], "Problem": "Prove whether or not $ n^x \\minus{} n$ is divisible by $ x$, for any n\r\n\r\nedit: composite number", "Solution_1": "$ n^x\\minus{}n$ is divisible by x, then x and n are relatively prime.", "Solution_2": "What do you mean? The simplest example occurs precisely when $ x | n$.", "Solution_3": "$ n^x\\minus{}n\\equal{}0 \\pmod{n}$ would also be true if x is prime !?\r\n(fermats little theorem)\r\n\r\n...\r\ni think this is what Bachukas meant", "Solution_4": "I seem to have read your problem wrong. Do you mean the following:\r\n\r\n[i]Do there exist composite $ x$ such that $ n^x \\equiv n \\bmod x$ for all $ n$?[/i]\r\n\r\nWhen $ (n, x) \\equal{} 1$, the answer is given by the [url=http://en.wikipedia.org/wiki/Carmichael_number]Carmichael numbers[/url]." } { "Tag": [ "inequalities", "trigonometry", "function", "inequalities proposed" ], "Problem": "Find all solutions $ (x,y)$ , satisfy\r\n\r\n$ 20\\text{sin} x \\minus{} 21 \\text{cos} x\\equal{}81y^2\\minus{}18y\\plus{}30$", "Solution_1": "hello, solving the given equation for $ y$ we get\r\n$ y_{1,2}\\equal{}\\frac{1}{9}\\pm\\frac{1}{9}\\sqrt{\\minus{}29\\plus{}20\\sin(x)\\minus{}21\\cos(x)}$ for real solutions it must be\r\n$ \\minus{}29\\plus{}20\\sin(x)\\minus{}21\\cos(x)\\geq0$, but we get $ \\minus{}29\\plus{}20\\sin(x)\\minus{}21\\cos(x)\\equal{}0$ by discussing the function\r\n$ f(x)\\equal{}\\minus{}29\\plus{}20\\sin(x)\\minus{}21\\cos(x)$, from here we get all real solutions as\r\n$ x_1\\equal{}\\minus{}\\arctan(\\frac{20}{21})\\plus{}\\pi\\plus{}2k_1\\pi$, $ y_1\\equal{}\\frac{1}{9}$\r\n$ x_2\\equal{}\\minus{}\\arctan(\\frac{20}{21})\\plus{}\\pi\\plus{}2k_2\\pi$, $ y_2\\equal{}\\frac{1}{9}$\r\nSonnhard.", "Solution_2": "no need to solve anything dr. :wink: \r\nJust observe that the RHS has minimum valu=29 whereas the LHS has maximum value=29=$ \\sqrt {20^2 \\plus{} 21^2}$." } { "Tag": [ "function", "algorithm", "search", "algebra", "polynomial", "logarithms" ], "Problem": "how about if we all put our ideas to find a formula for prime numbers.i know it is difficult but ,in my opinion it has an exact formula. \r\n\r\n [color=green]1[/color],[color=red]2[/color],[color=red]3[/color],4,[color=red]5[/color],6,[color=red]7[/color],8,9,10,[color=red]11[/color],12,[color=red]13[/color],14,15,16,[color=red]17[/color],18,[color=red]19,[/color]20,21,22,[color=red]23[/color],24,25,26,27,28,[color=red]29[/color],30,[color=red]31[/color],32,33,34,35,36,[color=red]37[/color],38,39,40,[color=red]41[/color],42,[color=red]43[/color],44,45,46,[color=red]47[/color],48,49,50,51,52,[color=red]53[/color],54,55,56,57,58,,,,,,\r\ni think 1 is also a prime number, but another kind of prime i will say why, i am sleepy now,", "Solution_1": "f(n) = [3n] ??\r\n\r\nmaybe off by 1 or so", "Solution_2": "... Good Luck with that\r\n\r\nTry $ f(n) = 6n \\pm 1$, although that doesn't get 2 as a prime, and it's not technically a function, but other than that :lol:", "Solution_3": "http://mathworld.wolfram.com/PrimeFormulas.html\r\nhttp://en.wikipedia.org/wiki/Formula_for_primes\r\n\r\nPlease try to remember that very smart people have been studying number theory for centuries, and if an easily-computable formula for the primes existed, someone would've found it before the creation of this topic. Stop being so arrogant to think that you don't need to look up any of the previous progress made on a given problem.\r\n\r\nAnd there are a lot of very good number-theoretic reasons [b]not[/b] to consider $ 1$ a prime. Among them, $ 1$ is a [url=http://mathworld.wolfram.com/Unit.html]unit[/url].", "Solution_4": "DiscreetFourierTransform, $ 6n \\pm 1$ is not always prime for $ n \\in \\mathbb{N}$. Try $ n=8$. However, all primes can provably be represented in that form. In other words, IF a number is prime, THEN it can be represented as $ 6n \\pm 1$, but not vice versa.\r\n\r\nA related result, finding an exact formula for the number of primes less than a given number, is easy... that is, once you've proven the Riemann Hypothesis, one of the seven hardest math problems known to mankind ;)", "Solution_5": "t0rajir0u i know what you are saying, i know that loads of smart people have been studying number theory for centuries,but that doesnt mean we have to stop searching, and lay back. i am not arrogant, and please try not to stop others from working on it,you dont have a 100% proof that primes doesnt have one exact formula do you ?\r\n\r\n[color=green]1[/color],[color=red]2[/color],[color=red]3[/color],4,[color=red]5[/color],6,[color=red]7[/color],8,9,10,[color=red]11[/color],12,[color=red]13[/color],14,15,16,[color=red]17[/color],18,[color=red]19,[/color]20,21,22,[color=red]23[/color],24,25,26,27,28,[color=red]29[/color],30,[color=red]31[/color],32,33,34,35,36,[color=red]37[/color],38,39,40,[color=red]41[/color],42,[color=red]43[/color],44,45,46,[color=red]47[/color],48,49,50,51,52,[color=red]53[/color],54,55,56,57,58,,,,,,\r\nne wayz this is my idea, i might be wrong but..\r\n\r\n1+1=[color=red]2[/color]\r\n2+3=[color=red]5[/color]\r\n3+4=[color=red]7[/color]\r\n5+6=[color=red]11[/color]\r\n6+7=[color=red]13[/color]\r\n8+9=[color=red]17[/color]\r\n9+10=[color=red]19[/color]\r\n11+12=[color=red]23[/color]\r\n14+15=[color=red]29[/color]\r\n15+16=[color=red]31[/color]\r\n18+19=[color=red]37[/color]\r\n21+22=[color=red]43[/color]\r\n23+24=[color=red]47[/color]\r\n26+27=[color=red]53[/color]\r\n............\r\n............\r\n............\r\n\r\ni think there is a pattern here, every prime is made from adding consecutive numbers before it, except 2, do you notice the pattern?", "Solution_6": "[quote=\"mathnerd314\"]DiscreetFourierTransform, $ 6n \\pm 1$ is not always prime for $ n \\in \\mathbb{N}$. Try $ n=8$. However, all primes can provably be represented in that form. In other words, IF a number is prime, THEN it can be represented as $ 6n \\pm 1$, but not vice versa.\n\nA related result, finding an exact formula for the number of primes less than a given number, is easy... that is, once you've proven the Riemann Hypothesis, one of the seven hardest math problems known to mankind ;)[/quote]\r\n\r\nNot quite correct, we can express all numbers as such: 6n+1, 6n+2, 6n+3, 6n+4, 6n+5 6n+6. Now, when testing for primes, we don't choose 6n+2 because that implies divisibility by 2. 6n+3 implies divisibility by 3. 6n+4 implies divisibility by 2 again. And 6n+6 is just 7n, divisibility by 7. The one ones who survived that test are 6n+1 and 6n+5, and the 6n+5 for some integer n, is 6n-1 for some other integer n. That is why 6n +/- 1 works alright. We can easily get better functions with a larger co efficient of n, but it is not very efficient at all. \r\n\r\nBy the way: Solving the Riemann Hypothesis only IMPLIES that an EFFICIENT algorithm exists, exploiting a result of the RH. Finding an efficient algorithm does not prove RH, and that infact ARE exact results for the Prime Counting Function, but they are not efficient.", "Solution_7": "[quote=\"binomial_4eva\"]i think there is a pattern here, every prime is made from adding consecutive numbers before it, except 2, do you notice the pattern?[/quote]\n\nAny odd number $ 2k+1$ can be trivially represented in the form $ k+(k+1)$. It doesn't mean anything. Instead of searching for a pattern for the primes $ p_{k}$ you're now searching for a pattern for the numbers $ \\frac{p_{k}-1}{2}$, which doesn't do anything to change the difficulty of the problem.\n\nHere's another example of what I'm talking about:\n\n$ 5 = \\sqrt{24+1}$\n$ 7 = \\sqrt{24 \\cdot 2+1}$\n$ 11 = \\sqrt{24 \\cdot 5+1}$\n$ 13 = \\sqrt{24 \\cdot 7+1}$\n\nAgain, instead of searching for a pattern for $ p_{k}$ you might now be tempted to search for a pattern for $ \\frac{p_{k}^{2}-1}{24}$. Doesn't change the difficulty of the problem. In fact, with these calculations alone you might be tempted to guess that the coefficients are themselves always prime. But\n\n$ 17 = \\sqrt{24 \\cdot 12+1}$\n\n[quote=\"binomial_4eva\"]you dont have a 100% proof that primes doesnt have one exact formula do you ?[/quote]\r\n\r\nDid you read the links I posted or not?\r\n\r\nThere [b]are[/b] exact formulas for the primes. Calculations with any of them are [b]less efficient than searching.[/b]", "Solution_8": "[quote=\"Gib Z\"]Not quite correct, we can express all numbers as such: 6n+1, 6n+2, 6n+3, 6n+4, 6n+5 6n+6. Now, when testing for primes, we don't choose 6n+2 because that implies divisibility by 2. 6n+3 implies divisibility by 3. 6n+4 implies divisibility by 2 again. And 6n+6 is just 7n, divisibility by 7. The one ones who survived that test are 6n+1 and 6n+5, and the 6n+5 for some integer n, is 6n-1 for some other integer n. That is why 6n +/- 1 works alright. We can easily get better functions with a larger co efficient of n, but it is not very efficient at all.[/quote]\n\nAgain, you have merely proven that IF a number is prime, THEN it satisfies that formula. Binomial_4eva's claim is that all numbers satisfying that formula are prime; that is, IF a number satisfies that formula, THEN it is prime. That claim is false. Again, try $ 6 \\cdot 8+1$.\n\nAnd you have a typo: 6n+6 is not 7n, it's 6(n+1).\n\n[quote=\"binomial_4eva\"]t0rajir0u i know what you are saying, i know that loads of smart people have been studying number theory for centuries,but that doesnt mean we have to stop searching, and lay back. i am not arrogant, and please try not to stop others from working on it,you dont have a 100% proof that primes doesnt have one exact formula do you?\n...\ni think there is a pattern here, every prime is made from adding consecutive numbers before it, except 2, do you notice the pattern?[/quote]\r\n\r\nWell, for a start, I notice that you conveniently skipped over 10+11=21 and 12+13=25. So it's obvious that this \"pattern\" doesn't always work. And by the way, do you know the reason the pattern seemed to work? When you add two consecutive numbers, you get an odd number. And all primes, \"except 2\", are odd.\r\n\r\nBy the way, we do have a 100% proof that you can't find a polynomial that always generates primes. Did you read the Wikipedia article?\r\n\r\nOops, t0rajir0u beat me to it.", "Solution_9": "[quote=\"mathnerd314\"]In other words, IF a number is prime, THEN it can be represented as $ 6n \\pm 1$, but not vice versa.[/quote] Unless it's 2 or 3, of course :wink: \n\n[quote=\"mathnerd314\"]A related result, finding an exact formula for the number of primes less than a given number, is easy... that is, once you've proven the Riemann Hypothesis, one of the seven hardest math problems known to mankind ;)[/quote]\n[quote=\"Gib Z\"]By the way: Solving the Riemann Hypothesis only IMPLIES that an EFFICIENT algorithm exists, exploiting a result of the RH. Finding an efficient algorithm does not prove RH, and that infact ARE exact results for the Prime Counting Function, but they are not efficient.[/quote]\r\n\r\nSo, some more correct information: there is a very good approximate formula for the number of primes less than $ x$ for large values of $ x$. This is given by the Prime Number Theorem. However, people are still interested in how large the error for that approximation is. The Riemann Hypothesis is equivalent to the \"best possible\" error term. (Why \"best possible\"? Because we're approximating something discrete, the distribution of the primes, with continuous functions -- there always has to be some error.) Thus, RH implies very strong conditions on the distribution of the primes.\r\n\r\nDeterministic, polynomial-time algorithms are known for primality testing. It was not proven that this was the case until 2002, by Agarwal et al. (working with some undergraduates, I believe). Their original algorithm ran in time 12th degree in the length of the prime -- this is polynomial, but very slow. They believed that if they Riemann Hypothesis was true, they could substantially increase the speed of the algorithm. More generally, any regularity condition on the distribution of the primes is likely to allow more efficient prime testing algorithms. However, it might be that the speed of the best algorithms for primality testing do not depend on RH to any significant degree.", "Solution_10": "[quote]Well, for a start, I notice that you conveniently skipped over 10+11=21 and 12+13=25. So it's obvious that this \"pattern\" doesn't always work. And by the way, do you know the reason the pattern seemed to work? When you add two consecutive numbers, you get an odd number. And all primes, \"except 2\", are odd.[/quote]\r\n\r\nmay be the formula has something that skips 10+11=21 and 12+13=25, who knows.\r\n\r\nby the way this is good to know that \"When you add two consecutive numbers, you get an odd number. And all primes, \"except 2\", are odd\" i didnt know this.", "Solution_11": "[quote=\"JBL\"]So, some more correct information: there is a very good approximate formula for the number of primes less than $ x$ for large values of $ x$. This is given by the Prime Number Theorem. However, people are still interested in how large the error for that approximation is. The Riemann Hypothesis is equivalent to the \"best possible\" error term. (Why \"best possible\"? Because we're approximating something discrete, the distribution of the primes, with continuous functions -- there always has to be some error.) Thus, RH implies very strong conditions on the distribution of the primes.[/quote]\n\nYes, I was just trying to explain everything in terms that binomial_4eva would understand. (For anyone interested, the prime number theorem gives an estimate of $ \\frac{x}{\\ln x}$ primes less than or equal to x. A few slightly better approximations exist.)\n\n[quote=\"JBL\"]Deterministic, polynomial-time algorithms are known for primality testing. It was not proven that this was the case until 2002, by Agarwal et al. (working with some undergraduates, I believe). Their original algorithm ran in time 12th degree in the length of the prime -- this is polynomial, but very slow. They believed that if they Riemann Hypothesis was true, they could substantially increase the speed of the algorithm. More generally, any regularity condition on the distribution of the primes is likely to allow more efficient prime testing algorithms. However, it might be that the speed of the best algorithms for primality testing do not depend on RH to any significant degree.[/quote]\n\nInteresting point. Does that mean that the Riemann Hypothesis is somehow inextricably related to P=NP?\n\n[quote=\"binomial_4eva\"]may be the formula has something that skips 10+11=21 and 12+13=25, who knows.[/quote]\r\n\r\nWhat's your point? As t0rajir0u stated, using formulas to find prime numbers is pointless. Why exactly would this \"formula\" skip those numbers? You have to invent something to explain that, and then you'll have to invent something else, ad infinitum.\r\n\r\nOh, and by the way, would somebody please edit binomial_4eva's first post, because it's stretching the forum tables. Thanks.", "Solution_12": "For those of you who actually want to know a function that generates primes, and are familiar with the floor function, heres one :\r\n\r\n$ \\pi(m) = \\sum_{j=2}^{m}\\left\\lfloor{(j-1)!+1 \\over j}-\\left\\lfloor{(j-1)! \\over j}\\right\\rfloor \\right\\rfloor.$", "Solution_13": "[quote=\"mathnerd314\"][quote=\"JBL\"]Deterministic, polynomial-time algorithms are known for primality testing. It was not proven that this was the case until 2002, by Agarwal et al. (working with some undergraduates, I believe). Their original algorithm ran in time 12th degree in the length of the prime -- this is polynomial, but very slow. They believed that if they Riemann Hypothesis was true, they could substantially increase the speed of the algorithm. More generally, any regularity condition on the distribution of the primes is likely to allow more efficient prime testing algorithms. However, it might be that the speed of the best algorithms for primality testing do not depend on RH to any significant degree.[/quote]\n\nInteresting point. Does that mean that the Riemann Hypothesis is somehow inextricably related to P=NP?[/quote]\r\n\r\nNo. The problem of primality testing is in P regardless of the status of the RH and P vs. NP. However, it is possible that the speed of the \"best possible\" polynomial deterministic algorithm for primality testing depends on the error term in the Prime Number Theorem, and thus on the RH (but this says nothing about NP).\r\n\r\nThe fate of factoring integers is more closely related to P vs. NP (although not so closely related as NP complete problems), but I don't know if the Riemann Hypothesis is thought to have any implications for easy factoring.", "Solution_14": "My deepest apologies i just realised what I posted before was the formula for the prime counting function. For a function that generates primes, look below:'\r\n\r\n$ f(n) = 2+(2n! \\,\\operatorname{mod}(n+1))$ for all non negative integer n.", "Solution_15": "I'm feeling very skeptical about this thread. You're not going to find a polynomial expression that returns a prime value for all integral inputs.", "Solution_16": "[quote=\"binomial_4eva\"]by the way this is good to know that \"When you add two consecutive numbers, you get an odd number. And all primes, \"except 2\", are odd\" i didnt know this.[/quote]\n\nI'm not sure how you'd expect to get anywhere with this problem without knowing that any prime greater than $ 2$ is odd. Even numbers are, by definition, divisible by $ 2$, so no even number greater than $ 2$ can be prime because it has a divisor that is not $ 1$ or itself. \n\nWhat kind of formula are you expecting, exactly? It's known that a polynomial one simply does not exist. If you're envisioning any kind of closed form in terms of elementary functions, then I suggest you [b]read the links.[/b]\n\n[quote=\"Gib Z\"]My deepest apologies i just realised what I posted before was the formula for the prime counting function. For a function that generates primes, look below:'\n\n$ f(n) = 2+(2n! \\,\\operatorname{mod}(n+1))$ for all non negative integer n.[/quote]\r\n\r\nIn the links I posted, as was the other one. And it doesn't really help, since this function returns $ 2$ for all composite $ n$, so it amounts to a (very slow) prime-checking algorithm.", "Solution_17": "what about this one:\r\n\r\nhttp://www.primenumbersformula.com/default.htm\r\n\r\nOr Ghandi's Formula that i've explained about it in my Blog before:\r\n\r\nhttp://www.mathlinks.ro/weblog_entry.php?t=131537\r\n\r\nand wow i didn't know about the formulas in wikipedia [b]t0rajir0u[/b] :) they are really cool, Prime numbers are really CooOl :lol:", "Solution_18": "Several points.\r\n\r\n1)I like to point that that there [b]are[/b] prime producing functions.\r\nSome are a consequence of \"Wilson's theorem\". Notice that this theorem tells us a necessary and sufficient condition for a number to be a prime. However, it is computationally impossible. For example, to see if $ n=101$ is prime we need to consider $ 100!$ which is really big, even for fast computers. So it really isnt efficient.\r\n\r\n2)One $ 1$ isn't a prime. One reason why not is that the uniqueness part of the fundamental theorem of arthimetic will be violated.\r\n\r\n3)There is [b]no[/b] non-constant polynomial to produces only primes. Try to prove it.\r\n\r\n4)Wait a minute! What about $ f(n)=n^{2}+n+41$ for $ n\\geq 0$? Doesnt that contradict #3 :rotfl:", "Solution_19": "I laughed when I read this:\r\n\r\n[quote=\"binomial_4eva\"]by the way this is good to know that \"When you add two consecutive numbers, you get an odd number. [/quote]\r\n\r\nIf you have two consecutive integers, one must be $ 0 \\text{ (mod 2)}$, and the other must be $ 1 \\text{ (mod 2)}.$ Adding them together yields another $ 1 \\text{ (mod 2)}$, which is also known as [b][u]odd[/u][/b].\r\n\r\nThat was just a fancy way of saying odd + even = odd, if you [u]still[/u] don't understand what I'm saying.", "Solution_20": "[quote=\"davidyko\"]I laughed when I read this:\n\n[quote=\"binomial_4eva\"]by the way this is good to know that \"When you add two consecutive numbers, you get an odd number. [/quote]\n\nIf you have two consecutive integers, one must be $ 0 \\text{ (mod 2)}$, and the other must be $ 1 \\text{ (mod 2)}.$ Adding them together yields another $ 1 \\text{ (mod 2)}$, which is also known as [b][u]odd[/u][/b].\n\nThat was just a fancy way of saying odd + even = odd, if you [u]still[/u] don't understand what I'm saying.[/quote]\r\n\r\nYeah, I typed that thinking that binomial_4eva would recognize it as tautological, but apparently not. I laughed when I read that too :D\r\n\r\n$ n^{2}+n+41$... nice :rotfl:. I'm told that $ x^{2}-79x+1601$ is more deceptive :D", "Solution_21": "Ok, I'll try to give some constructive comments.[b] [color=red]I expect you to read this[/color][/b].\r\n\r\nFirst, I see you're interested in math. That's good. Now you need a good place to start. Fortunately, the world is full of [b][color=blue]resources[/color][/b]. One of the things you'll need to realize is that it sometimes helps to start from the basics and build your way up. It's good that you have the interest going, but you won't get too far unless you understand the basics. Pick up a copy of [b][color=blue]AoPS 1[/color][/b]. Read, read, read. Read Wikipedia. [b][color=blue]Wikipedia [/color][/b]is a god of all sorts. Mathworld is great but it's a bit more technical.\r\n\r\nThe AoPS forum is an excellent resource when you learn how to use it. You need to get a feel of who knows what they're talking about, and [b][color=blue]if they try to help you, take it[/color][/b]. If someone gives you a link to read, it's for your own good to read it and try to understand it. You can start by going all the way back to the beginning of the thread and reading all of t0ra's links.\r\n\r\nNext, you need to get a feel of the mathematical method. It's a lot like the scientific method. There are a few keys.\r\n1. [b][color=blue]Relevance, relevance, relevance[/color][/b]. Anything you spend time on in mathematics must have some sort of relevance. It should either connect concepts you know, or work towards a focused goal in terms of a concept you don't know. Questions out of the blue are generally not worth the thought, unless you can think of some sort of relevance. This is by no means something easy; it's something you have to get a feel of. But know that.\r\n2. [b][color=blue]Identify[/color][/b]. When approached with an unknown question, you need to be able to find the root problem. Suppose I ask for the solution to $ (j+456)(x-1)^{19}+(j_{2}^{7}2)(x-1)^{3}+(j_{3}!)(x-1)=42$. What's wrong with that? It's ugly. All the factors with $ j$ in them are different, and there's no reason why you need to have $ x-1$ everywhere. The problem is $ ay^{19}+by^{3}+y=42$. THAT's the problem. If you do this, it will prevent you from asking unnecessary questions and wasting your time.\r\n3. [b][color=blue]Research.[/color][/b] Millions of mathematicians have walked the earth. What we know today is the result of their collective work, and what we don't know today is what none of them could figure out. Their work can help you.\r\n4. [b][color=blue]Build[/color][/b]. Start with easier things, and combine them to form harder ideas. There is a natural flow of things. You need to catch that drift.\r\n5. [b][color=blue]Question[/color][/b]. Is it true? Must it be true? Why is it true? What does it mean if it is true? What kinds of interpretations of it are there? What does it suggest/imply?\r\n6. [b][color=blue]Explore[/color]. [/b]You're already doing this, but without focus. When you want to work towards something, you really should have a goal. Especially in the beginning, when there are huge patches of void, empty space. As these spaces fill up, random work can cause you to run into something. Until then, [b][color=blue]stay focused[/color][/b].\r\n\r\nI invite you to get started. Best of luck.\r\n\r\n[color=green]On a side-note, I think it's time that this thread be locked, unless anyone else has any [b]constructive[/b] comments on calculating primes.[/color]", "Solution_22": "[quote=\"The Zuton Force\"]On a side-note, I think it's time that this thread be locked, unless anyone else has any constructive comments on calculating primes.[/quote]\n\nHmmm, I noticed that nobody has mentioned the Sieve of Eratosthenes (or however it's spelled... :blush:)\n\n[quote=\"Elemennop\"]f(n) = [3n] ?? \n\nmaybe off by 1 or so[/quote]\r\n\r\n:rotfl: I hope you meant that as a joke... because any integer can be expressed as \"3n off by 1 or so\".", "Solution_23": "[quote=\"sylow_theory\"]Some are a consequence of \"Wilson's theorem\". Notice that this theorem tells us a necessary and sufficient condition for a number to be a prime. However, it is computationally impossible. For example, to see if $ n=101$ is prime we need to consider $ 100!$ which is really big, even for fast computers. So it really isnt efficient.[/quote]Actually, one needn't ever calculate numbers larger than 10,000 while applying the Wilson's Theorem-based primality test to 101: since it's a calculation $ \\mod 101$, you just reduce after each multiplication and you'll never face numbers larger than $ (101-1)^{2}= 10000$. (If you use positives as well as negatives, I guess you can actually get this down to $ 50^{2}$.) The actual problem is the [i]number[/i] of computations necessary, not their size: if $ p$ is, say, 8 (or 200) digits long, the largest number one potentially faces is about 16 (or 400) digits -- not actually that big for a computer (or a very determined elementary schooler with good handwriting) to handle. The problem is that you need to make $ p$ multiplications -- when $ p = 101$, this is not so bad; when $ p$ is 8 digits, though, you're straining my PC for several minutes at least, and when $ p$ has 200 digits we don't stand a chance.\r\n\r\n[i]Post has been edited to remove some faulty information.[/i]", "Solution_24": "[quote=\"JBL\"]\nClose: his name was Erastothenes.[/quote]\r\n\r\nhttp://en.wikipedia.org/wiki/Eratosthenes\r\n\r\nFunny...I thought it was Erasthostenes for some time...", "Solution_25": "[quote=\"The Zuton Force\"]Ok, I'll try to give some constructive comments.[b] [color=red]I expect you to read this[/color][/b].\n\nFirst, I see you're interested in math. That's good. Now you need a good place to start. Fortunately, the world is full of [b][color=blue]resources[/color][/b]. One of the things you'll need to realize is that it sometimes helps to start from the basics and build your way up. It's good that you have the interest going, but you won't get too far unless you understand the basics. Pick up a copy of [b][color=blue]AoPS 1[/color][/b]. Read, read, read. Read Wikipedia. [b][color=blue]Wikipedia [/color][/b]is a god of all sorts. Mathworld is great but it's a bit more technical.\n\nThe AoPS forum is an excellent resource when you learn how to use it. You need to get a feel of who knows what they're talking about, and [b][color=blue]if they try to help you, take it[/color][/b]. If someone gives you a link to read, it's for your own good to read it and try to understand it. You can start by going all the way back to the beginning of the thread and reading all of t0ra's links.\n\nNext, you need to get a feel of the mathematical method. It's a lot like the scientific method. There are a few keys.\n1. [b][color=blue]Relevance, relevance, relevance[/color][/b]. Anything you spend time on in mathematics must have some sort of relevance. It should either connect concepts you know, or work towards a focused goal in terms of a concept you don't know. Questions out of the blue are generally not worth the thought, unless you can think of some sort of relevance. This is by no means something easy; it's something you have to get a feel of. But know that.\n2. [b][color=blue]Identify[/color][/b]. When approached with an unknown question, you need to be able to find the root problem. Suppose I ask for the solution to $ (j+456)(x-1)^{19}+(j_{2}^{7}2)(x-1)^{3}+(j_{3}!)(x-1)=42$. What's wrong with that? It's ugly. All the factors with $ j$ in them are different, and there's no reason why you need to have $ x-1$ everywhere. The problem is $ ay^{19}+by^{3}+y=42$. THAT's the problem. If you do this, it will prevent you from asking unnecessary questions and wasting your time.\n3. [b][color=blue]Research.[/color][/b] Millions of mathematicians have walked the earth. What we know today is the result of their collective work, and what we don't know today is what none of them could figure out. Their work can help you.\n4. [b][color=blue]Build[/color][/b]. Start with easier things, and combine them to form harder ideas. There is a natural flow of things. You need to catch that drift.\n5. [b][color=blue]Question[/color][/b]. Is it true? Must it be true? Why is it true? What does it mean if it is true? What kinds of interpretations of it are there? What does it suggest/imply?\n6. [b][color=blue]Explore[/color]. [/b]You're already doing this, but without focus. When you want to work towards something, you really should have a goal. Especially in the beginning, when there are huge patches of void, empty space. As these spaces fill up, random work can cause you to run into something. Until then, [b][color=blue]stay focused[/color][/b].\n\nI invite you to get started. Best of luck.\n\n[color=green]On a side-note, I think it's time that this thread be locked, unless anyone else has any [b]constructive[/b] comments on calculating primes.[/color][/quote]\r\n\r\nWow that was really nice :) i hope it'll help anyone in doing anything in mathematics... :) \r\n\r\nOOk the links that i gave are really nice read them, specially i like Ghandi's Formula:\r\n\r\n$ p_{k+1}=\\left\\lfloor 1-\\log_{2}\\left(-\\frac{1}{2}+\\sum_{d\\: |\\: P_{k}}\\frac{\\mu(d)}{2^{d}-1}\\right)\\right\\rfloor$\r\n\r\nwhere $ p_{k+1}$ is the $ k+1$'th prime and $ P_{k}=p_{1}p_{2}\\cdots p_{k}$ :roll:", "Solution_26": "[quote=\"K81o7\"][quote=\"JBL\"]\nClose: his name was Erastothenes.[/quote]\n\nhttp://en.wikipedia.org/wiki/Eratosthenes\n\nFunny...I thought it was Erasthostenes for some time...[/quote]\r\n\r\nHaha, that's what I get for being lazy -- I just put it into my search engine, assuming that if it was wrong it would suggest that I should have used the alternate spelling. (In fact, both the correct spelling and my (wrong) spelling must be fairly common.)", "Solution_27": "atleast i can make u all laugh with my stupidity :blush:", "Solution_28": "It's nice to think about math problems but this problem you are looking at may look deceptively easy, but it is very difficult (at least difficult enough to not belong in this forum :lol: ) ... \r\n\r\nIn any case, I'm locking this topic." } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Prove that if n \\geq 5, then the interval [n!, (n + 1)!] contains a multiple of n3.", "Solution_1": "It was equal to the conclusion that (n+1)! - n! >= n^3\r\nor n*n!>n^3\r\nIt can be prove by mathematical induction.", "Solution_2": "Something similar:\r\n\r\nProve that if n > 9 then there is a perfect cube between n and 3n.", "Solution_3": "if n = 10,between 10 and 30,27 is a cube.\r\nif n = k,m^3 is between k and 3k\r\n if ( k+1 != m^3 ){\r\n m^3 is between k+1 and 3(k+1)\r\n It is true.\r\n }\r\n if( k + 1 == m^3 ){\r\n ( m+ 1 )^3 > m^3 = k + 1\r\n ( m + 1 )^3 < 3(k+1)=3m^3 + 3 can be prove easliy\r\n }" } { "Tag": [ "pigeonhole principle" ], "Problem": "Hello everybody!!!\r\nCan someone help me with pegeons?\r\nWhat do you know about them?\r\nI feed them every day from my hand, so\r\nIs it safe? how about virus H5N1(ptichij grip)?\r\nThey are wild!", "Solution_1": "Do not keep your pigeon outside!! Ptiji grip is transported by birds wich are sick with Ptiji grip. Pigeon is a good creature. I like pigeons too. In math there is also a kind problem solving method called Pigeonhole Pricinple." } { "Tag": [], "Problem": "If $ f(x) \\equal{} 2$ for all real numbers $ x$, what is the value of $ f(x \\plus{} 2)$?", "Solution_1": "Since $ x$ is real, $ x\\plus{}2$ must also be real, meaning $ f(x\\plus{}2)\\equal{}\\boxed{2}.$", "Solution_2": "Its a troll.", "Solution_3": "yup, its a troll\n", "Solution_4": "bruh i was so confused when i was playing. i was like that must be rly hard. Then I come here and read it carefully :wallbash: " } { "Tag": [ "analytic geometry", "geometry", "number theory proposed", "number theory" ], "Problem": "Let n>=5 be a positive integer and a_1, b_1, a_2, b_2, . . . , a_n, b_n be integers satisfying the following two conditions :\r\n(i) the pairs (a_i, b_i) are all distinct for i=1, 2, . . ., n;\r\n(ii) |a_i*b_(i+1) - a_(i+1)*b_i | =1 for i = 1, 2, . . . , n, \r\nwhere (a_(n+1), b_(n+1)) = (a_1,b_1).\r\nShow that exits i, j with 1<= i, j <=n such that \r\n1 4 is useful since the conclusion does not hold for M_1(1,0), M_2(0,1), M_3(-1,0) and M_4(0,-1).\r\n\r\nI hope there are no 10000 subcases to study...\r\n\r\nPierre." } { "Tag": [ "limit", "induction", "algebra unsolved", "algebra" ], "Problem": "[u][i]Problem [/i][/u]\r\n\r\n [i] Let $ a ,b , A , B$ be positives real numbers and the sequence $ (x_{n})$ \n\n\n defined by : $ x_{1} \\equal{} a ; x _{2} \\equal{}b$\n\n\n $ x_{n\\plus{}1} \\equal{} A\\sqrt[3]{x_{n}^{2}} \\plus{} B\\sqrt[3]{x_{n\\minus{}1}^{2}}$ $ (n \\equal{}2,3,4,......)$ \n\n\n\n Find $ \\lim_{n\\rightarrow\\infty}x_{n}$\n\n\n\n\n Love An Forever\n\n Hu\u1ef3nh V\u00f5 Ph\u01b0\u01a1ng An[/i]", "Solution_1": "The limit is $ (A+B)^3$.\r\n\r\nLet $ (y_n)$ be the terms of $ (x_n)$ (in order) that are less than or equal to $ (A+B)^3$, and let $ (z_n)$ be the terms of $ (x_n)$ (in order) that are greater than or equal to $ (A+B)^3$. It is enough to show that each of these sequences converges to $ (A+B)^3$, if it has any elements.\r\n\r\nEvidently if $ (y_n)$ is not finite, it is bounded above by $ (A+B)^3$, and by simple induction, it is bounded below by the increasing sequence given by\r\n\\begin{align*}\r\nr_1 &=r_2 = \\min(y_1,y_2) \\\\\r\nr_{n+1} &= (A+B) \\sqrt[3]{r_{n-1}^2} , \\end{align*}\r\nwhich evidently converges to $ (A+B)^3$, since all its terms are positive. Therefore $ (y_n)$ converges to $ (A+B)^3$.\r\n\r\nSimilarly, $ (z_n)$ is not finite, then is bounded below by $ (A+B)^3$, and bounded above by the decreasing sequence given by\r\n\\begin{align*}\r\ns_1 &= s_2 = \\max(z_1, z_2) \\\\\r\ns_{n+1} &= (A+B) \\sqrt[3]{s_{n-1}^2}, \\end{align*}\r\nwhich also converges to $ (A+B)^3$. Hence $ (z_n)$ also converges to $ (A+B)^3$. Therefore $ (x_n)$ converges to $ (A+B)^3$, as desired. $ \\blacksquare$", "Solution_2": "[i] Wrong solution, I afraid\n\n Please try again :maybe: \n\n I think Patrick , Darij will help me :lol: [/i]\r\n\r\n [i]This problem is from VMO 1990 [/i]", "Solution_3": "Please show me where my solution is wrong. Perhaps I am misreading the problem?\r\n\r\nI do not see this problem on the [url=http://www.kalva.demon.co.uk/vietnam/viet90.html]1990 Vietnam Math Olympiad[/url] (is this what you meant?).", "Solution_4": "[i]Sorry\n \n\nIt ' s from Vietnam TST 1990[/i] :blush:", "Solution_5": "Ah! I see. :) \r\n\r\nWell, other people got the same answer as I did [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=716703#716703]here[/url], and using similar methods." } { "Tag": [ "symmetry", "algorithm", "combinatorics proposed", "combinatorics" ], "Problem": "What is the simplest way to determine if its correct? If all of the 3x3 grids contains all values from 1-9, is it possible that it is wrong in another way. I was thinking that as long as all the 3x3 grid contains only one value from 1-9, it is nearly impossible that it is wrong in another way. \r\n\r\nIs there an easy way to determine if two puzzles are related by the symmetry and even better, is there a normal form (a distinct element for each orbit)?", "Solution_1": "[quote=\"LeoYard\"]What is the simplest way to determine if its correct?[/quote]\r\n\r\nIf you solve a Sudoku by hand, then the simplest way is to check each box, line, and column. Of course, if you use a computer program to solve (or to assist you in the solving) then the checking will be instantaneous.", "Solution_2": "[quote=\"LeoYard\"]I was thinking that as long as all the 3x3 grid contains only one value from 1-9, it is nearly impossible that it is wrong in another way. [/quote]\r\n\r\nActually it is possible to be incorrect, for example put all the 1 of the upper three 3x3 grids in the first row.", "Solution_3": "Use the algorithm which is used in MathCad.\r\nIt is named as \"nearest neighbour way to solve\".\r\nThanks by now for replying." } { "Tag": [ "geometry", "circumcircle", "perpendicular bisector", "cyclic quadrilateral", "geometry proposed" ], "Problem": "The triangle $ABC$ is such that $\\angle BAC=30^{\\circ},\\angle ABC=45^{\\circ}$. Prove that if $X$ lies on the ray $AC$, $Y$ lies on the ray $BC$ and $OX=BY$, where $O$ is the circumcentre of triangle $ABC$, then $S_{XY}$ passes through a fixed point.\n\n[i]Emil Kolev [/i]", "Solution_1": "The condition $\\angle ABC = 45^\\circ$ is irelevant, it is sufficient (and necessary) that the angle $\\angle CAB = 30^\\circ.$ This makes the triangle $\\triangle COB$ equilateral. Let q be a distance greater than or equal to the distance of the circumcenter O from the side CA. Points $X, X' \\in CA$ such that OX = OX' = q are intersections of CA with a circle $\\mathcal C_1$ centered at O and with radius q, while points $Y, Y' \\in BC$ such that BY = BY' = q are intersections of BC with a circle $\\mathcal C_2$ centered at B and with the same radius q. Since the circles $\\mathcal C_1 \\cong \\mathcal C_2$ are congruent and since CB = CO, powers of the point C to these 2 circles are equal, $CX \\cdot CX' = CY \\cdot CY'.$ This means that the quadrilateral XX'YY' is cyclic. The perpendicular bisector of the side YY' is fixed, namely a normal to BC at B, and the perpendicular bisector of the side XX' is also fixed, identical with the perpendicular bisector of CA. Thus the circumcenter P of the cyclic quadrilateral XX'YY' is fixed and the perpendicular bisectors of its remaining sides XY', X'Y and of its diagonals XY, X'Y' all pass through its fixed circumcenter P.", "Solution_2": "yetti your solution proves that if OX=OX'; the lines pass through a fixed point, but if it is not; your solution doesn't work i think." } { "Tag": [ "algebra", "polynomial", "vector", "superior algebra", "superior algebra unsolved" ], "Problem": "How do you prove that $ \\mathbb{Q}(\\sqrt [3]{2})$ does not contain in $ \\sqrt [3]{3}$?\r\n\r\nThe standard method of finding an irreducible polynomial with $ \\sqrt[3]{2}\\plus{}\\sqrt[3]{3}$ as a zero looked like it would get very messy.\r\n\r\nWhat else should I try? I am not even sure if it is true, but it seems very intuitive.", "Solution_1": "It's not terribly messy. Let $ t \\equal{} \\sqrt[3]{2} \\plus{} \\sqrt[3]{3}$; then\r\n\r\n$ t^3 \\equal{} 5 \\plus{} 3 \\sqrt[3]{12} \\plus{} 3 \\sqrt[3]{18} \\implies$\r\n$ t^3 \\minus{} 5 \\equal{} 3 \\left( \\sqrt[3]{12} \\plus{} \\sqrt[3]{18} \\right) \\implies$\r\n$ (t^3 \\minus{} 5)^3 \\equal{} 27 \\left( 30 \\plus{} 18 \\sqrt[3]{12} \\plus{} 18 \\sqrt[3]{18} \\right) \\equal{} 27 \\left( 30 \\plus{} 6(t^3 \\minus{} 5) \\right)$.", "Solution_2": "The first equality on the last line is not true.\r\n\r\nEDIT: sorry, it is obviously true", "Solution_3": "But do you know an easy way of showing that $ t^9\\minus{}15t^6\\minus{}87t^3\\minus{}125$ is irreducible?", "Solution_4": "This might be useful\r\n\r\n$ F({\\sqrt[q]{e}}\\equal{}F(({\\sqrt[q]{f}}) \\implies f\\equal{}e^tc^q$ where char $ F$= $ 0$ and $ e,f \\in F$.", "Solution_5": "What is c?", "Solution_6": "$ c$ is an element of $ F$.", "Solution_7": "[quote=\"Abel\"]This might be useful\n\n$ F({\\sqrt [q]{e}} \\equal{} F(({\\sqrt [q]{f}}) \\implies f \\equal{} e^tc^q$ where char $ F$= $ 0$ and $ e,f \\in F$.[/quote]\r\nI have never seen that before and I have no idea why that is true. I liked the t0rajir0u's approach of finding a polynomial but I need help proving that it is irreducible.", "Solution_8": "The minimal polynomials of $ u: = 2^{1/3}, v : = 3^{1/3}$ over $ \\mathbb{Q}$ are $ x^3 - 2, x^3 - 3$, respectively and therefore both $ \\mathbb{Q}(u), \\mathbb{Q}(v)$ are extensions of degree $ 3$ over $ \\mathbb{Q}$. If $ \\mathbb{Q}(u)$ contained $ v$ then $ \\mathbb{Q}(u)$ would be a $ 3$-dimensional $ \\mathbb{Q}$-vector space containing the 3-dimensional $ \\mathbb{Q}$ vector space $ \\mathbb{Q}(v) \\Leftrightarrow \\mathbb{Q}(u) = \\mathbb{Q}(v)$. But then $ u, v$ have to be roots of the same irreducible polynomial over $ \\mathbb{Q}$, which they are not.", "Solution_9": "[quote=\"puuhikki\"]But then $ u, v$ have to be roots of the same irreducible polynomial over $ \\mathbb{Q}$[/quote]\r\n\r\nWhy that?\r\n\r\n darij", "Solution_10": "[quote=\"puuhikki\"]$ \\mathbb{Q}(u) \\equal{} \\mathbb{Q}(v)$. But then $ u, v$ have to be roots of the same irreducible polynomial over $ \\mathbb{Q}$[/quote]\r\nthis step is wrong" } { "Tag": [ "algebra", "polynomial", "complex analysis", "function", "real analysis", "real analysis unsolved" ], "Problem": "Find $f:R->R$ satisfies \r\n$f(1+f(x))=1+x$ and $f(f(x))=x \\forall x \\in R$", "Solution_1": "Hello,\r\n[quote=\"kiemkhach\"]Find $f:R->R$ satisfies $f(1+f(x))=1+x$ and $f(f(x))=x \\forall x \\in R$[/quote]\r\nAre you sure that you are not forgetting some condition. For example, if it is a polynomial, it is very easy (almost trivial to proof) that $f(x)=x$ or $f(x) = -x$. If it has to be continuous, I think that it has to be also $f(x)=\\pm x$. Now, if it does not have to be continues (actually, it will just have one discontinuity) there would be a complete class of holomorphic functions that will do it, as $f(x)=\\frac1x$. If they don't have to be continuous, well man, you will get a huge family of functions.", "Solution_2": "$-x$ is not a solution.\r\n\r\nA very large family of functions (not continuous) that satisfy the conditions: Let $g(x)$ be an involution of the set $[0,1).$ Define $f(x)$ by the relation $f(x)=g(x)$ if $0\\le x<1$ and $f(x+n)=f(x)+n$ for all integers $n.$", "Solution_3": "Ups,\r\n\r\n[quote=\"jmerry\"]$-x$ is not a solution.\n\nA very large family of functions (not continuous) that satisfy the conditions: Let $g(x)$ be an involution of the set $[0,1).$ Define $f(x)$ by the relation $f(x)=g(x)$ if $0\\le x<1$ and $f(x+n)=f(x)+n$ for all integers $n.$[/quote]\r\n\r\nYeah, sorry, it seems like in some moment while I was writting the first condition scape of my mind. Jeje, as we would say in my country \"Que pelada!\" (Translated literally means \"what a peeled!\", and usually means that you are ashamed for the silly mistake you did!).\r\n\r\nBest regards," } { "Tag": [ "Princeton", "college", "function", "calculus", "derivative", "integration", "complex analysis" ], "Problem": "This is #8 in Ch.1 of Complx Analysis (vol. II of the Princeton Lectures in Analysis):\r\n\r\n8. Suppose U and V are open sets in the complex plane. Prove that if f : U \u2192 V\r\nand g : V \u2192 C are two functions that are differentiable (in the real sense, that is,\r\nas functions of the two real variables x and y), and h = g \u25e6 f, then\r\n\\[ \\frac {\\partial{h}}{\\partial{z}} \\equal{} \\frac {\\partial{g}}{\\partial{z}} \\frac {\\partial{f}}{\\partial{z}} \\plus{} \\frac {\\partial{g}}{\\partial{\\overline{z}}} \\frac {\\partial{\\overline{f}}}{\\partial{z}}\r\n\\]\r\nand\r\n\\[ \\frac {\\partial{h}}{\\partial{\\overline{z}}} \\equal{} \\frac {\\partial{g}}{\\partial{z}} \\frac {\\partial{f}}{\\partial{\\overline{z}}} \\plus{} \\frac {\\partial{g}}{\\partial{\\overline{z}}} \\frac {\\partial{\\overline{f}}}{\\partial{\\overline{z}}}\r\n\\]", "Solution_1": "I believe this is only true if you further assume that f and g are differentiable in the complex plane - how else is derivation with respect to z defined?", "Solution_2": "It's a formalism, sometimes called Wirtinger calculus, the \"derivatives\" with respect to $ z$ and $ \\bar{z}$ defined in terms of partial derivatives in $ x$ and $ y$ so that you get the right result for analytic and anti-analytic functions (so use the C-R equations). It's a very useful notation, because most of the time you can take formulas that are usually stated for functions $ u(x,y)$ and just replace $ x$ with $ z$, $ y$ with $ \\bar{z}$, $ \\partial x$ with $ \\partial z$, $ dx$ with $ dz$ and so on. It works for many changes of variables, Green's formula (but you might need to introduce a constant), etc. Stokes formula in the plane is\r\n$ \\int_U \\frac {\\partial u}{\\partial \\bar{z}} \\, dx \\, dy \\equal{} \\frac {1}{2i} \\oint_{\\partial U} u(z) \\, dz$.\r\n\r\nIt's not necessary to use it, but it's convenient, and when dealing with functions like for example $ |f|^{2p} \\equal{} f^p \\bar{f}^p$, $ f$ analytic, thinking in terms of Wirtinger calculus usually simplifies things a lot. It's mentioned in every textbook I've ever seen, but it's never explained why it is a good tool.\r\n\r\nAs for the given question, I don't see what possible issue you can have with this if you understand the definitions. What is the problem with rigor :?:", "Solution_3": "Frankly I can see many \"handwavy\" solutions to such a problem. But to answer your question, I've done some initial computations (just plugging the said identities in) and the answer doesn't seem to agree...it looks like\r\n\\[ \\frac {\\partial{h}}{\\partial{z}} \\equal{} \\frac {\\partial{g}}{\\partial{f}} \\frac {\\partial{f}}{\\partial{z}} \\plus{} \\frac {\\partial{g}}{\\partial{\\overline{f}}} \\frac {\\partial{\\overline{f}}}{\\partial{z}}\r\n\\]\r\nand its counterpart would fit the bill better, and it is more intuitive.", "Solution_4": "It is understood that in the original formula that $ \\frac {\\partial g}{\\partial z}$ and $ \\frac {\\partial g}{\\partial \\bar{z}}$ are to be evaluated in $ f(z)$. I think you have deduced the correct formula, but you are using incorrect notation in your answer. When one writes $ \\frac {\\partial g}{\\partial f}$, typically $ f$ would have a double meaning as a variable and a function (like one writes y = g(x) or u = h(x,y)). In this case we are not even intending $ g$ to be differentiated against one of its variables in the usual sense, so writing $ \\frac {\\partial g}{\\partial f}$ is very misleading.", "Solution_5": "So basically I interpreted my $ \\frac{\\partial{g}}{\\partial{f}}$ as $ \\frac{\\partial{g}}{\\partial{(u\\plus{}iv)}}$ and then evaluated that in $ u$ and $ v$ using the given equations - is that equivalent to what you're saying it means?", "Solution_6": "I have no idea what we're talking about anymore. I am not a fan of the accepted differential operator-notation. Maybe I have confused you with poor english. I am saying that the original formula is this:\r\n\\[ \\frac {\\partial{h}}{\\partial{z}}(w) \\equal{} \\frac {\\partial{g}}{\\partial{z}}(f(w)) \\cdot \\frac {\\partial{f}}{\\partial{z}}(w) \\plus{} \\frac {\\partial{g}}{\\partial{\\overline{z}}}(f(w)) \\cdot \\frac {\\partial{\\overline{f}}}{\\partial{z}}(w), \\quad w \\in U\r\n\\]\r\nSince for example $ \\frac {\\partial{g}}{\\partial{z}}$ is a function from $ V$ to $ C$, it is understood that in the formula it should be evaluated in $ f(w) \\in V$. Note also that when either $ g$ or $ f$ is analytic the formula looks like the usual chain rule in one variable.\r\n\r\nI am also saying that $ \\frac {\\partial{g}}{\\partial{z}}(f(w))$ should never be written as $ \\frac {\\partial{g}}{\\partial{f}}$ in this context, because even though we are using the notation $ \\frac {\\partial}{\\partial z}$ this operator is NOT a true derivative - the $ z$ has a special meaning.", "Solution_7": "I believe Princeton's notation is wrong and prefer the formulation from \"Function Theory of One Complex Variable\" by Greene and Krantz: Let $ f$ and $ g$ be $ C^1$ functions and assume $ f\\circ g$ is defined. Prove the following version of the chain rule with $ w \\equal{} g(z)$:\r\n\\[ \\frac {\\partial}{\\partial z} \\left(f\\circ g\\right) \\equal{} \\equal{} \\frac {\\partial{f}}{\\partial{w}}\\frac {\\partial{g}}{\\partial{z}} \\plus{} \\frac {\\partial{f}}{\\partial{\\overline{w}}}\\frac {\\partial{\\overline{g}}}{\\partial{z}}.\r\n\\]\r\nHere, I'm assuming $ f\\circ g \\equal{} f(g,\\overline{g})$. For example, let $ f \\equal{} z^2 \\plus{} 2\\overline{z}$ and $ g \\equal{} 3z \\minus{} 4\\overline{z}$ then $ \\overline{g} \\equal{} 3\\overline{z} \\minus{} 4z$ and $ f\\circ g \\equal{} (3z \\minus{} 4\\overline{z})^2 \\plus{} 2(3\\overline{z} \\minus{} 4z)$ and therefore\r\n\\[ \\frac {\\partial}{\\partial z}\\left(9z^2 \\minus{} 24z\\overline{z} \\plus{} 16\\overline{z}^2 \\plus{} 6\\overline{z} \\minus{} 8z\\right) \\equal{} 18z \\minus{} 24\\overline{z} \\minus{} 8 \\equal{} \\frac {\\partial{f}}{\\partial{w}}\\frac {\\partial{g}}{\\partial{z}} \\plus{} \\frac {\\partial{f}}{\\partial{\\overline{w}}}\\frac {\\partial{\\overline{g}}}{\\partial{z}}\r\n\\]\r\nwhere:\r\n\\[ \\frac {\\partial f}{\\partial w} \\equal{} \\frac {\\partial f}{\\partial z}\\Bigg|_{z \\equal{} g(z)}\r\n\\]\r\n\r\n\\[ \\frac {\\partial f}{\\partial \\overline{w}} \\equal{} \\frac {\\partial f}{\\partial \\overline{z}}\\Bigg|_{\\overline{z} \\equal{} \\overline{g(z)}}\r\n\\]\r\nThe authors give the suggestion: Write out everything in real and imaginary parts and use the real variable chain rule from Calculus. Note if both are analytic then the partial reduces to the ordinary chain rule since by definition $ \\frac{\\partial f}{\\partial \\overline{z}}\\equal{}0$." } { "Tag": [ "modular arithmetic", "Combinatorial Number Theory", "number theory", "prime numbers", "IMO", "IMO 1994", "combinatorics" ], "Problem": "Show that there exists a set $ A$ of positive integers with the following property: for any infinite set $ S$ of primes, there exist [i]two[/i] positive integers $ m$ in $ A$ and $ n$ not in $ A$, each of which is a product of $ k$ distinct elements of $ S$ for some $ k \\geq 2$.", "Solution_1": "Beautiful problem. Let's look at what it says for a moment. A version of Ramsey's Theorem says this: if we color the $k$-element subsets of $\\mathbb N$ with two colors (it works for a fintie number of colors, of course, but we're interested in this case here), then there is an infinite $k$-homogemous set, i.e. an infinite set all of whose $k$-element subsets bear the same color. This problem says that this sort of result is not necesarily true any more if we try to do this for all $k\\ge 2$ at the same time: if we color all the subsets of $\\mathbb N$ which have at least two elements with two colors, then there need not exist an infintie set which is $k$-homogenous for all $k$. Here, of course, the set of primes represents $\\mathbb N$, and the product $p_{i_1}\\cdot\\ldots\\cdot p_{i_k}$ represents the set $\\{\\i_1,\\ldots,i_k\\}$. If a product of distinct primes (we don't work with non-square-free numbers) belongs to the set we want to construct, we'll say the corresponding set is red, while if it does not we'll say the corresponding set is blue. That's why from now on we'll work in this setting, using subsets of $\\mathbb N$ instead of square-free positive integers.\r\n\r\nWe now prove that given $k\\ge 2$ and some $x\\in\\mathbb N$, we can find a two-coloration of the $k$-element subsets of $\\mathbb N$ such that $x$ belongs to no infinite $k$-homogenous set. This is almost obvious. We can simply color all the $k$-element sets which contain $x$ in red, and the rest of them in blue. We'll call this procedure $P(x,k)$. Now, in order to obtain what we want, we can apply the procedures $P(k,k)$ for all $k\\ge 2$. In other words, in the two-coloring of the subsets of $\\mathbb N$ which have at least $2$ elements that we obtain, $2$ cannot belong to any infinite $2$-homogenous set, $3$ cannot belong to any infinite $3$-homogenous set, and, in general, $n$ cannot belong to any infinite $n$-homogenous set. In other words, there are no infinite subsets of $\\mathbb N$ which are $k$-homogenous for all $k\\ge 2$.", "Solution_2": "Maybe another solution to this beautiful problem:\r\n\r\nThe set $ A$ of the square-free positive integers is defined as follows: A product $ n$ of $ k$ distinct primes $ p_{1},\\ldots,p_{k}$ is a member of $ A$ if and only if $ n\\equiv 1\\pmod{k+1}$, i.e. $ p_{1}\\cdot\\ldots\\cdot p_{k}\\in A\\Leftrightarrow p_{1}\\cdot\\ldots\\cdot p_{k}\\equiv 1\\pmod{k+1}$ for all $ k$ and all primes $ p_{i}$.\r\n\r\nNow, let $ S$ be an infinite set of primes and let $ p_{1}$ be the smallest and $ p_{2}$ be the second-smallest prime in $ S$. Let furthermore $ p$ be any prime with $ p>p_{2}$.\r\n\r\nLets now consinder $ S$ modulo $ p$. By the pigeonhole principle, some remainder modulo $ p$ must appear infinitely many times in $ S$, i.e. there exists an integer $ a$, so that $ q\\equiv a\\pmod p$ for infinitely many $ q\\in S$. Obviously, $ p\\nmid a$.\r\n\r\nTake now $ p-1$ distinct primes $ q_{1},\\ldots,q_{p-1}\\in S$ and let $ m=q_{1}\\cdot\\ldots\\cdot q_{p-1}$. Then $ m\\in A$ because by Fermat's little theorem, $ m = q_{1}\\cdot\\ldots\\cdot q_{p-1}\\equiv a^{p-1}\\equiv 1\\pmod p$.\r\n\r\nOn the other hand, we have $ p>p_{2}>p_{1}$ by definition, so $ p_{2}\\not\\equiv p_{1}\\pmod p$. Hence, at least one of the numbers $ p_{1}\\cdot a^{p-2}$ and $ p_{2}\\cdot a^{p-2}$ is not congruent to $ 1$ modulo $ p$. Thus, at least one of the numbers $ n_{1}= p_{1}\\cdot q_{1}\\cdot \\ldots \\cdot q_{p-2}$ and $ n_{2}= p_{2}\\cdot q_{1}\\cdot \\ldots \\cdot q_{p-2}$ is not congruent to $ 1$ modulo $ p$ and hence ist not a member of $ A$. $ \\Box$", "Solution_3": "Another solution: \r\nInterpretating A as a set of finite subsets of N, we say that $ \\{n_{1},n_{2},\\ldots,n_{k}\\}\\in A$, where $ n_{1}< n_{2}< \\ldots n_{k}$, if and only if $ n_{2}\\le n_{1}+k$.\r\nGiven an infinite set of naturals, just take the first $ d$ elements, where d is the difference between the two least elements: this set belongs to A.\r\nNow take the first element, not the second, an then all the next d-1 elements: this set does not belong to A.", "Solution_4": "And another one :D (but I'm not really sure if it is correct...):\r\nWe will construct two disjoint sets $ A, B$ consisting of only squarefree numbers such that for every infinite set of primes $ S$ there exist $ m \\in A$ and $ n \\in B$ such that $ m$ and $ n$ are product of the same number ($ \\geq 2$) of elements of $ S$. We define:\r\n$ p_{1}p_{2}...p_{n}\\in A$ iff $ n$ has exactly two prime divisors from $ p_{1}, p_{2}, ..., p_{n}$ and $ p_{1}p_{2}...p_{n}\\in B$ iff $ n$ has exactly one prime divisor from the same set (we don't care about the other numbers). Now let $ S=\\{q_{1}, q_{2}, ...\\}$. Consider the number $ m=q_{1}q_{2}q_{i_{1}}q_{i_{2}}...q_{i_{s}}$, $ n=q_{1}q_{i_{1}}...q_{i_{s}}q_{i_{s+1}}$, where $ s=q_{1}q_{2}-2$ and $ q_{i}$ are different elements of $ S$. From the definition we immidiately get that $ m \\in A$ and $ n \\in B$.\r\n\r\nI think that there can be very many significantly distinct solutions to this problem.", "Solution_5": "I think I have another construction: \r\nIf $ p_{i}$ is the $ i$-th prime, and if $ n \\equal{} p_{t_{1}}*p_{t_{2}}*...*p_{t_{m}}$, for some primes $ p_{t_{i}}$, assign to the number $ n$ the set $ (t_{1}, t_{2} ... t_{m})$ (clearly any squarefree number $ n$ can be written uniquely in such a form). Define the set $ A$ as follows: a number $ n$ is in $ A$ iff the assigned set $ (t_{1}, t_{2}, ... t_{m})$ is such that $ m | t_{1} \\plus{} t_{2} \\plus{} ... \\plus{} t_{m}$. Now suppose to the contrary this set does not satisfy the condition: then say an infinite set $ S$ of primes, $ {s_{1}, s_{2} ... }$ (where $ s_{i}$ refers to the prime $ p_{s_{i}}$, is such, so that for any $ k$, either [1] all $ k$-element subsets of $ S$ are in $ A$ OR [2] all $ k$ element subsets of $ S$ are not in $ A$.\r\nIf [2] occurs, then for any $ k$-element subset of $ S$, the sum of the elements in this $ k$-element subset is not divisible by $ k$. But this cannot happen, since using well-known Erdos result, in any $ 2k \\minus{} 1$ numbers, some $ k$ have sum divisible by $ k$, and $ S$ is an infinite set. \r\nSince [2] never occurs, that means [1] must occur for every $ k$. This means for any fixed $ k$, any $ k$-element subset $ S$ has sum divisible by $ k$. Choose one such set, $ (t_{1}, t_{2} ... t_{k})$. Now replace $ t_{1}$ by another element from $ S$ - since the subset still has sum $ 0$ mod $ k$, this means that all elements apart from $ (t_{2} .. t_{k})$ are the same as $ t_{1}$ in $ Z_{k}$. Doing the same thing, but with $ t_{i}$ for other values of $ i$ (ie, replacing $ t_{i}$ by some other element in infinite subset $ S$) we find that elements of $ S$ have the same value in $ Z_{k}$. Now pick any two elements $ s_{1}, s_{2}$ from $ S$. This means $ s_{2} \\minus{} s_{1}$ is divisible by $ k$, for any $ k$. But this is clearly a contradiction, since we can choose $ k$ to be arbitrarily large.", "Solution_6": "Let $p_1 < p_2 < \\cdots$ be the ordered sequence of positive prime numbers. Now let $A$ be the set of all positive integers which are the product of $k$ distinct prime numbers one of which is $p_k$ for each $k \\ge 2$. Consider an arbitrary set $S$ of primes and let $p_i$ be the minimal prime number in $S$. Now consider a sequence of $i$ distinct prime numbers $a_1, a_2, \\dots, a_i$ such that $a_j \\in S$ and $a_j > p_i$ for each $1 \\le j \\le i$. By the definition of $A$, it follows that $p_i \\cdot a_1 \\cdot a_2 \\cdots a_{i-1} \\in A$ and $a_1 \\cdot a_2 \\cdots a_i \\not \\in A$ and $A$ satisfies the conditions required by the problem.", "Solution_7": "I don't understand the problem. Is k fixed or does it vary depending on what S is?", "Solution_8": "@viperstrike $k$ varies depending on $S$.\n\nSorry to resurrect; I was wondering about the validity of a possible construction.\n\nDenote the set of all primes as $\\{p_1,p_2,p_3,\\ldots\\}$. For integral $i \\ge 2$, let $B_i = \\{ \\prod_{j_k \\in D} p_{j_k} \\mid |D| = i \\text{ and } j_k \\equiv j_{m} \\pmod {2^{i-2}} \\text{ for } 1 \\le k < m \\le i\\}$. Similarly, let $C_i = \\{\\prod_{j_k \\in E} p_{j_k} \\mid |E| = i \\text { and } \\exists k,m \\text{ such that } 1 \\le k < m \\le i \\text{ and } j_k \\not \\equiv j_m \\pmod {2^{i-1}} \\}$.\n\nLet $A_i = B_i \\cap C_i$. Take $A = \\bigcup_{i \\ge 2} A_i$. \n\nThe thought behind this was that if there exists at least one even-indexed prime and one odd-indexed prime in $S$, then since there are an infinite number of at least one of these classes represented in $S$, WLOG even-indexed primes, we can take a product of an even-indexed prime with an odd-indexed prime, which is in $A$, and an even-indexed prime with an even-indexed prime, which is not in $A$. This would correspond to the addition of the set $A_1$ to $A$. Motivation for larger $i$ works similarly. Notice also that the only set contributing $i$-factor products to $A$ is $A_i$.\n\nFor arbitrary $S$, let the two smallest elements be $p_a < p_b$ and let $n$ be the smallest natural number such that $b < 2^n$. Then $a$ and $b$ are different $\\pmod {2^n}$. We then proceed through the following $n$-step iteration:\n\nStep $1$: If not all indices in $S$ are odd and not all indices are even, then there exists at least one even-indexed $p_{2c}$ and one odd-indexed $p_{2d-1}$, the product of which is in $A_1$ and therefore in $A$. WLOG there are infinitely many odd-indexed primes in $S$. Then there exist $p_{2d-1}$ and $p_{2e-1}$ such that $p_{2d-1}p_{2e-1} \\not \\in A$. Done.\n\nStep $2$: Assume the assumption of step 1 is not true. If not all indices in $S$ are $a \\pmod 4$ and not all indices in $S$ are $a+2 \\pmod 4$, then there exists at least one index which is $a \\pmod 4$ and one which is $a+2 \\pmod 4$, and one of these classes contributes an infinite number of indices to $S$. Assume the first. Then we can take $a \\equiv c \\equiv d-2 \\pmod 4$ so that the product $p_a p_c p_d$ is in $A_2$ and therefore in $A$. Take $a \\equiv c \\equiv e \\pmod 4$, so that $p_a p_d p_e \\not \\in A$. Done.\n\nStep $3$: Assume the assumptions of steps $1$ and $2$ are not true. Proceed similarly to step 2, but work $\\pmod 8$.\n\n$\\vdots$\n\nStep $n$: Assume the assumptions of all previous steps are not true. Then all indices are the same $\\pmod {2^{n-1}}$. Since $a,b < 2^n$, it automatically follows that there exist two indices which are not the same $\\pmod {2^n}$, namely $a$ and $b$. There must exist an infinite number of indices in $S$ corresponding to $a \\pmod {2^{n}}$ or to $a+2^{n-1} \\pmod{2^{n}}$. WLOG assume the former. Let $b < i_1 < i_2 < \\cdots < i_{n-1} < i_n$ be such that $a \\equiv i_1 \\equiv i_2 \\equiv \\cdots \\equiv i_{n-1} \\equiv i_n \\pmod {2^{n}}$. Then $p_{a}p_{b}p_{i_1} \\cdots p_{i_{n-1}} \\in A_{n}$ and therefore is in $A$ but $p_{a}p_{i_1}p_{i_2}\\cdots p_{i_{n-1}} p_{i_n} \\not \\in A$. Done; this iteration handled the final case.", "Solution_9": "I had the same idea as Mc Teague but the resulting solution worded in a quite different way.\nlet $\\{p_i\\}$ the set of all prime numbers, where $p_i < p_{i+1}$.\nLet $A_k = \\{p_{i_1} p_{i_2}.....p_{i_{k}}, k >=2 $where all $i_j$ are not all congruent to the same value $mod (k)\\}$\nAt last, let $A =\\cup_{k>=2} A_k$.\n\nBy contradiction : We suppose that for any $m \\in A_k$ for any $k >=2$, there exists at least one prime divisor not in S.\nTherefore, all prime divisor in S have an indice congruent to the same value $mod(k)$. From here, it's easy to get a contradiction : take $p_i$ and $p_j$ two prime divisors of S and let k big enough, you'll have $k>i$ and $k>j$ and $i \\equiv j [k]$, therefore $i=j$ : S has only one element, absurd.\n\nSo there must exist $k$ and $m \\in A$ where $m$ is a product of $k$ disctinct elements of S. \n\nlet $k$ such a number. As $S$ as infinitely many primes, we can find infinitely many primes for which all indices are congruent to the same value $mod (k)$. Take $n$ the product of $k$ such numbers of S. $n \\notin A$ but n is a product of k disctinct element of S : done.\n\n", "Solution_10": "It boils down to prove that there exists a countable collection $\\mathcal{F}$ of subsets of $\\mathbb{N}$ with the following property. For every infinite subset $D$ of $\\mathbb{N}$ there exists $k\\ge 2,$ such that $D$ contains a $k$-element subset of $D$ which is a member of $\\mathcal{F}$ as well as a $k$-element subset of $D$ which is not a member of $\\mathcal{F}.$\n\n[i]Proof.[/i] Arrange all 2-element subsets of $\\mathbb{N}$ as $A_1,A_2,\\dots.$ Map to each $A_i$ a family $\\mathcal{F}_i$ of $i+1$-element subsets of $\\mathbb{N}$ as follows\n$$\\mathcal{F}_i := \\{A_i\\cup X : X\\subset \\mathbb{N}\\setminus A_i\\,,\\, |X|=i-1 \\}$$\nSo, each set in $\\mathcal{F}_i$ contains exactly $i+1$ elements. Let $\\mathcal{F}:=\\bigcup_{i=1}^{\\infty} \\mathcal{F}_i.$ This is the wanted collection of subsets. Indeed, let $D$ be an infinite set of numbers. Take any $x,y\\in D.$ Let $A_n=\\{x,y\\}.$ Take any $A,B\\subset D\\setminus\\{x,y\\}$ such that $|A|=n-1, |B|=n.$ Now $A_n\\cup A \\in \\mathcal{F}_n$ and $\\{x\\}\\cup B\\notin \\mathcal{F}_n,$ which means $A_n\\cup A\\in \\mathcal{F}$ and $\\{x\\}\\cup B\\notin \\mathcal{F},$ since the only subsets in $\\mathcal{F}$ with $n+1$ elements are those in $\\mathcal{F}_n$ \n\n[b][u]Comment[/u][/b]. It was ISL 1994, N3, but it's a pure combinatorial one, just dressed as NT. I saw it given at [url=https://artofproblemsolving.com/community/c6h2978219]Germany 2010 - Problem 3.[/url] Something about motivation. One may try first to find a collection of, say, 2-element sets, because it is the simplest way. It means you have to connect some of numbers in $\\mathbb{N}$ with edges, that is, the edges will be the required sets. The imposed requirement just says that there should be no infinite set of vertices that form a clique or an anti-clique. Unfortunately, the infinite version of Ramsey's theorem shows, that it's impossible. The same happens if we try to choose subsets with same size $k>2,$ because of a version of Ramsey's theorem for $k$-uniform hypergraphs. This means that we are doomed by trying to create a collection of same size sets. We should vary $k$ by choosing bigger and bigger sets. " } { "Tag": [ "geometry", "topology", "Princeton", "college", "Putnam", "Columbia", "number theory" ], "Problem": "I was wondering from what colleges top quantitative companies recruit from. I know several of the administrators here worked at DE Shaw, so it would be nice if they could recall the list of colleges that were recruited. Specific colleges, please. Do they usually recruit physics/ math majors, or business majors?\r\n\r\nHow hard is it to get interviews in general for top companies (not only quants, but scientific companies, other financial groups, etc) from an above-average college (like University of Michigan)? Or do companies snub all non-Ivies / non-elite tech schools grads?\r\n\r\nHow does recruiting work anyways? \r\n\r\n\r\nThanks for answering the numerous questions.", "Solution_1": "im only in high school, so i don't know much about what you're asking, but on Jane Street Capital's website (the company Sandor Lehoczky works for) has a listing of colleges they recruit from on their website, and they also mention that you can still get an interview if you go to a different college, you just have to e-mail them with your resume. http://www.janestcapital.com/careers.html at the bottom of the page is the list.", "Solution_2": "[quote=\"1234567890\"]I was wondering from what colleges top quantitative companies recruit from. I know several of the administrators here worked at DE Shaw, so it would be nice if they could recall the list of colleges that were recruited. Specific colleges, please. Do they usually recruit physics/ math majors, or business majors?\n\nHow hard is it to get interviews in general for top companies (not only quants, but scientific companies, other financial groups, etc) from an above-average college (like University of Michigan)? Or do companies snub all non-Ivies / non-elite tech schools grads?\n\nHow does recruiting work anyways? \n\nThanks for answering the numerous questions.[/quote]\r\n\r\nUniv of Michigan is certainly on the 'will interview' list in some disciplines, though I admit it would be harder to get an interview from there than it would be from MIT. As for what schools are on the definite list, it's pretty much the usual suspects. Look at lists of top math/CS schools, and that will give you the list of schools Shaw/Jane Street would definitely pay attention to.\r\n\r\nIf you want to be a quant, don't major in business. Physics, CS, Math. And learn how to code.\r\n\r\nThe typical path to getting hired is:\r\n\r\n1) Send in a resume, or have a headhunter send in a resume.\r\n2) Have a phone interview.\r\n3) Have an in-house interview.\r\n4) Get hired.\r\n\r\nYour college/awards/test scores/etc are probably most important in clearing those first couple hurdles. Once you are in house, your performance there can overcome other areas that might not be as good as other applicants.", "Solution_3": "I actually want to attend University of Michigan in math. I think their math program is decent. So is it very hard to even get an interview?\r\n\r\nDo firms look for coursework/GPA heavily?", "Solution_4": "math, cs, and physics were the exact subjects i plan on majoring when i go to college! i didn't realize physics was helpful for being a trader or quant, but i guess i'm in all the better position already.", "Solution_5": "[quote=\"1234567890\"]I actually want to attend University of Michigan in math. I think their math program is decent. So is it very hard to even get an interview?\n\nDo firms look for coursework/GPA heavily?[/quote]\r\n\r\nThey look at coursework and GPA - GPA is an indicator of willingness to work.\r\n\r\nIt is tough to get an interview regardless of where you go. Make sure you excel at your coursework, and learn how to program.", "Solution_6": "[quote=\"ffdbzathf\"]math, cs, and physics were the exact subjects i plan on majoring when i go to college! i didn't realize physics was helpful for being a trader or quant, but i guess i'm in all the better position already.[/quote]\r\n\r\nIt's not so much that physics is helpful in trading, but that people who master physics have shown an ability to model phenomena very well. It's this ability that financial firms are interested in, not the actual physics.", "Solution_7": "sorry, i may just ask a silly question.. :? \r\nbut i what to know is that quants = quantitative finance??\r\nif yes, then why dont just study a major called quantitative finance?\r\n( actually i am not sure if this major exists or not...... :P )", "Solution_8": "[quote=\"ffdbzathf\"]math, cs, and physics were the exact subjects i plan on majoring when i go to college! i didn't realize physics was helpful for being a trader or quant, but i guess i'm in all the better position already.[/quote]\r\n\r\nMajoring in three subjects like that inevitably comes at the expense of breadth in any one subject.", "Solution_9": "[quote=\"asjsd\"]sorry, i may just ask a silly question.. :? \nbut i what to know is that quants = quantitative finance??\nif yes, then why dont just study a major called quantitative finance?\n( actually i am not sure if this major exists or not...... :P )[/quote]\r\n\r\nSome places do have a major similar to that, but most top-notch hedge funds will place their bets on the top physics/CS/math graduates over 'quantitative finance'. The former is all but guaranteed to be much more challenging. The hedge funds aren't hiring for financial knowledge - they're hiring for modeling/programming ability. The finance can be learned relatively quickly.", "Solution_10": "blahblahblah, i disagree. it solely depends on the work you're willing to put in. i know people who have had triple majors and learned each topic in a great amount of depth.", "Solution_11": "At the end of next summer, I'll basically have finished a degree in math (that is, I won't have enough credits, but I'll have taken a majority of the math courses offered at my university)\r\n\r\nThat will include:\r\n\r\n5 courses in analysis\r\n4 courses in algebra\r\n2 courses in number theory\r\n2 courses on differential equations\r\n1 course on algebraic topology\r\n1 course on numerical analysis\r\n1 course on problem solving\r\n2 courses in statistics\r\n2 courses in computer science\r\n3 lower division math courses\r\n\r\nThis will have taken me 2 full years, including summers (I'm allowed to take a couple of courses while I hold NSERCs, which are the equivalents of REUs over here)\r\n\r\nIf I wanted to take similarly many courses in CS (my university is very strong in CS), I could easily take another 3 semesters doing so, and I would still be only two thirds of the way towards your hypothetical triple major.", "Solution_12": "blah^3, what field of math are you interested in.... and how did you take that many courses in just 2 years? How much math per semester? Just curious.", "Solution_13": "at certain colleges, they have programs where you can skip some mandated requirements and go straight to working towards your major, such as Harvard's advanced standing program where you use AP grades of 5 to replace many topics not necessary for your major. Also, math, physics, and CS have some overlap in their courseload, mainly math and physics.", "Solution_14": "I'm most interested in analysis, but I like algebra and number theory as well.\r\n\r\nWell, I got a full year's worth of credits by completing the IB Diploma, so I don't have to take any arts courses or experimental sciences or whatever. I had 12 math credits coming in to university, so I basically got to skip the whole mind numbing calculus/linalg/discrete math introductory sequence. So I've basically been taking 4 math courses a semester and 1 'non-math' course. I also took a lot of courses without prerequisites, and found that it was a good thing, because it requires you to really focus (my GPA in courses I took without prereqs is actually higher than in courses I took with prereqs). I'll also be taking two grad courses in each of the next two semesters (analysis, algebra, algebraic number theory, functional analysis)\r\n\r\nI chose universities based on money. I turned down Princeton for a considerably less prestigious school with an excellent math department, but certainly not one that compares with Princeton's. So I figured that to make up for that, I would have to take as many math courses as I could and challenge myself as much as I can, and that's what I've done so far. In future semesters, I plan on taking a few more economics courses and CS courses, but I'll still take at least 3 math courses every semester.", "Solution_15": "[quote=\"blahblahblah\"]I chose universities based on money. I turned down Princeton for a considerably less prestigious school with an excellent math department, but certainly not one that compares with Princeton's.[/quote] \r\n\r\nI've been meaning to ask you about that. I thought Princeton's financial aid is about as good as it comes in the Ivy League. So did you get an \"offer you couldn't refuse\" from your current university, in the form of REALLY complete coverage of all your out-of-pocket costs? Do Canadian students have more actual expense if admitted to Princeton than do American students? \r\n\r\nMaybe I should start a separate thread about this in the College Forum, in the interest of staying on topic.", "Solution_16": "I'll reply here, and it can be split later.\r\n\r\nAnyways, as I remember, Princeton offered me about 33000 (USD) a year in financial aid, so it would have cost me about 11000-12000 a year out of my own pocket (my parents weren't willing to help out much). After exchange rates and everything (they've gotten significantly better in my favor since then), it would have been about 15000-17000 per year, and that, of course, would rise a little each year. So I was looking at paying roughly 70000 for a degree, minus whatever earnings I could have made in the summer. I absolutely loved Princeton when I visited, and that's a bargain for the quality of university we're talking about here, but...\r\n\r\nI was the top applicant at my local university (Simon Fraser University). Tuition there is about 2500 (CDN) per semester - my scholarship is worth 4250 per semester over 8 semesters. The rest goes back to me. The math department there has achieved some fame for some of the computational work they've done (record calculations of pi, among other things) as well as their affiliation with the Maple CAS. I also won a 5000 scholarship as one of the 13 top all-around students in my province and a 4000 scholarship from the federal government (a Millenium Excellence Award, for those who know what that means). All told, I think I took in about 55000 in scholarships. I also am pretty much guaranteed to have an NSERC research grant every summer (I just completed my first one) which is worth 6000-8000 over 4 months, depending on your supervisor and the year you're in. So over the course of my degree (which I will probably complete in three years), I will make roughly 80000 and only have to pay about 20000 in tuition, which is a rather nice sum of money for someone just completing their undergrad. I also do not plan to work immediately after - I will probably just go to grad school, so assuming that I get into the grad school of my choice, I will have sacrificed very little and gained very much, tangibly speaking.\r\n\r\nAll finances aside, I wouldnt've gone to SFU if I felt that I was seriously compromising the quality of my education. Now, I don't think very much of it as a liberal arts university, but it is quite strong in mathematics and computer science (it has been top 10 on the Putnam before, and does very well in the ACM on a year to year basis, both events which I have (Putnam) or will (ACM) be involved in). My professors have been excellent, and because of my IB Diploma, I have been free to take only the courses I like (whereas at Princeton, I would have had to take a very large number of non-mathematics courses). I also have access to all of the grad courses at the University of British Columbia, which has a better grad program than SFU (and some intriguing grad courses on mathematical finance). I do regret that I haven't been able to pursue interests in English lit and philosophy and whatnot, but I think that's a small price to pay for a lot of the benefits that I've gotten.\r\n\r\nI would have enjoyed the experience of going to Princeton and interacting with a much brighter student body, as well as moving away from home and taking a more balanced set of courses, but I feel that for me, it's an indulgence that wasn't worth the price.\r\n\r\nI'll also note that my family is fairly well off. The other Canadians I visited Princeton with had financial aid awards as high as 40000 a year, which really makes it an easier choice (they also didn't apply to many local universities and hence didn't have any big canadian scholarships to turn down).", "Solution_17": "Blahblahblah, \r\n\r\nI just wonder \r\n\r\n1. if your parents were willing to help with that 11-12g per year, would you make the same choice? or\r\n\r\n2. if your parents were quite wealthy that you are not qualified for any need based financial aid from Princeton. Your parents would provide full amount (44-45g per year for 4 years) to you regardless of Princeton or Simon Fraser, you could either pocket their money (total about 200g) and go to Simon Fraser, or you could spend all the money to go to Princeton, what would you do?\r\n\r\nHow about other bright students on this forum? How would you all make your choice(s)?", "Solution_18": "If there was no debt (or a very small amount) associated with going to Princeton, I would have gone there in a second. I thought that it was an incredible opportunity, but more than that, it would be a once in a lifetime experience. If I ever have kids, I will definitely encourage them to go to the best school they can get into.", "Solution_19": "I am very sympathetic to the situation of students whose parents don't pay out the full calculated \"family contribution\" for their children's education, because I was in that situation in my day. My conclusion then was to attend my state's flagship university, which had a low \"list price.\" My conclusion now is that I will pay out every penny, and then some, up to the maximum limit of my creditworthiness, for my children's education, and certainly fully pay out what each desirable college calculates that I can afford to pay. I admire blahblahblah's resourcefulness in coming up with a good plan B--obviously one has to pay a certain price in terms of effort and achievement in the high school years to have multiple channels for covering college costs. \r\n\r\nBTW, I would split this thread myself, except I haven't got moderation powers in this forum. If the Careers in Mathematics moderator chooses to split off these last few posts to a new topic \"Paying for College\" in the College Forum, I'll happily moderate it there from then on.", "Solution_20": "[quote=\"1234567890\"]I actually want to attend University of Michigan in math. I think their math program is decent. So is it very hard to even get an interview?\r\n\r\nDo firms look for coursework/GPA heavily?[/quote]\r\n\r\nHey, so are you in college yet? are you going to mich now?" } { "Tag": [ "Pythagorean Theorem", "geometry" ], "Problem": "The base length of an isosceles triangle is 24 feet, and the side length is 20 feet. What is the number of feet in the altitude to the base?", "Solution_1": "You can draw this:\r\n\r\n[asy]draw((0,0)--(24,0)--(12,16)--cycle);\nlabel(\"$12$\",(6,0),S);\nlabel(\"$12$\",(18,0),S);\ndraw((12,0)--(12,16));\nlabel(\"$20$\",(6,8),NW);\nlabel(\"$20$\",(18,8),NE);[/asy]\r\n\r\nBy Pythagorean Theorem, the height is $ \\sqrt{20^2\\minus{}12^2}$ or $ \\boxed{16}$.", "Solution_2": "Dang I suck. I didn't see that." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "What is the maximal number of knights that can be placed on the usual 8x8 chessboard so that each of then threatens at most 7 others?", "Solution_1": "[hide]\n60\nfirst fill the boards with knights\nin order to avoid the situation that the center knight threatens 8 other knights, we just need to take the 4 center knights away\n[/hide]" } { "Tag": [ "parameterization", "calculus", "integration", "function", "algebra", "linear equation", "calculus computations" ], "Problem": "Can anyone help me with this problem?\r\n\r\nConsider the following method of solving the general linear equation of 1st order: \r\n\r\ny' + p(t) y = g(t)\r\n\r\nIf g(t) is not everywhere zero, assume that the solution is of the form y = A(t) exp [- integral of p(t) dt]\r\n\r\nwhere A is now a function of t. By substituting for y in the given differential equation, [b]show [/b]that A(t) must satisfy the condition \r\n\r\nA'(t) = g(t) exp [integral p(t) dt]\r\n\r\nHow can I show this? \r\n\r\nThanks.", "Solution_1": "I think it will help to you:\r\n\r\n$I(t)=e^{\\int p(t)dt}$ is integrating factor and if we product the both side of the equation then it changs in the form $\\frac{d(yI)}{dt}=Ig(t)$ and integrate both side of this last equation with respect to $t$ , and then solve the resulting equation for the $y$. The general solution is $y=\\frac{\\int I(t)g(t)+c}{I(t)}$ where $c$ is the constant of integration." } { "Tag": [], "Problem": "Find the (x,y,z) that satisfy\r\n(x+y)(x+y+z)=72\r\n(x+z)(x+y+z)=96\r\n(y+z)(x+y+z)=120", "Solution_1": "[hide]Add all three equations:\n$2(x+y+z)^{2}= 288$\nHence, $x+y+z = \\pm12$. Thus, plugging in,\n$x+y = \\pm 6$\n$x+z = \\pm 8$\n$y+z = \\pm 10$\n\nUsing $x+y+z = \\pm 12$, subtract all three equations from this one to get:\n$z = \\pm 6$\n$y = \\pm 4$\n$x = \\pm 2$\n$(x,y,z) = (\\pm 2,\\pm 4,\\pm 6)$[/hide]\r\n\r\nEDIT: Wasn't this in the AOPS II Book?", "Solution_2": "[hide=\"i didnt see vishal s solution\"] let $k=x+y+z$. summing the three equations we get $2k(x+y+z)=288=2k^{2}$ thus $k=\\pm12$.\nso $(k-z)k=72\\Rightarrow z=\\frac{k^{2}-72}{k}=\\frac{144-72}{\\pm 12}=\\pm 6$\nsimilarly $y=\\frac{144-96}{\\pm 12}=\\pm 4$\nand $x=\\frac{144-120}{\\pm 12}=\\pm 2$. [/hide]" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "Construct the semicircle $ h$ with the diameter $ AB$ and the midpoint $ M$. Now construct the semicircle $ k$ with the diameter $ MB$ on the same side as $ h$. Let $ X$ and $ Y$ be points on $ k$, such that the arc $ BX$ is $ \\frac{3}{2}$ times the arc $ BY$. The line $ MY$ intersects the line $ BX$ in $ D$ and the semicircle $ h$ in $ C$.\r\nShow that $ Y$ ist he midpoint of $ CD$.", "Solution_1": "let $ O$ be center of $ k$ and $ a \\equal{} < YOB$, \r\n\r\nwe have $ < XMY \\equal{} < XMB \\minus{} < YMB \\equal{} 1/2( < XOB \\minus{} < YOB) \\equal{} a/4$,(because $ BX$ is $ 3/2$ times \r\n\r\n$ BY$). therfore $ < CDB \\equal{} 90 \\minus{} a/4$ (since $ < DXM \\equal{} < BXM \\equal{} 90$).\r\n\r\nand since $ MC \\equal{} MB$ then $ 2 < DCB \\plus{} < CMB \\equal{} 90$ .\r\n\r\nso $ < DCB \\equal{} 90 \\minus{} 1/2( < CMB) \\equal{} 90 \\minus{} a/4$(because $ < CMB \\equal{} < XMB \\equal{} a/2$).\r\n\r\nso $ < CDB \\equal{} < DCB$ and since $ BY$ is orthogonal to $ CD$ then $ YC \\equal{} YD$." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Think of a triangle ABC with $ \\angle$B=30.\r\n\r\nThe orthocenter of ABC is H.\r\n\r\nThe center of gravity(?) of ABH is G. (EDIT : G=the intersection point of the three medians)\r\n\r\nCG meets with AB at M and meets with AH at N.\r\n\r\nShow that AMN a equilateral triangle\r\n\r\nThanks.", "Solution_1": "[quote=\"bjh7790\"]Think of a triangle ABC with $ \\angle$B=30.\n\nThe orthocenter of ABC is H.\n\nThe center of gravity(?) of ABH is G. (EDIT : G=the intersection point of the three medians)\n\nCG meets with AB at M and meets with AH at N.\n\nShow that AMN a equilateral triangle\n\nThanks.[/quote]\r\n Let Q be the midpoint of $ AB, AH\\cap BC \\equal{} \\{J\\}$.Draw a line d which pass through C and parallel to $ QJ. d\\cap AH \\equal{} \\{N'\\}$\r\nWe have $ \\angle N'CJ \\equal{} \\angle BJQ \\equal{} 30^o \\equal{} 1/2\\angle HCJ$ then $ \\frac {HN'}{JN'} \\equal{} \\frac {HC}{JC} \\equal{} 2 \\equal{} \\frac {HG}{GQ}$\r\nSo $ GN'//QJ$ we get $ G,N',C$ are collinear. \r\n$ \\Rightarrow N\\equiv N'$ and $ MC//QJ$\r\nSince AQJ is the equilateral triangle we have q.e.d" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "let $a_{i}>0$ such that $a_{1}+a_{2}+...+a_{n}=n$. Prove that for each $r\\in N$,${{a_{1}^{a_{1}}a_{2}}^{a_{2}}...a_{n}}^{a_{n}}\\ge a_{1}^{r}a_{2}^{r}...a_{n}^{r}$", "Solution_1": "i'm waiting...\r\nif nobody has a solution then its ok but i will never post mine (until i change my mind) :P .", "Solution_2": "[quote=\"nayel\"]let $a_{i}>0$ such that $a_{1}+a_{2}+...+a_{n}=n$. Prove that for each $r\\in N$,${{a_{1}^{a_{1}}a_{2}}^{a_{2}}...a_{n}}^{a_{n}}\\ge a_{1}^{r}a_{2}^{r}...a_{n}^{r}$[/quote]\r\n\r\nI don't understand why $r$ has to be a natural number. The inequality is true for any $r \\in \\mathbb{R}^{+}$.", "Solution_3": "please dont post your solution. how was the problem?", "Solution_4": "[quote=\"nayel\"]please dont post your solution. how was the problem?[/quote]\r\n\r\nDon't post my solution? Sure. The problem was easy to me though.", "Solution_5": "which formula did you use?", "Solution_6": "[quote=\"nayel\"]which formula did you use?[/quote]\r\n\r\nIf you mean inequality, I used just Jensen's Inequality.", "Solution_7": "[quote=\"weiquan\"][quote=\"nayel\"]which formula did you use?[/quote]\n\nIf you mean inequality, I used just Jensen's Inequality.[/quote]\r\n\r\ncan you state the inequality?", "Solution_8": "blablablasdjfakljsklafjeiwfjk;lsdjzkl;fjwksdjzfoweoijfiowkjsdoif", "Solution_9": "[quote=\"nayel\"][quote=\"weiquan\"][quote=\"nayel\"]which formula did you use?[/quote]\n\nIf you mean inequality, I used just Jensen's Inequality.[/quote]\n\ncan you state the inequality?[/quote]\r\n\r\n$\\frac{1}{n}\\sum_{i=1}^{n}{f\\left(a_{i}\\right)}\\geq f\\left(\\frac{1}{n}\\sum_{i=1}^{n}{a_{i}}\\right)$ if $f(x)$ is convex over the range of $\\{a_{i}\\}$ and the inequality is reversed when $f(x)$ is concave. Now why [i]can't[/i] I go offline whenever I like? I can't possibly stay online [i]forever[/i]. And besides I did not go offline yet contrary to what you said." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "Let\n\n$f\\left( a;\\;b;\\;c\\right) =\\left| \\frac{\\left| b-a\\right| }{\\left| ab\\right| }+\\frac{b+a}{ab}-\\frac{2}{c}\\right| +\\frac{\\left| b-a\\right| }{\\left| ab\\right| }+\\frac{b+a}{ab}+\\frac{2}{c}$\n\nProve that\n$$\nf\\left( a,b,c\\right) =4\\max \\left( \\frac{1}{a}, \\frac{1}{b}, \\frac{1}{c}\\right) .\n$$\n\nFor any $n$ numbers $a_{1},$ $a_{2},$ $...,$ $a_{n},$ we use the notation $ \\max \\left( a_{1}, a_{2}, \\ldots, a_{n}\\right)$ for the greatest of the numbers $a_{1}, a_{2},\\ldots a_{n}.$", "Solution_1": "Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url] :)", "Solution_2": "No one solved this.", "Solution_3": "Hello, anyone has a solution! :maybe:", "Solution_4": "LaTeX back then was different. I think this is the updated latex:\n$ f(a,b,c)=4\\max(\\frac{1}{a}, \\frac{1}{b}, \\frac{1}{c})$\n[color=#f00]Amir: I just fixed it, thanks.[/color]", "Solution_5": "Use $|x-y|+x+y = 2\\max\\left(x,y\\right)$ two times then\n\n$$f(a,b,c) = 2\\max \\left( \\frac{\\left| b-a\\right| }{\\left| ab\\right| }+\\frac{b+a}{ab}, \\frac{2}{c}\\right) = 2\\max \\left( \\left| \\frac{1}{a}-\\frac{1}{b} \\right| + \\frac{1}{a}+\\frac{1}{b}, \\frac{2}{c}\\right) = 2\\max \\left( 2\\max \\left( \\frac{1}{a}, \\frac{1}{b} \\right), \\frac{2}{c}\\right) = 4\\max \\left( \\frac{1}{a}, \\frac{1}{b}, \\frac{1}{c}\\right)$$" } { "Tag": [ "geometry" ], "Problem": "In triangle $ABC$, let $K$ be midpoint of $AB$ and $L$ be on $AC$ such that $AL=LC+CB$. \r\n\r\nProve: $\\angle KLB$ is right if and only if $AC=3CB$", "Solution_1": "[hide]Let $E$ be a point in the extension of $AC$ such that $CE=CB$\n\n$AL = LC+CB \\Rightarrow AL=LC+CE \\Rightarrow AL=LE$\n\nSo, $L$ is the midpoint of $AE$\nAlso, $K$ is the midpoint of $AB$. So $KL \\parallel BE$\n\nLet $F$ be the symmetric point of $E$ w.r.t. $C$\nThe circle with diameter $FE$ passes through $B \\Rightarrow BF \\perp BE$\n\n\nLet $k$ be the positive real number such that $AC = k\\cdot BC$\n\n$AE = AC+BC \\Rightarrow AE = (k+1)BC$\n$AF = AC-BC \\Rightarrow AF = (k-1)BC$\n$AL=\\frac{AE}{2}\\Rightarrow AL = \\frac{k+1}{2}BC$\n\n$KL \\perp BL \\Leftrightarrow BE \\perp BL \\Leftrightarrow F \\equiv L \\Leftrightarrow k-1=\\frac{k+1}{2}\\Leftrightarrow k=3$\n\n\n[color=brown]Note \nWe have $BE \\perp BL \\Rightarrow F \\equiv L$, because only one perpendicular line is drawn to $AC$ through the point $B$. \nAnd the points $F,L$ are on $AC$[/color][/hide]", "Solution_2": "[hide=\"Part 1\"]If: $\\measuredangle KLB=90^\\circ$\n\nIt is enough to prove $BC=CL$\n\nDraw a parallel to $KL$ through $B$, it intersects $AC$ at $X$.\n\nThen we get $BC=CX$ and $\\measuredangle XBC= \\measuredangle CXB= \\measuredangle KLA$\n\nAlso: $\\measuredangle BCL= 2 \\measuredangle XBC$.\n\nWe can see that $BCL$ is isosceles then: $\\boxed{BC=CL}$ $\\mathbb{QED}$[/hide]\n\n[hide=\"Part2\"]If: $AC=3CB$\n\n$X$ is defined the same way.\n\nWe have: $AL=2LC=2BC=2CX$, then $\\measuredangle CBL= 90^\\circ$\n\n$BC//KL$: $\\measuredangle CBL= \\boxed{\\measuredangle KLB= 90^\\circ}$ $\\mathbb{QED}$[/hide]", "Solution_3": "[b][u]Very nice and ... easy problem ![/u][/b]\r\n\r\n[hide=\"Here are two proofs and an easy extension.\"][color=darkblue][b]Proof 1 (metric).[/b]\n\n$\\{\\begin{array}{c}LA=\\frac{b+a}{2}\\\\\\\\ LC=\\frac{b-a}{2}\\end{array}$ $\\Longrightarrow$ $\\{\\begin{array}{c}4b\\cdot BL^{2}=[(2a-b)(a+b)^{2}+2c^{2}(b-a)]\\\\\\\\ 4b\\cdot KL^{2}=[a(a+b)^{2}-ac^{2}]\\end{array}\\ .$\n\n$\\boxed{\\ LK\\perp LB\\ }$ $\\Longleftrightarrow$ $LK^{2}+LB^{2}=KB^{2}$ $\\Longleftrightarrow$ $4b\\cdot LK^{2}+4b\\cdot LB^{2}=4b\\cdot (\\frac{c}{2})^{2}$ $\\Longleftrightarrow$\n\n$a(a+b)^{2}-ac^{2}+(2a-b)(a+b)^{2}+2c^{2}(b-a)=bc^{2}$ $\\Longleftrightarrow$\n\n$(3a-b)[(a+b)^{2}-c^{2}]=0$ $\\Longleftrightarrow$ $\\boxed{\\ b=3a\\ }\\ .$\n\n[b]Proof 2 (synthetic).[/b] Denote the point $S\\in (BK)$ for which $\\overline{BS}=2\\cdot\\overline{SK}\\ .$ Prove easily that $\\overline{AB}=3\\cdot \\overline{SB}$ and $\\frac{\\overline{AB}}{\\overline{AK}}=2=-\\frac{\\overline{SB}}{\\overline{SK}}\\ ,$ i.e. the points $A,S$ are (harmonical) conjugate w.r.t. the pair of the points $\\{B,K\\}\\ .$ Therefore, $b=3a$ $\\Longleftrightarrow$ $LA=2a$ and $LC=a$ $\\Longleftrightarrow$ $LS\\parallel BC$ and the ray $[LB$ is the bisector of the angle $\\widehat{CLS}$ $\\Longleftrightarrow$ $LK\\perp LB\\ .$ \n\n[b]An easy extension.[/b] Let $ABC$ be a triangle for which exists the point $L\\in (AC)$ so that $2\\cdot LA=(k-1)\\cdot (LC+CB)\\ ,$ where $k>1\\ .$ Denote the point $K\\in (AB)$ for which $\\frac{KA}{KB}=\\frac{k-1}{2}\\ .$ Prove that $LK\\perp LB$ $\\Longleftrightarrow$ $b=ka\\ .$\n\n[b]Remark.[/b] For $k: =3$ obtain the proposed problem.[/color][/hide]" } { "Tag": [ "geometry", "real analysis", "real analysis unsolved" ], "Problem": "Prove that the intersection of 2 neighborhoods $N_1$ and $N_2$ of a point x is as well a neighborhood of point x", "Solution_1": "Would \"neighborhood\" be a better word here than \"area\"?", "Solution_2": "Fixed, thank you" } { "Tag": [ "algebra", "system of equations", "algebra unsolved" ], "Problem": "Find all values of $ a$ and $ b$ for which the system of equations \r\n $ xyz \\plus{} z \\equal{} a$\r\n$ xyz^2 \\plus{} z \\equal{} b$\r\n$ x^2 \\plus{} y^2 \\plus{} z^2 \\equal{} 4$\r\nhas only one real solution.", "Solution_1": "Let $ a$ and $ b$ be real numbers satisfying the condition.\r\nObserve that if$ x$ or $ y$ is not zero, and $ (x,y,z)$ is a solution, then $ ( \\minus{} x, \\minus{} y,z)$ will be another solution.\r\nHence, we have $ x \\equal{} y \\equal{} 0$.\r\nThen it follows that $ a \\equal{} b \\in \\{ \\minus{} 2,2\\}$.\r\nAnd now we have to check which value of $ a,b$ that satisfies the condition.\r\n\r\n(i) $ a \\equal{} b \\equal{} 2$ doesn't work since we have $ (x,y,z) \\equal{} (\\frac {\\sqrt {5} \\plus{} 1}{2},\\frac {\\sqrt {5} \\minus{} 1}{2},1)$ be a solution.\r\n(ii) $ a \\equal{} b \\equal{} \\minus{} 2$ doesn't have real solutions.\r\n\r\nFinally, there doesn't exist any value of $ a,b$ which satisfies the condition.\r\n :)" } { "Tag": [], "Problem": "In the next three days, 2000 people will move to Florida. How\nmany people, on average, will move to Florida each hour?\nExpress your answer to the nearest whole number.", "Solution_1": "There are 72 hours in 3 days. On average, $ \\frac {2000}{72}$, approximately equal to 28, migrate every hour.\r\n\r\nEdited: Added a 3.", "Solution_2": "You mean 72 hours in 3 days." } { "Tag": [ "calculus", "integration", "logarithms", "calculus computations" ], "Problem": "Evaluate $ \\int_1^e \\frac{1\\minus{}x(e^x\\minus{}1)}{x(1\\plus{}xe^x\\ln x)}\\ dx$.", "Solution_1": "hello \r\n$ \\int_{1}^{e}\\frac{1-x(e^{x}-1)}{x(1+xe^{x}\\ln x)}\\ dx$ \r\nit can be simplified to \r\n$ =\\int_{1}^{e} \\frac{x+1}{x}- \\frac{e^{x}(ln(x)(x+1)+1)}{1+xe^{x}\\ln x} \\,dx$ \r\n$ =e -ln(e^{e+1}+1)$ \r\nthank u", "Solution_2": "That's correct. :)" } { "Tag": [ "probability", "geometry", "AMC", "AIME", "MATHCOUNTS", "calculus", "number theory", "\\/closed" ], "Problem": "The Art of Problem Solving Spring Schedule is available [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url]. The classes we offer this fall and winter are listed below.\r\n[b]Note[/b]: We have added a second session of the Algebra 1 class, so you can enroll in either the Wednesday night class or the Thursday night class. The two offerings are the same.\r\n\r\n\r\n[b]Introductory level: Grades 6-10[/b]\r\n\r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=41]Algebra 1[/url]. Session one: Wednesdays starting Feb 24. Session 2: Thursdays starting Feb 25. \r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=1]Introduction to Counting & Probability[/url]. Mondays starting March 1.\r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=2]Introduction to Number Theory[/url]. Fridays starting March 5.\r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=42]Algebra 2[/url]. Tuesdays starting Feb 23.\r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=10]Introduction to Geometry[/url]. Wednesdays starting March 3.\r\n[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=5]Advanced MATHCOUNTS/AMC 8[/url]. Fridays starting Feb 12.\r\n\r\n[b]Intermediate level: Grades 8-12[/b]\r\n\r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=4]Algebra 3[/url]. Thursdays starting March 4.\r\n[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=31]Intermediate Number Theory[/url]. Tuesdays starting March 2.\r\n*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]Precalculus[/url]. Mondays starting March 8.\r\n[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=8]Special AIME Problem Seminar[/url]. Saturday-Sunday Mar 6-7.\r\n\r\nClasses marked with a (*) have an accompanying textbook. If you plan to enroll in one of these courses, you should do so well before the start of the course, so that you receive the text before the class starts. \r\n\r\nAll of our courses offer full transcripts for each class, so if you're going to miss a class or two, you'll still have access to everything that happened in class.\r\n\r\nIf you have any questions about a specific class, please contact us at classes@artofproblemsolving.com.", "Solution_1": "What will the Intermediate Number Theory seminar help with (as in level of contests)?", "Solution_2": "Hard AMC 12, AIME, beginning USAMO.", "Solution_3": "[quote]Classes marked with a (*) have an accompanying textbook. If you plan to enroll in one of these courses, you should do so well before the start of the course, so that you receive the text before the class starts.[/quote]\r\n\r\nUhh, I don't think Intermediate Number Theory has a book (unless you were planning to surprise us? :O).", "Solution_4": "Thanks; I fixed it.", "Solution_5": "when do the signups need to be in for intro to counting and number theory?", "Solution_6": "We recommend signing up for classes with textbooks at least 2 weeks before they start, so you can make sure to receive the text and have time to work a little on the first couple chapters.", "Solution_7": "Thanks!! i will enroll!", "Solution_8": "Im enrolled in Intro 2 Number Theroy... :) :) :)", "Solution_9": "[quote]Im enrolled in Intro 2 Number Theroy... \n\n[/quote]\r\nThat class is fun!!! I took it in the fall.\r\n\r\nAnother question: If i'm going to enroll in advanced mathcounts, should I do Intro to geometry, beause I have took Number theory, and Intro to counting and probabilty and I took geometry in school??? :?: \r\n\r\nCould someone please help me?\r\nThanks in advance :)", "Solution_10": "Take a look at the diagnostic test for the geometry course [url=http://www.artofproblemsolving.com/Classes/IntroGeom/PostTest.pdf]here[/url]. If you can handle most of the problems on this, then you probably don't need the Intro Geometry course.", "Solution_11": "how many classes are there in intro to counting and probability???", "Solution_12": "Umm... if you click the \"Online Classes\" button then it gives an in-depth explanation on all of the classes. You can find length of Intro to C&P there.", "Solution_13": "[quote=\"Letti\"]how many classes are there in intro to counting and probability???[/quote]12", "Solution_14": "May I just ask... what's the difference between Algebra 3 and Precalculus?", "Solution_15": "Take a look at the course descriptions [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=4]here[/url] and [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]here[/url] to see the difference between the two courses.", "Solution_16": "what about calculus?", "Solution_17": "Also for calculus, look [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=40]here[/url]. Seems like a long description.\r\n\r\nBy the way, is the Calculus textbook close to finishing? I remember it is supposed to come out in late January, and I'm really curious what it's going to be like.", "Solution_18": "We're hoping to get it to the printer in the next month.", "Solution_19": "are you gonna offer a class for calculus though?", "Solution_20": "We do; it starts every fall.", "Solution_21": "I was looking at the online classes and came upon Olympiad Geometry and I am interested in that. However it has no future date. Are you planning to do it or just abandon it? If you are doing it, can you do it in the summer? Thank you!", "Solution_22": "We haven't decided when we will next offer the Olympiad Geometry. It's unlikely to be summer, since many of the students who might be interested in that class will likely be in some summer program.", "Solution_23": "oh dang! well what would the cost be of that class?", "Solution_24": "[quote=\"JITTT\"]oh dang! well what would the cost be of that class?[/quote]\r\n\r\nWe haven't decided. I think last time we offered it, the cost was around $ \\$250$.", "Solution_25": "Im enrolled in Algebra 2", "Solution_26": "Since the Special AIME Seminar is sort of a wrap-up class, how many questions is it going to have? Does it also discuss the test-taking strategies and message board problems?", "Solution_27": "It doesn't have message board problems, but there will be some discussion of test-taking strategies. The number of problems depends on how well the students are doing in the class. We do as many as we can get through in the 5 hours of class.", "Solution_28": "When will Olympiad Geometry be offered?", "Solution_29": "We haven't decided yet. Most likely, it will be in spring, 2011.", "Solution_30": "That is a long time away. Would WOOT be too for me? I have not taken a formal AIME yet, but looking at the old AIMES, I can do almost all geometry problems. I also did 5/6 on the Are You Ready? for Olympiad Geometry. Now besides the geometry, everything else is average. Would you recommend Precalculus or AIME Problem Series?", "Solution_31": "The Precalculus is much more thorough than the AIME Problem Series; in general, if you have the time, the subject classes will give you a much broader and deeper exposure than the Problem Series classes.\r\n\r\nIf you keep working at math between now and the fall, it's likely you'll be ready for WOOT, though I'd want to hear more about your math background before I'd say that for sure. That said, you can learn plenty between now and September, so if you're anywhere close to ready for WOOT now, you'll likely be ready by September.", "Solution_32": "Thank you sir, I think I will take Precalculus." } { "Tag": [], "Problem": "What were your favorite problems and what were the hardest problems from this year's contest?", "Solution_1": "Not many of the problems were very original, but it was a good written test none the less. Some of the problems had too much computation, such as the one asking for the number of nines in $ 999999899999^2$ and the one asking for the number of partitions of $ 9$. The former could be simplified by rewriting the given number as $ (10^{12} \\minus{} 10^5 \\minus{} 1)^2$, but it was still fairly annoying from there. I don't think there was any non-brute-force way to do the latter. \r\nThere was not much of a difficulty gap in my opinion. I think that most of the problems were at essentially the same difficulty level, resulting in a number of very low scores and a number of very high scores (three-way tie for first place). Maybe the tie-breakers should have been a little more difficult because both the first and second place finishers got both of them correct, so the first place plaque was given at the judges' discretion.", "Solution_2": "I thought it was a nicely written test. Do you guys know when the results will be posted online?" } { "Tag": [ "function", "vector", "superior algebra", "superior algebra solved" ], "Problem": "Let $A$ be a ring and $a\\in A$ a non-invertible element. Denote by $X_a=\\{x\\in A: ax=1\\}$. Prove that $X_a=\\Phi$ or $X_a$ is infinite.", "Solution_1": "Simple algebraic manipulation shows that if $ax=1$ but $a$ is not left invertible then the elements of the fom $\\alpha_n=xa^{n+1}-a^n,\\ n\\ge 0$ are all distinct (we take $a^0=1$), and satisfy $a\\alpha_n=0,\\ \\forall n$. This means that we can take our infinite set of right inverses to be $x+\\alpha_n$.", "Solution_2": "I foud a proof for it in a book. Here it is: \r\n\r\nAssume that $X_a\\neq \\Phi$and $X_a$ is finite, and choose $b$ one of it's elements. Define $f: X_a\\to X_a$, $f(x)=xa-1+b$ this function well defined: $a(xa-1+b)=ab=1$, is injective: $f(x)=f(y)\\Rightarrow ax=ay\\Rightarrow a(x-y)=0\\Rightarrow ba(x-y)=0\\Rightarrow x=y$. \r\n\r\nSince $X_a$ is finite it is surjective, hence $\\exists c$ s.t. $b=f(c)=ca-1+b\\Rightarrow ca=1$ hence $a$ has leftinverse, hence it is invertible.", "Solution_3": "Can anyone give me an example in which that set is infinite?", "Solution_4": "Try taking $A$ to be the ring of endomorphisms of an infinite-dimensional vector space (over any field), and $a$ a surjective but not injective endomorphism." } { "Tag": [ "conics", "ellipse", "search", "geometry", "ratio", "analytic geometry", "trigonometry" ], "Problem": "Two different ellipses are given. One focus of the first ellipse coincides with one focus of the second ellipse. Prove that the ellipses have at most two points in common.", "Solution_1": "There's a [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=646639421&t=29201]very old thread[/url] in the Geometry section where this is proved. I think posts #7 and 8 in that topic add up to a proof. \r\n\r\nThe idea is that given two ellipses $ \\mathcal E_i,\\ i\\equal{}1,2$ with common focus $ F$, (a) all their points of intersection lie in one of the four regions into which the directrices of $ \\mathcal E_i$ corresponding to $ F$ cut the plane, and (b) for all the intersection points $ A$, the ratio between the distances from $ A$ to those two directrices is the same. It follows that all the intersection points are collinear, so, since they all belong to an ellipse, there can't be $ 3$ (or more) of them.", "Solution_2": "I have a very interesting solution, different and shorter than the official one. It came to me on the train, on my way home. :)\r\n\r\nGiven 3 points $ A,B,C$ and a focus $ F$, we want to prove that there is only one ellipse containing $ A,B,C$ and having one focus $ F$.\r\nConsider the circles $ \\mathcal{A},\\mathcal{B},\\mathcal{C}$ of centers $ A,B,C$ and rays $ FA,FB,FC$ respectively. There exists at most one circle $ \\mathcal{D}$ which contains all other three circles and is tangent to them. If one such circle exists, then its center is the second focus of the ellipse, so the ellipse is uniquely determined. If such circle does not exist, then there is no ellipse with the required proprieties. \r\n\r\nSo in our problem, if the two ellipses have 3 points in common and one focus in common, then they are not different anymore. :)", "Solution_3": "Here's a more calculative approach. \r\nThe equation in polar coordinates $ (r,\\theta)$ of an ellipse with a focus at the origin and the big axis aligned with $ \\theta\\equal{}0$ is $ r\\equal{}\\frac{b^2}{a\\minus{}c\\cos\\theta}$\r\nThus, the equations for the 2 ellipses can be written as $ r\\equal{}\\frac{b_1^2}{a_1\\minus{}c_1\\cos(\\theta\\minus{}\\theta_1)}$ and $ r\\equal{}\\frac{b_2^2}{a_2\\minus{}c_2\\cos(\\theta\\minus{}\\theta_2)}$. \r\nTherefore, the a point of intersection $ (r,\\theta)$ satisfies $ \\frac1{b_1^2}(a_1\\minus{}c_1\\cos(\\theta\\minus{}\\theta_1))\\equal{}\\frac1{b_2^2}(a_2\\minus{}c_2\\cos(\\theta\\minus{}\\theta_2))$, which is equivalent to $ \\cos(\\theta\\minus{}\\theta_0)\\equal{}\\alpha$ for some constants $ \\theta_0$ and $ \\alpha$. The last equation has at most 2 solutions in $ \\theta$.", "Solution_4": "Another way is to apply the transformation which maps one of the ellipses onto the circle: the common focus is mapped to the center. From this the question becomes obvious.", "Solution_5": "Why is it *obvious*? \r\n\r\nEllipses and circles can have four points of intersection in principle.", "Solution_6": "Just adding the geogebra picture :D [geogebra]29df07b3e987dbe200c2eb9b47d17de706915fcd[/geogebra]", "Solution_7": "Let $\\mathcal E_1$ be an ellipse with foci $F$ and $F_1$ and $\\mathcal E_2$ be an ellipse with foci $F$ and $F_2$. Suppose for the sake of contradiction that three intersection points $P_1$, $P_2$, $P_3$ exist. Then \\[P_iF_1 - P_iF_2 = (P_iF_1 + P_iF) - (P_iF_2 + P_iF)\\quad\\text{for }i=1,2,3,\\] so $P_1$, $P_2$, and $P_3$ all lie on the same branch of a hyperbola with foci $F_1$ and $F_2$. But such a branch can only intersect an ellipse in two places, contradiction." } { "Tag": [ "number theory proposed", "number theory" ], "Problem": "Find all triples $ (a,b,c)$ of positive integers such that $ a^2\\plus{}2^{b\\plus{}1}\\equal{}3^c$.", "Solution_1": "I think it is an old problem . \r\n$ 3^{c}\\equiv (\\minus{}1)^{c}(\\mod 4)$\r\nTherefore $ 4|a^2\\minus{}(\\minus{}1)^{c}$\r\nIt gives $ c\\equiv 0(\\mod 2)$ ,then exist an integer m such that $ c\\equal{}2m$\r\n$ (3^m\\minus{}a)(3^m\\plus{}a)\\equal{}2^{b\\plus{}1}$\r\nExist $ (x,y)\\in N_0$ such that :\r\n$ 3^m\\minus{}a\\equal{}2^x,3^m\\plus{}a\\equal{}2^y,x\\plus{}y\\equal{}b\\plus{}1$\r\n$ \\Rightarrow 2.3^m\\equal{}2^x\\plus{}2^y$\r\nIt follows that $ x\\equal{}1$ ,then we have : \r\n$ 3^m\\equal{}1\\plus{}2^{y\\minus{}1}$\r\nThis is catalant equation ,it have only solutions : \r\n$ (m,y)\\equal{}(1,2),(2,4)$", "Solution_2": "[quote=\"TTsphn\"]I think it is an old problem . \n$ 3^{c}\\equiv ( \\minus{} 1)^{c}(\\mod 4)$\nTherefore $ 4|a^2 \\minus{} ( \\minus{} 1)^{c}$\n [/quote]\r\nSorry , but just currious why could you find out $ 4|a^2\\minus{}(\\minus{}1)^c$ in paragraph I quoted.I think your result only happens if b+1 is even", "Solution_3": "[quote=\"nkht-tk14\"][quote=\"TTsphn\"]I think it is an old problem . \n$ 3^{c}\\equiv ( - 1)^{c}(\\mod 4)$\nTherefore $ 4|a^2 - ( - 1)^{c}$\n [/quote]\nSorry , but just currious why could you find out $ 4|a^2 - ( - 1)^c$ in paragraph I quoted.I think your result only happens if b+1 is even[/quote]\r\nIt is clearly$ a\\equiv b (\\mod c)$ so $ c|a - b$\r\nBut his proof is not suitable for a contes (catalan equation is very hard)\r\nWho can solve by an other way", "Solution_4": "Catalant equation is not very hard in the case : \r\n\\[ |2^a\\minus{}3^b|\\equal{}1\r\n\\]\r\nThis is one of problems in my school contest . \r\nYou can find it in our forum :)", "Solution_5": "[quote=\"TTsphn\"]Catalant equation is not very hard in the case :\n\\[ |2^a \\minus{} 3^b| \\equal{} 1\n\\]\nThis is one of problems in my school contest . \nYou can find it in our forum :)[/quote]\r\nYes,thanks ,of course in that case,it is not really hard,I remember that case was found first by a France", "Solution_6": "Good solutions.\r\n\r\nThe latest equations can solve also with graph, $ y \\equal{} 2^{x\\minus{}1} \\plus{} 1$ and $ y \\equal{} 3^x$ :lol:", "Solution_7": "Seeing modulo $4$ we have $a^2\\equiv_4(-1)^c$, therefore $a$ is odd and $c$ is even. So put $c=2d$. Then we have $(3^d-a)(3^d+a)=2^{b+1}$, which means $3^d+a$ and $3^d-a$ are powers of $2$ whose difference is $2a$, and since $2\\parallel2a$, we have $3^d-a=2$, therefore we have $3^d=2^{b-1}+1$. Now consider the following cases:\nCase 1: if $b=1$, then we have $3^d=2$, which is impossible.\nCase 2: if $b=2$, then we have $d=1$, $c=2$ and $a=1$, from where we obtain $(a,b,c)=(1,2,2)$, which is a solution.\nCase 3: if $b\\ge3$, seeing modulo $4$ we have $(-1)^d\\equiv_41$, therefore $d$ is even. So put $d=2e$. Then we have $(3^e-1)(3^e+1)=2^{b-1}$, which means $3^e+1$ and $3^e-1$ are powers of $2$ whose difference is $2$, and since $2\\parallel 2$, we have $3^e-1=2$, which means $e=1$, $d=2$, $c=4$, $b=4$ and $a=7$, from where we obtain $(a,b,c)=(7,4,4)$, which is a solution.", "Solution_8": "Considering $\\mod{4}$,$a$ is odd and $c$ is even.Let $c=2c'(c'\\in \\mathbb N)$,then $(3^{c'}+a)(3^{c'}-a)=2^{b+1}$.For $m>n\\ge 0$,\n\\[\\left\\{\\begin{array}{cc}\n3^{c'}+a=2^m \\\\\n3^{c'}-a=2^n \\\\\n\\end{array}\\right.\\]\n$\\rightarrow 2a=2^n(2^{m-n}-1)$.Since $a$ is odd,$n=1$ and $2\\cdot 3^{c'}=2^m+2\\Leftrightarrow 3^{c'}=2^{m-1}+1$.\n\nIf $m=1$,then $3^{c'}=2$ which is absurd.Thus $m\\ge 2$.Considering $\\mod{3}$,$m$ is even.\n\n[b]Case1:[/b]$m=2$\n$c'=1\\rightarrow c=2,a=1,b=2$ which satisfies the condition.\n\n[b]Case2:[/b]$m=4$\n$c'=2\\rightarrow c=4,a=7,b=4$ which satisfies the condition.\n\n[b]Case3:[/b]$m>4$\nBy [b]Zsigmondy's theorem[/b],\u2203$p$(prime) s.t. $p|2^{m-1}+1,p\\nmid 2+1=3$ which is absurd.\n\nFrom case$1,2,3$,the answer is\n$\\boxed{(a,b,c)=(1,2,2),(7,4,4)}\\blacksquare$ :coool:", "Solution_9": "my solution is here;\n\neasily, a is odd so, a^2 = 1(mod 4)\n\nalso, 2^(b + 1) = 0(mod 4)\n\nso, 3^c = 1(mod 4)\n\nc = 2m;\n\na^2 + 2^(b + 1) = (3^m)^2;\n\n2^(b + 1) = (3^m)^2 - a^2 = (3^m + a)(3^m - a);\n\n2^x = 3^m + a , 2^y = 3^m - a;\n\nso, 2*3^m = 2^y(2^(x - y) + 1);\n\ny = 1, 3^m = 2^(x - 1) + 1;\n\nlet p = x - 1;\n\n3^m = 2^p + 1;\n\nif p = 1, m = 1;\n\nx = 2 , c = 2 , a = 1\n\n(a, b, c) = (1, 2, 2)\n\nif p > 1,\n\n3^m = 1(mod 4)\n\nm = 2n;\n\n2^p = (3^n + 1)(3^n - 1)\n\n3^n + 1 = 2^e , 3^n - 1 = 2^f;\n\n2^e - 2^f = 2;\n\n2^f(2^(e - f) - 1) = 2;\n\nf = 1, n = 1 , m = 2, c = 4, a = 7\n\nso, (a, b, c) = (1, 2, 2),(7, 4, 4) these are the only answer", "Solution_10": "good problem I solved it with a great fun hard but nice", "Solution_11": "@fighter\nWhy don't you use $\\LaTeX$? $\\LaTeX$ make mathematical expression be easy to see. :D", "Solution_12": "please give me a tutorial also turn my answer into latex", "Solution_13": "easy just find that $c$ is even and factor", "Solution_14": "@fighter\nI don't know tutorial of $\\LaTeX$.I've studied $\\LaTeX$ by seeing other people's code.It is easy.You just enclose mathematical expression with '$' mark.Try by all means!!! :D ", "Solution_15": "@fighter\nYou can click other person's mathematical formula and see the code.\n\nOr see here.\nhttp://www.artofproblemsolving.com/wiki/index.php/LaTeX\n\n\nLaTeX makes your solution beautiful. :-D ", "Solution_16": "First of all notice that $a^2+2^{b+1}+1\\equiv0\\text{ or }1\\pmod 4$ and $3^c\\equiv(-1)^c\\pmod 4$ thus $c$ is even and $a\\equiv1\\text{ or }3\\pmod 4$\nSo now let $c=2k$ for some $k\\in\\mathbb{Z}^+$\nIf $b=1$ we have that $a^2+4=3^c$ however $a^2+4\\equiv1\\text{ or }2\\pmod 3$ while $3^c\\equiv0\\pmod 3$ which is a contradiction. From now on assume that $b\\ge2$\nUsing $c=2k$ we can transform in the original equation into $a^2+2^{b+1}=(3^k)^2\\Longrightarrow2^{b+1}=(3^k-a)(3^k+a)$ \nTherefore $3^k-a=2^x\\text{ and }3^k+a=2^y\\text{ such that }x1$\nNow notice that by the Catalan conjecture the only solution for $3^k-2^{b-1}=1$ for $k,b-1>1$ is $k=2\\text{ and }b-1=3\\Longrightarrow c=4\\text{ and }b=4$\nPlugging this in yields $a^2+2^5=3^4\\Longrightarrow a^2=49\\Longleftrightarrow a=7$ thus $(a,b,c)=(7,4,4)$ is the only other solution.\n\nSo, to sum up $\\boxed{(a,b,c)=(1,2,2)\\text{ and }(7,4,4)\\text{ are the only solutions for }a,b,c\\in\\mathbb{Z}^+}$ $\\blacksquare$." } { "Tag": [ "group theory", "abstract algebra", "superior algebra", "superior algebra unsolved" ], "Problem": "The well-known $ 15$-puzzle consists of a shallow box filled with $ 16$ small squares in $ 4 \\times 4$ array. The bottom right corner square is removed, and the other squares are labeled as ;(all in this order) $ 1,2,3,4$ in first row ,$ 5,6,7,8$ in second row, $ 9,10,11,12$ in third row and $ 13,14,15$ in the end row.By sliding the squares around (without lifting them up),Show that the set of possible permutations that can be obtained with the bottom right square blank is precisely $ A_{15}$. (There is no known easy proof that all elements in $ A_{15}$ must occur.)", "Solution_1": "My solution for the easy part (a permutation must lie in A_15)\r\n\r\nWe label the blank as 16. Then the required permutation is an element in $ S_{16}$ with 16 fixed. \r\nThere are two types of moves: \r\nType A: Horizontal, in which the two numbers in the corresponding transposition differ by 1.\r\nType B: Vertical, in which the two numbers in the corresponding transposition differ by 4.\r\nWe want to show that there are even number of both types of transpositions.\r\n\r\nFor type A, consider this:\r\nAny required permutation is a product $ (16 ,a_1)(a_1,a_2),...,(a_n,16)$\r\nmeaning the blank moves to $ a_n$, then $ a_{n \\minus{} 1}$,...,then $ a_1$, then move back the the original square. Note that the sum $ 16 \\minus{} a_n \\plus{} ... \\plus{} a_2 \\minus{} a_1 \\plus{} a_1 \\minus{} 16 \\equal{} 0$. In this sum, a type A move accounts for 1 or -1, and a type B move accounts for 4 or -4. Since the total sum is even, there are even number of type A moves.\r\n\r\nFor type B, consider this:\r\nA type B move shifts a \"level\" to another. (There are 4 levels, Level 1 = {1,2,3,4}, Level 2 = {5,6,7,8}, Level 3 = {9,10,11,12} and Level 4 = {13,14,15,16}). \r\nWe start at the fourth level. Again write a required permutation as a product $ (16 ,a_1)(a_1,a_2),...,(a_n,16)$. In which the total sum of \"level changes\" = 0 (since we go back to the original block eventually). Since a type A move doesn't affect the level, and a type B move changes the level by 1 or -1, there must be even number of type B moves.\r\n\r\nThere is a brute force way for the other half, that is to show that $ (123)$,...,$ (13,14,15)$ all can be done. Since it lies in $ A_{15}$ and it is obviously a subgroup, all 3-cycles can then be generated and thus $ A_{15}$.", "Solution_2": "[quote=\"ehsan2004\"]The well-known $ 15$-puzzle consists of a shallow box filled with $ 16$ small squares in $ 4 \\times 4$ array. The bottom right corner square is removed, and the other squares are labeled as ;(all in this order) $ 1,2,3,4$ in first row ,$ 5,6,7,8$ in second row, $ 9,10,11,12$ in third row and $ 13,14,15$ in the end row.By sliding the squares around (without lifting them up),Show that the set of possible permutations that can be obtained with the bottom right square blank is precisely $ A_{15}$. (There is no known easy proof that all elements in $ A_{15}$ must occur.)[/quote]\r\n\r\nHere's a proof that any element $ s \\in A_{15}$ must occur. \r\n\r\nGiven $ s$, you can move 2 to the top left corner, then 3 to the right of 2, then 4 to the right of 3. Finally move 1 to the square below 2:\r\n\r\n[code]2 3 4 *\n1 * * *\n* * * *\n* * * *\n[/code]\n\nNow move the empty square to the top right position. Move 4 right, 3 right, 2 right and 1 up. This completes the first row. In this way, you can complete the second row as well:\n\n[code]1 2 3 4\n5 6 7 8\n* * * *\n* * * *\n[/code]\n\nNow move 13 to the square below 5, and hope that 9 does not lie below 13:\n\n[code]1 2 3 4\n5 6 7 8\n13* * *\n* * * *\n[/code]\n\nIn that case, move 9 to the right of 13, move the empty square to below 13, then move 13 down, 9 left. This completes the column for 9 and 13. If 9 were below 13, we're stuck temporarily - but fortunately not for long. Move the empty square to the right of 13, then perform: RULLDRRULDRULDLURDRUL, and now 9 and 13 are in place.\n\nSimilarly, move 10 and 14 in place. This leaves us with:\n\n[code]1 2 3 4\n5 6 7 8\n9 10 * *\n13 14 * *\n[/code]\r\n\r\nMove 11 to its rightful position and the empty square to the bottom right. Now the remaining two squares must be in their correct position - otherwise, it'll be an odd permutation. :)" } { "Tag": [ "geometry", "circumcircle", "geometric transformation", "reflection", "symmetry", "homothety", "perpendicular bisector" ], "Problem": "Consider $\\triangle{ABC}$ with circumcircle $\\omega$, and let $K$ be an interior point (to triangle). Let $AK$, $BK$ and $CK$ meet $\\omega$ at $A_1$, $B_1$ and $C_1$, respectively.\r\nNow, denote by $A_2$, $B_2$ and $C_2$ the reflection of $A_1$, $B_1$ and $C_1$, respectively, in the midpoints of $BC$, $CA$ and $AB$.\r\nProve that the circumcircle of $\\triangle{A_2B_2C_2}$ passes through a fixed point (when $K$ varies).\r\n\r\nNote: It\u2019s quite obvious that such a fixed point is $H$ - the orthocenter of $\\triangle{ABC}$, considering the trivial case: $K=O$.", "Solution_1": "I think this is from 2003 year. It is not easy(for me).", "Solution_2": "Note that the perpendicular bisector of $A_2H$ is the image of line $AK$ through a symmetry centered at $F$, the midpoint of $OH$, also the center of the nine point circle (see figure, here $AA'$ is a diameter). Thus, the perpendicular bisectors of $A_2H,B_2H,C_2H$ are concurrent, hence the conclusion.\r\nI used well known facts, like $H$ is symmetric to $A'$ with respect to the midpoint of $BC$...", "Solution_3": "1. Thanks to Enescu for a cute solution!\r\n\r\n2. Thanks to moderator for notification in the source :D , it's much better now :P !", "Solution_4": "Great solution, [b]enescu[/b]. I have another solution, not so ellegant. First note that $HA_2BC$ can be inscribed in a circle. The same goes for $HB_2AC$ and $HC_2AB$. Take an inversion with pole $H$ and power that invariates circle $\\omega$. Let $X'$ be the image of $X$ in this inversion. Clearly we have $B', C'$ and $A_2'$ on a line. The same goes for $A',C',B_2'$ and $A',B',C_2'$. Now we have to show that $A_2',B_2' and C_2'$ are on a line. Some anglechasing shows that $\\angle A_2'HB' = \\angle KAC$ and $\\angle A_2'HC' = 180-\\angle KAB$. Thus $\\frac{A_2'B'}{A_2'C'} = \\frac {HB'sin(KAC)}{HC'sin(KAB)}$. (And the same for $B_2'$ and $C_2'$). Multiply these 3 identities and use [b]Menealaos Theorem[/b] for triangle $A'B'C'$ and the desired result follows easily.", "Solution_5": "enescu,i didn't understood what the quefitient of your homotety is." } { "Tag": [], "Problem": "Ok, basically this is a game where I start by listing a food.\r\n\r\nThe next person must list a new food that starts with the last letter of my food.\r\n\r\nYou 1 point for each new food that you list.\r\n\r\nIf someone catches a repeated food, that person receives 10 points!\r\n\r\nREADY?? I will list the first food in bold.\r\n\r\n[b]Oatmeal[/b]", "Solution_1": "It is like the Geography game. \r\n\r\n[b]Lasagna[/b] (if that is how you spell it)\r\n\r\nPS Do you lose points for mispelled words? If so, don't make me lose points because I didn't know.\r\n\r\nEDIT: I misspelled misspell. :rotfl: :rotfl: (not on purpose, though)", "Solution_2": "no lost points for mispellings :lol: don't worry I am generous!\r\n\r\n[b]Apple[/b] unfortunately, this is the only one I could think of.\r\n\r\nBy the way, if the food is a compound word, go with the last lettter.", "Solution_3": "EGG\r\n\r\nMy message is too small, so I am typing this to make it longer.", "Solution_4": "Guacamole\r\n\r\nis this message too small?", "Solution_5": "enchilada\r\n\r\nAnybody keepin score?", "Solution_6": "[b]anchovy[/b]", "Solution_7": "[b]Yoghurt[/b]", "Solution_8": "[b]tuna[/b]\r\n\r\nwill this include drinks and brand-name snacks?", "Solution_9": "Avocado\r\n\r\n_", "Solution_10": "Orange\r\n\r\nThis message is most certainly NOT too small", "Solution_11": "[b]Egg Salad :P[/b]", "Solution_12": "[b]Donut[/b]", "Solution_13": "[b]Turkey[/b]", "Solution_14": "[b]Yeast[/b]", "Solution_15": "liquified human brains served fear factor style in a real skull.", "Solution_16": "@abcak, I would [i]drink it[/i], if it was some smart brain liquid... :rotfl: :rotfl: \r\n[b]\nAn impaled scorpion dipped in garlic and tomato sauce, freshly caught and poisoned from the desert[/b]\r\n\r\nI WILL [u]NEVER [/u] EAT THAT^^^^^^^\r\n\r\nNEVER", "Solution_17": "[b]Tandoori Chicken[/b]", "Solution_18": "[b]noisette[/b] (look it up)", "Solution_19": "[b]edible calculators[/b]", "Solution_20": "[b]Swine flu[/b]", "Solution_21": "ok maybe we should get back on track...", "Solution_22": "[quote=\"abcak\"]liquified human brains served fear factor style in a real skull.[/quote]\r\n\r\nThere was a case where some workers inhaled liquified pig brains and got a neurotic disorder. Read about it [url=http://www.startribune.com/lifestyle/health/15089906.html]here[/url].", "Solution_23": "How did you know that :huh: :huuh: :what?: \r\n\r\nwow, and in my state too.", "Solution_24": "Our Life Science teacher told us. It took him 30 mins to comment about it and it wasn't even on topic (I think we were studying parts of a cell or something).", "Solution_25": "Ukranian Pooridge", "Solution_26": "eel pot pie", "Solution_27": "eggplant :D", "Solution_28": "tarantula (fried)", "Solution_29": "[b]artichoke[/b]" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Dera Mathlinkers,\r\nlet ABC be a triangle, (1) the Euler's circle of ABC,\r\nX a point, PQR the X-pedal triangle of ABC, (2) the circumcircle of PQR, X* the isogonal of X wrt ABC\r\nand (3) the Euler's circle of X*BC.\r\nProuve that (1), (2) and (3) are concurrent.\r\nSincerely\r\nJean-Louis", "Solution_1": "[b]This problem is equivalent to the following:[/b]\r\n$ \\mbox{Let triangle ABC and point D. The four Euler's circles of triangles BCD, ACD, ABD, ABC and} \\\\\r\n\\mbox{the circumcircle of D \\minus{} pedal triangle of ABC are concurent (have a common point E).} \\\\\r\n\\mbox{The point E trrough and the circle circumscribed of A \\minus{} pedal triangle of } \\triangle{BCD},\\dots$", "Solution_2": "See also:\r\nhttp://mathworld.wolfram.com/Nine-PointCircle.html" } { "Tag": [ "topology" ], "Problem": "Let $ X$ be a connected, normal topological space with more than a single point, prove that $ X$ is uncountable . ? \r\n\r\nAny idea is highly appreciated . Thanks", "Solution_1": "What is \"normal\" for you? Does \"normal\" imply $ T_2$ ? If not then every indiscrete space is normal and connected, hence your claim would be false.\r\nIf \"normal\" implies hausdorff then every normal space is in particular a tychonoff space and has therefore for every $ x\\in X$ and every closed subset $ \\emptyset\\neq A\\subseteq X$ not containing $ x$ a continouos mapping $ f: X\\to[0,1]$ with $ f(x)\\equal{}0$ and $ f(A)\\equal{}\\{1\\}$. Since $ X$ is connected, $ f(X)$ must be connected too. Thus $ f(X)\\equal{}[0,1]$ and so $ |X|\\geq |[0,1]|$. (the required $ x$ and $ A$ exist becauce $ X$ has more than one point and is hausdorff)" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Find the maximum value of A = (x+y+z)*(1/x+1/y+1/z) with 2<= x,y,z <= 4 .\r\nHelp me with this problem , please !", "Solution_1": "Does this help you? ;)\r\n\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?t=19244\r\n\r\n Darij" } { "Tag": [ "function" ], "Problem": "If f(x) = 3x^2 - Bx + 4 and f(-1) = 12, what is the value of B?", "Solution_1": "[hide]Plugging in $-1$ in the function we have $3+B+4=12$. Solve for B and get $\\boxed{5}$.[/hide]", "Solution_2": "[quote=\"Interval\"]If f(x) = 3x^2 - Bx + 4 and f(-1) = 12, what is the value of B?[/quote]\r\n\r\nugh no offense but i think u have been here at AoPS long enough to learn and use latex. im getting sick and tired of looking at \"^\" symbols.", "Solution_3": "[quote=\"Interval\"]If f(x) = 3x^2 - Bx + 4 and f(-1) = 12, what is the value of B?[/quote]\r\n\r\n[hide]\n$12=3(-1)^{2}+B+4$\n$12=3+4+B$\n$B=5$.\n\nA different story:\n\nyou meant $(3x)^{2}$?\n\n$12=(-3)^{2}+B+4$\n$12=9+B+4$\n$B=-1$[/hide]", "Solution_4": "anirudh, try doing the first one again. You subtracted wrong.", "Solution_5": "If f(x) = 3x^2 - Bx + 4 and f(-1) = 12, what is the value of B?\r\n\r\n[hide]\n\n$f(-1) = 3(-1)^{2}+B+4 = 12$\n\n$7+B = 12$\n\n$B = 5$\n\n$f(x) = 3x^{2}-5x+4$\n\n$f(-1) = 3+5+4 = 12$\n\n[/hide]", "Solution_6": "[quote=\"ckck\"]anirudh, try doing the first one again. You subtracted wrong.[/quote]\r\n\r\nIt was a typo :blush:" } { "Tag": [ "algebra", "polynomial", "algebra unsolved" ], "Problem": "Find all polinomials $ P(x)$ with real coefficients, such that \r\n\r\n$ P(\\sqrt {3}(a \\minus{} b)) \\plus{} P(\\sqrt {3}(b \\minus{} c)) \\plus{} P(\\sqrt {3}(c \\minus{} a)) \\equal{} P(2a \\minus{} b \\minus{} c) \\plus{} P( \\minus{} a \\plus{} 2b \\minus{} c) \\plus{} P( \\minus{} a \\minus{} b \\plus{} 2c)$\r\nfor any $ a$,$ b$ and $ c$ real numbers", "Solution_1": "Let $ Q(a,b,c)\\equal{}P(\\sqrt{3}(a\\minus{}b))\\plus{}P(\\sqrt{3}(b\\minus{}c))\\plus{}P(\\sqrt{3}(c\\minus{}a))$, $ Q'(a,b,c)\\equal{}P(2a\\minus{}b\\minus{}c)\\plus{}P(\\minus{}a\\plus{}2b\\minus{}c)\\plus{}P(\\minus{}a\\minus{}b\\plus{}2c)$\r\n\r\nLet $ P(x)\\equal{}a_nx^n\\plus{}a_{n\\minus{}1}x^{n\\minus{}1}\\plus{}...\\plus{}a_1x\\plus{}a_0$.\r\n\r\nSo the terms of degree $ k$ of $ Q(a,b,c)\\equal{}a_k(\\sqrt{3})^k[(a\\minus{}b)^k\\plus{}(b\\minus{}c)^k\\plus{}(c\\minus{}a)^k]$\r\nthat of $ Q'(a,b,c)\\equal{}a_k[(2a\\minus{}b\\minus{}c)^k\\plus{}(\\minus{}a\\plus{}2b\\minus{}c)^k\\plus{}(\\minus{}a\\minus{}b\\plus{}2c)^k]$\r\n\r\nSo if $ a_k \\neq 0$, then\r\n\r\n$ (\\sqrt{3})^k[(a\\minus{}b)^k\\plus{}(b\\minus{}c)^k\\plus{}(c\\minus{}a)^k]\\equal{}(2a\\minus{}b\\minus{}c)^k\\plus{}(\\minus{}a\\plus{}2b\\minus{}c)^k\\plus{}(\\minus{}a\\minus{}b\\plus{}2c)^k$.\r\nClearly it's possible for $ k\\equal{}0$ it's true. Put $ a\\equal{}b\\equal{}\\minus{}c$ we know it's not possible for $ k \\ge 1$.\r\n\r\nSo $ P(x)\\equal{}C$, where $ C$ is any real constant , are the only possible polynomials.", "Solution_2": "that's not the answer... try again...", "Solution_3": "[quote=\"stephencheng\"] ...\n$ (\\sqrt {3})^k[(a \\minus{} b)^k \\plus{} (b \\minus{} c)^k \\plus{} (c \\minus{} a)^k] \\equal{} (2a \\minus{} b \\minus{} c)^k \\plus{} ( \\minus{} a \\plus{} 2b \\minus{} c)^k \\plus{} ( \\minus{} a \\minus{} b \\plus{} 2c)^k$.\nClearly it's possible for $ k \\equal{} 0$ it's true. Put $ a \\equal{} b \\equal{} \\minus{} c$ we know it's not possible for $ k \\ge 1$.\n\n...[/quote]\r\n\r\nYou are wrong , because the equality holds for $ k \\equal{} 0,k \\equal{} 1,k \\equal{} 2$ . Let's suppose now $ k > 2$\r\nFor $ a \\equal{} 1,b \\equal{} 0,c \\equal{} 0$ you obtain $ (\\sqrt 3)^k(1 \\plus{} ( \\minus{} 1)^k) \\equal{} 2^k \\plus{} 2( \\minus{} 1)^k$, and only $ k \\equal{} 4$ may be possible . You have to check this case. \r\n:cool:", "Solution_4": "$ P(\\sqrt {3}s) \\plus{} P(\\sqrt {3}t) \\plus{} P(\\sqrt {3}u) \\equal{} P(s \\minus{} u) \\plus{} P(t \\minus{} s) \\plus{} P(u \\minus{} t)$ for $ s,\\ t,\\ u\\in{\\mathbb{R}}.$\r\n\r\n[color=red]Last edit[/color]\r\n\r\n$ s\\plus{}t\\plus{}u\\equal{}0.$", "Solution_5": "[quote=\"kunny\"]$ P(\\sqrt {3}s) \\plus{} P(\\sqrt {3}t) \\plus{} P(\\sqrt {3}u) \\equal{} P(s \\minus{} u) \\plus{} P(t \\minus{} s) \\plus{} P(u \\minus{} t)$ for $ s,\\ t,\\ u\\in{\\mathbb{R}}.$[/quote]\r\n\r\nThere is a mistake here. This equation is only true when $ s \\plus{} t \\plus{} u \\equal{} 0$\r\nHere is my solution:\r\n\r\nDenote $ x \\equal{} a \\minus{} c, y \\equal{} b \\minus{} a, z \\equal{} c \\minus{} b$, which implies $ x \\plus{} y \\plus{} z \\equal{} 0$\r\nThe first equation is equivalent to\r\n$ P(\\sqrt{3}x) \\plus{} P(\\sqrt{3}y) \\plus{} P(\\sqrt{3}z) \\equal{} P(x \\minus{} y) \\plus{} P(y \\minus{} z) \\plus{} P(z \\minus{} x)$\r\n\r\nAssume that $ deg P \\equal{} n$. Compare the coefficients of degree n follows\r\n$ (\\sqrt{3})^n(x^n \\plus{} y^n \\plus{} z^n) \\equal{} (x \\minus{} y)^n \\plus{} (y \\minus{} z)^n \\plus{} (z \\minus{} x)^n (*)$.\r\n\r\nLet $ x \\equal{} \\minus{}1, y \\equal{} 0, z \\equal{} 1$ which takes\r\n$ (\\sqrt{3})^n[(\\minus{}1)^n \\plus{} 1^n] \\equal{} 2.(\\minus{}1)^n \\plus{} 2^n (1)$\r\n\r\nLet $ x \\equal{} 1, y \\equal{} 0, z \\equal{} \\minus{}1$ which takes\r\n$ (\\sqrt{3})^n[(\\minus{}1)^n \\plus{} 1^n] \\equal{} 2 \\minus{} 2^n (2)$\r\n\r\nCombine (1) and (2) gives us\r\n$ 2^n \\plus{} 2.(\\minus{}1)^n \\equal{} 2 \\minus{} 2^n,$\r\n$ 2^n \\equal{} 1 \\plus{} (\\minus{}1)^n (3)$\r\n\r\nIt's easy to verify that $ n \\equal{} 0$ and $ n \\equal{} 2$ are only values satisfying (3).\r\n- If $ n \\equal{} 0$, it follows that $ P(x) \\equal{} C$. Indeed this polynomial satisfies the first equation\r\n- If $ n \\equal{} 2$, which follows that $ P(x) \\equal{} t_2x^2 \\plus{} t_1x \\plus{} t_0$. This polynomial satisfies the first equation\r\n\r\nAnd the solution is $ P(x) \\equal{} C$ and $ P(x) \\equal{} ax^2 \\plus{} bx \\plus{} c$\r\n\r\nP/S: My English is very poor :(", "Solution_6": "[quote=\"ll931110\"]There is a mistake here. This equation is only true when $ s \\plus{} t \\plus{} u \\equal{} 0$[/quote]\n\nYou are right, I forgot to type the condition.\nMy solution is almost same as yours.\n\n[quote=\"ll931110\"] P/S: My English is very poor :([/quote]\r\n\r\nWe can understand your English well. :)", "Solution_7": "The answer is $ P(x) \\equal{} a_4x^{4} \\plus{} a_2x^{2} \\plus{} a_1x \\plus{} a_0$.", "Solution_8": "$ p(x)\\equal{}ax^4\\plus{}bx^2\\plus{}cx\\plus{}d$ is the correct answer... \r\n\r\nll931110: check at $ (2)$, you must have $ 2\\plus{}2^n$ and not $ 2\\plus{}(\\minus{}2)^n$...", "Solution_9": "[quote=\"campos\"]$ p(x) \\equal{} ax^4 \\plus{} bx^2 \\plus{} cx \\plus{} d$ is the correct answer... \n\nll931110: check at $ (2)$, you must have $ 2 \\plus{} 2^n$ and not $ 2 \\plus{} ( \\minus{} 2)^n$...[/quote]\r\n\r\nSorry. I didn't check when I posted it.\r\n\r\nIt's easy to verify that n is even. Set $ n \\equal{} 2k$, where k is an integer\r\nCompare the coefficient of degree n follows\r\n$ 3^k(x^{2k} \\plus{} y^{2k} \\plus{} z^{2k}) \\equal{} (x \\minus{} y)^{2k} \\plus{} (y \\minus{} z)^{2k} \\plus{} (z \\minus{} x)^{2k}$ for all x, y, z satisfying $ x \\plus{} y \\plus{} z \\equal{} 0$\r\n\r\nPut $ x \\equal{} 2, y \\equal{} z \\equal{} \\minus{}1$ gives\r\n$ 4^k \\plus{} 2 \\equal{} 2.3^k$, and we could verify the roots of this equation is $ k \\equal{} 0,1,2$. So we have the answer is\r\n$ P(x) \\equal{} ax^4 \\plus{} bx^2 \\plus{} cx \\plus{} d$" } { "Tag": [ "floor function" ], "Problem": "For how many positive integers $n$ is $\\left(2003+\\frac{1}{2}\\right)^n +\\left(2004+\\frac{1}{2}\\right)^n$ an integer?", "Solution_1": "I only found 3 that work ($0,1,3$), but I wasn't using a formula; there might be more.", "Solution_2": "binomial expansion is $(a\\pm b)^n=\\sum_{k=0}^{n}(\\pm 1)^k\\binom{n}{k}a^{n-k}b^{k}$\r\n\r\n$\\left(2003+\\frac12\\right)^n+\\left(2004+\\frac12\\right)^n=\\left(2004-\\frac12\\right)^n+\\left(2004+\\frac12\\right)^n$\r\n\r\nif $n$ is even, last term of sum of both expansion is $2\\cdot\\left(\\frac12\\right)^n$ which is not integer for $n\\geq 2$, so $n$ cannot be even (except for zero, of course).\r\n\r\nassume $n$ is odd. then last term of sum of both binomials is zero so the penultimate term is $2\\cdot(\\pm 1)^{n-1}\\binom{n}{n-1}2004^1\\left(\\frac12\\right)^{n-1}$. because $2004=2^2\\cdot 501$, $n-1\\leq 3$. only odds less than 4 are $n=1,3$\r\n\r\nanswer $n=0,1,3$", "Solution_3": "Well, say $t_n = (2003 + \\frac{1}{2})^n + (2004 + \\frac{1}{2})^n$. For now, say $2003 + \\frac{1}{2} = 2004 - \\frac{1}{2} = a$ and $2004 + \\frac{1}{2} = b$.\r\n\r\nSo, $t_n(a+b) = (a^n + b^n)(a+b) = a^{n+1} + ab^n + a^n b + b^{n+1}$ \r\n\r\n$= a^{n+1} + b^{n+1} + ab(a^{n-1} + b^{n-1}) = t_{n+1} + ab t_{n-1}$\r\n\r\nSince we have numerical values for $a$ and $b$, we know $a+b = 2004 - 1/2 + 2004 + 1/2 = 4008 = 501(2^3)$ and also $ab = (2004 - \\frac{1}{2})(2004 + \\frac{1}{2}) = 2004^2 - \\frac{1}{4} = (501^2)(2^4) - (2^{-2})$\r\n\r\nUsing these, we get:\r\n$t_{n+1} = t_n(a+b) - ab t_{n-1} = t_n(501(2^3)) - t_{n-1}((501^2)(2^4) - (2^{-2}))$\r\n\r\nSince the $t_{n-1}$ term is being divided by 4 everytime we generate a new term $t_{n+1}$, we must be aware of the greatest power of 2 that divides a given $t_i$. So, $t_0 = a^0 + b^0 = 2$ and $t_1 = a^1 + b^1 = 501(2^3)$. So, we know the greatest powers of 2 that divide $t_0$ and $t_1$ are 1 and 3, respectively. So, lets define $[x]$ to be the greatest power of 2 that divides $x$. So, \r\n$[t_{n+1}] =[t_n(501(2^3)) - t_{n-1}((501^2)(2^4) - (2^{-2}))]$\r\n\r\n$= \\min(3 + [t_n], 4 + [t_{n-1}], -2 + [t_{n-1}] ) = \\min(3 + [t_n], -2 + [t_{n-1}] )$\r\n\r\nNow we are more concerned with the new sequence $f_n = [t_n]$ where $f_0 = 1$ and $f_1 = 3$ and $f_{n+1} = \\min(3 + f_n, -2 + f_{n-1} )$. When $f_n$ is always negative, we can have no more integers in our sequence. So, say for some $j$, $f_j$ and $f_{j-1}$ are negative, then $f_{j+1} = \\min(3 + f_j, -2 + f_{j-1} )$. Since $-2 + f_{j-1}$ is always negative due to the fact that $ f_{j-1}$ is and the minimum must be less than or equal to $-2 + f_{j-1}$, $f_{j+1}$ must always be negative. So, if we have 2 consecutive $f_i$'s negative, then all remaining are negative due to line of infinite descent. So all we need do is find the first consecutive pair of negatives.\r\n$f_0 = 1$\r\n\r\n$f_1 = 3$\r\n\r\n${f_2 = \\min(3 + 3, -2 + 1} ) = -1$\r\n\r\n${f_3 = \\min(3 + -1, -2 + 3} ) = 1$\r\n\r\n${f_4 = \\min(3 + 1, -2 + -1} ) = -3$\r\n\r\n${f_5 = \\min(3 + -3, -2 + 1} ) = -1$\r\n\r\nSo, we have 2 consecutive and we are done with only $0,1$ and $3$ producing positive powers of 2 and thus being integers.", "Solution_4": "Heh, I got it right by guess-and-test. :P\r\n\r\nNice formulae, though.", "Solution_5": "hehe, guess-and-test this one :D\r\n\r\nFor how many positive integers $n$ is $\\left(2005!-\\frac{1}{2}\\right)^n +\\left(2005!+\\frac{1}{2}\\right)^n$ an integer? ;)", "Solution_6": "Well, you know there are immediately two, 0 and 1. The other one can be found out.", "Solution_7": "[quote=\"236factorial\"]Well, you know there are immediately two, 0 and 1. The other one can be found out.[/quote]\r\n\r\nhmm yea, that was a bit too obvious, :spam: :?: :rotfl:", "Solution_8": "Refer to my old post for bits and pieces of this if need be.\r\nSay $a = 2005! - \\frac{1}{2}$ and $b = 2005! + \\frac{1}{2}$. So $a+b = 2(2005!)$ and $ab = (2005!)^2 - \\frac{1/4}$. So, we need to know how many powers of 2 are in 2005. We can easily find this by finding the sum $\\displaystyle\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{2005}{2^i}\\right\\rfloor$. This sum is just $1002 + 501 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 1997$. So, we get a recursion and other cool stuff but it comes down to $f_{n+1} = \\min(1998 + f_n,-2 + f_{n-1})$ With $f_0 = 1$ and $f_1 = [a+b] = 1998$. Sorry I gotta go for now." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "If $ a,b,c,d$ are distinct and not equal to $ 2$,show that $ abcd=5(a+b+c+d)+2(abc+bcd+cda+dab)$,if the points ${{{{ (\\frac{a^{4}}{a-2},\\frac{a^{3}-5}{a-2}}),(\\frac{b^{4}}{b-2},\\frac{b^{3}-5}{b-2}}),(\\frac{c^{4}}{c-2},\\frac{c^{3}-5}{c-2}}),(\\frac{d^{4}}{d-2},\\frac{d^{3}-5}{d-2}})$ are collinear.", "Solution_1": "since $ a,b,c,d$ are distinct, the slope (m) of the line passing through these four points is not equal zero.\r\nLet the line passing through these four points be$ L: y \\equal{} mx \\plus{} c$\r\nlet $ a \\equal{} x_1$, $ b \\equal{} x_2$, $ c \\equal{} x_3$, $ d \\equal{} x_4$,\r\nthen$ \\frac {{x_i}^3 \\minus{} 5}{x_i \\minus{} 2} \\equal{} \\frac {{x_i}^4}{x_i \\minus{} 2}m \\plus{} c$\r\nwe have $ 0 \\equal{} m{x_i}^4 \\minus{} {x_i}^3 \\plus{} cx_i \\minus{} 2c \\plus{} 5$........(*)$ i \\equal{} {1,2,3,4}$\r\nso $ a,b,c,d$ are the roots of (*).\r\nthen $ a \\plus{} b \\plus{} c \\plus{} d \\equal{} \\frac {1}{c}$, $ abc \\plus{} bcd \\plus{} acd \\plus{} abd \\equal{} \\minus{} \\frac {c}{m}$,\r\n$ abcd \\equal{} \\frac {5 \\minus{} 2c}{m} \\equal{} 5(\\frac {1}{c}) \\plus{} 2( \\minus{} \\frac {c}{m}) \\equal{} 5(a \\plus{} b \\plus{} c \\plus{} d) \\plus{} 2(abc \\plus{} bcd \\plus{} cda \\plus{} dab)$" } { "Tag": [], "Problem": "A sequence is defined by $ u_{1}\\equal{}8$ and $ u_{n\\plus{}1}\\equal{}u_{n} \\plus{} 3.$\r\n\r\nFind the value of $ N$ such that $ \\sum_{n\\equal{}1}^{2N} u_{n} \\minus{} \\sum_{n\\equal{}1}^{N} u_{n} \\equal{} 1256$.", "Solution_1": "[hide=\"Solution\"]From the recursive equation $ u_n \\equal{} u_{n \\minus{} 1} \\plus{} 3$, we can find the closed equation $ u_n \\equal{} 8 \\plus{} 3(n \\minus{} 1)$. So,\n\n$ \\sum_{n \\equal{} 1}^{2N} u_n \\minus{} \\sum_{n \\equal{} 1}^{N} u_n \\equal{} (8 \\plus{} 11 \\plus{} ... \\plus{} u_{2N}) \\minus{} (8 \\plus{} 11 \\plus{} ... \\plus{} u_N)$\n\n$ \\equal{} u_{N \\plus{} 1} \\plus{} u_{N \\plus{} 2} \\plus{} ... \\plus{} u_{2N}$\n\n$ \\equal{} (8 \\plus{} 3N) \\plus{} (8 \\plus{} 3(N \\plus{} 1)) \\plus{} ... \\plus{} (8 \\plus{} 3(2N \\minus{} 1))$\n\n$ \\equal{} 8N \\plus{} 3\\frac {N(3N \\minus{} 1)}{2} \\equal{} 1256$\n\n$ 16N \\plus{} 9N^2 \\minus{} 3N \\equal{} 2512$\n\n$ 9N^2 \\plus{} 13N \\minus{} 2512 \\equal{} 0$\n\n$ (N \\minus{} 16)(9N \\plus{} 157) \\equal{} 0$\n\n$ \\therefore \\boxed{N \\equal{} 16}$[/hide]", "Solution_2": "Thanks modularmarc101." } { "Tag": [ "Princeton", "college", "geometry", "search", "LaTeX", "Science Olympiad" ], "Problem": "Not anything specific unless you want. :P \r\n\r\nNorthern Jersey for me.", "Solution_1": "Central Jersey.", "Solution_2": "north jersey!!!!!!!!!!!! :w00t: :winner_first: :winner_first:", "Solution_3": "I live near NYC too. :wink:", "Solution_4": "Smack dab in the middle...\r\nPhew, it's chilly...", "Solution_5": "Northern New Jersey, Morris County more specifically.", "Solution_6": "Northern New Jersey.", "Solution_7": "northern new jersey for me :) i live RIGHT next to the george washington bridge its like a 5 minute walk away from the house haha", "Solution_8": "northern new jersey for me :) i live RIGHT next to the george washington bridge its like a 5 minute walk away from the house haha", "Solution_9": "northern new jersey for me :) i live RIGHT next to the george washington bridge its like a 5 minute walk away from the house haha", "Solution_10": "Central Jersey. Near Princeton.", "Solution_11": "Princeton. Like right next to the University.", "Solution_12": "Central Jersey!!! XD", "Solution_13": "Woot me too!!!\n\nMe lives near Princeton because my mom works at the university :O", "Solution_14": "[quote=\"Chuthecommie\"]Woot me too!!!\n\nMe lives near Princeton because my mom works at the university :O[/quote]\nThat's so cool! I live near the Princeton area too! :w00t:", "Solution_15": "[quote=\"JSGandora\"][quote=\"dingdong162\"]Go nyc jersey-ians!!!![/quote]\nWhat's NYC? I doubt that it stands for New York City...[/quote]\n\nSome of the New York metropolitan area is in New Jersey.", "Solution_16": "Yeah, like the part where filetmignon821, BOGTRO, and I live.\n\nA very important part that also includes the great AAST.\n\nAlso, BOO THOMAS GROVER AND COMMUNITY FOR WINNING SCIENCE OLYMPIAD AND MATHCOUNTS. BOOOOOOOOO!!!!!!!!!!!!", "Solution_17": "[quote=\"Chuthecommie\"]\nSince you are in Grover, I am obliged to say this out of loyalty to my science olympiad team: GROVER S.O. IS WORSE THAN MONTY S.O.[/quote]\nWhich is why we placed 2nd and you guys placed (checking...) 4th. :)", "Solution_18": "[quote=\"JSGandora\"][quote=\"Chuthecommie\"]\nSince you are in Grover, I am obliged to say this out of loyalty to my science olympiad team: GROVER S.O. IS WORSE THAN MONTY S.O.[/quote]\nWhich is why we placed 2nd and you guys placed (checking...) 4th. :)[/quote]\n\nYea, we were the school that sat next to you guys and I was one of the Pink People...yeah\n\nIt's funny that we did not receive a top-6 finish and we were sitting between Thomas Grover and Community. Ironic, no?", "Solution_19": "central, more specifically mercer county", "Solution_20": "central monmouth county", "Solution_21": "@middlesex, WW-P, Community Middle School", "Solution_22": "bergen county\nwere all the asians live LOL", "Solution_23": "middlesex county", "Solution_24": "[quote=ironpanther29]middlesex county[/quote]\n\nSAME!\n\nDo you guys all go to the Princeton Campus?", "Solution_25": "[quote=ironpanther29]middlesex county[/quote]\n\nsame", "Solution_26": "reeee i was born in Princeton and nj is my favorite state", "Solution_27": "NOT bergen county", "Solution_28": "NJ even looks cool\nis fat in some parts and skinny in others", "Solution_29": "Bergen :)\n" } { "Tag": [], "Problem": "A house worth $ \\$9000$ is sold by Mr. A to Mr. B at a $ 10\\%$ loss. Mr. B sells the house back to Mr. A at a $ 10\\%$ gain. The result of the two transactions is:\r\n$ \\textbf{(A)}\\ \\text{Mr. A breaks even} \\qquad\\textbf{(B)}\\ \\text{Mr. B gains }\\$900 \\qquad\\textbf{(C)}\\ \\text{Mr. A loses }\\$900\\\\\r\n\\textbf{(D)}\\ \\text{Mr. A loses }\\$810 \\qquad\\textbf{(E)}\\ \\text{Mr. B gains }\\$1710$", "Solution_1": "[hide=\"Hehe, nice title.\"]\nThe cost that Mr. A sells the house for is\n$ 9000(.9)\\equal{} 8100$\nThe cost that Mr. B sells the house back at is\n$ 9000*(.9)*(1.1) \\equal{} 8910$\nThus, the answer is\n\n$ \\boxed{\\text{D}}$[/hide]" } { "Tag": [ "geometry", "inradius", "inequalities" ], "Problem": "Let $r$ be the inradius of an acute triangle. Prove that the sum of the distances from the orthocenter to the sides of the triangle does not exceed $3r$.", "Solution_1": "Ehasan2004, i think you posted this problem once before. Anyway, it is easy to see that $HA' =2R cosB cosC,$ etc... So the inequality becomes \\[ cosA cosB +cosB cosC +cosC cosA\\leq\\frac{3r}{2R}. \\] but since \r\n $cosA cosB +cosB cosC +cosC cosA=\\frac{p^{2}+r^{2}-4R^{2}}{R^{2}}$ the inequality reduces to a well-known one. ;)", "Solution_2": "[quote]Ehasan2004, i think you posted this problem once before[/quote]\r\n\r\nWell ,because I post it long time ago and I want to see the Iranians solution ;)" } { "Tag": [ "calculus", "calculus computations" ], "Problem": "find $y(x)$ from this differential equation\r\n\\[ (2+2x \\sqrt{y})x dx + (2+x^2 \\sqrt{y})y dy =0 \\]\r\nI'm just learning differential eqution about three weeks and just know separable, finding integrate factor, excact diff, bernoulli, and rucatti. Is any of this method works?", "Solution_1": "sorry for this confused. I miss typo it, I've edit it. Just give me a hint, if it can be solved with the method that I've metioned above.", "Solution_2": "anybody solve this please? I'm stuck. I've working on it 5 days ago.Still, I don't get the answer.come on I'm askking for hint please...\r\nOr this problem use advance tehnique :?" } { "Tag": [ "complex numbers", "complex analysis" ], "Problem": "What is the formal defenition of this.. I was told they couldn't be ordered but I could think of a few ways to do it so I think I may have misinterpreted what \"ordered\" meant..\r\n\r\nthanks for any help", "Solution_1": "The actual statement is that the complex numbers cannot be given an order that makes them an ordered field. Of course you can invent a total order for the set of complex numbers - lexicographical order, for example - but you can't make that order cooperate with the arithmetic.\r\n\r\nA field $F$ is an [b]ordered field[/b] iff it is a field an satisfies the following properties:\r\n\r\nThere exists a set $P\\subset F,$ informally known as the positive elements of $F.$\r\n\r\nFor all $x\\in F,$ exactly one of the following is true: $x\\in P,\\,-x\\in P,$ or $x=0.$\r\n\r\nIf $x,y\\in P,$ then $x+y\\in P.$\r\n\r\nIf $x,y\\in P,$ then $xy\\in P.$\r\n\r\nWe can prove the following theorem: in any ordered field, if $x\\ne 0,$ then $x^2\\in P.$ This is simple: either $x$ or $-x$ is in $P$, and either way, $x^2=(-x)^2\\in P.$\r\n\r\nThis proves that $1\\in P,$ since $1=1^2.$\r\n\r\nBut in the complex numbers, $-1$ is also a square, being $i^2,$ hence positive. That violates the trichotomy axiom.\r\n\r\nOne can also prove that no field of finite characteristic can be an ordered field." } { "Tag": [ "invariant", "combinatorics proposed", "combinatorics" ], "Problem": "Is it possible to cover a $ 6\\times 6$ square perfectly, only using long (4x1) and T-shaped tetromino's?\r\n\r\nI can't find an example nor a good invariant.", "Solution_1": "This solution works for every (4k+2)*(4k+2) square!!!\r\n\r\n1. EVEN:\r\nwe paint the square like a chess. And then the number of black and white 1*1 squares are equal, so we have that the number of T-shaped tetromino's is even.\r\n\r\n2. ODD:\r\nwe paint 1*1 squares with 4 colors.\\[ color(i,j)\\equiv (i\\plus{}j)(mod 4 )\\]\r\nthe number of the 1*1 squares with the same color is are 8,9,10,9.\r\n\r\ni) if we cover the 6*6 square with only long 4*1, we have 9, 1*1 squares from each color.\r\nii) if we replace a T-shaped tetromino instead of a long 4*1, one of the colors have 1 more and the opposite color have one less, because each T-shaped tetromino covers 2 of a color and none of the opposite color. \r\n\r\nfrom i and ii we have that the number of the T-shaped tetromino's is odd.\r\n\r\nBut the number of T-shaped tetromino's can't be even and odd at the same time, contradiction!!! :lol:", "Solution_2": "[quote=\"sinaj\"]ii) if we replace a T-shaped tetromino instead of a long 4*1, one of the colors have 1 more and the opposite color have one less, because each T-shaped tetromino covers 2 of a color and none of the opposite color.[/quote]Can you please clarify this? What is \"the opposite color\" when you have 4 colors?", "Solution_3": "is the meaning of tetromino a T-shape domino with 4 squares?", "Solution_4": "A tetromino is any domino with 4 squares. A T-shaped tetromino is the T-shaped domino with 4 squares.", "Solution_5": "To be letter-perfect,[color=red] a domino has $ 2$ squares[/color] (the initial \"d\" from \"deuce\"), so Solomon Golomb introduced the term [color=blue]polyomino[/color].\r\n\r\nTherefore a tetromino is a polyomino with $ 4$ squares, a pentomino one with $ 5$ squares, etc.", "Solution_6": "Now the problem is still unsolved. :P", "Solution_7": "We don't need an invariant for this one. First we check that a I-piece cannot cover a corner, for, if say, we place a horizontal I-piece on the top left square then clockwise we need I-pieces to cover the rest of the corner squares. It's easy to see that you end up trying to fill a 2x2 square, which is impossible.\r\n\r\nThen we need T-pieces to cover each corner. But placing one of them completely determines how to place the other ones. After placing the four of them, we have 2 forced I-pieces, and we end up with a 4x4 square with corners removed, which is clearly impossible to cover.", "Solution_8": "How about a 10x10? \r\n14x14?", "Solution_9": "[quote=\"Peter\"]How about a 10x10? \n[/quote]\r\n[asy]size(150);\npen p=blue+linewidth(1.5);\nD((0,0)--(0,10)--(4,10)--(4,6)--(2,6)--(2,4)--(4,4)--(4,0)--(0,0),p);\nD((1,0)--(1,1)--(2,1)--(2,3)--(3,3)--(3,4),p);\nD((0,0)--(10,0)--(10,10)--(4,10),p);\nD((1,2)--(1,8),p);\nD((4,4)--(10,4),p);\nD((0,7)--(1,7)--(1,8)--(3,8)--(3,9)--(4,9),p);\nD((4,6)--(10,6)--(6,6)--(6,4),p);\nD((1,5)--(10,5),p);\nD((4,1)--(3,1)--(3,2)--(1,2)--(1,3)--(0,3),p);\nD((1,10)--(1,9)--(2,9)--(2,7)--(3,7)--(3,6),p);[/asy]\r\n\r\nSo obviously sinaj's approach doesn't work. :maybe:" } { "Tag": [ "probability", "combinatorics proposed", "combinatorics" ], "Problem": "In a class of $n$ students, what is the probability of existence of two students, whose birth dates differ by $k$ days?", "Solution_1": "Do you mean [i]exactly[/i] $k$ days, or $\\leq k$ days?", "Solution_2": "k is between 1 to 30 or about 365 days?" } { "Tag": [ "trigonometry", "algebra solved", "algebra" ], "Problem": "Prove that there exists a natural number n such that\r\nsin n>2003/2004", "Solution_1": "More generally, you may prove that the sequence {sin(n)/ n is a positive integer} is dense in [0,1].\r\n\r\nPierre.", "Solution_2": "Yes, Pierre, how can you prove that the sequence sin n is dense in \r\n0 \\leq x \\leq 1? (Ive asked a teacher who said that it might not be true)", "Solution_3": "Ok, I give you the steps, I left the details :\r\n\r\nLet a>0 be an irrational number.\r\nLet U(n) = {an} = an - [an], where {.} denotes the decimal part, and [.] denotes the integer part.\r\nLet E = { U(n) / n > 0}.\r\n\r\n1) Use that a is irrational to deduce that if n,m are two distinct integers then U(n) =/= U(m).\r\n\r\n2) Use the pigeon-hole principle to deduce that, for each integer N > 0 there exists p,q in {1,...,N+1} such that p =/= q and 0 < U(p) - U(q) < 1/N.\r\n\r\n3) Deduce from above that there exists a non-zero integer k = p-q (not necessarly positive) such that 0 < U(k) < 1/N.\r\n\r\nNow let x be in [0,1[ and e > 0 be given. Let I = ]x-e,x+e[.\r\nLet N > 0 be an integer such that 1/N < e. Use 3) to get k.\r\n\r\n4) If k > 0.\r\nLet F = { {ka}, 2*{ka}, ...., [1/{ka}]*{ka} }.\r\nProve that I \\cap F contains at least one element, and deduce that I \\cap E contains at least one element.\r\n\r\n5) If k < 0.\r\nLet r = -k. Then 1 - 1/n < U(r) < 1. Let U(r) = 1-d.\r\nLet G = {1,2,..., [1/d]}. For i in G, we have {i*r*a} = 1 - i*d.\r\nAs in 4), deduce that I \\cap E contains at least one element.\r\n\r\n6) Deduce that E is dense in [0,1].\r\n\r\nLet x be in [-1,1], and w in [0,2 \\pi [ such that sin(w) = x. Let t = w/(2 \\pi ), thus t is in [0,1]. Let e > 0 and I = ]x-e,x+e[.\r\n\r\n7) Use 6) for a = 1/(2 \\pi ). Deduce that there exists a positive integer n such that |2 \\pi *{n/(2 \\pi } - w| < e.\r\n\r\n8) Note that, for every real u,v we have |sin(u) - sin(v)| \\leq |u-v|.\r\nUse it to deduce that | sin(2 \\pi *{n/(2 \\pi )}) - x| < e.\r\n\r\n9) Note that 2 \\pi *{n/(2 \\pi )} = n - 2 \\pi *[n/(2 \\pi )], and deduce that |sin(n) - x| < e.\r\nThus { sin(n) / n > 0 is an integer} is dense in [-1;1] (and therefore in [0,1] ).\r\n\r\nPierre.", "Solution_4": "Let k \\in R\\Q, the sequence (e^(i \\pi nk))_{n \\in Z} is dense in \r\nS^1={ z \\in C; |z|=1}", "Solution_5": "You are right Moubi, and in fact this latter result is easier to prove.\r\nBut, the original problem ask for a natural number, thus the work is a little bit difficult...\r\n\r\nPierre." } { "Tag": [ "calculus", "integration", "real analysis", "trigonometry", "calculus computations" ], "Problem": "Solve\r\n\r\n$y(t)+\\int_{0}^{t}e^{t-v}y(v)dv=\\sin(t)$", "Solution_1": "With that convolution in there, I would highly recommend a transform method.\r\n\r\n[hide=\"Fixed below.\"]Laplace transforms, with $Y(s)$ being the Laplace transform of $y(t):$\n\n$Y(s)+\\frac1{s-1}Y(s)=\\frac{s}{s^{2}+1}$\n\n$sY=\\frac{s(s-1)}{s^{2}+1}$\n\n$Y=\\frac{s-1}{s^{2}+1}$\n\n$y(t)=\\sin t-\\cos t.$[/hide]", "Solution_2": "OR\r\n\r\n$\\int_{0}^{t}e^{-v}y(v)dv=e^{-t}(\\sin t-y)$ \r\n\r\nDifferentiate and you get that\r\n\r\n$y'=\\cos t-\\sin t$. It follows that $y(t)=\\sin t-\\cos t$.", "Solution_3": "Your answers are wrong, two guys.\r\nIsn't $y(0)=0?$", "Solution_4": "Now that kunny mentions it ... I had the Laplace transforms of sine and cosine reversed. Let's try this again.\r\n\r\nLaplace transforms, with $Y(s)$ being the Laplace transform of $y(t):$\r\n\r\n$Y(s)+\\frac1{s-1}Y(s)=\\frac{1}{s^{2}+1}$\r\n\r\n$sY=\\frac{s-1}{s^{2}+1}$\r\n\r\n$Y=\\frac{s-1}{s(s^{2}+1)}=\\frac{A}{s}+\\frac{Bs+C}{s^{2}+1}$\r\n\r\n$Y=\\frac{s+1}{s^{2}+1}-\\frac1s$\r\n\r\n$y(t)=\\cos t+\\sin t-1.$" } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "$ \\mathcal F$ is a family of 3-subsets of set $ X$. Every two distinct elements of $ X$ are exactly in $ k$ elements of $ \\mathcal F$. It is known that there is a partition of $ \\mathcal F$ to sets $ X_1,X_2$ such that each element of $ \\mathcal F$ has non-empty intersection with both $ X_1,X_2$. Prove that $ |X|\\leq4$.", "Solution_1": "Let $ |X_1| \\equal{} m$, $ |X_2| \\equal{} n$ and $ |X| \\equal{} \\ell$. ($ \\ell \\equal{} m \\plus{} n$)\r\n\r\nFor each pair $ \\{a,b\\}$ from $ X_1$ we have $ p$ different subsets $ A_i$\r\n\r\nThen there are in total $ {m\\choose 2}\\cdot p$ different subsets $ A_i$ (beacuse each $ A_i$ cuts both $ X_1$ and $ X_2$)\r\n\r\nAlso there is $ {n\\choose 2}\\cdot p$ (beacuse of $ X_2$)\r\n\r\nSo we have at least $ {m\\choose 2}\\cdot p \\plus{} {n\\choose 2}\\cdot p$ subsets $ A_i$:\r\n\r\n$ k \\geq {m\\choose 2}\\cdot p \\plus{} {n\\choose 2}\\cdot p \\geq {\\ell ^2 \\minus{} \\ell\\over 4}\\cdot p$\r\n\r\n-------------------------------------------------------------------------------------------------\r\n\r\nOn the other side each pair $ \\{a,b\\}$ from $ X$ is ''connected'' with $ p$ diffrent subsets $ A_i$\r\nand each subset $ A_i$ is connected with exactly $ 3$ pairs $ \\{a,b\\}$, so we have\r\n\\[ {\\ell \\choose 2}\\cdot p \\equal{} 3k\r\n\\]\r\n-------------------------------------------------------------------------------------------------\r\n\r\nJoining both equations we get $ \\ell\\leq 4$." } { "Tag": [ "analytic geometry", "graphing lines", "slope", "function", "rotation", "conics", "ellipse" ], "Problem": "Consider a graph of a line segment with slope zero from (0,0) to (1,0). This line segment contains all real x coordinates from 0 to 1. Now consider a sloped line from (0,0) to (1,1). This line segment is longer. On a graph of real numbers, if you move either way, inevitably you'll either increase or decrease your coordinate numbers since it's imposible to equally space coordinates you can't list. So if a line is longer, wouldn't it contain more points? Since the first line contains all the real x coordinates, doesn't this mean that there are more than one y coordinate for each x coordinate? That would mean that the graph of a sloped line isn't a function, right? I think my logic is wrong with the spacing, but i don't understand how a line can be longer without containing more points, especially when there are an infinite amount of points.", "Solution_1": "Infinity's a wierd concept, because some infinite quantities are greater than other infinite quantities, and some infinite quantities are the same as other infinite quantities. For example, there are the same number of even numbers as integers, because you can multiply every integer by two. \r\n\r\nIn your question, the infinite quantities are the same. If you graphed them side by side, you could move from the horizontal line a certain distance up such that you hit the other line. Therefore every point on the bottom line corresponds to a point on the top line, and vice versa.", "Solution_2": "Hello, filletwho!\r\n\r\nAs RC-7th pointed out, Infinity is a mind-boggling concept.\r\nWe are simple carbon-based lifeforms and cannot completely understand it.\r\n\r\nConsider the dilemma of a shepherd who has no number system.\r\nHow can tell if all his sheep have returned to the pen?\r\n\r\nAt the beginning of the day, he releases the sheep one at a time.\r\nAs each sheep leaves the pen, he places a pebble in his pocket.\r\n\r\nAnd the end of the day, he removes a pebble for each sheep that returns.\r\nIf he has any pebbles left over, he knows there are some sheep missing.\r\n[Note: He may not know that there are \"three\" sheep missing, but he knows\r\nthat there are as many sheep missing as there are pebbles in his pocket.]\r\n\r\nThis is a primitive method for determining which group is larger (sheep or pebbles).\r\n\r\n\r\nWe are equally baffled when comparing infinite sets.\r\nWe have no \"number system\" to handle them.\r\n\r\nWhich has more points:\r\nthe line segment from (0,0) to (0,1) or the line segment from (0,0) to (0,2) ?\r\n\r\nSince we can't really count them, we will pair them up (like pebbles and sheep).\r\n\r\nEvery point on (0,1) has a coordinate, some real number assigned to it.\r\nEvery point on (0,2) has a coordinate, too.\r\n\r\nWe will pair them up like this:\r\nthe point $x$ on (0,1) will correspond to the point $2x$ on (0,2).\r\n\r\nImagine that we can (somehow!) pair up ALL of them . . .\r\nWill there be any \"left over\"? (Will one set be larger than the other?)\r\n\r\nThe answer is NO.\r\nFor every point $x$ on (0,1), there is a corresponding point $2x$ on (0,2).\r\nFor every point $y$ on (0,2), there is a corresponding point $\\frac{y}{2}$ on (0,1).\r\n\r\nThere is a [i]one-to-one correspondence[/i] between the points in the two intervals.\r\nNeither set will \"run out of number\" first.\r\n\r\nWe are forced to conclude that the two sets have [i]the same number of points[/i].\r\n\r\nThis is contrary to our Common Sense, of course.\r\nBut then, Common Sense does not apply to concepts like \"Infinity\" or \"forever\".", "Solution_3": "Cantor proved that number of decimal numbers that are bigger than zero but smaller than one is greater than the number of counting numbers. By doing so, he showed that there are different sizes of infinity. \r\n\r\nAssume that the above-mentioned sets are the same size.\r\n\r\nLet every counting number $a_n$ correspond to a number $0.d_{n1}d_{n2}...$. After $d_{n9}$, we go on to $d_{n10},d_{n11}...$ or something like that.\r\n\r\nWhat about $0.x_{1}x_{2}x_{3}...$, where $x_{1}$ is something not equal to $d_{11}$, $x_{2}$ is not equal to $d_{22}$, and so on?\r\n\r\nClearly it doesn't correspond to any $a_n$ because it difffers from every $a_{n}$ by at least one digit.\r\n\r\nWe have a contradiction, proving that there are more decimal numbers between 0 and 1 than there are counting numbers.", "Solution_4": "With that logic, there are more points in the segment from (0,0) to (1,1) because there is a greater distance between each set of corresponding points.\r\n\r\nThat makes a lot more sense to me--one infinity can be greater than another.\r\n\r\nFor instance, the limit of x as x -> :inf: is greater than the limit of (x/2) as x -> :inf: \r\n\r\n(well, technically, neither limit exists...)", "Solution_5": "No, that's not true at all. If you can find a bijection between two sets, they have to have the same size. (That's the definition.)", "Solution_6": "One infinity can only be larger when speaking with respect to the orders of sets. X and X/2 are both unbounded as X increases, and the value of X is always greater than that of X/2 for every X>0, but that doesn't make the limit larger.", "Solution_7": "good call. sorry about that...", "Solution_8": "Infinity from different perspective.\r\n\r\nImagine a circle viewed from a point on the line normal to the center.\r\nThat is you are looking straight on at the circle.\r\n\r\nThen draw east-west & north-south diagonals.\r\nNext count all of the points in each direction.\r\n(It may require a more that a few seconds, but the result is infinity.)\r\n\r\nNext rotate the plane of the circle until you have an ellipse, such that\r\nthe east-west diagonal is 10 times longer that the north-south diagonal.\r\n\r\nIf points cannot warp, stretch, elongate, contract, or change, then when\r\nyou recount the points on each diagonal you will have 10 times as many\r\nin the east-west direction as in the north-south direction.\r\n\r\nActually, you haven't changed anything except your viewpoint. No matter\r\nhow the circle in observed there is only an infinity of points. Just as you\r\nchange you perspective it appears as if the values alter. \r\n\r\nTen infinities or 20 or 100 infinities do not change infinity.\r\nIt would seem likely that 20 times Zero is a lot bigger that 2 times Zero.\r\nThus, an infinite infinities are not larger that 1 infinity.\r\n\r\nAt infinity ALL LINE SEGMENTS are the same length.\r\n\r\nYou just just need to view your coordinate system from the necessary\r\npoint of view to get the correct count.", "Solution_9": "I still hold that one infinity can be larger than another infinity.\r\n\r\nFirst of all, it can be shown that $\\lim_{x\\rightarrow c}\\frac{a}{b} = \\frac{\\lim_{x\\rightarrow c} a}{\\lim_{x\\rightarrow c} b}$ \r\n\r\nTherefore $\\lim_{x\\rightarrow + \\infty}\\frac{x^3}{x^2} = \\frac{\\lim_{x\\rightarrow + \\infty}{x^3}}{\\lim_{x\\rightarrow + \\infty}{x^2}}$ \r\n\r\n\r\nand we can simplify\r\n\r\n$\\lim_{x\\rightarrow + \\infty}\\frac{x^3}{x^2} = \\lim_{x\\rightarrow + \\infty}{x} $\r\n\r\nand, obviously, $\\lim_{x\\rightarrow + \\infty}{x}> 1$\r\n\r\nSo therefore :\r\n$\\lim_{x\\rightarrow + \\infty}{x^3} / \\lim_{x\\rightarrow + \\infty}{x^2} > 1$\r\n\r\nwhich leads to:\r\n$\\lim_{x\\rightarrow + \\infty}{x^3} > \\lim_{x\\rightarrow + \\infty}{x^2}$\r\n\r\nEven though ${\\lim_{x\\rightarrow + \\infty}{x^3} = \\infty_1}$ and ${\\lim_{x\\rightarrow + \\infty}{x^2} = \\infty_2}$\r\n\r\nso $\\infty_1>\\infty_2$\r\n\r\n[size=150]\nNot all infinities are created equal :P [/size]", "Solution_10": "With cantor's argument, you can show that there are indescribable real numbers. :D :P", "Solution_11": "[quote=\"boy_21\"]First of all, it can be shown that $\\lim_{x\\rightarrow c}\\frac{a}{b} = \\frac{\\lim_{x\\rightarrow c} a}{\\lim_{x\\rightarrow c} b}$ [/quote]\r\n\r\nThis isn't true at all, only under certain conditions. One of those conditions is that both the numerator and denominatory [i]not[/i] approach infinity or 0. What your argument amounts to saying is that \"different functions approach infinity at different speeds.\" This doesn't tell you anything about sizes of infinities. There are in fact differently-sized infinities, but they mean something entirely different that how you're treating them. It's possible to define systems in which there are infinities more like the one that you are trying to use, but such systems aren't the standard real numbers that we're used to working with. For example, they admit of \"infinitesimal numbers,\" positive numbers which are less than $\\frac{1}{n}$ for all positive integers n -- something that we don't have in the \"normal\" reals." } { "Tag": [ "MATHCOUNTS" ], "Problem": "Why is my TI-84 plus silver edition so slow with graphing? All the memory is cleared, yet it graphs much slower than a normal TI-84.", "Solution_1": "There is no \"normal\" T-84. It is the TI-84+, which is what I have. I don't have much experience with the silver but I find that it tends to graph slowly only for exceedingly complex functions. Does it graph slow for something like $ y\\equal{}2x$? I don't know what would cause it go slow.", "Solution_2": "change the graph speed. The difference between a 84+ and 84+ silver is only that silver has more memory. The difference between 83+ silver and 84+ silver I don't know.", "Solution_3": "How do you change the graph speed?", "Solution_4": "[quote=\"JRav\"]There is no \"normal\" T-84. It is the TI-84+.[/quote]\r\n\r\nThen what's the point on naming it \"plus?\" Is it just a cheap scam to get people to think they are buying a \"better\" product? Or is this common practice and the \"normal\" version of a TI-89 is a TI-89 plus? :huh:", "Solution_5": "TI-83$ \\Rightarrow$TI-83+$ \\Rightarrow$TI-84+$ \\Rightarrow$TI-84+ Silver", "Solution_6": "If you're in parametric mode, it might be the Tstep. Use a higher value.", "Solution_7": "The Silver Edition has a way of lagging in all aspects of calculations. I find that deleting [i]most[/i] of your applications (the ones that are useless to you) will speed up the overall process.", "Solution_8": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]\r\nI overclocked mine.", "Solution_9": "[quote=\"7h3.D3m0n.117\"]The Silver Edition has a way of lagging in all aspects of calculations. I find that deleting [i]most[/i] of your applications (the ones that are useless to you) will speed up the overall process.[/quote]\nYeah, true. But I deleted all of my apps, and it's still really slow. \n\n[quote=\"#H34N1\"]I overclocked mine.[/quote]\r\nIt seems you can overclock all the TIs except the TI-84, how did you do it?", "Solution_10": "[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]\r\nUnmounting some random chip. My friend unlocked mine for me. I'll ask him when I go back to school tomorrow.\r\n\r\nEDIT: Nvm he overclocked my other calculator.", "Solution_11": "It runs fast enough for me. I like it more than java on my computer (for programs). I can go up to higher numbers, and it's less work. I have 1 app left though (Mirage OS lol).", "Solution_12": "I accidentally deleted all my applications. I also accidentally messed up all my programs too when they started going crazy (this factoring program was written incredibly inefficiently, testing all numbers up to the number instead of all primes up to the floor of the square root), which were only useful for random mathcounts stuff.\r\nMy calculator is fast :)", "Solution_13": "I noticed my TI-89 Titanium executes programs much slower than my TI-83 plus does. :(", "Solution_14": "[quote=\"serialk11r\"]I accidentally deleted all my applications. I also accidentally messed up all my programs too when they started going crazy (this factoring program was written incredibly inefficiently, testing all numbers up to the number instead of all primes up to the floor of the square root), which were only useful for random mathcounts stuff.\nMy calculator is fast :)[/quote]\r\nHow would you do that? To tes tall numbers to the square root use a for() loop\r\nlike for (a,1,sqrt(a),[1])\r\nThis means it takes variable A, starts at 1, goes to $ \\sqrt{a}$ and increments 1. If it's increment is 1, the last part is optional, if you want to increase it by any number besides 1, you'll have to put another num there.\r\nOf course you can also use an if statement" } { "Tag": [ "probability" ], "Problem": "Three cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a Jack, the second card is a Queen, and the third card is a King?", "Solution_1": "Simple probability.\r\n\r\n$ \\frac{4}{52} \\times \\frac{4}{51} \\times \\frac{4}{50}\\equal{}\\boxed{\\frac{8}{16575}}$.", "Solution_2": "Since it never specified whether the cards were drawn after one another or drawn at the same time, I think this problem should be modified.", "Solution_3": "It says FIRST and SECOND. That gives order and that is why it should be $52 \\times 51 \\times 50$", "Solution_4": "This problem, at least in my opinion, is a little too brute-forcey." } { "Tag": [ "trigonometry", "inequalities", "inequalities proposed" ], "Problem": "$a,b,c\\in\\mathbb{R^{+}}, a+b+c=abc$ prove:\r\n\r\n$\\frac{1}{\\sqrt{1+a^{2}}}+\\frac{1}{\\sqrt{1+b^{2}}}+\\frac{1}{\\sqrt{1+c^{2}}}\\leq\\frac{3}{2}$", "Solution_1": "very easy\r\n\r\nagain the famous cosine problem\r\n\r\nlet $x=\\tan^{-1}a,y=\\tan^{-1}b,z=\\tan^{-1}c$\r\nby hypothesis:\r\n$x,y,z>0,x+y+z=\\pi$ we wish to show:\r\n$\\cos{x}+\\cos{y}+\\cos{z}\\leq \\frac{3}{2}$ which is well known and true by Jensen.\r\n\r\nso we are done. :)", "Solution_2": "Ok,here is a first step to the solution:The first relation is rewritten as: \\[\\frac1{ab}+\\frac1{bc}+\\frac1{ac}=1\\] .Because of this,we may take $\\frac1{ab}=\\frac{x}{s}$ etc..with $s=x+y+z$.And after solving the sistem we easily get $c=\\sqrt{\\frac{xs}{yz}}$ so \\[c^{2}+1=\\frac{x(x+y+z)+yz}{yz}=\\frac{(x+y)(x+z)}{yz}\\] therefore our initial inequality becomes \\[\\sum \\sqrt{\\frac{yz}{(x+y)(x+z)}}\\leq\\frac32\\] .Can someone finish it from here?", "Solution_3": "[quote=\"andyciup\"] \\[\\sum \\sqrt{\\frac{yz}{(x+y)(x+z)}}\\leq\\frac32\\] .Can someone finish it from here?[/quote]\r\n\r\nCauchy.\r\n\r\n$\\sum \\sqrt{\\frac{yz}{(x+y)(x+z)}}\\leq \\sqrt{\\sum yz \\sum \\frac{1}{(x+y)(x+z)}}$\r\n$=\\sqrt{\\frac{2\\sum xy \\sum x}{(x+y)(y+z)(z+x)}}$\r\n$=\\sqrt{2 \\frac{\\sum x^{2}y+3xyz}{\\sum x^{2}y+2xyz}}$\r\n\r\n\r\n$\\sqrt{2 \\frac{\\sum x^{2}y+3xyz}{\\sum x^{2}y+2xyz}}\\leq \\frac{3}{2}$\r\n$\\iff 8(\\sum x^{2}y+3xyz) \\leq 9(\\sum x^{2}y+2xyz)$\r\n$\\iff 6xyz \\leq \\sum x^{2}y$ which is true by AM-GM, and this completes the proof.", "Solution_4": "[quote=\"andyciup\"] \\[\\sum \\sqrt{\\frac{yz}{(x+y)(x+z)}}\\leq\\frac32\\] .Can someone finish it from here?[/quote]\r\n\r\nAlso AM-GM . \r\n$2\\sqrt{\\frac{yz}{(x+y)(x+z)}}\\leq\\frac{y}{x+y}+\\frac{z}{x+z}$ \r\n\r\n :wink:" } { "Tag": [ "geometry", "circumcircle", "geometry unsolved" ], "Problem": "Let $ ABC$ be an equilateral and $ C$ its incircle. If $ D$ and $ E$ are points on $ AB$ and $ AC$, respectively, such that $ DE$ is tangent to $ C$.\r\nShow that $ \\frac{AD}{DB} \\plus{} \\frac{AE}{EC} \\equal{} 1$", "Solution_1": "Just use algebraic manipulation: $ AD^2 \\plus{} AE^2 \\equal{} (a\\minus{}AD)^2 \\plus{} (a\\minus{}AE)^2$ where $ a$ is half the side of the triangle. Expand and we shall see that the equation to be proven is equivalence to this.", "Solution_2": "Denote M, N and P the contact points of AB, AC and DE with the incircle, then AM = a, MD = DP = x, EN = EP = y.\r\nThe relation to prove is equivalent to a^2 - a(x + y) - 3xy = 0. \r\nCalculating DE from the triangle ADE with .<), 88, 102, and of course 144 and factors thereof\r\n\r\nand of course 137 :D (because I like 137?)" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "prove $ lim_{x \\minus{} > 6} 3 \\equal{} 3$ via an epsilon-delta proof\r\n\r\n-- my attempt --\r\n$ Lim_{x \\minus{} > 6} 3 \\equal{} 3$\r\n$ |3 \\minus{} 3| < \\varepsilon$\r\n$ |0| < \\varepsilon$\r\nbut i need to find \r\n$ |x \\minus{} 6| < k*{ \\varepsilon}$, and [b]i can't see how i'd do that.[/b] \r\nThen, let $ \\delta \\minus{} k \\varepsilon$\r\nSince all steps are reversible, whenever $ 0 < |x \\minus{} 6| < \\delta$, then $ |3 \\minus{} 3| < \\varepsilon$ and $ 3$ is the limit.", "Solution_1": "youre overthinking it\r\n\r\nall epsilon delta proofs start with (even if they dont say it)\r\n\r\nlet $ \\epsilon >0$\r\n\r\nthen we want to find a $ \\delta>0$ s.t.\r\n\r\n$ |x\\minus{}6|<\\delta \\to |3 \\minus{} 3| < \\epsilon$\r\n\r\nbut of course such a delta exists, since 3 is constant no matter what we choose for x\r\n\r\nso let $ \\delta \\equal{} 1$\r\n\r\nthen we know when $ |x\\minus{}6|<\\delta\\equal{}1$ we have $ |3\\minus{}3|\\equal{}0 < \\epsilon$\r\n\r\ntherefore $ \\lim_{x\\to 6} 3 \\equal{} 3$" } { "Tag": [ "function", "inequalities", "limit", "Functional Analysis", "triangle inequality", "topology", "real analysis" ], "Problem": "Show that if X is a separable Banach space there exists an isometric inclusion, T, between X and the space l^(infinity), the space of real valued functions with the sup norm, which is also a Banach space. Hahn Banach could be used, but I am not sure how to get it. Much thanks!", "Solution_1": "The right trick here is a direct construction. Let $ x_1,x_2,\\cdots$ be a dense sequence, and define $ f(x)\\equal{}(\\|x\\minus{}x_1\\|,\\|x\\minus{}x_2\\|,\\cdots)$. By the triangle inequality, $ \\|f(x)\\minus{}f(y)\\|\\equal{}\\sup |\\,\\|x\\minus{}x_n\\|\\minus{}\\|y\\minus{}x_n\\| \\,|\\le \\sup \\|x\\minus{}y\\|$. Since we can choose a subsequence $ x_{n(k)}$ converging to $ y$, we also get that $ \\sup |\\,\\|x\\minus{}x_n\\|\\minus{}\\|y\\minus{}x_n\\| \\,|\\ge \\lim_k |\\,\\|x\\minus{}x_{n(k)}\\|\\minus{}\\|y\\minus{}x_{n(k)}\\| \\,|\\equal{}\\|x\\minus{}y\\|$ and $ f$ is an isometry.\r\n\r\nThis actually proves a stronger result; we didn't use any norm properties, so in fact any separable metric space has an isometric injection into $ \\ell^\\infty$." } { "Tag": [], "Problem": "I found this problem and now I'm starring at it with lot's of what if's. I'm working on this problem but feel free to beat me at it It's not like I'm going back to school to take Algebra II again.\r\n\r\nA * [(B + C)(D - E) - F(G*H) ] / J = 10\r\n\r\nKnowing that each variable is a unique, single-digit, nonzero number, and that C - B = 1, and H - G = 3, what is the number ABCDEFGHJ, where each letter is a digit? For example, if A = 4, B = 2, and C = 7, ABC would equal 427.", "Solution_1": "Harder than it looks...\r\n\r\n[hide=\"partial solution\"]Since C-B=1, then B is one of 1,2,3,4,5,6,7,8 and C is one of 2,3,4,5,6,7,8,9 AND B+C is one of 3,5,7,9,11,13,15,17\n\nSince H-G=3, H is one of 9,8,7,6,5,4 and G is one of 6,5,4,3,2,1 AND HG is one of 54,40,28,18,10,4\n\nWe can make a little distribution of the possible values of B, C, G, and H:\n\nA| \nB| 1,2,3,4,5,6,7,8\nC| 2,3,4,5,6,7,8,9\nD| \nE| \nF| \nG| 6,5,4,3,2,1\nH| 9,8,7,6,5,4\nJ| \n\nWe also know that $A\\left(\\left(B+C\\right)\\left(D-E\\right)-FGH\\right)=10J$[/hide]\r\n\r\n\r\nAnd I thought I knew Intermediate Algebra! :P", "Solution_2": "This was supposebly \"simple\" algebra :|" } { "Tag": [ "algebra", "polynomial", "inequalities", "algebra proposed" ], "Problem": "A polynomial $ P(x)$ has degree at most $ 2k$, where $ k= 0, 1, ...$. Given that for an integer $ i$, the inequality $ -k\\leq i\\leq k$ implies $ |P(i)|\\leq 1$. Prove that for all real numbers $ x$, with $ -k\\leq x\\leq k$, the following inequality holds: $ |P(x)|< (2k+1){2k\\choose k}$", "Solution_1": "$ (*)$ if $ deg(P)=2k$\r\nwe have $ P(x)=\\sum_{i=1}^{2k}Q_{i}(x)P(i)$ where $ Q_{i}(x)=\\prod_{j=1 \\\\ j\\neq i}^{2k}\\frac{x-i}{j-i}$.\r\nso $ |p(x)|\\le \\sum_{i=1}^{2k}|Q_{i}(x)|$\r\nand we have for $ x\\in [-k,k]$ :$ |Q_{i}(x)|\\le |Q_{i}(2k+1)|=\\frac{(2k)!}{(2k+1-i)\\prod_{j=1 \\\\ j\\neq i}^{2k}|j-i|}\\le \\frac{(2k)!}{(2k+1-i)(k!)^{2}}$\r\nthen $ |p(x)|\\le{2k\\choose k}\\sum_{i=1}^{2k}\\frac{1}{2k+1-i}<(2k+1){2k\\choose k}$\r\n___________________________\r\n$ (*)$ if $ deg(P)=s<2k$\r\nwe have $ P(x)=\\sum_{i=1}^{s}Q_{i}(x)P(i)$ where $ Q_{i}(x)=\\prod_{j=1 \\\\ j\\neq i}^{s}\\frac{x-i}{j-i}$.\r\nso $ |p(x)|\\le \\sum_{i=1}^{s}|Q_{i}(x)|$\r\nand we have for $ x\\in [-k,k]$ :$ |Q_{i}(x)|\\le |Q_{i}(2k+1)|=\\frac{(2k)!}{(2k-s)!(2k+1-i)\\prod_{j=1 \\\\ j\\neq i}^{s}|j-i|}\\le \\frac{(2k)!}{ (2k-s)!(2k+1-i)(s!)^{2}}$\r\nand we have $ (2k-s)!(2k+1-i)(s!)^{2}>(k!)^{2}$\r\nthen $ |p(x)|<(2k+1){2k\\choose k}$" } { "Tag": [ "algebra unsolved", "algebra" ], "Problem": "Let $a_{1},a_{2},\\cdots a_{N}$ be distinct positive integers that do not contain a $9$ in their decimal representation.\r\n\r\nProve that $\\frac{1}{a_{1}}+\\cdots+\\frac{1}{a_{n}}\\leq \\frac{1}{30}$", "Solution_1": "[quote=\"Yosh...\"]Let $a_{1},a_{2},\\cdots a_{N}$ be distinct positive integers that do not contain a $9$ in their decimal representation.\n\nProve that $\\frac{1}{a_{1}}+\\cdots+\\frac{1}{a_{n}}\\leq \\frac{1}{30}$[/quote]\r\n\r\nThis is obviously wrong : take $N=1$ and $a_{1}=1$\r\n\r\nI think that the problem is to show $\\frac{1}{a_{1}}+\\cdots+\\frac{1}{a_{n}}< 30$ \r\n\r\nLet $A=\\{$ posive integers without any $9$ in their decimal representaion $\\}$\r\nLet $A_{k}=\\{x\\in A, 10^{k-1}\\leq x<10^{k}\\}$\r\n\r\nLet $S_{k}=\\sum_{x\\in A_{k}}\\frac{1}{x}$\r\n\r\n$S_{k+1}=\\sum_{x\\in A_{k}}(\\frac{1}{10x}+$ $\\frac{1}{10x+1}+\\frac{1}{10x+2}+\\frac{1}{10x+3}+$ $\\frac{1}{10x+4}+\\frac{1}{10x+5}+\\frac{1}{10x+6}$ $+\\frac{1}{10x+7}+\\frac{1}{10x+8})<9\\sum_{x\\in A_{k}}\\frac{1}{10x}=\\frac{9}{10}S_{k}$\r\nSo $S_{k+1}10.\r\n\r\nusually the problems involving complex numbers are actualy very simple, requiring more geometry and intuition than algebra.", "Solution_12": "In answer to the orginal question in this post, see our Frequently Asked Questions, AIME #1:\r\nhttp://www.unl.edu/amc/f-miscellaneous/faq.shtml#aime\r\n\r\nSteve Dunbar\r\nMAA Director, American Mathematics Competitions" } { "Tag": [ "probability" ], "Problem": "[b]Hi. This is based on a math game called \"24.\" What you have to do is with four numbers, perform 3 different operations (only multiplication, division, addition, and subtraction) with all numbers, to equal 24. All parts must be whole. You can only use each number once, unless it is listed twice. The numbers don't have to be used in the order listed. Do not hide it since it is a game with points. Just type it, don't put in the \"hide=Answer\" stuff. Do not post once somebody has, it doesn't matter and you will not recieve points. If no one posts the answer in 3 days, I will post a new one. I mean exactly 3 days. The time is at the top of posts. If it's 3 days and 1 minute, it won't count. So if you solve one that is more than 3 days old, good for you but it doesn't count. :D :D First person to 5 points earns the right to post 1 problem, whenever they want, as long as an answer has been posted and I have not made a new one yet (Write in the post that you earned it). And for people who are not new: PEMDAS no longer counts. If your are don't worry. At no point can the number be a decimal of fraction\n\n[/b][b]ex.[/b] the numbers are 2,4, 44 and 6. Notice how you can get 24 by doing $4\\times6$. However, thats only using two numbers. The correct answer would be $44\\div2+6-4=24$\r\n\r\n[b]The first set of numbers are...[/b]\r\n[b]8, 20, 2, and 18 (remember, the numbers do not have to be used in this order)[/b]", "Solution_1": "Here is the answer to the first one:\r\n\r\n$2\\cdot18-20+8=24$\r\n\r\nAre you going to post several more after they have been answered?", "Solution_2": "I will kind of make this into a game so the scores are:\r\n\r\nb-flat: 1", "Solution_3": "[b]Round 2.[/b]\r\nHere are the next numbers. [b]Look at the top post for instructions.[/b]\r\n\r\n[b]3, 5, 57, & 10 [/b](they don't have to be used in order)\r\n\r\nP.S. If you have looked at the instructions look again because I added some stuff.", "Solution_4": "[b]57/3 + 10 - 5 = 24[/b]", "Solution_5": "ok the score is:\r\n\r\n[b]b-flat: 1\ndogseatcheese: 1[/b]\r\n\r\n[b]Next numbers[/b]. Look back for instructions\r\n[b]4. 4. 5. & 8[/b]\r\n\r\nRemember, you cannot use a number twice, unless it is listed twice, so you have to use 4 twice.", "Solution_6": "4*5+8-4=24 :lol:", "Solution_7": "Here's another:\r\n8*5-4*4=24 :D", "Solution_8": "another:\r\n\r\n8*(4+4-5) :)", "Solution_9": "Next numbers????? i was the frst 1 2 answer =P(i don't like dah smiley)", "Solution_10": "You could make it the first person to reply gets 1 point, and if the second person posts a completely different (but still correct) solution, they get 1 point also. Like Eru posted a completely different but correct solution so he/she should also get a point. When are you posting the next problems?", "Solution_11": "SRY. I''ve had 2 many school projects these days.s Well, I don't believe Eru would be right because I'm pretty sure in the game, PEMDAS doesn't matter. However [b]I will make a new rule that if you solve it with PEMDAS you get half of a point[/b]. If there are multiple soulutions without PEMDAS both will get a point\r\n\r\n[b]b-flat: 1 \ndogseatcheese: 1\nNaonao: 1\nEru: .5[/b]\r\n\r\n[b]Next number 5, 13, 20, 3[/b]", "Solution_12": "13*3-20+5 ;) ;)", "Solution_13": "[b]b-flat: 1 \ndogseatcheese: 1 \nNaonao: 1\nrobinhe: 1\nEru: .5 [/b]\r\n\r\nCome on...some 1 who has already\r\nanswered don't leave-No one has even reached 2\r\n[b]\nNext #'s\n3, 3, 9, 7 [/b]", "Solution_14": "Here I am, with 2. It is hard to find multiple solutions, and if someone gets there before you, you can't get any point. It doesn't use PEMDAS.\r\n\r\n$\\frac93\\cdot7+3=24$", "Solution_15": "There's no way to make 24 using those. I'm pretty sure.\r\n23 I also got, 24 is next to impossible.\r\nBy the way, check out my game!\r\nLink's in my sig!", "Solution_16": "[b]Hey. Due to the new \"Kleiban's 24\", which has took all the attention away from mine, this whole 24 forum can be deleted by a moderator, because many people aren't replying anyway. And if a mod doesn't delete it, I'm not replying to anything or posting, so do whatever you want in thid forum.[/b]\r\n\r\n[b]:bomb: :mad: :furious: THANKS KLEIBAN :furious: :mad: :bomb: [/b]\r\n\r\nyes, I show anger through emoticons.", "Solution_17": "[quote=\"R49917S\"][b]Hey. Due to the new \"Kleiban's 24\", which has took all the attention away from mine, this whole 24 forum can be deleted by a moderator, because many people aren't replying anyway. And if a mod doesn't delete it, I'm not replying to anything or posting, so do whatever you want in thid forum.[/b]\n\n[b]:bomb: :mad: :furious: THANKS KLEIBAN :furious: :mad: :bomb: [/b]\n\nyes, I show anger through emoticons.[/quote]\r\n\r\n? You don't need someone to delete this. I would have responded to this 24, except i couldnt' find a solution to your numbers :lol:", "Solution_18": "[quote=\"ch1n353ch3s54a1l\"][quote=\"R49917S\"][b]Hey. Due to the new \"Kleiban's 24\", which has took all the attention away from mine, this whole 24 forum can be deleted by a moderator, because many people aren't replying anyway. And if a mod doesn't delete it, I'm not replying to anything or posting, so do whatever you want in thid forum.[/b]\n\n[b]:bomb: :mad: :furious: THANKS KLEIBAN :furious: :mad: :bomb: [/b]\n\nyes, I show anger through emoticons.[/quote]\n\n? You don't need someone to delete this. I would have responded to this 24, except i couldnt' find a solution to your numbers :lol:[/quote]\r\n\r\nI couldn't find an answer either so I just gave up on it.", "Solution_19": "[b]Okay. Have 6 other people say that they would answer and I will continue. Otherwise, make your own or something.[/b]", "Solution_20": "i will continue, sure it's a fun game", "Solution_21": "I started a new topic because this is getting way to confusing.\r\nSome of your nubmers are off, and your rules are hard to understand.", "Solution_22": "[quote=\"R49917S\"][b]Hi. This is based on a math game called \"24.\" What you have to do is with four numbers, perform 3 different operations (only multiplication, division, addition, and subtraction) with all numbers, to equal 24. All parts must be whole. You can only use each number once, unless it is listed twice. The numbers don't have to be used in the order listed. Do not hide it since it is a game with points. Just type it, don't put in the \"hide=Answer\" stuff. Do not post once somebody has, it doesn't matter and you will not recieve points. If no one posts the answer in 3 days, I will post a new one. I mean exactly 3 days. The time is at the top of posts. If it's 3 days and 1 minute, it won't count. So if you solve one that is more than 3 days old, good for you but it doesn't count. :D :D First person to 5 points earns the right to post 1 problem, whenever they want, as long as an answer has been posted and I have not made a new one yet (Write in the post that you earned it). And for people who are not new: PEMDAS no longer counts. If your are don't worry. At no point can the number be a decimal of fraction\n\n[/b][b]ex.[/b] the numbers are 2,4, 44 and 6. Notice how you can get 24 by doing $4\\times6$. However, thats only using two numbers. The correct answer would be $44\\div2+6-4=24$\n\n[b]The first set of numbers are...[/b]\n[b]8, 20, 2, and 18 (remember, the numbers do not have to be used in this order)[/b][/quote]\r\n\r\n$24$", "Solution_23": "2*18-20+8\r\n\r\nCan I post the next problem?", "Solution_24": "tz. You apparently have not read the previous page that says I have forever abandoned this post (well at least making problems). Type anything you want about anything. (For example-Duck are neat.", "Solution_25": "[quote=\"R49917S\"] Type anything you want about anything. (For example-Duck are neat.[/quote]\r\nwouldn't that be just spamming?", "Solution_26": "Duck are neat", "Solution_27": "oh well.. i liked this 24.", "Solution_28": "Penguins walk wobilly. Elephants have huge ears.", "Solution_29": "Please stop! If R49917S really doesn't want to continue, then leave this topic alone. Thank you. :)" } { "Tag": [ "inequalities", "inequalities solved" ], "Problem": "Let a,b,x,y,z be positive real numbers.\r\nShow that\r\n\r\n x/(ay+bz) + y/(az+bx) + z/(ax+by) \\geq 3/(a+b)", "Solution_1": "Just a remark: This is a generalization of Nesbitt's inequality, for a=b we get his inequality.", "Solution_2": "it is a very easy application of a derived version of Cauchy-Schwartz inequality (sometimes called Cauchy-Buniakowski):\r\n\r\n\\sum ai / bi \\sum aibi \\geq ( \\sum ai)2 , for all positive reals ai, bi." } { "Tag": [ "probability", "geometry", "number theory", "\\/closed" ], "Problem": "What would be the next course to take at AoPS after the Into to Algebra course that began in March 2009 (assuming the student did fine in the course)? I am confused by the new names of the AoPS Algebra courses. Thank you, from the mom of a current Intro to Alg. student.", "Solution_1": "Introduction to Algebra is the new Algebra 1 + Algebra 2 courses, so the next Algebra course would be Algebra 3 (the old Intermediate Algebra).", "Solution_2": "[quote=\"westiepaw\"]What would be the next course to take at AoPS after the Into to Algebra course that began in March 2009 (assuming the student did fine in the course)? I am confused by the new names of the AoPS Algebra courses. Thank you, from the mom of a current Intro to Alg. student.[/quote]\r\n\r\nI would recommend going through Intro to Number Theory and Intro to Counting & Probability (in either order) after Intro Algebra. Then, Intro to Geometry, and then Algebra 3 (formerly Intermediate Algebra). After that, the rest of the Intermediate classes in basically any order." } { "Tag": [ "geometry", "3D geometry", "number theory", "least common multiple", "MATHCOUNTS", "calculus", "trigonometry" ], "Problem": "1.Find all those 5-digit positive whole numbers that are square and cube numbers also.\r\n \r\n\r\n 2. How many 4-digit numbers are there in which the sum of the digits is 5?\r\n\r\n 3 .Find a 6-digit number that gets 6 times greater when you move its last 3 digits to the front of the number without changing the order of these three digits.\r\n \r\n 4.How many 3-digit numbers are there in which the digits are such that if you create all the different 3-digit numbers from them, and you add these numbers, the sum is divisible by 111?\r\n\r\ncan someone else post some problems -_-", "Solution_1": "ummm...what's with the title of this thread?", "Solution_2": "[hide]\n1) If a number is to be a perfect square and cube, it must be a perfect sixth power (6=LCM of (2,3)). The only numbers which are five digit sixth powers are 5^6=15625 and 6^6=46656. Thus, there are two numbers. \n\n2) After a little bit of counting, you'll figure out that there are six partitions of 5 which can be expressed as 4 or fewer numbers (if fewer, then add enough zeroes to the end of the partition until it has four digits). These partitions are:\n\n5+0+0+0\n4+1+0+0\n3+1+1+0\n2+1+1+1\n2+2+1+0\n3+2+0+0\n\nIf you know about combinatorics, you should know that each of four numbers, if placed in the thousands place of a four digit number, adds six more different numbers to the total # of possibilities (provided all the numbers are distinct, and none are zero). But here, this is not the case. For the first partition, we have (4!-(3x6))/3!=1 combo.\n\nFor the second, we have (4!-(2x6))/2!=6 combos.\nFor the third, we have (4!-(1x6))/2!=9.\n4th=4!/3!=4.\n5th=(4!-(1x6))/2!=9.\n6th=(4!-(2x6))/2!=6.\n\nThis is a total of 1+6+9+4+9+6=35 numbers.\n\n3) Umm, I'm still working on it...\n\n4) As long as the three digits are not zero, all possible arrangements added up will be divisible by 111, since their sum will be 222(a+b+c), where a,b, and c are the digits. Otherwise, you'll run into complications. \n\nThe total possible of such numbers is 9x9x9=729. [/hide]", "Solution_3": "1.[hide]These are just powers are 6... 5^6=15625 and 6^6=46656 are the only powers of 6 that are 5 digits.[/hide]\n\n2.[hide]The ways to make the sum of the digits 5 are as follows: 5+0+0+0, 4+1+0+0 (can be repeated 6 times, 4100, 1400,4010, 4001, etc...), 3+2+0+0 (also 6 times), 3+1+1+0 (9 times), 2+2+1+0, (9 times) 2+1+1+1 (4 times). 9+9+6+6+4+1=35.[/hide]\n\n3.[hide]600000a+60000b+6000c+600d+60e+6f=100000d+10000e+1000f+100a+10b+c\n599900a+59990b+5999c=99400d+9940e+994f\n\nClearly, the unit's digits of 5999c and 994f must be the same. We want 994f as big as possible and 599900 as small as possible so we make f=7, and c=2. That puts 9 in the tens digits for the left side and 8 for the right side...\nEdit: This is not a good method, read natmc's method.[/hide]\n\n4.[hide]You have 3 digits, a, b, and c. You can have the numbers 100a+10b+c, 100a+10c+b... etc... and if you sum all of those you get 222a+222b+222c. Note that this will only work if none of the digits are 0. That leaves us with 9^3=729 numbers in which it works.[/hide]", "Solution_4": "[hide=\"1\"]\nmust be $x^6$ so, $5^6=15625$ and $6^6=26656$\n[/hide]\n[hide=\"2\"]\npartitions for 5\n5000 -1 number\n4100 -6 numbers\n3200 -6 numbers\n3110 -9 numbers\n2210 -9 numbers\n2111 -4 numbers\n$1+6+6+9+9+4=35$\n[/hide]\n[hide=\"3\"]\n$n=1000x+y$\n\n$6(1000x+y)=1000y+x$\n\n$6000x+6y=1000y+x$\n\n$5999x=994y$\n\n$\\frac{x}{y}=\\frac{994}{5999}=\\frac{142}{857}$\n\n$x=142$\n\n$y=857$\n\n$n=142857$\n[/hide]\n[hide=\"4\"]\nYou go through all cases and figure out it cannot contain a 0.\nnumbers $x,y,z$\n\n$xyz$ $xzy$ $yxz$ $yzx$ $zxy$ $zyx$\n\n$\\sum=200(x+y+z)+20(x+y+z)+2(x+y+z)={2}\\cdot{111}(x+y+z)$\n\nnumbers $x,x,y$\n\n$xxy$ $xyx$ $yxx$\n\n$\\sum=100(2x+y)+10(2x+y)+(2x+y)=111(2x+y)$\n\nnumbers $x,x,x$\n\n$\\sum=111x$\n\nnumbers $x,y,0$\n\n$xy0$ $x0y$ $yx0$ $y0x$\n\n$\\sum=200(x+y)+10(x+y)+(x+y)=211(x+y)$\n\nnumbers $x,0,0$\n\n$\\sum=100x$\n\nanswer ${9}\\cdot{9}\\cdot{9}=729$\n[/hide]", "Solution_5": "[quote=\"eryaman\"]ummm...what's with the title of this thread?[/quote]\r\n\r\nYeah, someone should change the title, because it will confuse people, and stop them from coming to see the problems, since I think we can assume that most people in Mathcounts aren't learning calculus or trig yet.", "Solution_6": "The title of this thread has been changed by moderator rcv.\r\n\r\nG-UNIT: The moderator requests that you use appropriate, meaningful titles.", "Solution_7": "[hide=\"1.\"]Must be powers of 6:\n5^6=15625\n6^6=46656\ntwo.[/hide]\n[hide=\"2.\"]\nPossible combinations for the digits:\n1,1,1,2\n1,2,2,0\n2,3,0,0\n3,1,1,0\n4,1,0,0\n5,0,0,0\n\n1st one: 4 combinations\n2nd one: 9 combinations\n3rd one: 6 combinations\n4th one: 9 combinations\n5th one: 6 combinations\n6th one: 1 combination\n9+9+6+6+4+1=[b]35[/b][/hide]", "Solution_8": "sorry about that, i didnt really think my choice of title mattered, i simply wrote down the first title i could think off... :)" } { "Tag": [ "Vieta", "algebra unsolved", "algebra" ], "Problem": "Find all real solutions of the system:\r\n\r\n$ x^2\\plus{}2yz\\equal{}x,$\r\n$ y^2\\plus{}2zx\\equal{}y,$\r\n$ z^2\\plus{}2xy\\equal{}z.$", "Solution_1": "[hide]Case I: one of $ x,y,z$ is $ 0$: \n\nWLOG, let $ x\\equal{}0$. \n\nThen, $ yz\\equal{}0$, $ y^2\\equal{}y$, $ z^2\\equal{}z$. \n\nThus, $ y,z \\in \\{ 0,1 \\}$ and either $ y\\equal{}0$ or $ z\\equal{}0$. \n\nTherefore, the only solutions in this case are: $ (0,0,0)$, $ (1,0,0)$, $ (0,1,0)$, $ (0,0,1)$. \n\nCase II: $ x,y,z \\neq 0$: \n\nThen, multiplying the three equations by $ x$, $ y$, $ z$ respectively yields: \n$ x^3\\minus{}x^2 \\equal{} y^3\\minus{}y^2 \\equal{} z^3\\minus{}z^2 \\equal{} \\minus{}2xyz \\equal{} K$ for some non-zero $ K \\in \\mathbb{R}$. \n\nThus, $ x,y,z$ are solutions of $ f(t) \\equal{} t^3\\minus{}t^2\\minus{}K \\equal{} 0$. \n\nIf $ x,y,z$ are distinct, by Vieta's $ xyz \\equal{} K \\equal{} \\minus{}2xyz$ which is impossible for $ x,y,z \\neq 0$. \n\nWLOG, let $ y \\equal{} z$. \n\nThen: \n$ y^2 \\plus{} 2xy \\equal{} y \\Rightarrow y\\equal{}1\\minus{}2x$ \n$ x^2 \\plus{} 2y^2 \\equal{} x \\Rightarrow x^2 \\plus{} 2(1\\minus{}2x)^2 \\minus{} x \\equal{} (3x\\minus{}1)(3x\\minus{}2) \\equal{} 0$ \n\nThus, $ x \\equal{} \\frac{1}{3} \\Rightarrow y \\equal{} z \\equal{} \\frac{1}{3}$ OR $ x \\equal{} \\frac{2}{3} \\Rightarrow y \\equal{} z \\equal{} \\minus{}\\frac{1}{3}$ \n\nTherefore, the solutions in this case are: $ \\left(\\frac{1}{3},\\frac{1}{3},\\frac{1}{3}\\right)$, $ \\left(\\frac{2}{3},\\minus{}\\frac{1}{3},\\minus{}\\frac{1}{3}\\right)$, $ \\left(\\minus{}\\frac{1}{3},\\frac{2}{3},\\minus{}\\frac{1}{3}\\right)$, $ \\left(\\minus{}\\frac{1}{3},\\minus{}\\frac{1}{3},\\frac{2}{3}\\right)$. [/hide]", "Solution_2": "And another solytion" } { "Tag": [], "Problem": "I can't get go2pdf to work on my computer, so I won't be able to include images in my solutions file. Is it OK to submit images as a separate file?", "Solution_1": "Try CutePDF:\r\n\r\nhttp://www.cutepdf.com/Products/CutePDF/writer.asp\r\n\r\nAnother option is to print out you solutions and hand-draw them.\r\n\r\nYou can also just make JPG files and include those the same way the instructions describe including PDF files. \r\n\r\nWe need to have a single document we can print and hand to graders." } { "Tag": [], "Problem": "Which of the following statements are true?\r\n\r\nI. The set {-1,0,1} is closed under addition\r\nII. The set {-1,0,1} is closed under multiplication \r\nIII. The set of rational numbers is closed under addition\r\nIV. The set of rational numbers is closed under multiplication", "Solution_1": "The first two questions can be answered making a table to see if all the possible sums/products of elements of the set belong to the set. If so, we have closure.\r\n[hide=\"I.\"] 1+1=2 by my definition, and 2 does not belong to {-1,0,1}.\nThe set {-1,0,1} is not closed under addition.\nOf course, we could define an operation that set would be closed under, but I would not use the name addition.[/hide]\n[hide=\"II.\"]All nine possible products are in {-1,0,1}. The set {-1,0,1} is closed under multiplication [/hide]\nFor the other two problems, it is a question of deciding if every time two rational number are added/multiplied the result is necessarily also rational.\n[hide=\"III.\"] The set of rational numbers is closed under addition[/hide]\n[hide=\"IV.\"] The set of rational numbers is closed under multiplication[/hide]" } { "Tag": [ "algebra", "polynomial", "Rational Root Theorem" ], "Problem": "I know you can use the Rational Root Theorem to look for rational roots of a polynomial, but how would you find the roots if there are no rational roots, if all the roots are irrational or imaginary?\r\n\r\nThanks.", "Solution_1": "Sometimes an approximation is good enough, say through Newton's Method.", "Solution_2": "[quote=\"tornado.adv4\"]I know you can use the Rational Root Theorem to look for rational roots of a polynomial, but how would you find the roots if there are no rational roots, if all the roots are irrational or imaginary?\n\nThanks.[/quote]\r\n\r\nThat's a good question, although usually on AMC there are rational roots.\r\n\r\nI had a similar question :blush: \r\n\r\nAs JRav said, sometimes approximation is good enough.\r\n\r\nMaybe Intermediate Algebra can help, or ask at class if you're taking Algebra 3", "Solution_3": "[quote=\"tornado.adv4\"]I know you can use the Rational Root Theorem to look for rational roots of a polynomial, but how would you find the roots if there are no rational roots, if all the roots are irrational or imaginary?\n\nThanks.[/quote]Depends. If you are looking for the roots in a 'nice' form, you should know by now polynomials degree of 5 or higher are not 'solvable' by radicals due to Galois Theory. There are other conditions that allow it to be 'solvable' but that's beyond the scope of this subforum.\r\n\r\nJRav suggests a good way of finding an approximation.\r\n\r\nEDIT: Descartes Rule of Signs is helpful if you want a rough idea about the nature of the roots. Other interesting topics are like Horner's method, Sturm sequences, etc." } { "Tag": [ "ratio", "Gauss" ], "Problem": "What are some of the differences between American and foreign Universities? Would some of you students ever consider attending a foreign University?\r\n\r\nWhat do you all think of the proposed changes to the British University system?", "Solution_1": "I'm afraid you'll have to provide me a little background (or some good links) about the British system.\r\n\r\nCanadian schools are cheap, and they have some good ones (McGill and U. Toronto pop to mind).", "Solution_2": "Under the currect system all British citizens (and I think even EU citizens) pay the same tuition for any British Univeristy. I think it's around $2000 US a year. Tony Blair wants to make it so that British Universities can charge more, but the students don't have to pay until after they graduate and their annual income becomes at least equal to the national average.\r\n\r\nTo us Americans this seems like a dream situation! But some of the British are fighting the proposal vehemently.", "Solution_3": "Which of them sounds like a dream situation - the old or the new format? (Though I'd wager there are many who'd answer 'both'.)\r\n\r\nAnd what are the grounds on which the complaining British object?", "Solution_4": "I think the main objection from some British is that they see this as a first step in the Americanization of the system. They fear that the tuitions will skyrocket, and that the elite Universities it will become exclusively for the rich.\r\n\r\nBlair's argument is that in order to reach Britain's stated goal of having 50% of it's population go to University, they have to find a way to bring in more revenue. He also claims that the system as it is can't compete with the American system and that Britain is losing many of it's best professors to American Universities that can pay them better.", "Solution_5": "Heh when it comes to cost no one...er, country can beat Japan: 5K for all Japanese government-built universities (I apologize for that, I mistook \"You can go to any college if you don't be choosy\" for \"You can go to any college no matter what\" and applied it to all people...although in Japan most people are middle-classes and not a lot of people are really rich so I'm guessing most could afford it)\r\n\r\nTokyo University is still very cheap and it's very good too, around five grand (the best school in Japan). There are two other private great school in Japan, Waseda (around 20K) and Keiou (around 30K) depending on what you study. (All rates are annual and only includes the fees for the studying, does not include stuff like the living cost or transportation cost)\r\n\r\n\r\nIn general Japanese schools have this huge entrance exams which everybody studies for like maniacs (a.k.a. the Examination War) and then they slack off for the rest of the school until the end in which they start studying for The War again and this starts happening as soon as preschool :shocked:\r\n\r\nThis is the reason contributing to \"you can at least go to somewhere for college\" is the decreasing number of Japanese children. We have so much seniors and too little children such that we have 2:1 ratio for seniors versus children. Pretty soon we might perish...(I'm just having a shot in the dark here, don't take it seriously folks :wink:)\r\n\r\nIn fact the total number of total seats available in Japan exceeds the number of children in Japan thus this is possible.", "Solution_6": "What is the nature and set up of the entrance exams in Japan? Are they like the SATs, or like AP Exams, or something totally different?", "Solution_7": "Well there is this \"Central Exam\" (even though it's pronounced \"center exam\" I believe this is grammatically correct) but most college/university have their own tests and that is perhaps the only factor which will decide whether you get into that college/university or not. There is no scholarships, (maybe) retakes, very little looking over the other stuff (unless you have something big like a criminal record or an IMO winner :wink:). Basically if you fail the war you'll have to come back next year, even if you were just sick the day you took the test.\r\nOn the other hand there's no ridiculous requirement like \"swim 100m\" but at the same time there are no requirement...kinds extreme huh?\r\n\r\nI wonder how much Russian university cost...", "Solution_8": "I mean what subjects do the exams usually cover - are they \"aptitude\" tests like the SAT or are they subject tests like the AP exams? And what is the level of difficulty of the tests.", "Solution_9": "Most colleges in other countries actually usually have an extremely simple process, compared to the U.S. In many countries, you just take a test(similar to the SAT/ACT), and submit your grades. Those may be the only considerations! \n\nGauss, the test usually depends on the country: each have their own test, and it usually covers all the material learned in high school only." } { "Tag": [ "inequalities", "trigonometry", "geometry proposed", "geometry" ], "Problem": "$1.\\blacktriangleright\\ \\triangle ABC\\Longrightarrow$ $\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+\\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\\right)\\le 12R^{2}\\left(a^{2}+b^{2}+c^{2}\\right)\\ .$\r\n\r\n$2.\\blacktriangleright\\ \\sqrt 3\\le\\boxed{\\frac{4R+r}{p}\\le\\sqrt{3+4\\cdot \\frac{R-2r}{r}}}\\ .$\r\n[hide=\"Hint.\"]Apply the first inequality in the (acute) triangle $A'B'C\\ (\\sqrt a,\\sqrt b,\\sqrt c)\\ ,$ where $\\cos A'=\\frac{p-a}{\\sqrt{bc}}$ a.s.o. and $R'=\\sqrt{\\frac{pR}{4R+r}}$ [/hide]", "Solution_1": "we have \r\n$12R^{2}(a^{2}+b^{2}+c^{2})\\ge\\frac{4}{3}(a^{2}+b^{2}+c^{2})^{2}\\ge(a^{2}+b^{2}+c^{2})^{2}+b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}$\r\nwe have the first becasue $a^{2}+b^{2}+c^{2}\\le 9R^{2}$\r\nthus we got the stronger inequality and completed the solution! :D", "Solution_2": "Indeed, the first inequality is a trivial consequence of the identity\r\n\\[OH^{2}=9R^{2}-a^{2}-b^{2}-c^{2}. \\]" } { "Tag": [ "algebra solved", "algebra" ], "Problem": "Problem2\r\n\r\ni)Let n positive integer (n>= 0) find the sign of n^6 +5n^5.sin(n) +1\r\n\r\nii)For wich positive integer n (n>= 0)\r\n\r\n(n^2 +5n.cos(n) +1)/(n^6+5n^5.sin(n)+1) >= 10^(-4) \r\n\r\nholds ? \r\n\r\nWithout calcultrice", "Solution_1": "I suppose \"angle n\" (in sin(n)) is in degrees ?", "Solution_2": "Yes", "Solution_3": "NO\r\n\r\ni) I think \"angle n\" in sin(n) is in radians, otherwise the problem loses significance.\r\n1) if n=1,2,3: sin n>0 ==> n^6+5n^5.sin(n)+1>0\r\n2) if n=4: sin 4=-sin(4-pi)>-sin 0.9 > -0.9+0.9^3/3!-0.9^5/5!>-4/5 ==> \r\n n^6+5n^5.sin(n)+1>4^5(4+5.sin 4)+1>4^5.0+1>0\r\n3) if n>5: n^6+5n^5.sin(n)+1=n^5(n+5.sin n)+1>n^5(n-5)+1>0\r\n\r\n----------\r\nPrepare yourself to boredom.\r\n\r\nii) Let us determine sign of n^2 +5n.cos(n) +1\r\n1) if n=1: n^2+5n.cos(n)+1=2+5cos(1)>2\r\n2) if n=2: \r\n n^2+5n.cos(n)+1=5+10cos(2)=5(1+2cos(2))=10(cos(pi/3)+cos(2))=20.cos(1-pi/6).cos(1+pi/6)=20.cos(1-pi/6).sin(pi/3-1)>20.cos(pi/6).2/pi.(pi/3-1)>10(\\sqrt 3).1/2.1/25>1/5\r\n3) if n=3: n^2+5n.cos(n)+1=10+15cos(3)<0\r\n4) if n=4: n^2+5n.cos(n)+1=17+20cos(4)=17-20cos(4-pi)>17-20cos(pi/4)>5/2\r\n5) if n=5: n^2+5n.cos(n)+1=26+20cos(5)>26\r\n6) if n\\geq 6: n^2-5n10-4.\r\nf(2)>(1/5)/(2^6+5*2^5+1)>10-4\r\nf(3)<0\r\nf(4)>(5/2)/(4^6+1)>10-4\r\nf(5)>26/(5^6+1)>10-4\r\nf(6)>36/(6^6+1)>10-4\r\nf(7)>49/(7^6+5.7^5+1)>10-4\r\nIf n=8 then \r\n n^2+5n.cos(n)+1=65+40cos(8)=65-40sin(8-5pi/2)>65-40.(8-5pi/2)>65-40*0.15=59\r\n n^6+5n^5.sin(n)+1<8^6+5.8^5+1<500000\r\n f(8)>59/500000>10-4\r\nIf n=9 then\r\n n^2+5n.cos(n)+1=82+45cos(9)=82-45cos(3pi-9)<82-45cos(pi/6)<45\r\n n^6+5n^5.sin(n)+1=9^5(9+5sin(9))+1=9^5(9+5sin(3pi-9))+1>\r\n >9^5(9+5.(0.4).2/pi)+1>9^5(9+5.(0.4).2/4)+1=9^5.10+1=590491\r\n f(9)<45/590491<10-4\r\nif n=10 then\r\n n^2+5n.cos(n)+1=101+50cos(10)=101-50cos(10-3*pi)<\r\n <101-50cos(3/5)<101-50(1-(3/5)^2/2)=60\r\n n^6+5n^5.sin(n)+1=10^5(10+5sin(10))+1=10^5(10-5sin(10-3pi))+1>\r\n >10^5(10-5(10-3pi))+1>10^5(10-5.3/5)+1>70000\r\n f(10)<10-4\r\nif n=11 then\r\n cos(11)>0 ==> n^2+5n.cos(n)+1>122\r\n n^6+5n^5.sin(n)+1=11^5(11+5sin(11))+1=11^5(11-5*cos(11-7pi/2))+1<\r\n <11^5(11-4)+1=1127357\r\n f(11)>10-4\r\nif n=12 then\r\n n^2+5n.cos(n)+1=145+60cos(12)<145+24=205\r\n 12+5sin(12)=12-5sin(4pi-12)>12-5(4pi-12)>9 ==> n^6+5n^5.sin(n)+1>2239489\r\n f(12)<10-4\r\n\r\nif n\\geq 13 then\r\n f(n)<(n^2+5n+1)/(n^6-5n^5+1)<10-4\r\n (if you will ask me \"why?\", I kill you)" } { "Tag": [ "trigonometry", "inequalities", "geometry", "circumcircle", "trig identities", "Law of Sines", "inequalities unsolved" ], "Problem": "Prove that for every acute triangle $ABC$ we have\r\n\r\n$\\sin A \\sin 3A + \\sin B \\sin 3B + \\sin C \\sin 3C \\leq 0$", "Solution_1": "No solutions? :(", "Solution_2": "It's really not difficult and It's a lemma of $\\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$\r\nAny one post a solution,My solution use deriveate and I think It has other way.", "Solution_3": "[color=blue][b]Problem.[/b] Let ABC be an acute-angled triangle. Prove the inequality\n\n$\\sin A\\sin\\left(3A\\right)+\\sin B\\sin\\left(3B\\right)+\\sin C\\sin\\left(3C\\right)\\leq 0$.[/color]\r\n\r\n[i]Solution.[/i] By a well-known formula from trigonometry,\r\n\r\n$\\sin\\left(3A\\right)=3\\sin A-4\\sin^3 A$.\r\n\r\nThus, $\\sin A\\sin\\left(3A\\right)=\\sin A\\left(3\\sin A-4\\sin^3 A\\right)=3\\sin^2 A-4\\sin^4 A$. Similarly, $\\sin B\\sin\\left(3B\\right)=3\\sin^2 B-4\\sin^4 B$ and $\\sin C\\sin\\left(3C\\right)=3\\sin^2 C-4\\sin^4 C$. Thus, the inequality in question,\r\n\r\n$\\sin A\\sin\\left(3A\\right)+\\sin B\\sin\\left(3B\\right)+\\sin C\\sin\\left(3C\\right)\\leq 0$,\r\n\r\nis equivalent to\r\n\r\n$\\left(3\\sin^2 A-4\\sin^4 A\\right)+\\left(3\\sin^2 B-4\\sin^4 B\\right)+\\left(3\\sin^2 C-4\\sin^4 C\\right)\\leq 0$.\r\n\r\nIf R is the circumradius of triangle ABC, then the extended law of sines yields $\\sin A=\\frac{a}{2R}$, $\\sin B=\\frac{b}{2R}$ and $\\sin C=\\frac{c}{2R}$, so that this inequality rewrites as\r\n\r\n$\\left(3\\left(\\frac{a}{2R}\\right)^2-4\\left(\\frac{a}{2R}\\right)^4\\right)+\\left(3\\left(\\frac{b}{2R}\\right)^2-4\\left(\\frac{b}{2R}\\right)^4\\right)+\\left(3\\left(\\frac{c}{2R}\\right)^2-4\\left(\\frac{c}{2R}\\right)^4\\right)$\r\n$\\leq 0$.\r\n\r\nAfter multiplication with $4R^4$ and simplification, this becomes\r\n\r\n$\\left(3R^2a^2-a^4\\right)+\\left(3R^2b^2-b^4\\right)+\\left(3R^2c^2-c^4\\right)\\leq 0$.\r\n\r\nThis is equivalent to\r\n\r\n$3R^2\\left(a^2+b^2+c^2\\right)-\\left(a^4+b^4+c^4\\right)\\leq 0$,\r\n\r\ni. e. to\r\n\r\n$3R^2\\left(a^2+b^2+c^2\\right)\\leq a^4+b^4+c^4$.\r\n\r\nNow, let's apply the Conway substitution (cf. http://www.mathlinks.ro/Forum/viewtopic.php?t=6523 post #4): Since the triangle ABC is acute-angled, we have $b^2+c^2>a^2$, $c^2+a^2>b^2$ and $a^2+b^2>c^2$; thus, the numbers $x=\\frac12\\left(b^2+c^2-a^2\\right)$, $y=\\frac12\\left(c^2+a^2-b^2\\right)$ and $z=\\frac12\\left(a^2+b^2-c^2\\right)$ are all positive. Of course, we have $y+z=\\frac12\\left(c^2+a^2-b^2\\right)+\\frac12\\left(a^2+b^2-c^2\\right)=a^2$ and similarly $z+x=b^2$ and $x+y=c^2$, and thus $a^2+b^2+c^2=\\left(y+z\\right)+\\left(z+x\\right)+\\left(x+y\\right)=2\\left(x+y+z\\right)$.\r\n\r\nAlso, from $y+z=a^2$, $z+x=b^2$ and $x+y=c^2$, it follows that $\\left(y+z\\right)^2=\\left(a^2\\right)^2=a^4$ and similarly $\\left(z+x\\right)^2=b^4$ and $\\left(x+y\\right)^2=c^4$. \r\n\r\nFinally, since the circumradius R of triangle ABC can be found using the formula $R=\\frac{abc}{4F}$, where F is the area of triangle ABC, we have\r\n\r\n$R^2=\\left(\\frac{abc}{4F}\\right)^2=\\frac{a^2b^2c^2}{16F^2}=\\frac{\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)}{16F^2}$.\r\n\r\nBy the Heron formula for the area F of triangle ABC, we have\r\n\r\n$16F^2=\\left(a+b+c\\right)\\left(b+c-a\\right)\\left(c+a-b\\right)\\left(a+b-c\\right)$;\r\n\r\non the other hand, using the equations $x=\\frac12\\left(b^2+c^2-a^2\\right)$, $y=\\frac12\\left(c^2+a^2-b^2\\right)$ and $z=\\frac12\\left(a^2+b^2-c^2\\right)$ and performing some calculations (there is a synthetic proof, too, but I'm too lazy to write it down), we can find\r\n\r\n$4\\left(yz+zx+xy\\right)=\\left(a+b+c\\right)\\left(b+c-a\\right)\\left(c+a-b\\right)\\left(a+b-c\\right)$.\r\n\r\nThus, $16F^2=4\\left(yz+zx+xy\\right)$. Hence,\r\n\r\n$R^2=\\frac{\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)}{16F^2}=\\frac{\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)}{4\\left(yz+zx+xy\\right)}$.\r\n\r\nCombining all the above results, we get\r\n\r\n$a^2+b^2+c^2=2\\left(x+y+z\\right)$\r\n$a^4=\\left(y+z\\right)^2$; $b^4=\\left(z+x\\right)^2$; $c^4=\\left(x+y\\right)^2$;\r\n$R^2=\\frac{\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)}{4\\left(yz+zx+xy\\right)}$.\r\n\r\nWith these formulas, the inequality that we must prove,\r\n\r\n$3R^2\\left(a^2+b^2+c^2\\right)\\leq a^4+b^4+c^4$,\r\n\r\nbecomes\r\n\r\n$3\\frac{\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)}{4\\left(yz+zx+xy\\right)}\\cdot 2\\left(x+y+z\\right)\\leq \\left(y+z\\right)^2+\\left(z+x\\right)^2+\\left(x+y\\right)^2$.\r\n\r\nThis simplifies to\r\n\r\n$3\\frac{\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)\\left(x+y+z\\right)}{2\\left(yz+zx+xy\\right)}\\leq \\left(y+z\\right)^2+\\left(z+x\\right)^2+\\left(x+y\\right)^2$.\r\n\r\nUpon multiplication with 2 (yz + zx + xy), this becomes\r\n\r\n$3\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)\\left(x+y+z\\right)\\leq 2\\left(yz+zx+xy\\right)\\left(\\left(y+z\\right)^2+\\left(z+x\\right)^2+\\left(x+y\\right)^2\\right)$.\r\n\r\nBut this inequality follows from\r\n\r\n$2\\left(yz+zx+xy\\right)\\left(\\left(y+z\\right)^2+\\left(z+x\\right)^2+\\left(x+y\\right)^2\\right)-3\\left(y+z\\right)\\left(z+x\\right)\\left(x+y\\right)\\left(x+y+z\\right)$\r\n$=yz\\left(y-z\\right)^2+zx\\left(z-x\\right)^2+xy\\left(x-y\\right)^2\\geq 0$.\r\n\r\nSo the proof is complete.\r\n\r\n[[b]EDIT:[/b] Now I see that this inequality has already been proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=22866 , posts #10 and #14.]\r\n\r\n Darij", "Solution_4": "Since\r\n$R^2=\\frac{a^2b^2c^2}{16S^2}=\\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$,\r\nusing the substitution $a^2=\\frac{y+z}{2}$ etc., the inequality\r\n$\\displaystyle \\frac{a^4+b^4+c^4}{a^2+b^2+c^2} \\ge 3R^2$\r\nbecomes as follows\r\n$xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \\ge 0.$" } { "Tag": [ "quadratics", "modular arithmetic", "probability", "algebra", "linear equation", "number theory" ], "Problem": "Problem 2.18 is:\r\n\r\n\"Is 9409 prime?\" \r\n\r\nThere is a hint, # 52, that gives the expression:\r\n\r\n(100-x)^2 = 10000-200x+x^2\r\n\r\nNow, this is great and I'm still working on the problem, but what I find confusing is how can one tell from the problem that algebra would be useful from the problem where it's not.\r\nFor example, most of the problems in this Challenge section are done with no algebra or arithmetic at all, just words. So, out of nowhere, comes this problem where this quadratic is given as a hint. \r\nProblem 2.14 is similar, since the hint involves a linear equation, but there's nothing I can see from the problem that would lead me to believe algebra would be useful.\r\nI realize I should just let go and trust that understanding will dawn at some point, but if anyone has any insight on this, I would be most appreciative.\r\n\r\nThanks,\r\nEd Beaugard", "Solution_1": "hello, solve the equation $ 10000\\minus{}200x\\plus{}x^2\\equal{}9409$ for $ x$.\r\nSonnhard.", "Solution_2": "Algebra is useful because there are many factorization techniques that can be applied if you know that the number is of a special form, and factorizations are usually useful in number theory.\r\n\r\nFor example, [url=http://www.mathlinks.ro/viewtopic.php?t=150527]PEN E 4[/url] asks to prove that $ 1280000401$ is composite. By recognizing the special form $ 1280000401 \\equal{} 20^7 \\plus{} 20^2 \\plus{} 1$ you may use the factorization $ a^7 \\plus{} a^2 \\plus{} 1 \\equal{} (a^2 \\plus{} a \\plus{} 1)(a^5 \\minus{} a^4 \\plus{} a^2 \\minus{} a \\plus{} 1)$.\r\n\r\nIt's all a matter of familiarity with the technique.", "Solution_3": "Not only is 9409 not prime, it is a perfect square. $ 97^2\\equal{}9409$.", "Solution_4": "...which is EXACTLY what the hint implied?\r\n\r\nIn general, when there's a question that says \"Is ??? prime?\", about 70% of the time (unless it's in a school textbook) it will be composite. :)", "Solution_5": "I really appreciate the replies and patience with my Newbie questions. \r\nI found ffao interesting, since he(I assume) relates factorization in algebra to factorization in number theory, which is exactly what I was wondering about, that is, when would one turn to algebra to solve these kinds of problems.\r\n\r\nThanks again.", "Solution_6": "[quote=\"AIME15\"]...which is EXACTLY what the hint implied?\n\nIn general, when there's a question that says \"Is ??? prime?\", about 70% of the time (unless it's in a school textbook) it will be composite. :)[/quote]\r\n\r\nLol , I like your way of thinking :lol:", "Solution_7": "So, the key to recognizing when to apply algebra, such as special factorizations, to number theory problems is simply familiarity with the technique? Or is there a certain set of factorizations that we should learn to test on all prime/composite problems?", "Solution_8": "Also, note 9409 ends in a 9.\r\nAnd the hint says 100^2-200x+x^2\r\nSo x must be 3 mod 10, otherwise you wouldn't have a 9 as the units digit.", "Solution_9": "Why can't it be $ \\equiv 7 \\pmod 10$?", "Solution_10": "If 9409 fits 10000 - 200x + x^2 you'll know that it is composite because then it is also a perfect square. If it doesn't then the hint was pretty much useless.. :P", "Solution_11": "Generally, in my opinion, it would be helpful in the book if there were indications about when to use algebra and when not to on particular problems.\r\n\r\nThanks,\r\nEd Beaugard", "Solution_12": "But you aren't given these hints in real problems...", "Solution_13": "When and when not to use algebra to solve real problems are something we should know, I agree, but I'm talking about the textbook itself(which I like a great deal), where we're supposed to learn this. It's just something that could be thought about that might make a superb book even better.\r\n\r\nVery sincerely,\r\nEd Beaugard", "Solution_14": "I think you have a good point. For instance, Introduction to Counting and Probability emphasizes that many times algebra and counting can be used together, but doesn't discuss too much about how and when. Perhaps it would be useful if a section or two were dedicated to explaining what clues to look for when approaching a counting problem that may require algebra.", "Solution_15": "Maybe someone could start a topic or forum about this, if there's enough interest." } { "Tag": [], "Problem": "in soal ro javab bedin.man ke khodam tv mibinam.shayad ham ba doostam beram biroon.gahi cinema raftam ham ba doostan hal mide :roll:", "Solution_1": "\nIn the filming of the movie, you can watch the TV show. Maybe I have done a lot of films like this?\nrough translation", "Solution_2": "@aasta.sharma the guy is iranian not indian", "Solution_3": "Ya ik I plugged into google translate... XD\nAlso u spelled my name wrong :mad: \nLol", "Solution_4": "What is this?", "Solution_5": "a dead post (and forum), that's what. i can't believe i'm the only poster on here for 5 years." } { "Tag": [ "LaTeX", "\\/closed" ], "Problem": "The character \u22ef is not properly displayed on IE 6 for some reason ...", "Solution_1": "I have IE6. What is that character (if I see it, I'll then know what is it)?", "Solution_2": "This was a note for myself ... no need to reply ;)" } { "Tag": [ "AMC", "USA(J)MO", "USAMO", "USAMTS" ], "Problem": "I've know that the USAMO is probably much harder than the USAMTS in terms of how strictly it's graded, but does anyone know how much stricter? In other words, would a 5 on a USAMTS problem always get a 7 on a USAMO problem? Obviously a USAMTS problem wouldn't appear on the USAMO, but just compare the scoring.", "Solution_1": "You need to have a complete, rigorous solution to get a 5 on a USAMTS problem, so a 5 on USAMTS would probably get a 7 on USAMO.\r\n\r\nOn the other hand, one could receive as much as 3/5 on a USAMTS problem without coming close to solving it. On the USAMO, this would probably result in 1-2 points if even that much.", "Solution_2": "[quote=\"le00327146\"]I've know that the USAMO is probably much harder than the USAMTS in terms of how strictly it's graded, but does anyone know how much stricter? In other words, would a 5 on a USAMTS problem always get a 7 on a USAMO problem? Obviously a USAMTS problem wouldn't appear on the USAMO, but just compare the scoring.[/quote]\r\n\r\nFrom my own experience (which is not much) I would say no, since USAMTS graders allow a certain degree of handwaving which the USAMO graders wouldn't. I do believe I received a couple of fives which probably would've reached a ~1/7 if the problem was on the USAMO.\r\n\r\nHere's my weak attempt to match up the two grading scales (what constitutes a hole/minor is probably different for the two contests): \r\n\r\n0 - No work, or completely trivial work \r\n*0/5 = nothing presented that is credible. \r\n*1/5 = evidence of understanding the problem\r\n*2/5 = some good ideas but not much more;\r\n\r\n1-2 - Progress on the problem, but not completely solved \r\n*3/5 = larger holes in a promising solution;\r\n\r\n3-4 - All steps are present, but may lack clarity. (These scores are very rare.) \r\n\r\n5-6 - Complete solution with minor errors \r\n*4/5 = minor shortcomings in the solution;\r\n\r\n7 - Perfect solution \r\n*5/5 = clear, concise, correct solution;", "Solution_3": "[quote=\"azjps\"]\n3-4 - All steps are present, but may lack clarity. (These scores are very rare.) \n[/quote]\r\n\r\nSo does this mean you have a correct, valid solution, but it's messy or hard to understand? Because that sounds like an awfully harsh point deduction for having the correct steps.", "Solution_4": "Basically, you [b]will not[/b] get a 3-4 on a one part problem. Even 2 and 5 are pretty rare, but much less so. On a two part problem, depending on the relative difficulties of the parts, it'll either be 3/4 or 2/5. On 3+ part problems, the middle points are much easier to obtain", "Solution_5": "I did USAMO last year, and I thought I had adequately solved two of them, but I ended up getting a one one both :| . But I'm doing fine in USAMTS.", "Solution_6": "Did you do USAMTS last year? I'd be surprised if you haven't learned anything about writing solutions in the 5 months between the last USAMO and the beginning of this year's USAMTS.", "Solution_7": "These are interesting answers. Having been through MOSP, I think I can say that I am familiar with the USAMO grading process, which favors very concise, to-the-point solutions. Many USAMTS graders, however, seem to like lots of elaboration. There have been instances where I have gotten 4 on USAMTS problems that I am quite certain I would have gotten 7 for on the USAMO. Of course, the same problems would not appear on the USAMO, but still... Basically, I think USAMO graders are much more likely to understand and like unusual arguments. On the other hand, they can be very hard on incomplete proofs, which doesn't seem to bother USAMTS graders as much (not sure about this part, though).", "Solution_8": "As I see it, USAMTS is aimed at those students who are just being introduced to proofs. There are more problems that you can do by utter brute-force, the use of a computer is even allowed, and partial credit is given out gratuitously. The only thing that really distinguishes a nice, elegant solution is the commending of solutions.", "Solution_9": "[quote=\"MellowMelon\"]Did you do USAMTS last year? I'd be surprised if you haven't learned anything about writing solutions in the 5 months between the last USAMO and the beginning of this year's USAMTS.[/quote]\r\n\r\nWell, no I didn't do USAMTS last year, but I don't feel like I learned much... :|", "Solution_10": "[quote=\"jb05\"]These are interesting answers. Having been through MOSP, I think I can say that I am familiar with the USAMO grading process, which favors very concise, to-the-point solutions.[/quote]\r\n\r\nSo I guess this means the shorter the better, and points will be taken off for brute-force?", "Solution_11": "I wish you luck on brute-forcing a USAMO question (just kidding). Shorter indicates being more concise, but without losing any rigorosity, of course. \r\n\r\nLast year the only (literally only) proof writing experience I had was this, and I somehow got a 12 on the USAMO ... still don't know how that happened ...", "Solution_12": "I didn't have any proof-writing experience for USAMO last year :D but then, I only got a 4...", "Solution_13": "[quote=\"azjps\"]but without losing any rigorosity, of course. \n[/quote]\r\n\r\nFor future reference, there's no such word as \"rigorosity.\" There is \"rigorism\" or more simply just \"rigor.\"", "Solution_14": "To go slightly off-topic... is \"rigorize\" a word? Probably not, but I use it a lot.", "Solution_15": "[url=http://dictionary.reference.com/browse/rigorize]Woah .. its not?[/url] [url=http://en.wiktionary.org/wiki/rigorize]Or maybe?[/url]\r\n\r\n :lol: I was too lazy to check if \"rigorisity\" was a word (it sounded like one), but I use 'rigorize' all the time. Apparently AoPS shows up in the top google results for 'rigorize'\r\n\r\nEdit: this post is off-topic ..", "Solution_16": "According to Merriam-Webster's Collegiate Dictionary, rigorize is not a word. Of course, I bet if you looked at one of those huge, 80 dollar dictionaries, you might find it. Also, it doesn't really matter if it's a word if people can understand it." } { "Tag": [ "function", "inequalities", "complex analysis", "complex analysis unsolved" ], "Problem": "Suppose that f is analytic on complex plane except for a finite number of singularity and z*f(z) goes zero as z goes to infinity.\r\nProve that there exits M>0 such that |z^2*f(z)| < M for all z such that |z|>= M", "Solution_1": "Hint: $ f$ minus the sum of $ f$'s singular parts at these singularities is an entire function. Consider $ z$ times this entire function, using the fact that each singular part is is $ O(1/|z|)$ at $ \\infty$.", "Solution_2": "Thanks for your help. I have tried the problem according to your suggestion and I did not get it, could you please help me more", "Solution_3": "Now that I think of it, the problem is badly posed. A stronger result is easier to prove: Suppose $ f$ is analytic in $ \\{|z|>R\\}$ for some $ R > 0$. Show that if $ zf(z)\\rightarrow 0$ at $ \\infty$, then $ f(z) \\equal{} O(1/|z|^2)$ as $ z\\rightarrow \\infty$. Hint: Consider the Laurent expansion of $ f$ in $ \\{|z|>R\\}$." } { "Tag": [ "quadratics", "number theory proposed", "number theory" ], "Problem": "Show that the following equations have an infinite number of simultaneous solutions. \r\n\r\n $ a^4 \\plus{} b^4 \\plus{} c^4 \\equal{} 2d^4$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2d^2$ with $ \\gcd(a,b,c,d) \\equal{} 1$ and $ a,b,c,d \\in \\mathbb{N}$", "Solution_1": "\\[ \\frac{a^4\\plus{}b^4\\plus{}c^4}{2}\\equal{}\\left ( \\frac{a^2\\plus{}b^2\\plus{}c^2}{2} \\right )^2\\]\r\n\r\n\\[ a^4\\minus{}2a^2(b^2\\plus{}c^2)\\plus{}(b^2\\plus{}c^2)^2\\equal{}0\\]\r\nsolve quadratic in $ a$.\r\n\\[ a^2\\equal{}\\frac {2(b^2\\plus{}c^2)\\pm \\sqrt{2^2(b^2\\plus{}c^2)^2\\minus{}4(b^2\\plus{}c^2)^2}}{2}\\equal{}b^2\\plus{}c^2\\]\r\nQED", "Solution_2": "There is a sign error in your second equation. It should be\r\n\r\n$ a^4 \\minus{} 2a^2(b^2 \\plus{} c^2) \\plus{} (b^2 \\minus{} c^2)^2 \\equal{} 0$\r\n\r\nso your solution is incomplete", "Solution_3": "thanks for the correction..\r\n\\[ a^{2}\\equal{}\\frac{2(b^{2}\\plus{}c^{2})\\pm\\sqrt{2^{2}(b^{2}\\plus{}c^{2})^{2}\\minus{}4(b^{2}\\minus{}c^{2})^{2}}}{2}\\]\r\n\\[ \\equal{}b^{2}\\plus{}c^{2}\\pm \\sqrt{4b^2c^2}\\equal{}b^2\\plus{}c^2\\pm 2bc\\equal{}(b\\pm c)^2\\]", "Solution_4": "[quote=\"phymax\"]thanks for the correction..\n\\[ a^{2} \\equal{} \\frac {2(b^{2} \\plus{} c^{2})\\pm\\sqrt {2^{2}(b^{2} \\plus{} c^{2})^{2} \\minus{} 4(b^{2} \\minus{} c^{2})^{2}}}{2}\\]\n\n\\[ \\equal{} b^{2} \\plus{} c^{2}\\pm \\sqrt {4b^2c^2} \\equal{} b^2 \\plus{} c^2\\pm 2bc \\equal{} (b\\pm c)^2\\]\n[/quote]\r\n\r\nBut this is not finished !\r\n\r\nIf $ a^2\\equal{}(b\\pm c)^2$, you get $ \\frac{a^4\\plus{}b^4\\plus{}c^4}2\\equal{}\\left(\\frac{a^2\\plus{}b^2\\plus{}c^2}2\\right)^2$ but you are not sure that $ \\frac{a^2\\plus{}b^2\\plus{}c^2}2$ is a perfect square ($ d^2$)\r\n\r\nfor example : $ b\\equal{}2$, $ c\\equal{}1$ and $ a\\equal{}b\\minus{}c\\equal{}1$ imply $ \\frac{a^2\\plus{}b^2\\plus{}c^2}2\\equal{}3$ and $ \\frac{a^4\\plus{}b^4\\plus{}c^4}2\\equal{}9$ but there is no integer $ d$ such that $ a^2\\plus{}b^2\\plus{}c^2\\equal{}2d^2$\r\n\r\nIMHO", "Solution_5": "So close, but there are solutions.\r\n\r\nE.G $ (a,b,c,d) \\equal{} (3,5,8,7)$\r\n\r\nYou just have to show there are infinitely many.\r\n\r\nSo far you have shown that\r\n$ a \\pm b \\pm c \\equal{} 0$ and $ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2d^2 \\implies a^4 \\plus{} b^4 \\plus{} c^4 \\equal{} 2d^4$", "Solution_6": "[quote=\"PhilG\"]Show that the following equations have an infinite number of simultaneous solutions. \n $ a^4 \\plus{} b^4 \\plus{} c^4 \\equal{} 2d^4$,$ a^2 \\plus{} b^2 \\plus{} c^2 \\equal{} 2d^2$ with $ \\text{gcd}(a,b,c,d) \\equal{} 1$ and $ (a,b,c,d) \\in \\mathbb{N}^4$[/quote]\r\n$ \\displaystyle \\left(\\sum_{cyc}{a^2}\\right)^2 \\equal{} 2\\left(\\sum_{cyc}{a^4}\\right)$ so $ \\displaystyle \\prod_{cyc}{(a \\plus{} b \\minus{} c)} \\equal{} 0$. So it is enough to set $ c: \\equal{} a \\plus{} b$ and show that the equation $ a^2 \\plus{} ab \\plus{} b^2 \\equal{} d^2$ iff $ \\left(\\frac {2a \\plus{} b}{2d}\\right)^2 \\plus{} 3\\left(\\frac {b}{2d}\\right)^2 \\equal{} 1$ has infinitely many coprime solution in $ \\mathbb{Z}^3$. But it is straghforward and well known that the elliptic curve $ X^2 \\plus{} 3Y^2 \\equal{} 1$ has infinitely many points in $ \\mathbb{Q}^2$ (if more details are necessary look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150614]here[/url] for all class solutions).", "Solution_7": "Thank you for completing the problem :) \r\n\r\nThe result can also be seen as an application of the multiplicative property of quadratic forms. I.e.\r\n\r\n$ (x^2 \\plus{} 3y^2)(z^2 \\plus{} 3t^2) \\equal{} (xz \\minus{} 3yt)^2 \\plus{} 3(xt \\plus{} yz)^2$\r\n\r\nWe require a solution to $ u^2 \\plus{} 3v^2 \\equal{} 4d^2$. \r\nTo get one just take any odd integers $ x,y$ and take $ d$ from $ x^2 \\plus{} 3y^2 \\equal{} 2d$\r\nUsing the multiplicative identity we find $ u \\equal{} x^2 \\minus{} 3y^2, v \\equal{} 2xy \\implies u^2 \\plus{} 3v^2 \\equal{} 4d^2$\r\n\r\nIn fact we can repeat the multiplication of solutions to get higher powers for the quadratic form which can be used to show that there are an infinite number of solutions for $ a^4 \\plus{} b^4 \\plus{} c^4 \\equal{} 2d^{2n}$ for any $ n \\in \\mathbb{N}$" } { "Tag": [], "Problem": "GREECE ARE IN THE FINAL BEATING GERMANY :D :D :D \r\nFRIDAY THE CUP AGAINST SPAIN", "Solution_1": "Tha to paroume xalaraa ! :lol: \r\n\r\n\r\nMas duskolepsan ligo sth fash twn omilwn, alla h ellhnikh omada den exei deiksei pros to paron ola ths ta opla :wink:", "Solution_2": ":mad: :mad: spain-greece 1-0 instead of 1-5 gongrats to them but htey were awful :wink: \r\nnext time we will win the future is us" } { "Tag": [ "LaTeX" ], "Problem": "I'm sorry if this was asked/answered before.\r\n\r\n\r\nLaTeX => pdf. \r\nIn the folder, the pdf file is created (I can open it by double-clicking it), but\r\nwhen I press F5 after pressing ctrl+F7, it says \"Cannot excuse the command\" and does not show the pdf file.\r\n\r\n\r\nThank you.", "Solution_1": "You have given no information about what program you are referring to, but on the assumption you are using Windows and TeXnicCenter, then see http://www.artofproblemsolving.com/Forum/topic-42360.html", "Solution_2": "Thank you very much.\r\n\r\nHowever, now after I press F5, \r\nadobe reader opens up, but it doesn't seem like it's opening the document.\r\n\r\nIs this still wrong path?\r\nC:\\Program Files\\Adobe\\Acrobat 7.0\\Reader\\AcroRd32.exe\r\n\r\n\r\n\r\n\r\nOnce again, thank you very much.", "Solution_3": "The path looks ok, but check that all the other commands are set as at http://en.wikipedia.org/wiki/TeXnicCenter", "Solution_4": "It works now!\r\nThank you very much!!" } { "Tag": [ "function" ], "Problem": "Let $ k$ = a real number greater than $ 1$.\r\n\r\nGiven $ y \\equal{} k^x \\minus{} (k^2)^x$, what is the maximum value of the function?", "Solution_1": "hello, setting $ k^x\\equal{}t$ so we get\r\n$ f\\left(\\frac{\\ln(t)}{\\ln(k)}\\right)\\equal{}\\minus{}\\left(t\\minus{}\\frac{1}{2}\\right)^2\\plus{}\\frac{1}{4}\\le\\frac{1}{4}$\r\nthe maximum will be obtained for $ x\\equal{}\\minus{}\\frac{\\ln(2)}{\\ln(k)}$.\r\nSonnhard." } { "Tag": [], "Problem": "\u03a4\u03c1\u03b9\u03b1 \u03c3\u03b7\u03bc\u03b5\u03b9\u03b1$ A,B,C$ \u03bc\u03b5\u03c4\u03b1\u03b2\u03b1\u03bb\u03bf\u03bd\u03c4\u03b1\u03b9 \u03c3\u03b5 \u03c4\u03b5\u03c4\u03b1\u03c1\u03c4\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03bf \u03ba\u03b5\u03bd\u03c4\u03c1\u03bf\u03c5 $ O(0,0)$ \u03ba\u03b1\u03b9 \u03b1\u03ba\u03c4\u03b9\u03bd\u03b1\u03c2 $ 2$.\r\n\u0391\u03bd $ G$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03c4\u03bf \u03b2\u03b1\u03c1\u03c5\u03ba\u03b5\u03bd\u03c1\u03bf \u03c4o\u03c5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03c5 $ ABC$ (\u03b1\u03ba\u03bf\u03bc\u03b7 \u03ba\u03b1\u03b9 \u03b5\u03ba\u03c6\u03c5\u03bb\u03b9\u03c3\u03bc\u03b5\u03bd\u03bf\u03c5) \u03bd\u03b1 \u03b2\u03c1\u03b5\u03b8\u03b5\u03b9 \u03c4\u03bf $ (OG)_{min}$\r\n(\u0398\u03b1 \u03c0\u03c1\u03bf\u03c4\u03b9\u03bc\u03bf\u03c5\u03c3\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03b5 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03c5\u03c2)", "Solution_1": "$ 9|z_G|^2\\equal{}|z_1\\plus{}z_2\\plus{}z_3|^2\\equal{}12\\plus{}8(a\\plus{}b\\plus{}c)$ \u03cc\u03c0\u03bf\u03c5 $ a\\equal{}z_1/z_2$ \u03ba\u03b1\u03b9 \u03ba\u03c5\u03ba\u03bb\u03b9\u03ba\u03ac\r\n\u03bf\u03bc\u03c9\u03c2 $ 0\\equal{}0$ kai oxi toy 2\r\n\u03a1\u03bf\u03b4\u03bf\u03bb\u03c6\u03b5 \u03c4\u03bf \u03b2\u03b9\u03b2\u03bb\u03b9\u03bf \u03c3\u03bf\u03c5 \u03c4\u03bf \u03be\u03b5\u03ba\u03bf\u03ba\u03b1\u03bb\u03b9\u03b6\u03c9 \u03b1\u03ba\u03bf\u03bc\u03b7 \u039f\u03bc\u03bf\u03bb\u03bf\u03b3\u03c9 \u03bf\u03c4\u03b9 \u03b1\u03c1\u03ba\u03b5\u03c4\u03b1 \u03b1\u03c0 \u03c4\u03b1 \u03b8\u03b5\u03bc\u03b1\u03c4\u03b1 \u03c0\u03bf\u03c5 \u03b4\u03b9\u03b1\u03c0\u03c1\u03b1\u03b3\u03bc\u03b1\u03c4\u03b5\u03c5\u03b5\u03c3\u03b1\u03b9 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b5\u03bd\u03c4\u03c5\u03c0\u03c9\u03c3\u03b9\u03b1\u03ba\u03b1 .......\u03c3\u03c4\u03bf \u03c4\u03b5\u03bb\u03bf\u03c2 \u03c4\u03bf\u03c5 \u03b8\u03b1 \u03c4\u03b1 \u03c0\u03bf\u03c5\u03bc\u03b5 \u03a3\u03c5\u03b3\u03c7\u03b1\u03c1\u03b7\u03c4\u03b7\u03c1\u03b9\u03b1..", "Solution_4": "\u03a1\u03bf\u03b4\u03bf\u03bb\u03c6\u03b5 \u03b4\u03b5\u03bd \u03c4\u03bf \u03b5\u03b3\u03c1\u03b1\u03c8\u03b1 \u03ba\u03b1\u03bb\u03b1 \u03b8\u03b5\u03bb\u03b5\u03b9 $ 8Re(z_1/z_2)\\plus{}...$", "Solution_5": "\u039a\u03b1\u03bb\u03b7\u03c3\u03c0\u03b5\u03c1\u03b1\r\n\r\n\u0395\u03b9\u03bd\u03b1\u03b9 \u03bc\u03b9\u03b1 \u03c0\u03bf\u03bb\u03c5 \u03c9\u03c1\u03b1\u03b9\u03b1 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7, \u03b2\u03b1\u03b6\u03c9 \u03bc\u03b9\u03b1 \u03bb\u03c5\u03c3\u03b7 \u03b1\u03bb\u03bb\u03b1 \u03b4\u03b5\u03bd \u03b5\u03b9\u03bc\u03b1\u03b9 100% \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03bf\u03c0\u03bf\u03c4\u03b5 \u03b4\u03b9\u03bf\u03c1\u03b8\u03c9\u03c3\u03c4\u03b5 \u03bc\u03b5 \u03b1\u03bd \u03b5\u03c7\u03c9 \u03bb\u03b1\u03b8\u03bf\u03c2\r\n\r\n\u0391\u03bd $ z_1, z_2, z_3, z_G$ \u03bf\u03b9 \u03b5\u03b9\u03ba\u03bf\u03bd\u03b5\u03c2 \u03c4\u03c9\u03bd \u03ba\u03bf\u03c1\u03c5\u03c6\u03c9\u03bd \u03c4\u03bf\u03c5 \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf\u03c5 \u03ba\u03b1\u03b9 \u03c4\u03bf\u03c5 \u03b2\u03b1\u03c1\u03c5\u03ba\u03b5\u03bd\u03c4\u03c1\u03bf\u03c5 \u03c3\u03c4\u03bf \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf \u03b5\u03c0\u03b9\u03c0\u03b5\u03b4\u03bf, \u03c4\u03bf\u03c4\u03b5:\r\n$ |z_G| \\equal{} \\frac {1}{3}|z_1 \\plus{} z_2 \\plus{} z_3| \\equal{} \\frac {\\sqrt {2}}{3}\\sqrt {6 \\plus{} \\sum{ab} \\plus{} \\sum{\\sqrt {4 \\minus{} a^2}\\sqrt {4 \\minus{} b^2}}}$\r\n\u03bf\u03c0\u03bf\u03c5 $ a \\equal{} Re(z_1), b \\equal{} Re(z_2), c \\equal{} Re(z_3)$\r\n\u039f\u03bc\u03c9\u03c2 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b7 \u03c3\u03c5\u03bd\u03b1\u03c1\u03c4\u03b7\u03c3\u03b7 $ f(a,b,c) \\equal{} \\sum{ab} \\plus{} \\sum{\\sqrt {4 \\minus{} a^2}\\sqrt {4 \\minus{} b^2}}$ \u03b5\u03b9\u03bd\u03b1\u03b9 \u03ba\u03bf\u03b9\u03bb\u03b7 \u03c9\u03c2 \u03c0\u03c1\u03bf\u03c2 \u03c4\u03b9\u03c2 3 \u03bc\u03b5\u03c4\u03b1\u03b2\u03bb\u03b7\u03c4\u03b5\u03c2 \u03ba\u03b1\u03b9 \u03b5\u03c0\u03bf\u03bc\u03b5\u03bd\u03bf\u03c2 \u03c0\u03b1\u03b9\u03c1\u03bd\u03b5\u03b9 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf \u03c3\u03c4\u03b1 \u03b1\u03ba\u03c1\u03b1 \u03c4\u03b7\u03c2, \u03bf\u03c0\u03bf\u03c4\u03b5 \u03ba\u03bf\u03b9\u03c4\u03b1\u03b6\u03bf\u03bd\u03c4\u03b1\u03c2 \u03c3\u03c4\u03b1 \u03b1\u03ba\u03c1\u03b1 \u03b5\u03c5\u03ba\u03bf\u03bb\u03b1 \u03b2\u03b3\u03b1\u03b6\u03bf\u03c5\u03bc\u03b5 \u03bf\u03c4\u03b9 \u03b7 \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03b7 \u03c4\u03b9\u03bc\u03b7 \u03c4\u03b7\u03c2 \u03b5\u03b9\u03bd\u03b1\u03b9 \u03bf \u03b1\u03c1\u03b9\u03b8\u03bc\u03bf\u03c2 $ 4$\r\n\u0391\u03c1\u03b1 \u03c4\u03bf \u03b6\u03b7\u03c4\u03bf\u03c5\u03bc\u03b5\u03bd\u03bf \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf \u03b9\u03c3\u03bf\u03c5\u03c4\u03b5 \u03bc\u03b5 $ {|z_G|}_{min} \\equal{} \\frac {\\sqrt {2}}{3}\\sqrt {6 \\plus{} 4} \\equal{} \\frac {2\\sqrt {5}}{3}$\r\n\u03bc\u03b5 \u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03bf\u03c4\u03b1\u03bd \u03c4\u03b1 2 \u03b1\u03c0\u03bf \u03c4\u03b1 3 \u03c3\u03b7\u03bc\u03b5\u03b9\u03b1 \u03c4\u03b1\u03c5\u03c4\u03b9\u03b6\u03bf\u03bd\u03c4\u03b1\u03b9 \u03bc\u03b5 \u03c4\u03b7\u03bd \u03c4\u03bf\u03bc\u03b7 \u03c4\u03bf\u03c5 \u03c4\u03b5\u03c4\u03b1\u03c1\u03c4\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03bf\u03c5 \u03bc\u03b5 \u03c4\u03bf\u03bd $ x'x$ \u03ba\u03b1\u03b9 \u03c4\u03bf 3\u03bf \u03b5\u03b9\u03bd\u03b1\u03b9 \u03b7 \u03c4\u03bf\u03bc\u03b7 \u03c4\u03bf\u03c5 \u03c4\u03b5\u03c4\u03b1\u03c1\u03c4\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03bf\u03c5 \u03bc\u03b5 \u03c4\u03bf\u03bd \u03b1\u03be\u03bf\u03bd\u03b1 $ y'y$ \u03b7 \u03b1\u03bd\u03c4\u03b9\u03c3\u03c4\u03c1\u03bf\u03c6\u03b1 (\u03b5\u03c7\u03bf\u03c5\u03bc\u03b5 \u03c5\u03c0\u03bf\u03b8\u03b5\u03c3\u03b5\u03b9 \u03c7\u03b2\u03c4\u03b3 \u03bf\u03c4\u03b9 \u03c4\u03bf \u03c4\u03b5\u03c4\u03b1\u03c1\u03c4\u03bf\u03ba\u03c5\u03ba\u03bb\u03b9\u03bf \u03bf\u03c1\u03b9\u03b6\u03b5\u03c4\u03b1\u03b9 \u03b1\u03c0\u03bf \u03c4\u03bf\u03c5\u03c2 \u03b4\u03c5\u03bf \u03b8\u03b5\u03c4\u03b9\u03ba\u03bf\u03c5\u03c2 \u03b7\u03bc\u03b9\u03b1\u03be\u03bf\u03bd\u03b5\u03c2).\r\n\r\n\u0394\u03b9\u03bf\u03c1\u03b8\u03c9\u03c3\u03c4\u03b5 \u03bc\u03b5 \u03b1\u03bd \u03b5\u03c7\u03c9 \u03ba\u03b1\u03bd\u03b5\u03b9 \u03bb\u03b1\u03b8\u03bf\u03c2 \u03b3\u03b9\u03b1\u03c4\u03b9 \u03b4\u03b5\u03bd \u03b5\u03b9\u03bc\u03b1\u03b9 \u03c3\u03b9\u03b3\u03bf\u03c5\u03c1\u03bf\u03c2 \u03b3\u03b9\u03b1 \u03bf\u03c3\u03b1 \u03b3\u03c1\u03b1\u03c6\u03c9.", "Solution_6": "\u03a3\u03c5\u03b3\u03bd\u03ce\u03bc\u03b7 \u03c0\u03b1\u03c1\u03b1\u03c3\u03cd\u03c1\u03b8\u03b7\u03ba\u03b1 \u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2", "Solution_7": "\u039b\u03bf\u03b9\u03c0\u03bf\u03bd Nick \u03b4\u03c9\u03c3\u03b1\u03bc\u03b5 \u03c0\u03b1\u03c1\u03bf\u03bc\u03bf\u03b9\u03b5\u03c2 \u03bb\u03c5\u03c3\u03b5\u03b9\u03c2 \u03bc\u03bf\u03bd\u03bf \u03c0\u03bf\u03c5 \u03b5\u03b3\u03c9 \u03b2\u03c1\u03b7\u03ba\u03b1 \u03c4\u03bf \u03b5\u03bb\u03b1\u03c7\u03b9\u03c3\u03c4\u03bf \u03c4\u03bf\u03c5 $ 8Re(z_1/z_2)\\plus{}...$ $ 8$ (\u03b4\u03b5\u03c2 \u03c4\u03bf\u03bd \u03c4\u03c1\u03bf\u03c0\u03bf \u03c4\u03bf\u03c5 \u03a1\u03bf\u03b4\u03bf\u03bb\u03c6\u03bf\u03c5) \u03bc\u03b5 \u03c4\u03b7\u03bd \u03b9\u03c3\u03bf\u03c4\u03b7\u03c4\u03b1 \u03c3\u03c4\u03bf \u03b5\u03ba\u03c6\u03c5\u03bb\u03b9\u03c3\u03bc\u03b5\u03bd\u03bf \u03c4\u03c1\u03b9\u03b3\u03c9\u03bd\u03bf......\u03bf\u03c0\u03c9\u03c2 \u03c0\u03c1\u03bf\u03b1\u03bd\u03b5\u03c6\u03b5\u03c1\u03b5\u03c2\r\n\u0397 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03ba\u03b1\u03c4\u03b1\u03c3\u03ba\u03b5\u03c5\u03b1\u03c3\u03c4\u03b7\u03ba\u03b5 \u03c3\u03c4\u03b7\u03bd \u03c0\u03c1\u03bf\u03c3\u03c0\u03b1\u03b8\u03b5\u03b9\u03b1 \u03bd\u03b1 \u03b2\u03b3\u03b5\u03b9 \u03b1\u03c3\u03ba\u03b7\u03c3\u03b7 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u0393' \u03ba\u03b1\u03c4\u03b5\u03c5\u03b8\u03c5\u03bd\u03c3\u03b7 .\u039c\u03bf\u03bd\u03bf \u03c0\u03bf\u03c5 \u03b2\u03b3\u03b7\u03ba\u03b5 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u0394' \u03ba\u03b1\u03c4\u03b5\u03c5\u03b8\u03c5\u03bd\u03c3\u03b7............\r\n\u0391\u03bd \u03ba\u03b1\u03c0\u03bf\u03b9\u03bf\u03c2 \u03b5\u03c7\u03b5\u03b9 \u03bb\u03c5\u03c3\u03b7 \u03bc\u03b5 \u03bb\u03c5\u03ba\u03b5\u03b9\u03b1\u03ba\u03b5\u03c2 \u03b1\u03c5\u03c3\u03c4\u03b7\u03c1\u03b1 \u03b3\u03bd\u03c9\u03c3\u03b5\u03b9\u03c2 \u03bc\u03b5 \u03b5\u03bd\u03b4\u03b9\u03b1\u03c6\u03b5\u03c1\u03b5\u03b9.....", "Solution_8": "[quote=\"r_boris\"]\u03a3\u03c5\u03b3\u03bd\u03ce\u03bc\u03b7 \u03c0\u03b1\u03c1\u03b1\u03c3\u03cd\u03c1\u03b8\u03b7\u03ba\u03b1 \u03bc\u03b5 \u03c4\u03bf\u03c5\u03c2 \u03bc\u03b9\u03b3\u03b1\u03b4\u03b9\u03ba\u03bf\u03cd\u03c2[/quote]\r\n\r\n\u03a0\u03a1\u039f\u039a\u0395\u0399\u039c\u0395\u039d\u039f\u03a5 \u039d\u0391 \u0395\u039e\u0399\u039b\u0395\u03a9\u0398\u03a9 \u03b3\u03b9\u03b1 \u03c4\u03b7\u03bd \u03c7\u03b1\u03b6\u03bf\u03bc\u03ac\u03c1\u03b1 \u03c0\u03bf\u03c5 \u03b2\u03b9\u03ac\u03c3\u03c4\u03b7\u03ba\u03b1 \u03bd\u03b1 \u03c0\u03c9 \u03b4\u03af\u03bd\u03c9 \u03c3\u03c4\u03bf \u03c3\u03c5\u03bd\u03b7\u03bc\u03bc\u03ad\u03bd\u03bf \u03bc\u03b9\u03b1 \u039a\u0391\u0398\u0391\u03a1\u0391 \u0393\u0395\u03a9\u039c\u0395\u03a4\u03a1\u0399\u039a\u0397 \u039b\u03a5\u03a3\u0397" } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "Let $ a,b,c$ be the length of the sides of an triangle. Prove that $ \\sum_{cyc}\\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ca}{ab+bc-b^{2}}\\leq 3$ and $ \\sum_{cyc}\\sqrt{\\frac{a-b+c}{9a+b-c}}\\leq 1$", "Solution_1": "We have that $ \\sum_{cyc}\\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ac}{ab+bc-b^{2}}=-12+\\sum_{cyc}\\frac{(a+c)^{2}+4ac-3b^{2}}{ab+bc-b^{2}}$. Now we use $ 4ac\\leq (a+c)^{2}$ so we are left to prove $ \\sum_{cyc}\\frac{2(a+c)^{2}-3b^{2}}{ab+bc-b^{2}}\\leq 15$.This is equivalent after we use $ (a+c)^{2}-b^{2}=(a+b+c)(a+c-b)$ to $ 2\\left(\\sum_{cyc}a \\right)\\left(\\sum_{cyc}\\frac{1}{a}\\right) \\leq 15+\\sum_{cyc}\\frac{a}{b+c-a}$.Now if $ a,b,c$ are sidelengths of a triangle we have $ a=y+z$,$ b=z+x$,$ c=x+y$ with $ x,y,z\\geq 0$.The inequality becomes \r\n$ 4\\left(\\sum_{cyc}x\\right)\\left(\\sum_{cyc}\\frac{1}{x+y}\\right)\\leq 15+\\sum_{cyc}\\frac{x+y}{2z}$.We substract $ 18$ from each side and after grouping we reach $ \\sum_{cyc}\\frac{(x-y)^{2}}{(x+z)(y+z)}\\leq \\sum_{cyc}\\frac{(x-y)^{2}}{4xy}$.The last is easy to prove by SOS.", "Solution_2": "[quote=\"Vorelin Tancu\"]Let $ a,b,c$ be the length of the sides of an triangle. Prove that $ \\sum_{cyc}\\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ca}{ab+bc-b^{2}}\\leq 3$[/quote]\r\nLet $ a=y+z,$ $ b=x+z$ and $ c=x+y.$\r\nHence, $ \\sum_{cyc}\\frac{a^{2}+b^{2}+c^{2}-4ab-4bc+6ca}{ab+bc-b^{2}}\\leq 3\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{y^{2}+z^{2}-4x^{2}+xy+xz}{(y+z)x}\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\frac{y^{2}+z^{2}-2x^{2}}{(y+z)x}+\\sum_{cyc}\\frac{y+z-2x}{y+z}\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\left(\\frac{(z-x)(x+z)}{(y+z)x}-\\frac{(x-y)(x+y)}{(y+z)x}\\right)+\\sum_{cyc}\\left(\\frac{z-x}{y+z}-\\frac{x-y}{y+z}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}\\left(\\frac{(x-y)(x+y)}{(x+z)y}-\\frac{(x-y)(x+y)}{(y+z)x}\\right)+\\sum_{cyc}\\left(\\frac{x-y}{x+z}-\\frac{x-y}{y+z}\\right)\\geq0\\Leftrightarrow$\r\n$ \\Leftrightarrow\\sum_{cyc}(x-y)^{2}(xz+yz-xy)(x+y)z\\geq0.$\r\nLet $ x\\geq y\\geq z.$ Hence, $ S_{x}\\geq0$ and $ S_{y}\\geq0,$\r\nwhere $ S_{x}=x(y+z)(xy+xz-yz)$ and $ S_{y}=y(x+z)(xy+yz-xz).$\r\nId est, it remains to prove that $ S_{y}+S_{z}\\geq0.$\r\nBut $ S_{y}+S_{z}\\geq0\\Leftrightarrow x^{2}(y-z)^{2}+2y^{2}z^{2}+y^{2}xz+z^{2}xy\\geq0.$", "Solution_3": "And the other inequality? :D\r\n\r\n$ \\sum_{cyc}\\frac{1}{3a^{2}+\\left( b-c\\right)^{2}}\\geq \\frac{9}{\\left( a+b+c\\right)^{2}}$", "Solution_4": "[quote=\"Vorelin Tancu\"]Let $ a,b,c$ be the length of the sides of an triangle. Prove that $ \\sum_{cyc}\\sqrt{\\frac{a-b+c}{9a+b-c}}\\leq 1$[/quote]\r\nLet $ a=y+z,$ $ b=x+z$ and $ c=x+y.$\r\nThen $ \\sum_{cyc}\\sqrt{\\frac{a-b+c}{9a+b-c}}\\leq 1\\Leftrightarrow\\sum_{cyc}\\sqrt{\\frac{x}{4x+5y}}\\leq1,$ which known:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=150022", "Solution_5": "That's great! How about the last?", "Solution_6": "[quote=\"Vorelin Tancu\"]And the other inequality?\n$ \\sum_{cyc}\\frac{1}{3a^{2}+\\left( b-c\\right)^{2}}\\geq \\frac{9}{\\left( a+b+c\\right)^{2}}$[/quote]\r\nThe same way:\r\nit's equivalent to following inequality:\r\n$ \\sum_{cyc}\\frac{1}{x^{2}+xy+y^{2}}\\geq\\frac{9}{(x+y+z)^{2}},$ which known:\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=6364" } { "Tag": [], "Problem": "How many squares of any size appear in a 3x5 grid of unit squares?\r\n\r\nHow would you do these types of problems?\r\n\r\nIs there any quick method?", "Solution_1": "Methodically count the number of each possible size of square:\r\n\r\n[hide=\"like this\"]\nNumber of $1\\times1$ squares: 15\nNumber of $2\\times2$ squares: 8\nNumber of $3\\times3$ squares: 3\n\nTotal:26[/hide]", "Solution_2": "You are forgetting squares of noninteger lengths...", "Solution_3": "If we can make squares that have sides that are not parallel to the gridlines, we will have more squares. If not, we are finished. Usually the problem will specify, but if it does not, we assume sides do NOT have to be parallel to the gridlines.", "Solution_4": "i just counted them all (its pretty fast)\r\nbut you can consider (for positive sides only)\r\ndimension 1 * dimension 2 + dimension1 * 1dimension 2-1 + dimension1-2 + dimension 2 -2. until one dimension = zero. then add them all up\r\nso...\r\n3x5=15 squares with side lengths 1\r\n2x4=8 squares with side lengths 2\r\n1x3=3 squares with side lengths 3\r\n\r\nand so on...\r\nif there are square root lengths, then thats 8 for root 2 and 4 for root 5.", "Solution_5": "[quote]How many squares of any size [b]appear[/b] in a 3x5 grid of unit squares? \n[/quote]\r\n\r\nI would argue that unless the sides of the squares were part of the grid, then the squares do not actually 'appear'. \r\n\r\nBut if we do count other squares, then there are an extra 12 making the total $\\boxed{38}$" } { "Tag": [ "algorithm", "geometry", "perimeter", "analytic geometry", "conics", "ellipse", "search" ], "Problem": "Find a point P on equilateral triangle ABC such that (AP+BP+CP)/(AB+BC) is minimized, and prove your answer (preferebly, dont use the first fermat point algorithm)", "Solution_1": "(you mean on the perimeter of the triangle right?) AB and BC are constant, WLOG on AB, then AP+PB is constant, then minimize CP, pretty obvious the shortest distance is when C is P is the midpoint of AB, if you move it away from there, then by pythagorean theorem, it will be greater\r\n\r\nif it is inside the triangle...not sure...a messy cooridinate solution could work...", "Solution_2": "P is inside the triangle\r\nthe solution (my solution at least) doesnt use coordinates but i suppose it could be done that way\r\n\r\n[hide=\"hint\"]use shearing (i think thats what the transformation is called[/hide]", "Solution_3": "[hide=\"Original thought\"]Let $\\triangle ABC$ be the equilateral triangle, and let $D, E, F$ be the midpoints of $AB, BC, CA$ respectively. If $P$ is the minimal point for $\\triangle ABC$, then it also must be the minimal point for $\\triangle DEF$, since they are similar and all the corresponding quotients for the two are the same. Therefore, $P$ must be inside $\\triangle DEF$. Now take the midpoints of $DE, EF, FD$ and apply the same reasoning. Continue to infinity and you'll conclude that $P$ must be the centroid of $\\triangle ABC$, because that's the only point that belongs to all the triangles, no matter how small they are.[/hide]", "Solution_4": "[hide=\"why that doesnt work\"] Its the right answer for equilaterals but the reasoning is incorrect. How do I know it has to be inside all of the triangles? it just has to be in the same relative position for the triangles.[/hide]", "Solution_5": "[hide]it seems that it is the orthocenter, but i don't know how to prove it...i tried it for a right triangle as well, and i don't think shearing is the right word...that is for sheep :lol: \n\nanother thing that might work would somehow have ellipses with the verticies as foci to show that it is a minimum sum of distances\n[/hide]", "Solution_6": "one quck thing; its very difficult for the general triangle (search up first fermat point if you want to know). Use the fact that each pair of sides in the equilateral triangle are the same", "Solution_7": "[quote=\"Scrambled\"]it just has to be in the same relative position for the triangles.[/quote]\r\n\r\nRight, and since it's given that $P$ should be inside the initial triangle, it must be inside all of them in order to be in the same relative position for all.", "Solution_8": "[quote=\"Farenhajt\"][quote=\"Scrambled\"]it just has to be in the same relative position for the triangles.[/quote]\n\nRight, and since it's given that $P$ should be inside the initial triangle, it must be inside all of them in order to be in the same relative position for all.[/quote]\r\nuh, I dont exactly understand you\r\nfirst of all, if K isnt in the triangle formed by the midpoints of the original triangle, that just means that its not the point for the centroid triangle; it doesnt imply anything about the original triangle" } { "Tag": [ "induction" ], "Problem": "Prove that if $n$ is a positive integer number then:\r\na) $1.3.5...(2n-1)\\sqrt[n]{\\frac{1}{n}}$\r\n$\\Rightarrow 1+\\left(1-\\frac{1}{2}\\right)+\\left(1-\\frac{1}{3}\\right)+...+\\left(1-\\frac{1}{n}\\right)> \\frac{n}{\\sqrt[n]{n}}$\r\n$\\Rightarrow n-\\frac{n}{\\sqrt[n]{n}}>\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{n}$, or:\r\n$1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{n}\\sqrt[n]{n+1}$\r\n$\\Rightarrow 2+\\left(1+\\frac{1}{2}\\right)+\\left(1+\\frac{1}{3}\\right)+...+\\left(1+\\frac{1}{n}\\right)>\\sqrt[n]{n+1}$\r\n$1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{n}>n\\left(\\sqrt[n]{n+1}-1\\right)$\r\nThe probelm has been almost proved almost", "Solution_8": "There are some problem is the same type:\r\n[b]Problem 1[/b]:\r\n$\\frac{1}{1^{1}}+\\frac{1}{2^{2}}+...+\\frac{1}{n^{2}}>\\sqrt[n+1]{\\frac{n+1}{2}}$", "Solution_9": "[b]Problem 2[/b] \r\n$\\sqrt[n]{n}<1+\\frac{1}{\\sqrt{n}}$\r\n[b]Problem 3[/b]\r\n$1+\\frac{1}{\\sqrt[3]{2}}+\\frac{1}{\\sqrt[3]{3}}+...+\\frac{1}{\\sqrt[3]{n}}>\\frac{3}{2}\\left(\\sqrt[3]{n+1}-1\\right)$\r\n[b]Problem 3[/b]\r\n$\\frac{1!2!+2!3!+...+n!(n+1)!}{n.\\sqrt[n]{(1!)^{2}...(n!)^{2}}}\\geq 2\\sqrt[2n]{n!}$\r\n[b]Problem 4[/b]\r\n$\\sqrt[n]{1+\\frac{\\sqrt[n]{n}}{n}}+\\sqrt[n]{1-\\frac{\\sqrt[n]{n}}{n}}<2$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ x,y,z$ be non-negative real numbers such that $ x\\plus{}y\\plus{}z\\equal{}3$. Prove that $ \\sqrt[10]{x^{3}}\\plus{}\\sqrt[10]{y^{3}}\\plus{}\\sqrt[10]{z^{3}}\\geq xy\\plus{}yz\\plus{}zx$.", "Solution_1": "Please, someone can solve the problem" } { "Tag": [ "percent", "geometry", "percent increase" ], "Problem": "The side of a square is increased by $ 50\\%$. By what percent is the area of the square increased?", "Solution_1": "try to giver urself an example so let's say that the side is 4. when the side is 4, the area is 16. then u increase 4 by 50% making it 6. that area is 36, so the percent increase from 16 to 36 is 225%.", "Solution_2": "First thing here, the answer is incorrect. Secondly, when you formally prove something, you have to prove that it works for any side length $ x$, not just by giving a specific example.\r\n\r\nWell, you can't just assume a definite side length of the square. In fact, this rule works out for any side length $ x$ of the square. Note that when you increase the side length by $ 50$ percent, you are multiplying the side length by 1.5, hence multiplying the original area by $ (1.5)^2 \\equal{} 2.25$. So if you let $ x$ be the side length of the square, $ x^2$ the area, then $ 2.25x^2 \\minus{} x^2 \\equal{} 1.25x^2$, which means that the area of the square is increased by $ \\boxed{125}$ percent." } { "Tag": [ "geometry", "3D geometry", "analytic geometry", "Euler", "IMO Shortlist", "combinatorics" ], "Problem": "Unit cubes are made into beads by drilling a hole through them along a diagonal. The beads are put on a string in such a way that they can move freely in space under the restriction that the vertices of two neighboring cubes are touching. Let $ A$ be the beginning vertex and $ B$ be the end vertex. Let there be $ p \\times q \\times r$ cubes on the string $ (p, q, r \\geq 1).$\r\n\r\n[i](a)[/i] Determine for which values of $ p, q,$ and $ r$ it is possible to build a block with dimensions $ p, q,$ and $ r.$ Give reasons for your answers.\r\n[i](b)[/i] The same question as (a) with the extra condition that $ A \\equal{} B.$", "Solution_1": "I am having trouble understanding the solution from IMO Compendium. Aren't we trying to look for hamiltonian cycles rather than eulerian ones?\r\nThanks\r\n\r\nSolution by IMO Compendium:\r\nLet us set a coordinate system denoting the vertices of the block. The\r\nvertices of the unit cubes of the block can be described as {(x, y, z) | 0 \u2264\r\nx \u2264 p, 0 \u2264 y \u2264 q, 0 \u2264 z \u2264 r}, and we restrict our attention to only\r\nthese points. Suppose the point A is fixed at (a, b, c). Then for every other\r\nnecklace point (x, y, z) numbers x \u2212 a, y \u2212 b, and z \u2212 c must be of equal\r\nparity. Conversely, every point (x, y, z) such that x \u2212 a, y \u2212 b, and z \u2212 c are\r\nof the same parity has to be a necklace point. Consider the graph G whose\r\nvertices are all such points and edges are all diagonals of the unit cubes\r\nthrough these points. In part (a) we are looking for an open or closed\r\nEuler path, while in part (b) we are looking for a closed Euler path.\r\nNecklace points in the interior of the (p, q, r) box have degree 8, points on\r\nthe surface have degree 4, points on the edge have degree 2, and points\r\non the corner have degree 1. A closed Euler path can be formed if and\r\nonly if all vertices are of an even degree, while an open Euler path can be\r\nformed if and only if exactly two vertices have an odd degree. Hence the\r\nproblem in part (a) amounts to being able to choose a point A such that\r\n0 or 2 corner vertices are necklace vertices, whereas in part (b) no corner\r\npoints can be necklace vertices. We distinguish two cases.\r\n (i) At least two of p, q, r, say p, q, are even. We can choose a = 1, b = c =\r\n 0. In this case none of the corners is a necklace point. Hence a closed\r\n Euler path exists.\r\n(ii) At most one of p, q, r is even. However one chooses A, exactly two\r\n necklace points are at the corners. Hence, an open Euler path exists,\r\n but it is impossible to form a closed path.\r\nHence, in part (a), a box can be made of all (p, q, r) and in part (b) only\r\nthose (p, q, r) where at least two of the numbers are even.", "Solution_2": "Solution from [i]Twitch Solves ISL[/i]: For (a), the answer is always yes; whereas for (b), the answer is only if at least two of $p$, $q$, $r$ are even.\n\nLet the box $\\mathcal B$ be located in space spanning $(0,0,0)$ to $(p,q,r)$. Then the string can move from $(x,y,z)$ to $(x \\pm 1, y \\pm 1, z \\pm 1)$. So, the string is going to visit only points of two parity classes in ${\\mathbb F}_2 \\times {\\mathbb F}_2 \\times {\\mathbb F}_2$, say $(\\varepsilon_x \\bmod 2, \\varepsilon_y \\bmod 2, \\varepsilon_z \\bmod 2)$ and $(1+\\varepsilon_x \\bmod 2, 1+\\varepsilon_y \\bmod 2, 1+\\varepsilon_z \\bmod 2)$. However, every unit cube has a string through it, so we find that the string visits every point of this form in $\\mathcal B$. Let us denote the set of points by $V$ (which depends on the choice of $(\\varepsilon_x, \\varepsilon_y, \\varepsilon_z)$).\nHence the question is asking when we can have a Eulerian path or circuit on the resulting graph $G$ (whose vertex set is $V$ and whose edges are the string).\nHowever, the degrees of vertices of $G$ are always even [i]except[/i] for the corners of $\\mathcal B$, i.e.\\ the eight points \\[ \t\\Big\\{ (0,0,0), \\; (p,0,0), \\; (0,q,0), \\; (0,0,r), \\; \t(p,q,0), \\; (p,0,r), \\; (0,q,r), \\; (p,q,r) \\Big\\}. \\] [list] \t[*]We can always ensure there are at most \ttwo corners in $V$ \tby choosing $\\varepsilon_x$, $\\varepsilon_y$, $\\varepsilon_z$ randomly; \tthe expected number of corners is $\\frac14 \\cdot 8 = 2$. \t[*]We can ensure there are zero corners \tif at least two of $(p,q,r)$ are even; \totherwise we cannot. [/list]\nHence by the classical theorem on Eulerian paths and circuits, (a) is possible always, (b) is possible when at least two of $(p,q,r)$ are even.\n" } { "Tag": [ "geometry", "incenter" ], "Problem": "I finally got a letter from the awesome people who make Olymon. Aparently, i made [b]many[/b] mistakes. Anyways, it had the questions for september, and all the solution up till september for 2006 (I think). I know we can't discuss the questions, but could i get a hint for the following question, thanks. :) \r\n\r\n454. Let ABC be a non-isosceles triangle with circumcentre O, incentre I and orthocentre H. Prove that the angle OIH exceeds 90 degrees.", "Solution_1": "I am not sure if hints can be given out, please ask Mr. Barbeau if this is allowed. Just in case, this has to be done. :!:", "Solution_2": "[quote=\"rem\"]I am not sure if hints can be given out, please ask Mr. Barbeau if this is allowed. Just in case, this has to be done. :!:[/quote]\r\noh, sorry wrong question. And wrong wording :blush: . Could you explain the following question to me, thanks.\r\n\r\nLet a and b be positive integers and let $u=a+b$ and $v=lcm(a,b)$. Prove that\r\n$gcd(u,v)=gcd(a,b)$", "Solution_3": "This is from a problem set that is not due until October 31, so it should not be discussed. The topic is locked.", "Solution_4": "Problem is Equivalent to Saying that $I$ lies inside the circle with diameter $HO$ .\r\n\r\nStronger result is true that is $I$ lies inside the circle with diameter $HG$.This is the best possible bound ie $I$ can be anywhere inside this Circle.", "Solution_5": "Just got the corrected solutions back the day before yesterday.\r\n\r\nHow did you guys do?" } { "Tag": [ "calculus", "integration", "function", "calculus computations" ], "Problem": "Determine\r\n\r\n\\[\\int_{0}^{1}\\left( \\sum_{n=1}^{\\infty}\\frac{1}{(x+n)^{3}}\\right)\\, dx \\]", "Solution_1": "The functions are positive so you just switch the integral and the summation sign? The sum even converges uniformly..." } { "Tag": [], "Problem": "how many 6 digit numbers are there such that out of each 2 digit, the one to the right is smaller?", "Solution_1": "Don't know, but is it:\r\n\r\n\r\nThe first digit could be $9,8,7,6,5$, second $8,7,6,5,4$, third $7,6,5,4,3$, fourth $6,5,4,3,2$, fifth $5,4,3,2,1$, and sixth $4,3,2,1,0$, so would it just be $5^6$?\r\n\r\nEDIT: Misunderstood question.\r\n\r\nWell, since each two digit should have a number and a number smaller, the first out of the two digits can be a number between $1-9$ and then the second has to be a number between $0-8$ that goes for the other 4 digits too, so would it be $\\boxed{9^6}$?", "Solution_2": "The hard thing is that you have to be careful about the numbers you change. Right now all I know is that the smallest is 543210 and the largest 987654.", "Solution_3": "I think my method makes sense though because for every place value, the digit is one less than the one to the left and the one to the left can not be $0$, so it should be $9^3$.", "Solution_4": "$9^6$ doesn't seem right because $9^6/(9 \\cdot 10^5) = .59049$, and the answer is definitely $<.5$", "Solution_5": "[quote=\"236factorial\"]The hard thing is that you have to be careful about the numbers you change. Right now all I know is that the smallest is 943210 and the largest 987654.[/quote]\r\n??? Isn't the smallest 543210?\r\nHmm I got one answer as [hide]$\\binom{9}{5}=126$ :?:[/hide]", "Solution_6": "I got a similar answer to 10000th User's.\r\n\r\nThere are $\\binom{10}{6}=210$ ways to pick 6 distinct numbers (if the order does not matter.)\r\nFor each possible set of 6 numbers, there's only one way to line up 6 numbers to satisfy the given condition.\r\n\r\nExample: If the set was $\\{1,2,3,5,7,9\\}$, then there's only one possibility:$975321$\r\n\r\n\r\nSo I think the answer is $\\boxed{210}$...", "Solution_7": "[quote=\"10000th User\"][quote=\"236factorial\"]The hard thing is that you have to be careful about the numbers you change. Right now all I know is that the smallest is 943210 and the largest 987654.[/quote]\n??? Isn't the smallest 543210?\n[/quote]\r\n\r\nDuhhhh :blush:", "Solution_8": "[quote=\"jli\"]how many 6 digit numbers are there such that out of each 2 digit, the one to the right is smaller?[/quote]\r\n\r\nOnce you pick the digits, they must be arranged decreasingly, so $\\binom{10}{6}$ is right.", "Solution_9": "Say it is an 8 digit number...\r\n\r\nCall a skip when it doesnt include a number\r\n\r\nFor example, if the skip is on 7, there wont be a 7 in the number\r\n\r\nCASE 1: if the first digit is 8\r\n\r\nThere is only one way to arrang it.\r\n\r\nCASE 2: if the first digit is 9\r\n\r\nThere has to be exactly one skip.\r\n\r\nThe number of ways to place the skip is ${8\\choose 1}=8$\r\n\r\nSo the total is 9...\r\n\r\nis this true?", "Solution_10": "If you're still talking about digits in descending order, that's right.", "Solution_11": "[quote=\"frt\"]I got a similar answer to 10000th User's.\n\nThere are $\\binom{10}{6}=210$ ways to pick 6 distinct numbers (if the order does not matter.)\nFor each possible set of 6 numbers, there's only one way to line up 6 numbers to satisfy the given condition.\n\nExample: If the set was $\\{1,2,3,5,7,9\\}$, then there's only one possibility:$975321$\n\n\nSo I think the answer is $\\boxed{210}$...[/quote]\r\n\r\noh...i didnt see your post\r\n\r\nyes. this is correst" } { "Tag": [ "geometry", "inradius", "ratio", "perimeter" ], "Problem": "In a triangle, two altitudes are $ 1.2$ and $ 1.4$ times longer than the third, respectively. What's the ratio between the shortest altitude and the inradius?", "Solution_1": "[hide=\"Answer\"]In a triangle, the product of an altitude and the side it touches is constant. Thus, if the shortest altitude has length $ a$ and the side it touches has length $ x$, then the other sides have length $ x/1.2=5x/6$ and $ x/1.4=5x/7$.\n\nNow, equate two different area formulas. Note that $ r$ is the inradius and $ p$ is the perimeter:\n\\begin{eqnarray*}\nA=\\frac{xa}{2} &=& \\frac{rp}{2}\\\\\n\\displaystyle xa &=& r(x+\\frac{5x}{6} +\\frac{5x}{7})\\\\\na &=& r(\\frac{107}{42})\n\\end{eqnarray*}\n\nSo $ a: r = 107: 42$[/hide]", "Solution_2": "[hide=\"General formula\"]We can derive the formula connecting the inradius and the altitudes. If the area of the triangle is $ P$ and its semiperimeter $ s$, then\n\n$ 2s\\equal{}a\\plus{}b\\plus{}c$\n\n$ 2\\cdot{P\\over r}\\equal{}{2P\\over h_a}\\plus{}{2P\\over h_b}\\plus{}{2P\\over h_c}$\n\n$ \\boxed{{1\\over r}\\equal{}{1\\over h_a}\\plus{}{1\\over h_b}\\plus{}{1\\over h_c}}$\n\nNow $ {1\\over r}\\equal{}{1\\over h}\\plus{}{1\\over 1.2h}\\plus{}{1\\over 1.4h}\\iff {h\\over r}\\equal{}1\\plus{}{5\\over 6}\\plus{}{5\\over 7}\\equal{}{107\\over 42}$[/hide]" } { "Tag": [], "Problem": "I will be attending a Scholar School in Canberra early next year for the Chemistry Olympiad (I hope to get selected :D ). They will be using:\r\n\r\n- Silberberg, Martin; Chemistry: The Molecular Nature of Matter and Change, 3rd edition, ISBN0-07-239681-4\r\n- Morrison and Boyd; Organic Chemistry, 6th edition, ISBN 0-13-643669-2\r\n\r\nwith us during the Scholar School but I was wondering if there are other books that I may use as well to strengthen and add to my current knowledge so I might a better chance of getting into the IChO Team for 2008 (or 2009).", "Solution_1": "I find \"Chemistry: Molecules, Matter, and Change\" from Peter Atkins a very good book. Take also a look at some other Organic Chemistry books, like Solomons or Carey. If spectroscopy is also part of the program, then I would strongly advise \"Introduction to Spectroscopy\" from Pavia, Lampman and Kriz (Brooks /Cole) - it is the book I use in my classes. Also, after studying the general chemistry books I would advise you to take a look at an Inorganic Chemistry book (like the one from Atkins).\r\nYou can also use this forum to improve your skills. \r\n\r\nIn any way, I wish you good look!" } { "Tag": [ "geometry", "3D geometry", "sphere", "topology", "real analysis", "real analysis unsolved" ], "Problem": "H is infinite-dim. Why unit sphere S is weakly dense in unit ball B?\r\n( I was given a hint that in fact, every x in B is weak sequential limit of points in S) Can someone explain? Thanks.", "Solution_1": "First, any infinite orthonormal sequence tends weakly to zero- can you show this?\r\n\r\nNow, given $x$, let $V$ be the orthogonal complement of $x$. The intersection of $x+V$ with $S$ is a sphere of radius $\\sqrt{1-\\|x\\|^{2}}$ in $V$ centered at $x$; find an infinite orthonormal set in $V$, scale appropriately, and add $x$.", "Solution_2": "any open set is unbounded, so every open set that contains a point of the unit ball cuts the unit sphere.", "Solution_3": "[quote=\"alekk\"]any open set is unbounded, so every open set that contains a point of the unit ball cuts the unit sphere.[/quote]\r\n\r\nI am not sure alekk's method works: what if this open set is not \"connected\", i.e.: part of it is in unit ball, part of it outside unit ball. Then it can be unbounded and do not intersect unit sphere.", "Solution_4": "well, I should have said that each open set contains a line ($\\{x+\\lambda y : \\lambda \\in \\mathbb{R}\\}$)" } { "Tag": [ "algebra", "polynomial", "function", "induction", "superior algebra", "superior algebra unsolved" ], "Problem": "Let $ A$ be a finite commutative ring with the property that the function $ f: A\\minus{}>A,f(x)\\equal{}x^2$ is injective.Prove that $ \\forall n \\in N^{*}$ and $ \\forall a \\in A$ the polynomial $ P(X)\\equal{}X^{2n}\\plus{}a$ is reducible in $ A$.", "Solution_1": "Since $ A$ is finite, $ f$ must also be surjective, so every element of $ A$ is a square. By induction, we can write $ \\minus{} a \\equal{} b^{2n}$ for some $ b\\in A$. Then $ P(x) \\equal{} X^{2n} \\minus{} b^{2n} \\equal{} (X^n \\minus{} b^n)(X^n \\plus{} b^n)$.", "Solution_2": "The induction doesn't work like that. For example take $ A\\equal{}\\mathbb F_4$ (field with $ 4$ elements), $ a \\neq 0,1$ and $ n\\equal{}3$. Then for any $ b$ it is $ b^6 \\equal{}0,1$.", "Solution_3": "[quote=\"ZetaX\"]The induction doesn't work like that. For example take $ A \\equal{} \\mathbb F_4$ (field with $ 4$ elements), $ a \\neq 0,1$ and $ n \\equal{} 3$. Then for any $ b$ it is $ b^6 \\equal{} 0,1$.[/quote]\r\n\r\nYou're right, I got carried away. We just need $ \\minus{}a\\equal{}b^2$, then $ X^{2n}\\minus{}b^2\\equal{}(X^n\\minus{}b)(X^n\\plus{}b)$.", "Solution_4": "NICE!!!...thank's :D" } { "Tag": [ "logarithms" ], "Problem": "Find all values of x such that: $2\\log_{9}(15-x)-\\log_{3}(x-4)=3\\log_{27}(x+5)$.", "Solution_1": "[hide]\\begin{eqnarray*}2log_{9}(15-x)-log_{3}(x-4)&=&3log_{27}(x+5)\\\\ log_{9}((15-x)^{2})-log_{3}(x-4)&=&\\\\ log_{3^{2}}((15-x)^{2})-log_{3}(x-4)&=&\\\\ log_{3}(15-x)-log_{3}(x-4)&=&log_{3}(x+5)\\\\ log_{3}(\\frac{15-x}{x-4})&=&log_{3}(x+5)\\\\ \\frac{15-x}{x-4}&=&x+5\\\\ 15-x&=&(x+5)(x-4)\\\\ &=&x^{2}+x-20\\\\ 0&=&x^{2}+2x-35\\\\ 0&=&(x+7)(x-5) \\end{eqnarray*}\n\n-7 does not work because you cannot take the log of a negative number ($-7-4$), so the only answer is $x=5$.[/hide]", "Solution_2": "[quote=\"Ignite168\"]Find all values of x such that: $2log_{9}(15-x)-log_{3}(x-4)=3log_{27}(x+5)$.[/quote]\r\n[hide]\n$\\log_{9}(15-x)^{2}-\\log_{3}(x-4)=\\log_{27}(x+5)^{3}$\n$\\log_{3}\\frac{(15-x)^{2}}{x-4}=\\log_{3}(x+5)$\n$x^{2}+x-20=x^{2}-30x+225$\n$31x=245$\n$x=\\frac{245}{31}$[/hide]", "Solution_3": "[quote=\"bpms\"][quote=\"Ignite168\"]Find all values of x such that: $2log_{9}(15-x)-log_{3}(x-4)=3log_{27}(x+5)$.[/quote]\n[hide]\n$\\log_{9}(15-x)^{2}-\\log_{3}(x-4)=\\log_{27}(x+5)^{3}$\n$\\log_{3}\\frac{(15-x)^{2}}{x-4}=\\log_{3}(x+5)$\n$x^{2}+x-20=x^{2}-30x+225$\n$31x=245$\n$x=\\frac{245}{31}$[/hide][/quote]\r\nYou messed up on your second line. $log_{9}((15-x)^{2})-log_{3}(x-4)$ is not equal to $log_{3}(\\frac{(15-x)^{2}}{x-4})$ because the two logarithms have different bases. That must be changed first.", "Solution_4": "[quote=\"The QuattoMaster 6000\"][quote=\"bpms\"][quote=\"Ignite168\"]Find all values of x such that: $2log_{9}(15-x)-log_{3}(x-4)=3log_{27}(x+5)$.[/quote]\n[hide]\n$\\log_{9}(15-x)^{2}-\\log_{3}(x-4)=\\log_{27}(x+5)^{3}$\n$\\log_{3}\\frac{(15-x)^{2}}{x-4}=\\log_{3}(x+5)$\n$x^{2}+x-20=x^{2}-30x+225$\n$31x=245$\n$x=\\frac{245}{31}$[/hide][/quote]\nYou messed up on your second line. $log_{9}((15-x)^{2})-log_{3}(x-4)$ is not equal to $log_{3}(\\frac{(15-x)^{2}}{x-4})$ because the two logarithms have different bases. That must be changed first.[/quote]\nOopsies.\n[hide]\n$\\log_{3}\\frac{|15-x|}{x-4}=\\log_{3}x+5$\n$|15-x|=x^{2}+x-20$\nAssume the LHS is positive.\n$15-x=x^{2}+x-20$\nx=-7, 5\nIf the LHS is negative\n$x-15=x^{2}+x-20$\n$x^{2}=5$\n$x=\\sqrt{5}$, 5.[/hide]", "Solution_5": "[quote=\"bpms\"]\nOopsies.\n[hide]\n$\\log_{3}\\frac{|15-x|}{x-4}=\\log_{3}x+5$\n$|15-x|=x^{2}+x-20$\nAssume the LHS is positive.\n$15-x=x^{2}+x-20$\nx=-7, 5\nIf the LHS is negative\n$x-15=x^{2}+x-20$\n$x^{2}=5$\n$x=\\sqrt{5}$, 5.[/hide][/quote] And so 3 roots are extraneous roots and the answer is $x=5$." } { "Tag": [], "Problem": "Who will win the NL West? I think Padres. I love the Dodgers, but don't think they have much of a chance :( I will only post for NL East, AL East, and AL West (and the wild cards) because the Centrals both have dominant teams.", "Solution_1": "i voted D-Backs.", "Solution_2": "isn't one baseball topic enough? there are about six now. that's a bit much. no wonder why everyone now how a 1000 posts. :D \r\nanywho, go d-backs.", "Solution_3": "furious, i think you are right. I will not make new posts for the AL teams :) maybe next week when this topic has all the votes it will get" } { "Tag": [ "AMC", "USA(J)MO", "USAMO" ], "Problem": "Do they just choose 20 finalists from the 900 people (that qualified from local) that take the national exam? \r\nIn addition, how are the 3 sections scored? Thanks in advance!", "Solution_1": "Scott said this in the other USNCO topic on this forum:\r\n\r\n[quote]For everyone interested in how they grade the national exam, lets [i][sic][/i] run through the basics. \n\n1) If you're not making above a 55/60 on the Multiple Choice, you may have some problems qualifying \n2) If you're answering none of the higher-level (d through g) questions in the Free Response, you may have some problems qualifying \n3) If you're absolutely terrible at labs, you'll have NO PROBLEM getting in. [/quote]\r\n\r\nTo answer the first part of your question, yes. I don't know how exactly the three sections are scored, but I will say (in agreement with Scott) that the lab receives very little weight. And by very little weight, I mean [b]very little weight[/b]. I [i]completely[/i] blew the lab, but still somehow qualified for camp. The logic is that most high school chem students have very little lab expertise, I suppose. As far as the relative weights of MC/FR, I don't know. Try to do well on both, I guess.\r\n\r\nTo practice, take the practice tests they offer. Many of the questions (especially multiple choice) repeat themselves by necessity; there are certain skills they have to test, and there are only so many ways to ask for the $E$ of a cell...", "Solution_2": "So that means that there is no semi-finalist thing for the Chemistry Olympiad like the USAMO is for the math olympiad, right? Anyways, thanks for the advice. I am currently studying with Mortimer's book, and I hope to be ready by test time :D.", "Solution_3": "Think of it this way: The local qualifier is the AMC, and the USNCO national exam is the USAMO.\r\n\r\nAnd of course, both olympiads have the finals (the team selection test).", "Solution_4": "Hm.. the practice national exams seem tough, but do-able. Are the cutoffs for camp usually very high? I took a practice on the 2005 National Exam, and I got a 52/60 on the multipe choice, and explained the Free-response questions fairly well (messed up on part of the Organic chem problems) . Would that give me a good chance to be invited to the camp?", "Solution_5": "Wow, I read in the introduction topic that you made the USNCO camp in the same year you took AP Chem, inteluser. That's amazing!" } { "Tag": [ "algebra", "polynomial" ], "Problem": "1. Express the five distinct roots of 1 in terms of radicals.\r\n2. Prove that the three perpendiculars dropped to the three sides of an equilateral triangle from any point inside the triangle have a constant sum.\r\n3. For all $ n > 1$, show that $ (n \\minus{} 1)^2$ is a factor of $ n^{n \\minus{} 1} \\minus{} 1$", "Solution_1": "What do you mean by the five distinct roots of 1? If you mean the fifth roots of unity, then that would require knowledge of $ \\sin36^\\circ$, etc in terms of radicals.", "Solution_2": "[quote=\"boo\"]1. Express the five distinct roots of 1 in terms of radicals.\n2. Prove that the three perpendiculars dropped to the three sides of an equilateral triangle from any point inside the triangle have a constant sum.\n3. For all $ n > 1$, show that $ (n \\minus{} 1)^2$ is a factor of $ n^{n \\minus{} 1} \\minus{} 1$[/quote]\r\n[hide=\"2\"] Okay, so call the equilateral triangle to be ABC, the point on the interior to be P, and the perpindiculars to AB, BC, and AC to be PX, PY, and PZ respectively. Also, say that AB=BC=AC=x.\n\\[ [ABC] \\equal{} [PAB] \\plus{} [PBC] \\plus{} [PAC]\n\\]\n\n\\[ [ABC] \\equal{} \\frac {PX(AB)}{2} \\plus{} \\frac {(PY)(BC)}{2} \\plus{} \\frac {(PZ)(AC)}{2}\n\\]\n\n\\[ [ABC] \\equal{} \\frac {PX(x) \\plus{} PY(x) \\plus{} (PZ)(x)}{2}\n\\]\n\n\\[ [ABC] \\equal{} (\\frac {x}{2})(PX \\plus{} PY \\plus{} PZ)\n\\]\n\n\\[ PX \\plus{} PY \\plus{} PZ \\equal{} \\frac {2[ABC]}{x}\n\\]\nSince x and [ABC] are constant, we have that PX+PY+PZ is constant. \n[/hide]\n[hide=\"3\"] Start by dividing $ n^{n \\minus{} 1} \\minus{} 1$ by $ n \\minus{} 1$. It is quite easy to see that this quotient is $ n^{n \\minus{} 2} \\plus{} n^{n \\minus{} 3}... \\plus{} 1$. We have to show that $ n \\minus{} 1$ is a factor of this polynomial. Look at the polynomial $ \\bmod (n \\minus{} 1)$. So, we obtain that $ n^{n \\minus{} 2} \\plus{} n^{n \\minus{} 3}... \\plus{} 1\\equiv 1 \\plus{} 1 \\plus{} 1...1\\equiv n \\minus{} 1\\equiv 0\\bmod (n \\minus{} 1)$. Thus, n-1 is a factor of the polynomial and it is proven. \n[/hide]" } { "Tag": [ "trigonometry", "algebra solved", "algebra" ], "Problem": "What is,\r\n\r\n$\\sin \\frac{\\pi}{7}\\cdot \\sin \\frac{2\\pi }{7}\\cdot ... \\cdot \\sin \\frac{6 \\pi }{7}=?$", "Solution_1": "$\\sin \\frac{\\pi}{7}\\cdot \\sin \\frac{2\\pi }{7}\\cdot ... \\cdot \\sin \\frac{6 \\pi }{7}=\\frac{7}{2^{6}}.$\r\nPosted before.", "Solution_2": "I tried searching it before for a long time, and still cant find it. Can you please post a link.", "Solution_3": "Let \\[P(x)=\\frac{x^{n}-1}{x-1}=\\prod_{k=1}^{n-1}(x-exp(\\frac{2\\pi k i}{n})).\\]\r\nConsider \\[P(1)=\\prod_{k=1}^{n-1}(1-cos\\frac{2\\pi k}{n}-isin\\frac{2\\pi k}{n})=\\prod_{k=1}^{n-1}2sin\\frac{\\pi k}{n}(-i)exp\\frac{\\pi ki}{n}.\\]\r\nBecause $sin\\frac{\\pi k}{n}>0$ we get \r\n\\[\\prod_{k=1}^{n-1}sin\\frac{\\pi k}{n}=\\frac{|P(1)|}{2^{n-1}}=\\frac{n}{2^{n-1}}.\\]", "Solution_4": "See also http://www.mathlinks.ro/Forum/viewtopic.php?t=137400", "Solution_5": "Thank you guys" } { "Tag": [], "Problem": "as apart of their preschool science project , mrs. joy pribble's class is holding snail races . the first snail, zippy , is known to move at a rate of 5 inches per hour .the second snail,lightning, is known to move at 4.5 inches per hour . if the snails race along a straight path, and if zippy finishes the race 0.25 hour before lightning , determine the time it takes lightning to run the race\r\n\r\na) 3 hours b) 2.5 hours c) 2.75 hours d) 3.5 hours e) none of these", "Solution_1": "Let the time lighting takes equal $ t$. Then $ t\\equal{}\\frac9{10}t\\plus{}.25$ \\[ t\\equal{}2.5\\]The answer is $ B$." } { "Tag": [ "modular arithmetic", "induction" ], "Problem": "Prove that every positive integer that can be written using only 1 as each digit in base 9 is a triangular number:\r\n\\begin{eqnarray*} 1_9 &=& 1 \\\\ 11_9 &=& 1 + 2 + 3 + 4 \\\\ 111_9 &=& 1 + 2 + \\cdots + 13 \\\\ 1111_9 &=& 1 + 2 + \\cdots + 40 \\\\ & \\vdots & \\end{eqnarray*}", "Solution_1": "[hide]\n$\\underbrace{11\\ldots 11_9}_{n \\text { ones}} = 1+9+9^2+\\cdots+9^{n-1}$\n\n$=\\frac{9^n-1}{9-1}$\n\n$=\\frac{9^n-1}{8}$\n\nWe wish to show that $\\frac{9^n-1}{8}$ can be written in the form $\\frac{1}{2}(k)(k+1)$ for some positive integer $k$.\n\n$\\Longleftrightarrow 9^n-1=4k(k+1)$\n\n$9^n-1=4k^2+4k$\n\n$\\iff 4k^2+4k-(9^n-1)=0$\n\n$\\Rightarrow k=\\frac{-4\\pm\\sqrt{16+4(4)(9^n-1)}}{8}$\n\n$=-\\frac{1}{2}\\pm \\frac{1}{2}\\sqrt{9^n}$\n\n$=\\frac{1}{2}(-1\\pm 3^n)$\n\nSince we want $k>0$, we will take $k=\\frac{1}{2}(-1+3^n)$.\n\nBecause $3^n\\equiv 1\\pmod {2}$, $k$ must be an integer, as claimed. $\\Box$\n[/hide]", "Solution_2": "[quote=\"MCrawford\"]Prove that every positive integer that can be written using only 1 as each digit in base 9 is a triangular number:\n\\begin{eqnarray*} 1_9 &=& 1 \\\\ 11_9 &=& 1 + 2 + 3 + 4 \\\\ 111_9 &=& 1 + 2 + \\cdots + 13 \\\\ 1111_9 &=& 1 + 2 + \\cdots + 40 \\\\ & \\vdots & \\end{eqnarray*}[/quote][hide]Theorem: \\[ \\underbrace{11\\ldots11_9}_{n\\text{ ones}}=1+2+\\ldots+\\sum_{i=0}^{n-1}3^i. \\]The proof will be by induction upon the number of digits. Consider $n=1$ as the base case. If $n=1$, then, as demonstrated, \\[ 1_9=1=\\sum_{i=0}^{1-1}3^i. \\]\nNow, for the inductive hypothesis, suppose the theorem is true for some integer $k$. That is, \\[ \\underbrace{11\\ldots11_9}_{k\\text{ ones}}=1+2+\\ldots+\\sum_{i=0}^{k-1}3^i. \\] Then \\[ \\underbrace{11\\ldots11_9}_{(k+1)\\text{ ones}}=\\underbrace{11\\ldots11_9}_{k\\text{ ones}}+9^k. \\] Notice that \\begin{eqnarray*}9^k&=&\\frac{(3^k)(3^k+1)}{2}+\\frac{(3^k)(3^k-1)}{2}\\\\ &=&\\sum_{j=1}^{3^k}j+(3^k)\\left(\\sum_{i=0}^{k-1}3^i\\right)\\\\ &=&\\sum_{j=1}^{3^k}\\left(\\sum_{i=0}^{k-1}3^i+j\\right)\\end{eqnarray*} Thus, \\[ \\underbrace{11\\ldots11_9}_{(k+1)\\text{ ones}}=\\underbrace{11\\ldots11_9}_{k\\text{ ones}}+\\sum_{j=1}^{3^k}\\left(\\sum_{i=0}^{k-1}3^i+j\\right). \\] But from the inductive hypothesis, \\[ \\underbrace{11\\ldots11_9}_{k\\text{ ones}}=1+2+\\ldots+\\sum_{i=0}^{k-1}3^i. \\] Therefore, \\[ \\underbrace{11\\ldots11_9}_{(k+1)\\text{ ones}}=1+2+\\ldots+\\sum_{i=0}^{k-1}3^i+\\sum_{j=1}^{3^k}\\left(\\sum_{i=0}^{k-1}3^i+j\\right). \\] Note that the final double sum is equivalent to \\[ \\left(\\sum_{i=0}^{k-1}3^i\\right)+1+\\left(\\sum_{i=0}^{k-1}3^i\\right)+2+\\ldots+\\left(\\sum_{i=0}^{k-1}3^i\\right)+3^k. \\] Indeed, \\[ \\underbrace{11\\ldots11_9}_{(k+1)\\text{ ones}}=1+2+\\ldots+\\sum_{i=0}^{k}3^i, \\] and so the proof is complete.[/hide]", "Solution_3": "[hide]I just noticed something, any number that starts with a 1 and has the rest of its digits as zeros is a triangular number. In base 9, for every 10 numbers in base ten, there is +1 in base 9, therefore all number that consist of 1 only end in the same number of 1's as the quotient of the original number divided by ten. So, all numbers that consist only of 1 in base 9 will be triangular. I'm sorry that the wording is confusing.[/hide]", "Solution_4": "[quote=\"maniacman384\"]\nI just noticed something, any number that starts with a 1 and has the rest of its digits as zeros is a triangular number.[/quote]\r\n\r\nLast time I checked, 1000 was a triangle number, not in base 10 nor base 9..." } { "Tag": [ "function", "floor function", "algebra proposed", "algebra" ], "Problem": "Suppose $f : \\mathbb{N} \\longrightarrow \\mathbb{N}$ is a function that satisfies $f(1) = 1$ and\r\n$f(n + 1) =\\{\\begin{array}{cc} f(n)+2&\\mbox{if}\\ n=f(f(n)-n+1),\\\\f(n)+1& \\mbox{Otherwise}\\end {array}$\r\n\r\n$(a)$ Prove that $f(f(n)-n+1)$ is either $n$ or $n+1$.\r\n$(b)$ Determine$f$.", "Solution_1": "Let $\\phi=\\frac{1+\\sqrt{5}}{2}$, so that we have the identities $\\phi+1=\\phi^2$ and $\\phi-1=\\frac{1}{\\phi}$. Then $f(n)=\\lfloor\\phi n\\rfloor$.\r\n\r\nWe prove this by strong induction. The base case is given. So suppose $f(n)=\\lfloor\\phi n\\rfloor$ for all $n\\leq k$, so that $f(f(k)-k+1)=f(\\lfloor\\phi k\\rfloor -k+1)=\\lfloor\\phi(\\lfloor\\phi k\\rfloor-k+1)\\rfloor$. It is easy to prove that this value is either $k$ or $k+1$.\r\n\r\nTo establish the inductive hypothesis, if $f(f(k)-k+1)=k$ (so that $f(k+1)=\\lfloor\\phi k\\rfloor+2$) then\r\n$k<\\phi(\\lfloor\\phi k\\rfloor-k+1)0 , a+b+c=1 and n integer find the max of\r\n$ a^n+b^n+c^n-\\frac{1}{3^n} $", "Solution_1": "$1-\\frac{1}{3^n}$", "Solution_2": "which inequality did u use?", "Solution_3": "Please siuhochung explain your answer analitically", "Solution_4": "I don't think that value can be attained since $a$, $b$ and $c$ are positive (not nonnegative). However that value should be the supremum.", "Solution_5": "Maybe. We wait siuhochung's analitical solution", "Solution_6": "it's obvious that there is only a supremum. The result is immediate because $(a+b+c)^n \\ge a^n + b^n + c^n$. Equality \"holds\" when a->1, b->0, c->0.", "Solution_7": "Supremum means that there is a value that the RHS could not take it? Someone explain to me because my English are not good", "Solution_8": "[quote=\"silouan\"]Supremum means that there is a value that the RHS could not take it? Someone explain to me because my English are not good[/quote]\r\n1. not really can't take it. It varies from case to case\r\n2. Supremum is a mathematical term. http://mathworld.wolfram.com/Supremum.html", "Solution_9": "Silouan, mabye you should post this type of inequalities in the pre-olympiad forum?" } { "Tag": [ "trigonometry", "calculus", "calculus computations" ], "Problem": "Solve $ \\frac {d^2y}{dt^2} \\minus{} 2\\frac {dy}{dt} \\plus{} 5y \\equal{} e^{ \\minus{} t}\\cos 2t$.", "Solution_1": "I feel like something similar was recently posted. Use undetermined coefficients and \"guess\" $ y(t)\\equal{}e^{\\minus{}t}(A\\cos(2t)\\plus{}B\\sin(2t))$ and try to solve for the coefficients. If that doesn't work, try $ y(t)\\equal{}te^{\\minus{}t}(A\\cos(2t)\\plus{}B\\sin(2t))$.", "Solution_2": "yeah, it was [url=http://www.mathlinks.ro/viewtopic.php?t=242997]Differential Equation 8[/url].", "Solution_3": "Oh, the problem looks almost just like Today's Differential Equation 8, but the problelm 13 is easy to solve than problem 8. :wink:" } { "Tag": [ "articles", "ratio", "search", "geometry" ], "Problem": "http://mathcircle.berkeley.edu/mpgeo.pdf\r\n\r\nI'm getting very confused about mass points. In the article above, I don't understand the postulates at all... what is meant by $nP+mQ+kR$, for example? (3) \r\n\r\nAlso....\r\n\r\n-how do you do #6 in the Basics section?\r\n-Angle Bisectors #1: why can you put on the weights suggested by the hint?\r\n-Angle Bisectors #3... how do you prove concurrency with mass points?\r\n-Angle Bisectors #5: how do you do it?\r\n\r\nThanks to anyone who answers these questions, I'm really very confused after reading this article :(", "Solution_1": "[hide=\"6\"]\nIf we assign weight $1$ to each vertex of the original quadrilateral, the weights of the midpoints are $2$.\nThen, we can observe that the diagonals of the new quadrilateral (whose vertices are the midpoints of the original quadrilateral) bisect each other, so it is a parallelogram.\n[/hide]", "Solution_2": "mass points are easy if you just accept the postulates as true, i.e. if you place the masses to balance the side ratio, and you add the masses to the point in the middle", "Solution_3": "Isn't mass point a simple variation of Menelaus theorem?", "Solution_4": "I'm going to learn these right before the Aime. I just might need em. :)", "Solution_5": "[quote=\"Altheman\"]mass points are easy if you just accept the postulates as true, i.e. if you place the masses to balance the side ratio, and you add the masses to the point in the middle[/quote]\r\n\r\nWell all I know about mass points is that if you have a point $C$ on line segment $\\overline{AB}$, then $w_A\\cdot AC = w_B \\cdot CB$, and $w_C = w_A+w_B$. :( And I don't really understand what is meant by any of the postulates...", "Solution_6": "$\\textsc{Anyone}?$", "Solution_7": "Here, I drew some images of basic layman's mass points instead of all the fancy wording and lettering.\r\n\r\n[url=http://img.photobucket.com/albums/v734/Elemennop/massp1.jpg]1st Mass Points Info[/url]\r\n[url=http://img.photobucket.com/albums/v734/Elemennop/massp2.jpg]2nd Mass Points Info[/url]", "Solution_8": "amm I also didn't under stand this--Angle Bisectors #1: why can you put on the weights suggested by the hint? thanks. I search the forum and it didn't help at all.", "Solution_9": "As far as I know, \"Mass Point\" has no use if you are farmiliar with Ceva, Menelaus and Area Ratios.", "Solution_10": "Yeah, but sometimes Mass Point geometry is still useful even if you know Ceva, Menelaus and area ratios. For instance, there's a case where you have to use Ceva/Menelaus twice, whereas you only need one operation if you use Mass Points.", "Solution_11": "so maybe help me? :lol: thanks", "Solution_12": "[quote=\"frt\"]For instance, there's a case where you have to use Ceva/Menelaus twice, whereas you only need one operation if you use Mass Points.[/quote]\r\nIn my experiences, the problems that involves more than on Ceva or Menelaus can be easily solved by Area Ratios." } { "Tag": [ "induction", "modular arithmetic", "geometric series" ], "Problem": "Prove the following by induction:\r\n$ 10^{n \\plus{} 1} \\minus{} 10(n \\plus{} 1) \\plus{} n|81, n>\\equal{}0$", "Solution_1": "Here's my failed attempt. I wrote the 2nd step of induction like this:\r\n$ 10 \\cdot 10^k \\minus{} 10(k \\plus{} 1) \\plus{} k|81$\r\nThen did the following with the last step:\r\n$ \\begin{array}{l}\r\n 10^{k \\plus{} 2} \\minus{} 10(k \\plus{} 2) \\minus{} k \\plus{} 1 \\equal{} \\\\ \r\n 10^2 \\cdot 10^k \\minus{} 10(k \\plus{} 1) \\minus{} 10 \\plus{} k \\plus{} 1 \\equal{} \\\\ \r\n (81 \\plus{} 9) \\cdot 10^k \\minus{} 10(k \\plus{} 1) \\minus{} 10 \\plus{} k \\plus{} 1 \\equal{} \\\\ \r\n 81 \\cdot 10^k \\plus{} 9 \\cdot 10^k \\plus{} 10 \\cdot 10^k \\minus{} 10(k \\plus{} 1) \\plus{} k \\minus{} 9 \\\\ \r\n \\end{array}$\r\nSo, now I have $ 81 \\cdot 10^k|81$ and $ 10 \\cdot 10^k \\minus{} 10(k \\plus{} 1) \\plus{} k|81$ but don't know what to do with $ 9 \\cdot 10^k \\minus{} 9$", "Solution_2": "Err, I guess you meant $ 81 \\mid 10^{n\\plus{}1} \\minus{} 10(n\\plus{}1) \\plus{} n$?\r\n\r\n[hide=\"With induction\"]\nFor $ n \\equal{} 0$, we get $ 81 \\mid 0$, which is true.\n\nSuppose $ 10^n \\minus{} 10n \\plus{} n \\minus{} 1 \\equal{} 10^n \\minus{} 9n \\minus{} 1 \\equal{} 81k$ (case $ n\\minus{}1$). Then $ 10^{n\\plus{}1} \\minus{} 90n \\minus{} 10 \\equal{} 81(10k)$, and also $ 10^{n\\plus{}1} \\minus{} 9n \\minus{} 10 \\equal{} 10^{n\\plus{}1} \\minus{} 10(n\\plus{}1) \\plus{} n \\equal{} 81(10k\\plus{}n)$. This shows that $ 81 \\mid 10^{n\\plus{}1} \\minus{} 10(n\\plus{}1) \\plus{} n$ (case $ n$), so we're done.\n[/hide]\n\n[hide=\"Without induction\"]\nSubstitute $ m \\equal{} n\\plus{}1$. So for positive integers $ m$ we want to prove $ 81 \\mid 10^m \\minus{} 9m \\minus{} 1$.\n\nSince $ \\frac{10^m\\minus{}1}{10\\minus{}1} \\equal{} 1 \\plus{} 10 \\plus{} 100 \\plus{} \\ldots \\plus{} 10^{m\\minus{}1}$ by geometric series, $ 10^m \\minus{} 1 \\equal{} 9(1 \\plus{} 10 \\plus{} \\ldots \\plus{} 10^{m\\minus{}1})$. Rewriting the original problem, we want to show $ 81 \\mid 9(1 \\plus{} 10 \\plus{} \\ldots \\plus{} 10^{m\\minus{}1}) \\minus{} 9m$, equivalent to showing that $ 9 \\mid 1 \\plus{} 10 \\plus{} \\ldots \\plus{} 10^{m\\minus{}1} \\minus{} m$. But notice that $ 1 \\plus{} 10 \\plus{} \\ldots \\plus{} 10^{m\\minus{}1} \\equiv 1 \\plus{} 1 \\plus{} \\ldots \\plus{} 1 \\equiv m \\pmod{9}$. Since $ a \\equiv b \\pmod{m} \\Leftrightarrow m \\mid a \\minus{} b$ by definition, we're done.\n[/hide]\n\n[hide=\"To answer your previous post\"]\nIt turns out that $ 9 \\cdot 10^k \\minus{} 9$ is always divisible by $ 81$. This fact inspired and is explained by the \"Without induction\" solution in this post. To summarize, realize that showing this is equivalent to showing that $ 10^k \\minus{} 1$ is divisible by $ 9$. Then look at $ 10^k \\minus{} 1$ modulo $ 9$.\n[/hide]" } { "Tag": [], "Problem": "Hello, can someone help me on this problem:\r\n\r\nIf $ \\frac{2y\\plus{}2z\\minus{}x}{a} \\equal{} \\frac{2z\\plus{}2x\\minus{}y}{b} \\equal{} \\frac{2x\\plus{}2y\\minus{}z}{c}$ then show that \r\n$ \\frac{x}{2b\\plus{}2c\\minus{}a} \\equal{} \\frac{y}{2c\\plus{}2a\\minus{}b} \\equal{} \\frac{z}{2a\\plus{}2b\\minus{}c}$\r\n\r\nHints, anyone?", "Solution_1": "[b]Lemma:[/b] If $ \\frac{a}{b} \\equal{} \\frac{c}{d}$ then $ \\frac{a}{b} \\equal{} \\frac{c}{d} \\equal{} \\frac{a \\plus{} ck}{b \\plus{} dk}$ for any $ k$.", "Solution_2": "I think I got it:\r\n\r\nLet $ \\frac{2y\\plus{}2z\\minus{}x}{a} \\equal{} (A)$\r\n$ \\frac{2z\\plus{}2x\\minus{}y}{b} \\equal{} (B)$\r\n$ \\frac{2x\\plus{}2y\\minus{}z}{c} \\equal{} (C)$ \r\n\r\nAdd $ (B)$ and $ (C)$ then subtract $ (A)$\r\n\r\n$ \\frac{2z\\plus{}2x\\minus{}y\\plus{}2x\\plus{}2y\\minus{}z}{b\\plus{}c} \\equal{} \\frac{4x\\plus{}y\\plus{}z}{b\\plus{}c} \\equal{} \\frac{2(4x\\plus{}y\\plus{}z)}{2(b\\plus{}c)}$\r\n\r\n$ \\frac{8x\\plus{}2y\\plus{}2z}{2b\\plus{}2c} \\equal{} \\frac{8x\\plus{}2y\\plus{}2z\\minus{}(2y\\plus{}2z\\minus{}x)}{2b\\plus{}2c\\minus{}a} \\equal{} \\frac{9x}{2b\\plus{}2c\\minus{}a}$\r\n\r\nSimilarly, add $ (A)$ and $ (B)$ and then subtract $ (C)$\r\n\r\n$ \\frac{2y\\plus{}2z\\minus{}x\\plus{}2z\\plus{}2x\\minus{}y}{a\\plus{}b} \\equal{} \\frac{4z\\plus{}x\\plus{}y}{a\\plus{}b} \\equal{} \\frac{8z\\plus{}2x\\plus{}2y}{2a\\plus{}2b}$\r\n\r\n$ \\frac{8z\\plus{}2x\\plus{}2y\\minus{}(2x\\plus{}2y\\minus{}z)}{2a\\plus{}2b\\minus{}c}$ = $ \\frac{9z}{2a\\plus{}2b\\minus{}c}$\r\n\r\nOnce again, add $ (A)$ and $ (C)$ and subtract $ (B)$\r\n\r\nSince each of these fractions is equal to the sum of numerators and denominators:\r\n\r\n$ \\frac{9x}{2b\\plus{}2c\\minus{}a} \\equal{} \\frac{9y}{2a\\plus{}2c\\minus{}b} \\equal{} \\frac{9z}{2a\\plus{}2b\\minus{}c}$\r\n\r\nOr $ \\frac{x}{2b\\plus{}2c\\minus{}a} \\equal{} \\frac{y}{2a\\plus{}2c\\minus{}b} \\equal{} \\frac{z}{2a\\plus{}2b\\minus{}c}$", "Solution_3": "We can assume x+y+z=1 and a+b+c=1, because any scale of them would also work.\r\nSo we have $ \\frac {2 \\minus{} 3x}{a} \\equal{} \\frac {2 \\minus{} 3y}{b} \\equal{} \\frac {2 \\minus{} 3z}{c} \\equal{} n$\r\nWe need to show $ \\frac{x}{2\\minus{}3a}\\equal{}\\frac{y}{2\\minus{}3b}\\equal{}\\frac{x}{2\\minus{}3c}$\r\nThen $ n \\equal{} n(a \\plus{} b \\plus{} c) \\equal{} na \\plus{} nb \\plus{} nc \\equal{} 6 \\minus{} 3(x \\plus{} y \\plus{} z) \\equal{} 3$\r\nSo $ a \\equal{} \\frac {2 \\minus{} 3x}{3}$, etc.\r\nSubbing all of them into the second, we get 1/3=1/3=1/3, clearly true." } { "Tag": [ "probability" ], "Problem": "Some board games come with special dice. One game in particular uses an 8-sided die with the numbers 1 through 8 printed on the sides. What is the probability of rolling a number other than a 1, 2, or 5 with this die?", "Solution_1": "[hide]5/8, there are 8 numbers and you want to roll 5 of them[/hide]", "Solution_2": "[hide]1-P(1)-P(2)-P(5) = 5/8[/hide]", "Solution_3": "[hide]\nwanted outcome=5\npossible outcome=8\nso the answer is obviously 5/8.[/hide]", "Solution_4": "[quote=\"colts18\"]Some board games come with special dice. One game in particular uses an 8-sided die with the numbers 1 through 8 printed on the sides. What is the probability of rolling a number other than a 1, 2, or 5 with this die?[/quote]\r\n\r\nIf you want to roll a 1, 2, or 5, then the probability would be $\\frac{3}{8}$. \r\n\r\nBut if you don't want those numbers, then the probablilty would obviously be $\\frac{5}{8}$" } { "Tag": [ "limit" ], "Problem": "Sa se arate ca nu exista functii continue neconstante $f:R\\to R$ cu proprietatea:\\[\r\nf(x + \\sqrt 2 ) = f(x + \\sqrt 3 ) = f(x),\\forall x \\in R.\r\n\\]", "Solution_1": "-----------------------------------------------------------------------------------------------------------\r\n[b][color=blue]\nS\u0103 se arate c\u0103 nu exist\u0103 func\u0163ii continue $ f: \\mathbb{R} \\to \\mathbb{R}$, neconstante cu proprietatea:\n$ f\\left( {x \\plus{} \\sqrt 2 } \\right) \\equal{} f\\left( {x \\plus{} \\sqrt 3 } \\right) \\equal{} f\\left( x \\right),\\forall x \\in \\mathbb{R}$\n[/color][/b]\r\n------------------------------------------------------------------------------------------------------------\r\n[b][u]Solu\u0163ie:[/u][/b]\r\nDin ipoteze rezult\u0103 c\u0103 func\u0163ia are period\u0103 pe $ \\sqrt 2$ dar \u015fi pe $ \\sqrt 3$. \r\nSe poate demonstra c\u0103 aceast\u0103 func\u0163ie are ca perioad\u0103 orice num\u0103r de forma\r\n\\[ m\\sqrt 2 \\plus{} n\\sqrt 3 ,\\forall n,m \\in \\mathbb{Z}\r\n\\]\r\nDeci avem:\r\n\\[ f\\left( x \\right) \\equal{} f\\left( {x \\plus{} m\\sqrt 2 \\plus{} n\\sqrt 3 } \\right),\\forall m,n \\in \\mathbb{Z},\\forall x \\in \\mathbb{R}\r\n\\]\r\n\u00een particular avem:\r\n\\[ f\\left( 0 \\right) \\equal{} f\\left( {m\\sqrt 2 \\plus{} n\\sqrt 3 } \\right),\\forall m,n \\in \\mathbb{Z}\r\n\\]\r\nDac\u0103 se define\u015fte mul\u0163imea $ A \\equal{} \\left\\{ {m\\sqrt 2 \\plus{} n\\sqrt 3 \\left| {m,n \\in \\mathbb{Z}} \\right.} \\right\\}$ aceasta este dens\u0103 in $ \\mathbb{R}$, adic\u0103: \r\n$ \\forall a \\in \\mathbb{R},\\exists \\left( {x_n } \\right)_{n \\in \\mathbb{N}}$ astfel \u00eenc\u00e2t $ x_n \\in A$ \u015fi $ \\mathop {\\lim }\\limits_{n \\to \\infty } x_n \\equal{} a$\r\n\r\nDin $ f\\left( 0 \\right) \\equal{} f\\left( {m\\sqrt 2 \\plus{} n\\sqrt 3 } \\right),\\forall m,n \\in \\mathbb{Z}$ avem: $ f\\left( {x_n } \\right) \\equal{} f\\left( 0 \\right)$ \u015fi deci $ \\mathop {\\lim }\\limits_{n \\to \\infty } f\\left( {x_n } \\right) \\equal{} f\\left( a \\right) \\equal{} f\\left( 0 \\right)$\r\n\r\n[u]In concluzie[/u] func\u0163iile c\u0103utate sunt func\u0163ii constante." } { "Tag": [ "trigonometry", "limit", "calculus", "calculus computations" ], "Problem": "Prove that the series $ \\sum_{i \\equal{} 1}^{\\infty} \\frac{\\sin i}{i}$ converge.", "Solution_1": "http://en.wikipedia.org/wiki/Dirichlet%27s_test", "Solution_2": "Set $ a_{k} \\equal{} \\sin k$ and $ b_{k} \\equal{} \\frac {1}{k}$\r\n\t\r\nLet $ S_{k} \\equal{} \\sum_{i \\equal{} 1}^{k}\\sin i \\equal{} \\frac {1}{2\\sin \\frac {1}{2}}\\left( \\cos \\frac {1}{2} \\minus{} \\cos \\left( k \\plus{} \\frac {1}{2}\\right) \\right)$\r\n\t\r\n$ \\left| S_{k}\\right| \\leq \\frac {1}{2\\sin \\frac {1}{2}}\\Rightarrow \\lim_{n\\rightarrow \\infty }S_{n}b_{n} \\equal{} 0$\r\n\t\r\nNow,\r\n\t\t\r\n$ \\left| \\sum_{i \\equal{} 1}^{\\infty }\\frac {\\sin i}{i}\\right|$\r\n\t\r\n$ \\equal{} \\lim_{n\\rightarrow \\infty }\\left| \\sum_{i \\equal{} 1}^{n}\\frac {\\sin i}{i}\\right|$\r\n\t\r\n$ \\equal{} \\lim_{n\\rightarrow \\infty }\\left| \\sum_{i \\equal{} 1}^{n}a_{i}b_{i}\\right|$\r\n\t\r\n$ \\equal{} \\lim_{n\\rightarrow \\infty }\\left| S_{n}b_{n} \\minus{} \\sum_{i \\equal{} 1}^{n \\minus{} 1}S_{i}\\left( b_{i \\plus{} 1} \\minus{} b_{i}\\right) \\right|$\r\n\t\r\n$ \\leq \\sum_{i \\equal{} 1}^{\\infty }\\left| S_{i}\\left( b_{i} \\minus{} b_{i \\plus{} 1}\\right) \\right| \\leq \\frac {1}{\\sin \\frac {1}{2}}\\sum_{i \\equal{} 1}^{\\infty }\\left( \\frac {1}{i} \\minus{} \\frac {1}{i \\plus{} 1}\\right) \\equal{} \\frac {1}{\\sin \\frac {1}{2}}$\r\n\t\r\nHence the given series converge.", "Solution_3": "Converges, and its value is $ \\tfrac{{\\pi \\minus{} 1}}\r\n{2}$.\r\n\r\nProve ... is a fun challenge. :rotfl:", "Solution_4": "the last challenge is easy if you use C\u00e8saro means.\r\n\r\nif I'm not mistaken, using arithmetic means of Fourier series (or C\u00e8saro sums), we have $ \\dfrac{\\pi\\minus{}x}{2} \\equal{} \\sum_{n\\equal{}1}^{\\infty}\\dfrac{\\sin{nx}}{n}$\r\nSo, take $ x\\equal{}1$.\r\n\r\nThis example is due to Fejer, in the begin of early 20. \r\n\r\n^^" } { "Tag": [ "limit", "calculus", "calculus computations" ], "Problem": "Study $ \\lim_{n\\to\\infty} a_n$ such that $ na_{n \\plus{} 1} \\equal{} (n \\plus{} 1)a_n \\minus{} \\max(a_n,\\ n^2)\\ (n \\equal{} 1,\\ 2,\\ 3,\\ \\cdots).$", "Solution_1": "$ na_{n \\plus{} 1} \\equal{} (n \\plus{} 1)a_n \\minus{} \\max(a_n,\\ n^2) \\leq (n \\plus{} 1)a_n \\minus{} a_n \\equal{} na_n$, hence $ a_{n \\plus{} 1} \\leq a_n$, so the sequence is decreasing. Moreover, for $ n$ such that $ a_1 < n^2$ we have $ na_{n \\plus{} 1} \\equal{} (n \\plus{} 1)a_n \\minus{} n^2 \\leq a_1 \\plus{} na_1 \\minus{} n^2$, hence $ a_{n \\plus{} 1} \\leq a_1/n \\plus{} a_1 \\minus{} n$, therefore the sequence is lower unbounded. Finally $ \\lim_{n \\to \\infty} a_n \\equal{} \\minus{} \\infty$." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Hi all, I am stuck trying to simplify or evaluate the following two sums, N is odd:\r\n\r\n\\sum\\limits_{i=0}^{(N-1)/2}{N\\choose i}(N-i)\r\n\r\n\\sum\\limits_{i=0}^{(N-1)/2}{N\\choose i}{N-i\\choose 2}\r\n\r\n(also see the pdf-file attached)\r\n\r\nThe problem would be much easier were the sums taken over i=0 to N instead of (N-1)/2. I would really appreciate you help!\r\n\r\nBest,\r\nserge", "Solution_1": "\\[ \\sum_{i=0}^{(N-1)/2}\\binom N i \\cdot (N-i)=\\\\ =N\\sum_{i=0}^{(N-1)/2}{\\binom N i}-\\sum_{i=0}^{(N-1)/2}i\\cdot\\binom N i\\\\ =N\\sum_{i=0}^{(N-1)/2}{\\binom N i}-\\sum_{i=1}^{(N-1)/2}i\\cdot\\binom N i\\\\ =N\\sum_{i=0}^{(N-1)/2}{\\binom N i}-N\\sum_{i=1}^{(N-1)/2}\\binom {N-1}{i-1}\\\\ =N\\sum_{i=0}^{(N-1)/2}{\\binom N i}-N\\sum_{j=0}^{(N-3)/2}\\binom {N-1}{j}\\\\ =\\frac N2 \\left( \\sum_{i=0}^{(N-1)/2}{\\binom N i}+\\sum_{i=(N+1)/2}^{N}{\\binom N {N-i}} -\\sum_{j=0}^{(N-3)/2}\\binom {N-1}{j} -\\sum_{j=(N+1)/2}^{N-1}\\binom {N-1}{N-1-j} \\right)\\\\ =\\frac N2 \\left( \\sum_{i=0}^{N}{\\binom N i} -\\sum_{j=0}^{N-1}\\binom {N-1}{j} + \\binom{N-1}{(N-1)/2}\\right)\\\\ = \\frac N2 \\left( 2^N -2^{N-1} + \\binom{N-1}{(N-1)/2}\\right)\\\\ = \\frac N2 \\left( 2^{N-1} + \\binom{N-1}{(N-1)/2}\\right) \\]\r\n\r\nThe other one could be done similarly." } { "Tag": [ "probability" ], "Problem": "My friend thought up a problem and gave it to me, but...\r\n\r\nThree people, A,B, and C are running a race. The probability that A finishs before B is 60%. The probability that B finishes before C is 90%. What is the probability that C finishes before A?\r\n\r\nI thought that the question as is was impossible, but he thinks that since there is a set number of cases, the probability can be calculated. Doing it that way, however, I got -.08% :? Without at least defining bounds for this race, is it solvable? And if not, would giving the track bounds make it solvable?\r\n\r\n\r\n*Finishing before doesn't necessarily mean directly before. If it went A then C then B, A finishes before B", "Solution_1": "[hide=\"solution\"]\n\nthere are only three ways C can finish before A:\n\ncases: CAB, CBA, BCA (leftmost finish first)\n\n$CAB = 0.1 * 0.6 = 0.06$\n$CBA = 0.4 * 0.1 = 0.04$\n$BCA = 0.9 * 0.4 = 0.36$\n\nadd them all, we get 0.46. the probability that C beats A is $ \\frac{0.46}{total}$ =\n$ \\frac{0.46}{1.42} = \\frac{23}{71}$[/hide]\r\n\r\ni hope that is somewhat right.", "Solution_2": "This is a very interesting problem. I don't believe that reddevil's answer is correct right now. And I think the problem needs to specify that the respective placements of the individuals are independent of one another (or if they aren't, how they are related).\r\n\r\n[hide=\"Hint\"]Try a Venn-Diagram, or Inclusion-Exclusion.[/hide]" } { "Tag": [], "Problem": "$ x\\plus{}\\frac{1}{y}\\equal{}y\\plus{}\\frac{1}{z}\\equal{}z\\plus{}\\frac{1}{x}$\r\n\r\nProve that $ x\\equal{}y\\equal{}z$ or $ (xyz)^2\\equal{}1$\r\n\r\n :maybe:", "Solution_1": "If $ (x, y, z) \\equal{} \\left(3, 1, \\frac{1}{3}\\right)$\r\n\r\n$ (xyz)^2\\equal{}1^2\\equal{}1$.\r\n\r\nBut, $ \\left(x\\plus{}\\frac{1}{y}, y\\plus{}\\frac{1}{z}, z\\plus{}\\frac{1}{x}\\right) \\equal{} \\left(4, 4, \\frac{2}{3}\\right)$ :D", "Solution_2": "You are wrong.\r\nIt's not if and only if, it's just if.\r\nIf x+(1/y)=... then (xyz)^2=1 or...", "Solution_3": "from given condition we have $ x\\minus{}y \\equal{} \\frac{y\\minus{}z}{yz} ; y\\minus{}z \\equal{} \\frac{z\\minus{}x}{xz} ; x\\minus{}z \\equal{}\\frac{y\\minus{}x}{xy}$ \r\nmultiply all of them together gives $ (x\\minus{}y)(y\\minus{}z)(z\\minus{}x) \\equal{} \\frac{(x\\minus{}y)(y\\minus{}z)(z\\minus{}x)}{x^2y^2z^2}$\r\nif $ x\\minus{}y \\equal{}0$ then $ y\\minus{}z \\equal{} 0 \\minus{}> x\\equal{}y\\equal{}z$\r\nif $ x\\minus{}y <> 0$ and $ y\\minus{}z <> 0$ and $ z\\minus{}x <> 0$ then $ (xyz)^2 \\equal{}1$", "Solution_4": "Nice solution :omighty:", "Solution_5": "Hi, HTA, Bachukas :) \r\n\r\nIs the problem correct? \r\nDid my post read?", "Solution_6": "And have you read mine?\r\nYou were going in a wrong direction.", "Solution_7": "Well done HTP ,very elegant solution !", "Solution_8": "[quote=\"HTA\"]from given condition we have $ x \\minus{} y \\equal{} \\frac {y \\minus{} z}{yz} ; y \\minus{} z \\equal{} \\frac {z \\minus{} x}{xz} ; x \\minus{} z \\equal{} \\frac {y \\minus{} x}{xy}$ \nmultiply all of them together gives $ (x \\minus{} y)(y \\minus{} z)(z \\minus{} x) \\equal{} \\frac {(x \\minus{} y)(y \\minus{} z)(z \\minus{} x)}{x^2y^2z^2}$\nif $ x \\minus{} y \\equal{} 0$ then $ y \\minus{} z \\equal{} 0 \\minus{} > x \\equal{} y \\equal{} z$\nif $ x \\minus{} y < > 0$ and $ y \\minus{} z < > 0$ and $ z \\minus{} x < > 0$ then $ (xyz)^2 \\equal{} 1$[/quote]\r\n\r\nSure, nice & elegant.. :D :roll: :clap:", "Solution_9": "[quote=\"Bugi\"]\nYou are wrong.\nIt's not if and only if, it's just if.\nIf x+(1/y)=... then (xyz)^2=1 or...\n\nAnd have you read mine?\nYou were going in a wrong direction.[/quote]\r\nOops :(\r\nI misunderstood a problem.\r\n\r\n$ \\color{red}\\times\\color{black}\\boxed{x\\plus{}\\frac{1}{y}\\equal{}y\\plus{}\\frac{1}{z}\\equal{}z\\plus{}\\frac{1}{x} \\color{blue}\\iff\\color{black} x\\equal{}y\\equal{}z \\ \\text{or} \\ (xyz)^2\\equal{}1}$\r\n\r\n$ \\color{red}\\bigcirc\\color{black}\\boxed{x\\plus{}\\frac{1}{y}\\equal{}y\\plus{}\\frac{1}{z}\\equal{}z\\plus{}\\frac{1}{x} \\color{blue}\\Longrightarrow\\color{black} x\\equal{}y\\equal{}z \\ \\text{or} \\ (xyz)^2\\equal{}1}$\r\n\r\nThank you :)" } { "Tag": [ "Asymptote", "limit", "algebra", "polynomial", "LaTeX" ], "Problem": "I have this fraction i have to find all the limits for:\r\n\r\n$f(x)=\\frac{x^3+1}{x^2+1}$\r\n\r\nI found out (and checked with my calculator) that there was no vertical limit, but for this fraction i can't find any diagonal limits (as this is the only one left). My calculator also tells me there is a diagonal limit.\r\n\r\nThx in advance for your help.", "Solution_1": "You mean oblique asymptotes? Try reading what your textbook has to say.", "Solution_2": "you just divide it as much as you can, and the quotient is the asymptote, and the remainder drops out since the degree of the remainder is higher on the bottom, so it will approach 0", "Solution_3": "In this case it might work, but generally an oblique asymptote is defined like this:\r\n\r\nIf there exist both $k=\\lim_{x\\rightarrow\\infty}{f(x)\\over x}$ and $n=\\lim_{x\\rightarrow\\infty}({f(x)-kx})$, then $y=kx+n$ is an oblique asymptote. Of course, \"$\\infty$\" can be either $+\\infty$ or $-\\infty$, and those asymptotes can be different.", "Solution_4": "The problem is i can't divide it any further than this, unless i'm doing it wrong:\r\n\r\n$(x^3+1): (x+1)=x^2-x+1$\r\n\r\nAnd if i then try to find the x's for this 2nd degree equation:\r\n\r\n$x^2-x+1=0$\r\n\r\n$\\frac{1\\pm\\sqrt{-3}}{2}$\r\n\r\nThis means the 2nd degree equation doesn't intersect with $x=0$.\r\n\r\nAnd it is the same i end up with if i try with the denominator:\r\n\r\n$x^2+1$\r\n\r\nSo i can't split it up more than this:\r\n\r\n$f(x)=\\frac{(x^2-x+1)(x+1)}{x^2+1}$\r\n\r\nAm i doing anything wrong in these calculation or did i just miss the point on how to continue from here?", "Solution_5": "I'm a bit confused as to what you're doing here. As $x\\rightarrow\\infty$, only the leading terms in the numerator and denominator really matter. Here you have\r\n\r\n$y=\\frac{x^{3}+1}{x^{2}+1}\\approx\\frac{x^{3}}{x^{2}}=x\\quad\\text{as}\\quad x\\rightarrow\\infty$", "Solution_6": "Oh yeah, that makes perfect sense to do it this easy instead of following the book exactly :D\r\n\r\nThx for the help.", "Solution_7": "Oh, BTW: Found out it wasn't enough to find what $f(x)$ goes towards when it goes towards $x$, i have to find the 'oblique asymptotes' as Chess64 called it.", "Solution_8": "use long division for polynomials", "Solution_9": "Then i will have to ask you to show me step by step how to do it. I have tried that with this fraction but it didn't work well for me.", "Solution_10": "ok, here is the long division\r\n\r\n[code]\n x\n ----------\nx^2+1| x^3+1\n -(x^3+x)\n ----------\n 1-x\n[/code]\r\n\r\nnow we now know by our division that $f(x)=\\frac{x^3+1}{x^2+1}=x+\\frac{1-x}{x^2+1}$\r\nnow look at the last term, as x gets very large, the whole fraction approaches 0 because there is $x^2$ on the bottom, and $x$ on the top (i.e. the bottom has a higher degree than the top)\r\n\r\nso our answer is $h(x)=x$\r\n\r\na simple definition of oblique asympote:\r\n\r\nif we have $f(x)=h(x)+g(x)$ where for large $x$, $g(x)$ approaches $0$, then our oblique asymptote is $h(x)$, make sure you understand this and can replicate this, since this seems like a standard school problem that you will see in your class again", "Solution_11": "OK, so long division is just to divide the whole fraction without splitting it into fractions first. I've never learned that, but i can surely believe it is true - why wouldn't it...\r\n\r\nYes, i have learned about the $f(x)=h(x)+g(x)$ though it was described in another way. And $g(x)$ is usally, if not always, the remaining fraction. So when $x$ becomes a great number it is reduced to almost 0 if it is a degree bigger in the denominator (it either has to be bigger in the denominator or the same degree as the nominator). That's the reason why the asymptote is found in $h(x)$.\r\n\r\nGreat, think i got the point now. Thx a lot everyone, this really helped ;)", "Solution_12": "Well, \r\n the problem was not formulated in the best way, but finally is was solved. Please, read and learn what was written in posts 4 and 11. These are basic results to all similar problems. \r\n\r\n [u] Babis[/u]", "Solution_13": "Number 4 wasn't too easy, but i think i got the point now. I guess the biggest problem with this forum for me is that english isn't my main language. I know this is how it is for many of you ppl, but i guess you expierence this often everytime a new member joins. You might be good in english, but when i studied english we read texts without math words. Guess that is what makes many the many problems with us foreigners :)", "Solution_14": "BTW: How do you make that unlimited sign using LaTeX amcavoy made?", "Solution_15": "[quote=\"Sporally\"]BTW: How do you make that unlimited sign using LaTeX amcavoy made?[/quote]\r\n\r\n\\infty", "Solution_16": "Testing:\r\n\r\n$\\infty$\r\n\r\nOK, thx." } { "Tag": [ "calculus", "function", "inequalities", "inequalities unsolved" ], "Problem": "Question: Prove that \r\n$ \\frac {a}{b \\plus{} 2c \\plus{} 3d} \\plus{} \\frac {b}{c \\plus{} 2d \\plus{} 3a} \\plus{} \\frac {c}{d \\plus{} 2a \\plus{} 3b} \\plus{} \\frac {d}{a \\plus{} 2b \\plus{} 3c}\\ge \\frac {2}{3}$, given that $ a,b,c,d > 0$.\r\nAnswer: Using calculus, LHS is a convex function of $ a,b,c,d$. So minimum occurs when $ a \\equal{} b \\equal{} c \\equal{} d$, so $ LHS\\ge 4\\cdot \\frac {a}{6a} \\equal{} \\frac {2}{3}$.\r\n\r\nI have another question:\r\n\r\nGiven that $ x,y,z\\ge \\minus{} 1$, show that $ \\frac {1 \\plus{} x^2}{1 \\plus{} y \\plus{} z^2} \\plus{} \\frac {1 \\plus{} y^2}{1 \\plus{} z \\plus{} x^2} \\plus{} \\frac {1 \\plus{} z^2}{1 \\plus{} x \\plus{} y^2}\\ge 2$.", "Solution_1": "Your first one is easy by Cauchy-Schwartz:\r\n\\[ \\sum_{cyc}\\frac{a}{b\\plus{}2c\\plus{}3d}\\geqslant\\frac{(a\\plus{}b\\plus{}c\\plus{}d)^2}{4(ab\\plus{}ac\\plus{}ad\\plus{}bc\\plus{}bd\\plus{}cd)}\\stackrel{!}{\\geqslant}\\frac{2}{3}\\mbox{,}\\]\r\nwhere the last one is eqivalent to\r\n\\[ \\frac{a\\plus{}b\\plus{}c\\plus{}d}{4}\\geqslant\\sqrt{\\frac{ab\\plus{}ac\\plus{}ad\\plus{}bc\\plus{}bd\\plus{}cd}{6}}\\mbox{,}\\]\r\nwhich is MacLaurin's inequality. :lol:", "Solution_2": "[quote=\"knol\"]\nI have another question:\n\nGiven that $ x,y,z\\ge \\minus{} 1$, show that $ \\frac {1 \\plus{} x^2}{1 \\plus{} y \\plus{} z^2} \\plus{} \\frac {1 \\plus{} y^2}{1 \\plus{} z \\plus{} x^2} \\plus{} \\frac {1 \\plus{} z^2}{1 \\plus{} x \\plus{} y^2}\\ge 2$.[/quote]\r\nIt is JBMO proposed by Panaiotopol.\r\nUse the AM-GM inequality on $ x,y,z$ at the denominators to acquire $ \\frac{1\\plus{}x^2}{2},\\frac{1\\plus{}y^2}{2},\\frac{1\\plus{}z^2}{2}$ and then use CS to kill it.\r\nsee also here:\r\nhttp://gbas2010.wordpress.com/2009/12/07/inequality-8laurentiu-panaitopol-jbmo-2003/" } { "Tag": [], "Problem": "What is the largest integer $ n$ for which $ \\binom{8}{3} \\plus{} \\binom{8}{4} \\equal{} \\binom{9}{n}$?", "Solution_1": "By Pascal's Identity, $ \\binom{8}{3}\\plus{}\\binom{8}{4}\\equal{}\\binom{9}{4}$. However, we want the largest value of $ n$, and since $ 9\\minus{}4\\equal{}5>4$, the answer is $ \\boxed{5}$." } { "Tag": [ "inequalities", "inequalities unsolved" ], "Problem": "someone needs help for this question. Personally, I am not interested in this type of inequality. So I came here to ask for help. He told me that he is secondary-school pupil. So please find an elementary proof for the problem.\r\n\r\nGiven $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:\r\n\r\n$2(a+b+c) + \\frac{1}{abc} \\geq 7$\r\n\r\nThanks in advance.", "Solution_1": "$It is very easy$ :D \r\n\r\nBy $A(m)\\geq G(m)$ we have that $(3=a^2+b^2+c^2\\geq 3sqrt[3]{{abc}})$\r\n=> $abc\\leq 1$ \r\n$By A(m)\\geq G(m)$\r\n$2(a+b+c) + \\frac{1}{abc} =a+b+c+a+b+c+ \\frac{1}{abc}\\geq 7sqrt[7]{{abc}}\\geq 7$ because $abc\\leq 1$ .\r\n\r\n$Q-E-D$", "Solution_2": "[quote=\"vardanch\"]$It is very easy$ :D \n\nBy $A(m)\\geq G(m)$ we have that $(3=a^2+b^2+c^2\\geq 3sqrt[3]{{abc}})$\n=> $abc\\leq 1$ \n$By A(m)\\geq G(m)$\n$2(a+b+c) + \\frac{1}{abc} =a+b+c+a+b+c+ \\frac{1}{abc}\\geq 7sqrt[7]{{abc}}\\geq 7$ because $abc\\leq 1$ .\n\n$Q-E-D$[/quote]\r\n\r\nif $abc\\leq 1$ then $7\\sqrt[7]{abc}\\leq 7$ and not $\\geq 7$", "Solution_3": "[quote=\"ptung\"]Given $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:\n\n$2(a+b+c) + \\frac{1}{abc} \\geq 7$\n\nThanks in advance.[/quote]\r\n\r\nUnfortunately, this inequality is not true. For example, take $a = b = \\sqrt{9/10}$, $c = \\sqrt{12/10}$. Then $a^2 + b^2 + c^2 = 3$ and\r\n\\[ 2(a + b + c) + \\frac{1}{abc} \\approx 6.999924 < 7. \\]", "Solution_4": "I ask Mr.ptung for help , that problem is wrong but this one is right\r\nGiven a,b,c positive numbers, a^2+b^2+c^2=3.prove that\r\n5(a+b+c)+3/(abc) >= 18\r\n\r\n\r\nI have some problem like this (certainly they're true ) \r\n\r\n[color=blue] I'm waiting for a solution for secondary school like me [/color]", "Solution_5": "just put: Find the infinum value of:\r\n\r\n$\\frac{\\frac{1}{abc} - 1}{3 - a - b - c}$", "Solution_6": "$PROBLEM$\r\n\r\n[quote=\"ptung\"]someone needs help for this question. Personally, I am not interested in this type of inequality. So I came here to ask for help. He told me that he is secondary-school pupil. So please find an elementary proof for the problem.\n\nGiven $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:\n\n$2(a+b+c) + \\frac{1}{abc} \\geq 7$\n\nThanks in advance.[/quote]\r\n\r\n$SOLUTION$\r\nBy $A(m)\\geq G(m)$ we have $abc\\leq 1$.\r\n$1)$ If $a^2+b^2+c^2\\leq a+b+c$ => $2(a+b+c)+1/abc\\geq 2(a^2+b^2+c^2)+1/abc\\geq 3*2+1=7$\r\n$2)$ If $a^2+b^2+c^2\\geq a+b+c$ => $a+b+c\\leq 3$\r\nBy $A(m)\\geq G(m)$ we have that for positive real x,y,z\r\n$sqrt(x)+sqrt(y)+sqrt(z)\\leq sqrt(3*(x+y+z))$ => $sqrt(a)+sqrt(b)+sqrt(c)\\leq sqrt(3*(a+b+c))\\leq a+b+c$ =>\r\n\r\n$2(a+b+c)+1/abc\\geq 2(sqrt(a)+sqrt(b)+sqrt(c))+1/abc$\r\n By $A(m)\\geq G(m)$\r\n\r\n$2(sqrt(a)+sqrt(b)+sqrt(c))+1/abc=sqrt(a)+sqrt(a)+sqrt(b)+sqrt(b)+sqrt(c)+sqrt(c)+1/abc\\geq 7$\r\n\r\n$Q-E-D$", "Solution_7": "$Now$ $I$ $will$ $proof$ $that$ \r\n$\\sqrt x + \\sqrt y + \\sqrt z \\le \\sqrt{3(x + y + z)}$\r\nif x,y,z are positive real numbers.\r\nBy $A(m)\\geq G(m)$ we have tha $x+y\\geq \\sqrt xy$ $\\Rightarrow$ $x+y+z\\geq \\sqrt xy+\\sqrt yz+\\sqrt xz$ $\\Rightarrow$\r\n$3(x+y+z)\\geq (\\sqrt x + \\sqrt y + \\sqrt z)^2$ => $\\sqrt x + \\sqrt y + \\sqrt z \\le \\sqrt{3(x + y + z)}$", "Solution_8": "[quote=\"vardanch\"]\n$2)$ If $a^2+b^2+c^2\\geq a+b+c$ => $a+b+c\\leq 3$\nBy $A(m)\\geq G(m)$ we have that for positive real x,y,z\n$sqrt(x)+sqrt(y)+sqrt(z)\\leq sqrt(3*(x+y+z))$ => $sqrt(a)+sqrt(b)+sqrt(c)\\leq sqrt(3*(a+b+c))\\leq a+b+c$\n[/quote]\r\n\r\nThe last inequality is false; it should be $\\sqrt{3(a+b+c)} \\ge a+b+c$.\r\n\r\nA proof of this inequality was posted here:\r\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86904]www.artofproblemsolving.com/Forum/viewtopic.php?t=86904[/url]", "Solution_9": "[quote=\"nsato\"][quote=\"vardanch\"]\n$2)$ If $a^2+b^2+c^2\\geq a+b+c$ => $a+b+c\\leq 3$\nBy $A(m)\\geq G(m)$ we have that for positive real x,y,z\n$sqrt(x)+sqrt(y)+sqrt(z)\\leq sqrt(3*(x+y+z))$ => $sqrt(a)+sqrt(b)+sqrt(c)\\leq sqrt(3*(a+b+c))\\leq a+b+c$\n[/quote]\n\nThe last inequality is false; it should be $\\sqrt{3(a+b+c)} \\ge a+b+c$.\n\nA proof of this inequality was posted here:\n[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86904]www.artofproblemsolving.com/Forum/viewtopic.php?t=86904[/url][/quote]\r\n\r\nWhat is FALSE????\r\n\r\nI proved that $\\sqrt x + \\sqrt y + \\sqrt z \\le \\sqrt{3(x + y + z)}$.\r\nWhat is wrong in that proof???", "Solution_10": "just as nsato said... if \\[ a+b+c\\leq 3 \\] then $\\sqrt{3(a+b+c)} \\geq a+b+c$ and not $\\leq a+b+c$ and... He already gave you a counter example...\r\nHe didn't say that your proof of \\[ \\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\leq \\sqrt{3(a+b+c)} \\] was wrong...", "Solution_11": "[quote=\"ptung\"]someone needs help for this question. Personally, I am not interested in this type of inequality. So I came here to ask for help. He told me that he is secondary-school pupil. So please find an elementary proof for the problem.\n\nGiven $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:\n\n$2(a+b+c) + \\frac{1}{abc} \\geq 7$\n\nThanks in advance.[/quote]\r\nLet b=c and a=1,1, this inequality is wrong.\r\n\r\nWe can prove $5(a+b+c) + \\frac{3}{abc} \\geq 18$ by mixing variable.\r\nA great question : Find all positive real number k satisfying the following inequality is true : $k(a+b+c)+\\frac{1}{abc} \\ge 3k+1$", "Solution_12": "[quote=\"yookon88\"]just as nsato said... if \\[ a+b+c\\leq 3 \\] then $\\sqrt{3(a+b+c)} \\geq a+b+c$ and not $\\leq a+b+c$ and... He already gave you a counter example...\nHe didn't say that your proof of \\[ \\sqrt{a}+\\sqrt{b}+\\sqrt{c}\\leq \\sqrt{3(a+b+c)} \\] was wrong...[/quote]\r\n\r\nOK. I understand my mistake. :D" } { "Tag": [], "Problem": "Hello! Is anyone there?", "Solution_1": "hello, lol ur a big eragon fan", "Solution_2": "hello, is there any point in this?", "Solution_3": "lol hello? dude this is a spam post, and must be removed by the moderator of the board!", "Solution_4": "OK, if not hello, then ... hi everybody, how do you think about math?", "Solution_5": "Spam post ... isn't good to say everything here.. we discute about mathematics..." } { "Tag": [ "ratio", "geometric series" ], "Problem": "I am wondering if there is a quick way to sum an exponential series of base b from 0 to n.\r\n\r\nb^0 + b^1 + b^2 + ... + b^n\r\n\r\nA few things:\r\n\r\nWhen b=1, the sum is n (e.g., 1+1+1=3)\r\nWhen b=2, the sum is 2^n-1 (e.g., 1+2=3=2^2-1)\r\nWhen b=3, the sum is [3^(n+1)/2], where [x] is the greatest integer not exceeding x (the floor function).\r\n\r\nMaybe I'm getting out of my league here, but I'm curious :)\r\n\r\nThanks in advance.", "Solution_1": "If b != 1 we have :\r\n\r\n :Sigma:(k=0..n) b^k = (b^(n+1)-1)/(b-1)\r\n\r\n\r\nIt can be proved a lot of ways, the simplest being perhaps induction.", "Solution_2": "Wow...\r\n\r\nI didn't think it would be so simple.", "Solution_3": "[quote=\"belenos\"]If b != 1 we have :\n\n :Sigma:(k=0..n) b^k = (b^(n+1)-1)/(b-1)\n\n\nIt can be proved a lot of ways, the simplest being perhaps induction.[/quote]\r\n\r\nNote that this is just using the formula for geometric series, with 1st term (t_0) equal to 1 and common ratio b.\r\n\r\nA geometric series is a series of the form\r\nt_0, t_0*r, t_0*r^2, t_0*r^3, ...\r\n\r\nIn this case, t_0 is the first term and r is the ratio between any two consecutive terms.\r\n\r\nNow, the sum of this series t_0, t_0*r, t_0*r^2, t_0*r^3, ... is simply t_0*(1-r^(n+1)/(1-r) if r is not 1, or n*t_1 if r is 1.\r\n\r\nPlugging in t_0 = 1 and r = b into this yields the formula that belenos gave.", "Solution_4": "For a finite sum, the nicest way to prove it is probably this:\r\n\r\nLook at 1 + b + b^2 + ... + b^n = S (what you are looking for), and compare it to b + b^2 + ... + b^(n + 1) = b*S. Do you see how to go on from there?" } { "Tag": [ "limit", "factorial", "algebra", "binomial theorem", "binomial coefficients" ], "Problem": "Find $1.01^{100}$ to the nearest tenth[i] without using a calculator[/i].\r\n\r\n[hide=\"read story behind it\"]This is a great example of the usefulness of problem-solving in everyday life. My dad asked me to go on the computer and calculate that because he needed it to figure out something regarding stock. Instead, I thought for a moment and used problem-solving :yup: [/hide]", "Solution_1": "[hide=\"Hint\"] Find a pattern. Do $1.01^2$, $1.01^3$, etc. [/hide]\r\nThat might not help, actually. You might as well use the binomial theorem.", "Solution_2": "[quote=\"lotrgreengrapes7926\"][hide=\"Hint\"] Find a pattern. Do $1.01^2$, $1.01^3$, etc. [/hide]\nThat might not help, actually. You might as well use the binomial theorem.[/quote]\r\n\r\nNope, see if you can find a much slicker way ;)", "Solution_3": "[hide]2.7? It's very close to $e$...[/hide]", "Solution_4": "[hide]\n$e=\\displaystyle\\lim_{x\\to\\infty}(1+\\frac{1}{x})^x$ which will give you infinite digits of e.\n\n$(1+\\frac{1}{100})^{100}=2.7...$ (two digits of $e$)\n\nI double-checked with a calculator and for each 10 you multiply to the denominator and the exponent, you get another digit of $e$.[/hide]\r\n\r\nedit:\r\neryaman beat me. ;)", "Solution_5": "[quote=\"calc rulz\"][quote=\"lotrgreengrapes7926\"][hide=\"Hint\"] Find a pattern. Do $1.01^2$, $1.01^3$, etc. [/hide]\nThat might not help, actually. You might as well use the binomial theorem.[/quote]\n\nNope, see if you can find a much slicker way ;)[/quote]\r\nActually, using Binomial Theorem is not that bad.\r\nIf you didn't know $\\lim_{x\\to \\infty} \\left(1+\\frac{1}{x}\\right)^x =e$, then this would be one of the best approaches.", "Solution_6": "[quote=\"frt\"]Actually, using Binomial Theorem is not that bad.\nIf you didn't know $\\lim_{x\\to \\infty} \\left(1+\\frac{1}{x}\\right)^x =e$, then this would be one of the best approaches.[/quote]\r\nOh. :rotfl:", "Solution_7": "Wouldn't binomial theorem depend on a large factorial (without a calculator)?", "Solution_8": "[quote=\"surge\"]Wouldn't binomial theorem depend on $100!$ (without a calculator)?[/quote]Well, for the first couple of values most of it cancels out with the denominator...makes it easier but it is still kind of tedious.", "Solution_9": "[quote=\"surge\"]Wouldn't binomial theorem depend on a large factorial (without a calculator)?[/quote]\r\n\r\nI agree... you'd still have to calculate around 50 binomial coefficients, and it would be a lot of multiplication to do without a calculator.", "Solution_10": "Wouldn't you need only a few binomial terms to get an accurate enough answer?", "Solution_11": "Yeah, you only need to calculate the first 4 or 5 terms.\r\n(Remember, we only need to approximate it to the nearest tenth.)\r\n\r\nAfter the first 5 terms, each term becomes so small that they have no effect on the tenth (or hundredth) term.", "Solution_12": "oh wow, I didn't realize it was to the nearest 10th... that makes it way easier... :blush: :blush:" } { "Tag": [], "Problem": "$10$ friends rented $3$ apartments, each with maximum occupation of $4$ people. \r\n\r\n[b]1[/b]. In how many ways the friends could organize themselves, such that:\r\n\r\n[b]a)[/b] $2$ apartments have $4$people each and the other only $2$ people.\r\n\r\n[b]b)[/b] $2$ apartments have $3$people each and the other $4$ people.\r\n\r\n[b]\n2.[/b] Assume that exist $2$ antagonist people who cannot stay in the same group. How many ways could we organize the group of friends?\r\n\r\nThanks in advance. :lol:", "Solution_1": "uhh... there can't be 4 people in a group of three.", "Solution_2": "[quote=\"nutz_for2.718281828\"]uhh... there can't be 4 people in a group of three.[/quote]\r\n\r\nSorry, you are right. I just corrected the number of people." } { "Tag": [ "puzzles" ], "Problem": "A fun little puzzle I found in a book of mine.\r\n\r\nMy house number is a positive integer.\r\nThese three statements concerning my house number are true:\r\n\r\nIf it is a multiple of 3, then it is a number from 50 to 59 inclusive.\r\nIf it is not a multiple of 4, then it is a number from 60 to 69 inclusive.\r\nIf it is not a multiple of 6, then it is a number from 70 to 79 inclusive. \r\n\r\nWhat is my house number?", "Solution_1": "[hide][size=200]76[/size]\n\nSize 29! :D[/hide]", "Solution_2": "[hide]It would be 76. If the number was divisible by 3, it would have to be either 51, 54, or 57. It can't be 51 or 57, because if it isn't a multiple of 6, it has to be between 70 and 79. It isn't 54 because if it isn't a multiple of 4, it must be between 60 and 69. If it isn't a multiple of 4, then it must be between 60 and 69. If it is divisible by 3, it must be between 50 and 59, so 60, 63, 66, and 69 are out. If it isn't divisible by 3, it isn't divisible by 6 either, and it must be between 70 and 79. That means that the number has to be between 70 and 79, but not a multiple of 6 or 3, and it has to be a multiple of 4. That means that the only number possible is 76[/hide]" } { "Tag": [], "Problem": "In triangle $ ABC$, $ AB$ is congruent to $ AC$, the measure of angle $ ABC$ is $ 72^{\\circ}$ and segment $ BD$ bisects angle $ ABC$ with point $ D$ on side $ AC$. If point $ E$ is on side $ BC$ such that segment $ DE$ is parallel to side $ AB$, and point $ F$ is on side $ AC$ such that segment $ EF$ is parallel to segment $ BD$, how many isosceles triangles are in the figure shown?\n\n[asy]draw((0,0)--(5,15)--(10,0)--cycle,linewidth(1));\ndraw((0,0)--(8,6)--(6.5,0)--(9.25,2.25),linewidth(1));\nlabel(\"B\",(0,0),W);\nlabel(\"A\",(5,15),N);\nlabel(\"D\",(8,6),E);\nlabel(\"E\",(7,0),S);\nlabel(\"F\",(9,3),E);\nlabel(\"C\",(10,0),E);[/asy]", "Solution_1": "[quote=\"GameBot\"]In triangle $ ABC$, $ AB$ is congruent to $ AC$, the measure of angle $ ABC$ is $ 72^{\\circ}$ and segment $ BD$ bisects angle $ ABC$ with point $ D$ on side $ AC$. If point $ E$ is on side $ BC$ such that segment $ DE$ is parallel to side $ AB$, and point $ F$ is on side $ AC$ such that segment $ EF$ is parallel to segment $ BD$, how many isosceles triangles are in the figure shown?\n\n[asy]draw((0,0)--(5,15)--(10,0)--cycle,linewidth(1));\ndraw((0,0)--(8,6)--(6.5,0)--(9.25,2.25),linewidth(1));\nlabel(\"B\",(0,0),W);\nlabel(\"A\",(5,15),N);\nlabel(\"D\",(8,6),E);\nlabel(\"E\",(7,0),S);\nlabel(\"F\",(9,3),E);\nlabel(\"C\",(10,0),E);[/asy][/quote]\r\n\r\nABC, BED, DEF, FEC, DEC, BDC, ABD.\r\n\r\n7 triangles." } { "Tag": [], "Problem": "Prove that there is a power of 2 who's last 1000 digits are composed entirely of 1's and 2's.\r\n\r\nhelp me on this one :blush: \r\n\r\nplease?", "Solution_1": "Anyone? Any help would be appreciated..." } { "Tag": [ "probability" ], "Problem": "Two different integers are randomly selected from the set of integers greater than 0 and less than 10 . What is the probability that their they have no common prime factors? Express your answer as a common fraction. \r\n\r\nThe answer and solution that I don't agree with\r\n[hide]There are 9*(9-1)= 72 possibilities.\nThe good ones are listed below:\n{{1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {1,7}, {1,8}, {1,9}, {2,1}, {2,3}, {2,5}, {2,7}, {2,9}, {3,1}, {3,2}, {3,4}, {3,5}, {3,7}, {3,8}, {4,1}, {4,3}, {4,5}, {4,7}, {4,9}, {5,1}, {5,2}, {5,3}, {5,4}, {5,6}, {5,7}, {5,8}, {5,9}, {6,1}, {6,5}, {6,7}, {7,1}, {7,2}, {7,3}, {7,4}, {7,5}, {7,6}, {7,8}, {7,9}, {8,1}, {8,3}, {8,5}, {8,7}, {8,9}, {9,1}, {9,2}, {9,4}, {9,5}, {9,7}, {9,8}}. \nThere are 54 of them. So the answer is: 54/72=3/4 if simplification is needed. [/hide]\n\n[hide=\"I got\"]$\\frac{5}{6}$ which I'm not sure if it's right[/hide]", "Solution_1": "[hide]\n9C2 or 36 ways to choose 2 different integers.\n1 works with anything so 8 possibilities.\n2 works with the odd numbers (don't count 1 as we already did it) so 4 more possibilities.\n3 works with 4, 5, 7, and 8 so 4 more possibilities\n4 works with 5, 7, 9 so 3 more possibilities\n5 works with 6, 7, 8, and 9 so 4 more possibilities\n6 works with 7 so 1 more possibility\n7 works with 8 and 9 so 2 more possibilities\n8 works with 9 so 1 more possibility\n8+4+4+3+4+1+2+1 = 27\nAnswer is $\\frac{27}{36}$ or $\\boxed{\\frac{3}{4}}$.[/hide]\n[hide=\"What I think you did wrong\"]\nYou might've counted 4 and 6, 6 and 8, and 6 and 9.[/hide]", "Solution_2": "Actually I worked backwards and counted the pairs that did have a common prime factor but I only got 6 pairs\r\n2,4\r\n2,6\r\n2,8\r\n3,6\r\n3,9\r\n4,8\r\n\r\nbut I forgot 3 you said\r\n\r\n4,6\r\n6,8\r\n6,9\r\n\r\nthanks", "Solution_3": "Since those numbers are small, its easier. Try numbers from 30 to 40. \r\n\r\n[hide=\"hint to solution\"] How many have two in common, three, five? [/hide]", "Solution_4": "[quote=\"Ignite168\"][hide]\n9C2 or 36 ways to choose 2 different integers.\n1 works with anything so 8 possibilities.\n2 works with the odd numbers (don't count 1 as we already did it) so 4 more possibilities.\n3 works with 4, 5, 7, and 8 so 4 more possibilities\n4 works with 5, 7, 9 so 3 more possibilities\n5 works with 6, 7, 8, and 9 so 4 more possibilities\n6 works with 7 so 1 more possibility\n7 works with 8 and 9 so 2 more possibilities\n8 works with 9 so 1 more possibility\n8+4+4+3+4+1+2+1 = 27\nAnswer is $\\frac{27}{36}$ or $\\boxed{\\frac{3}{4}}$.[/hide]\n[hide=\"What I think you did wrong\"]\nYou might've counted 4 and 6, 6 and 8, and 6 and 9.[/hide][/quote]\r\n\r\nYou mean $\\frac{3}{4}$ right", "Solution_5": "[quote=\"mathgeniuse^ln(x)\"][quote=\"Ignite168\"][hide]\n9C2 or 36 ways to choose 2 different integers.\n1 works with anything so 8 possibilities.\n2 works with the odd numbers (don't count 1 as we already did it) so 4 more possibilities.\n3 works with 4, 5, 7, and 8 so 4 more possibilities\n4 works with 5, 7, 9 so 3 more possibilities\n5 works with 6, 7, 8, and 9 so 4 more possibilities\n6 works with 7 so 1 more possibility\n7 works with 8 and 9 so 2 more possibilities\n8 works with 9 so 1 more possibility\n8+4+4+3+4+1+2+1 = 27\nAnswer is $\\frac{27}{36}$ or $\\boxed{\\frac{3}{4}}$.[/hide]\n[hide=\"What I think you did wrong\"]\nYou might've counted 4 and 6, 6 and 8, and 6 and 9.[/hide][/quote]\n\nYou mean $\\frac{3}{4}$ right[/quote]\r\n\r\nThat's what I wrote? :?", "Solution_6": "Oh, right, on my computer I only saw a box, not a number itself, sorry." } { "Tag": [], "Problem": "Fie A o matrice patratica cu elemente intregi si pentru care exista k natural nenul cu $A^k=I_n$. Ce valori poate lua cel mai mare divizor comun al elementelor matricii $A-I$?", "Solution_1": "Obviously, I can't fully understand the whole statement, but I think I got idea.\r\nI don't see where you wrote $A\\neq I$, otherwise the answer is clear...", "Solution_2": "ahh... ce mirat am fost cand am cautat un post si am vazut \"O problema frumoasa-pentru Kuba, de la Vali Vitejie\" I was like :o:o\r\n:))\r\ndin pacate doar astazi [acuma] am dat peste ea :) mersi harazi :)", "Solution_3": "am exclus $A=I_n$.. dar.. nu am reusit mai mult decat 2 :? (care oricum era destul de evident pt $A=-I_n$..." } { "Tag": [ "linear algebra", "matrix", "group theory", "abstract algebra", "linear algebra unsolved" ], "Problem": "$A \\in GL_n(\\frac{\\mathbb{Z}}{p \\mathbb{Z}})$, and $n \\leq p$.\r\nProve that $ord(A) \\neq p^2$, where $ord(A)$ is the least integer $k \\geq 2$ such that $A^k=Id$.", "Solution_1": "The exact same problem was posted and solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53879]here[/url].", "Solution_2": "I found yesterday a weird proof, but I prefer grobber's simple one.\r\nAnyway, here's how it goes:\r\n\r\nlemma:\r\nevery upper triangular matrix $T$ with $1$ on the main diagonal \r\nsatisfies: $T^p=1$.\r\nProof:\r\nOne can work in an algebraic closure of $\\mathbb{F}_p$. In this extension,\r\none can suppose that $T$ is in its Jordan form. Each jordan cell is of the \r\nform $M=I+J$. Conclusion follows because $(I+J)^p=I$ if $n \\leq p$.\r\n\r\nNow, consider a matrix $A$, such that $A^{p^2}=I$. Then $A$ generates a subgroup $H$\r\nof $GL_n(F_p)$ of order $p^2$. A classical theorem says that $H$ is a subgoup of \r\na $p$-sylow $S$ of $GL_n(F_p)$. But every Sylow subgoups are conjugated and the subgroup \r\n$U$ of upper triangular matrices with $1$ on the main diagonal is a Sylow subgroup. Hence \r\n$A$ is similar to a matrix $T \\in U$. But $T^p=Id$, so $A^p=I$. Hence $ord(A) \\neq p^2$." } { "Tag": [ "search", "complex numbers", "superior algebra", "superior algebra solved" ], "Problem": "Hi, \r\n\r\nis the additive group of the complex numbers isomorphic with the additive group of the reals?\r\n\r\nI have no idea, I find this a weird problem?\r\n\r\nThx", "Solution_1": "I think they are. Just regard them as vectorial spaces over $Q$. They have the same dimension thus they are isomorphic but this easily implies they are isomorphic as groups.", "Solution_2": "harazi: They are both of infinte dimensions over Q? I don't believe that property implies isomorphism.\r\n\r\nThey are indeed isomorphic, but concrete relation cannot be constructed, since the proof of isomorphism relies on the axiom of choice.", "Solution_3": "I did not say that the fact they have infinite dimension over $Q$ implies isomorphism. Read more carefully the post and you'll se I said the same dimension, which is the cardinality of R in fact. I know it uses axiom of choice, but if you want to find another solution without using it, so be it. I think mathematics is more interesting if we assume the axiom of choice. ;)", "Solution_4": "I think I didn't communicate clearly, but I meant that since the proof of isomorphism relies on the axiom of choice, if we were able to construct a direct relation, then it would show the axiom is redundant. So even though they are isomorphic, it maybe a waste of time to try to search for a mapping. I too like that axiom.\r\nSorry T_______________T" } { "Tag": [ "ARML", "AMC", "AMC 12", "USA(J)MO", "USAMO" ], "Problem": "Our school is considering getting up a group to go to 2004 National MAO Convention. How does National MAO rate with respect to other renown competitions such as AMC or ARML,in terms of organization, quality of problems, etc.?", "Solution_1": "what's national MAO.\r\nI know it's a math contest but specifically what is it?", "Solution_2": "It's Mu Alpha Theta \r\n\r\ntheir website is http://www.mualphatheta.org/", "Solution_3": "[quote=\"msmswatson\"]Our school is considering getting up a group to go to 2004 National MAO Convention. How does National MAO rate with respect to other renown competitions such as AMC or ARML,in terms of organization, quality of problems, etc.?[/quote]\r\n\r\nI have attended 8 national MA :theta: conventions as a student and organizer. The quality of the problems varies because there are different test writers, though the test-writing has become more organized and more centralized for most recent conventions. The problems are not as difficult as AMC12 problems. Basically, if you took the first 15 problems on the AMC12 and extended that level of difficulty to x-problem tests, you would have something close to MA:theta:. Of course, the quality even varies from test to test since there are different test writers (who have to come up with thousands of problems overall).\r\n\r\nI personally recommend the convention for students who are anywhere below USAMO qualification level. The convention is also a very social environment and I think it's good for students to try to meet as many of their peers nationally as possible.\r\n\r\nI enjoy going to the conventions and will probably be in Huntsville this summer for the convention.", "Solution_4": "[quote=\"MCrawford\"]\nI enjoy going to the conventions and will probably be in Huntsville this summer for the convention.[/quote]\r\n\r\nIs that Huntsville, Texas? or another huntsville?", "Solution_5": "Huntsville, Alabama" } { "Tag": [ "function", "real analysis", "real analysis unsolved" ], "Problem": "Prove that the class of functions mapping $[0, 2\\pi]$ to $\\mathbb{C}$ with finite Holder norm is complete under the norm: Holder norm + sup norm.", "Solution_1": "[quote=\"mli\"]Prove that the class of functions mapping $[0, 2\\pi]$ to $\\mathbb{C}$ with finite Holder norm is complete under the norm: Holder norm + sup norm.[/quote]For a Cauchy sequence in $C^\\alpha$, find a subsequence that converges under the sup norm. Then show that the limit is Hoelder and the convergence occurs in the Hoelder norm." } { "Tag": [ "videos", "Support" ], "Problem": "I can't hear any of the videos...", "Solution_1": "Make sure your sound is turned up and the sound on the video is turned up. \r\n\r\nCan you hear sound on videos from YouTube and other sources?\r\n\r\nAnyone else having sound problems?", "Solution_2": "I doubt this has something to do with the vids, but I could only hear sounds when I watch vids. Opposite..." } { "Tag": [ "combinatorics proposed", "combinatorics" ], "Problem": "Consider the tiling of the plane by hexagonal cells. For $n\\in\\mathbb{N}$, denote by $T_n$ the triangular-shaped board of side $n$, i.e. containing $\\frac{n(n+1)}2$ cells.\r\nWe are interested in non-overlapping tilings of $T_n$ by tiles consisting of three adjacent cells. As those tiles can have three different shapes, let's call $I$-tiles thoses consisting of three hexagons 'in a row', $\\Delta$-tiles those of shape $T_2$ and $U$-tiles those of u-shape (like a hydrogen molecule if you want). Prove:\r\n\r\n(1) There is no tiling of $T_n$ consisting only of $I$-tiles.\r\n(2) There is no tiling of $T_n$ consisting only of $\\Delta$-tiles.\r\n(3) There is no tiling of $T_n$ consisting only of $U$-tiles.", "Solution_1": "I remember Thurston wrote an article called \"Conway's Tiling Groups\", which showed a technique that solved these kinds of problems.\r\n\r\nIn searching for this article, I came across these reference:\r\n[url=http://faculty.plattsburgh.edu/don.west/tiling/]faculty.plattsburgh.edu/don.west/tiling/[/url]\r\n[url=http://www.claymath.org/library/senior_scholars/stanley_ardila_tilings.pdf]www.claymath.org/library/senior_scholars/stanley_ardila_tilings.pdf[/url]\r\n\r\nEdit: Two more references:\r\n[url=http://www.math.ubc.ca/~elissa/tilinggrps/Tilinggrps.pdf]www.math.ubc.ca/~elissa/tilinggrps/Tilinggrps.pdf[/url]\r\n[url=http://www.mat.puc-rio.br/~nicolau/publ/papers/resenhas.pdf]www.mat.puc-rio.br/~nicolau/publ/papers/resenhas.pdf[/url]" } { "Tag": [ "trigonometry", "quadratics", "algebra", "polynomial", "analytic geometry", "graphing lines", "slope" ], "Problem": "1)Find the sum of all values of x that satisfy the following equation $2 \\tan^{-1}x-\\tan^{-1}(\\frac{1}{2})=\\frac{\\pi}{4}$\r\n\r\n2)A regular n-gon has 350 diagonals. Find n.\r\n\r\n3)Find $\\sum_{k=1}^{\\infty} \\frac{k}{5^k}$\r\n\r\n4) ABCD is a quadrilateral with right angles at A and C. Points E and F are on AC. Also DE and BF are perpendicular to AC. AE=3, DE=5, and CE=7, then find BF. \r\n\r\n5) Find all roots to $(x+5)(x+6)(x+7)(x+8)=-1$", "Solution_1": "2:\n\n[hide]\n\nthe number of diagonals in an n-gon is C[n,2] - n.\n\nC[n,2]=n(n-1)/2-n=700\n\nsimplifying into a quadratic, and solving for the positive value for n gives n=28.\n\n[/hide]\n\n\n\n3:\n\n[hide]\n\nS=1/5 + 2/25 + 3/125...\n\n5S = 1 + 2/5 + 3/25 ...\n\n\n\n 5S - S = 1 + 1/5 + 1/25 .. = 5/4\n\n 4S = 5/4\n\n S = 5/16\n\n[/hide]", "Solution_2": "1.\n\n[hide]\n\nLet arctan x be :theta: and let arctan 1/2 be :beta:.\n\n\n\ntan :theta: = x, so tan 2:theta: = (2x)/(1-x^2)\n\n\n\nTaking the tangent of both sides of the given equation... we get\n\n\n\ntan(2:theta - :beta:) = 1\n\n(tan 2:theta: - tan :beta:)/(1 + tan :theta: tan :beta:) = 1\n\n\n\ntan 2:theta: - tan :beta: = 1 + tan :theta: tan :beta:\n\n(2x)/(1-x^2) - 1/2 = 1 + 1/(1-x^2)\n\n\n\nSimplifying results in a polynomial whose roots sum to 3.\n\n\n\ni bet i made a mistake.\n\n\n\n[/hide]", "Solution_3": "Need help with 4 (I previously said 5 by accident)\n\n[hide]CD is sqrt74, AD is sqrt34, and those are the only sides I've found. ABCD is cyclic, so I drew a circle around it. But I'm not too sure where to go from here. I thought maybe Ptolemy would help, but I get the equation asqrt74 + bsqrt34 = 10e, where a is side AB, b is side BC, and e is DB. I used some trig on the right triangles to find all of the angles. Also, minor arc BC is 62, minor arc CD is 118, minor arc DA is 72, and minor arc BA is 108. Is there some trig identity I don't know about that would help me here? If only I knew what FC and BC were...\n\n\n\nHmm since all the angles in triangles BFC and DEC are the same, they are similar. Also, ABF and DEA are similar.[hide][/hide][/hide]", "Solution_4": "5: Set x+6.5=a. Then you get a biquadratic equation in a :D :D :D .\r\n\r\nMisha", "Solution_5": "For 5: Hint: Producut of 4 consecutive integers +1 is always a square..\n\n\n\n[hide]That is the expression is (x^2+13x+41)^2 = 0\n\n\n\nHow do you see that right away:\n\n\n\nthink of (x+5)(x+8) as (x^2+13x+40)\n\nand (x+6)(x+7) as (x^2+13x +42) \n\n\n\nThink of 40 as (41-1) and 42 as (41+1) and the expression of x^2+13x+41 as y\n\n\n\nso the expression now is (y-1)(y+1) +1 = 0 or y^2=0 or y=0\n\nYou can solve x^2+13x+41=0 easily. (x= (-13+- sqrt(5))/2 etc..)\n\n [/hide]", "Solution_6": "mathkid wrote:1.\n[hide]\nLet arctan x be :theta: and let arctan 1/2 be :beta:.\n\ntan :theta: = x, so tan 2:theta: = (2x)/(1-x^2)\n\nTaking the tangent of both sides of the given equation... we get\n\ntan(2:theta - :beta:) = 1\n(tan 2:theta: - tan :beta:)/(1 + tan :theta: tan :beta:) = 1\n\ntan 2:theta: - tan :beta: = 1 + tan :theta: tan :beta:\n(2x)/(1-x^2) - 1/2 = 1 + 1/(1-x^2)\n\nSimplifying results in a polynomial whose roots sum to 3.\n\ni bet i made a mistake.\n\n[/hide]\n\n\n\nBe careful....", "Solution_7": "[quote]Need help with 4 (I previously said 5 by accident) [/quote]\r\n\r\nNF - You are almost there.\r\nSay assume AC is X asis and ED is y-axis You know cooridnates of A (3,0), D and C now.. you know slope (and thus equation) of line AB, and and also of line BC.. just find out where they meet (actually you need the y value only).\r\n\r\nHope this helps.", "Solution_8": "[quote=\"Gyan\"][quote]Need help with 4 (I previously said 5 by accident) [/quote]\n\nNF - You are almost there.\nSay assume AC is X asis and ED is y-axis You know cooridnates of A (3,0), D and C now.. you know slope (and thus equation) of line AB, and and also of line BC.. just find out where they meet (actually you need the y value only).\n\nHope this helps.[/quote]\r\n\r\nGood, but can you do this without coordinates?", "Solution_9": "Isn't 1 out of AHSIMC?", "Solution_10": "True... in fact first 3 problems are from AHSMIC, for the #2, #3 I changed the numbers", "Solution_11": "[quote]Good, but can you do this without coordinates?[/quote]\r\nSorry did not see it but answer is obviously yes. Once you know that the lines are 'known' rest is easy. You can see similar traingles ( the two right triangles on one side of AC are similar to the ones on the other side of AC..) multiply two numbers and diivid by third - some thing like that ((7*3)/5) - Hope that helps.", "Solution_12": "beta wrote:1)Find the sum of all values of x that satisfy the following equation \n\n2)A regular n-gon has 350 diagonals. Find n.\n\n3)Find \n\n4) ABCD is a quadrilateral with right angles at A and C. Points E and F are on AC. Also DE and BF are perpendicular to AC. AE=3, DE=5, and CE=7, then find BF. \n\n5) Find all roots to \n\n\n\n[hide]2. From any of the n vertices, we can choose n-3 other points to form a diagonal (we exclude the chosen vertex and the two adjacent ones, which when connected form the sides). Since we counted each diagonal twice - once from each vertex, the total number of vertices is n(n-3)/2 = 350\n\n\n\nn^2 - 3n = 700\n\nn^2 - 3n - 700 = 0\n\n(n-28)(n+25) = 0\n\nn = 28, -25\n\n\n\nBut n > 2, so 28.\n\n\n\n3. 1/5 + 2/25 + 3/125 + ... = (1/5 + 1/25 + 1/125 + ...) + (1/25 + 1/125 + ...) + (1/125 + 1/625 + ...) + ... = 1/4 + 1/20 + 1/100 + ... = 5/16.\n\n\n\n5. (x+5)(x+6)(x+7)(x+8) = -1\n\n[(x+5)(x+8)][(x+6)(x+7)] = -1\n\n[x^2 + 13x + 40][x^2 + 13x + 42] = -1\n\nLet y = x^2 + 13x + 40\n\ny(y+2) = -1\n\ny^2 + 2y + 1 = 0\n\n(y+1)^2 = 0\n\ny = -1\n\n\n\nx^2 + 13x + 40 = -1\n\nx^2 + 13x + 41 = 0\n\nx = [-13 + sqrt(5)]/2 or [-13 - sqrt(5)]/2\n\n\n\n(It makes sense that there are two roots, since in order for (x+5)...(x+8) to be negative, one or three terms must be negative, so either (x+5) < 0, (x+6), ... > 0 or (x+5), ... (x+7) < 0, (x+8) > 0.\n\n\n\nBah, I can't do 4. Uh, yet.[/hide]", "Solution_13": "A big hint for people who are having trouble doing #4:\n\n\n\n[hide]Obviously ABCD is cyclic, so draw the circumcircle of ABCD with center at midpt. BD. And extend DE such that it intersect the circle at another point G.........[/hide]" } { "Tag": [ "MATHCOUNTS" ], "Problem": "In countdown round, what is the fastest amount of time to answer one question. I believe it was 45 seconds by Bobby Shen. \r\n\r\nThis is what I have heard going around on the Today Show with the non-stop talking ladies. My teacher let me watch it during class and we had to mute it because it was getting way too annoying.\r\n\r\n\r\n :idea: thinking is a good thing....", "Solution_1": "I'm assuming you're implying what the least amount of time in which a question in countdown round was answered correctly? If so, then it is WAY less than 45 seconds (this is actually the max :wink: ). \r\nIt even could be less than a second (David Yang) if you don't count the seconds after you press the buzzer.\r\n\r\nAlso, what in the world was your last portion about?\r\n\r\n(AOPS23 is not a scammer -- I traded with him and I got my requested tests: I'm beginning to doubt if he's a multi or not, too -- he's been on a long time [other multis die out very quickly]. My point is: stop being mean to him just because he copied someone a bit and is a bit suspicious -- we'd all be guilty if he actually wasn't a scammer or a multi [sort of like how whites discriminated against black when blacks are awesome])\r\n\r\nThere! I'm done with my speech! lol...", "Solution_2": "Multiple times, people have buzzed in insanely quickly. Of course, this is because many problems are quite easy.", "Solution_3": "[quote=\"myruggy89\"]In countdown round, what is the fastest amount of time to answer one question. I believe it was 45 seconds by Bobby Shen. [/quote]\r\n\r\n\r\n45 seconds is not the fastest time. It is the slowest time. 45 seconds is too much time to do a count down problem.\r\nFast time in national count down is about 2 - 3 seconds. Most people barely finishing reading the problems.", "Solution_4": "[quote=\"tiger21\"][quote=\"myruggy89\"]In countdown round, what is the fastest amount of time to answer one question. I believe it was 45 seconds by Bobby Shen. [/quote]\n\n\n45 seconds is not the fastest time. It is the slowest time. 45 seconds is too much time to do a count down problem.\nFast time in national count down is about 2 - 3 seconds. Most people barely finishing reading the problems.[/quote]\r\n\r\nMax Schindler vs. Bobby Shen (this year) had some pretty quick problems (first 3 especially).", "Solution_5": "thanks Akhil0422!", "Solution_6": "Yes, as you've heard before, people at the National Mathcounts competition are super fast. However, at my chapter(where I actually won the Countdown Round), we pretty much all, me included, stunk. 75% of the time no one even answered, and I ended up getting through some of the rounds by default, or by getting just 1 right So obviously, the chapter competition was NOTHING like nationals, and my goal is to not only go to nationals by 8th grade, but also to make it to the countdown round. Also, how can I improve my reaction time for the problems? I'm always losing the \"For the Win' game.", "Solution_7": "For maybach:\r\n\r\nA.) don't revive an old thread like that...\r\nB.) it kinda depends on your reading speed, so read more books, i guess...", "Solution_8": "[hide=\"A long quote\"]\n[quote=\"AIME15\"]If you've ever had to go against Bobby Shen in a countdown, then you know what it feels like to be in a CD, when the problem is displayed and the buzzer goes off before you can read more than 3 words. \n\nHow does he do it? \n\nGeneral rule: Read the last sentence first. \n\nReading the last sentence tells you exactly what you need to know before you trudge through the problem. Many countdown problems are loaded with useless information that you don't need to know. If you know what you need to know, you can sift through that much faster. \n\nGeneral rule: Learn number tricks. \n\nObviously I can't explain all of them here, but you should be comfortable with multiplying a large number by 11 in under 1.5 seconds, etc. \n\nGeneral rule: Buzz in early. \n\nYou need some practice to get this done correctly; however, when used optimally, allows you to buzz in about 4 seconds before you actually have the answer, and just finish off the arithmetic during this time. This can save you lots of time and can make the difference in a countdown match. Or, if the countdown reader doesn't know the rules, such as in the North California State Countdown Round, you can just buzz in if the question is long. This rule is not a general rule. \n\nGeneral rule: Say your answer correctly. \n\nA countdown moderator will not (or at least, should not) accept an answer of \"One thousand...I meant ten thousand!\". If your answer is \"367887654\", make sure you say it as \"three hundred sixty-seven million, eight hundred eighty-seven thousand, six hundred fifty-four\" and not \"three billion, six hundred seventy-eight thousand, oh shoot, I meant...\" Take the extra 0.25 seconds to make sure you know how to say your answer. \n\nGeneral rule: \"Answer the stinking question.\" \n\nThis is copied from my MATHCOUNTS coach (Mr. Lomas). Many times a problem will ask for \"inches\" when everything is in \"feet\", or things like that. Not answering the question is giving away free points to your opponent. Again, take the extra 0.25 seconds to make sure you're answering the question.[/quote]\n[/hide]" } { "Tag": [ "topology", "real analysis", "real analysis theorems" ], "Problem": "I am unable to follow the proof of the following theorem, could sm1 explain:\r\n\r\nLet $ \\{Q_{1} , Q_{2} ... \\}$ be a countable collection of nonempty sets in $ \\mathbb{R}^{n}$ such that \r\n1) $ Q_{k \\plus{} 1} \\subseteq Q_{k} \\forall k$\r\n2) Each set $ Q_{k}$ is closed and $ Q_{1}$ is bounded.\r\n\r\nthen\r\n$ \\cap_{k \\equal{} 1}^{\\infty} Q_{k}$ is closed and nonempty.", "Solution_1": "Some1 Please answer. \r\nThanks", "Solution_2": "Since $ Q_i$ are closed $ A : \\equal{} \\cap_{i \\equal{} 1}^\\infty Q_i$ is closed because any intersection of closed sets is closed in any topology. It remains to show that $ A \\neq \\emptyset$. Assume the contrary. In the sequel we work in the topology induced on $ Q_1$ by the natural topology on $ \\mathbb R^n$. $ Q_1$ is compact since it is a closed bounded subset of $ \\mathbb R^n$. Taking complements in $ A \\equal{} \\emptyset$ we get $ Q_1 \\equal{} \\cup_{i \\equal{} 1}^\\infty Q_i^c$. The sets $ Q_i^c$ form an open cover of $ Q_1$ so by compactness there must exist a finite subcover. Thus $ Q_1 \\equal{} \\cup_{i \\equal{} 1}^n Q_i^c$ and taking complements once again we see that $ \\emptyset \\equal{} Q_1^c \\equal{} \\cap_{i \\equal{} 1}^n Q_i \\equal{} Q_n$ which is a contradiction.", "Solution_3": "Thanks sir", "Solution_4": "I have not looked at the above solution, so here's my solution for some practice with my definitions. Here's my first formal solution to a topology problem. Please give some constructive criticism. :)\r\n\r\n[i]Proof.[/i] Since $ Q_k$ is closed, $ Q_k^c$ is open, so $ \\bigcap_{k \\equal{} 1}^{\\infty}Q_k \\equal{} \\left(\\bigcup_{k \\equal{} 1}^{\\infty}Q_k^c\\right)^c$ is closed. Since $ Q_1$ is bounded and closed in $ \\mathbb{R}^n$, there exists an n-cell $ S$ such that $ Q_1\\subseteq S$. n-cells are compact, and since $ Q_1$ is closed, $ Q_1$ is compact. Similarly, each $ Q_k$ is compact. Now assume to the contrary that $ Q_1\\cap\\bigcap_{k \\equal{} 2}^{\\infty}Q_k \\equal{} \\emptyset$ even though each finite intersection of the $ Q_i$ is nonempty. Then is follows that $ Q_1\\subseteq \\left(\\bigcap_{k \\equal{} 2}^{\\infty}Q_k\\right)^c \\equal{} \\bigcup_{k \\equal{} 2}^{\\infty}Q_k^c$. Since $ \\{Q_k^c\\}_{k \\equal{} 2}^{\\infty}$ is an open cover of $ Q_1$ and $ Q_1$ is compact, $ Q_1\\subseteq\\bigcup_{k \\equal{} 1}^nQ_{a_k}^c$ for some finite $ \\{Q_{a_k}^c\\}_{k \\equal{} 1}^n$. But this implies that $ Q_1\\cap\\bigcap_{k \\equal{} 1}^nQ_{a_k} \\equal{} \\emptyset$. This is clearly a contradiction. And the conclusion follows. $ \\Box$" } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "Hi, I need help finding the x and y intercepts for problems like these..\r\n\r\n1.Passes through points (-1,5) and (2,-1)\r\n\r\n2.Slope -4 and x-intercept 7.\r\n\r\n3. slope 2/3 and passes through point (5,7)", "Solution_1": "common and easy.\r\n\r\nput them in the highschool section! :)", "Solution_2": "Please don't post the same topic under two sections. Decide which section the topic would be best under and post it there.\r\n\r\nOk, here goes. \r\n\r\n1. First, find the slope using $\\frac{y_2-y_1}{x_2-x_1}=\\frac{-6}{3}=-2$\r\nThen use the point-slope form of a linear equation.\r\n$y-y_1=m(x-x_1)$ where m is the slope, and $y_1, x_1$ are a point that the line passes through.\r\n\r\n$y-5=-2(x+1)$\r\n$y=-2x+3$\r\n\r\nThen, find the intercepts by pluggin in 0 for x and y and finding the value.\r\nWhen $y=0$, we find $x=\\frac{3}{2}$ When $x=0$, $y=3$" } { "Tag": [ "number theory", "prime numbers" ], "Problem": "What is the arithmetic mean of all prime numbers between 20 and 30?", "Solution_1": "The prime numbers between $ 20$ and $ 30$ are $ 23$ and $ 29$. Thus, the arithmetic mean of all prime numbers between $ 20$ and $ 30$ is $ \\frac{23\\plus{}29}{2}\\equal{}\\frac{52}{2}\\equal{}\\boxed{26}$" } { "Tag": [ "symmetry" ], "Problem": "dollar/>find a,b,c wich are the length sides of an right-angled triangle and are verifing the next relation: $ \\frac{1}{a}\\plus{}\\frac{2}{b\\plus{}c}\\equal{}\\frac{6}{a\\plus{}b\\plus{}c}$)", "Solution_1": "$ a \\equal{} 4$, $ b \\equal{} 3$, $ c \\equal{} 5$", "Solution_2": "[quote=\"akech\"]$ a \\equal{} 4$, $ b \\equal{} 3$, $ c \\equal{} 5$[/quote]\r\nHow did you get that?\r\nDid you start off with multiplying $ abc$ on both side?", "Solution_3": "[quote=\"DonkeyKong\"][quote=\"akech\"]$ a \\equal{} 4$, $ b \\equal{} 3$, $ c \\equal{} 5$[/quote]\nHow did you get that?\nDid you start off with multiplying $ abc$ on both side?[/quote]\r\n[hide=\"Solution\"] \nMultiply $ a\\plus{}b\\plus{}c$ on both sides to obtain that \\[ \\frac{a\\plus{}b\\plus{}c}{a}\\plus{}2(\\frac{a\\plus{}b\\plus{}c}{b\\plus{}c})\\equal{}6\\] \\[ 1\\plus{}\\frac{b\\plus{}c}{a}\\plus{}2\\plus{}\\frac{2a}{b\\plus{}c}\\equal{}6\\] \\[ \\frac{b\\plus{}c}{a}\\plus{}\\frac{2a}{b\\plus{}c}\\equal{}3\\] \\[ \\frac{b\\plus{}c}{a}\\equal{}x\\] \\[ x\\plus{}\\frac{2}{x}\\equal{}3\\] \\[ x^2\\plus{}2\\minus{}3x\\equal{}0\\] \\[ (x\\minus{}1)(x\\minus{}2)\\equal{}0\\] \\[ x\\equal{}1, x\\equal{}2\\] \\[ x\\equal{}1\\rightarrow a\\equal{}b\\plus{}c\\] which is impossible since $ b\\plus{}c>a$ in a right triangle. Next, \\[ x\\equal{}2\\rightarrow 2a\\equal{}b\\plus{}c\\] If a is the hypotenuse, then we have that \\[ 4a^2\\equal{}b^2\\plus{}c^2\\plus{}2bc\\rightarrow 4(b^2\\plus{}c^2)\\equal{}(b^2\\plus{}c^2)\\plus{}2bc\\rightarrow b^2\\plus{}c^2\\equal{}\\frac{2bc}{3}nx) -> 1/2 and then use cos2x=2cos2x-1 to get a contradiction.\r\n\r\nIn order for the limit to be 0 we must have the value of 2nx get close to k.pi with k in Z, that is 2nx/pi must be 'almost' integer. Not sure how to finish the proof, but I guess it works... :D", "Solution_4": "This is another good one coming from old times. I remember that it appeared a couple more times on the forum though I couldn't find exactly where. Also Myth posted a very similar (though a bit harder) problem at\r\n[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33765[/url]. Anyway, I don't remember seeing a complete solution and this thread comes quite close. So can someone finish it? :) \r\n\r\nThe problem is to find all $x$ such that the sequence $x_n=\\sin(2^nx)$ converges.\r\n\r\nAgain, if nobody finds a solution in 2 weeks, I'll post mine. If this problem seems too easy, try to solve the version posted by Myth.", "Solution_5": "This one I can solve, but the one posted by Myth is really awful. I will wait 2 weeks to see your proof for Myth's problem, fedja... :(" } { "Tag": [ "probability" ], "Problem": "Which is more likely: rolling a pair of dice $ 100$ times and getting a double six at least once or rolling four dice $ 3600$ times and getting a quadruple six at least once?", "Solution_1": "The probability of getting a double six is $ \\frac16\\times{\\frac16} = \\frac {1}{36}$. Thus, the probability of not getting at least one of these in $ 100$ tries is $ (\\frac {35}{36})^{100}$, which means that the probability of one showing up is $ 1 - ({\\frac {35}{36})^{100}}$. In a similar way the probability of rolling at least one quadruple six is $ 1 - (\\frac {1295}{1296})^{3600}$. After computing this with my TI-89, I found that $ 1 - (\\frac {35}{36})^{100}\\approx.94022$ and $ 1 - (\\frac {1295}{1296})^{3600}\\approx.93789$. Thus, the double six showing up once is more likely." } { "Tag": [ "inequalities" ], "Problem": "For $a,b,c$ are positive numbers. Find minimum of expression:\r\n$A=\\frac{x^{2}}{(2y+3z)(2z+3y)}+\\frac{y^{2}}{(2x+3z)(2z+3x)}+\\frac{z^{2}}{(2x+3y)(2y+3x)}$", "Solution_1": "It's not hard. By the [b]AM-GM Inequality[/b] we have:\r\n$\\sum\\frac{x^{2}}{(2y+3z)(2z+3y)}\\ge\\sum\\frac{x^{2}}{\\left[\\frac{5(y+z)}{2}\\right]^{2}}=\\frac{4}{25}\\sum\\left(\\frac{x}{y+z}\\right)^{2}\\ge\\frac{4}{25}\\cdot\\frac{\\left(\\sum\\frac{x}{y+z}\\right)^{2}}{3}\\ge\\frac{4}{25}\\cdot\\frac{\\left(\\frac{3}{2}\\right)^{2}}{3}=\\frac{3}{25}$" } { "Tag": [ "geometry", "inequalities" ], "Problem": "Now that the Math Jam is over, it's fair game. Here are mine: I'm proud of my solution to #5, but I think I was way too wordy with #1. Feel free to critique as much as you like! I found it interesting that they downplayed geometry in this round - usually there is at least one hard geometry problem (#3 was fairly straightforward), and sometimes two.", "Solution_1": "Nice, looks like you'll get a 25. :) \r\n\r\nI never thought of assuming that m>n>p>q, though. How did you think of that?", "Solution_2": "Posted in solafidefarms' thread.\r\nWow, he beat you to it by 3 minutes...", "Solution_3": "[quote=\"1234567890\"]Nice, looks like you'll get a 25. :) \n\nI never thought of assuming that m>n>p>q, though. How did you think of that?[/quote]\r\nIt's a nice trick with symmetric equations - once you assume some sort of ordering, you can play with inequalities that set bounds on what some of the variables can be. I find that a symmetric equation is always a prompt to at least try ordering the variables, because more often than not some good can come of it.", "Solution_4": "[quote=\"1234567890\"]Nice, looks like you'll get a 25. :) \n\nI never thought of assuming that m>n>p>q, though. How did you think of that?[/quote]\r\n\r\nI don't think you can assume that - although you can assume $m \\ge n \\ge p \\ge q$. In fact the solution (2,2,2,6) does have some equal terms.", "Solution_5": "[quote=\"Philip_Leszczynski\"][quote=\"1234567890\"]Nice, looks like you'll get a 25. :) \n\nI never thought of assuming that m>n>p>q, though. How did you think of that?[/quote]\n\nI don't think you can assume that - although you can assume $m \\ge n \\ge p \\ge q$. In fact the solution (2,2,2,6) does have some equal terms.[/quote]\r\nI think he probably meant that, and merely left out the equalities for speed. It is important to note that distinction, though." } { "Tag": [ "geometry", "perimeter", "MATHCOUNTS" ], "Problem": "In $ \\triangle ABC$, $ D$ is a point on $ \\overline{AB}$ such that $ CD \\equal{} 6$ and $ DB \\equal{} 8$. If $ \\angle CAD \\equal{} \\angle BCD$, how many units are in the perimeter of $ \\triangle ACD$.", "Solution_1": "[geogebra]ef28ceeb4636592176dacbc2fdafb2ddee765746[/geogebra] \r\n\r\nOne possible valid representation:\r\n\r\nCAD ~ BCD\r\nAD = 4.5\r\nAC = 7.5\r\nPerimeter is 18.\r\n\r\nIn mathcounts, since proof is unnecessary I will leave it to here." } { "Tag": [ "inequalities", "algorithm", "algebra", "polynomial" ], "Problem": "Is there some relatively fast algorithm to solve [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118813]this problem[/url] or at least how can I find an algorithm that finds bigger and bigger values of the expression. \r\n\r\nOr in general, if we have been given $n$ integers $A=\\{a_{1},a_{2},\\cdots,a_{n}\\}=\\{1,2,\\cdots,n\\}$, is there an algorithm to maximize the value of $P(b_{1},\\cdots,b_{n})$ where $P$ is a polynomial with respect $b_{1},\\cdots,b_{n}$ with coefficients chosen in the set $A$?", "Solution_1": "If you want progressively larger values, you could try this: http://en.wikipedia.org/wiki/Simulated_annealing" } { "Tag": [ "calculus", "integration", "algebra", "polynomial", "function", "calculus computations" ], "Problem": "The problem is this:\r\n\r\nIntegrate $\\int{\\frac{{x}}{\\sqrt{x-1}}dx}$\r\n\r\nTo evaluate, do we need to use u-subsitution or no? If so, can someone show me how to get started or if not, how would we evaluate it?\r\n\r\nThe second problem is\r\n\r\nIntegrate $2\\pi\\int{x^{2}\\sqrt{x+1}}dx$\r\n\r\nThis I have no idea how to evaulate...\r\n\r\nany help is appreciated!", "Solution_1": "You could do both of these by u-substitutions, to make the thing under the integral a little bit nicer, but the substitution is by no means necessary in either case. For the first one, try either $u = x-1$ or $u = \\sqrt{x-1}$. You'll probably want to express $\\frac 1{\\sqrt{(\\cdot)}}$ as $(\\cdot)^{-\\frac12}$ if you use the first substitution.\r\n\r\nA similar example:\r\nSuppose we wanted to integrate\r\n$\\int (2x-1) \\sqrt{x+2}dx$\r\nMethod 1: I could do a u-substitution, $u = x+2$ to transform this integral to $\\int (2(u-2)-1) \\sqrt u du = \\int 2u\\sqrt u-5 \\sqrt u du = \\int 2u^{\\frac32}-5u^{\\frac{1}{2}}du = \\frac45u^{\\frac52}-\\frac{10}3 u^{\\frac32}+c$ (which I must then convert back to $x$)\r\n\r\nor\r\nMethod 2: I could substitute $u = \\sqrt{x+2}$, $u^{2}-2 = x$, $2udu = dx$ so the integral becomes $\\int (2(u^{2}-2)-1)\\cdot u \\cdot 2udu$ which is then a big polynomial thing which I can easily integrate.", "Solution_2": "Integrate $\\int{\\frac{{x}}{\\sqrt{x-1}}dx}$\r\n\r\nSo if we let $u = x-1$, $du = dx$\r\n\r\nTherefore, would we just integrate the following, and substitute $x-1$ for $u$ when we're done? I don't think that'd work because we still need to get rid of the $x dx$ in the integral..\r\n\r\n$\\int{\\frac{{1}}{\\sqrt{u}}du}$", "Solution_3": "Yeah, you killed off the $x$ in the numerator -- what happened to it? It should have been substituted by $u+1$, right?\r\n\r\n\r\nAlso, something I was going to add at the end of my last post, but you responded to quickly:\r\n\r\nI could also do this by parts, although it is much more work than necessary. The substitution in Method 1 is so simple that you might think of the idea (breaking the $2x-1$ into a sum of powers of $x+2$, which is exactly what the substitution did for us) with without actually making the substitution.", "Solution_4": "True - but even with this substitution, how does it help us? We still have a square root in the denominator.\r\n\r\n$\\int{\\frac{{u+1}}{\\sqrt{u}}du}$", "Solution_5": "Try writing $\\frac 1{\\sqrt{u}}$ as a power of $u$ and then multiplying.", "Solution_6": "I'll do the first one out for you\r\n\r\n$\\int{\\frac{x}{\\sqrt{x-1}}}\\,dx$\r\n\r\nLet $u=x-1$\r\n\r\nThen the integral is transformed to\r\n$\\int{\\frac{u+1}{u^{\\frac{1}{2}}}\\,du}$\r\n\r\nWe can separate the numerator, so the whole integral\r\n$=\\int{u^{\\frac12}}\\,du+\\int{u^{-\\frac{1}{2}}}\\,du$\r\n\r\n$=\\frac{2u^{\\frac32}}{3}+2u^{\\frac12}+constant$\r\n\r\n$=u^{\\frac12}\\left(\\frac{2u}{3}+2\\right)+constant$\r\n\r\nFill in the $u$s again:\r\n\r\n$=\\sqrt{x-1}\\left(\\frac{2(x-1)}{3}+2\\right)+constant$\r\n\r\n$=\\frac23\\sqrt{x-1}(x+2)+constant$\r\n\r\nand thats it...I hope :oops:", "Solution_7": "you can also try to use\r\n\r\n$\\frac{x}{\\sqrt{x-1}}=\\sqrt{x-1}+\\frac{1}{\\sqrt{x-1}}$\r\n :)", "Solution_8": "[quote=\"shyong\"]you can also try to use\n\n$\\frac{x}{\\sqrt{x-1}}=\\sqrt{x-1}+\\frac{1}{\\sqrt{x-1}}$\n :)[/quote]\r\n\r\nnice i never would have thought about that, I would have done it w/ u-sub.", "Solution_9": "In teaching a calculus class, I would always show them the substitution, and do what coffeym did. \r\n\r\nAs for what shyong did: note that he needed to write the numerator of the fraction as $x=(x-1)+1,$ then pull that fraction apart into the sum of two terms. coffeym did the exact same step - only he did so after he had already made the substitution, and for him was much easier to see that $u+1$ should be pulled apart into the sum of two different things. \r\n\r\nDon't think that a substitution will necessarily be the last word, or that it will solve the problem all by itself. In many cases, you perform substitutions - espicially \"little\" substitution where $u$ is a linear function of $x$ and $du$ is a constant times $dx$ - just to streamline the presentation of the problem and to make it easier to see what to do next. And that's exactly what coffeym did.\r\n\r\nThere's nothing wrong with what shyong said. The only problem with it is what maokid7 said: \"i would never have thought of that.\" Doing the substitution first makes that piece of algebra easier to find." } { "Tag": [ "email" ], "Problem": "Here's the story.\r\n\r\nAbout a month ago, there was some discussion on the Etc. forum about an \"AoPS Movie\" or an \"AoPS Story.\" I'm pretty sure that no one was being serious about it, but about two weeks ago I decided to write a little dialogue involving some members of AoPS, inspired by Fierytycoon's \"Interesting Problem: Christmas Holidays\" post.\r\n\r\nToday I posted the second part of my work-in-progress, which was met with mixed reviews. Somebody called it \"lovely.\" somebody called it \"interesting\". But it apparently offended some people, so I had to change it around a little.\r\n\r\nI have now come to the realization that:\r\n1) I need an editor before I post anything to ensure that this doesn't happen again.\r\n2) I need to be more careful and be aware that people may take something in the story too personally\r\nand\r\n3) I need to only include people who specifically say \"I would like my pseudonym to be a part of this.\" Note that I say pseudonym, because you could be [i]anybody[/i] on the internet, and that's fine by me.\r\n\r\nIf you \"would like your pseudonym to be a part of this\", please contact me via PM, replying here, or posting a comment in my supposedly slanderous LiveJournal.\r\n\r\nFrom now on, I will post any announcements regarding The Saga under this thread. I will repost links to Parts I and II, and it is up to The Moderator to decide what to do with the old thread, called \"'The AoPS Saga' book\". However, if it were up to me, I would leave it as it not only provides some insightful comments that will help me, but will also show us what we did wrong in making comments to each other. \r\n\r\nFeel free to comment on the story whenever you like. However, please keep your comments free of any expletives, or anything that hints at expletives (including asterisks, filler words, etc.). In this way, we will create a community of constructive comments and criticism. Hooray for alliteration!\r\n\r\nYour loyal servant and author, \r\nmathfanatic", "Solution_1": "-edited-\r\n\r\nIt's a sad, sad world when things come to this. But as of today, September 1st, 2003, I have chosen to make all previous references to \"The AoPS Saga\" no longer publically viewable. This comes as a result of certain complaints from AoPS Community members about my writings.", "Solution_2": "First off I apologize for using asterisks and segregating gay people.\r\nI think this story is humourous and entertaining. Still, I think the stripper scene is unnecessary because\r\n1.) As mentioned by several people, there are parents reading the material on AoPS as well as things linked to it, and I think some parents would be reluctant to allow their child/ren logging in to AoPS if they see what we wrote, even if it was a joke\r\n2.) Even though you say \"asterisk swearing\" is a bad thing, you used it in your story too, which is a contradiction. Furthermore, you used it in my line, which is another contradiction since you're telling me not to use asterisk swearing.\r\n3.) I have been recommended to keep away from the following topics: religion, sexuality, politics and I suggest you do the same because anyone could be reading your story.\r\n4.) I don't think materials which needs a disclaimer shouldn't be posted.\r\n5.) Keep in mind that this is a \"kids\" site, meaning there are people under 18, maybe even 13 for whom this material is completely inappropriate\r\n6.) This part of the story along with the rest is currently in violation of the Livejournal Term of Service Clause XVI: Member Conduct subordinate 1:\"Upload, post or otherwise transmit any content that is in LiveJournal.com's opinion to be unlawful, harmful, threatening, abusive, harassing, tortious, defamatory, vulgar, obscene, libelous, invasive to another's privacy (up to, but not excluding any address, email, phone number, or any other contact information with out the written consent of the owner of such information), hateful, or racially, ethnically or otherwise objectionable; \" which can result in the enforcement of Clause XV: Journal Content subordinate 2:\"Should any Content be reported to LiveJournal.com as being offensive or inappropriate, LiveJournal.com might call upon the author to retract, modify, or protect (by means of private and friends only settings) the Content in question within a reasonable amount of time, as set forth by the LiveJournal.com staff. Should the author fail to meet such a request from LiveJournal.com staff, LiveJournal.com has the full authority to terminate any such reported and verified offensive account holding such inappropriate content. LiveJournal.com, however, is under no obligation to restrict or monitor journal Content in any way; \"\r\n\r\nP.S. If it's not too much trouble, I candidate for the post of editor for this story.\r\n\r\nRegards,\r\nTare", "Solution_3": "Feel free to include me if you find a way. I know that I haven't been around here for very long though", "Solution_4": "WOW Tare... you were [size=200][b][i][u]REALLY BORED!!![/u][/i][/b][/size]", "Solution_5": "you must be bored replying like that", "Solution_6": "LoL", "Solution_7": "No, I'm not bored. I'm [b]serious[/b].", "Solution_8": "[quote=\"Syntax Error\"]you must be bored replying like that[/quote]\r\n\r\nHey... I only did it like 4 times or so with all the bold and italics and stuff...\r\n\r\nAnd to Tare: sorry.", "Solution_9": "In that case, I have a new, new policy.\r\n\r\nNo more.", "Solution_10": "I completely agree with tare, I would be banned from this site if my parents saw some of the stuff that was going on, and I don't want that to happen. Also, I would be ok with being included in mathfanatics new and improved (and appropriate) story. I'm done now.", "Solution_11": "o bleh...im not in mathfanatic's story...(or am i...i havent read the second chapter yet)...\r\n\r\nHowever, I agree with the cussing discussion...Moderators, lets get rid of it or at least enforce it.", "Solution_12": "I can't enforce it unless you agree to stop the swearing. I can edit it out later, but I would much prefer it if all of you just stopped that.", "Solution_13": "Ftr (for the record), I only sweared (well, wrote asterisks) three times (counting the one forced by mathfanatic) and I've already deleted them. The only swearing involving the AoPS book is the swearing by teratomato, me, and mathfanatic (well, it's me who's saying it, but I didn't want to; he wrote it as [b]my [/b]line...*sigh*) and the only one left is the one by teratomato.\r\nand I stopped swearing when my [color=red] burning anger [/color]cooled off.", "Solution_14": "Yeah guys, is it really that hard to control your swearing on an online forum? Seriously, I mean, its not as if you instintively say it out of anger, as could happen in real life. And yeah, I like talking about girls and dancing and fashion and stuff like that, but when it gets down to stripping, well...that's just not cool.\r\n\r\nWell, even if no one pays any note of my words, at least they'll pay attention to ripply's, simply cause she is ripply.", "Solution_15": "now we've got three peeps out there with an almost identical pear :)", "Solution_16": "lol...I want a peach though!!", "Solution_17": "That can be arranged... :twisted: (but it'd take a while, so be patient; it won't be ready today because of reasons mentioned earlier, but I might get it in tomorrow.)", "Solution_18": "spiffin...that works...", "Solution_19": "I agree with Syntax Error. Please stick to the topic, and if the topic no longer exists, then you can't stick to anything, therefore you shouldn't post here. Especially 2-person conversations. Thank you.", "Solution_20": "nuff said.", "Solution_21": "well, its not exactly a 2-way since we're inviting everyone to join. Personally, I think it's their fault that they havent joined in. Of course it's not completely right for us to just keep posting by ourselves, but nobody's keeping u guys from sayin stuff too...", "Solution_22": ":P Nostalgia. \r\n\r\nHow many 1st/2nd generation AoPSers are still around?", "Solution_23": "Wow, this is one old thread. From before I even joined :o. The funny thing is I can't imagine little freshmen mathfanatic and MithsApprentice...\r\n\r\nI think a lot of the earliest AoPSers are still \"active\" but just not posting with the same frequency that they once did (I fall into this category).\r\n\r\nEDIT: On another note, I think this thread should be locked.", "Solution_24": "I'm totally inactive, but Jeffrey pointed this out to me. scary stuff...the ripply-FyTy joke, not so much, but Tare and I were out of control. (Tare more though!)", "Solution_25": "Math fanatic you're a legend. ", "Solution_26": "What is this?", "Solution_27": "[quote=ns0631]Math fanatic you're a legend.[/quote]\n\nyay 15 year bump", "Solution_28": "/adding to feed", "Solution_29": "[quote=MathNinja234]/adding to feed[/quote]\n\nYou can bookmark it next time instead of posting." } { "Tag": [ "analytic geometry", "graphing lines", "slope" ], "Problem": "For the simultaneous equations \n\\[ 2x\\minus{}3y\\equal{}8\\]\n\\[ 6y\\minus{}4x\\equal{}9\\]\n\n$\\textbf{(A)}\\ x=4,y=0 \\qquad\n\\textbf{(B)}\\ x=0,y=\\dfrac{3}{2}\\qquad\n\\textbf{(C)}\\ x=0,y=0 \\qquad\\\\\n\\textbf{(D)}\\ \\text{There is no solution} \\qquad\n\\textbf{(E)}\\ \\text{There are an infinite number of solutions}$", "Solution_1": "[quote=\"Smartguy\"]For the simultaneous equations\n\\[ 2x \\minus{} 3y \\equal{} 8\n\\]\n\n\\[ 6y \\minus{} 4x \\equal{} 9\n\\]\n(A) $ x \\equal{} 4,y \\equal{} 0$\n(B) $ x \\equal{} 0,y \\equal{} \\frac {3}{2}$\n(C) $ x \\equal{} 0,y \\equal{} 0$\n(D) there is no solution\n(E) there are an infinite number of solutions[/quote]\r\n\r\nThe graphs of the lines have the same slope so there is no solution.", "Solution_2": "Rather, we have 6y - 4x = 9 implies 4x - 6y = -9 implies 2x - 3y = -9/2 = 8, which is FALSE- no solution. D.", "Solution_3": "$ 2x \\minus{} 3y \\equal{} 8$\r\n$ y \\equal{} \\frac {2}{3}x \\minus{} \\frac {8}{3}$\r\n\r\n$ 6y \\minus{} 4x \\equal{} 9$\r\n$ y \\equal{} \\frac {2}{3}x \\plus{} \\frac {3}{2}$\r\n\r\nThus, they are parallel lines and there is no solution or empty set because they have the same slope. D is the correct answer.", "Solution_4": "[hide=\"Solution\"]\nWe rearrange the second equation to get\n$ 2x\\minus{}3y \\equal{} \\frac{\\minus{}9}{2}$, which is not the same as the first equation. Therefore, the answer is $ \\boxed{(D)}$.\n[/hide]", "Solution_5": "To make a rigorously complete solution, there is no solution because the slopes are equal [i]and[/i] the y-intercepts are different." } { "Tag": [ "calculus", "integration", "derivative", "modular arithmetic", "real analysis", "real analysis unsolved" ], "Problem": "Let $m$ and $n$ be two natural numbers. Calculate:\r\n\r\n$\\int_{0}^{\\infty}\\frac{d^{m}}{dx^{m}}(e^{-x^{2}})\\frac{d^{n}}{dx^{n}}(e^{-x^{2}})dx.$", "Solution_1": "Hmm. I can solve the case where $m-n \\equiv 1 \\pmod 2$, but I'm still trying to figure out the case where $m \\equiv n \\pmod 2$." } { "Tag": [ "LaTeX", "trigonometry", "real analysis", "real analysis unsolved" ], "Problem": "I need help in the following problem :\r\nStudy the convergence of the serie associated to the sequence sin(sqrt(n)*(-1)^n/(n^a+ (-1)^n)) where \"a\" is a real (there are some trivial cases about it)\r\nSorry for not using Latex(it's not my computer) , and just to remind you * stands for product and ^ for power ; thank you :)", "Solution_1": "Here it is in LaTeX: \\[(-1)^{n}\\sin\\frac{\\sqrt{n}}{n^{a}+(-1)^{n}}\\] Converges when $a>1/2$ by the alternating series test (may requires some work). I'd say it diverges for $a\\le 1/2$, but do not have a proof.", "Solution_2": "(-1)^n is after the bracket of sin , and with leibniz ciriterion we can only say that if the serie converges a must be > 1/2 , don't mess up things !!:)" } { "Tag": [ "geometry", "circumcircle", "geometry proposed" ], "Problem": "let ABC be a triangle(angle A=90).draw the circumcircleof ABC and name it (C).\r\nAH is perpendicular to BC.circle(C')is tangent to BC in point D.and is tangent to AH and circle(C).\r\nprove that:angle BAD=angle DAH.", "Solution_1": "It isn't clearly ! I think that there are two circles C'.", "Solution_2": "I hope I correctly understood your problem:\r\n\r\n[color=blue][b]Problem.[/b] Let ABC be a triangle with < BAC = 90\u00b0, and let k be the circumcircle of triangle ABC. Let H be the foot of the perpendicular from the point A to the line BC. Let a circle k' touch the segments BH and AH and the circle k, and let D be the point where this circle k' touches the segment BH. Prove that < BAD = < DAH.[/color]\r\n\r\nSince < BAC = 90\u00b0, the segment BC is a diameter of the circle k. On the other hand, the assertion < BAD = < DAH is equivalent to the assertion that the line AD bisects the angle BAH. And thus your problem is equivalent to the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=23606 .\r\n\r\n[b]Ashegh, please try to write your problems in a more readable way.[/b] It is a real pain to read your posts; even if your English is not too good, you can certainly care about putting spaces between words!\r\n\r\n darij", "Solution_3": "I suppose WLOG $H\\in [OB]$ and note $k'=C(I,x)$ and the midpoint $O$ of the side $[BC]$. I apply the generalizationed Pitagora's theorem to the side $[IO]$ in the triangle $OIH$:\r\n${IH=x\\sqrt 2 ,\\ OH=\\frac {b^2-c^2}{2a},\\ OI=\\frac a2 -x,\\ m(\\widehat {IHO})=135^{\\circ}}\\Longrightarrow $\r\n$(a^2-2ax)^2=8a^2x^2+(b^2-c^2)^2+4ax(b^2-c^2)\\Longrightarrow$\r\n\r\n$(b^2+c^2)^2-4a^3x=4a^2x^2+(b^2-c^2)^2+4ax(b^2-c^2)\\Longrightarrow $\r\n\r\n$a^2x^2+2ab^2x-b^2c^2=0\\Longrightarrow x=HD=\\frac {b(a-b)}{a}$.\r\n\r\n$DB=HB-HD=\\frac {c^2}{a} -\\frac {b(a-b)}{a}=a-b\\Longrightarrow \\frac {DB}{DH}=\\frac ab ;$\r\n\r\n$\\frac {AB}{AH}=c\\cdot \\frac {a}{bc}=\\frac ab\\Longrightarrow \\frac {DB}{DH}=\\frac {AB}{AH}\\Longrightarrow \\widehat {BAD}\\equiv \\widehat {HAD}.$\r\n\r\n[b]Generalization (Neculai Roman). [/b]\r\n\r\nLet $k=C(O,R)$ be the circumcircle of the triangle $ABC$ with ${A\\ge 90^{\\circ}}$.\r\nI note the point $H\\in OC,\\ AH\\perp OC$. Let $k'=C(I,r)$ be the circle which\r\nis interior tangent to the circle $k$ and which is tangent to the lines $BC,\\ AH$.\r\nI note $D\\in BC\\cap k'$. Then $\\widehat {BAD}\\equiv \\widehat {HAD}.$\r\n\r\n[b]Indication. [/b]I note the point $E\\in AH\\cap k$. Apply the Cassey's theorem to the degenerate circles $A,\\ C,\\ E$ and the circle $k$.", "Solution_4": "now i can prove it with inversion :D :D" } { "Tag": [ "geometry", "circumcircle", "trigonometry", "geometric transformation", "homothety", "geometry proposed" ], "Problem": "In triangle $ABC$, let $R$ be the circumradius, and $r_{a}$ the exradius opposite $A$. Show that\r\n\r\n$\\frac{r_{a}}{R}=(-\\cos A+\\cos B+\\cos C+1)$", "Solution_1": "Let X, Y, Z be the tangency points of the excircle $(I_{a})$ with BC, CA, AB. Inversion in the excircle $(I_{a})$ carries the triangle vertices A, B, C into the midpoints A', B', C' of the excircle chords YZ, ZX, XY and the circumcircle (O) into the 9-point circle (O') of $\\triangle XYZ.$ Thus the radius of (O') is $R' = 2r_{a}.$ The inversion center $I_{a}$ is the homothety center of (O') and (O). The homothety coefficient of (O'), (O) is equal to the power of the inversion center $I_{a}$ to the inverted circle (O') divided by the power of inversion $r_{a}^{2},$ hence\r\n\r\n$\\frac{r_{a}}{2R}= \\frac{R'}{R}= \\frac{O'I_{a}^{2}-R'^{2}}{r_{a}^{2}}= \\frac{H'I_{a}^{2}-r_{a}^{2}}{4r_{a}^{2}}$\r\n\r\nwhere H' is the orthocenter of the $\\triangle XYZ$ and $2O'I_{a}= H'I_{a}.$ According to a well-known formula, the orthocenter-circumcenter distance $H'I_{a}$ of the $\\triangle XYZ$ is given by\r\n\r\n$H'I_{a}^{2}= 9r_{a}^{2}-(x^{2}+y^{2}+z^{2}) = r_{a}^{2}[9-4 (\\sin^{2}X+\\sin^{2}Y+\\sin^{2}Z)]$\r\n\r\nwhere x = YZ, y = ZX, z = XY and $\\angle Y \\equiv \\angle XYZ =\\frac{ \\angle B}{2},$ $\\angle Z \\equiv \\angle YZX = \\frac{\\angle C}{2}$ (because $XY \\perp CI_{a},\\ YZ \\perp AI_{a},\\ ZX \\perp CI_{a}$) and $\\angle X \\equiv \\angle ZXY = 180^\\circ-(\\angle Y+\\angle Z) = 90^\\circ+\\frac{\\angle A}{2}.$ Therefore,\r\n\r\n${\\frac{r_{a}}{R}= 4-2 \\cos^{2}\\frac{A}{2}-2 \\sin^{2}\\frac{B}{2}}-2 \\sin^{2}\\frac{C}{2}=$\r\n\r\n$= 4-(1+\\cos A)-(1-\\cos B)-(1-\\cos C) = 1-\\cos A+\\cos B+\\cos C$\r\n\r\n[hide=\"Comment\"]The formula $HO^{2}= 9R^{2}-(a^{2}+b^{2}+c^{2})$ is usually proved using medians and centroid (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101687]Orthic triangle[/url], post #6, or Coexter & Greitzer, Geometry Revisited, problem 1.7.2), but I prefer the straightforward use of the cosine theorem:\n\n$HO^{2}= AO^{2}+AH^{2}-2AO \\cdot AH\\ \\cos \\widehat{OAH}=$\n\n$= R^{2}+4R^{2}\\cos^{2}A-4R^{2}\\cos A \\cos (B-C) =$\n\n$= 5R^{2}-4R^{2}\\sin^{2}A+4R^{2}\\cos(B+C) cos (B-C) =$\n\n$= 5R^{2}-4R^{2}\\sin^{2}A+2R^{2}(\\cos 2B+\\cos 2C) =$\n\n$= 9R^{2}-(4R^{2}\\sin^{2}A+4R^{2}\\sin^{2}B+4R^{2}\\sin^{2}C) =$\n\n$= 9R^{2}-(a^{2}+b^{2}+c^{2})$\n\nThere are other ways, for example, if N is the 9-pont circle center, AN is the A-median of $\\triangle AHO,$\n\n$AN^{2}= \\frac{2AO^{2}+2AH^{2}-HO^{2}}{4}$\n\n$HO^{2}= 2AO^{2}+2AH^{2}-4AN^{2}$\n\nIf C' the midpoint of AB and F the foot of the C-altitude, power of A to the 9-point circle (N) is\n\n$AN^{2}-\\frac{R^{2}}{4}= AF \\cdot AC' = \\frac{h_{c}c}{2 \\tan A}= \\frac{|\\triangle ABC|}{\\tan A}$\n\n$4AN^{2}= R^{2}+\\frac{4|\\triangle ABC|}{\\tan A}= 2bc \\cos A = R^{2}+b^{2}+c^{2}-a^{2}$\n\n$HO^{2}= 2R^{2}+8R^{2}\\cos^{2}A-(R^{2}+b^{2}+c^{2}-a^{2}) =$\n\n$= 10R^{2}-2a^{2}-(R^{2}+b^{2}+c^{2}-a^{2}) = 9R^{2}-(a^{2}+b^{2}+c^{2})$[/hide]" } { "Tag": [ "geometry", "3D geometry", "vector", "rotation", "analytic geometry" ], "Problem": "What is the side length of the biggest square that can fit inside a cube of side length n?", "Solution_1": "Yuck. (More to follow later, hopefully.)", "Solution_2": "Okay, so I've gotten one step so far: any square that can fit in can fit in with its center at the center of the cube. Let's just define axes from the center of the square, parallel to the edges of the cube. The square's outermost extents are the same in the +x and -x directions, the +y and -y directions, and the +z and -z directions. Any point in the cube is at least as close to three sides of the cube which make a corner as the center is. These distances from the center of the square determine the maximal extent of the square. If we move it to the center of the cube, these distances all increase.\r\n\r\nI hope that was clear -- it definitely is true, even if I haven't explained it well. Now I \"just\" need to find the longest length for two equal perpendicular vectors emanating from the center of a cube such that their endpoints are still within the cube.", "Solution_3": "n!!!!!!! \r\n\r\nIf the cube side length is n, and a square length could be as long as the cube side length, it would be n.", "Solution_4": "How do you know it can't be longer?", "Solution_5": "I agree with JBL, i have a feeling that it is longer than n, but I have no idea as to what it is.", "Solution_6": "It is definitely longer than n.\r\n\r\nThe original problem asked for the smallest positive integer n so a square of side length n+1 could fit in a cube side length n. Both problems are exactly the same - the hardest thing seems to be able to find what the optimal arrangement of the square is.", "Solution_7": "Does anyone have a resolution to this problem?", "Solution_8": "[hide]\ni'm fairly certain we could put the diagonal of the square along a space diagonal of the cube, and tilt the square so it fits\nthat would give $ \\frac{n\\sqrt{6}}{2}$ as the sidelength\n[/hide]", "Solution_9": "No, definitely not: if you arrange a cube so that its space diagonal is vertical, no matter how you rotate it it will always be substantially taller than it is wide. (Your suggestion could only work if two of the space diagonals of a cube were perpendicular.)", "Solution_10": "It's effectively a lower bound, because it doesn't show whether there are bigger squares:\r\n\r\n[hide]\nside length of the square = $ s$\n\nAs the square is maximum, it must contain some sort of symmetry. Assume it touches 4 edges of the cube. The minimal distance between the corner of the cube and the corner of the square is $ x$. For a cube with length 1, by Pythagoras, we have:\n\n$ (x\\sqrt2)^2 \\plus{} 1 \\equal{} s^2$\n$ 2(1\\minus{}x)^2 \\equal{} s^2$\n\nCompare to get:\n\n$ 2x^2 \\plus{} 1 \\equal{} 2(1\\minus{}x)^2$\n$ x \\equal{} \\frac14$\n\nTherefore, substitute in the first equation to get:\n\n$ (x\\sqrt2)^2 \\plus{} 1 \\equal{} s^2$\n$ 2x^2 \\plus{} 1 \\equal{} s^2$\n$ 1/8 \\plus{} 1 \\equal{} s^2$\n$ s \\equal{} \\frac{3\\sqrt2}4$\n\nThis can be further generalized to $ s \\equal{} \\frac{3n\\sqrt2}4$.\n[/hide]", "Solution_11": "Let's use JBL's idea and try to find two longest orthogonal vectors whose coordinates are less than $ 1$ in absolute value. Suppose that our vectors are $ (a,b,c)$ and $ (x,y,z)$. Then $ ax\\plus{}by\\plus{}cz\\equal{}0$ so two of the three products $ ax,by,cz$ must have the same sign. Thus, without loss of generality, we may assume that $ a,x,b,y$ are positive. Since $ |cz|\\le 1$, we can say that $ ax\\plus{}by\\le 1$. Now we essentially reduced our problem to finding two vectors in the unit square whose scalar product is $ \\le 1$ and maximizing the minimum of their lengths. Note that $ (1.0.5)$ and $ (0.5,1)$ are two such vectors and the following picture shows that if we try increase the length, we must also decrease the angle, thus increasing the scalar product\r\n[asy]size(100);\nD((0,0)--(0,1)--(1,1)--(1,0)--cycle);\nD((0,0));\nD(CR((0,0),sqrt(5)/2,0,90),red);\nD((1,0.5)--(0,0)--(0.5,1),black);[/asy]\r\nSo, $ (0.5,1,1)$ and $ (1,0.5,\\minus{}1)$ is the optimal pair." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "[b]1[/b].$\\ ABC$ is a triangle. Assume that exist $\\ X, Y, Z$ be points on the sides $\\ BC, CA, AB$ respectively such that $\\ AX=BY=CZ$ and $\\ BX=CY=AZ$. Prove that: $\\ ABC$ is an equilateral triangle.\r\n[b]2[/b].$\\ ABC$ is a triangle. $\\ M$ be the point on the incircle of $\\ ABC$, $\\ H, K, L$ respectively the projections of $\\ M$ on $\\ BC, CA, AB$. Determine the positions of $\\ M$ so that $\\ (MH+MK+ML)$ attains its greates value and the least value.", "Solution_1": "Nobody has solution? :D", "Solution_2": "[quote=\"April\"][b]1[/b].$ \\ ABC$ is a triangle. Assume that exist $ \\ X, Y, Z$ be points on the sides $ \\ BC, CA, AB$ respectively such that $ \\ AX \\equal{} BY \\equal{} CZ$ and $ \\ BX \\equal{} CY \\equal{} AZ$. Prove that: $ \\ ABC$ is an equilateral triangle.\n[/quote]\r\n\r\nEdit: go to http://www.mathlinks.ro/Forum/viewtopic.php?p=558585#558585", "Solution_3": "problem 1. - http://www.mathlinks.ro/Forum/viewtopic.php?p=558585#558585" } { "Tag": [ "real analysis", "real analysis solved" ], "Problem": "Prove that there are not existing an infinite real sequence {x_n} that lx_nl \\leq 0,6666666 for every n in N \r\nand lx_n - x_m l \\geq 1/n*(n+1) + 1/m*(m+1) for every m,n in N.", "Solution_1": "Let k = 0,6666666 \r\nconsider seqence :x_1,x_2,..,x_n and each term belongs to [-k,k]\r\nBy pigeon-hole principle we have:\r\n2k/(n-1) \\geq |x_i-x_j| \\geq 1/i*(i+1) + 1/j*(j+1) \\geq 2/n*(n+1)\r\nwhen n gets to infinity we have contradiction!" } { "Tag": [ "calculus", "probability", "probability and stats" ], "Problem": "I'm currently a high school student (10th grade) who's not really taking any really difficult math courses right now. I have a tutor in pre-calculus/calculus purely because I'm interested. We're still on pre-calculus but we're moving at a pretty quick rate and after single-variable calculus I plan on going on to multi-variable,etc. This notes aren't a problem here b/c the tutor basically provides notes via lecture/textbook. \r\n\r\nHowever, I want to teach myself Statistics too and not just the basics, I mean the whole way with the calculus stuff too. The problem is that the only books at my local library are readinsg like \"Statistics De-mystified\" and \"Idiot's Guide to Statistics\". I checked some of them out and these types of readings are really difficult for me to follow because they don't really explain why in-depth but they just list facts. \r\n\r\nStrangely enough, it's much easier for me to follow college-texts and such. I really don't have the money to be spending a fortune even on a used copy. Any ideas on how to go about teaching myself something like this/what to do about texts? \r\n\r\nI just think it's going to be really difficult but it's something I'm truly interested in. I can't afford a tutor right now either.\r\n\r\nEdit: Actually, if there are any accomplished statisticians here who have lectures or outlines and such already prepared and would like to send them to me, that would be amazing.", "Solution_1": "You can't really understand much of statistics if you don't know probability. A good book on probability is \"A First Course on Probability\" by Sheldon Ross. \r\n\r\nHere's an online text on probability, but I can't vouch for it because I've never used it: \r\nhttp://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/amsbook.mac.pdf" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "$ a,b,c \\geq 0$\r\n$ a \\leq b\\plus{}c$\r\n\r\nProve that :\r\n\r\n$ (b\\plus{}c)\\plus{}1\\plus{}|a\\minus{}1|\\geq |b\\minus{}1|\\plus{}|c\\minus{}1|\\plus{}a$", "Solution_1": "You could just divide into cases of values for expressions under absolute braces. There are 8 cases, some impossible because of the condition, and other easily solved.", "Solution_2": "[quote=\"Bugi\"]You could just divide into cases of values for expressions under absolute braces. There are 8 cases, some impossible because of the condition, and other easily solved.[/quote]\r\n\r\ni know this solution ... i want a nice solution for it ...... thanks you :)" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $a,b,c$ be nonegative real numbers, prove that \\[2\\leq\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}\\leq3.\\]", "Solution_1": "q: for non-negative reals a,b,c prove:\r\n2 <= a/((b^2 -bc + c^2)^(1/2)) + b/((c^2 -ca + a^2)^(1/2)) + c/((a^2 -ab + b^2)^(1/2)) <= 3.\r\na: as for the right inequality, take a = 4, b = 1, c = 0 then lhs = 4.5 > 3 = rhs. \r\nthe left inequality's status needs to be checked.\r\n____________________________________________________________", "Solution_2": "[quote=\"pvthuan\"]Let $a,b,c$ be nonegative real numbers, prove that \\[2\\leq\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}.\\] [/quote]\r\nJensen kills it.", "Solution_3": "Can you post your solution, argady? Thank you!", "Solution_4": "q: for non-negative reals a,b,c:\r\n2 <= a/((b^2 - bc + c^2)^(1/2) + b/((c^2 -ca + a^2)^(1/2)) + c/((a^2 -ab + b^2)^(1/2)).\r\na: if say a = 0 then the rhs = b/c + c/b >= 2 by cauchy and \"=\" occurs when b = c. \r\nso consider case a, b, c > 0. then by a fairly well-known inequality\r\n\r\n2 < (a^2/(b^2 + c^2))^(1/2) + (b^2/(c^2 + a^2))^(1/2) + (c^2/(a^2 + b^2))^(1/2) < (a^2/(b^2 - bc + c^2))^(1/2) + (b^2/(c^2 - ca + a^2))^(1/2) + (c^2/(a^2 - ab + b^2))^(1/2) = rhs. and we're done.\r\n\r\n________________________________________________________", "Solution_5": "[quote=\"pvthuan\"]Let $a,b,c$ be nonegative real numbers, prove that \\[2\\leq\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}\\leq3.\\] [/quote]\r\nThe right part is certainly wrong,because \r\n$\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}$ can tends to $\\infty$...\r\nDoes $a,b,c$ should to be three sides of an triangle?", "Solution_6": "[quote=\"ehku\"]Can you post your solution, argady? Thank you![/quote]\n[quote=\"pvthuan\"]Let $a,b,c$ be nonegative real numbers, prove that \\[2\\leq\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}.\\] [/quote]\r\nSince, it's a homogeneous inequality let's assume that $a+b+c=1.$\r\n$f(x)=\\frac{1}{\\sqrt{x}}$ is convex function. Hence, \r\n$\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}\\geq\\frac{1}{\\sqrt{a(b^{2}-bc+c^{2})+b(a^{2}-ac+c^{2})+c(a^{2}-ab+b^{2})}}.$\r\nId est, remain to prove that\r\n$\\frac{1}{\\sqrt{a(b^{2}-bc+c^{2})+b(a^{2}-ac+c^{2})+c(a^{2}-ab+b^{2})}}\\geq2.$\r\nBut $\\frac{1}{\\sqrt{a(b^{2}-bc+c^{2})+b(a^{2}-ac+c^{2})+c(a^{2}-ab+b^{2})}}\\geq2\\Leftrightarrow$\r\n$\\Leftrightarrow1\\geq4\\cdot\\sum_{cyc}(a^{2}b+a^{2}c-abc)\\Leftrightarrow(a+b+c)^{3}\\geq4\\cdot\\sum_{cyc}(a^{2}b+a^{2}c-abc)\\Leftrightarrow$\r\n$\\Leftrightarrow15abc+\\sum_{cyc}(a^{3}-a^{2}b-a^{2}c+abc)\\geq0,$ which true by Schur.", "Solution_7": "[quote=\"pvthuan\"]Let $a,b,c$ be nonegative real numbers, prove that \\[2\\leq\\frac a{\\sqrt{b^{2}-bc+c^{2}}}+\\frac b{\\sqrt{c^{2}-ca+a^{2}}}+\\frac c{\\sqrt{a^{2}-ab+b^{2}}}\\] [/quote]\r\nMy solution\r\n[hide] \\[{2a\\sqrt{b^{2}-bc+c^{2}}}\\le a^{2}+b^{2}+c^{2}-bc \\le a^{2}+b^{2}+c^{2}\\] [/hide]", "Solution_8": "The left part is quite simple. I am wondering about \\[\\frac a{\\sqrt{a^{2}-ab+b^{2}}}+\\frac b{\\sqrt{b^{2}-bc+c^{2}}}+\\frac c{\\sqrt{c^{2}-ca+a^{2}}}\\leq3.\\] What a pity! It is wrong. Try $(0,1,\\tfrac14)$", "Solution_9": "[quote=\"pvthuan\"]The left part is quite simple. I am wondering about \\[\\frac a{\\sqrt{a^{2}-ab+b^{2}}}+\\frac b{\\sqrt{b^{2}-bc+c^{2}}}+\\frac c{\\sqrt{c^{2}-ca+a^{2}}}\\leq3.\\] [/quote]\r\nMr. Thuan, it is true and here is my solution for it.\r\nTo prove it, we will prove the stronger inequality \\[\\sum\\frac{a^{2}}{a^{2}-ab+b^{2}}\\leq 3\\] Or, \\[\\sum\\frac{1}{x^{2}-x+1}\\leq 3\\] with $x,y,z>0$ and $xyz=1$\r\nWe will peove the following Lemma:\r\nLemma. \\[\\sum\\frac{1}{x^{2}+x+1}\\geq 1\\] with $x,y,z>0$and$xyz=1$\r\nProof.\r\nSince $x,y,z>0$and$xyz=1$, then there exists $m,n,p>0$ such that $x=\\frac{np}{m^{2}}, y=\\frac{pm}{n^{2}}, z= \\frac{mn}{p^{2}}$. And our Lemma becomes \\[\\sum\\frac{m^{4}}{m^{4}+m^{2}np+n^{2}p^{2}}\\geq 1\\] Applying Cauchy-swharz Inequality, we have \\[\\sum\\frac{m^{4}}{m^{4}+m^{2}np+n^{2}p^{2}}\\geq \\frac{(m^{2}+n^{2}+p^{2})^{2}}{%Error. \"summ\" is a bad command.\n^{2}n^{2}+%Error. \"summ\" is a bad command.\n^{2}np}\\geq 1\\] Our Lemma is proven. Now, applying this Lemma by replacing $x,y,z$ with $\\frac{1}{x^{2}}, \\frac{1}{y^{2}},\\frac{1}{z^{2}}$, we have\r\n\r\n$\\sum\\frac{x^{4}}{x^{4}+x^{2}+1}\\geq 1$\r\n\r\nOr, \r\n\r\n$\\sum\\frac{2(x^{2}+1)}{x^{4}+x^{2}+1}\\leq 4 (*)$\r\n \r\nNotes that $2(x^{2}+1) =(x^{2}+x+1)+(x^{2}-x+1)$ and $x^{4}+x^{2}+1 =(x^{2}+x+1)(x^{2}-x+1)$, we have \r\n\r\n$(*) \\Leftrightarrow \\sum\\frac{1}{x^{2}-x+1}+\\sum\\frac{1}{x^{2}+x+1}\\leq 4$\r\n\r\nNow, applying the Lemma again, we will have\r\n\r\n$\\sum\\frac{1}{x^{2}-x+1}\\leq 3$\r\n\r\nQED. :)" } { "Tag": [ "inequalities", "function", "LaTeX", "inequalities solved" ], "Problem": "my freind faisl gave me a nice congecture and i find a solution aexp[n+1]\\aexp[n]+bexp(n) +bexp(n+1)\\cexp(n)+bexp(n) +cexp(n+1)\\cexp(n) +bexp(n) >= a+b+c\\2 (E)\r\n assume that we have a+b+c=1 then (E) become a\\(1+(b\\a)exp(n) ) +b\\(1+(b\\c)exp(n) ) +c\\(1+(c\\a)exp(n)) >=0.5 so applying jensen inegualitie to function 1\\(1+(x)exp(n) then we have\r\n\r\n(E)>=1\\(1+(ab\\a+bc\\b+ca\\a)exp(n)=0.5\r\n\r\n\r\n[/hide][/list][/code][/quote][/u][/i]", "Solution_1": "[quote=\"fermat3\"]my freind faisl gave me a nice congecture and i find a solution aexp[n+1]\\aexp[n]+bexp(n) +bexp(n+1)\\cexp(n)+bexp(n) +cexp(n+1)\\cexp(n) +bexp(n) >= a+b+c\\2 (E)\n assume that we have a+b+c=1 then (E) become a\\(1+(b\\a)exp(n) ) +b\\(1+(b\\c)exp(n) ) +c\\(1+(c\\a)exp(n)) >=0.5 so applying jensen inegualitie to function 1\\(1+(x)exp(n) then we have\n\n(E)>=1\\(1+(ab\\a+bc\\b+ca\\a)exp(n)=0.5\n\n\n[/hide][/list][/code][/quote][/u][/i][/quote]\r\n\r\n\r\nuse latex please. nobody will untrstand a word...\r\n :(", "Solution_2": ":stretcher: :what?: \r\nUse Latex please.Thank you.", "Solution_3": "I think he wanted to say that \\[ \\sum \\frac{a^{n+1}}{a^n+b^n} \\geq \\frac{a+b+c}{2} , \\] but I'm not too sure.", "Solution_4": "The function $\\frac1{1+x^n}$ is nor concave nor convex. Jensen doesn't work here :(", "Solution_5": "Well I guess it can't be solved in general (as it's false) as already mentioned [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83149]here.[/url] :P", "Solution_6": "fermat3, why do you dont use latex in all your posts!!!! :mad:" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $a,b,c>0$. Show that:\r\n$\\frac{a^4}{a^4+\\sqrt[3]{(a^6+b^6)(a^3+c^3)^2}}+\\frac{b^4}{b^4+\\sqrt[3]{(b^6+c^6)(b^3+a^3)^2}}+\\frac{c^4}{c^4+\\sqrt[3]{(c^6+a^6)(c^3+b^3)^2}} \\le 1$", "Solution_1": "I think that the following is very suggestive\r\n\\[ \\frac x{x+y+z}+\\frac y{x+y+z}+\\frac z{z+x+y}=1. \\]", "Solution_2": "This quation is easy than I think.\r\n :D This is my solution. \r\nBy Holder's we have that \r\n$\\sqrt[3]{((a^2)^3+(b^2)^3){((c^2)^3/2)+(a^2)^3/2}^2} \\geq (a^2)(c^2)+ (a^2)(b^2)$ \r\n\r\n\r\nIt's beautiful ? :D", "Solution_3": "oh\r\ni think we can replace x^3=a\r\nthen it is easier to sovle", "Solution_4": "I guessed this when I solved the problem:\r\n\\[ \\frac{a^4}{a^4+\\sqrt[3]{\\cdots}} \\le \\frac{a^2}{a^2+b^2+c^2} \\]\r\nBut Holder's give solution only, while I tried other ones (because I don't like Holder's). Can... ?", "Solution_5": "See\r\nhttp://www.mathlinks.ro/Forum/viewtopic.php?p=490540#490540\r\nIn this page,you will see quite general of this problem :) :)", "Solution_6": "[quote=\"Lovasz\"]Let $a,b,c>0$. Show that:\n$\\frac{a^4}{a^4+\\sqrt[3]{(a^6+b^6)(a^3+c^3)^2}}+\\frac{b^4}{b^4+\\sqrt[3]{(b^6+c^6)(b^3+a^3)^2}}+\\frac{c^4}{c^4+\\sqrt[3]{(c^6+a^6)(c^3+b^3)^2}} \\le 1$[/quote]\r\n\r\nThis is a question of a competition called Olimpic 30/4 in VN.\r\nFirstly, I prove this:$(a_1+b_1)(a_2+b_2)(a_3+b_3) \\geq(\\sqrt[3]{a_1a_2a_3}+\\sqrt[3]{b_1b_2b_3})^3$\r\n\r\nWe have:\r\n$\\sqrt[3]{a^6+b^6)(c^3+a^3)(c^3+a^3)}\\geq (a^2c^2+b^2a^2)$\r\n$\\frac{a^4}{a^4+\\sqrt[3]{a^6+b^6)(c^3+a^3)^2}}\\leq\\frac{a^2}{a^2+b^2+c^2}$\r\n\r\n\r\nPS: I'm very bad in English, so please forgive if I have any mistakes", "Solution_7": "[quote=\"detectivehien\"]\nThis is a question of a competition called April 4th Olympiad in VN.\nFirstly, I prove this:$(a_1+b_1)(a_2+b_2)(a_3+b_3) \\geq(\\sqrt[3]{a_1a_2a_3}+\\sqrt[3]{b_1b_2b_3})^3$\n[/quote]\r\nAnd this just is Mink's triangle inequality. \r\nVery simple solution :) ." } { "Tag": [ "AMC", "AIME", "quadratics", "geometry", "MATHCOUNTS", "AIME II", "algebra" ], "Problem": "Are we allowed to have scotch tape, scissors, or carbon paper on the AIME?", "Solution_1": "Only scratch paper, graph paper, rulers, compasses, and protractors are permitted, unless there was a change of rules this year.", "Solution_2": "[quote=\"davidyko\"]Only scratch paper, graph paper, rulers, compasses, and protractors are permitted, unless there was a change of rules this year.[/quote]\r\nIt doesn't say clothes are allowed either, better not bring those.", "Solution_3": "Fine, \"No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.\" - 2006 AIME II booklet sitting on my desk.\r\nSo unless your clothes are your mathematical aids...(in which case that would be quite strange).", "Solution_4": "[quote=\"davidyko\"]Fine, \"No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.\" - 2006 AIME II booklet sitting on my desk.\nSo unless your clothes are your mathematical aids...(in which case that would be quite strange).[/quote]\r\nHeh, you never know :D\r\nI'm just wondering, I especially think that they wouldn't care about carbon paper", "Solution_5": "Can anyone clarify?\r\nI don't see what the big deal would be, especially with carbon paper, but if the MAA says no, there must be a good reason.\r\nAlso, add construction paper to my list.", "Solution_6": "[quote=\"davidyko\"]Fine, \"No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.\" - 2006 AIME II booklet sitting on my desk.\nSo unless your clothes are your mathematical aids...(in which case that would be quite strange).[/quote]\r\nIn my middle school days one team at a regional math meet came in with a team t-shirt that had the quadratic equation written on the back ... that caused quite an uproar.", "Solution_7": "Why would you need carbon paper?\r\n\r\nI guess you could take it because it wouldn't exactly be um...mathematical aids?\r\n\r\n(Unless you scratched formulas into construction paper)", "Solution_8": "Maybe to help with a geometry problem, but you could probably do it with a compass, protractor, and ruler anyway.", "Solution_9": "[quote=\"randomdragoon\"][quote=\"davidyko\"]Fine, \"No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.\" - 2006 AIME II booklet sitting on my desk.\nSo unless your clothes are your mathematical aids...(in which case that would be quite strange).[/quote]\nIn my middle school days one team at a regional math meet came in with a team t-shirt that had the quadratic equation written on the back ... that caused quite an uproar.[/quote]\r\n:rotfl:\r\nLast year at state mathcounts this one team had shirts with the distance formula on the back :D", "Solution_10": "Please, can someone in charge confirm or deny whther these can be used?", "Solution_11": "Just to add to this, such a discussion interestingly came up last year when scissors turned out to be very useful in solving #10 on last year's AIME I. From what I remember, people were talking as if scissors were allowed, although most people did not think of bringing them.", "Solution_12": "Read the Teachers Manual: on page 6 is a facsimile of the AIME cover page. On item 3 it says:\r\n\"No aids other than scratch paper, graph paper, ruler, compass and protractor are allowed. In particular, calculators and computers are not allowed.\"\r\n\r\nThe rules say exactly what the rules mean, the rules mean exactly what the rules say.\r\n\r\nSteve Dunbar\r\nDirector, American Mathematics Competitions" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Prove that for any $ n \\in {\\mathbb N}^* - \\{ 1 \\}$ we have: $\\sqrt[p+1]{{log_p}^p {\\left ( {p+1} \\right )}^p} + \\sqrt[p+1]{{log_{p+1}}^p {\\left ( {p} \\right )}^{p+1}} > 2{[p(p+1)]}^{\\frac {p}{2p+2}}$.\r\n\r\n :ninja:", "Solution_1": "Nice application of log rules and AM-GM (if my solution is correct). \r\n\r\nThe left expression is equivalent to\r\n\r\n$ (p*log_{p}{(p+1)})^{\\frac{p}{p+1}} + ((p+1)*log_{p+1}{(p)})^{\\frac{p}{p+1}}.$\r\n\r\nSince p is a natural number not equal to 1, both of those terms are positive, so we can apply AM-GM.\r\nBy AM-GM, this thing is greater than or equal to \r\n\r\n$ 2 [p (p+1) log_{p}{(p+1)}*log_{p+1}{(p)}]^{\\frac{p}{2p+2}}$\r\n\r\nWe use the fact that $log_{a}{(b)}*log_{b}{(a)} = 1$ to finally simplify it to \r\n\r\n$ 2 [p(p+1)]^{\\frac{p}{2p+2}}.$\r\n\r\nTo show that the two quantities cannot be equal, we use the fact that the equality case of AM-GM is only when all terms are equal. Therefore we would have\r\n\r\n$ p*log_{p}{(p+1)} = (p+1)*log_{p+1}{(p)}$\r\n\r\nwhich is equivalent to\r\n\r\n$\\frac{log_{p}{(p+1)}}{log_{p+1}{(p)}} = \\frac{p}{p+1}$\r\n\r\nwhich is further equivalent to \r\n\r\n$log_{p}{(p+1)}*log_{p}{(p+1)} = \\frac{p}{p+1}$\r\n\r\nThe left is irrational and the right is rational for natural p, so the equality case is impossible. So, we are left with\r\n\r\n$\\sqrt[p+1]{{log_p}^p {\\left ( {p+1} \\right )}^p} + \\sqrt[p+1]{{log_{p+1}}^p {\\left ( {p} \\right )}^{p+1}} > 2{[p(p+1)]}^{\\frac {p}{2p+2}}.$ QED", "Solution_2": "nice :roll:", "Solution_3": "Well i saw your sollution and it's ok.\r\nHere is my sollution:\r\n\r\n$\\sqrt[p+1]{{log_p}^p {\\left ( {p+1} \\right )}^p} + \\sqrt[p+1]{{log_{p+1}}^p {\\left ( {p} \\right )}^{p+1}}$ = \r\n= $ \\left ( \\sqrt[p+1]{{log_p} {\\left ( {p+1} \\right )}^p} \\right )^p + \\left ( \\sqrt[p+1]{{log_{p+1}}{\\left ( {p} \\right )}^{p+1}} \\right )^p $$(*)$\r\n\r\nHere we use the following inequality : $ \\frac {x^n + y^n}{2} \\geq \\left ( \\frac {x+y}{2} \\right )^n $ which can be prooved easy by Jensen. So we get:\r\n\r\n$ (*) \\geq 2 \\left ( \\frac {{\\sqrt[p+1]{log_p (p+1)^p} + {\\sqrt[p+1]{log_{p+1}p^{p+1}}}}}{2} \\right )^p $ $(**)$ and here\r\nusing AM-GM we get: $(**) \\geq 2 \\left ( \\sqrt {\\sqrt[p+1]{p(p+1)log_p {(p+1)} log_{(p+1)}p}} \\right )^p = 2 \\left [ p(p+1) \\right ]^{\\frac {p}{2(p+1)}}$ and for prooving that the inequality is strict i prooved like u did.\r\n\r\nq.e.d." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "Consider all possible sequences of 2004 balls, white, black and red on a line.\r\nI will say transformation to the following process: We choose 200 consecutive balls, the first $100$ balls at the left($in$ $the$ $same$ $order$) are ubicated inmediately on the right of the other $100$ balls, we say that two sequences of balls are $similar$ if one can be obtained from the other with transformations, find the number of sequences of balls which are $not$ $similar$ :read:", "Solution_1": "find all OR find how many", "Solution_2": "[quote=\"OHO\"]find all OR find how many[/quote]\r\nFind how many", "Solution_3": "nobody$?$, please. Its a very nice and dificult $problem$." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Prove that every symmetric matrix has orthogonal eigenvectors.", "Solution_1": "This is the finite-dimensional [url=http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices]spectral theorem[/url]." } { "Tag": [ "combinatorics unsolved", "combinatorics" ], "Problem": "$ m$ and $ n$ are positive integers. In a $ 8 \\times 8$ chessboard, $ (m,n)$ denotes the number of grids a Horse can jump in a chessboard ($ m$ horizontal $ n$ vertical or $ n$ horizontal $ m$ vertical ). If a $ (m,n) \\textbf{Horse}$ starts from one grid, passes every grid once and only once, then we call this kind of Horse jump route a $ \\textbf{H Route}$. For example, the $ (1,2) \\textbf{Horse}$ has its $ \\textbf{H Route}$. Find the smallest positive integer $ t$, such that from any grid of the chessboard, the $ (t,t\\plus{}1) \\textbf{Horse}$ does not has any $ \\textbf{H Route}$.", "Solution_1": "Can anyone help?\r\nIt seems that t=3 can be easily proved \r\nbut t=2 seems like a hard case.", "Solution_2": "If I'm not mistaken, $ t \\equal{} 2$ is handled by noticing that only two moves can be made from $ (1,1), (1,7), (7,1), (7,7)$. The moves from these points form a cycle along with $ (3,4), (4,3), (5,4), (4,5)$. We can make similar cycles by looking at translates to $ (1,2), (2,1), (2,2)$. We only have two endpoints on the path, so we can cut at most two of the cycles, not enough." } { "Tag": [ "function", "limit", "integration", "real analysis", "real analysis unsolved" ], "Problem": "Let $ f: [0,1] \\rightarrow R$ be a integrable function with the property that for all $ x \\in [0,1)$ we have:\r\n\r\n $ \\lim_{n\\to\\infty}{n \\cdot \\int_{x}^{x\\plus{}\\frac{1}{n}}{f(t)}dt}\\equal{}0.$\r\n\r\nProve that for all $ a,b \\in (0,1)$ we have:\r\n\r\n $ \\int_{a}^{b}{f(t)}dt\\equal{}0.$", "Solution_1": "let $ F(x) \\equal{} \\int_{0}^{x}f(t)dt$\r\nthen we have $ n[F(x \\plus{} \\frac {1}{n}) \\minus{} F(x)] \\equal{} n\\int_{x}^{x \\plus{} \\frac {1}{n}}f(t)\\rightarrow F'(x) \\equal{} f(x)$", "Solution_2": "Since we weren't given that $ f$ is continuous, that isn't quite the way to manage it. Close, but we should rearrange the argument slightly.\r\n\r\nDefine $ F(x)$ as you have. \r\n\r\n\r\n$ F'(x)\\equal{}\\lim_{n\\to\\infty}{n \\cdot \\int_{x}^{x\\plus{}\\frac{1}{n}}{f(t)}dt}\\equal{}0$ for all $ x.$\r\n\r\nHence (by the Mean Value Theorem) $ F(a)\\equal{}F(b)$ for all $ a$ and $ b.$\r\n\r\n$ \\int_a^bf(t))\\,dt\\equal{}\\int_0^bf(t)\\,dt\\minus{}\\int_0^af(t)\\,dt\\equal{}F(b)\\minus{}F(a)\\equal{}0$ and we're done." } { "Tag": [ "inequalities", "function", "induction", "inequalities unsolved" ], "Problem": "Hello,I have the following problem.\r\nIf $f:R->R$ is a convex function and $a_{k+1}\\ge a_{k}$ for $k=1,2,...,n-1$ then prove that \r\n $a_{1}f(a_{n})+a_{n}f(a_{n-1}))+...+a_{2}f(a_{1})\\leq f(a_{1})a_{n}+f(a_{n})a_{n-1}+...+f(a_{2})a_{1}$", "Solution_1": "Are you sure the function isn't concave instead? Is the sign backwards? Subtracting the RHS from the LHS yields\r\n\r\n$(a_1-a_{n-1})f(a_n) + \\cdots + (a_2-a_n)f(a_1) \\ge 0$\r\n\r\nBut $a_i > a_j$ for each $i > j$, which means that the coefficients of $f(a_n)$ and $f(a_1)$ are both negative, but $f$ evaluated at the endpoints can be made arbitrarily large without losing the convexity of $f$, making the LHS arbitrarily small.\r\n\r\nIf $f$ is made concave, the problem looks like it can be solved with smoothing; I'll think about it.", "Solution_2": "$\\sum{a_{i}f(a_{i+1})} \\ge \\sum {a_{i+1}f(a_{i})}$\r\nIn case $n=3$\r\n$a_1f(a_2)+a_2f(a_3)+a_3f(a_1) \\ge a_2f(a_1)+a_3f(a_2)+a_1f(a_3)$\r\n$<->a_1(f(a_2)-f(a_3))+a_2(f(a_3)-f(a_1))+a_3(f(a_1)-f(a_2)) \\ge 0$\r\n$<->(a_1-a_2)(f(a_2)-f(a_3))+(a_3-a_2)(f(a_1)-f(a_2)) \\ge 0$\r\n$<->\\frac{f(a_3)-f(a_2)}{a_3-a_2} \\ge \\frac{f(a_2)-f(a_1)}{a_2-a_1}$\r\nIt's true because f is a convex function!\r\nAnd the ineq is easy to prove,use induction!", "Solution_3": "[quote=\"hungkhtn\"]$\\sum{a_{i}f(a_{i+1})} \\ge \\sum {a_{i+1}f(a_{i})}$\nIn case $n=3$\n$a_1f(a_2)+a_2f(a_3)+a_3f(a_1) \\ge a_2f(a_1)+a_3f(a_2)+a_1f(a_3)$\n$<->a_1(f(a_2)-f(a_3))+a_2(f(a_3)-f(a_1))+a_3(f(a_1)-f(a_2)) \\ge 0$\n$<->(a_1-a_2)(f(a_2)-f(a_3))+(a_3-a_2)(f(a_1)-f(a_2)) \\ge 0$\n$<->\\frac{f(a_3)-f(a_2)}{a_3-a_2} \\ge \\frac{f(a_2)-f(a_1)}{a_2-a_1}$\nIt's true because f is a convex function!\nAnd the ineq is easy to prove,use induction![/quote]\r\n\r\nAh, but this is the reverse of hardsoul's original statement, which, for $n=3$, asks us to prove\r\n\r\n$a_1f(a_3) + a_3f(a_2) + a_2f(a_1) \\ge a_3f(a_1)+a_2f(a_3)+a_1f(a_2)$", "Solution_4": "Only take $f(x)=x^2$ abd $x_1=1,x_2=2,x_3=3$ we have:\r\n$x_1f(x_2)+x_2f(x_3)+x_3f(x_1) \\ge x_1f(x_3)+x_3f(x_2)+x_2f(x_1)$\r\nAnd the sign of hardsoul reverse.", "Solution_5": "Sorry,the sign is really reversed.", "Solution_6": "You can see a similiar problem in Iran 1999 first round.", "Solution_7": "Sorry, does anyone know how to finish the proof by an induction?", "Solution_8": "Finally I proved it :P .\r\nFor $n=3$ it is proved .\r\nSuppose that it's true for n i.e \r\n$x_1\\cdot f(x_2)+...+x_n\\cdot f(x_1)\\ge x_2\\cdot f(x_1)+...+x_1f(x_n)$\r\nNow take $x_{n+1}\\ge x_n\\ge x_{n-1}\\ge ...\\ge x_1$\r\nThen adding $x_{n+1}f(x_1)-x_{n}f(x_1)\\ge x_1f(x_{n+1})-x_1f(x_n)$ equivalent to $\\frac{f(x_{n+1})-f(x_n)}{x_{n+1}-x_n}\\ge \\frac{x_{n+1}-f(x_1)}{x_{n+1}-x_1}$ we complete the induction." } { "Tag": [ "probability" ], "Problem": "Ralph and Waldo each simultaneous hold out some fingers on one hand. Ralph always holds out a prime number of fingers; Waldo always holds out an odd number. What is the probability that the sum of the number of fingers they hold out is even? Express your answer as a common fraction.", "Solution_1": "WHen prime guy holds out 2 sum is odd, when 3 or 5, even, so it's 2/3", "Solution_2": "well, since waldo always holds out a odd number, that means to get an even sum, ralph has to hold out an odd number as well\r\n\r\ntherefore, ralph holding out 3 or 5 fingers would get us an even sum\r\n\r\nthe probability of it being an even sum would be $ \\frac {2 \\times 3} {3 \\times 3} \\equal{} \\frac {6} {9} \\equal{} \\frac {2} {3}$\r\n\r\nedit: dang, beaten to it, but mine is more detailed :P" } { "Tag": [ "integration", "trigonometry", "limit", "real analysis", "real analysis unsolved" ], "Problem": "Find $\\displaystyle\\int^\\frac{\\pi}{2}_0 \\sin x\\,dx$ only bu using Riemann metode.", "Solution_1": "$\\int_{0}^{\\frac{\\pi}2}\\sin{x}dx=\\lim_{n\\to\\infty}\\frac{\\pi}{2n}\\sum_{k=1}^{n}\\sin{\\frac{k\\pi}{2n}}=\\lim_{n\\to\\infty}\\frac{\\pi}{4n}\\frac{\\sum_{k=1}^{n}(\\cos{\\frac{2k-1}{4n}\\pi}-\\cos{\\frac{2k+1}{4n}\\pi})}{\\sin{\\frac{\\pi}{4n}}} =\\lim_{n\\to\\infty}\\frac{\\pi}{4n}\\frac{\\cos{\\frac{\\pi}{4n}}-\\cos{(1+\\frac{1}{2n})\\frac{\\pi}{2}}}{\\sin\\frac{\\pi}{4n}}=\\lim_{n\\to\\infty}(\\frac{\\frac{\\pi}{4n}}{\\tan\\frac{\\pi}{4n}}-\\frac{\\pi}{4n})=1$." } { "Tag": [ "calculus", "integration", "arithmetic sequence", "number theory unsolved", "number theory" ], "Problem": "1. Show that there exists some integral power of 2 equal to 2002...\r\n\r\n2. Given any n+1 integers in [1,2n], show that we can find one element that divides another.\r\n\r\n3. How many primes are there among the positive integers (written in the usual base 10) such that the prime begins with 1, ends with 1, and only contains 0's and 1's for digits?\r\n\r\n4. Let A be any set of 20 distinct integers chosen from the arithmetic progression 1, 4, 7, ..., 100. Prove that there must be two distinct integers in A whose sum is 104.", "Solution_1": "2. Let us write every integer $ i$ from $ 1$ to $ 2n$ in the form $ 2^{k_{i}}l_{i}$ where $ i$ is odd and $ k$ is a non-negative integer. Then, because in the set $ \\{1,2,...,2n\\}$ there are only $ n$ odd integers, then in the set $ \\{l_1,l_2,...,l_{2n}\\}$ there are only $ n$ distinct integers, therefore by Pigeon-Hole Principle among the $ n+1$ integers $ x_1,x_2,...,x_{n+1}$ selected from the set $ \\{1,2,...,2n\\}$ we will find distinct integers $ a,b\\in\\{x_1,x_2,...,x_{n+1}\\}$ such that $ l_a=l_b$ and therefore, if ${ a=2^{k_{a}}l_a,b=2^{k_b}}l_b$ and wolog $ k_a[hide]15+pi?????[/hide]", "Solution_2": "[color=cyan]Well, the form of your answer isn't bad . . . the actual numbers involved are different, though. Reread the question and take each of your steps very carefully. Going slowly and not making silly mistakes is the most important part of this problem.[/color]", "Solution_3": "O man, I can't believe i messed up again.\n\n\n\nOK, renewed answer:---->[hide]30+4pi[/hide]\n\nhmm, better be rite...", "Solution_4": "[color=cyan]That sounds right.[/color]", "Solution_5": "alright!\r\nniceness....", "Solution_6": "Explanation please?", "Solution_7": "Ok, if u draw the boundaries of the path of the circle, you see that it is a triangle around the original equilateral triangle, except that at the corners, its rounded (do u see?)\r\n\r\nNow, from each of the three vertices, we draw two perpendiculars to both sides of the \"triangle\", which is really the path of the circle\r\n\r\nWat we are left with is three rectangles, 2x5, and 3 sections of circles, each of which is 120 degrees. \r\n\r\nSo the area is 3*2*5+3*(1/3)*4pi, which is 30+4pi", "Solution_8": "Oh, ok. That was pretty simple. Thanks.", "Solution_9": "Any time man...." } { "Tag": [ "function", "calculus", "integration", "calculus computations" ], "Problem": "In how many ways one can prove that (for example) that $ f(x) \\equal{} 2x$ is integrable on $ x\\in[0,1]$ ?", "Solution_1": "Explicitly evaluating the integral from the definition would be one way. Also, we know $ f$ is bounded on $ [0,1]$ so if you prove that $ \\sup\\{L(f,P)\\}\\equal{}\\inf\\{U(f,P)\\}$ where $ P$ is a partition of $ [0,1]$, then that's the definition of being integrable, so there's another way." } { "Tag": [ "calculus", "integration", "trigonometry", "logarithms", "complex numbers", "calculus computations" ], "Problem": "Okay friends, here is an exercise:\r\nShow that $ \\int\\limits_0^\\pi {\\frac {{a \\minus{} b\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\equal{} \\frac{\\pi}{a}$, where $ a$ and $ b$ are real numbers with $ a \\neq 0$.", "Solution_1": "hello, we have $ \\int\\frac{a\\minus{}b\\cos(x)}{a^2\\plus{}b^2\\minus{}2ab\\cos(x)}\\,dx\\equal{}\\frac{x\\minus{}2\\arctan\\left(\\frac{(a\\minus{}b)\\cot(\\frac{x}{2})}{a\\plus{}b}\\right)}{2a}\\plus{}C$.\r\nSonnhard.", "Solution_2": "Can you please show how you obtained that solution?", "Solution_3": "For $ |r| < 1,$\r\n\r\n$ \\sum_{k \\equal{} 0}^{\\infty}r^k\\cos kx \\equal{} \\text{Re}\\,\\sum_{k \\equal{} 0}^{\\infty}r^ke^{ikx} \\equal{} \\text{Re}\\,\\left(\\frac {1 \\minus{} re^{ \\minus{} ix}}{1 \\minus{} 2r\\cos x \\plus{} r^2}\\right) \\equal{} \\frac {1 \\minus{} r\\cos x}{1 \\minus{} 2r\\cos x \\plus{} r^2}.$\r\n\r\nCase 1: Suppose that $ |b| < |a|.$\r\n\r\n$ \\sum_{k \\equal{} 0}^{\\infty}\\left(\\frac ba\\right)^k\\cos kx \\equal{} \\frac {1 \\minus{} \\frac ba\\cos x}{1 \\minus{} 2\\frac ba\\cos x \\plus{} \\left(\\frac ba\\right)^2} \\equal{} \\frac {a(a \\minus{} b\\cos x)}{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}$\r\n\r\nThen $ \\int_0^{\\pi}\\frac {a \\minus{} b\\cos x}{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}\\,dx \\equal{} \\frac1a\\int_0^{\\pi}\\sum_{k \\equal{} 0}^{\\infty}\\left(\\frac ba\\right)^k\\cos kx\\,dx$\r\n\r\n$ \\equal{} \\sum_{k \\equal{} 0}^{\\infty}\\frac {b^k}{a^{k \\plus{} 1}}\\int_0^{\\pi}\\cos kx\\,dx \\equal{} \\frac {\\pi}{a}$\r\n\r\nsince the integral is zero except for $ k \\equal{} 0$ and in that case is the integral of a constant.\r\n\r\nCase 2: Suppose that $ |b| > |a|.$\r\n\r\nThis is a little more problematical.\r\n\r\n$ \\sum_{k \\equal{} 0}^{\\infty}\\left(\\frac ab\\right)^k\\cos kx \\equal{} \\frac {b(b \\minus{} a\\cos x)}{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}$\r\n\r\nWith some algebraic maneuvering, we get that\r\n\r\n$ \\frac{a\\minus{}b\\cos x}{a^2\\plus{}b^2\\minus{}2ab\\cos x}\\equal{}\\frac1a\\minus{}\\frac1a\\sum_{k \\equal{} 0}^{\\infty}\\left(\\frac ab\\right)^k\\cos kx.$\r\n\r\nPut that in and discover that the integral is zero.\r\n\r\nCase 3: $ |a|\\equal{}|b|\\ne 0$\r\n\r\nIn this case, this is simply $ \\int_0^{\\infty}\\frac1{2a}\\,dx\\equal{}\\frac{\\pi}{2a}.$\r\n\r\n=============================\r\n\r\nakech's original statement was not quite right. It should have been\r\n\r\n$ \\int_0^{\\pi}\\frac{a\\minus{}b\\cos x}{a^2\\plus{}b^2\\minus{}2ab\\cos x}\\,dx\\equal{} \\begin{cases}\\frac{\\pi}a,&|a|>|b|\\\\\\frac{\\pi}{2a},&|a|\\equal{}|b|\\ne 0\\\\0,&|a|<|b|\\end{cases}$\r\n\r\nwith the further note that this is meaningless if $ a\\equal{}b\\equal{}0.$", "Solution_4": "$ \\int\\limits_0^\\pi {\\frac {{a \\minus{} b\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\\\\r\n\\equal{} \\int\\limits_0^\\pi {\\frac {1}{2a}\\frac {{2a^2 \\minus{} 2ab\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\\\\r\n\\equal{} \\int\\limits_0^\\pi {\\frac {1}{2a}\\frac {{a^2 \\minus{} b^2 \\plus{} a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\\\\r\n\\equal{} \\int\\limits_0^\\pi {\\frac {a^2 \\minus{} b^2}{2a}\\frac {{1}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\plus{} \\pi$\r\nfor the remaining part, write \r\n$ a^2 \\plus{} b^2 \\equal{} c \\\\\r\n\\minus{} 2ab \\equal{} d$\r\nThen,\r\n$ \\frac {1}{c \\plus{} dcos x} \\\\\r\n\\equal{} \\frac {sec^2{\\frac {x}{2}}}{(c \\minus{} d) tan^2{\\frac {x}{2}} \\plus{} (c \\plus{} d)}$\r\nput value of c and d.\r\ncorresponding to different cases you will have to break $ ln{|{Denominator}|}$\r\n\r\n\r\nmathematica:\r\n[code]\nif a^3 b != a b^3, ans=(\\Pi (1 + Sign[a + b]/Sign[a - b]))/(2 a) [/code]", "Solution_5": "Thank you, Dr. Merryfield for pointing out my error and also for the complete solution.\r\nMy partial solution was for the case $ |a| > |b|$:\r\nConsider the integral:\r\n$ I \\equal{} \\frac {1}{2} \\int \\log (a^2 \\plus{} b^2 \\minus{} 2ab \\cos x) dx$\r\n$ \\equal{} x\\log a \\plus{} \\frac {1}{2} \\int \\log \\left(1 \\plus{} ( \\minus{} \\frac {b}{a})^2 \\plus{} 2( \\minus{} \\frac {b}{a}) \\cos x\\right) dx$\r\n$ \\equal{} x\\log a \\plus{} \\frac {1}{2} \\int \\log (1 \\plus{} ( \\minus{} \\frac {b}{a})e^{ix}(1 \\plus{} ( \\minus{} \\frac {b}{a})e^{ \\minus{} ix}dx$\r\n$ \\equal{} x\\log a \\minus{} \\sum_{n \\equal{} 1}^{\\infty} (\\frac {b}{a})^{n}\\frac {\\sin xn}{n^2} \\plus{} C$. \r\n\r\n\r\n$ \\frac {dI}{da} \\equal{} \\frac {x}{a} \\plus{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {b^{n}}{a^{n \\plus{} 1}}\\frac {\\sin xn}{n}$ \r\n\r\nAlso, $ I \\equal{} \\frac {1}{2} \\int \\log (a^2 \\plus{} b^2 \\minus{} 2ab \\cos x) dx \\implies \\frac {dI}{da} \\equal{} \\int {\\frac {{a \\minus{} b\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx$\r\n\r\nHence $ \\int {\\frac {{a \\minus{} b\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\equal{} \\frac {x}{a} \\plus{} \\sum_{n \\equal{} 1}^{\\infty}\\frac {b^{n}}{a^{n \\plus{} 1}}\\frac {\\sin xn}{n} \\plus{} C$\r\n\r\nThus, $ \\int\\limits_0^\\pi {\\frac {{a \\minus{} b\\cos x}}{{a^2 \\plus{} b^2 \\minus{} 2ab\\cos x}}} dx \\equal{} \\frac {\\pi}{a}$.", "Solution_6": "Another perspective: $ a^2\\plus{}b^2\\minus{}2ab\\cos \\theta\\equal{}(b^2\\\\minus{}2ab\\cos\\theta\\plus{}a^2\\cos^2\\theta)\\plus{}a^2\\sin^2\\theta$\r\n$ \\equal{}(\\minus{}b\\plus{}a\\cos\\theta)^2\\plus{}(a\\sin\\theta)^2\\equal{}|ae^{i\\theta}\\minus{}b|^2$. Introduce complex numbers.\r\n\r\nConsider the complex integral $ I\\equal{}\\int_{|z|\\equal{}a}\\frac1{z\\minus{}b}\\,dz$. This can be parametrized as $ \\int_0^{2\\pi}\\frac1{ae^{i\\theta}\\minus{}b}\\cdot iae^{i\\theta}\\,d\\theta$, which becomes in real form\r\n$ I\\equal{}\\int_0^{2\\pi}\\frac{i(a\\cos\\theta\\minus{}b\\minus{}ia\\sin\\theta)(a\\cos\\theta\\plus{}ia\\sin\\theta)}{a^2\\plus{}b^2\\minus{}2ab\\cos \\theta}\\,d\\theta$\r\n$ \\equal{}a\\int_0^{2\\pi}\\frac{b\\sin\\theta\\plus{}i(a\\minus{}b\\cos\\theta)}{a^2\\plus{}b^2\\minus{}2ab\\cos \\theta}\\,d\\theta$\r\n\r\nThe imaginary part of $ I$ is $ a\\int_0^{2\\pi}\\frac{a\\minus{}b\\cos\\theta}{a^2\\plus{}b^2\\minus{}2ab\\cos \\theta}\\,d\\theta$, $ 2a$ times the integral we want to compute. We have a simple pole of residue $ 1$, so $ I\\equal{}2\\pi i$ if that pole is inside the circle, $ I\\equal{}0$ if the pole is outside, and $ I\\equal{}\\pi i$ as a principal value if it's on the circle." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "I am not a spammer, at least, this is the way I use to think about myself, and thus I will not open a new thread for the following problem from today's DeMO exam:\r\n\r\nLet Q(n) denote the sum of the digits of a positive integer n. Prove that $Q\\left(Q\\left(Q\\left(2005^{2005}\\right)\\right)\\right)=7$.\r\n\r\n[[b]EDIT:[/b] Since this post was split into a new thread, I comment:\r\n\r\nThe problem is completely analogous to the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 , with the only difference that you have to consider the number $2005^{2005}$ instead of $4444^{4444}$.]\r\n\r\n Darij", "Solution_1": "But, Darij, won't the method be the same? So we're just left to make some computations", "Solution_2": "I think Darij knew that when posting ;) \r\nBut the DeMO is going down and down... they are not able to find new problems (or to take them from a little bit more unknown olympiads)" } { "Tag": [], "Problem": "Prove\r\nThe sum of the 3rd powers of three consecutive numbers from Z is divisible by 9.", "Solution_1": "Let\u00b4s expand this cubics:\r\n\r\n(a-1)^3+a^3+(a+1)^3=a^3-3a^2+3a-1+a^3+a^3+3a^2+3a+1=3a^3+6a=3a(a^2+2). And now we have to prove that 3/a(a^2+2).\r\n\r\n1)a=0(mod3) and so is proved\r\n2)a=1(mod3) -> a^2=1(mod3) -> a^2+2=3(mod3) and so is proved\r\n3)a=2(mod3) -> a^2=4=1(mod3) and so is proved too." } { "Tag": [ "counting", "derangement", "search" ], "Problem": "Here's a problem from my math homework that my math teacher couldn't answer (just parts d and e) So if anyone knows how to answer d or e, it would be greatly appreciated\r\n\r\n26. The ten cards, ace to ten in hearts are removed from a deck of cards, shuffled, and placed in a row. The ace to ten of clubs are also removed, shuffled, and placed in a row just below the first cards.\r\n\r\na) How many different pairs (of ten cards) can be made? (answer: (10!)2)\r\nb) How many of the pairings in 26a have the same card in the third position? (answer: (10!)(9!) )\r\nc) How many of the pairings in 26a have the same card in the fourth position? (answer: (10!)(9!) )\r\nd) How many of the pairings in 26a have exactly one of the pairs the same? (any position) (answer: ???)\r\ne) How many of the pairings in 26a have at least of of the pairs being the same? (answer: ???)\r\n\r\nAnswers and explainations for parts d and e would be appreciated.", "Solution_1": "What do you mean by part a? I would say that there are C(20, 2) pairs of cards, 10 pairs by the poker definition. What exactly do you mean? How many orderings? I think you need to explain to me what you are doing better, because I'm not really sure what we are counting.", "Solution_2": "Sorry, I just copied it from the book word-for-word.\r\nYou will end up with two rows of ten cards each, with the second row right below the first. The first row only contains hearts (Ace to ten), and the second only clubs (Ace to ten). There would be 10! ways to arrange each row, for a total of (10!)2 permutations.", "Solution_3": "Okay -- and we consider a heart and club to be paired when they are directly above (below) each other?", "Solution_4": "correct", "Solution_5": "For part d, you have several things to do:\n\n\n\nFirst, [hide]you have to pick the pair that will stay the same.[/hide]\n\n\n\nSecond, (and this is the hard part), \n\n[hide]you have to make sure none of the other pairs is still together.[/hide] Now, I have a question: If we were playing with only 4 cards, {A, 2, 3, 4}, and I had\n\n\n\nA 2 3 4\n\nA 2 3 4\n\n\n\nas one arrangement and\n\n\n\nA 2 4 3\n\nA 3 4 2\n\n\n\nas another arrangement, do the 44 pairs count as the same pair? I assume that is what \"any position\" means, but I just wanted to check.\n\n\n\nThis second bit is rather hard. The problem is known as \"derangements,\" and you could try a search for that word on this forum in order to find out about it. I personally would start by looking at how many derangements of 1 card, 2 cards, 3 cards, and so on, there are, and trying to find a pattern.\n\n\n\n\n\nFor part e, try using the principle of inclusion-exclusion.\n\n\n\n\n\nI still feel like the question isn't well-posed, though -- it leaves me with too many \"what exactly do they mean?\" feelings.", "Solution_6": "[quote=\"picounts\"]Sorry, I just copied it from the book word-for-word.\nYou will end up with two rows of ten cards each, with the second row right below the first. The first row only contains hearts (Ace to ten), and the second only clubs (Ace to ten). There would be 10! ways to arrange each row, for a total of (10!)2 permutations.[/quote]\r\n\r\nThats kinda weird anyway... \r\n\r\nLets simplify this to hearts and clubs with 2h, 3h, 4h being 3 hearts and 2c,3c,4c being 4 clubs.... then if you lay it out like this:\r\n\r\n2h 3h 4h\r\n2c 3c 4c\r\n\r\nIt's different from\r\n\r\n2h 4h 3h\r\n2c 4c 3c \r\n\r\nBy how you've defined it. This might be what you meant, but its kind of odd.", "Solution_7": "Yes, I agree -- this definition of pairings is very strange. Are you really convinced your answer to the first question is correct? A much more \"usual\" answer to a question like this would be 10!.", "Solution_8": "Yeah, pretty sure. For example, with 3 cards instead of 10 we woud have (3!)2:\r\n\r\nAd 2d 3d\r\nAc 2c 3c\r\n\r\nAd 2d 3d\r\nAc 3c 2c\r\n\r\nAd 2d 3d\r\n2c Ac 3c\r\n\r\nAd 2d 3d\r\n2c 3c Ac\r\n\r\nAd 2d 3d\r\n3c Ac 2c\r\n\r\nAd 2d 3d\r\n3c 2c Ac\r\n\r\n\r\nAd 3d 2d\r\nAc 2c 3c\r\n\r\nAd 3d 2d\r\nAc 3c 2c\r\n\r\nAd 3d 2d\r\n2c Ac 3c\r\n\r\nAd 3d 2d\r\n2c 3c Ac\r\n\r\nAd 3d 2d\r\n3c Ac 2c\r\n\r\nAd 3d 2d\r\n3c 2c Ac\r\n\r\n\r\nAnd so on", "Solution_9": "I understand what you are saying -- however, the conventional definition of pairing doesn't rely on what order your pairs are in. Thus,\r\n\r\nA 2\r\n2 A\r\n\r\n2 A\r\nA 2\r\n\r\nwould normally be considered to be the same. It makes the later questions much nicer, and it makes them make more sense.", "Solution_10": "Wow, I feel slow today....\r\n\r\nSo, to get the number of ways of counting we want, can't we just multiply the \"nicer\" way of counting by 10P10=10! ?" } { "Tag": [ "combinatorics theorems", "combinatorics" ], "Problem": "What are combinatorics? like, the definition of them and what r they classified as and stuff.", "Solution_1": "look it in the dicktionary\r\n\r\ntry wikipedia.com", "Solution_2": "[quote=\"Tecnogram888\"]What are combinatorics? like, the definition of them and what r they classified as and stuff.[/quote]" } { "Tag": [ "AMC", "AIME", "probability", "pigeonhole principle" ], "Problem": "9. A fair coin is to be tossed ten times. Let $i/j$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i + j$", "Solution_1": "consider the string of H's and T's, say that there are $h$ H's, then there are $10-h$ T's, and $11-h$ spots for the H's to go for a good configuration, so we choose $h$ of those $11-h$ spots to place the H's, and giving all the possible numbers of H's in a summation:\r\n\r\n$\\frac{ \\sum_{h=0}^{5}\\binom{11-h}{h} }{2^{10}}=\\frac{9}{64} \\implies m+n=73$", "Solution_2": "[hide]5 heads: We must distribute these 5 heads among 6 slots, so there are 6 possibilities (ie, HTHTHTHTHT, HTHTHTHTTH, HTHTHTTHTH, HTHTTHTHTH, HTTHTHTHTH, THTHTHTHTH).\n\n4 heads: 4 heads are to be placed into 7 slots. So there are $\\binom74=35$ ways.\n\n3 heads: $\\binom83=56$ ways\n\n2 heads: $\\binom92=36$ ways\n\n1 head: 10 ways\n\nNo heads: 1 way\n\nSo in total, we have \\[ \\binom65+\\binom74+\\binom83+\\binom92+\\binom{10}1+\\binom{11}0=6+35+56+36+10+1=144 \\] good possibilities. There are $2^{10}$ total ways, so the probability in question is \\[ \\frac{144}{2^{10}}=\\frac{9}{64}\\implies m+n=9+64=\\boxed{73} \\][/hide]", "Solution_3": "athlet, whered you get the 11 from? i thought it was 10 tosses? \r\nk8, where did you get those 6 possibilities? why are they the only ones? \r\nhow come you guys didnt take into account the probability of getting a TTTTTTTTTT, HTTTTTTTTT, THTTTTTTTT, etc. 2 heads, 3 heads, etc.?", "Solution_4": "[quote=\"Tecnogram888\"]athlet, whered you get the 11 from? i thought it was 10 tosses? \nk8, where did you get those 6 possibilities? why are they the only ones? \nhow come you guys didnt take into account the probability of getting a TTTTTTTTTT, HTTTTTTTTT, THTTTTTTTT, etc. 2 heads, 3 heads, etc.?[/quote]\r\n\r\nWhen he counts 11 slots, he's basically counting the possible strings of heads and tails. For example,\r\n\r\n_T_ T_ T_ T_ T_ and you can place 5 heads in one of 6 slots. There are 6, not 5 slots (endpoints count! :D )\r\n\r\nThe rest follows. For example, if there are 5 heads, then we calculate (6 c 5). And if there are 4 heads, there would be 6 tails, so 7 spots to put them (7 c 4) etc. all the way down to no heads.\r\n\r\nThe reason why we can't have 6 heads, is because then we'll have 4 tails, and by Pigeonhole, 2 of these heads must be beside each other...an impossibility. So we consider when there are 5 or fewer heads.\r\n\r\nAnd as for the last possibility...they did take into account the probability of getting all tails etc. All tails is represented by (11 c 0) = 1. etc. And as for 1 head, 2 heads, etc, both have considered the possible strings containing those numbers of heads, combinatorially of course.\r\n\r\nHope this helps.", "Solution_5": "[hide=\"Another Solution\"]After 2 coins are chosen, the possibilities are HT, TH, and TT. The only significant characteristic of each possibility is the outcome of the last flip, because it determines our possibilities for the outcomes of the next flip. So after two coins are chosen, we can write the number of possibilities as 2T, 1H, where the letter denotes the outcome of the last flip.\n\nAfter 3 coins are chosen the number of possibilities ending in T is equal to the total number of possibilites after 2 coins are chosen, because we can add a T to any possibility without breaking any restrictions. The number of possibilities ending in H is equal to the number of possibilites ending in T after 2 coins are chosen, because we can only add an H to possibilities ending in T.\n\n2: 2T, 1H\n3: 3T, 2H\n4: 5T, 3H\n5: 8T, 5H\n6: 13T, 8H\n7: 21T, 13H\n8: 34T, 21H\n9: 55T, 34H\n10: 89T, 55H\n\nIf you notice that these are all Fibonacci numbers, then you can save some time.\n\nSo the probability is $\\frac{89+55}{2^{10}}=\\frac{9}{64}\\Rightarrow\\frac{i}{j}=73$[/hide]" } { "Tag": [], "Problem": "How many ways can $ 5$ distinct paperback books and $ 1$ hardcover book be arranged on a shelf if the hardcover must be the rightmost book on the shelf?", "Solution_1": "Since the placement of the hardcover book is set, we need to find the ways to arrange $ 5$ books in a row, which is $ 5!\\equal{}\\boxed{120}$." } { "Tag": [ "\\/closed" ], "Problem": "I actually planned to go to todays math jam, but couldn't get thru after trying for about 40 minutes :( Suppose the classroom site was too busy for a lowly dialup connection to get through to :)", "Solution_1": "We're still going - try now!", "Solution_2": "I've been trying continuously for the last hour. About 10 minutes ago, I actually got through to the page, but now the Java applet won't load.", "Solution_3": "Weird - has anyone else had this problem? Maybe you need to update your Java?", "Solution_4": "I got reapetedly kicked out and finally gave up.", "Solution_5": "What sort of internet connection do you have?", "Solution_6": "I have wireless.", "Solution_7": "That explains it. Many wireless connections are fragile - lots of very brief disconnections. This is no problem if you are surfing the net, in which you don't have to stay connected continuously, but our classroom requires you stay connected contininuously (without interruptions, no matter how small). Your wireless has lots of tiny interruptions - you lose contact with the classroom each time.", "Solution_8": "oh, now i see why i get kicked out once or twice an hour,", "Solution_9": "Also check ur CPU processing percentage in the task manager. I rmr once not being able to get in just b/c I was running at 100% and didn't know it.", "Solution_10": "Nah my java thingy is fine.. and I have been in once before (though I remember it took a while then too). It must just be the speed of my modem (my friend gets in fine)", "Solution_11": "well if you have a pentium 2 or lower java is very slow i know because i have one. and it takes like a century to use maple 9 on them.", "Solution_12": "Na, I don't have one of those. I tried disabling my firewall, tried downloading a different internet browser, and updating java. Nothing works. It just says it can't load the applet.. if anyone knows anything about Java , this is what it said.. (a tiny bit got cut off, cos I copied it into messenger and then later into here)\r\n\r\n[quote=\"Java thing\"]\nload: class com.diginet.digichat.client.DigiChatApplet not found. java.lang.ClassNotFoundException: com.diginet.digichat.client.DigiChatApplet at sun.applet.AppletClassLoader.findClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.applet.AppletClassLoader.loadClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.applet.AppletClassLoader.lo\n\nCaused by: java.io.IOException: open HTTP connection failed. at sun.applet.AppletClassLoader.getBytes(Unknown Source) at sun.applet.AppletClassLoader.access$100(Unknown Source) at sun.applet.AppletClassLoader$1.run(Unknown Source) at java.security.AccessController.doPrivileged(Native Method) ... 10 more \n\nException: java.lang.ClassNotFoundException: com.diginet.digichat.client.DigiChatApplet 29/02/2004 13:19:05 sun.plugin.util.PluginLogger log INFO: Exception: java.lang.ClassNotFoundException: com.diginet.digichat.client.DigiChatApplet Finding information ... 29/02/2004 13:19:05 sun.plugin.util.PluginLogger log \n[/quote]", "Solution_13": "I asked the folks at Digichat and they asked that you run a tracert. Here's how:\r\n\r\n[quote]Open up a command line (for windows, start > run > cmd), then type \"tracert \", where host is replaced with the actual host. This will tell us if they can establish a connection to our servers.[/quote]\r\n\r\nLet me know what that yields and we'll see if we can get you connected.", "Solution_14": "Cool! So, uh.. this is probably obvious, but whats the host? :D", "Solution_15": "i believe the host would be artofproblemsolving.com, doing a trace route just shows you what servers/junction points you pass when trying to connect to a certain host, and how long it takes.", "Solution_16": "Oh.. well here we go...\r\n\r\nMicrosoft Windows XP [Version 5.1.2600]\r\n(C) Copyright 1985-2001 Microsoft Corp.\r\n\r\nC:\\Documents and Settings\\Stephen>tracert artofproblemsolving.com\r\n\r\nTracing route to artofproblemsolving.com [65.36.154.75]\r\nover a maximum of 30 hops:\r\n\r\n 1 2079 ms 1729 ms 128 ms e0.nas5.akl1.maxnet.net.nz [210.55.230.147]\r\n 2 146 ms 139 ms 149 ms fa0-0-0-2.br01.akl1.maxnet.net.nz [210.55.230.15\r\n8]\r\n 3 372 ms 375 ms 246 ms 39.ATM3-0.r4akl2.wcom.net.nz [203.110.31.105]\r\n 4 152 ms 139 ms 139 ms f0-0.r0akl2.wcom.net.nz [203.21.28.1]\r\n 5 153 ms 182 ms 182 ms 64.Serial3-0.GW7.SYD2.ALTER.NET [203.166.42.237]\r\n\r\n 6 174 ms 171 ms 170 ms 427.at-6-0.XR2.SYD2.ALTER.NET [210.80.33.193]\r\n 7 160 ms 171 ms 139 ms so-6-0-0.TR1.SYD2.ALTER.NET [210.80.51.249]\r\n 8 331 ms 332 ms 332 ms 0.so-5-0-0.IR1.LAX12.Alter.Net [210.80.49.149]\r\n 9 317 ms 325 ms 354 ms POS2-0.IR1.LAX9.ALTER.NET [137.39.31.222]\r\n 10 827 ms 709 ms 624 ms 0.so-5-2-0.TL1.LAX9.ALTER.NET [152.63.0.146]\r\n 11 949 ms 857 ms 981 ms 0.so-1-0-0.XL1.LAX9.ALTER.NET [152.63.115.141]\r\n 12 1092 ms 772 ms 429 ms POS6-0.BR2.LAX9.ALTER.NET [152.63.114.241]\r\n 13 513 ms 493 ms 375 ms 204.255.169.138\r\n 14 504 ms 461 ms 321 ms bur-core-01.inet.qwest.net [205.171.13.9]\r\n 15 1742 ms * 2551 ms iah-core-02.inet.qwest.net [205.171.205.26]\r\n 16 1556 ms 1320 ms 385 ms iah-core-01.inet.qwest.net [205.171.31.1]\r\n 17 398 ms 418 ms 407 ms ewr-core-02.inet.qwest.net [205.171.8.206]\r\n 18 386 ms 396 ms 418 ms ewr-edge-01.inet.qwest.net [205.171.17.38]\r\n 19 399 ms 386 ms 397 ms 65.115.225.238\r\n 20 400 ms 386 ms 396 ms linux1.artofproblemsolving.com [65.36.154.75]\r\n\r\nTrace complete.\r\n\r\nC:\\Documents and Settings\\Stephen>", "Solution_17": "[quote=\"TripleM\"]Oh.. well here we go... [/quote]\r\n\r\nHi, TripleM, I for one would love to see you participating in math jams, so I hope your technical problems get resolved. It does seem that some of those hops are rather long (more than a second, in one instance I noted). I used to live in Taiwan (till July 2001), and in those days trans-Pacific connections in the north Pacific were sometimes painfully slow. I had a lot of problem in Taiwan, too, with local modem pool latency. \r\n\r\nNow back in North America, usually connecting to North American sites through a local Internet service provider with a local broadband connection to a good Internet backbone, I usually enjoy high speed of connection, but I do notice that when I type messages here on AoPS (as I am doing right now) that sometimes the characters don't show up on the screen right away. I DON'T seem to have that issue on other sites at which I type to reply to PHPBB boards. So I'm still investigating what speed is workable for connecting to AoPS. We do get the math jams here, and stay on them successfully. \r\n\r\nI suppose/hope that these days there is a good fiber optic cable between New Zealand and North America. But perhaps sometimes just the extra router connections along the way may be adding latency to your connection. Best wishes to you for getting connected and participating in the math jams.", "Solution_18": "check out http://www.edgehill.ac.uk/mltr/lessons/viewfaq.htm it has some instructions on fixing Java problems, i have a feeling that your problem is cause 2 or 3 at that site, but just go through it and see if anything helps.", "Solution_19": "My bad - I should have read what I posted more closely. Try host5.digichat.com for the host.", "Solution_20": "Ahha!\r\n\r\nTracing route to digichat5.digi-net.com [64.235.52.50]\r\nover a maximum of 30 hops:\r\n\r\n 1 134 ms 117 ms 117 ms e0.nas2.akl1.maxnet.net.nz [210.55.230.151]\r\n 2 141 ms 150 ms 139 ms fa0-0-0-2.br01.akl1.maxnet.net.nz [210.55.230.15\r\n8]\r\n 3 679 ms 149 ms 139 ms 39.ATM3-0.r4akl2.wcom.net.nz [203.110.31.105]\r\n 4 147 ms 128 ms 117 ms f0-0.r0akl2.wcom.net.nz [203.21.28.1]\r\n 5 152 ms 149 ms 150 ms 64.Serial3-0.GW7.SYD2.ALTER.NET [203.166.42.237]\r\n\r\n 6 163 ms 149 ms 149 ms 327.at-3-0-0.XR1.SYD2.ALTER.NET [210.80.33.189]\r\n\r\n 7 646 ms 139 ms 246 ms so-3-0-0.TR1.SYD2.Alter.Net [210.80.48.133]\r\n 8 281 ms 300 ms 321 ms 0.so-3-1-0.IR1.LAX12.Alter.Net [210.80.49.161]\r\n 9 281 ms 311 ms 321 ms POS2-0.IR1.LAX9.ALTER.NET [137.39.31.222]\r\n 10 291 ms 311 ms 300 ms 0.so-5-2-0.TL1.LAX9.ALTER.NET [152.63.0.146]\r\n 11 282 ms 300 ms 300 ms 0.so-5-0-0.CL1.LAX15.ALTER.NET [152.63.115.197]\r\n\r\n 12 292 ms 300 ms 289 ms POS6-0.GW1.LAX15.ALTER.NET [152.63.115.209]\r\n 13 302 ms 300 ms 311 ms wcgGigELAX-gw.customer.alter.net [157.130.42.202\r\n]\r\n 14 282 ms 300 ms 300 ms anhmca1wcx2-pos5-0.wcg.net [64.200.140.69]\r\n 15 282 ms 311 ms 289 ms lsanca1wcx1-pos9-0.wcg.net [64.200.240.210]\r\n 16 * * * Request timed out.\r\n 17 * * * Request timed out.\r\n 18 * * lsanca9lca1-premianetworks-gige.wcg.net [64.200.139.98]\r\nreports: Destination net unreachable.\r\n\r\nSeems to be some problem there indeed..", "Solution_21": "Unfortunately, it appears that one of the routers between you and the digichat servers is bad. Sadly, this is not on either your end or ours, so we're pretty powerless to fix it. Hopefully the problem will be resolved in the near future and you'll be able to join our Math Jams in the future." } { "Tag": [ "modular arithmetic", "number theory unsolved", "number theory" ], "Problem": "Prove that if $p$ is a prime and $p^2+2$ is a prime then $p^3+2$ is a prime!", "Solution_1": "[quote=\"pavel25\"]Prove that if $p$ is a prime and $p^2+2$ is a prime then $p^3+2$ is a prime![/quote]\r\n\r\nbecause $p$ must be $3$. Indeed, if $p,p^2+2$ are both primes then if $(p,3)=1$ then $p^2+2\\equiv0\\pmod{3}$ which cotract! So $p=3$ and $p^3+2=29$ is prime!", "Solution_2": "[quote=\"pluricomplex\"][quote=\"pavel25\"]Prove that if $p$ is a prime and $p^2+2$ is a prime then $p^3+2$ is a prime![/quote]\n\nbecause $p$ must be $3$. Indeed, if $p,p^2+2$ are both primes then if $(p,3)=1$ then $p^2+2\\equiv0\\pmod{3}$ which cotract! So $p=3$ and $p^3+2=29$ is prime![/quote]\r\nthanks pluricomplex! and i have to add that $p=1$ too!", "Solution_3": "but i dont think of adding$p=1$." } { "Tag": [], "Problem": "Find the value of the sum: \r\n\\[ \\sum_{i=1}^{2001} [\\sqrt{i}] \\]\r\nwhere $[ {x} ]$ denotes the greatest integer which does not exceed $x$.", "Solution_1": "[hide]\nNote that for terms\n$i \\in [1,3], [\\sqrt i ] = 1$\n$i \\in [4,8], [\\sqrt i ] = 2$\n$i \\in [9,15], [\\sqrt i ] = 3$\nSo, $[\\sqrt i ] = n$ for $2n+1$ integers of $n$.\nThus we can express the sum as $\\sum_{k=1}^{43} (n(2n+1)) + \\sum_{k=44^2}^{2001} 44$\n$2 \\sum n^2 + \\sum n + 66 \\cdot 44$\n$\\dfrac{43 \\cdot 44 \\cdot 87}{3} + \\dfrac{43 \\cdot 44}{2} + 66 \\cdot 44$\n$\\boxed{58718}$\n[/hide]", "Solution_2": "[hide]\nI have an another approach.We can consider on cartesian coordinates.The value $\\sum_{i=1}^{2001} [\\sqrt i]$ denotes the number of lattice point $(x>0,y>0)$ below $y=\\sqrt x$.We can calculate the number of lattice point by considering the number of lattice point $(y>0,x>0)$ below $x=y^2$.Namely, $44\\cdot 2001-\\sum_{y=1}^{44} (y^2-1)=\\boxed{58718}$." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Let $f(x)=2x-sinx$. Prove that : there exists a real $b$ and functions $h(x)$and $g(x)$ satisfies the coditions \r\n1. $g(x)=bx+h(x)$\r\n2. $h(x)$is periodic \r\n3. $f(g(x))=x$ for all x", "Solution_1": "Obviosly b=1/2 and \r\n(1) $h(2x-sinx)=\\frac{sin x}{2}$.\r\nBecause $2x-sinx$ increase in $(0,2\\pi )$, then exist unique function$h: (0,4\\pi)\\to (-\\frac{1}{2},\\frac{1}{2})$, suth that (1). Obviosly$h(x)$ had period $4\\pi.$" } { "Tag": [ "superior algebra", "superior algebra solved" ], "Problem": "for any set $X$, let $P(X)=\\{G|G \\subseteq X\\}$ be the power set of $X$. For any $G$ and $H$ in $P(X)$ define.\r\n$G+H=\\{x|x \\in G \\cup H, x \\notin G \\cap H \\}$ and $G\\cdot H= G \\cap H$\r\n\r\nunder these two operations how can I show $P(X)$ is a ring with unity?", "Solution_1": "Nice, I never saw that before :)\r\n\r\nzero is the empty set, unity is whole $X$\r\nto prove the other necessary things: draw some graphic!", "Solution_2": "It's just a construction of a Boolean ring from a Boolean algebra. Another way to look at it is as being the direct product $\\mathbb Z_2^X$." } { "Tag": [ "geometry", "similar triangles", "geometry unsolved" ], "Problem": "[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14821[/img]\r\n\r\n :(", "Solution_1": "This problem is coming from the past.\r\n\r\n$ \\bullet$ When the triangles $ \\bigtriangleup AO_{1}B$ and $ \\bigtriangleup A'O_{1}B'$ are directly similar, we have that $ \\frac {O_{1}A}{O_{1}A'} \\equal{} \\frac {AB}{A'B'}$ $ ,(1)$ and $ \\frac {O_{1}B}{O_{1}B'} \\equal{} \\frac {AB}{A'B'}$ $ ,(2)$\r\n\r\nFrom $ (1),$ we conclude that the point $ O_{1},$ lies on the [b][size=100]Appolonius circle[/size][/b] $ (C_{1}),$ with diameter the segment $ MM',$ \r\n\r\nwhere $ M,\\ M',$ defined points on the line segment $ AA',$ such that $ \\frac {MA}{MA'} \\equal{} \\frac {M'A}{M'A'} \\equal{} \\frac {AB}{A'B'}$ $ ($ $ M,$ between $ A$ and $ A'$ $ ).$\r\n\r\nSimilarly from $ (2),$ we conclude that the point $ O_{1},$ lies also on the [b][size=100]Appolonius circle[/size][/b] $ (C_{2}),$ with diameter the segment $ NN',$\r\n\r\nwhere $ N,\\ N',$ defined points on the line segment $ BB',$ such that $ \\frac {NB}{NB'} \\equal{} \\frac {N'B}{N'B'} \\equal{} \\frac {AB}{A'B'}$ $ ($ $ N,$ between $ B$ and $ B'$ $ ).$\r\n\r\nThere are two points $ O_{1}$ and $ O'_{1},$ with the property as the problem states, as the intersection points of the circles $ (C_{1})$ and $ (C_{2}).$\r\n\r\n$ \\bullet$ By the same way, in the case of inversely similar triangles $ \\bigtriangleup AO_{2}B$ and $ \\bigtriangleup B'O_{2}A',$\r\n\r\nwhen we have that $ \\frac {O_{2}A}{O_{2}B'} \\equal{} \\frac {AB}{A'B'}$ $ ,(3)$ and $ \\frac {O_{2}B}{O_{2}A'} \\equal{} \\frac {AB}{A'B'}$ $ ,(4)$ \r\n\r\nthe point $ O_{2},$ lies on the [b][size=100]Appolonius circles[/size][/b] $ (C'_{1}),\\ (C'_{2}),$ with respect to the segments $ AB',\\ BA'$ respectively.\r\n\r\nThere also two points $ O_{2}$ and $ O'_{2},$ with the property as the problem states and the solution is completed.\r\n\r\nKostas vittas." } { "Tag": [], "Problem": "If a*b=22 and a+b=12,\r\nwhat is absolute value of b-a?\r\nA)2 sqrt 66\r\nB)2 sqrt 14\r\nC)5 sqrt 66\r\nD)2 sqrt 10\r\nE)5 sqrt 14", "Solution_1": "[quote=\"iamhe\"]If a*b=22 and a+b=12,\nwhat is absolute value of b-a?\nA)2 sqrt 66\nB)2 sqrt 14\nC)5 sqrt 66\nD)2 sqrt 10\nE)5 sqrt 14[/quote]\r\n\r\n[hide=\"generic\"]\n$ a\\plus{}b\\equal{}12$\n$ (a\\plus{}b)^2\\equal{}144$\n$ a^2\\plus{}2ab\\plus{}b^2\\minus{}4ab\\equal{}144\\minus{}88$\n$ (a\\minus{}b)^2\\equal{}56$\n$ |a\\minus{}b|\\equal{}|b\\minus{}a|\\equal{}2\\sqrt{14}$\n\n[/hide]" } { "Tag": [ "geometry", "geometry proposed" ], "Problem": "Let $ H$ be the orthocenter of an acute triangle $ ABC$ whose edgesare $ BC\\equal{}a;CA\\equal{}b;AB\\equal{}c$\r\nProve that\r\n$ \\frac{HA^2\\plus{}HB^2\\plus{}HC^2}{a^2\\plus{}b^2\\plus{}c^2} \\leq (cotA.cotB)^2\\plus{}(cotB.cotC)^2\\plus{}(cotCcotA)^2$", "Solution_1": "I posted it before! Please solve it!" } { "Tag": [ "trigonometry" ], "Problem": "The value of $ \\tan 10^{\\circ}$ is denoted by $ p$. Find, in terms of $ p$, the value of \r\n\r\n(i) $ \\tan 55^{\\circ}$,\r\n\r\n(ii) $ \\tan 5^{\\circ}$,\r\n\r\n(iii) $ \\tan \\alpha$, where $ \\theta$ satisfies the equation $ 3 \\sin(\\theta \\plus{} 10^{\\circ}) \\equal{} 7 \\cos (\\theta \\minus{} 10^{\\circ})$.\r\n.", "Solution_1": "[hide=\"i\"]$ \\tan 55^{\\circ}\\equal{}\\tan(10\\plus{}45)^{\\circ}\\equal{}\\frac{\\tan 10\\plus{}\\tan 45}{1\\minus{}\\tan 45\\tan 10}\\equal{}\\frac{p\\plus{}1}{1\\minus{}p}$[/hide]\n\n[hide=\"ii\"]$ \\frac{2\\tan 5}{1\\minus{}\\tan^2 5}\\equal{}p\\Rightarrow 2\\tan 5\\equal{}p\\minus{}p\\tan^2 5$\n$ \\Rightarrow \\tan 5\\equal{}\\frac{\\minus{}2\\plus{}\\sqrt{4\\plus{}4p^2}}{2p}\\equal{}\\frac{\\minus{}1}{p}\\plus{}\\frac{\\sqrt{1\\plus{}p^2}}{p}$[/hide]" } { "Tag": [ "number theory", "prime factorization" ], "Problem": "How many positive factors does the number 900 have?", "Solution_1": "Since $ 900\\equal{}2^2\\cdot3^2\\cdot5^2$, there are $ (2\\plus{}1)^3\\equal{}\\boxed{27}$ factors.", "Solution_2": "[hide=\"To find the number of factors of a number\"]\nYou find the prime factorization of that number and multiply the number of exponents + 1 for each prime number. Yeah, I know it sounds confusing, but lemme give you an example. To find the number of factors for the number $20$, you factors it with primes, so $20=2^2\\cdot5^1$. Any factor(s) of $2^{2}5^{1}$ must be in the form $2^{x}5^{y}$, where $2\\ge{x}\\ge0$ and $1\\ge{y}\\ge0$, and they must be integers. Therefore, $x$ could be $0,1,2$ and $y$ could be $0,1$. We multiply them to get $\\boxed{6}$ factors. This is how we determine that $20$ has $6$ factors.\n[/hide]\n\nNow, we know that $900=2^{2}3^{2}5^{2}$, so we multiply by $3\\cdot3\\cdot3=3^{3}=\\boxed{27}$.\n\nHope I helped!\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\n[color=transparent]Please rate, like, comment, or subscribe to my YouTube channel at youtube.com/user/DAMENAOCL12[/color]\n\n[size=50]teehee[/size]", "Solution_3": "Thanks! Forgot how to find the number of factors but I know now! I know you probably won't see this because it has been 7 years though\n", "Solution_4": "[quote=Greenacid]Thanks! Forgot how to find the number of factors but I know now! I know you probably won't see this because it has been 7 years though[/quote]\n\nLol" } { "Tag": [ "geometry", "incenter", "analytic geometry", "linear algebra", "matrix", "trigonometry", "function" ], "Problem": "I, G, H are incenter, gravity center (medicenter) and orthocenter of triangle ABC respectively (there are no two equal sides).\r\na) Is it posible IH to be parallel to some of the sides of a triangle?\r\nb) Is it possible triangle IGH to be equaliteral?", "Solution_1": "$ \\text{1). In barycentric coordinate I(a; b; c) and G(1; 1; 1)} \\Longrightarrow \\\\\r\n\\text{IG line equation: }\\begin{equation*} \\begin{vmatrix} x & y & z \\\\\r\na & b & c \\\\\r\n1 & 1 & 1 \\end{vmatrix} \\end{equation*} = 0 \\Longleftrightarrow (b - c).x + (c - a)y + (a - b).z = 0 \\text{ and BC line equation: x = 0.}$\r\n$ \\text{ 2). }\\boxed{{IG}\\|{BC}}\\Longleftrightarrow$\r\n$ \\Longleftrightarrow \\begin{cases} \\text{(b - c).x + (c - a)y + (a - b).z = 0} \\\\\r\nx = 0 & \\text{have solution: }y^2 + z^2\\ne{0} \\\\\r\nx + y + z = 0 \\end{cases} \\end{equation*} \\Longleftrightarrow$\r\n$ \\Longleftrightarrow \\begin{equation*} \\begin{cases} \\text{(c - a)y + (a - b).z = 0} \\\\\r\ny + z = 0 \\end{cases} \\end{equation*} \\text{have solution: } y^2 + z^2\\ne{0} \\Longleftrightarrow$ $ \\begin{equation*} \\begin{vmatrix} c - a & a - b \\\\\r\n1 & 1 \\end{vmatrix} \\end{equation*} = 0$ \r\n$ \\Longleftrightarrow \\boxed{ b + c = 2.a.}$\r\n$ \\text{3). Have: H(tgA; tgB; tgC), and barycentric equation the line IH: }$\r\n$ \\begin{equation*} \\begin{vmatrix} x & y & z \\\\\r\na & b & c \\\\\r\ntgA & tgB & tgC \\end{vmatrix} \\end{equation*} = 0 \\Longleftrightarrow \\\\\r\n\\Longleftrightarrow(b.tgC - c.tgB).x + (c.tgA - a.tgC)y + (a.tgB - b.tgA).z = 0\\text{ and: } \\\\\r\n\\boxed{{HI}\\|{BC}}\\Leftrightarrow{c.tgA - a.tgC = a.tgB - b.tgA}\\Longleftrightarrow\\boxed{(c + b).tgA = a.(tgB + tgC).}$", "Solution_2": "Thank you a lot, but the condition of a) were wrong. I edited it. It should be IH to be parallel to some of the sides.\r\n\r\nFor b) I've an idea. If it is possible give an answer may we from:\r\n$ AI^{2} \\plus{} BI^{2} \\plus{} CI^{2} \\equal{} AH^{2} \\plus{} BH^{2} \\plus{} CH^{2}$\r\nTo get ABC is equaliteral? If yes it is impossible, because it is impossible IG=GH for non equaliteral triangle. Please if you can - verify this hypothesis.", "Solution_3": "[b][color=darkblue]Prove easily that $ IH\\parallel BC$ $ \\Longrightarrow$ the projection of $ A$ on $ BC$ belongs to $ (BC)\\ !$\nThus, $ \\boxed {IH\\parallel BC}$ $ \\Longleftrightarrow$ $ \\frac {2p}{a} \\equal{} \\tan B\\tan C$ $ \\Longleftrightarrow$ $ a\\sin B\\sin C \\equal{} 2p\\cos B\\cos C$ $ \\Longleftrightarrow$ $ abc \\equal{} 8R^2p\\cos B\\cos C$ $ \\Longleftrightarrow$ $ 4Rpr \\equal{} 8R^2p\\cos B\\cos C$ $ \\Longleftrightarrow$ $ \\frac rR \\equal{} 2\\cos B\\cos C$ $ \\Longleftrightarrow$ $ \\frac rR \\equal{} \\cos (B \\minus{} C) \\minus{} \\cos A$ $ \\Longleftrightarrow$ $ 2\\cos A \\plus{} \\cos B \\plus{} \\cos C \\equal{}$ $ 1 \\plus{} \\cos (B \\minus{} C)$ $ \\Longleftrightarrow$ $ \\cos^2\\frac A2 \\minus{} \\sin^2\\frac A2 \\plus{} \\cos\\frac {B \\plus{} C}{2}\\cos\\frac {B \\minus{} C}{2} \\equal{}$ $ \\cos^2\\frac {B \\minus{} C}{2}$ $ \\Longleftrightarrow$ $ \\sin^2\\frac {B \\plus{} C}{2} \\minus{} \\cos^2\\frac {B \\plus{} C}{2} \\equal{}$ $ 2\\sin\\frac B2\\sin\\frac C2\\cos\\frac {B \\minus{} C}{2}$ $ \\Longleftrightarrow$ $ \\left(\\tan\\frac B2 \\plus{} \\tan\\frac C2\\right)^2 \\minus{} \\left(1 \\minus{} \\tan\\frac B2\\tan\\frac C2\\right)^2 \\equal{}$ $ 2\\tan \\frac B2\\tan\\frac C2\\left(1 \\plus{} \\tan\\frac B2\\tan\\frac C2\\right)$ $ \\Longleftrightarrow$ $ \\boxed {\\left(\\tan\\frac B2 \\plus{} \\tan\\frac C2\\right)^2 \\equal{} 1 \\plus{} 3\\tan^2\\frac B2\\tan^2\\frac C2}\\ .$ Denote $ \\tan\\frac B2 \\equal{} x$, $ \\tan\\frac C2 \\equal{} y\\ .$ Thus, $ IH\\parallel BC$ $ \\Longleftrightarrow$ exist $ x > 0$, $ y > 0$ so that $ (x \\plus{} y)^2 \\equal{} 1 \\plus{} 3x^2y^2$, i.e. exists $ x > 0$ so that the equation (in $ y$) $ (3x^2 \\minus{} 1)\\cdot y^2 \\minus{} 2x\\cdot y \\plus{} 1 \\minus{} x^2 \\equal{} 0$ has at least a positive root. It is possibly, for example $ \\frac {\\sqrt 3}{3} < x < 1\\ !$\n\nI hope that there are no any mistakes in the upper proof. Here are two remarks :\n\n$ \\boxed {IG\\parallel BC}$ $ \\Longleftrightarrow 3r \\equal{} h_a$ $ \\Longleftrightarrow$ $ 3ar \\equal{} ah_a$ $ \\Longleftrightarrow$ $ 3ar \\equal{} 2pr$ $ \\Longleftrightarrow$ $ 2p \\equal{} 3a$ $ \\Longleftrightarrow$ $ \\boxed {b \\plus{} c \\equal{} 2a}\\ .$[/color]\n\n[color=darkred][u]Exercise.[/u]$ \\boxed {IG\\perp BC\\Longleftrightarrow b \\equal{} c\\ \\ \\vee\\ \\ b \\plus{} c \\equal{} 3a}$ (it is interestingly).[/color][/b]", "Solution_4": "I've only one stupid question. How you get the condition IH||BC?", "Solution_5": "$ IH\\parallel BC\\Longleftrightarrow$ $ [BHC]\\equal{}[BIC]$ (the areas).", "Solution_6": "Very good idea. I think your proof is very good (if i don't have a mistake, too). And I also think there are infinitely many triangles with required property.", "Solution_7": "[b][color=darkblue]Here is the particular case $ \\|\\begin{array}{c} \\tan\\frac B2 = \\frac {r}{DB} = x = \\frac 23\\in(\\frac {\\sqrt 3}{3},1) \\\\\n \\\\\n\\tan\\frac C2 = \\frac {r}{DC} = y = \\frac {6 - \\sqrt {21}}{3} < 1\\end{array}$, where $ D\\in BC\\cap C(I,r)\\ .$[/color][/b]\r\n\r\n[b][color=darkred][u]Can someone to draw[/u] a triangle $ ABC$ for which $ \\frac {r}{10} = \\frac {DB}{15} = \\frac {DC}{12 + 2\\sqrt {21}}\\ ?$[/color]\n\n[color=darkblue]Please answer me if you\"ll obtain $ IH\\parallel BC\\ !$ Thank you a lot ![/color][/b]", "Solution_8": "I don't understood you in depth. But I think:\r\nBy your conditions we may find $ sin \\alpha$ and $ sin \\gamma$\r\nDo you want more details? I think this triangle is constructible. And the construction is easy.", "Solution_9": "$ \\boxed{IH\\|BC}\\Longleftrightarrow (b \\plus{} c).tgA \\equal{} (tgB \\plus{} tgC).a\\Longleftrightarrow \\\\\r\n\\Longleftrightarrow (b \\plus{} c).\\frac {tgB \\plus{} tgC}{1 \\minus{} tgB.tgC} \\plus{} (tgB \\plus{} tgC).a \\equal{} 0\\Longleftrightarrow tgB.tgC \\equal{} \\frac {a \\plus{} b \\plus{} c}{a} \\equal{} \\frac {h_a}{r}.$\r\n$ \\text{Similar of Virgil Nicula assertion!}$", "Solution_10": "Thank you Virgil Nicula and mihai miculita! I think you are right. In original statement of the problem were written to find dimension of the angles of the triangles for a) and b). But I think it is not a good idea, because in my opinion for a) we may have infinitely many triangles with this property. Am I right?", "Solution_11": "$ \\text{Using the sin and cos laws, have: }\\\\\r\n(b\\plus{}c).tgA\\equal{}a.(tgB\\plus{}tgC)\\Longleftrightarrow (b\\plus{}c).\\frac{sinA}{cosA}\\equal{}a.(\\frac{sinB}{cosB}\\plus{}\\frac{sinC}{cosC})\\Longleftrightarrow \\\\\r\n\\Longleftrightarrow \\dots \\Longleftrightarrow \\frac{b\\plus{}c}{b^2\\plus{}c^2\\minus{}a^2}\\equal{}a.(\\frac{1}{a^2\\plus{}c^2\\minus{}b^2}\\plus{}\\frac{1}{a^2\\plus{}b^2\\minus{}c^2})\\Longleftrightarrow \\\\\r\n\\Longleftrightarrow \\boxed{2.a^3.(b^2\\plus{}c^2\\minus{}a^2)\\equal{}(b\\plus{}c).(a^2\\plus{}c^2\\minus{}b^2).(a^2\\plus{}b^2\\minus{}c^2)}.$", "Solution_12": "What about b) ?", "Solution_13": "$ \\text{Have: } \\\\\r\n|IG|^2 \\equal{} \\frac {2}{9}.(p^2 \\plus{} r^2 \\minus{} 16.Rr),\\text{ } |GH|^2\\equal{}R^2\\minus{}\\frac{4}{9}.(a^2\\plus{}b^2\\plus{}c^2),\\\\\r\n|HI|^2 \\equal{} 2.r^2 \\minus{} 4.R^2.cosA.cosB.cosC \\equal{} 4.R^2 \\plus{} 4.Rr \\plus{} 3.r^2 \\minus{} p^2, \\text{ if }p \\equal{} \\frac {a \\plus{} b \\plus{} c}{2}.$", "Solution_14": "[color=darkblue][b]The relations $ \\left\\|\\begin{array}{c} 9\\cdot GI^2 \\equal{} p^2 \\plus{} 5r^2 \\minus{} 16Rr \\\\\n \\\\\nHI^2 \\equal{} 4R^2 \\plus{} 4Rr \\plus{} 3r^2 \\minus{} p^2 \\\\\n \\\\\n9\\cdot GH^2 \\equal{} 36\\cdot R^2 \\minus{} 8\\cdot \\left(p^2 \\minus{} r^2 \\minus{} 4Rr\\right)\\end{array}\\right\\|$ are well-known.\n\nTherefore, $ GI \\equal{} HI \\equal{} GH$ $ \\Longleftrightarrow$ $ \\left\\|\\begin{array}{c} 5p^2 \\equal{} 18R^2 \\plus{} 26Rr \\plus{} 11r^2 \\\\\n \\\\\n3p^2 \\equal{} 12R^2 \\plus{} r^2 \\plus{} 16Rr\\end{array}\\right\\|$ $ \\Longleftrightarrow$\n\n$ \\left\\|\\begin{array}{c} p^2 \\equal{} 3\\cdot\\left(2R^2 \\plus{} 2Rr \\minus{} 3r^2\\right) \\\\\n \\\\\n3R^2 \\plus{} Rr \\minus{} 14r^2 \\equal{} 0\\end{array}\\right\\|$ $ \\Longleftrightarrow$ $ \\left\\|\\begin{array}{c} R \\equal{} 2r \\\\\n \\\\\np \\equal{} 3r\\sqrt 3\\end{array}\\right\\|$ $ \\Longleftrightarrow$ $ a \\equal{} b \\equal{} c\\ .$\n\nIn conclusion, $ \\triangle\\ GHI$ is equilateral iff $ \\triangle\\ ABC$ is equilateral, i.e. $ G\\equiv H\\equiv I\\ .$[/b][/color]", "Solution_15": "Excellent job, Virgil Nicula! Very good.", "Solution_16": "Virgil Nicula, I've an idea - you use very interesting equalities for the three distances. How do you get them? May you find similar relations for the sides of the triangle by its angle bisectors? \r\n\r\nI ask you, because I've posted a problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=172431 and I'm interested is there a solution in elementary functions and if there is a solution how to find it and if there is no a solution - how to prove it.\r\n\r\nP.S. Just to know the triangle cannot be equaliteral because GIH>90 for every triangle. I found it after I found this problem on another forum and you had solved it." } { "Tag": [ "geometry", "geometry unsolved" ], "Problem": "Dear Mathlinkers,\r\nLet ABC be a triangle, Na the Nagel's point of ABC, (1) the incircle of ABC, I the center of (1), DEF the contact triangle of ABC, H' the orthocenter of DEF, and U, V, W the antipoles of D, E, F wrt (1).\r\nProuve : the isogonal of H' wrt the triangle UVW is the Nagel's point Na of ABC.\r\nSincerely\r\nJean-Louis", "Solution_1": "[b]Lemma:[/b] \"$ AU, BV, CL$ concur at Nagel point of triangle $ ABC$\". It's well-known so I leave proof for the reader :oops: \r\nApplying this lemma above, we will prove that $ AU$ and $ UH'$ are the isogonal lines wrt $ \\angle VUW. (*)$\r\n$ \\Leftrightarrow \\angle AUW\\plus{}\\angle VUH'\\equal{}180^o$\r\nBut $ \\angle AUW\\equal{}\\angle AUE\\plus{}\\angle EUW\\equal{}\\angle AUE\\plus{}\\angle IFE$\r\nAnd $ \\angle VUH'\\equal{}\\angle FUH'\\minus{}\\angle FUV\\equal{}\\angle FUH'\\minus{}\\angle IEF$\r\nThus we need to show $ \\angle AUE\\plus{}\\angle FUH'\\equal{}180^o$\r\nOn the other side, $ U$ is the antipole of $ D$ wrt $ (I)$ then $ U$ and $ H'$ are symmetric through the midpoint of $ EF$.\r\nDenote $ L$ the second intersection of $ AU$ and $ (I)$.\r\nThe well-known result about harmonic quadrilateral: $ UH'$ and $ UL$ are isogonal lines wrt $ \\angle FUE$\r\nWhich follow that $ \\angle AUE\\plus{}\\angle FUH'\\equal{}\\angle AUE\\plus{}\\angle EUL\\equal{}180^o$\r\nTherefore $ (*)$ is true. Similarly for $ BV, CL$ we completed the proof." } { "Tag": [ "limit", "integration", "trigonometry", "calculus", "real analysis", "function", "absolute value" ], "Problem": "Suppose $f$ is bounded on $[0,\\infty)$ and\r\n\r\n$\\lim_{a\\to\\infty}\\int_0^af(x)dx=L$ exists.\r\n\r\nProve that $\\lim_{t\\to0^+}\\int_0^{\\infty}e^{-tx}f(x)dx=L.$\r\n\r\nApplication: once you have proved this, use it in a computation of\r\n\r\n$\\int_0^{\\infty}\\frac{\\sin x}xdx.$", "Solution_1": "nice one.\r\nalso known as laplace transform", "Solution_2": "Tr: saying that this is related to the Laplace transform doesn't solve it. There's work to do.\r\n\r\nIf $\\int_0^{\\infty}|f(x)|dx$ is finite, the question isn't very interesting. For the theorem to gain you anything, you might as well assume the indefinite integral is conditionally convergent (as in the application.)\r\n\r\nAlso, the hypothesis \"$f$ is bounded\" is much stronger than we really need; it would be sufficient for $f$ to be locally (Lebesgue) integrable. That is, $\\int_a^b|f(x)|dx$ is finite for each finite $a$ and $b$ - that's enough.", "Solution_3": "$F(u)=-\\int_{u}^{+\\infty}f(t)dt$ tend to $0$ when $u$ tends to $+\\infty$\r\n\r\n$\\epsilon>0, \\exists A>0, \\forall u\\geq A, |F(u)|\\leq \\epsilon$\r\nFor $x\\geq 0$ and $A\\leq \\alpha < \\beta$\r\nIntegration by parts gives\r\n$\\int_{\\alpha}^{\\beta}e^{-xt}f(t)dt=[e^{-tx}F(t)]_{\\alpha}^{\\beta}+x\\int_{\\alpha}^{\\beta}e^{-xt}F(t)dt$\r\nThis implies\r\n(1) $|\\int_{\\alpha}^{\\beta}e^{-xt}f(t)dt|\\leq 3\\epsilon$\r\n\r\n(2) ${| \\int_{0}^{+\\infty}(e^{-tx}-1)f(t)dt|\\leq |\\int_{0}^A}(e^{-tx}-1)f(t)dt|+|\\int_{A}^{+\\infty}e^{-xt}f(t)dt|+|\\int_{A}^{+\\infty}f(t)dt|$\r\n\r\nthe first integral is bounded by $||f||_{\\infty}.A.(1-e^{-Ax})$ tends to $0$ when \r\nx tends to $0$\r\nthe second integral is less than $3\\epsilon$ \r\nthe third one is less than $\\epsilon$\r\n\r\nThe LHS of (2) tends to $0$ when x tends to $0$\r\n\r\nMathlinks great place to be for practice the Art of Problem Solving\r\nthank you Valentin Vornicu for this forum.", "Solution_4": "[quote=\"Moubinool\"]the first integral is bounded by $ ||f||_{\\infty}.A.(1-e^{-Ax}) $ tends to $ 0 $ when x tends to $ 0 $ [/quote]\r\nI made the comment that we don't really need $f$ to be bounded, just locally integrable, so I'd like not to use $||f||_{\\infty}$ if possible. Since $A$ has already been chosen above, this integral is bounded by $(1-e^{-Ax})\\int_0^A|f(t)|dt,$ and that does go to $0$ as $x\\to0.$\r\n\r\nThank you Moubinool, that works nicely. We don't need $f$ to be bounded; this works just fine for, say, $f(x)=x\\cos(x^3).$\r\n\r\n(Note that Moubinool reversed the usage of the letters $t$ and $x$.)\r\n\r\nNow, what about $\\int_0^{\\infty}\\frac{\\sin x}xdx$?\r\n\r\nSuggestion: can you differentiate $G(t)=\\int_0^{\\infty}e^{-tx}\\frac{\\sin x}xdx$?", "Solution_5": "[quote=\"Kent Merryfield\"]\nSuggestion: can you differentiate $G(t)=\\int_0^{\\infty}e^{-tx}\\frac{\\sin x}xdx$?[/quote]\r\n\r\nThis one is well known for Oral Examination X-ENS in France \r\nI let it for harazi, grobber, alekk good training \r\n\r\nHints: Show that G is derivable for $t \\geq 0$, find the expression of $G'$ without \r\nintegral symbol, take a primitive and use the continuity of $G$ at $0$, also limit \r\nof G when $t$ tends to $+\\infty$", "Solution_6": "I think that $G'(t)=-\\frac{1}{t^2+1}$. Anyway, I won't post the proof I found, since it is too long and ugly. There must be a nice solution for this monster.", "Solution_7": "Yes, for $t>0, G'(t)=-\\frac1{t^2+1}.$ As for the justification of that:\r\n\r\n$\\frac{G(t+h)-G(t)}h=\\int_0^{\\infty}\\frac{e^{-(t+h)x}-e^{-tx}}h\\cdot\\frac{\\sin x}xdx$\r\n\r\nWe now want to take the limit as $h\\to0$ and bring that limit inside the integral sign. By the Mean Value Theorem, $\\frac{e^{-(t+h)x}-e^{-tx}}h=e^{-sx}$ for some $s$ between $t$ and $t+h$. If we restrict $h$ to $|h|<\\frac t2$ then this is bounded in absolute value by $e^{-tx/2}.$ Using also the estimate $\\left|\\frac{\\sin x}x\\right|\\le 1$ we get that the integrand for the difference quotient is dominated by the integrable function $e^{-tx/2}.$ Then the interchange of limit and integral is justified by the Dominated Convergence Theorem. Once you've done that, the actual computation of $G'(t)$ is a routine calculus exercise.\r\n\r\nOne thing to note: because we're going to use the \"continuous version of Abel's Theorem\" proved above, we only need to do these calculations for $t>0$, not worrying about what happens when $t=0.$" } { "Tag": [], "Problem": "The equation $ x^3\\plus{}ax^2\\plus{}bx\\plus{}c\\equal{}0$ has three real roots.Prove that:\r\n $ 2(ab\\plus{}c)\\le\\ a^3\\plus{}4\\sqrt{2(a^2\\minus{}b)}$", "Solution_1": "you can post solution?It very nice :D" } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "For $a,b,c,d \\in \\mathbb{R}^{+}$ and $k \\in \\mathbb{R}$ prove that:\r\n\r\n$3(a^{k+4}+b^{k+4}+c^{k+4}+d^{k+4})+abcd (a^{k}+b^{k}+c^{k}+d^{k}) \\geq (a+b+c+d)(a^{k+3}+b^{k+3}+c^{k+3}+d^{k+3})$", "Solution_1": "Let $S_{k}(a,b,c)=a^{k}(a-b)(a-c)+b^{k}(b-a)(b-c)+c^{k}(c-a)(c-b)$.\r\n$3(LHS-RHS)$\r\n$=a S_{k+1}(b,c,d)+b S_{k+1}(a,c,d)+c S_{k+1}(a,b,d)+d S_{k+1}(a,b,c)$\r\n$+2(S_{k+2}(b,c,d)+S_{k+2}(a,c,d)+S_{k+2}(a,b,d)+S_{k+2}(a,b,c)) \\ge 0$.", "Solution_2": "Hello,\r\n\r\nThank you, that solution works. But how did you come up with it in the first place?\r\nAlso, how do I show that the whole expression indeed evaluates to $LHS-RHS$ without expanding everything and brute-force algebra? Is there a trick that I'm missing?" } { "Tag": [ "function", "MIT", "college" ], "Problem": "Cells of certain organisms have organelles that produce ATPs and use in the synthesis of organic substances from carbon dioxide. These organelles are:\r\n\r\na) the lysosomes\r\nb) the mitochondria\r\nc) the chloroplasts\r\nd) the Golgi system\r\ne) the nucleoli", "Solution_1": "Here's one way to go about the problem: which of the organelles definitely DON'T do what you're asking? Lysosomes are similar to vacuoles, which aid in the digestion of macromolecules. The Golgi apparatus aid in the processing and packaging of macromolecules. The nucleoli works with the transcription of ribosomal DNA.\r\n\r\nThat leaves you down to the mitochondria and the chloroplasts, both of which deal with ATP. Your goal, is to figure out which one fits. The key is that the question description gives you all the information you need-one of those two organelles should scream out at you, as its function is carry out the process which converts carbon dioxide into organic compounds.", "Solution_2": "I don't know :blush: \r\n\r\nI only copy and cut this question from my book. Do you know the answer ?\r\n\r\n :D", "Solution_3": "Yes, I know the answer. Just look at the last line of my first post. If you know what does that you've got the answer (and I already narrowed it down to mitochondria and chloroplast for you, so you don't have much work to do).", "Solution_4": "Well ... then ... the answer is C ? chloroplasts ? \r\n\r\nThanks. I think this is the answer. It's correct ?", "Solution_5": "yeah correct!!!\r\n\r\nthough both mit and chl prod energy,org matter in synth in chl only (i.e. frm CO2)", "Solution_6": "Oh :lol: \r\n\r\nThanks ;)", "Solution_7": "This is kinda related but i forgot what the fold inside the mitochondria was called, can anyone remind me?", "Solution_8": "[quote=\"eli140\"]This is kinda related but i forgot what the fold inside the mitochondria was called, can anyone remind me?[/quote]\r\n\r\nCristae (sing. crista). Just look at Wiki or one of the diagrams in your bio book (Campbell or whatnot).", "Solution_9": "Thanks :lol:" } { "Tag": [ "probability", "conditional probability" ], "Problem": "The digits from 1 to 6 are arranged to form a six-digit multiple of 5. What is the probability that the number is greater than 500,000? Express your answer as a common fraction.", "Solution_1": "Because it is a multiple of 5, the last digit must be 5. There are 5 other digits besides 5. The number is only greater than 500000 if the first digit is 6. Because there are 5 possible digits, and only 1 results in the number being greater than 500000, the answer is $ \\frac {1}{5}$.", "Solution_2": "hmm...\r\nwait\r\nif the first digit has to be a $ 6$, and the last digit has to be a $ 5$, then the remaining four digits can be arranged in $ 24$ ways\r\nthere are $ 6!$ ways to arrange every digit, so that's $ 720$\r\nso, the answer is $ \\frac {24} {720} \\equal{} \\frac {1} {30}$", "Solution_3": "No... because we are already GIVEN that the last digit is 5. It's conditional probability.", "Solution_4": "Don't divide by 6 because it already says its divisible by 5", "Solution_5": "In conditional probability, the probability of an event A given an event B is $ \\frac {P(A \\cap B)}{P(B)}$. We are looking for the probability that the number is greater than 500000 GIVEN that it is a multiple of 5. This is obviously different from the probability that the number is greater than 500000 AND is a multiple of 5, which is what you found, vallon22." } { "Tag": [ "algorithm", "number theory", "Euclidean algorithm" ], "Problem": "find the minimum integer value of $ n$ for which $ 3n \\minus{} 233$ and $ 8n \\plus{} 201$ will have a common positive divisor, other than $ 1$.", "Solution_1": "By the Euclidean Algorithm,\r\n$ \\gcd(3n\\minus{}233,8n\\plus{}201)$\r\n$ \\equal{}\\gcd(3n\\minus{}233,2n\\plus{}667)$\r\n$ \\equal{}\\gcd(n\\minus{}900,2n\\plus{}667)$\r\n$ \\equal{}\\gcd(n\\minus{}900,2467)$\r\n$ 2467$ is prime, so the smallest possible value of $ n$ is $ 900\\plus{}2467\\equal{}\\boxed{3367}$.", "Solution_2": "[b]@ OP:[/b] Remember to write \"positive\" integer next time :)" } { "Tag": [ "probability", "group theory", "expected value", "combinatorics unsolved", "combinatorics" ], "Problem": "In a simple graph on $ n$ points, the points have degrees $ 010$.\r\n$ p_n\\equal{}\\frac{10S_{n\\minus{}1}}{Pr_{n\\minus{}1}\\minus{}10}$. If $ n\\equal{}2$ we get $ p_2\\equal{}\\frac{10p_1}{p_1\\minus{}10}\\to p_1>10\\to p_2$ is not prime.\r\nIf $ p_1>2$, then $ 10S_{n\\minus{}1}$ is even and $ Pr_{n\\minus{}1}\\minus{}10$ odd, therefore $ 2|p_3, p_n>2$ give $ p_n$ is not prime.\r\nConsider $ n\\equal{}3$. If $ p_1\\equal{}2$, then $ p_3\\equal{}\\frac{10p_2\\plus{}20}{2p_2\\minus{}10}\\equal{}\\frac{5(p_2\\plus{}2)}{p_2\\minus{}5}$ give $ p_3$ is not prime.\r\nTherefore $ 4\\le n\\le 5$. Let $ Pr\\equal{}p_2*...p_{n\\minus{}1},S\\equal{}p_2\\plus{}...\\plus{}p_{n\\minus{}1}$, then $ p_n\\equal{}\\frac{5(S\\plus{}2)}{Pr\\minus{}5}$.\r\nIf $ 5\\not |Pr$, then must be $ p_n\\equal{}5, Pr\\equal{}S\\plus{}7$ and $ p_i,i\\equal{}2,..n\\minus{}1$ equal 2 or 3. It is easy to chek, that it had nod solution.\r\nTherefore $ 5|Pr$ or one of primes $ p_2,..,p_{n\\minus{}1}\\equal{}5$. Let $ a\\equal{}\\frac{Pr}{5}\\equal{}\\frac{p_2...p_{n\\minus{}1}}{5}, b\\equal{}S\\minus{}5\\equal{}p_2\\plus{}..p_{n\\minus{}1}\\minus{}5$, then $ p_n\\equal{}\\frac{b\\plus{}7}{a\\minus{}1}$. \r\nIf $ n\\equal{}4$, then must be $ a\\equal{}b\\equal{}p$ and $ p\\equal{}3$ or $ p\\equal{}5$.\r\nIt give solution $ (2,3,5,5)$.\r\nIf $ n\\equal{}5$ $ p_n\\equal{}\\frac{p\\plus{}q\\plus{}7}{pq\\minus{}1}\\ge 5$ have not solution.", "Solution_2": "in the question Does the product of primes means just a.b(i.e.product of two primes) OR a.b.c.d.....(product of any number of primes)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\nsorry if my english is not good :(", "Solution_3": "I think $ 2\\le n<6$ is not obviosly :roll: The reason is $ 2^6\\ge 10\\times 6$. But I think it is not logic :blush: \r\n\\[ 10(a_1\\plus{}a_2\\plus{}...\\plus{}a_n)\\equal{}a_1a_2...a_n,\\forall a_n\\ge ...\\ge a_2\\ge a_1\\ge 2\\]\r\nWith $ a_1,a_2,...,a_n$ are prime numbers. I think it is not easy to give $ n<6$\r\nI misunderstood something in your post. For expamle: why $ p_1>10 \\to p_2$ is not prime :maybe: Can you explain clearly? Thank you so much :wink:", "Solution_4": "Denote the numbers by $ p_{1},....,p_{n}$. We have $ \\prod p_{i} \\equal{} 10 \\sum p_{i}$, hence, WLOG, $ p_{1}\\equal{}2, p_{2}\\equal{}5$. Now we have:\r\n$ \\prod_{i\\equal{}3}^{n} p_{i} \\equal{} \\sum_{i\\equal{}3} p_{i} \\plus{}7$\r\nObviously (I sound like one of the most active members on this forum... :P ) n>3.\r\nIf $ n\\equal{}4: xy\\equal{}x\\plus{}y\\plus{}7 \\implies (x\\minus{}1)(y\\minus{}1)\\equal{}8 \\implies x\\equal{}2^{k}\\plus{}1, y\\equal{}2^{3\\minus{}k}\\plus{}1$. The only possible solutions are $ p_{3}\\equal{}3, p_{4}\\equal{}5$ (and of course vice versa).\r\nIf n>4:\r\nNone of the primes is bigger than 3. If WLOG $ p_{3} \\ge 5$, we have: $ (p_{3}\\minus{}1)(p_{4}\\minus{}1) \\ge 4$, or: $ p_{3}p_{4} \\ge p_{3}\\plus{}p_{4}\\plus{}3$, and than: $ p_{3}p_{4}p_{5} \\ge (p_{3}\\plus{}p_{4}\\plus{}3)(p_{5}) \\ge p_{3}\\plus{}p_{4}\\plus{}3\\plus{}p_{5}\\plus{}8$ (because: $ (x\\minus{}1)(y\\minus{}1) \\ge 9, x\\equal{}p_{3}\\plus{}p_{4}\\plus{}3 \\ge 10, y\\equal{}p_{5} \\ge 2$), a contradiction (If we use $ p_{i}p_{j} \\ge p_{i}p_{j}$ for the rest of the primes).\r\nHence, the only primes are 2 and 3. We need to solve $ 2^{k}3^{t}\\equal{}2k\\plus{}3t\\plus{}7$. If k=0 or t=0, we have an easy contradiction.\r\nI will show that t=0:\r\nIf $ p_{3} \\ge 3$, in the same way as above: $ p_{3}p_{4}p_{5}\\ge p_{3}\\plus{}p_{4}\\plus{}p_{5}\\plus{}5$\r\nIf n=5: we need to solve $ 3yz\\equal{}3\\plus{}y\\plus{}z\\plus{}7, z,y\\equal{}2,3$. It's simple to check that there are no solutions.\r\nIf n>5: $ p_{3}p_{4}p_{5}p_{6}\\ge (p_{3}\\plus{}p_{4}\\plus{}p_{5}\\plus{}5)(p_{6}) \\ge p_{3}\\plus{}p_{4}\\plus{}p_{5}\\plus{}5\\plus{}p_{6}\\plus{}10$, a contradiction. Hence, t=0, another contradiction.\r\n\r\nHence, the only solution is {2,3,5,5}" } { "Tag": [ "search", "geometry", "calculus", "derivative", "videos" ], "Problem": "We are having this debate in our (suburban) public elementary school. Should all children in the class be taught at the same level or should they be divided into, say, three levels and taught different material? I run a math club and it is quite apparent that the top 1/3 is just coasting with the current material. The counterpoint is that having all kids do the same material really pulls the kids not doing well by providing them with role-models.\r\n\r\n Are there any authoritative study of this? What does your school do?", "Solution_1": "[quote=\"rsinha\"] The counterpoint is that having all kids do the same material really pulls the kids not doing well by providing them with role-models.\n[/quote]\r\nJust one small remark: I doubt that somebody who does virtually no work in class but still gets excellent grades can really serve as a good \"role model\". Other kids may envy him but what is there to learn from looking at such a student? He would serve as a good role model if he had to struggle with the material and still came a winner but this would require the classroom material be ajusted according to the abilities of the top students in the class, which is something very few teachers would dare to do and which may be not the best idea for several other reasons. So, I am inclined to disagree with this particular objection. Of course there may be other arguments against separation and there are many alternative techniques of giving different workloads to different students.", "Solution_2": "I have a clear bias in this debate :) I think the kids should be given different material. However the administrators in charge of making this decision are convinced that this will adversely affect the overall class atmosphere and they are backed by a professor of math education whom they consulted for curriculum design. They see it as a policy decision of individual good vs. societal good. \r\n\r\n They do start separating the kids in higher grade but I work with these kids in my math club and I see first hand how much more they can learn and how excited they are with more challenging material. Given all the debate about US kids falling behind their peers, I find it crazy to hold these kids back. I can pass myself as a mathematician but this doesn't carry the same weight as a professor of math [i]education[/i]. So I am hoping for an authoritative study that I can cite in my argument.\r\n\r\nSorry if I sound frustrated.", "Solution_3": "[quote=\"rsinha\"]We are having this debate in our (suburban) public elementary school. Should all children in the class be taught at the same level or should they be divided into, say, three levels and taught different material? [/quote]\r\n\r\nWell, I can see where you are going. Our school are divided into different classes in subjects. For example, for math, we usually have GT in one class and Non-GT in the other. Then my teacher sits with the struggling GT people (usually the same people) and the higher, more, advanced GT people are doing their own work.", "Solution_4": "A search for \"ability grouping\" will turn up a bunch of links like [url=http://www.ericdigests.org/pre-927/grouping.htm]this one[/url]. But why bother reading them when you can just say \"Research shows...\" ? :) \r\n\r\nI would advise* against ability grouping. Chances are that it will correlate with race/ethnicity, which will sooner or later make a story in your local newspaper. :| More math-intensive extracurricular activities could be a better way to go.\r\n\r\n(*) having no knowledge or experience in the field :ninja:", "Solution_5": "Yeah in my old elementary days everyone learned practically the same thing. Even in the GT program, you just learn the same thing everyday...it was kinda boring :D \r\n\r\nGive students the option of choosing a normal class (easy and good for their moral but might be boring), challenging GT class (might drop morale). Make sure their parents know about it. There's nothing worse then a young student discouraged about math.", "Solution_6": "When I was in elementary school, once a week, a bunch of us would go to a separate room with a math specialist to talk about some more advanced material for about an hour or something. Then we'd get independent study work for the rest of the week so we wouldn't be slacking off while the rest of the class learned material that we already knew. I thought the system worked pretty well.", "Solution_7": "[quote=\"randomdragoon\"]When I was in elementary school, once a week, a bunch of us would go to a separate room with a math specialist to talk about some more advanced material for about an hour or something. Then we'd get independent study work for the rest of the week so we wouldn't be slacking off while the rest of the class learned material that we already knew. I thought the system worked pretty well.[/quote]\r\n\r\nI don't like the once a week thing. That means if you have a question you would have to ask it in 6 days, which could be unhelpful (other than posting on AoPS that is). The people who take geometry at my middle school have the once a week thing and they just hang out in a math class of Algebra II and do nothing, which is kinda wasting time. Preferablly, Id either go to a class specifically about geometry at a high school or something rather than learn with that much space between lessons and lectures.", "Solution_8": "Hmm, in middle school, I, along with a few other people, had an \"independent study\" from the second semester of 7th grade through the end of 8th grade. We never went to class at all, we worked from AoPS Vol. I, with a teacher there to help us if we needed it. I don't know if this is practical or not at your school, but that's what we did to remedy the problem.", "Solution_9": "[quote=\"davidyko\"]Hmm, in middle school, I, along with a few other people, had an \"independent study\" from the second semester of 7th grade through the end of 8th grade. We never went to class at all, we worked from AoPS Vol. I, with a teacher there to help us if we needed it. I don't know if this is practical or not at your school, but that's what we did to remedy the problem.[/quote]\r\n\r\nAww lucky! I wanted to take some AoPS courses... :D", "Solution_10": "i think that dividing the class into more and less advanced sections should be one of the top priorities of the school\r\nits completely stupid to discourage the top students from learning, even if it means taking away attention from some of the people who need help\r\nthis is one of the main reasons america is now falling way behind asia in math and science :( \r\n\r\npersonally, i get to take an AoPS class and work on the material during class\r\ni hadn't been able to do so before, and it has made me a happier, and smarter person. a few years ago, i was acting up in class cuz i had nothing to do :P\r\n[quote=\"mlok\"]\t\nA search for \"ability grouping\" will turn up a bunch of links like this one. But why bother reading them when you can just say \"Research shows...\" ? Smile\n\nI would advise* against ability grouping. Chances are that it will correlate with race/ethnicity, which will sooner or later make a story in your local newspaper. Neutral More math-intensive extracurricular activities could be a better way to go.\n\n(*) having no knowledge or experience in the field [/quote]\r\ni have to say that i reeeealy disagree with you\r\nif they write a story about it in the newspaper, just say, \"well i guess these people work harder cuz they got into the better class we didn't have any bias\"\r\nand not doing so will just discourage the top students\r\nif i had not lived in an area with lots of good math resources, i probably would have became dissillusioned with math, thinking it was too easy and boring. the only reason that i like math is that i was introduced to AoPS and harder math", "Solution_11": "[quote=\"daermon\"]i think that dividing the class into more and less advanced sections should be one of the top priorities of the school\nits completely stupid to discourage the top students from learning, even if it means taking away attention from some of the people who need help\nthis is one of the main reasons america is now falling way behind asia in math and science :( [/quote]\r\n\r\nI agree but I believe the reason why elementary schools do not seperate students is the lack of available resources. No extra rooms, no extra textbooks, no extra teachers, etc. cause advanced students to be holed up with for lack of better term, lesser students. Also, there can be a minimal amount of students say 3, that would require advanced classes but the school would not provide the extra work and money into providing the students their needs. Thats why in middle school, where there is a higher chance of advanced students, that there is the extra funding for advanced classes. Of course in high school then the classes seperate in to 4 different skill levels (academic, honors, GT, and AP)", "Solution_12": "This is interesting for me largely because I grew up with an rather different system. At the elementary school I went to in 1st - 4th grade (moved for 5th & 6th), I was in an 'accelerated' / 'gifted' program within a normal elementary school. In 1st and 2nd grade, all the students worked on the same material at the same time, but once we got to 3rd grade we could go at our own pace. This meant we read the book on our own, did the homework our teacher assigned us, asked for help if we didn't understand something, and took the chapter tests when we were ready. If you got a poor grade (C or below) you had to redo the chapter. This had the end result of letting me finish \"7th grade math\" in 4th grade (of course, I had to repeat it when I moved because no one believed me...). In my opinion, this is an excellent system - it allows students to work at their own level of ability, provides 'role models' (i.e. the students farther ahead), allows talented people to work ahead and actually enjoy themselves in math class. The 'redo-on-failure' part also ensures that people actually understand the material once they're done, unlike a traditional classroom setting where poor-performing students are just dragged along with the rest of the class because the teacher can't spare the class time to go back over material.\r\nNaturally, I'm a bit biased in favor of this system... :D", "Solution_13": "Everyone at my school doesn't even know what Art of Problem Solving is so there is next to no chance of us getting taught from their books...", "Solution_14": "[quote=\"Danbert\"]This is interesting for me largely because I grew up with an rather different system. At the elementary school I went to in 1st - 4th grade (moved for 5th & 6th), I was in an 'accelerated' / 'gifted' program within a normal elementary school. In 1st and 2nd grade, all the students worked on the same material at the same time, but once we got to 3rd grade we could go at our own pace. This meant we read the book on our own, did the homework our teacher assigned us, asked for help if we didn't understand something, and took the chapter tests when we were ready. If you got a poor grade (C or below) you had to redo the chapter. This had the end result of letting me finish \"7th grade math\" in 4th grade (of course, I had to repeat it when I moved because no one believed me...). In my opinion, this is an excellent system - it allows students to work at their own level of ability, provides 'role models' (i.e. the students farther ahead), allows talented people to work ahead and actually enjoy themselves in math class. The 'redo-on-failure' part also ensures that people actually understand the material once they're done, unlike a traditional classroom setting where poor-performing students are just dragged along with the rest of the class because the teacher can't spare the class time to go back over material.\nNaturally, I'm a bit biased in favor of this system... :D[/quote]\r\n\r\nI wished more schools offered these kind of policies. I guess an alternative to enhancing a youngster's education in elementary school is to enroll him/her into extracurricular programs, which I've seen in a few places. And of course, a number of problems arise, which include:\r\n1) Risk of overworking the student. Of course this would be a lot less if the student wasn't forced (like in most places in Asia. So there is a trade-off involved)\r\n2) Since the material taught in elementary school is not very broad, there will bound to be overlapping material --> waste of time for the student in his math class\r\n3) Sort of an extension to 2, many schools (especially elementary, at least it applies for the one I went to) are simply too rigid to let you skip grades in a specific course, let alone let you take an apititude test. And depending on the environment and the concerned student, being surrounded by a class of upper years (aka \"bigger kids\") may be intimidating, especially if intellect is not regarded highly in the particular student body.", "Solution_15": "[quote=\"rsinha\"]I have a clear bias in this debate :) I think the kids should be given different material. However the administrators in charge of making this decision are convinced that this will adversely affect the overall class atmosphere and they are backed by a professor of math education whom they consulted for curriculum design. They see it as a policy decision of individual good vs. societal good. \n\n They do start separating the kids in higher grade but I work with these kids in my math club and I see first hand how much more they can learn and how excited they are with more challenging material. Given all the debate about US kids falling behind their peers, I find it crazy to hold these kids back. I can pass myself as a mathematician but this doesn't carry the same weight as a professor of math [i]education[/i]. So I am hoping for an authoritative study that I can cite in my argument.\n\nSorry if I sound frustrated.[/quote]\r\npersonally, I find the situation infuriating. The amount of potential wasted through poor, inadequate, counter-productive, and generally faulty teaching and teaching methods is tragic. I've become very interested in education lately. \r\n\r\nI personally attended a very small Montessori school from age 4-10, and despite some issues with the teachers and some obvious improvements which could have been made, that time was one of the best educational experiences of my life. I know personally that, perhaps as a result of the Montessori program- although if it is, I would call it a benefitial result- I am only motivated to learn by my own curiosity and desire for knowledge. This has made my more conventional schooling extremely frustrating at times, since the authoritarian structure of middle school and high school demand that one does an excessive amount of things only because they are required.... and why is there this host of requirements? the answer seems to differ depending on the situation, but the most common answer I have been given is that students need to learn how to do stuff they don't want to do... how to do what they're told, and how to submit to authority. :mad: :mad: :mad: :mad: :mad: :mad: \r\n\r\nOh it makes me mad. \r\n\r\nKids don't like school. This seems to have been accepted as fact. But I really think it is more symptomatic of the failings of our system. I've suffered a baffling amount of worthless classes and work, sadistic or idiotic authority figures, and other bad things in my education, and the sad thing is, I consider myself extremely fortunate despite all this when I compare my education to others'. \r\n\r\nanyhow, as evidenced by my post, I'm thinking about eduction... I certainly don't have it worked out yet. But I think a key component of an improved educational system is differentiation between levels of abilities and PLACEMENT STUDENTS AMONG ACADEMIC PEERS, something it seems our society is loath to do for several reasons (a desire for normalcy being top-most).", "Solution_16": "[quote=\"plucesiar\"]\n3) Sort of an extension to 2, many schools (especially elementary, at least it applies for the one I went to) are simply too rigid to let you skip grades in a specific course, let alone let you take an apititude test. And depending on the environment and the concerned student, being surrounded by a class of upper years (aka \"bigger kids\") may be intimidating, especially if intellect is not regarded highly in the particular student body.[/quote]\r\n\r\nskipping grades doesn't do much to alleviate the underlying issue. The speed and depth of a course is much more important than the level of the material, IMO.", "Solution_17": "Our middle school deals with this in a fairly good manner.\r\n\r\nOut teams are divided based on math level (Pre-Algebra all the way to Pre-Calculus).\r\n\r\nBelow Pre-Algebra kids are placed on the \"Jaguar Team\" (lower than normal).\r\nPre-Algebra kids to Algebra kids are placed on the \"Panther Team\" (normal).\r\nBeyond Algebra kids are placed on the \"Tiger Team\" (advance).\r\n\r\nThis is for 8th grade.", "Solution_18": "I think the school absolutely must allow a student to learn what they're ready to learn, especially at the elementary level. I was extremely lucky to have an extremely cooperative elementary school that pretty much satisfied whatever needs I had. I really can't say what might have happened to my enthusiasm for math if that hadn't been the case. Certainly, there are students who would have been turned off to math at that point. And I really don't buy the tired old argument about enhancing the atmosphere of the classroom. It's irresponsible to hold a student back from their full potential just because the other students haven't reached that level. And on top of that, honestly, what good does it do to add in the kids who are bored and already know what they're doing to the class? Maybe in high school, they can help the other kids master it, but (a) most of them won't want to, and (b) elementary school kids are not math tutors. I was completely incapable of helping other students understand math the way I did back then. Instead of restricting the more talented students, invest in good teachers who can inspire enthusiasm in students.\r\n\r\nNow the disclaimer: I really don't have the authority to speak on this, as I'm not any kind of educator or anything. This is just my personal opinion, which is not based on any solid facts except my own (limited) experience.\r\n\r\n[quote=\"plucesiar\"]\nAnd depending on the environment and the concerned student, being surrounded by a class of upper years (aka \"bigger kids\") may be intimidating, especially if intellect is not regarded highly in the particular student body.[/quote]\r\n\r\nWhen I was in 4th grade, I skipped up to an 8th grade algebra class; since my elementary school was adjacent to a junior high, I could take it in a normal classroom, mixed in with the most advanced 8th grade kids. I didn't find it at all intimidating, actually. All the kids were really nice to me, and I think they thought it was pretty cool to have a 4th grader in their class.\r\n\r\nCoincidentally, that class was where I was first introduced to the concept of a \"math team\".", "Solution_19": "At my elementary school there wasn't a GT program, but in my opinion, if a students clearly isn't challenged at all, there should be some effort so they can stay interested. For example, when I was in fourth grade, my teacher had me go to a 5th grade room for math and I definitely learned a bunch that year (more than before at least). Next year, I (read: my parents) became more involved and I studied more at home.\r\n\r\nIn middle school and higher, there should definitely be classes offered that are ahead of grade level since there are so many more students, which means that more are going to be ahead of their grade level (as in enough to fill a classroom at least).\r\n\r\nOh, and while we're on the topic of elementary school math, has anyone heard of the Every Day Mathematics books? Well there are a few videos on youtube about them, this one has been particularly active:\r\n[url=http://youtube.com/watch?v=Tr1qee-bTZI]Youtube video[/url]", "Solution_20": "I think it's nice to have different levels, as that would encourage students to try harder, but the schools would be wary of trying to have different levels because that would take more time. Not all teachers would want to do that, though. My brother is in elementary school, and he is coasting through math very easily. If there was a 'advanced' level, I'm pretty sure it would encourage him to do more than he is right now and think more than he is. (I do know that several of the boys, including him, try to finish their math work in the fastest amount of time.)\r\n\r\nIn my old elementary school, we had an enrichment program. Once or twice each week, we'd be taken out of class to discuss something that extended the curriculm.\r\n\r\nI think my school did the Every Day Mathematics books. It's hard to remember." } { "Tag": [ "algebra proposed", "algebra" ], "Problem": "given n is positive integer ,\r\n consider sequence {x_n } :01$\r\n\\[\\lim_{n \\to+\\infty}e^{-nx}\\sum_{k=1}^{n-1}\\frac{(nx)^{k}}{k!}\\]\r\n\r\n\r\nI find the answer is $0$,when $x>1$;is $1$,when $0n-1 \\end{array}$\r\n\r\nThen you can multiply,\r\n$\\sum_{k=0}^{\\infty}a_{k}\\cdot \\sum_{n=0}^{\\infty}\\frac{(-nx)^{k}}{k!}= \\sum_{k=0}^{\\infty}b_{k}x^{k}$\r\n\r\nWhere $b_{k}= \\sum_{i=0}^{k}a_{k-i}\\cdot \\frac{(-n)^{i}}{i!}$", "Solution_2": "Hey,thank you,but I have try to do with that approach when I post the problem but get nothing :(", "Solution_3": "Consider a Poisson-distributed random variable $u$ with parameter $nx$; $\\frac{(nx)^{k}}{k!}$ is the probability that $u$ is exactly $k$, so this sum is the probability that $u$ is between $1$ and $n-1$.\r\n\r\nIf $x=0$, the sum is zero for all $n$.\r\nIf $01$, the probability that $u$ is less than $n$ is at most $\\frac1{t^{2}}$, with $t=\\frac{nx-n}{\\sqrt{nx}}=\\sqrt{n}\\frac{x-1}{\\sqrt{x}}$. This goes to zero as $n\\to\\infty$, so the limit is 0.\r\nIf $x=1$, we need stronger estimates. In this case, that's the central limit theorem or some variant. A Poisson distribution with large parameter looks approximately normal- and the normal is symmetric around its mean. The limit is $\\frac12$ in that case.\r\n\r\nThis approach says nothing about $x<0$.\r\n\r\nThe $x=1$ case has certainly appeared on the forum before, and it hasn't been long.", "Solution_4": "[quote=\"jmerry\"]\nThe $x=1$ case has certainly appeared on the forum before, and it hasn't been long.[/quote]\r\n\r\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=146401\r\n\r\nI think there is an older thread where myth solved it using non-probabilistic methods.", "Solution_5": "Thank you,That's very nice result." } { "Tag": [ "inequalities", "function", "ratio", "real analysis", "real analysis unsolved" ], "Problem": "In the following $a,b, A,B$ [b] are positive numbers[/b]. [b] Let $I: =[0,\\infty)$ , $f: I \\to {\\mathbb R}$ be such that $|f(x)|\\le Ae^{Bx}\\; ,\\; \\ ; \\forall x \\in I .$ \nVerify that the series $\\sum\\limits_{k=0}^{\\infty}\\frac{(ab)^k}{k!}f\\left(\\frac{k}{a}\\right)$ is convergent. [/b]\r\n[color=green][b]\nIf $f$ is assumed to be convex on $I$ prove $\\begin{array}{|c|} \\hline \\\\ \\displaystyle \\sum\\limits_{k=0}^{\\infty}\\frac{\\left((a+1)b\\right)^k}{k!}f\\left(\\frac{k}{a+1}\\right) \\le \\displaystyle e^b \\sum\\limits_{k=0}^{\\infty}\\frac{(ab)^k}{k!} f\\left(\\frac{k}{a}\\right) \\\\ \\\\ \\hline \\end{array}\\; .$\nShow that the equality is attained for any linear function.\n[/b][/color]", "Solution_1": "It seems to me RHS $e$ should be replace by $e^b$ \r\nCould you to check this professor LUPAS ?", "Solution_2": "[quote=\"Moubinool\"]It seems to me RHS $e$ should be replace by $e^b$ [/quote]\r\nRight ! Thanks, I make the necessary correction !", "Solution_3": "$|\\sum \\frac{(ab)^k}{k!}f(k/a)|\\leq A \\sum \\frac{(ab)^k}{k!}e^{Bk/a}$\r\nthe serie at the RHS is convergent since the ratio test is verify.\r\n-----------------------------------------------\r\n\r\nFor the inequality write $p/(a+1)=\\sum_{q=0}^{p}w(q)\\frac{p-q}{a}$ with \r\n$w(q)=\\frac{C_p^q}{(a+1)^p}$\r\n$w(0)+...+w(p)=1$\r\n \r\nUse convexity of f and arrange the RHS to get a \r\n\r\n(Le Barron Louis Augustin Cauchy (Polytechnicien) ) product\r\n\r\n$\\sum \\frac{((a+1)b)^p}{p!}f(p/(a+1)) \\leq \\sum_{p\\geq 0}\\sum_{q=0}^{p}\\frac{(ab)^{p-q}}{(p-q)!}f((p-q)/a)\\frac{b^q}{q!} = e^b\\sum \\frac{(ab)^k}{k!}f(k/a)$", "Solution_4": "[quote=\"Moubinool\"] ......(Le Barron Louis Augustin Cauchy (Polytechnicien) ) product........[/quote]\r\nNice !" } { "Tag": [ "geometry", "3D geometry", "algebra proposed", "algebra" ], "Problem": "Can we always find integer solutions to\r\n\r\n$a^n+b^n+c^n=d^n$ for $n \\neq 0$ in $\\mathbb{N}$", "Solution_1": "you've already solved this? wow, i'd like to see a solution plz... :D", "Solution_2": "Not quite no. I was hoping someone else might have a good idea. I know its true for squares, cubes and forth powers, but beyond that, it gets tougher!" } { "Tag": [ "LaTeX" ], "Problem": "Is it ok if my solutions are like 5 page per problem(I want to be throuogh[is that how u spell it??])...\r\nBecause even I get tired reading my solutions...\r\nbtw, what is the \"bad box\" thing when you compile, and how u get rid of them?????", "Solution_1": "That's a bit on the long side, but it won't cost you points.\r\n\r\nAs for the bad boxes, as long as you're happy with how it looks, then don't worry about it. LaTeX has a lot of ticky-tacky mystery warnings that aren't terribly important." } { "Tag": [ "algebra", "polynomial", "Ring Theory", "superior algebra", "superior algebra unsolved" ], "Problem": "Let R be a quotient ring of the polynomial ring C[X]. Prove that\r\nfor any maximal ideal M of R, the ring R/M is isomorphic with C", "Solution_1": "This is simply a consequence of the fundamental theorem of algebra. The only irreducible elements in $ \\mathbb{C}[x]$ are polynomials of degree one, the quotients by which are isomorphic to $ \\mathbb{C}$.", "Solution_2": "And the so-called Third Isomorphism Theorem." } { "Tag": [ "quadratics" ], "Problem": "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]\r\n\r\nFind all integers $ b$ and $ c$ such that the equation $ x^2 \\minus{} bx \\plus{} c \\equal{} 0$ has two real roots $ x_1, x_2$ satisfying $ x_1^2 \\plus{} x_2^2 \\equal{} 5$.", "Solution_1": "sorry, dont know latex.\r\n\r\nusing quadratic you get the two solutions to be x=(b+(sqrt(b^2-4c))/2, \r\nx=(b-(sqrt(b^2-4c))/2. squaring both yields b^2-2c=5. There are infinitely many solutions for b>1 or b<-1 and b is odd, however; b^2-4c>0. since we know b^2-2c=5, c<5/2. Thus, the only solutions are b=3, c=2 or b=-3, c=2.\r\n\r\nEDIT: :blush: sorry, i forgot the other two cases (-1, -2) and (1, -2).", "Solution_2": "You can skip a bit of work in that first part using Newton's sums:\r\n$ S_1\\equal{}b$ from Vieta's.\r\n$ S_2\\equal{}bS_1\\minus{}2c\\equal{}b^2\\minus{}2c\\equal{}5$.", "Solution_3": "Since the sum of the squares of the roots is 5 and the roots are real, the roots are between + and - sqrt(5) inclusive and so -5 <= c <= 5. So we have -5 <= c <= 2. Checking, I get (b, c) = (-3, 2) and (3, -2) as in lukezli's post and also (-1, -2) and (1, -2)." } { "Tag": [ "geometry", "parallelogram", "geometry proposed" ], "Problem": "In the parallelogram $ ABCD$ consider $ M$ on $ CD$ such that $ M \\not\\equal{}C$ and $ M \\not\\equal{}D$. Denote by $ N$ the intersection between $ AM$ and $ BC$. Prove that the triangles $ DMN$ and $ BCM$ have the same area.", "Solution_1": "$ \\frac {[\\triangle DMN]}{[\\triangle ADM]} \\equal{} \\frac {MN}{AM} \\ \\ , \\ \\ \\angle ADM \\plus{} \\angle BCM \\equal{} 180^{\\circ}$\n\n$ \\frac {[\\triangle ADM]}{[\\triangle BCM]} \\equal{} \\frac {AD \\cdot DM}{BC \\cdot MC} \\equal{} \\frac {DM}{MC}$\n\n$ \\frac {MC}{DM} \\equal{} \\frac {MN}{AM}\\Longrightarrow \\ \\frac {[\\triangle ADM]}{[\\triangle BCM]} \\equal{} \\frac{[\\triangle ADM]}{[\\triangle DMN]} \\Longrightarrow \\ [\\triangle BCM] \\equal{} [\\triangle DMN]$", "Solution_2": "[quote=\"moldovan\"][color=darkred]Let $ ABCD$ be a parallelogram. Consider the point $ M\\in CD \\minus{} \\{C,D\\}$ \n\nand the intersection $ N\\in AM\\cap BC$ . Prove that $ [DMN] \\equal{} [BCM]$ .[/color] [/quote]\r\n[color=darkblue][b][u]Proof[/u].[/b] $ [DNC]\\stackrel{(AD\\parallel NC)}{\\ \\ \\equal{} \\ \\ }[ANC]\\implies$ $ [DNC]\\pm [MNC] \\equal{} [ANC]\\pm [MNC]$ $ \\implies$\n\n $ [DMN] \\equal{} [ACM]\\stackrel{(AB\\parallel CM)}{\\ \\ \\equal{} \\ \\ }[BCM]\\implies$ $ [DMN] \\equal{} [BCM]$ .[/color]" } { "Tag": [ "superior algebra", "superior algebra unsolved" ], "Problem": "If you know some results about conjugacy classes\r\n or size of conjugacy classes in finite group, please write here.", "Solution_1": "I hear that they are disjoint.", "Solution_2": "On a less snarky note, what you probably want is the [url=http://en.wikipedia.org/wiki/Conjugacy_class#Conjugacy_class_equation]class equation[/url].\r\n\r\nThey also have an intimate connection with irreducible representations." } { "Tag": [], "Problem": "x,y,z\u5747\u4e3a\u6b63\u5b9e\u6570\u4e14$x+y+z=1$,\u6c42\u8bc1:\r\n$(1-x^2)^2+(1-y^2)^2+(1-z^2)^2 \\leq (1+x)(1+y)(1+z)$", "Solution_1": "see the picture[/list]", "Solution_2": "[img]http://www.mathlinks.ro/Forum/files/__________425.gif[/img]\u662f\u4ec0\u4e48\u610f\u601d\uff1f :?:", "Solution_3": "see here:\r\nhttp://www.mathlinks.ro/viewtopic.php?p=1697724#1697724", "Solution_4": "[quote=\"gsh91best\"][img]http://www.mathlinks.ro/Forum/files/__________425.gif[/img]\u662f\u4ec0\u4e48\u610f\u601d\uff1f :?:[/quote]\r\n\r\n\u6c42\u548c\u7b26\u53f7\u3002\u7edf\u8ba1\u5b66\u91cc\u9762\u5f88\u5e38\u7528\u7684\uff0c\u91cd\u70b9\u4e2d\u5b66\u7684\u9ad8\u4e09\u90fd\u8981\u6c42\u4f1a\u7528\u8fd9\u4e2a\u7b26\u53f7\u3002" } { "Tag": [ "geometry", "rectangle", "analytic geometry", "geometry unsolved" ], "Problem": "\"Given N distinct points in the unit square [0,1]x[0,1], including the origin (0,0) as one of the N points. Can you construct N rectangles, contained in the unit square, with sides parallel to the coordinate axes, pairwise non-intersecting, such that each of our N given points is the lower-left-hand corner of one of the rectangles, and such that the total area of the rectangles is at least 1/2?\r\n\r\nFor example, N=3 and the points are (0,0), (0.2,0.4) and (0.8,0.6). The three rectangles could be [0,1]x[0,0.39], [0.2,0.79]x[0.4,1], and [0.8,1]x[0.6,1]. Their areas are 0.39, 0.354 and 0.08, summing to 0.824, which is larger than 1/2.\r\n(We allow degenerate rectangles -- a point or a line segment -- but still do not allow them to intersect with other rectangles, even in a single point.)\r\n\r\nEither PROVE that such a construction is always POSSIBLE, or give a set of N distinct points (including the origin) for which it is IMPOSSIBLE.\"\r\n\r\n\r\nIt seems to me that \"worst case\" is when our N points are equally spaced along the main diagonal: (i/N, i/N) for i=0,1,...,N-1, then any solution will have area less than W = (N+1)/(2*N), which is slightly larger than 1/2. In any other situations we have a greater W. But a general solution is not known?!? (I cannot prove it entirely).\r\nI already check this for N = 2,3,4,5.\r\n\r\nDoes anyone has an ideea?", "Solution_1": "\u00a0\u2591\u064e7,\u00f4|7\u2567\u25cb\u25cb8\u25cb[12\u263a-||5D\u0637\u00e9", "Solution_2": "[quote=\"amir2\"]\u00a0\u2591\u064e7,\u00f4|7\u2567\u25cb\u25cb8\u25cb[12\u263a-||5D\u0637\u00e9[/quote]\r\n\r\nCan u be more specific, plz? :?" } { "Tag": [ "MIT", "college", "Columbia" ], "Problem": "i am thinking about going to University of Illinois because i hear it has a good engineering program. I live only three hours away from it, and the tuition is not terribly high, either. However, I know there are better schools for engineering that i could probably get into. \r\n\r\nMy question is: Would U of I be a good school to attend, or would it make a terrific difference if i went to a Ivy League-type school. Am i sacrificing anything by wanting to go there? and does it have a good math department too.\r\n\r\nI think i want to study mechanical engineering, but that isnt set in stone. my big things are math and physics.", "Solution_1": "I think it's probably not worth it to go to an Ivy League school, although you should probably apply anyway. The University of Illinois, assuming you like Champaign-Urbana, the environment, etc., is not much worse than the Ivy Leaguers and whatnot. Anyway, it's only undergrad (I'm assuming).", "Solution_2": "what about purdue and university of michigan? According to US News Report, michigan has the second best mechanical engineering school in the nation, while illinois has the sixth best. That difference is not that big of a deal, right? What makes that difference?", "Solution_3": "Hi, my name is Geehoon and I am currently a high school senior.\r\n\r\nYes, I did apply to and got accepted to University of Illinois at Urbana Champaign (along with many other schools), and had a chance to research the school a little bit.\r\n\r\nRankingwise, the engineering program is really awesome, I would say definitely top 20 in nation, if not top 10.\r\nIn terms of Environment, I have a friend there (he won IOI gold once) and it seems that he likes being there.\r\n\r\nIt will be a good choice for you especially since you live around there.\r\nIt is not that hard to get in, so it will be safer for you also.\r\n\r\nHowever, I will say these two things.\r\n\r\nFirst, you don't have to decide where youre going right now. What I mean by that is, apply to all the schools that you like, and see what happens. as JBL once mentioned, college admission is somewhat random and you cannot expect what is going to happen. For example, I did not get into Georgia Tech, but got into MIT and Caltech. If I were afraid of getting rejected and didnt apply to MIT/Caltech, I might have gotten into serious problem after getting rejected from Georgia Tech.\r\n\r\nSecond, don't worry about money so much. I'm closer to the poor than to the rich, but never hesitated to apply Ivy League schools. I see many people who do not apply to Ivy League schools simply because they are expensive. However, if you get into more than one school, you still have a chance to compare the financial aid packet from all schools that you got in, and obviously you have more options if you got into more schools.\r\n\r\n\r\nTo make the long story short, my opinion is:\r\n1. UIUC is wonderful school, especially their engineering.\r\n2. Apply Ivy leagues. It only widens your option.\r\n3. Worry about money after you get financial aid packet, not before.", "Solution_4": "thank you for your response. I am going to be a senior next fall, but i need to start thinking about these things now.\r\n\r\ni kind of like the fact that U of I is only 3 hours away. i have to consider travel costs as well.\r\n\r\nwhat is IOI?\r\nwhat school are you going to?", "Solution_5": "IOI = [url=http://olympiads.win.tue.nl/ioi/]International Olympiad in Informatics[/url]\r\n\r\nGeehoon is going to MIT next year.", "Solution_6": "UIUC is significantly better than most of the Ivies for Engineering, anyways.", "Solution_7": "\"most,\" probably. Columbia and Cornell are excellent, but the others are not.", "Solution_8": "But is it better than Purdue and University of Michigan?\r\n\r\nI have heard U of I has a pretty good math department too. If i can handle engineering i want to get a degree in mathematics.", "Solution_9": "My husband was a math major at U of I more than 20 years ago so this info is a little old but he thinks the quality of the science departments hasn't changed much. The math and physics departments were excellent. The intro classes were large but the upper level classes were small. Upon graduation he was accepted by U Chicago for math graduate studies so the U of I math dept is well-respected by other schools.", "Solution_10": "thank you all", "Solution_11": "What Geehoon said (post #4) is excellent advice for all the readers of this forum. My one comment about your specific question of whether or not you should study in your home state is that there is an advantage to seeing the rest of the world. I went to the same state university that most people from my high school went to. I lived close enough to it that I could take a bus to the campus and live at home. Later I went overseas and lived there for three years, after I had completed my undergraduate degree. Knowing what I know now as a forty-something-year-old, my advice is apply to all the schools you desire, including your state university, but if you can go somewhere far away at affordable cost, go to some place you have never lived in before to go to college, and prefer colleges, if you are accepted to more than one, that have students from all over the country and all over the world. You learn by meeting new kinds of people who didn't have the same kinds of life experiences you had when growing up. Changing your environment builds more flexibility of thinking and helps increase your creativity. \r\n\r\nGood luck in your applications. It sounds to me like your in-state choice is a very good choice indeed. Apply to various schools, and see what application results you end up with." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "find all integers like a ,b and c so that:\r\nb+c|bc , c+a|ac , b+a|ab .", "Solution_1": "no idea?\r\nthis is making me crazy.I hate this kind of problem in number theory.", "Solution_2": "Strange - maybe I misunderstood something but:\r\n\r\nlet $a = \\prod p_{i}^{\\alpha_{i}}$ , $b = \\prod p_{i}^{\\beta_{i}}$, $c = \\prod p_{i}^{\\gamma_{i}}$. Then our condition rewrites as:\r\n\r\n$\\alpha_{i}\\leq \\beta_{i}+\\gamma_{i}$ \r\n$\\beta_{i}\\leq \\alpha_{i}+\\gamma_{i}$\r\n$\\gamma_{i}\\leq \\alpha_{i}+\\beta_{i}$. \r\n\r\nso $\\alpha_{i}, \\beta_{i}, \\gamma_{i}$ have to be sides of triangle (maybe degenerated) and I suppose it's sufficient and necessary condition... :maybe:", "Solution_3": "I made a big mistake in saying the problem.so I edited it.excuse me." } { "Tag": [ "algebra", "polynomial", "superior algebra", "superior algebra unsolved" ], "Problem": "let $ \\varphi$ be a ring epimorphism $ \\varphi : R \\to S$ how to show that $ J(R)\\subseteq J(S)$ where $ J(R)$ is Jacobson radical", "Solution_1": "First, it should be $ \\varphi\\left(J\\left(R\\right)\\right)\\subseteq J\\left(S\\right)$.\r\nThen, what definition of Jacobson radical are you using? The $ J\\left(R\\right) \\equal{} \\left\\{x\\in R\\mid \\text{for every }r\\in R\\text{ there exists some }u\\in R\\text{ such that }u\\left(1 \\minus{} rx\\right) \\equal{} 1\\right\\}$ definition makes this completely obvious.\r\n\r\nAnd: problems of this kind belong into College Playground >> Superior Algebra, [b]not[/b] into the Olympiad Section >> Algebra board. In the Olympiad Section, algebra means number sequences, functional equations, polynomials, finite sums etc., but definitely not rings and ideals.\r\n\r\n darij" } { "Tag": [], "Problem": "Let $ f(x)=4x-x^{2}$ for $ x_{0}\\in\\mathbb{R}$ we consider the finite sequence $ x_{0}, x_{1}=f(x_{0}), x_{2}=f(x_{1}),....$ Prove that there exists many $ x_{0}$ such that $ x_{0},x_{1},x_{2},...$ consists of finitely many different values", "Solution_1": "$ f^{i}=fofof..(i\\ times)..ofof,J=]-\\infty;4]$\r\n$ f(J)=J\\Longrightarrow \\forall i\\in N^{*}, \\ f^{i}(J)=J$\r\n$ 4\\in J\\Longrightarrow \\exists a_{i}\\in R: \\ f^{i}(a_{i})=4$\r\n$ \\forall (i,j)\\in N^{*}^{2},\\ (we\\ suppose\\ i>j): \\ f^{i}(a_{j})=f^{i-j}of^{j}(a_{j})=f^{i-j}(0)=0\\neq 4=f^{i}(a_{i})$\r\nso, $ \\forall (i,j)\\in N^{*}^{2},\\ i\\neq j\\Longrightarrow a_{i}\\neq a_{j}$\r\nthen $ |\\{a_{i}|\\ i\\in N^{*}\\}|=+\\infty$\r\nand for $ i\\in N^{*}$, if we take $ x_{0}=a_{i}$\r\n$ {\\{x_{i}|\\ i\\in N\\}=\\{a_{i},f(a_{i}),f^{2}(a_{i}),...,f^{i-1}(a_{i}),f^{i}(a_{i}),f^{i+1}(a_{i}),f^{i+1}(a_{i}),......\\}=\\{a_{i},f(a_{i}),f^{2}(a_{i}),...,f^{i-1}(a_{i}),4,0\\}}$\r\nthen $ \\{x_{i}|\\ i\\in N\\}$ is finite" } { "Tag": [ "USAMTS", "email" ], "Problem": "I think that it was said about two weeks ago that scores will be coming in mid-December, but possibly late December. Are there any further updates on when we are going to get the scores? Is it more likely to be mid-December or late December? Thank you.", "Solution_1": "Mid-december is a few days after today...\r\n\r\nSo I think then", "Solution_2": "Hey guys maybe you should just wait.", "Solution_3": "I know but I'm a little nervous so I'm sort of impatient to get it over with...", "Solution_4": "We will make an announcement via this forum, the USAMTS website, and the email mailing list when the scores are available." } { "Tag": [], "Problem": "Compute the sum: $ 3^0\\plus{}3^1\\plus{}3^2\\plus{}...\\plus{}3^{2009}$.", "Solution_1": "Yep, you're right. :) I will try to progress. \r\n\r\n[b]Generalisation[/b]\r\n\r\nCompute the sum: $ n^0\\plus{}n^1\\plus{}...\\plus{}n^m$ where $ m$ and $ n$ are positive integers greater than $ 1$. \r\n\r\n[b]Solution[/b]\r\n\r\n$ S\\equal{}n^0\\plus{}n^1\\plus{}...\\plus{}n^m$\r\n\r\n$ n \\cdot S\\equal{}n^1\\plus{}n^2\\plus{}...\\plus{}n^{m\\plus{}1}$\r\n\r\n$ \\Rightarrow (n\\minus{}1) \\cdot S\\equal{}n^{m\\plus{}1}\\minus{}1$\r\n\r\n$ \\Rightarrow S\\equal{}\\frac{n^{m\\plus{}1}\\minus{}1}{n\\minus{}1}$", "Solution_2": "I think the answer is just 3^3000-1/3-1=3^3000-1/2", "Solution_3": "WELL OBVIOUSLY IT'S NOT AN INTEGER. I MEAN DUH!", "Solution_4": "I am afraid (3^3000-1)/2 is an integer Ace of Diamonds!", "Solution_5": "I am (actually not really) afraid that you did not use parentheses in the post that AoD responded to :wink:", "Solution_6": "The result is $ S\\equal{}\\frac{3^{2010}\\minus{}1}{2}$." } { "Tag": [ "calculus", "calculus computations" ], "Problem": "find the development/expansion in (-2) of $ \\sqrt[7]{x^2\\minus{}2}$;\r\nwhy does't it work (the first two terms are the same, the third is different) if I substitute $ x^2\\minus{}2$ with $ u\\plus{}2$ and then make the expansion in zero?", "Solution_1": "The substitution you want to use is $ u \\equal{} x \\plus{} 2,$ which means that $ x \\equal{} u \\minus{} 2$ and $ x^2 \\minus{} 2 \\equal{} 2 \\minus{} 4u \\plus{} u^2.$\r\n\r\nI get an expansion that starts out with\r\n\r\n$ \\sqrt[7]{2}\\left(1\\minus{}\\frac27u\\minus{}\\frac{17}{98}u^2\\plus{}O(u^3)\\right)$\r\n\r\n$ \\equal{}\\sqrt[7]{2}\\left(1\\minus{}\\frac27(x\\plus{}2)\\minus{}\\frac{17}{98}(x\\plus{}2)^2\\plus{}O((x\\plus{}2)^3)\\right).$", "Solution_2": "thanks, but why is not possible to substitute $ x^2\\minus{}2$ with u directily, maybe because you have to use linear powers?", "Solution_3": "The problem asks for a series expansion in terms of $ x\\plus{}2$; we have to get that into the picture." } { "Tag": [ "probability" ], "Problem": "A piggybank contains 2 quarters, 3 dimes, 4 nickels, and 5 pennies. One coin is removed at random. What is the probability that the coin is a dime or a nickel?", "Solution_1": "[quote=\"Interval\"]A piggybank contains 2 quarters, 3 dimes, 4 nickels, and 5 pennies. One coin is removed at random. What is the probability that the coin is a dime or a nickel?[/quote]\r\n[hide]There are 7 coins that are either a dime or a nickel and there are 14 coins over all so then the probability is 7/14 which is 1/2[/hide]", "Solution_2": "Did you reduce 7/14?", "Solution_3": "[quote=\"Interval\"]Did you reduce 7/14?[/quote]\r\nyes he did. If you can reduce a fraction,always do that. you always want a fraction in simplest form.", "Solution_4": "[quote=\"ckck\"][quote=\"Interval\"]Did you reduce 7/14?[/quote]\nyes he did. If you can reduce a fraction,always do that. you always want a fraction in simplest form.[/quote]\r\nUnless it tells you not to. (usually it'll say put it in a common fraction which is in lowest terms)", "Solution_5": "[quote=\"Quevvy\"][quote=\"ckck\"][quote=\"Interval\"]Did you reduce 7/14?[/quote]\nyes he did. If you can reduce a fraction,always do that. you always want a fraction in simplest form.[/quote]\nUnless it tells you not to. (usually it'll say put it in a common fraction which is in lowest terms)[/quote]\r\n\r\nand not in mixed numbers I believe", "Solution_6": "[hide]\nThere are $14$ coins in all. There are $4+3$ dimes and nickels. $7/14=1/2$[/hide]" } { "Tag": [], "Problem": "1. Lead storage batteries can be recharged, but why can't dry cells recharge? \r\n\r\n2. Why do boiling points increase as you go down the halogens, but decrease as you go down the alkali metals? \r\n\r\n3. Name a nonmetallic oxide that reacts with water to form a base and a metallic oxide that reacts with water to form an acid.", "Solution_1": "for # 3.............\r\n\r\n[hide]$Na_{2}O\\ +\\ H_{2}O\\ \\to \\ 2NaOH$\n\n$CO_{2}\\ +\\ H_{2}O\\ \\to \\ H_{2}CO_{3}$[/hide]", "Solution_2": "No.2:\r\n[hide]\nOn descending Group I, the boiling points decrease because the metallic bonding in the metals becomes longer and weaker (the metallic radius increases on descending).\n\nOn descending Group VII, the boiling points increase because there are more electrons available for intermolecular bonding. Thus, with a larger and more easily polarisable electron cloud, there is stronger and more extensive intermolecular van der Waal's forces of attraction.[/hide]", "Solution_3": "[quote=\"g1sk8terz\"]for # 3.............\n\n[hide]$Na_{2}O\\ +\\ H_{2}O\\ \\to \\ 2NaOH$\n\n[/hide][/quote]\r\n\r\nI tried that too! I just didn't realize that $Na_{2}O$ was nonmetallic... :blush: Well, is there a way you can tell?", "Solution_4": "A nonmetal oxide (like $\\text{SO}_2$) usually (meaning almost always) forms an acid in water, and a metal oxide (like $\\text{Na}_2\\text{O}$) almost always forms a base in water. Maybe (huge huge maybe) something like polonium oxide might form an acid ($\\text{H}_2\\text{PoO}_4$?), but I'm not sure about that.", "Solution_5": "Lead acid batteries are not fully rechargable. For example, during rechage, some of the electrons are energic enough to react and sepertate the water in the sulfuric acid into hydrogen and oxygen. They migrate away from their respective elecrodes and are lost to the battery. More water needs to be added to a battery to restore the orginial condition. \r\n\r\nTo address your question, sometimes the products of the redox reaction that releases energy in a battery happen (or are designed) to be reactants to consume far more energy than given off to restore (mostly) the material to the orginal condition. (Entropy will always get its share.)\r\n\r\nIn a lead acid battery, lead reacts to lead oxide, which can be readily converted back to lead.", "Solution_6": "For a metallic oxide that forms an acid, I think some sort of chromium oxide (VI?) would do it. Maybe something like:\r\n\r\n$2CrO_3 + H_2O \\rightarrow H_2Cr_2O_7$\r\n\r\nNo idea about a nonmetallic base." } { "Tag": [ "number theory solved", "number theory" ], "Problem": "(1) If n is a positive integer and d a divisor of 2n^2, prove that n^2+d cannot be a perfect square.\r\n\r\n(2) For positive integers m,n, prove that 3^m+3^n+1 can never be a perfect square.\r\n\r\n(3) If n is a positive integer > 11, prove that n^2-19n+89 is never a perfect square.", "Solution_1": "a) Let d*m= 2n 2 .\r\nThen m 2 *(n 2 +d) = n 2 ((m+1) 2 2 - 1)\r\nThus, if n 2 +d is a square so is (m+1) 2 2 - 1, from which it follows that (m+1) 2 2 = 0 (since 0 and 1 are twe only two squares differing by 1). A contradiction.\r\n\r\nb) An odd square is known to be equal to 1 mod[8].\r\nBut, according to the parities of m and n, it is easy to see that 3 m + 3 n is never equals to 0 mod[8].\r\n\r\nc) For n > 11, it is easy to verify that \r\n(n-9) 2 < n 2 -19n + 89 < (n-10) 2 , then it cannot be a square....\r\n\r\nPierre." } { "Tag": [ "geometry", "parallelogram", "circumcircle" ], "Problem": "$ ABCD$ is a parallelogram,$ O$ is the circumcenter of of $ ABC$ and $ BOB^{\\prime}$ is a diameter of this circle. Prove that $ DB^{\\prime}$ is perpendicular to $ AC$. :)", "Solution_1": "[b]lemma:$ XY \\bot MN \\iff XM^{2}-XN^{2}=YM^{2}-YN^{2}$[/b]\r\nnow according to the lemma its sufficient to show that:\r\n$ DA^{2}-DC^{2}=B'A^{2}-B'C^{2}$\r\nnow we have:\r\n$ BB'$ is a diameter so $ \\angle B'AB=90$ so $ BB'^{2}=B'A^{2}+BA^{2}\\Rightarrow B'A^{2}=BB'^{2}-AB^{2}$\r\nin a similar way we get that $ B'C^{2}=BB'^{2}-BC^{2}$\r\nhence we have:\r\n$ B'A^{2}-B'C^{2}=(BB'^{2}-AB^{2})-(BB'^{2}-BC^{2})=BC^{2}-AB^{2}=DA^{2}-DC^{2}$ so we have:\r\n$ B'A^{2}-B'C^{2}=DA^{2}-DC^{2}$ as wanted...", "Solution_2": "Wow really nice solution Babak :) \r\nmy solution was this:\r\nas $ BOB^{\\prime}$ is diameter we have $ CB^{\\prime}$ perpendicular to $ BC$ (or prpendicular to $ AD$) and $ AB^{\\prime}$ is perpendicular to $ AB$(or perpendicular to $ DC$)! as a consequence $ B^{\\prime}$ is the orthocenter of triangle $ ADC$ and $ DB^{\\prime}$ is perpendicular to $ AC$. :)", "Solution_3": "Let $E$ be midpoint of $AC$; since $O$ was the circumcenter and $AC$ a chord of the circle, $OE\\bot AC$, but $OE$ is midline of $\\Delta BB'D$, done!\n\nBest regards,\nsunken rock" } { "Tag": [ "geometry" ], "Problem": "dayereye delkhahe C1 az ro oose B ,C mosallase ABC migozarad\r\ndayereye C2 ra tori rasm mikonim ke bar C1 dar X va bar azlaae AB ,AC \r\naz mosallas momaas shavad.\r\n\r\nsabet konid nimsaaz e BXC az mahalle barkhorde nimsaaz haye mosallase ABC migozarad", "Solution_1": "ghablan [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59059]inja[/url]hal shode\r\n\r\nalbate man sheklhasho pak kardam :D" } { "Tag": [], "Problem": "What is the pressure in atmospheres on the body of a diver if he is 36 feet below the surface of the water when atmospheric pressure at the surface is 0.95 atm?", "Solution_1": "Are you referring to the pressure exerted by the water on the body, or the gas pressure inside the lungs of the diver?" } { "Tag": [ "limit", "trigonometry", "logarithms", "calculus", "derivative", "calculus computations" ], "Problem": "Evaluate of the folowing limit [b]without using l'Hopital's theorem[/b].\r\n\r\n\\[\\lim_{x\\rightarrow 0} \\frac{e^{x\\sin 2005x}-1}{x\\ln (1+x)}\\]", "Solution_1": "It is obviously $2005$ (replace each expression by its equivalent), but replacement by equivalents here is equivalent to knowledge of certain derivatives, which is equivalent to using L'Hopital... What I mean is that I doubt it is possible to do it without knowing that $\\lim_{x\\to 0}\\frac{\\log(1+x)}x=1$, but this is equivalent to saying that $\\log(1+x)'=1$ at $0$ (just by the definition of the derivative) and once we have a derivative, we are using L'Hopital. So I am really confused about what the problem asks to do :?", "Solution_2": "Thank you for your reply, fedja.\r\n\r\nThis problem is a practice for using the following basic definition of the limit.\r\n\r\n$\\lim_{h\\rightarrow 0}\\frac{e^h-1}{h}=1,\\lim_{x\\rightarrow 0}\\frac{\\ln (1+x)}{x}=1,\\lim_{\\theta \\rightarrow 0}\\frac{\\sin \\theta}{\\theta}=1$\r\n\r\nkunny" } { "Tag": [ "inequalities", "real analysis", "abstract algebra", "projective geometry", "geometry", "function", "trigonometry" ], "Problem": "What your favorite theorem?\r\nMy favorite is Sylow's First Theorem ... duh!", "Solution_1": "Well this thread is in the [b]Polls[/b] section so either make a poll or a mod is going to move this to G&FF or something. \r\n\r\nMy favorite theroem? I'm too tired to think so i'll just say Pythagorean Theorem (i probably didn't even spell it right)", "Solution_2": "Four Colors Theorem, for sure.", "Solution_3": "direchlet theorem", "Solution_4": "Is a theorem just something that can be proved?", "Solution_5": "Hack, you have to be kidding me.\r\nThis is way too hard and involves what you know, and what type of math you like.\r\n\r\nI think my favorite theorem is Fermat's Little Theorem. However, when I learn a new ,cooler one this will probably change.", "Solution_6": "what I liked most until now is Wedderburn's theorem :love:", "Solution_7": "Karamata's Weighted Inequality and Schur's Convexity.", "Solution_8": "Why isn't anyone saying Pappus's Theorum?", "Solution_9": "[quote=\"spix\"]what I liked most until now is Wedderburn's theorem :love:[/quote]\r\nYes, Wedderburn is nice.\r\nSome other nice ones are the residue theorem, local-to-global principles (especially Hasse-Minkowski), the main theorem of class field theory, Chebotarev's density theorem, the theorem of Kronecker-Weber and a lot of other quasi-cyclotomic stuff :D", "Solution_10": "Stewart's for the win.", "Solution_11": "what about Fermat's Last Theorem?", "Solution_12": "[quote=\"davidyko\"]Stewart's for the win.[/quote]\r\n\r\nI like Stewart's too.", "Solution_13": "Abel's Theorem", "Solution_14": "How about everyone actually posts the theorem when they declare them? (\"Declare\"? I've been programming too much)\r\nAnd the proof, if it's within reason.", "Solution_15": "The Chicken McNugget theorem states that for two relatively prime numbers m and n, the largest positive number you cannot make using an infinite amount of m and n is mn-m-n.", "Solution_16": "I don't really know many of them ... but for now my favourite is Binomial theorem :lol:", "Solution_17": "The Squeezing Theorem is a fun one to use :)", "Solution_18": "Here are my favorites:\r\nPtolemy\r\nPick's Theorem\r\nPythagorean Theorem\r\nFermat's Last Theorem\r\n\r\nMy criteria was to pick theorems that are very generalized and natural and in order to prove them you have to somehow break the natural symmetry of the problem.", "Solution_19": "I thought the Ptolemy Theorem was that the earth was the center of the galaxy or whatever..... :ninja: :huh: :|", "Solution_20": "[quote=\"dynamo729\"]I thought the Ptolemy Theorem was that the earth was the center of the galaxy or whatever..... :ninja: :huh: :|[/quote]\r\n....\r\n\r\nI really, really, really, hope that this wasn't a serious post.\r\n\r\nPtolemy did indeed support the false geocentric model, instead of the heliocentric one, but...", "Solution_21": "[quote=\"7h3.D3m0n.117\"]The Squeezing Theorem is a fun one to use :)[/quote]\r\n\r\n*squeeze* :what?:", "Solution_22": "that isn't ptolemy theorem :o .....but indeed...ptolemy said that the earth is in the center of tha galaxy and that is a sphere...but this is not math:))", "Solution_23": "The main part of galoin theory(like some one who gave the same choice i think)", "Solution_24": "You maybe mean \"Galois\".\r\nMost are lacking the more advanced really interesting stuff here.", "Solution_25": "The non-existence of the quintic formula using standard notation (addition, subtracton, multiplication, division, radicals)! (The Abel-Ruffini Theorem)", "Solution_26": "yes i meant 'galois\".it's a superb theorem i think.", "Solution_27": "Djikstra's Algorithm!", "Solution_28": "My favourite one is the Pythagorean Theorem :)", "Solution_29": "I like DeMoivre's theorem and the Cubic Formula.\r\n\r\n$ x^3\\plus{}ax^2\\plus{}bx\\plus{}c\\equal{}0$\r\n\r\n$ r_1 \\equal{} \\minus{}\\frac{a}{3} \\plus{} \\sqrt[3]{\\frac{\\minus{}2a^3 \\plus{} 9ab \\minus{} 27c \\plus{} \\sqrt{(2a^3\\minus{}9ab\\plus{}27c)^2 \\plus{} 4(\\minus{}a^2\\plus{}3b)^3}}{54}} \\plus{} \\sqrt[3]{\\frac{\\minus{}2a^3 \\plus{} 9ab \\minus{} 27c \\minus{} \\sqrt{(2a^3\\minus{}9ab\\plus{}27c)^2 \\plus{} 4(\\minus{}a^2\\plus{}3b)^3}}{54}}$\r\n\r\n$ r_2 \\equal{} \\minus{}\\frac{a}{3} \\minus{} \\frac{1\\plus{}i\\sqrt{3}}{2} \\sqrt[3]{\\frac{\\minus{}2a^3 \\plus{} 9ab \\minus{} 27c \\plus{} \\sqrt{(2a^3\\minus{}9ab\\plus{}27c)^2 \\plus{} 4(\\minus{}a^2\\plus{}3b)^3}}{54}} \\plus{} \\frac{\\minus{}1\\plus{}i\\sqrt{3}}{2} \\sqrt[3]{\\frac{\\minus{}2a^3 \\plus{} 9ab \\minus{} 27c \\minus{} \\sqrt{(2a^3\\minus{}9ab\\plus{}27c)^2 \\plus{} 4(\\minus{}a^2\\plus{}3b)^3}}{54}}$\r\n\r\n$ r_3 \\equal{} \\minus{}\\frac{a}{3} \\plus{} \\frac{\\minus{}1\\plus{}i\\sqrt{3}}{2} \\sqrt[3]{\\frac{\\minus{}2a^3 \\plus{} 9ab \\minus{} 27c \\plus{} \\sqrt{(2a^3\\minus{}9ab\\plus{}27c)^2 \\plus{} 4(\\minus{}a^2\\plus{}3b)^3}}{54}} \\minus{} \\frac{1\\plus{}i\\sqrt{3}}{2} \\sqrt[3]{\\frac{\\minus{}2a^3 \\plus{} 9ab \\minus{} 27c \\minus{} \\sqrt{(2a^3\\minus{}9ab\\plus{}27c)^2 \\plus{} 4(\\minus{}a^2\\plus{}3b)^3}}{54}}$" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a,b,c be the positive number satisfying ab+bc+ca=1.\r\nProve that:\r\n$ \\frac{ab}{a^2\\plus{}1}\\plus{}\\frac{bc}{b^2\\plus{}1}\\plus{}\\frac{ca}{c^2\\plus{}1}$\u2264$ \\frac{1}{16}(\\frac{a}{b}\\plus{}\\frac{b}{c}\\plus{}\\frac{c}{a}\\plus{}2(\\frac{b}{a}\\plus{}\\frac{c}{b}\\plus{}\\frac{a}{c})\\plus{}3)$", "Solution_1": "[quote=\"SUPERMAN2\"]Let a,b,c be the positive number satisfying ab+bc+ca=1.\nProve that:\n$ \\frac {ab}{a^2 + 1} + \\frac {bc}{b^2 + 1} + \\frac {ca}{c^2 + 1}$\u2264$ \\frac {1}{16}(\\frac {a}{b} + \\frac {b}{c} + \\frac {c}{a} + 2(\\frac {b}{a} + \\frac {c}{b} + \\frac {a}{c}) + 3)$[/quote]\r\n\r\nThis inequality equivolent to:\r\n$ \\frac {ab}{(a + b)(a + c)} + \\frac {bc}{(b + c)(a + b)} + \\frac {ac}{(a + c)(b + c)} \\le \\frac {1}{16}(\\frac { 2\\sum_{cyclic}{a^2b} + \\sum_{cyclic}{ab^2} + 3abc}{abc})$\r\nor\r\n${ 16(3abc + \\sum_{cyclic}{ab^2})abc} \\le ( 2\\sum_{cyclic}{a^2b} + \\sum_{cyclic}{ab^2} + 3abc)(\\sum_{cyclic}{a^2b} + \\sum_{cyclic}{ab^2} + 2abc)$\r\n\r\nLet's $ x = \\sum_{cyclic}{a^2b}$, $ y = \\sum_{cyclic}{ab^2}$, $ z = abc$ then our inequality become:\r\n\r\n$ (2x + y + 3z)(x + y + 2z) \\ge 48z^2 + 16yz$\r\nor\r\n$ 2x^2 + y^2 + 3xy + 7xz \\ge 42z^2 + 11yz$ \r\nSo it's left to prove\r\n$ 2x^2 + y^2 + 3xy + 7xz \\ge 42z^2 + 11yz$ \r\nwe have:\r\n$ 2x^2 \\ge 18z^2$ \r\n$ 7xz \\ge 21z^2$\r\n$ \\frac {2}{3}y^2 \\ge 2yz$\r\n$ \\frac {y^2}{3} \\ge 3z^2$\r\n$ 3xy \\ge 9yz$\r\nSo by summing this 5 inequalities we will have desired result.", "Solution_2": "[quote=\"SaYaT\"][quote=\"SUPERMAN2\"]Let a,b,c be the positive number satisfying ab+bc+ca=1.\nProve that:\n$ \\frac {ab}{a^2 + 1} + \\frac {bc}{b^2 + 1} + \\frac {ca}{c^2 + 1}$\u2264$ \\frac {1}{16}(\\frac {a}{b} + \\frac {b}{c} + \\frac {c}{a} + 2(\\frac {b}{a} + \\frac {c}{b} + \\frac {a}{c}) + 3)$[/quote]\n\nThis inequality equivolent to:\n$ \\frac {ab}{(a + b)(a + c)} + \\frac {bc}{(b + c)(a + b)} + \\frac {ac}{(a + c)(b + c)} \\le \\frac {1}{16}(\\frac { 2\\sum_{cyclic}{a^2b} + \\sum_{cyclic}{ab^2} + 3abc}{abc})$\nor\n${ 16(3abc + \\sum_{cyclic}{ab^2})abc} \\le ( 2\\sum_{cyclic}{a^2b} + \\sum_{cyclic}{ab^2} + 3abc)(\\sum_{cyclic}{a^2b} + \\sum_{cyclic}{ab^2} + 2abc)$\n\nLet's $ x = \\sum_{cyclic}{a^2b}$, $ y = \\sum_{cyclic}{ab^2}$, $ z = abc$ then our inequality become:\n\n$ (2x + y + 3z)(x + y + 2z) \\ge 48z^2 + 16yz$\nor\n$ 2x^2 + y^2 + 3xy + 7xz \\ge 42z^2 + 11yz$ \nSo it's left to prove\n$ 2x^2 + y^2 + 3xy + 7xz \\ge 42z^2 + 11yz$ \nwe have:\n$ 2x^2 \\ge 18z^2$ \n$ 7xz \\ge 21z^2$\n$ \\frac {2}{3}y^2 \\ge 2yz$\n$ \\frac {y^2}{3} \\ge 3z^2$\n$ 3xy \\ge 9yz$\nSo by summing this 5 inequalities we will have desired result.[/quote]\r\nYour solution is right but it is not very nice.\r\nHere is my sol.\r\nNote that:\r\n$ \\frac{ab}{a^2+1}=\\frac{ab}{(a+b)(a+c)}=\\frac{ab}{a+b} \\frac{1}{a+c}$\u2264$ \\frac{a+b}{4} \\frac{1}{4}(\\frac{1}{a}+\\frac{1}{c})$\r\n=> $ \\frac{ab}{a^2+1}$\u2264$ \\frac{1}{16}(a+b)(\\frac{1}{a}+\\frac{1}{c})$\r\nSimilarly,$ \\frac{bc}{b^2+1}$\u2264$ \\frac{1}{16}(b+c)(\\frac{1}{a}+\\frac{1}{b})$ and $ \\frac{ca}{c^2+1}$\u2264$ \\frac{1}{16}(c+a)(\\frac{1}{b}+\\frac{1}{c})$\r\nThen sum,we get LHS\u2264RHS.", "Solution_3": "just use a=cotA and so on where A,B,C are angles of triangle", "Solution_4": "[quote=\"Jaswinder\"]just use a=cotA and so on where A,B,C are angles of triangle[/quote]\r\nI have thought the same way to you,but when I do that,I can not solve the problem." } { "Tag": [ "geometry", "inradius" ], "Problem": "The inradius of $ \\Delta ABC$ is 1 and the sides are all integers. Prove that $ \\Delta ABC$ is right angled.", "Solution_1": "[quote=\"hell_ever\"]The inradius of $ \\Delta ABC$ is 1 and the sides are all integers. Prove that $ \\Delta ABC$ is right angled.[/quote]\r\n\r\n\r\n$ S\\equal{}pr$ --> $ S\\equal{}p$ $ a\\equal{}x\\plus{}y,b\\equal{}x\\plus{}z,c\\equal{}y\\plus{}z$ ----> $ S\\equal{}\\sqrt{xyz(x\\plus{}y\\plus{}z)}\\equal{}x\\plus{}y\\plus{}z$ ---> $ xyz\\equal{}x\\plus{}y\\plus{}z$ ---> \r\n$ 2(x\\plus{}y\\plus{}z)\\equal{}a\\plus{}b\\plus{}c$ and $ c\\equal{}y\\plus{}z$ integer ----> $ 2x$ -> integer ---> $ (2x)yz\\equal{}2(x\\plus{}y\\plus{}z)$ ---> $ yz$ integer ----> $ x,y$ integer ---> $ x,y,z$ integer ---> $ xyz\\equal{}x\\plus{}y\\plus{}z$< --->$ (x;y;z)\\equal{}(1;2;3)$ ---> $ a\\equal{}3,b\\equal{}4,c\\equal{}5$ right angled" } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "x^4 - 8y^4 = z^2\r\nAssuming x,y,z is the minimal solution,\r\nI understand that x and z are both odd, but how do I explain :\r\n- the remainder of x modulo 4 is 1, and\r\n- the remainder of z modulo 4 is 3?", "Solution_1": "Is that true?\r\n3^4-8x0^4=9^2\r\nDid you solve it in positive int or in int.", "Solution_2": "[quote=\"Muhigai\"]x^4 - 8y^4 = z^2\nAssuming x,y,z is the minimal solution,\nI understand that x and z are both odd, but how do I explain :\n- the remainder of x modulo 4 is 1, and\n- the remainder of z modulo 4 is 3?[/quote]\r\n\r\n $ x,y,z$ positive integer \r\n$ z^{2} \\equal{} x^{4} \\minus{} 8y^{4}$ ---> $ z^{4} \\equal{} x^{8} \\minus{} 16x^{4}y^{4} \\plus{} 64y^{8} \\equal{} (x^{4} \\plus{} 8y^{4})^{2} \\minus{} 32x^{4}y^{4}$\r\n\r\n$ \\minus{} 8(2xyz)^{4} \\equal{} [(x^{4} \\minus{} 8y^{4})^{2} \\minus{} 32x^{4}y^{4} \\minus{} (x^{4} \\plus{} 8y^{4})^{4}]2z^{4} \\equal{} [(x^{4} \\minus{} 8y^{4}){2} \\minus{} 32x^{4}y^{4} \\minus{} (x^{4} \\plus{} 8y^{4})^{4}][z^{4} \\plus{} z^{4}] \\equal{} [(x^{4} \\minus{} 8y^{4})^{2} \\minus{} 32x^{4}y^{4} \\minus{} (x^{4} \\plus{} 8y^{4})^{4}][z^{4} \\plus{} (x^{4} \\plus{} 8y^{4})^{2} \\minus{} 32x^{4}y^{4}] \\equal{} [(z^{2})^{2} \\minus{} 32x^{4}y^{4} \\minus{} (x^{4} \\plus{} 8y^{4})^{4}][z^{4} \\plus{} (x^{4} \\plus{} 8y^{4})^{2} \\minus{} 32x^{4}y^{4}] \\equal{} (z^{4} \\minus{} 32x^{4}y^{4})^{2} \\minus{} (x^{4} \\plus{} 8y^{4})^{4}$ ------>\r\n$ \\minus{} 8(2xyz)^{4} \\equal{} (z^{4} \\minus{} 32x^{4}y^{4})^{2} \\minus{} (x^{4} \\plus{} 8y^{4})^{4}$ ----->\r\n\r\n$ (x^{4} \\plus{} 8y^{4})^{4} \\minus{} 8(2xyz)^{4} \\equal{} (z^{4} \\minus{} 32x^{4}y^{4})^{2}$\r\n\r\n$ x_{1} \\equal{} x^{4} \\plus{} 8y^{4}$ $ y_{1} \\equal{} 2xyz$ $ z_{1} \\equal{} z^{4} \\minus{} 32x^{4}y^{4}$ ---> $ x \\equal{} y \\equal{} z \\equal{} 1$ --->\r\n$ x_{1} \\equal{} 9,y_{1} \\equal{} 2,z_{1} \\equal{} 31$", "Solution_3": "first of all, what do you mean by minimal solution? there are a number of ways to define minimal.", "Solution_4": "$ x,y,z$ are positive integers and \r\nwe assume there's a solution where $ x$ is minimal.\r\n\r\nThen you can show there are no solutions by the method of infinte descent.\r\n\r\n$ x \\equiv 1 \\bmod 4$\r\n$ z \\equiv 3 \\bmod 4$\r\n\r\nThis is the part that escapes me.\r\n\r\nEdit:\r\nOkay, this is the context : \r\n'We assume that there's a positive solution with minimal $ x$.\r\nThen $ (x,y) \\equal{} 1$\r\nif $ x, z$ are both even, $ y$ would be odd and $ z^2 \\equiv 8 \\bmod 16$, which is impossible.\r\nHence $ x, z$ are both odd.\r\n$ x^2 > z$ --> $ 8y^4 \\equal{} x^4 \\minus{} z^2 \\equal{} (x^2 \\minus{} z^)(x^2 \\plus{} z^)$\r\nSince $ x \\equiv 1 \\bmod 4$ and $ z \\equiv 3 \\bmod 4$, -> [b]What goes wrong if you negate this part?[/b]\r\n$ x^2 \\minus{} z \\equiv 2 \\bmod 4$ and $ x^2 \\plus{} z \\equiv 0 \\bmod 4$\r\nso $ m \\equal{} (x^2 \\minus{} z)/2$ is odd, $ n \\equal{} (x^2 \\plus{} z)/4$ is an integer......'" } { "Tag": [ "inequalities proposed", "inequalities" ], "Problem": "Given a,b,c >0 and a+b+c=3. Prove :\\[ \\sqrt{\\frac{a}{b^2\\plus{}1}}\\plus{}\\sqrt{\\frac{b}{c^2\\plus{}1}}\\plus{}\\sqrt{\\frac{c}{a^2\\plus{}1}} \\geq 6\\sqrt{\\frac{1}{(a^2\\plus{}1)(b^2\\plus{}1)(c^2\\plus{}1)}}\\]", "Solution_1": "[quote=\"phucvy\"]Given a,b,c >0 and a+b+c=3. Prove :\n\\[ \\sqrt {\\frac {a}{b^2 \\plus{} 1}} \\plus{} \\sqrt {\\frac {b}{c^2 \\plus{} 1}} \\plus{} \\sqrt {\\frac {c}{a^2 \\plus{} 1}} \\geq 6\\sqrt {\\frac {1}{(a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1)}}\\]\n[/quote]\r\nIT's wrong! :P \r\nLet $ a\\equal{}b\\equal{}0,c\\equal{}3\\equal{}>RHS \\leq LHS$ :P", "Solution_2": "[quote=\"2424\"][quote=\"phucvy\"]Given a,b,c >0 and a+b+c=3. Prove :\n\\[ \\sqrt {\\frac {a}{b^2 \\plus{} 1}} \\plus{} \\sqrt {\\frac {b}{c^2 \\plus{} 1}} \\plus{} \\sqrt {\\frac {c}{a^2 \\plus{} 1}} \\geq 6\\sqrt {\\frac {1}{(a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1)}}\\]\n[/quote]\nIT's wrong! :P \nLet $ a \\equal{} b \\equal{} 0,c \\equal{} 3 \\equal{} > RHS \\leq LHS$ :P[/quote]\r\n\r\nTake wrong value of $ (a,b,c)$. Please look at the statement of the problem carefully, [b]the condition is[/b] $ a,b,c>0$", "Solution_3": "[quote=\"mathVNpro\"][quote=\"2424\"][quote=\"phucvy\"]Given a,b,c >0 and a+b+c=3. Prove :\n\\[ \\sqrt {\\frac {a}{b^2 \\plus{} 1}} \\plus{} \\sqrt {\\frac {b}{c^2 \\plus{} 1}} \\plus{} \\sqrt {\\frac {c}{a^2 \\plus{} 1}} \\geq 6\\sqrt {\\frac {1}{(a^2 \\plus{} 1)(b^2 \\plus{} 1)(c^2 \\plus{} 1)}}\\]\n[/quote]\nIT's wrong! :P \nLet $ a \\equal{} b \\equal{} 0,c \\equal{} 3 \\equal{} > RHS \\leq LHS$ :P[/quote]\n\nTake wrong value of $ (a,b,c)$. Please look at the statement of the problem carefully, [b]the condition is[/b] $ a,b,c > 0$[/quote]\r\n@:Chao ban,voi bdt nay thi gt $ a,b,c>0$ hay $ a,b,c \\geq 0$ khong khac nhau! :P \r\nCo the cho $ a,b\\to 0,c\\to3$" } { "Tag": [ "calculus", "integration", "geometry", "real analysis", "real analysis unsolved" ], "Problem": "Following Evans' book \"PDE\":\r\n$ \\Phi (x,t)\\equal{}\\frac{1}{(4\\pi t)^{n/2} } e^{\\minus{}\\frac{|x|^2}{4t}}$,when $ t>0$. $ \\Phi \\equal{}0$ when $ t<0$. is fundamention solution to heat equation $ u_t\\equal{}\\Delta u$.\r\n\r\nFor $ r>0$, define \"heat ball\" around point (0,0) as: $ E(r)\\equal{} \\{ (y,s) \\in R^{n\\plus{}1}| s<\\equal{}0, \\Phi(y,\\minus{}s)>\\equal{} \\frac{1}{r^n} \\}$\r\n\r\nHow to show: \r\n\r\n$ \\iint_{E(1)} \\frac{|y|^2}{s^2} dyds\\equal{}4$\r\n\r\n(since $ \\Phi$ is $ \\infty$ at (0,0), so (0,0) is also in heat ball. the integrand would be singular?Maybe heat ball defination should be $ s<0$ instead of $ s<\\equal{}0$?)", "Solution_1": "I think that reading Evans with full understanding requires more knowledge of integration. For instance, you should know that $ s\\le 0$ is no different from $ s<0$ as far as integration is concerned, and that a singularity is not a death sentence for an integral. As for the integral, the substitution $ t=-s$, $ y=z\\sqrt{t}$ turns it into \r\n\\[{ \\int_{e^{-|z|^2/4}\\ge (4\\pi t)^{n/2}}|z|^2t^{n/2-1}\\,dz\\,dt =\\int_{\\mathbb R^n}|z|^2\\,dz \\int_0^{\\frac{1}{4\\pi}e^{-|z|^2/(2n)}}t^{n/2-1}dt\r\n=\\frac{2}{n(4\\pi)^{n/2}}\\int_{\\mathbb R^n}|z|^2e^{-|z|^2/4}}dz\\]\r\nThe last integral is the area of $ S^{n-1}$ times $ \\int_0^{\\infty} r^{n+1}e^{-r^2/4}dr$. It's quite possible that I made a mistake or a few here; it's getting late. :sleeping:", "Solution_2": "[quote=\"mlok\"]I think that reading Evans with full understanding requires more knowledge of integration. For instance, you should know that $ s\\le 0$ is no different from $ s < 0$ as far as integration is concerned, and that a singularity is not a death sentence for an integral. As for the integral, the substitution $ t = - s$, $ y = z\\sqrt {t}$ turns it into\n\\[{ \\int_{e^{ - |z|^2/4}\\ge (4\\pi t)^{n/2}}|z|^2t^{n/2 - 1}\\,dz\\,dt = \\int_{\\mathbb R^n}|z|^2\\,dz \\int_0^{\\frac {1}{4\\pi}e^{ - |z|^2/(2n)}}t^{n/2 - 1}dt = \\frac {2}{n(4\\pi)^{n/2}}\\int_{\\mathbb R^n}|z|^2e^{ - |z|^2/4}}dz\n\\]\nThe last integral is the area of $ S^{n - 1}$ times $ \\int_0^{\\infty} r^{n + 1}e^{ - r^2/4}dr$. It's quite possible that I made a mistake or a few here; it's getting late. :sleeping:[/quote]\r\n\r\nthanks. let me finish the last integral: by change variable: $ r^2/4=u$ it became:\r\n$ 2^{n+1} \\int_0^{\\infty} u^{n/2} e^{-u} du=2^{n+1} \\Gamma (n/2+1)$ by defination of gamma function. But $ S_{n-1}=\\frac{n \\pi^{n/2}}{\\Gamma (n/2+1)}$, ..." } { "Tag": [ "linear algebra", "matrix", "linear algebra unsolved" ], "Problem": "Let $f$ be a continious morphism of the additive groupes $f: \\mathbb{R} \\to M_n(\\mathbb{K})$. Prove that there exists $A \\in M_n(\\mathbb{K})$ such that $f(t)=e^{tA}$ for all $t \\in \\mathbb{R}$.", "Solution_1": "[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=homomorphism&t=28482[/url]" } { "Tag": [ "function", "integration" ], "Problem": "a rope of length $l$ and linear density function $\\lambda(x)$ where $x$ is the vertical position on the rope; holds up a block of mass $m$ What is the function $T(x)$ where $T$ is the tension of the rope at a vertical position $x$", "Solution_1": "At any point the tension is only due to the mass of the rope Below it , the upper part does not contribute to the tension.\r\n\r\nHence the tension due to the weight of the rope at a point at a distance x from the bottom is given by \r\n\r\n$T(x) =$ $\\int^{x}_{0}{\\lambda(x)g\\,dx}$\r\n\r\nHence ,the net tension at any point = tension due to rope+tension due to the overhanging mass which is equal to \r\n\r\n$T(x) =$ $mg$ $+$ $\\int^{x}_{0}{\\lambda(x)g\\,dx}$" } { "Tag": [ "geometry", "3D geometry", "prism" ], "Problem": "i was givin this in class (the picture is the work sheet) and my teacher wants us to find the most efficient way of filling the truck bed without going over 3375 in^3 \r\ni got 1800 rectangular prisms with the dimensions of 9.6x6x10 \r\np.s dont mind the diagram i did at the top.\r\n\r\nWOrk sh33t:\r\n[url=http://imageshack.us][img]http://img462.imageshack.us/img462/843/s3pg.jpg[/img][/url]", "Solution_1": "any got any bit of an answer lower than mine?", "Solution_2": "Well if you're not worry about surface area, it's easy to do a 12\" x 12\" x 12\" box. So are you trying yo conserve surface area AND volume? Which is the most important? Also, rule out cylinder and triangular prism.", "Solution_3": "yea got any answers lower than 1800 boxes?" } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let a,b,c be nonnegative real numbers. Prove that:\r\n$ a^2 \\plus{} b^2 \\plus{} c^2 \\plus{} \\frac {4018}{2000}\\sqrt [2009]{(abc)^{2000}} \\plus{} \\frac {1982}{2000}\\geq 2(ab \\plus{} bc \\plus{} ca)$", "Solution_1": "these numbers really threaten people in a scary way :D \r\nHowever, we have\r\n\r\n$ \\frac{4018(abc)^{\\frac{2000}{2009}}}{2000}\\plus{} \\frac{1982}{2000} \\ge 3 \\sqrt[3]{(abc)^2}$ \r\n\r\nNow we turn to the our well known inequality!\r\n\r\n$ a^2\\plus{}b^2\\plus{}c^2 \\plus{}3\\sqrt[3]{(abc)^2} \\ge 2(ab\\plus{}bc\\plus{}ca)$", "Solution_2": "we can use schur inequality :D" } { "Tag": [ "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Does there exsist positive constant $ C$ and $ \\alpha \\in (0,1]$ such that the following inequality \r\nholds for all $ x_i\\in [ \\minus{} \\pi,\\pi]$ and $ y_i\\in ( \\minus{} \\infty, \\plus{} \\infty), i\\equal{}1,2,3$:\r\n$ |1 \\minus{} cos(x_1y_1 \\plus{} x_2y_2 \\plus{} x_3y_3)|\\leq C |x_1 \\plus{} x_2 \\plus{} x_3|^\\alpha |y_1 \\plus{} y_2 \\plus{} y_3|^\\alpha$.", "Solution_1": "If $ x_1\\plus{}x_2\\plus{}x_3\\equal{}0$, but not all $ x_j\\equal{}0$, the right hand side is $ 0$ but the left hand side can be anything between $ 0$ and $ 2$. :roll:", "Solution_2": "thank you fedja,\r\nWhat if we add following condition: the product $ (x_1\\plus{}x_2\\plus{}x_3)(y_1\\plus{}y_2\\plus{}y_3)$ is not equal to zero." } { "Tag": [ "function", "algebra unsolved", "algebra" ], "Problem": "Find all functions $ f: \\mathbb R^2 \\to \\mathbb R$ such that it qualifies both conditions below:\ni) $ f(x,y)+f(y,z)+f(z,x)=x^2+y^2+z^2$, for all reals $ x,y,z$, and\nii) $ f(x^2+y-f(0,x),0)=x+f^2(y,0)-f(0,y^2-f(0,y))$, for all reals $ x,y$.", "Solution_1": "[quote=\"mathVNpro\"]Find all fuction $ f(x,y)$ such that it qualifies both condition below:\ni) $ f(x,y) \\plus{} f(y,z) \\plus{} f(z,x) \\equal{} x^2 \\plus{} y^2 \\plus{} z^2$, for all $ x,y,z$ real numbers\nii) $ f(x^2 \\plus{} y \\minus{} f(0,x),0) \\equal{} x \\plus{} f^2(y,0) \\minus{} f(0,y^2 \\minus{} f(0,y))$, for all $ x,y$ real numbers[/quote]\r\n\r\nLet $ g(x,y)\\equal{}f(x,y)\\minus{}y^2$\r\n\r\nThe two conditions may be writen :\r\nA1(x,y,z) : $ g(x,y)\\plus{}g(y,z)\\plus{}g(z,x)\\equal{}0$\r\nA2(x,y) : $ g(y\\minus{}g(0,x),0)\\equal{}x\\plus{}g^2(y,0)\\minus{}g(0,\\minus{}g(0,y))\\minus{}g^2(0,y)$\r\n\r\nA1(0,0,0) $ \\implies$ $ g(0,0)\\equal{}0$\r\nA1(x,0,0) $ \\implies$ $ g(x,0)\\equal{}\\minus{}g(0,x)$\r\n\r\nUsing this last property in A2, we get :\r\nA3(x,y) : $ g(y\\plus{}g(x,0),0)\\equal{}x\\plus{}g(g(y,0),0)$\r\n\r\nLet then $ h(x)\\equal{}g(x,0)$. A3 becomes :\r\nA4(x,y) : $ h(y\\plus{}h(x))\\equal{}x\\plus{}h(h(y))$ with $ h(0)\\equal{}0$\r\n\r\nA4(x,0) $ \\implies$ $ h(h(x))\\equal{}x$\r\nA4(0,x) $ \\implies$ $ h(x)\\equal{}h(h(x))$ \r\n\r\nSo $ h(x)\\equal{}x$ and $ g(x,0)\\equal{}x$\r\n\r\nA1(x,y,0) $ \\implies$ $ g(x,y)\\plus{}g(y,0)\\plus{}g(0,x)\\equal{}0$, so $ g(x,y)\\plus{}g(y,0)\\minus{}g(x,0)\\equal{}0$, and so $ g(x,y)\\equal{}x\\minus{}y$\r\n\r\nAnd so $ f(x,y)\\equal{}x\\minus{}y\\plus{}y^2$\r\n\r\nAnd it is easy to verify that this necessary condition works." } { "Tag": [ "vector", "function", "superior algebra", "superior algebra unsolved" ], "Problem": "Hello,\r\nI just \"finished\" a proof only to find a hole. The problem runs thus:\r\n\r\nProve that if $V$ is a finite dimensional vector space and $W$ is a subspace of $V$, then there exists an idempotent endomorphism of $V$ with image $W.$\r\n\r\nI (incorrectly) considered the \"endomorphism\" $\\phi$ defined a restriction on the\r\nidentity function $\\phi$ on $V$ where if $w\\in W,$ then $\\phi(w)=w$ and if $w\\in V\\backslash W,$ then $\\phi(w)=0.$ This, of course, falls apart when one notes that if you have two elements from $V,$ one of which is in $W$ and the other in $V \\backslash W$ (say $w\\in V$ and $v\\in V\\backslash W.$)\r\n\r\n$\\phi(w+v)\\neq \\phi(w)+\\phi(v)$ for all $w,v\\in V$ is not well defined for the operation. \r\n\r\nAny one have any other considerations?\r\n\r\n-JNG", "Solution_1": "Make $\\phi(x)=0$ not in all $V\\setminus W$, but on some (sum)complement of $W$." } { "Tag": [ "algebra", "polynomial", "linear algebra", "matrix", "linear algebra theorems" ], "Problem": "Suppose $ F \\equal{} C$, $ T\\in L(V)$, $ p \\in P(C)$, and $ a \\in C$. prove $ a$ is an eigenvalue of $ p(T)$ iff $ a \\equal{} p(S)$ for some eigenvalue $ S$ of $ T$.\r\n\r\nstarting with an eigenvalue of $ T$ and proving the implication is an easy 1 step solution. Going the other way should be easy but im a bit stuck. We can bring the eqn $ p(T)v \\equal{} p(S)v$, $ v \\in V$ to the form where $ (T \\minus{} SI)(very\\: messy\\: polynomial\\: in\\: T,S)v \\equal{} 0$ (LHS-RHS and factor out) so either one of those two linear transforms takes $ v$ to zero. If it is $ (T \\minus{} SI)$ then we are done, but I am not sure how to show this?", "Solution_1": "That argument won't work, because the inverse of $ p$ is multiple-valued. If $ p(S)$ is an eigenvalue, $ S$ might not be.\r\n\r\nHere's what I would do: Triangularize $ T$. The diagonal entries $ t_{ii}$ are the eigenvalues. The diagonal entries of $ p(T)$ are $ p(t_{ii})$, and are the eigenvalues of $ p(T)$.", "Solution_2": "[quote=\"jmerry\"] That argument won't work, because the inverse of p is multiple-valued. [/quote] I know very little about polynomials, so can you explain a bit more what you are saying here?\r\n\r\noh ok, the next question is to show why this is not necessarily true for the field of reals; will any 'un-diagonal' matrix do? how do we find an un-diagonal matrix? and how do we show p(T) does not have some $ S$ as an eigenvalue? or do we find a matrix with no eigenvalues and show $ p(T)$ has one?", "Solution_3": "An extremely simple counterexample in $ \\mathbb{R}$: $ T\\equal{}\\begin{bmatrix}0&1\\\\\\minus{}1&0\\end{bmatrix}$ has no eigenvalues (because they're $ \\pm i$). $ T^2\\equal{}\\minus{}I$ does have eigenvalues." } { "Tag": [ "vector", "geometry", "circumcircle", "analytic geometry", "search", "articles", "absolute value" ], "Problem": "Well I had a simple question on my Geometry test yesterday that asked to find the circumcenter of a triangle given the coordinates of the vertices. I know that it can be solved using some algebra, but I wondered if there was a more efficient method to reach the same answer. After a quick search, I found [url=http://en.wikipedia.org/wiki/Circumcenter]this Wikipedia article.[/url] I was wondering how one would:\r\n\r\n1. Go about subtracting $ P_2\\minus{}P_3$. My dad says that to subtract two vectors, you have to \"line them up\" and then find the distance between the two \"ends.\" (Not too specific, I know).\r\n\r\n2. Why do the have the absolute value signs when that part of the expression is already squared?", "Solution_1": "To add two vectors, stick the tail of one onto the head of another. It turns out that the components of the new vector are obtained by summing the corresponding components of the two vectors you added. I could try to go into more detail, but it'd probably be good to find an \"introduction to vectors\" website, maybe the Wikipedia article on vectors, and dig in.", "Solution_2": "The vector $ P_2 \\minus{} P_3$ is the vector that goes from the tip of $ P_3$ to the tip of $ P_2$ when you give $ P_2$ and $ P_3$ the same initial point. (This makes complete sense: it's what you have to add to $ P_3$ to get $ P_2$.) Algebraically, it's just the coordinatewise difference.\r\n\r\nFor the second question, the absolute values are lengths of vectors. If $ v$ is a vector, $ v^2$ doesn't mean anything, but $ |v|^2$ means the length of $ v$, squared." } { "Tag": [ "induction", "inequalities", "inequalities unsolved" ], "Problem": "Let $ x_{1},x_{2},\\cdots,x_{m},y_{1},y_{2},\\cdots,y_{n}$ be positive real numbers. Denote by $ X \\equal{} \\sum_{i \\equal{} 1}^{m}x,Y \\equal{} \\sum_{j \\equal{} 1}^{n}y.$ Prove that $ 2XY\\sum_{i \\equal{} 1}^{m}\\sum_{j \\equal{} 1}^{n}|x_{i} \\minus{} y_{j}|\\ge X^2\\sum_{j \\equal{} 1}^{n}\\sum_{l \\equal{} 1}^{n}|y_{i} \\minus{} y_{l}| \\plus{} Y^2\\sum_{i \\equal{} 1}^{m}\\sum_{k \\equal{} 1}^{m}|x_{i} \\minus{} x_{k}|$", "Solution_1": "Oh of course :blush: \r\nWell assume $ x_1\\ge x_2\\ge \\dots \\ge x_n$ and $ X\\ge Y$, then make an induction :\r\nif the inequality is true for $ m \\minus{} 1$, then we can prove for $ x \\equal{} x_1 \\plus{} x_2$ the following statement, which will solve our problem in matter of fact:\r\n$ X\\sum_{j \\equal{} 1}^{n} |x \\minus{} y_j| \\minus{} Y\\sum_{i \\equal{} 3}^{m} |x \\minus{} x_i|\\le X\\sum_{j \\equal{} 1}^{n} (|x_1 \\minus{} y_j| \\plus{} |x_2 \\minus{} y_j|) \\minus{}$ $ Y\\left(\\sum_{i \\equal{} 3}^{m} (|x_1 \\minus{} x_i| \\plus{} |x_2 \\minus{} x_i|) \\plus{} |x_1 \\minus{} x_2|\\right)$ (*).\r\n\r\nFirst $ LHS \\equal{} X\\sum_{j \\equal{} 1}^{n} |x \\minus{} y_j| \\minus{} Y\\sum_{i \\equal{} 3}^{m} |x \\minus{} x_i| \\equal{} \\sum_{j \\equal{} 1}^{n}\\sum_{i \\equal{} 1}^{m} x_i|x \\minus{} y_j| \\minus{} \\sum_{j \\equal{} 1}^{n}\\sum_{i \\equal{} 3}^{m} y_j|x \\minus{} x_j|$\r\n\r\n$ RHS \\equal{} X\\sum_{j \\equal{} 1}^{n} (|x_1 \\minus{} y_j| \\plus{} |x_2 \\minus{} y_j|) \\minus{}$ $ Y\\left(\\sum_{i \\equal{} 3}^{m} (|x_1 \\minus{} x_i| \\plus{} |x_2 \\minus{} x_i|) \\plus{} |x_1 \\minus{} x_2|\\right) \\equal{} \\sum_{j \\equal{} 1}^{n}\\sum_{i \\equal{} 1}^{m} (x_i|x_1 \\minus{} y_j| \\plus{} x_i|x_2 \\minus{} y_j|)$ $ \\minus{} \\sum_{j \\equal{} 1}^{n}\\sum_{i \\equal{} 3}^{m}\\left(y_j|x_1 \\minus{} x_i| \\plus{} y_j|x_2 \\minus{} x_i|) \\plus{} y_j|x_1 \\minus{} x_2|\\right)$\r\n\r\nand now using the inequality $ |xx_i \\minus{} y_jx_i| \\minus{} |y_jx \\minus{} y_jx_i|\\le$ $ x_i|x_1 \\minus{} y_j| \\plus{} x_i|x_2 \\minus{} y_j| \\minus{} y_j|x_1 \\minus{} x_i| \\minus{} y_j|x_2 \\minus{} x_i|$ which follows by verifying all cases - $ x_1\\ge x_2\\ge y_j \\cup x\\ge y_j$ etc. , we will take what we need.\r\n\r\n$ \\Rightarrow 2XY \\sum_{i \\equal{} 1}^m\\sum_{j \\equal{} 1}^n|x_i \\minus{} y_j|$ $ \\minus{} X^2 \\sum_{i \\equal{} 1}^n\\sum_{j \\equal{} 1}^n|y_i \\minus{} y_j| \\minus{} Y^2 \\sum_{i \\equal{} 1}^m\\sum_{j \\equal{} 1}^m|x_i \\minus{} x_j|\\ge$ $ 2XY \\sum_{i \\equal{} 2}^m\\sum_{j \\equal{} 1}^n|x'_i \\minus{} y_j| \\minus{} X^2 \\sum_{i \\equal{} 1}^n\\sum_{j \\equal{} 1}^n|y_i \\minus{} y_j| \\minus{} Y^2 \\sum_{i \\equal{} 2}^m\\sum_{j \\equal{} 1}^m|x'_i \\minus{} x'_j|\\ge 0$ by (*), where $ x'_2 \\equal{} x_1 \\plus{} x_2$ and $ x'_i \\equal{} x_i$ for $ i\\ge 3$...", "Solution_2": "Could you please show us where we can find $ \\sum y_j |x_1\\minus{}x_2|$? :)", "Solution_3": "Please ignore the previous post!", "Solution_4": "This is the shortest proof(not mine) for the hard problem\r\nI have to say this brilliant approach is owed to the official solution for a TST problem from Poland :)", "Solution_5": "[quote=\"hxy09\"]This is the shortest proof(not mine) for the hard problem\nI have to say this brilliant approach is owed to the official solution for a TST problem from Poland :)[/quote]\nSo this problem is from a Poland TST? If so, can you tell me which year? By the way, the solution you posted is very nice :)!", "Solution_6": "Well not all of us can be a genius who comes up with above brilliant approach...\n\nRelax the condition in the problem statement such that the $x_{i}$ and $y_{j}$ are nonnegative real numbers. \n\nLet $P(x_1, x_2, \\cdots, x_m, y_1, y_2, \\cdots, y_n)$ denote the desired inequality. Let $A = \\sum_{i = 1}^{m}\\sum_{j = 1}^{n}|x_{i} - y_{j}|$, $B = \\sum_{j = 1}^{n}\\sum_{l = 1}^{n}|y_{i} - y_{l}|$, and $C = \\sum_{i = 1}^{m}\\sum_{k = 1}^{m}|x_{i} - x_{k}|$.\n\nWe will write $(x)$ and $(y)$ to denote the numbers $x_1, x_2, \\cdots, x_m$ and $y_1, y_2, \\cdots, y_n$ respectively.\n\nNotice that if $m = 0$ or $n=0$, the LHS and the RHS of the desired inequality are both equal to zero. In what follows, we induct on $m+n$. \n\n[b]Lemma 1.[/b] If $x_m = 0$, we can delete $x_m$ to yield an equivalent desired inequality. In other words, $P(x_1, x_2, \\cdots, x_m, y_1, y_2, \\cdots, y_n)$ is implied by $P(x_1, x_2, \\cdots, x_{m-1}, y_1, y_2, \\cdots, y_n)$ if $x_m = 0$.\n\n[i]Proof.[/i] Note that $P(x_1, x_2, \\cdots, x_{m-1},y_1, y_2, \\cdots, y_n)$ is the following: \\[2XY(A-Y) \\ge X^2B+Y^2(C-2X).\\] This is easily seen to be equivalent to $2XYA \\ge X^2B + Y^2C$. $\\blacksquare$\n\nTime for the key idea of the proof.\n\n[b]Lemma 2.[/b] Suppose there are $d \\ge 3$ distinct values present among the numbers $(x),(y)$. Then we can find nonnegative reals $x_1', x_2', \\cdots, x_m', y_1', y_2', \\cdots, y_n'$ (henceforth abbreviated $(x'), (y')$) such that either there are at most $d-1$ distinct values present among them, or at least one of them is equal to zero, and such that $P((x'),(y'))$ implies $P((x),(y))$.\n\n[i]Proof.[/i] Let $a